Solutions to Kiritsis’ String Theory in a Nutshell

Alex Atanasov

Chapter 2: Classical String Theory

1. I don’t know what this question asks exactly given that 2.1.16 is an infinitesimal diffeomorphism.
9 “
We are still allowed to assume WLOG that τ runs from 0 to 1. For ξ infinitesimal, we have δe “ ξ e9 ` ξe
Bτ pξeq. So for a general epτ q define şτ 1 1
dτ epτ q
τ2 pτ q “ ş01 (1)
1 1
0 dτ epτ q
ş1 ´ ¯´1 ´ ¯´1
epτ q
Take L “ 0 dτ 1 epτ 1 q. Then e2 pτ2 pτ qq “ dτ
dτ
2
epτ q “ L epτ q “ L.
ş1 ş1
Note that we cannot get rid of this L, since it is invariant L “ 0 dτ epτ q “ 0 dτ2 epτ2 q

2. From analytic continuation, we have the functional equation for the Riemann zeta function:
πs
ζpsq “ 2s π s´1 sinp qΓp1 ´ sqζp1 ´ sq (2)
2
It is worth knowing that near s “ 0 we have ζp1 ´ sq “ ´ 1s ` γ and Γp1 ´ sq “ 1 ` γs. Expanding the right
hand side about s “ 0 gives
1 1?
ζpq “ ´ ´ 2π (3)
2 2
? log n
This gives ζp0q “ ´ 12 and ζ 1 p0q “ ´ 12 2π. Further, ζ 1 psq “ ´ 8
ř
n“1 ns .
´2 8
ř ` ř8 ˘
1
So we get 8
ś8
n“1 1 “ L´2ζp0q “ L and 2
ś
n“1 L2 “ L n“1 n “ exp 2 n“1 log n “ 2π.

3. For simplicity, we will work in the action with the einbein.

ż
1
dτ epe´2 Gµν x9 µ x9 ν ´ m2 q
2
The Euler-Lagrange equations for xµ is:
d B α β B ´ ´1 α β
¯
ν ν γ Gµν x9 ν
pe´1
G αβ x
9 x
9 q ´ e G αβ x
9 x
9 “ 2e´1
G µν x
: ` 2e´1
Bγ G µν x
9 x
9 ´ 2 e9 ´ e´1 Bµ Gαβ x9 α x9 β
dτ Bxµ Bxµ e2
1 1
Ñ Gµν x:ν ` pBγ Gµν ` Bν Gµγ ´ Bµ Gνγ qx9 ν x9 γ ´ Gµν x9 ν Bτ log e2
2 2
(4)
This last term looks particularly annoying, and is ignored by other authors. We have total freedom in
reparameterization of e, so we can WLOG set it equal to a (metric-dependent) constant by problem 1. Then
the term drops out and we get exactly the geodesic equations.
We could have done this explicitly as well:

d G x9 ν B
b :ν ` Bλ Gµν x9 ν x9 λ ´ 21 Bµ Gνλ x9 ν x9 λ
Gµν x d
a µν ´ µ Gαβ x9 α x9 β “ a ` Gµν x9 ν p´Gαβ x9 α x9 β q´1{2
dτ Gαβ x9 α x9 β Bx ´Gαβ x9 α x9 β dτ
(5)
ν ν α β 1 ν d α β
x ` Γαβ x9 x9 q ´ Gµν x9
“ Gµν p: logp´Gαβ x9 x9 q
2 dτ
And fix the parameterization so that x2 “ const and the last term vanishes.

4. We get the same as before, but now cannot drop the last term. Now the dots represent time derivatives.
1
xν ` Γναβ x9 α x9 β q ´ Gµν x9 ν BX 0 logp´Gαβ x9 α x9 β q
Gµν p: (6)
2

1
5. We get: ż b
´ mc dτ ´pG00 x9 0 x9 0 ` 2G0i x9 0 x9 i ` Gij x9 i x9 j q (7)

Taking τ “ x0 “ ct gives us our result. Further, we write G00 “ ´1 ´ 2φ

c2
where φ is the gravitational
potential. To first order then we get:
ż b ż
2
dt ´pG00 ` 2c G0i x9 ` c Gij x9 x9 q « dt ´mc2 ´ mφ ` mcG0i v i ` mGij v i v j
` ˘
´ mc ´1 i ´2 i j (8)

The last two terms in brackets are positive (kinetic) while the first two are negative (potential). This
explains why there is a - sign out front of the action.

a
6. The Lagrangian for a special relativistic particle
a in an electromagnetic field is ´mc2 1 ´ v 2 {c2 ´ eφ ` e~v ¨ A.
This has the Lorentz invariant form: ´m ´Gµν x9 µ x9 ν ` eAµ x9 µ . We get equations of motion as before: The
additional term gives the equations of motion:
e 9 e 9 ν “ e Fµν x9 ν
pAµ ´ Bµ Aν x9 ν q “ pBν Aµ ´ Bµ Aν qx (9)
m m m
don’t confuse e with the einbein.
If one coordinate is cyclic (neither the metric nor the vector potential depend on it), the corresponding
momentum is
BL Gµν x9 ν
“ m a ` eAµ (10)
B x9 µ ´Gµ ν x9 µ x9 ν

7. Ignoring the cosmological constant term (which is not reparameterization invariant), we note that any term
that involves the metric Gµν will require at least 2 xµ variables for it to be contracted with. Also, repa-
rameterization
a invariance requires that under dτ Ñ f 1 pτ qdτ we get L Ñ L{λ. The simplest such term is
´Gµν x9 x9 . Terms with more than 2 xµ s will be suppressed by powers of 1{`2s . Similarly, terms with more
µ ν

derivatives w.r.t. worldsheet coordinates will be less relevant in the IR.

8. Let’s set Gi0 “ 0 for simplicity. The Nambu-Goto action is:

ż b
´T dτ dσ pX9 ¨ X 1 q2 ´ pX9 2 qpX 1 2 q

Take τ “ ct, σ “ x and note T “ ρc2 with ρ the mass per unit length. Take X 0 “ ct and ~u “ X 1 i , ~v “ X9 i .
Appreciate that v gives us how that point of the string is moving, while u gives the direction parallel to the
string at that point (scaled according to σ’s parameterization). Inside the radical:
0 j 0 0 i j
pG00 X9 0 X 1 ` Gij X9 i X 1 q2 ´ pG00 X9 0 X9 0 ` Gij X9 i X9 j qpG00 X 1 X 1 ` Gij X 1 X 1 q
“ c´2 pGij ui v j q2 ´ pG00 ` c´2 Gij v i v j qpGij ui uj q

Take G00 “ ´1 ´ 2ϕ{c2 . Then the radical becomes:

c
1 2 1 p~u ¨ ~v q2
ˆ ˆ ˙˙
a
2 ´2 2 ´2 2 ´2 2 2 ´2 2
p~u ¨ ~v q2 ´2
u ´ c 2φu ` c p~u ¨ ~v q ´ c v u “ |u| 1 ´ c p2φ ` v ´ q “ |u| 1 ´ c ´φ ` v ´
u2 2 2 u2

2
 2
But note that v 2 ´ p~uu¨~v2q “ p~v ´ u¨v
u2
~uq2 . This is exactly the part of v transverse to u (the string itself). So
we can write this as ~vT , the transverse velocity.
ż ż
´2 1 2 1
´2
´ T dtdσ|u|p1 ` c φ ´ c vT q “ dtdσ|u|p´c2 ´ φ ` vT2 q (11)
2 2
ş
Note that ρ dσ|u| “ ρLs “ Ms . The first term is thus ´Ms c2 . The second terms is exactly the mass density
of the string interacting with the gravitational field, while the third (kinetic) is the motion of the transverse
components of the string. Note that the longitudinal excitations do not contribute.

9. Let’s work in lightcone gauge. We have B` B´ X “ 0 The vanishing of the stress-energy tensor gives us
X9 2 ` X 1 2 “ 0 and X9 ¨ X 1 2 “ 0. But at the endpoints we get X 1 “ 0 so that X9 2 “ 0 and the endpoints with
Neumann boundary conditions need to move at the speed of light.
?
10. The cosmological constant term gives the equation of motion δgδSab “ ´p T4π ab
` λ21 gab q ´g. But by reparame-
terization invariance we need Tab “ 0 so that λ1 must be 0.

11. It is quick to derive the current P µ under δX ν “ δ µν :

BL ?
µ
“ ´T ´gg αβ Bβ Xµ (12)
BpBα X q

Similarly under δX λ “ Mµν

λδ X with M λδ “ pδ λ δ δ ´ δ δ δ λ q. Then we have
δ µν µ ν µ ν

BL ?
λ
pδµλ δνδ ´ δµδ δνλ qXδ “ ´T ´gg αβ pXµ Bβ Xν ´ Xν Bβ Xµ q (13)
BpBα X q

12. Write:
i`s ÿ 1
X µ pτ, σq “ xµ ` `2s pµ τ ` ? pαn e´inσ ` ᾱn einσ qe´inτ
2 nPZ´t0u n
(14)
`s ÿ
X9 µ pτ, σq “ `2s pµ ` ? pαn e´inσ ` ᾱn einσ qe´inτ
2 nPZ´t0u

Now take the Fourier series (in σ) of the commutation relation:

δn`m 1 µν
tXnµ , X9 m
µ
u“ η (15)
2π T
The only nonzero terms are those we get when we pair each mode with its negative (in σ). Also note that
there is no τ dependence on the right-hand side, so we need to pair each τ mode with its negative. Let’s
look at xµ , the zero mode of X. We can only pair this with the other mode pµ and we necessarily have:
1
txµ , pν u “ η µν “ η µν (16)
2π`2s T
Similarly, we can only pair αn with α´n giving:

µ 2mδm`n µν
tαm , αnν u ` tᾱm
µ ν
, ᾱm u“ η (17)
2πi`2s T
By parity symmetry, both of these brackets should be the same. We get then that:
µ
tαm , αnν u “ tᾱm
µ ν
, ᾱm u “ ´iδm`n η µν (18)

13. For each coordinate on the n-torus, we have X i pτ, σ ` 2πq “ Xpτ, σq ` 2πni Ri . Then the corresponding
momenta have difference p ´ p̄ “ `22 ni Ri while the total momentum is quantized in multiples of p ` p̄ “ 2m
Ri .
i
s
Therefore we have: ˆ ˙
i 1 `s Ri
α0 “ ? mi ` ni (19)
2 Ri `s

3
14. We begin with a redefined pµ Ñ 2pµ as in the book.

µ `s ÿ
X 1 pτ, σq|σ“0 “ `2s ppµ ´ p̄µ q ` ? pαn ´ ᾱn qe´inτ
2 n
(20)
`s ÿ
X9 µ pτ, σq|σ“0 “ `2s ppµ ` p̄µ q ` ? pαn ` ᾱn qe´inτ
2 n

So then
`s ÿ ´inτ
X 1 ` λX9 “ `2s ppλ ` 1qpµ ` pλ ´ 1qp̄µ q ` ? e ppλ ` 1qαn ` pλ ´ 1qᾱn q “ 0
2 n
1´λ µ 1´λ µ
This gives pµ “ 1`λ p̄ and similarly αµ “ 1`λ ᾱn .
Further:
µ `s ÿ µ ´iπn ÿ 1 ` λ iπn ´inτ
X 1 pτ, σq|σ“π “ `2s ppµ ´ p̄µ q ` ? pαn e ´ ᾱnµ eiπn qe´inτ Ñ αnµ pe´iπn ´ e qe
2 n n
1´λ
(21)
`s
ÿ ÿ 1 ` λ iπn ´inτ
X9 µ pτ, σq|σ“π “ `2s ppµ ` p̄µ q ` ? pαnµ e´iπn ` ᾱnµ eiπn qe´inτ Ñ αnµ pe´iπn ` e qe
2 n n
1´λ

This gives:
1 ` λ iπn
p1 ` λqe´iπn ´ p1 ´ λq e “ 0 ñ sinpπnq “ 0 ñ n P Z. (22)
1´λ
The full equation is then
?
µ 2`2s pµ
µ i 2`s ÿ αnµ ´inτ
X “x ` ` e pcospσnq ` iλ sinpσnqq (23)
1 ´ λ p1 ´ λq n
nPZ´t0u

Clearly as λ Ñ 0 we recover Neumann boundary conditions. On the other hand as λ Ñ 8 we see that the
endpoint of the string is constrained to be unable to move and we indeed recover Dirichlet.

15. Looking at the DD solution:

µ ? ÿ
X 1 pτ, σq “ wµ ` 2`s αnµ e´inτ cospnσq (24)
nPZ

At the endpoints the momentum flow is

? ÿ
wµ ˘ 2`s αnµ e´inτ (25)
nPZ

16. In conformal gauge we have L “ 2T B` X µ B´ Xµ “ T2 pBτ ` Bσ qX µ pBτ ´ Bσ qX µ “ T 9 2

2 ppXq ´ pX 1 q2 q so that
9 and dσ ΠX9 ´ L “ T dσ ppXq
ş ş
Π “ T pXq 9 2 ` pX 1 q2 q ¨ Xq
9 as we needed.
2
For the closed string:

`2 ppµ ` p̄µ q `s ÿ `2s ppµ ´ p̄µ q `s ÿ

X9 “ s `? pαn e´inσ ` ᾱn einσ qe´inτ X1 “ `? pαn e´inσ ´ ᾱn einσ qe´inτ
2 2 n‰0 2 2 n‰0

Assuming no winding, we have p “ p̄. In the hamiltonian, the only contributions that will not vanish is
when each einσ is paired with e´inσ . So we can look at this mode-by-mode. Between the two of these, the
cross terms involving αn ᾱn e´2inτ will cancel. We will get:
8
T `2s ÿ 1 ÿ ÿ
ˆ ˆ 2π ˆ pαn α´n ` ᾱn ᾱ´n q ˆ 2 “ pα´n αn ` ᾱ´n ᾱn q “ pα´n αn ` ᾱ´n ᾱn q
2 2 n‰0
2 n‰0 n“1

The zero mode will contribute `4s p2 ˆ 2π ˆ T {2 “ 12 `2s p2 as required.

4
 For NN we again have p “ p̄
? ÿ ? ÿ
X9 “ 2`2s pµ ` 2`s αn cospnσqe´inτ X 1 “ ´i 2`s αn sinpnσqe´inτ
n‰0 n‰0

The zero mode gives 4`4s p2 ˆ π After squaring this, we can only pair cospnσq either with itself or cosp´nσq.
Pairing it with itself will give αn2 cos2 pnσqe´inτ which will be cancelled by the ´αn2 sin2 pnσqe´inτ obtained
from multiplying sinpnσq with itself. On the other hand, pairing cospnσq and sinpnσq with their negative
frequency counterparts and integrating gives two factors of παn α´n so that in total we get:
8
1 ÿ ÿ
`2s p2 ` α´n αn “ `2s p2 ` α´n αn (26)
2 n‰0 n“1

The exact same logic applies for DD except now only the difference term contributes. Instead of 2`2s pµ we
have wµ “ py ´ xqµ {π which must thus give zero mode px ´ yq2 {p2π`s q2 .
Lastly, for DN we have no zero-modes at all, only:
? ÿ αµ
X µ pσ, τ q “ xµ ´ 2`s k ´ikτ
e sinpkσq (27)
1
k
kPZ` 2

? ÿ µ ? ÿ
ñ X9 µ “ i 2`s αkµ e´ikτ sinpkσq, X 1 “ ´ 2`s αkµ e´ikτ cospkσq
k k

By the same reason as in DD and DN, the only terms that don’t cancel is when we pair each sinpkσq with
its negative and similarly for cos. We get
8
T 2 ÿ 1 ÿ ÿ
`s ˆ π ˆ 2 ˆ α´n αn “ α´n αn “ α´n` 1 αn´ 1 (28)
2 1
2 1 n“1
2 2
nPZ` 2 nPZ` 2

17. Immediately we have tLm , L̄n u “ 0. For tLm , Ln u we have:

1ÿ
tLm , Ln u “ tαm´k αk , αn´l αl u
4 k,l
1ÿ
“ αn´l αm´k tαk , αl u ` αm´k tαk , αn´l uαl ` αn´l tαm´k , αl uαk ` tαm´k , αn´l uαk αl
4 k,l
iÿ
“´ αn´l αm´k kδk`l ` αm´k αl kδk`n´l ` αn´l αk pm ´ kqδm´k`l ` αk αl pm ´ kqδm´k`n´l
4 k,l
iÿ
“´ αn`k αm´k k ` αm´k αn`k k ` loooooooooooooooooooooooooomoooooooooooooooooooooooooon
αn`m´k αk pm ´ kq ` αk αn`m´k pm ´ kq
4 k
kÑk`n
iÿ
“´ αm´k αn`k k ` αm´k αn`k pm ´ n ´ kq
2 k
1ÿ 1ÿ
“ ´i αm´k αn`k pm ´ nq Ñ ´ipm ´ nq αm`n´k αk “ ´ipm ´ nqLm`n
2 k 2 k1

The exact same logic applies to the conjugate charges.

5
Chapter 3: Quantization of Bosonic Strings
1. For simplicity we will ignore the µ index in our calculation first.
First consider rLm , Ln s with m ` n ‰ 0. Then expanding in terms of commutators: This is the same as
before, but now we must be careful about commutation:
1ÿ
rLm , Ln s “ r: αm´k αk :, : αn´l αl :s
4 k,l

Note that the indices m ´ k, k, n ´ l, l sum to n ` m, if any pairwise sum of them is equal to zero (necessary
for a nonvanishing commutator), then the other two will have sum equal to n`m. Then as long as m`n ‰ 0
αp , there will be no normal-ordering ambiguity and we will recover the standard commutation relations as
before.
So the remaining case to consider is n “ ´m. Take m positive WLOG. The logic of the question from last
chapter applies, but now we must be careful about the ordering of the αi outside of the commutator.
1ÿ
rLm , L´m s “ rαm´k αk , α´m´l αl s
4 k,l
1ÿ
“ αm´k αl kδk´m´l ` α´m´l αk pm ´ kqδm´k`l ` α´m´l αm´k kδk`l ` αk αl pm ´ kqδk`l (29)
4 k
1ÿ
“ αm´k αk´m k ` α´k αk pm ´ kq ` α´m`k αm´k k ` αk α´k pm ´ kq
4 k

We can split this into k ě 1 and k ď 1. The k “ 0 term is already in normal order. When k ě 1, the first,
third, and fourth terms of the sum are out of normal order. The first term has only m terms out of normal
order. Rearranging these gives the constant:
m
1 ÿ 1 mpm2 ´ 1q
kpk ´ mq “
4 k“1 4 6

The fourth term has all terms out of normal order and gives the formally infinite sum
8
ÿ
kpm ´ kq
k“1

The last term has all but the first m terms out of normal order, and so contributes the sum
8
ÿ 8
ÿ m
ÿ
p´m ` kqk “ ´ pm ´ kqk ` pk ´ mqk
k“m`1 k“1 k“1

The first part of this exactly cancels with the third term’s infinite contribution. The last part of this gives
exactly the same contribution as the first term.
ř
Now, for k ď ´1 only
ř the first two terms contribute. The first term contributes k pm ´ kqk while the second
term contributes k p´kqpm ´ kq which cancel. Thus the term left behind is exactly:
1 mpm2 ´ 1q mpm2 ´ 1q
2ˆ “ (30)
4 6 12
Note however that in fact our oscillators carry with them a µ index which we have ignored. If we incorporate
it, then each normal ordering of αiµ ανj will include a factor of η µν which would have to be summed over. This
will add in a copy of D to our final result for the normal ordering term.
Finally, we see that the normal ordering constant a must be equal to:
1ÿ i i ÿ
i 1 ÿ i i D´2 ÿ D´2
α´k αk Ñ α´k αki ` rαk , α´k s “: L0 : ` k “´ (31)
2 k k“0
2 ką0 2 24
lookmoon
ζp´1q

6
2. I believe that the treatment of the prior derivation of the central term was sufficiently careful, as I did not
need to use any zeta regularization to compute an infinite sum. I only used zeta regularization in calculating
the normal-ordering constant
3. Given that the Witt algebra is already given as an associative algebra, the commutator directly satisfies the
Jacobi identity, since pa ´ pb ´ cqq ` pb ´ pc ´ aqq ` pc ´ pa ´ bqq “ a ` b ` c “ 0. Adding a central term gives
1
rLa , rLb , Lc ss`rLb , rLc , La ss`rLc , rLa , Lb ss “ δa`b`c papa2 ´1qpb´cq`bpb2 ´1qpc´aq´cpc2 ´1qpa´bqq (32)
12
This is zero by algebra.
4. For the closed string, we have:
`s ÿ µ ´inσ
X9 µ pτ, σq “ `2s pµ ` ? pαn e ` ᾱnµ einσ qe´inτ
2 n‰0
µ `s ÿ µ ´inσ
X 1 pτ, σq “ ? pαn e ´ ᾱnµ einσ qe´inτ
2 n‰0

Taking X ` “ x` ` `s p` τ sets αn` , ᾱn` “ 0 for all n ‰ 0.

µ ? ÿ
X9 µ ` X 1 “ `2s pµ ` 2`s αnµ e´inσ e´inτ
n‰0
µ ? ÿ
X9 µ ´ X 1 “ `2s pµ ` 2`s ᾱnµ einσ e´inτ
n‰0

Let’s just look at the constraint pX9 ` X 1 q2 “ 0 and then the other constraint will give the same result for
the right-movers.
? ÿ ÿ
0 “ `4s p2 ` 2`3s p ¨ αn e´inpσ`τ q ` 2`2s αn ¨ αm e´ipn`mqpσ`τ q
n‰0 n,m‰0

The zero mode gives p2 “ 0. Noting that αn ¨ αm “ ´αn` αm ´ ´ α` α´ ` αi αi “ αi αi , we look at the

m n n m n m
remaining terms of each mode individually, so:
? ÿ i i ii
? ÿ
0 “ `s p ¨ αn ` 2 αm´n αm “ ´`s p` α´ ` `sop
lo moαon ` 2 αm´n αn

m i
m
α is transverse
? ? « ff
2 ÿ i 2 ÿ
ñ α´ “ α αi “ : i
αm´n αmi
: ´2aδn
`s p` m m´n m `s p` m
loooooomoooooon
2L0

5. Firstly, we see that L0 ´ L̄0 can only differ by an integer, otherwise there’s no combination of α´n ᾱ´m acting
on |pµ y that will give a physical state. Now lets say they differ by an integer n. Then α´n i ᾱi
´1 will be the
lowest-lying excitation at level pn ` 1, 1q. We see there are 24 of these that transform under SOp24q, so they
must give us a massless particle. We note also that we have exactly 24 excitations at levels pn ` k, kq for
1 ď k ă n, as the only way to get them is applying α´n´k i i
ᾱ´1´k . On the other hand, each of these has
mass-shell condition:
0 “ pL0 ´ aqα´n´k ᾱ´k |pµ y ñ `2s m2 “ 4pn ` k ´ aq
However if this is massless for some value of k, it will be massive for k ` 1, breaking Lorentz invariance.
Note that L ´ L̄0 generates translations along σ so this shows that any state should be invariant under
σ Ñ σ ` c.
6. Note that SOp25q acts on 25 ˆ 25 traceless symmetric tensors. Note that if we restrict to a subgroup
SOp24q that leaves one of the spatial direction fixed, the SOp25q representation breaks down into two SOp24q
representations: the symmetric tensor representation (including trace) on the 24 transverse directions, and
the vector representation in those directions as well. This is exactly what we have at level two. So, we see
we can arrange these two SOp24q rep’ns into the traceless symmetric SOp25q tensor rep.

7
 7. The generators (for the closed string) are:
ż 2π 8
ÿ
µν µ
J “T dσpX µ X9 ν ´ X ν X9 µ q “ xµ pν ´ xν pµ ´ i rα´n αnν ´ α´n
ν
αnµ ` p. . . qs
0 n“1

Upon computing the commutator rJ µν , J ρσ s the xµ pν ´ xν pµ will give no problems, and there will be no
cross terms between the right and left moving modes. So it is enough to look at the left movers. I’m gonna
pass on doing this computation...
8. For NN boundary conditions, αkµ is associated to the wavefunction cospkσq, σ P r0, πs. This has eigenvalue 1
under flip if k is even and ´1 if k is odd. Thus this αk must transform identically: Ωαkµ Ω´1 “ p´1qk αk . For
DD boundary conditions, we have sinpkσq, which has opposite eigenvalues, so instead we get p´1qk`1
9. This is a Lie algebra of dimension npn ´ 1q{2, which already looks promising. In the case of all θi equal, we
can pick basis so that the Rij are all 1. This is clearly sopnq. Now, take a diagonal unitary matrix γ (note
γ T “ γ). It clear that λ̃ij :“ γ 1{2 λij γ ´1{2 gives the right structure under transposition:

λ̃T “ γ ´1{2 λT γ 1{2 “ ´γ ´1{2 λγ 1{2 “ ´γ λ̃γ

But since λ̃ij is just a conjugation action on the λij , we will still have that the Lie algebra structure is
preserved, and maintain sopnq.
For the second part, again when all the θi “ 0, this is just the definition of the symplectic group and we have
λ “ ´ωλT ω ´1 “ ωλT ω for ω the canonical symplectic written in the px1 , p1 , x2 , p2 , . . . q basis. Now note that
the new symplectic form γ can be written as σ 1{2 ωσ ´1{2 with σ “ diagpeiθ1 , eiθ1 , eiθ2 , eiθ2 , . . . q. Then define
λ̃ “ σ ´1{2 λσ 1{2 and note that

λ̃T “ σ 1{2 λT σ ´1{2 “ ´σ 1{2 ωλωσ ´1{2 “ ´γ λ̃γ

as required. Again, conjugation action will preserve the Lie algebra structure, so this will remain spp2nq.
10. In the symmetric case, we have λT “ λ, so these are symmetric matrices of N indices. Naturally SOpN q
acts on these, and we see that they can be written as F b F for F the fundamental representation. This can
be decomposed as the trivial representation and the traceless symmetric representation.
In the anti-symmetric case with N even, I know that the symplectic group acts on RN . I’ll call this the
fundamental rep, and then note that tensoring it with its dual again gives an antisymmetric N ˆ N matrix
on which SppN q can act. This can be decomposed into the singlet and the skew-traceless antisymmetric
matrix.
11. Traceless means that any pair of indices contracted with η µν gives zero. Locally, we can pick the metric so
that only η`´ “ η´` “ 1 is nonzero. This means that Ti1 ...in “ 0 if any one i is set to ` with the other set
to ´. Thus we can have only T`¨¨¨` and T´¨¨¨´ nonzero.
12. The round metric is
4dzdz̄
ds2 “
p1 ` z z̄q2
The Lie derivative is:
LX gab “ X c Bc gab ` gac Bb X c ` gcb Ba X c (33)
Working with z, z̄ we get:
LX gzz “ 2gz z̄ Bz X z̄ “ 0
LX gzz “ 2gz z̄ Bz X z̄ “ 0 (34)
z z̄
LX gz z̄ “ pX Bz ` X Bz̄ qgz z̄ “ λpz, z̄qgz z̄
The first two equation shows us that X z , X z̄ must be holomorphic and anti-holomorphic respectively. We
want the function λ to be well-defined on the entire Riemann sphere and so the last equation gives us:
pX z z̄ ` X z̄ zq
´2 “ λpz, z̄q (35)
1 ` z z̄

8
 We see that X z , X z̄ cannot have any poles. Further, they cannot grow faster than z 2 , z̄ 2 respectively as
z Ñ 8 otherwise λ will blow up at the north pole. So our solutions space is spanned by Bz , zBz , z 2 Bz and
their conjugates.
Next, right away we can see that the only nonzero Christoffel symbols in the round metric are Γzzz and Γz̄z̄ z̄ .
Second, because T is traceless, by the previous problem we see it has only two components: Tzz and Tz̄ z̄ .
Now looking at ∇β Tαβ we see that this gives two equations:
g z z̄ ∇z̄ Tzz “ Bz̄ Tzz
(36)
g z z̄ ∇z Tz̄ z̄ “ Bz Tz̄ z̄
Note that there can be no Christoffel contribution. This simply asks for globally-defined holomorphic 2-
forms. Let’s look at Tzz . Around z “ 0, it must be a polynomial to avoid poles. Transforming to w “
dz dz 2
1{z, dw “ ´dz{z 2 ñ dw “ ´z 2 “ ´w´2 we get Tww pwq “ p dw q Tzz pwq. Note that the right hand side will
´2
only have poles at least as bad as w so we cannot have any global section of this vector bundle. Thus,
there are no Teichmuller parameters.
13. We can think of the torus as C{Λ. Note that scaling and rotation preserve the complex structure of the
fundamental parallelogram so WLOG we can pick Λ “ Z-spant1, τ u with τ P H. Thus we need vector fields
on C that respect the translation-invariance under Λ. Any translation-invariant holomorphic function is
zero, we can only have the constant vector fields Bz , Bz̄ .

We now look for holomorphic and anti-holomorphic traceless tensors. Again, Tzz and Tz̄ z̄ be translation-
invariant w.r.t the lattice, so again they must be constants. We get dz bdz and dz̄ bdz̄ as our two Teichmuller
deformations. As real tensors these are:
ˆ ˙ ˆ ˙
1 0 0 1
“ dx b dx ´ dy b dy, “ 2dx b dy,
0 ´1 1 0

14. Not sure exactly how they want us to calculate this. Let’s assume they are OK with Gauss-Bonnet. For the
disk with the flat metric, we have right away that the curvature R vanishes. The geodesic curvature at the
boundary is a constant, and is easily seen to be 1. Integrating this over the boundary of the disk gives 2π so
that χ “ 1.
Using the round metric, it is quick to see that the only contribution to Rµν “ Rzµzν ` Rz̄µz̄ν is for Rzz z
z̄ “
z z̄ z̄ 2
´Bz̄ Γzz and Rz̄ z̄z “ Bz Γz̄ z̄ giving Rz z̄ “ p1`|z|2 q . Tracing this gives R “ 1. Integrating this over half the
sphere gives 2π. The geodesic curvature vanishes on the great circle by symmetry, and we get χ “ 1 again.
15. Any such surface can be decomposed as a sphere with 2n holes connected to n handles. Let’s integrate the
scalar curvature over each piece individually. First the curvature integrated on the sphere with 2n disks
removed is equal to the curvature integrated on the Riemann sphere: 4π minus the curvature integrated on
2n disks. We have just done this in the previous problem, and we get 2π ˆ 2n. Lastly, the curvature on the
handles is just the same as the curvature on the sphere with two holes cut out, which we have just calculated
is 4π ´ 2 ˆ 2π “ 0. Thus the total curvature is just 2πp2 ´ 2nq, giving us χ “ 2 ´ 2n as required.

9
 ş
16. Our point particle action is S0 “ dτ epe´2 pBτ xq2 ´ m2 q. Let’s look at:
ż ż
DXDe ´S0 ş ş
e „ DXDeDbDc e´S0 ´ bpδF qc´i BF peq
Vgauge

Note we don’t need an α index on B, b, c because they just parameterize the continuous symmetry with no
discrete parameters:
pδτ1 τ qpτ q “ δpτ ´ τ1 q
Using Polchinski’s convention for coordinate transformation (he also has ´BA of Kiritsis),

δτ1 Xpτ q “ ´δpτ ´ τ1 qBτ X, δτ1 epτ q “ ´Bτ pδpτ ´ τ1 qepτ qq

and

rδτ1 δτ2 spτ q “ ´pδpτ ´τ1 qBτ δpτ ´τ2 q´δpτ ´τ2 qBτ δpτ ´τ1 qq Bτ ñ fττ13τ2 “ δpτ3 ´τ1 qBτ3 δpτ3 ´τ2 q´δpτ3 ´τ2 qBτ3 δpτ3 ´τ1 q

Then the BRST transformation is given by:

δ X “ icX9
9
δ e “ ipcXq
δ b “ BA (37)
δ c “ icc9
δ B “ 0

Now let’s take F peq “ e ´ 1. Then we get a ghost action

ż
α α
bA c δα F Ñ dτ1 dτ2 bpτ1 qcpτ2 qδτ2 p1 ´ epτ2 qq
ż ż ż
9
“ dτ1 bpτ1 qBτ1 dτ2 cpτ2 qrδpτ1 ´ τ2 qepτ1 qs “ ´ dτ bc

We enforce this constraint by integrating over B. Now we will have δ e “ 0 and δ b “ ipT X ` T gh q. Because
we are in Euclidean signature, we have p “ Bt X “ iX9 and similarly p “ ´ic (´i because the real time
9
Lagrangian has term ´ibc). Then the BRST current (equal to charge because we’re in 1D) is:
1 1 9 “ ´c 1 X9 ` 1 m2 “ c 1 pp2 ` m2 q “ cH.
QB “ pX δB X ` pb δB b “ ´cX9 2 ´ icip´ X9 ` m2 ´ bcq
2 2 2 2 2
Clearly Q2B “ 0. As before, the ghosts generate a two-state system. Our set of states is given by
|k µ , Òy , |k µ , Óy. Following convention, c raises and b lowers. QB |k, Óy “ 12 pk 2 ` m2 q |k, Òy so all states of
the form |k, Òy with k 2 ` m2 ‰ 0 are BRST exact. Similarly all states |k, Òy are BRST closed along with
all states of the form |k, Óy with k 2 ` m2 “ 0. So the closed states that are not exact are |k, Óy, |k, Òy with
k 2 ` m2 “ 0. We take only the states with b |ψy “ 0. The reason is that all states |k, Òy are physical, and
so we would need amplitudes between such states to be proportional to δpk 2 ` m2 q in order for the states
to decouple, but amplitudes cannot have such extreme singularities Don’t understand this. Appreciate
it, and its relation to Siegel gauge..

17. I believe these variations have Kiritsis taking c Ñ ´ic in his formalism. They also follow directly from
Polchinski’s formalism. Under a diffeomorphism δξ,ξ̄ X “ ´ξBX ´ ξ¯B̄X. These are two copies of the repa-
rameterization algebra developed in the previous problem, and so the commutation relations are the same.
We get, again in Polchinski’s formalism (37)
¯
δ X “ ipcBX ` c̄dXq
δ c “ ipcBc ` c̄B̄c̄q (38)
X gh
δ b “ ipT `T q

10
 I see no problem with Q2B giving zero when acting on the X and c fields.
¯ “ ip
δB pcBX ` c̄dXq cpBcqBX

c̄pB̄c̄q
 `   B̄X ´ cBp


cBX
 ` c̄B̄c̄Xq ´ c̄Bp
 cBX
 ` c̄B̄c̄Xqq


and the remaining terms die by the equations of motion. The c variation will always die because we’ve
already shown the transformations satisfy a Lie algebra with Bianchi identity.
It looks like the b field will be nontrivial. If one of the equations of motion is the T X ` T gh “ 0 then this
will be zero right away. Otherwise, we want to compute (WLOG in the holomorphic sector):
1 2
δB pT X `T gh q “ δB p pBXq2 `2bBc`Bbcq “ ir BXBpcBXq`2pT X `T gh qBc`2b BpcBcq`BpT X `T gh qBc`Bb cBcs
α α
The purely bc terms cancel. I’m left with
2
rpBXq2 Bc ` BXB 2 Xcs
α
I don’t know how to get rid of this. I can write it as a total derivative less something proportional to pBXq2 Bc
and perhaps note that this is just Bc times the stress tensor, which perhaps vanishes classically? At any rate,
there is no need to use d “ 26 here.

18. Integrating over σ will as usual pick out the zero mode. For T`` this gives us
ÿ ÿ
p2nb´n`m cn ` n ` mc´n bn`m q “ pm ´ nq : bn`m c´n :
n n

and similarly for the right-movers. To get the central charge we’ll have to proceed as before, noting that
only rLm , L´m s can give a nonzero central term. As before, we expect only a finite part of the infinite sum
to contribute to this. We thus take out the only terms of the sum with m, n having the same sign:
m
ÿ m
ÿ
pm ` nqbm´n cn , p´2m ` nqb´n c´m`n
n“1 n“1

Then our commutators of these finite terms give:

ÿ ÿ
pm ` kqp´2m ` k 1 qrbm´k ck , b´k1 c´m`k1 s “ pm ` kqp´2m ` k 1 qpbm´k cm`k1 ´ b´k1 ck qδk´k1
k,k1 k,k1
ÿm
“ pm ` kqp´2m ` kqpbm´k cm`k ´ b´k ck q
k“1

Looking at the non-normal-ordered part, this leaves:

m m
ÿ ÿ 1
bm´k c´m`k pm ` kqpk ´ 2mq Ñ pm ` kqpk ´ 2mq “ pm ´ 13m3 q
k“1 k“1
6

19. We have
BL BL 2 1 1
jB “ cBX ` cBc “ 1
pBXq2 c ` bcBc Ñ cT X ` bcBc “ cT X ` cT gh
BpB̄Xq BpB̄cq 2πα π 2

I don’t understand how other references include a 32 B 2 c.

20. Let’s do this for the open string, so we are then just calculating the holomorphic sector. We have:
8
ÿ 1 gh
QB “ : pLX
´m ` L´m ´ aδm,0 qcm :
m“´8
2

Note that the a is just from the X component of the theory since by definition Q contains the term : cT gh :
already in normal order.

11
 We now need to consider the total BRST charge Q ` Q̄. Then:
ÿ gh
Q2B “ prLX X gh gh X
m , Ln s ` rLm , Ln s ` pm ´ nqLm`n ` pm ´ nqLm`n ` 2amδm`n qc´m cm
n,m

This will vanish only if the commutators give no anomalous term. From previous exercises we see this is
only if:
dpm3 ´ mq pm ´ 13m3 q
` ` 2am “ 0
12 6
This happens exactly when d “ 26 and a “ 1.
21. We now have Q ` Q̄ that we need to be zero on states. Again, each of Q, Q̄ will no change the level, so their
sum will not either, and we have a (double) grading on the space which they will preserve.
Q “ Q0 ` Q1 , Q0 “ c0 pLX
0 ´ 1q, Q1 “ c´1 LX X
1 ` c1 L´1 ` c0 pb´1 c1 ` c´1 b1 q

and Q̄ is the conjugate of this.

At level zero we will again have Q0 “ c0 LX X
0 ` c̄0 L̄0 . We now have two copies of the Clifford algebra and our
Siegel gauge condition will make it so that we only consider states |Ó, Ó̄, p, p̄y. Now we need ppL0 ´ 1qc0 `
2 2 2 2
pL̄0 ´ 1qc̄0 q |Ó, Ó̄, p, p̄y “ pL0 ´ 1q |Ò, Ó̄, p, p̄y ` pL̄0 ´ 1q |Ó, Ò̄, p, p̄y so we need `s4p “ `s4p̄ “ 1 i.e. the total mass
is m2 “ ´4. We thus have the tachyon state.
Also because b0 |ψy “ 0 for any physical state, we will have tQ, b0 u |ψy “ 0 “ pL0 ´ aq |ψy so we have
L0 ´ 1 “ L̄0 ´ 1 “ 0 and this gives us the level-matching condition. So the next level we can have a state is
at p1, 1q.
As in the open string treatment, the most general such state has nine terms:
|ψ1 y “ pζ ¨ α´1 ζ̄ ¨ ᾱ´1 ` ζab ¨ α´1 b̄´1 ` ζac ¨ α´1 c̄´1
` b´1 ζba ¨ ᾱ´1 ` ξbb b´1 b̄´1 ` ξbc b´1 c̄´1
` c´1 ζca ¨ ᾱ´1 ` ξcb c´1 b̄´1 ` ξcc c´1 c̄´1 q |Ó, Ó̄, py
α20 ᾱ20
Let’s act on this with Q0 ` Q1 . First lets look at Q0 . On the αᾱ term it will give eigenvalue c0 2 ` c̄0 2
α2 ᾱ2
while something like the ζab αb̄ term it will give eigenvalue c0 20 ` c̄0 p 20 ´ 1q. This will be compensated by
the action of the c̄0 pb̄´1 c̄1 q (from Q1 ) on ζab α´1 b´1 . The exact same argument can be applied to any of
those four terms - there will always be one the four bc terms of Q1 ` Q̄1 that will give us the extra factor of
1 from its commutation relation with that term in |ψ1 y (it commutes with everything else).
2 2
So we get a term c0 `s4p |ψ1 y which then gives the p2 “ 0 constraint. The remaining term comes from the
c´1 LX X
1 ` c1 L´1 ` c.c. action. The c´1 L1 ` c1 L´1 will each annihilate everything except six terms, giving

ζ ¨ p c´1 ζ̄ ¨ ᾱ´1 ` ζab ¨ p c´1 b̄´1 ` ζac ¨ p c´1 c̄´1

(39)
` p ¨ α´1 ζba ¨ ᾱ´1 ` ξbb p ¨ α´1 b̄´1 ` ξbc p ¨ α´1 c̄´1 ` c.c.
and the conjugate of this will contribute the conjugate terms. For this to all be zero we need each of the
ζi ¨ p “ 0 as well as their conjugates. We also need ξbb “ ξbc “ ξcb “ 0. We also see that ζba “ ζab “ 0.
On the other hand the general form of an exact state is also given by (39) for the ζi and ξi arbitrary. Thus
all the terms involving c and/or c̄ are exact and so upon quotienting we get ζac “ ζbc “ ξcc “ 0. Lastly we
get the relation that we should identify ζi ζ̄j “ ζi ζ̄j ` pi ζj1 ` ζi1 p ie we project out any tensor of the form p b ζ 1
or ζ 1 b p. This is equivalent to identifying ζ – ζ 1 ` ξp and identically for ζ̄.
So we have eliminated everything except for ζ, ζ̄, each of which must be transverse to p and we identify ζ
differing by a longitudinal p component. This is 24 ˆ 24 parameters, as required.
22. If I have the Clifford algebra C`p2q, any vector v will have an orbit generated by 1, b0 , c0 , b0 c0 , so there can
be no irreducible representation of dimension greater than 4. Further, there is a vector v0 annihilated by
b. Consider v1 “ cv0 and assume it is distinct. Now bv1 “ bcv0 “ v0 ´ cbv0 “ v0 . So v1 and v0 span the
irreducible representation meaning that any irrep in fact has dimension 2. Thus, any higher dimensional
generalization would only be (probably direct or semidirect) extensions of this and the trivial irrep, and give
us no new information.

12
Chapter 4: Conformal Field Theory
1. We’ll do this directly. First observe:
d
|t“0 e´itPµ f pxq “ ´Bµ f
dt
d it µν
|t“0 e´ 2 ω Jµν f pxq “ ´ωνµ xν Bµ
dt (40)
d
|t“0 e´itD f pxq “ x ¨ Bf pxq annoying that there is no -
dt
d
|t“0 e´itKµ f pxq “ ´px2 Bµ ´ 2xµ px ¨ δqqf pxq
dt
x `x a µ 2 µ
The last one is exactly the first-order expansion of 1`2a¨x`a2 x2 . Note the dilatation and special conformal
generators are the negative of Di Francesco’s (SO ANNOYING OGM).
Now let’s do the commutator
rJµν , Pρ s “ ´Bρ pxµ Bν ´ Bν Bµ q “ ´pηµρ Bν ´ ηνρ Bµ q “ ´ipηµρ Bν ´ ηνρ Bµ q
rPµ , Kν s “ ´Bµ px2 Bν ´ 2xν x ¨ Bq “ ´p2xµ Bν ´ 2ηµν xλ Bλ ´ 2xν δµλ Bλ q “ 2iJµν ´ 2iηµν D
rJµν , Jρσ s “ ´ipηµρ Jνσ ´ ηµσ Jνρ ´ ηνρ Jµσ ` ηνσ Jµρ q Ð Everyone has done this one like 20 times
rJµν , Kρ s “ ´ipηµρ Kν ´ ηνρ Kµ q
rD, Kµ s “ xν ¨ Bν rx2 Bµ ´ 2xµ pxλ Bλ qs ´ rx2 Bµ ´ 2xµ x ¨ Bsxλ Bλ
“ 2
xν xν Bµ ´ 2xν ηµν px ¨ Bq ´ 
2x x2

µ px
 ¨ Bq ´  B 2x
µ ` µx B “ iKµ
¨

rD, Pµ s “ ´Bµ xλ Bλ “ ´Bµ “ ´iPµ

rJµν , Ds “ 0

The way we did the rJ, Ks commutator is by noting it should look the same as rJ, P s, since P is just
translation about the point at 8. The rJ, Ds commutator follows because rotation is scale invariant.

2. We see immediately that the Jµν can be mapped to the Mµν corresponding to a SOpp, qq subgroup of
SOpp ` 1, q ` 1q. The full group has:

rMµν , Mρσ s “ ´ipηµρ Mνσ ´ ηµσ Mνρ ´ ηνρ Mµσ ` ηνσ Mµρ q (41)

Note the commutation relations of J with P and K gives us:

ˆ ˙
1 1 1
rJµν , pKρ ˘ Pρ qs “ ´i ηµρ pK ˘ P qν ´ ηνρ pK ˘ P qµ
2 2 2

Writing these as Mρ,d`1 and Mρ,d respectively, we see that we get the second and fourth terms nonzero and
we get exactly (41). Note at this stage I didn’t need to do such linear combinations of K and P . That is
important for appreciating that we want:

rMµd , Mνd`1 s “ ´iηµν Mdd`1 “ ´iηµν Md,d`1 “ iηµν D

and we get exactly this:

1 1
rpK ´ P qµ , pK ` P qν s “ prKµ , Pν s ´ rPµ , Kν sq “ iηµν D
4 4
We needed that combination so that Jµν wouldn’t appear. As required rJµν , Ds “ rMµν , Md,d`1 s “ 0 for
µ P 0 . . . d ´ 1. I’m getting the wrong sign. Perhaps our friend’s convention is off.

13
3. This comes from noting that for f “ z ` pzq
ˆ ˙∆ ˆ ¯
˙∆
Bf Bf ¯ ¯ B̄
´ 1 “ p1 ` Bq∆ p1 ` B̄q∆ ´ 1 “ ∆B ` ∆
Bz Bz̄
ñ Φpzqp1 ´ p∆B ` ∆ ¯ B̄qq “ Φ1 pf pzq, f¯pz̄qq “ p1 ` B ` ¯B̄qΦ1 pzq
¯ B̄ ` B ` ¯B̄qqΦpzq “ Φ1 pzq
ñ p1 ´ p∆B ` ∆
¯ B̄ ` ¯B̄qΦpzq
ñ Φpzq ´ Φ1 pzq “ p∆B ` B ` ∆

How weird... think about this in terms of active/passive. Contrast with Di Francesco.

4. As in the 2-point greens function case, note that:

˜ ¸
N
ÿ
δ GN “ 0 ñ pzi qBzi ` ∆i Bpzi q ` c.c. GN “ 0
i“1

We can WLOG look at just the holomorphic sector (set ¯ “ ř 0) Now first set pzq “ 1. This directly gives
N N 2
ř
i G “ 0, as we wanted. Next, take pzq “ z. This gives
i Bř i pzi Bi ` ∆i qG “ 0. Finally, take  “ z to
2 N
get i pzi Bi ` 2zi ∆i qG “ 0 as desired. Note in all these cases, we are exactly performing the global SLp2q
transformations, so these Ward identities will always hold.

5. The first Ward identity tells us that the function can only depend on z12 , z23 . Then the next two can be
written as:
px1 B1 ` x2 B2 ` x3 B3 ` Σ∆i qf px12 , x23 q “ ppx1 ´ x2 qB12 ` px2 ´ x3 qB23 ` Σ∆i q f “ 0
px21 B1` x22 B2 ` x23 B3 ` Σ2xi ∆i qf px12 , x23 q “ px21 ´ x22 qB12 ` px22 ´ x23 qB23 ` Σ2xi ∆i f “ 0
` ˘

We can subtract out B23 to get the differential equation:

ˆ ˙ ÿ ˆ 2xi ˙
x1 ` x2
0“ ´ 1 x12 B12 ` ´ 1 ∆i Ñ px12 ` x23 qx12 B12 ` px12 ` x12 ` x23 q∆1 ` x23 p∆2 ´ ∆3 q
x2 ` x3 i
x2 ` x3
ñ 0 “ x212 B12 ` x23 x12 B12 ` 2x12 ∆12 ` x23 p∆1 ` ∆2 ´ ∆3 qf
` ˘

Now write f px12 , x23 q “ eg pu, x23 q with u “ log x1 2. This substitution gives the ODE:
ż log x12
u 1 u 2∆1 eu ´ x23 p∆1 ` ∆2 ´ ∆3 q
pe ` x23 qg puq ` 2∆1 e ` x23 p∆1 ` ∆2 ´ ∆3 q “ 0 ñ gpuq “ du
´8 eu ` x23

This integral can be done and gives:

C C
“
x∆
12
1 `∆2 ´∆3
px12 ` x23 q∆1 `∆3 ´∆2 x∆
12
1 `∆2 ´∆3 ∆1 `∆3 ´∆2
x13

We can do the same for B23 and get the general form:
λ123
∆1 `∆2 ´∆3 ∆1 `∆3 ´∆2 ∆2 `∆3 ´∆1
ˆ c.c.
x12 x13 x23

for λ123 , λ̄123 undetermined constants (call their product C123 ).

6. Again specialize to the holomorphic part. We see GN depends only on relative positions x12 , x13 , x14 . We
can WLOG take Gp4q to have the form:
f pz1 , z2 , z3 , z4 q
Gp4q pz1 , z2 , z3 , z4 q “ ∆12 ∆13 ∆14 ∆23 ∆24 ∆34
z12 z13 z14 z23 z24 z34

Here, because f is arbitrary, we have not made any assumptions. The Ward identities imply the following:

14
 • f depends only on the relative positions zij
ř ř ř
• iăj ∆ij “ ∆ with ∆ “ i ∆i and i zi Bi f “ 0
• ∆23 `
ř∆24 ` ∆34 “ 2∆1 , ∆13 ` ∆14 ` ∆34 “ 2∆2 , ∆12 ` ∆14 ` ∆24 “ 2∆3 , ∆12 ` ∆13 ` ∆23 “ 2∆4
and i zi2 Bi f “ 0

These give 4 constraints for the 6 ∆ij , so the system is underdetermined. The most symmetric solution is
given by:
1
∆ij “ ∆i ` ∆j ´ ∆
3
It remains to find the general form of f .

• The first ward identity gives us that it can only depend on the zi through zij .
• Further, it must transform trivially under dilatation, so we see that it can only depend on ratios of the
zij with an equal number of each zij in the numerator and denominator.
ij z zij
• Under special conformal transformations, each such ratio will transform as zkl Ñ zkl pzi ` zj ´ zk ´ zl q,
and more generally ź zi j ź zi j ÿ
a a a a
Ñ ˆ pzia ` zja ´ zka ´ zla q
a
zka la a
zk a l a a

The third Ward identity shows that f must transform trivially under these, and so f can only depend
on ratios where each zi appears an equal number of times in the numerator and denominator.

In total: we need ratios of zij with an equal number of zij in the numerator and denominator, and each zi
appears the same number of times in the numerator and denominator. All such ratios can be obtained as
rational functions of:
z12 z34 z14 z24
x :“ , y :“
z13 z24 z13 z24
But we see that y “ 1´x so in fact the most general such function is any function of x alone, as was required.

7. With conformal invariance (rescaling in particular), an infinite cylinder has no moduli, so you can set its
radius to be whatever you like and get the same theory.

8. Let’s perform the OPE within the correlator:

ÿ ∆ ´∆i ´∆j ∆¯ ´∆¯ ´∆
¯
xΦi pz1 qΦj pz2 qΦk pz3 qy “ z12` z̄12` i j Cij` xΦ` pz2 qΦk pz3 qy
`

By the orthonormality assumption of the OPE, we then have

δ`k Cijk pz12 q

xΦ` pz2 qΦk pz3 qy “ 2∆k 2∆¯ ñ xΦi pz1 qΦj pz2 qΦk pz3 qy “ ¯ k ∆i `∆j ´∆k ∆ ¯ `∆¯ ´∆
¯
2∆k 2∆
z23 z̄23 k z23 z̄23 z12 z̄12i j k

9. We assume that µ ! 1{r. The integral is in fact real, and we can approximate it by
1 2 ż 2π
p dp e´ 2 ppr cospθqq
ż8 ż8
du e´u
ż ż ż
2 cosppr cospθqq 1 1
d p “ dθ “ dθ “ dθΓp0, µ̃2 r2 cos2 pθqq
p2 ` m2 0 p2 ` µ 2 2 1 2 2
µ r cos2 pθq u 2 0
2

It is known that Γp0, q “ ´γ ´ log  so up to a constant (that can be absorbed into the redefinition of µ)
we get;
`2 1 `2
´ s p2πq logpµ2 |x ´ y|2 q “ ´ s logpµ2 |x ´ y|2 q
2π 2 2
10. By Stokes’ theorem its clear. Let Ω be any disk enclosing the origin:
ż ż ¿ ¿
dz
d2 z B̄B log |z|2 “ i dz ^ dz̄ B̄B log |z|2 “ ´i dzB log |z|2 “ ´i “ 2π
Ω Ω z
BΩ BΩ

15
 Alternatively we could put in a regulator and evaluate this directly:

µ2
ż ż ż
2 ¯ 2 2 2 z 2
d zB d logp|z| ` µ q “ d zB 2 “ d z
Ω Ω |z| ` µ2 Ω p|z|2 ` µ2 q2

As µ Ñ 0 this approaches 0 everywhere except for the origin. Taking |z| “ r and integrating in polar
coordinates (note d2 z “ 2dxdy “ 2rdrdθ):

µ2 r
ż8
2π ˆ 2 ˆ 2 2 2
“ 2π
0 pr ` µ q

as required.

11. We have:
ż ˆ ˙
1 2 ? ab 4π δS 1 1 c
d ξ ´g g Ba X Bb X ñ Tab “ ´? ´ 2 Ba XBb X ´ gab Bc XB X
4π`2s ´g δg ab `s 2

This is clearly traceless. Let’s specialize to the holomorphic sector to get T pzq “ ´ `12 : BXBX : and of course
s
this is the non-singular part of the BXpzqBXpwq OPE as z Ñ w.

12. The scaling dimensions of conserved currents don’t change.

1
ű
For a current to be conserved, we must that the surface operator 2πi dzJpzq is topological (independent of
contour). Applying dilatation z Ñ z{λ on this does not change the operator, so long as it does not pass any
operator insertions. So we have:
¿ ¿
1 1 z
dzJpzq ` c.c. “ d J 1 pz{λ, z̄{λq ` c.c.
2πi 2πi λ

And thus we get Jpz, z̄q “ λ´1 J 1 pz{λ, z̄{λq, and we get J has scaling dimension 1.
On the other hand for T µν , we have the conserved charge:
¿
Pν “ dnµ Tµν

Applying dilatation, we see from exponentiating the commutation relation for rD, Pν s that Pν “ Pν1 {λ so
1 dnµ 1
¿ ¿
Pν “ dnµ Tµν ` c.c. “ T pz{λ, z̄{λq ` c.c. “ Pν1 {λ
λ µν
λ loooooooooooooooomoooooooooooooooon
“Pν1 q

giving us that
Tµν pz, z̄q “ λ´2 Tµν
1
pz{λ, z̄{λq
so T properly has scaling dimension 2.

13.
N ż 1
ş ş
d2 zBX B̄X`i d2 z Xpzq i pi δ 2 pz´zi q
ř
ź
ipXpz,z̄q ´
x e y“ DXe 2π`2
s

n“1
1
ş 1 řN
d2 σd2 σ 1 JpσqJpσ 1 qGpσ,σ 1 q pi pj xXpzi qXpzj qy
“ 2πδpΣpi qe´ 2 “ 2πδpΣpi qe´ 2 i,j“1

Appreciate both the UV divergence (from coincident points in the correlator) and the IR divergence (from
the correlator going as a logarithm) will cancel (think Kosterlitz-Thouless/Mermin Wagner stuff here):
`2 2 `2
s p2i
ř
µ2 4 pΣpi q 2 4

Momentum conservation removes the IR, and if we normal-order the vertex operators within the product we
will not get the UV divergence.

16
14. By explicit calculation:

´3αpBXq2 pwq 4pBXq4

T pzqrpBXq4 spwq „ ` ` ...
pz ´ wq4 pz ´ wq2
´2α 4pBXB 2 Xqpwq 4pB 2 Xq2 pwq
T pzqrpB 2 Xq2 spwq „ ` `
pz ´ wq6 pz ´ wq3 pz ´ wq2
´3α 6pBXq pwq 6pBXB 2 Xqpwq 6pB 2 Xq2 pwq
2
T pzqrB 3 XBXspwq „ ` ` ` ` ...
pz ´ wq6 pz ´ wq4 pz ´ wq3 pz ´ wq2

where ` . . . are terms that are Oppz ´ wq´1 q or higher powers, which will not affect the non-primary terms.
We see that the combination:
α 3 2 2
pBXq4 ` B 3 XBX ´ pB Xq
2 4α
gives a primary operator of dimension 4. Along the way I noticed that there are no primary operators of
dimension 2 or 3 that are finite sums of products of derivatives of BX.
I can’t help but think that this might have something to do with the Schwarzian.

15. We look at:

? ? 8
2 ipXpwq 2 ÿ pipqn
:i BXpzq :: e :“i : BXpzq :: pXpwqqn : ` finite
`s `s n“0 n!
? 8
2 ÿ `2 n pipq2 : Xpwqn´1 : `s p 1
“i p´q s “? Vp pwq ` finite
`s n“0 2 z´w n! 2z´w

16. Directly:
ÿ piaqn pibqm
: X n pzq :: X m pwq :
n,m
n! m!

First lets look at when n “ m and say we contract everything. Then we need to contract all n Xpzq with
2
all n Xpwq. There are n! ways to do this, and each produces a factor of ´ `2 log |z ´ w|2 . The diagonal
components thus give the sum:
ÿ 1 ˆ ab`2 ˙n
s 2 2
log |z ´ w| “ |z ´ w|ab`s {2
n
n! 2

Now a more general term, say : Xpzqn :: Xpwqm : where we want to contract k ă n, m of them we must
choose k Xpzq and kXpwq to contract` ˘ the` Xpzq with and then figure out the order to contract those k
amongst themselves (k!), so we have nk ˆ m
˘ n!m!
k ˆ k! “ pm´kq!pn´kq!k! ways to do this. The contraction again
piaqn´k pibqm´k
gives the logk term as before, and now we have a remaining factor of pn´kq!pm´kq! : Xpzqn´k :: Xpwqm´k :.
For each k-contracted set which gives the logk term, we should therefore multiply it by:
8
ÿ piaqn´k pibqm´k
: Xpzqn´k :: Xpwqm´k :“ eiaXpzq`ibXpwq
m,n“k
pn ´ kq!pm ´ kq!

So the OPE is:

2 {2
: eiaXpzq :: eiaXpwq :“ |z ´ w|ab` eiaXpzq`ibXpwq

17. Directly: ˆ ˙
1 6
Bz JpzqBw Jpwq “ Bz Bw “´ ` finite
pz ´ wq2 pw ´ zq4
2
We have no pz´wq2
term, as would otherwise be required

17
18. The stress energy tensor is:
1 1 ´1 1 Bψpzq 1 1 Bψpwq
T pzq “ ´ : ψpzqBψpzq :ñ T pzqψpwq “ ´ ψpzqp q` “ ψpwq `
2 2 pz ´ wq2 2z´w 2 pz ´ wq2 pz ´ wq

so this shows that ψ is primary with weight 1{2.

19. I’ll instead have the notation w “ g ˝ f pzq. For T pzq “ pf 1 q2 T pf q ` tf, zu consider h “ g ˝ f . Then we have:

T pzq “ pf 1 q2 T pf q ` Cpf q “ pf 1 q2 ppg 1 q2 T pg ˝ f q ` Cpg ˝ f qq ` Cpf q “ ph1 q2 T phq ` pf 1 q2 Cpgq ` Cpf q

So we get the desired cocycle property:

Cphq “ pf 1 q2 Cpgq ` Cpf q

Now, we need Cpf q to have units of rzs´2 . The most naive guess is to let Cphq “ h2 , but this gives:

h2 “ pf 1 q2 g 2 ` f 2 g 1

If that last factor of g 1 were not there, we would be done. Instead we must think more deeply. We also need
the Schwarzian to include a term linear in the third derivative, and the only such term is a constant times
f 3 {f 1 . Let us look at how this transforms:

h3 1 2g
3 f 2g2 f 3
“ pf q ` 3 ` 1
h1 g1 g1 f

Now what stops us is the cross-term. The only terms that we can add to f 3 {f 1 that involve less than third
derivatives in  are f 2 , pf 1 q2 pf 2 {f 1 q2 .
There is one last term we could have built out of terms of order ď 3 that would give units of rzs´2 : pf 3 {f 2 q2 ,
however in the limit of an infinitesimal transformation z ` pzq, this would give p3 {2 q2 which is nonlinear
in , so this term cannot contribute. .
ph1 q2 “ pf 1 g 1 q2 has none of the properties we’d like, and adding it would break the term that pf 1 q2 multiplies
being proportional to Cpgq. Similarly, ´ adding
¯ f 2 would break the term that pf 1 q2 doesn’t multiply being
2
proportional to Cpf q. What is left is ff 1 . This transforms as:
ˆ ˙2 ˆ ˙2 ˆ ˙2
h2 1 2 g2 f2 2f 2 g 2
“ pf q ` `
h1 g1 f1 g1

The cross term is exactly of the form of the cross term in f 3 {f 1 , and so by appropriately subtracting:
˜ ˆ 2 ˙2 ¸
h3 3 h2 2 f3 3 f2 2
ˆ ˙ 3
ˆ ˙
1 2 g 3 g
´ “ pf q ´ ` 1 ´
h1 2 h1 g1 2 g1 f 2 f1

Another way to do this is to first look at the general nth derivative of the global conformal transformations
(the Möbius transformations). Note that:

az ` b ad ´ bc 1 n!p´cqn´1
g“ , g 1 pzq “ 2
“ ñ Bzn g “
cz ` d pcz ` dq pcz ` dq2 pcz ` dqn`1
In particular:
´2c 6c2
g 2 pzq “ , g 3 pzq “
pcz ` dq3 pcz ` dq4
The simplest combination of g 1 , g 2 , and g 3 that can give zero is:
2
pg 2 q2 ´ g 3 pzqg 1 pzq
3

18
 We want this to have units of rgs{rzs2 and to behave as 3 pzq to leading order when g “ z ` pzq. The only
way to do this (which fixes overall normalization and all) is to divide through by ´2{3pg 1 pzqq2 and get:

g3 3 g2 2
ˆ ˙
´ .
g1 2 g1
It is easy to check that this satisfies the cocycle property for composition, namely:

Bz2 2
ˆ ˙
tz3 , z1 u “ tz3 , z2 u ` tz2 , z1 u (42)
Bz1
Since for h “ g ˝ f we get:
˙2
h3 3 h2 2 f 3 f 2 f 1 g 2 pf 1 q3 g 3 3 f 2 g 1 ` pf 1 q2 g 2
ˆ ˙ ˆ 2 3 g2
1 2g
´ “ `3 ` ´ “ tf, zu`pf q ´ “ tf, zu`pf 1 q2 tg, f pzqu
h1 2 h1 f1 f 1g1 f 1g1 2 f 1g1 g1 2 g1
`z 1 ˘ Bz 1
` z1 ˘ Bz 1
20. I will use shorthand z for Bz and zz for B2 z
, also I will just write Γzzz , gz z̄ , g z z̄ as Γ, g, g ´1 respectively.
Now ˆˆ ˙ ˙ ˆ ˙ ˆ ˙2 ˆ 1 ˙
´1 1 ´1 1 1 1 ´1 z z z z
Γ“g Bg ñ Γ “g Bg “g B 1
g “ 1
Γ´ 1
z z z zz
So
ˆ ˙2 ˆ ˙3 ˆ 1 ˙ ˆ ˙4 ˆ 1 ˙2
1 2 z 2 z z z z
pΓ q “ 1
Γ ´ 2Γ 1 ` 1
z z zz z zz
ˆ ˙ «ˆ ˙ ˆ ˙2 ˆ 1 ˙ff ˆ ˙2 ˆ ˙3 ˆ 1 ˙ ˆ ˙2 ˆ 1 ˙2 ˆ ˙3 ˆ 1 ˙
1 1 z z z z z z z z z z z
BΓ “ B Γ´ 1 “ BΓ ´ Γ 1 `2 1 ´ 1
z1 z1 z zz z1 z zz z zz z zzz

To cancel out the Γ term we look at 2BΓ ´ Γ2 . We see this transforms as:
ˆ ˙2 ˆ ˙4 ˆ 1 ˙2 ˆ ˙3 ˆ 1 ˙ ˆ ˙2
12 z 2 z z z z z
1 1
2B Γ ´ Γ “ 1
p2BΓ ´ Γ q ` 3 1 ´2 1 “ p2BΓ ´ Γ2 ´ 2tz 1 , zuq
z z zz z zzz z1
So that
ˆ 1 ˙2 ˆ 1 ˙2
c z c 2 c c z c 2
Tzz ´ p2BΓ ´ Γ2 q “ pTz 1 z 1 ´ p2B 1 Γ1 ´ Γ1 qq ` tz 1 , zu ´ 2tz 1 , zu “ pTz 1 z 1 ´ p2B 1 Γ1 ´ Γ1 qq
24 z 24 12 24 z 24
c
So indeed T̂zz “ Tzz ´ 24 p2BΓ ´ Γ2 q transforms as a tensor.

21. We have:
¯ z z̄ “ ∇T̂zz “ g z z̄ B̄ T̂ “ ´ c z z̄ c
g B̄r2Bpg ´1 Bgq ´ pg ´1 Bgq2 s “ ´ g z z̄ 2B B̄pg ´1 Bgq ´ 2pg ´1 BgqB̄pg ´1 Bgq
“ ‰
´∇T
24 24
We can recognize this as:
c z z̄ c c c A
2g pBRz̄z ´ Γzzz Rz̄z q “ 2g z z̄ ∇z Rz̄ z̄ “ ∇z R “ BR “ ´ BR
24 24 24 24 2
so we have A “ ´c{12

22. The first part of the action is truly invariant. Let us look at how R changes under Weyl rescaling:

´2e´χ g ´1 B̄peχ g ´1 Bpeχ gqq “ e´χ pR ´ 2g ´1 B B̄χq “ e´χ pR ´ 2B B̄χq

? ?
Consequently: ´gR Ñ ´gpR ´ 2∇2 χq
So the action part will transform as:
ż ż
χ 1 ? 1 ?
SL pgαβ e , φq “ SL pgαβ , φq ´ d ξ gφ∇2 χ “ SL pgαβ , φq `
2
d2 ξ gg αβ Bα φBβ χ
48π 24π

19
23. The most general variation of the effective action is:
ż ż
1 ? 1
δ log Z “ ´ d2 ξ gpa1 R ` a2 qδφ ´ dξpa3 ` a4 K ` a5 na ∇a qδφ (43)
4π Σ 4π BΣ
The counterterms that we can introduce are:
ż ż
2 ?
d ξ gb1 ` dξpb2 ` kb3 q (44)
Σ BΣ

and the variation of the counterterm action:

ż ż
2 ? 1
d ξ gb1 δω ` dξpb2 ` b3 na Ba qδω
Σ 2 BΣ
So we can use this to set a2 , a3 , a5 “ 0. Further, we know the bulk integral’s variation is in fact:
ż ż ż
1 2 ? αβ 1 2 ? α c ? c
δ log Z “ ´δSef f “ d ξ g Tαβ δg “ ´ d ξ g Tα δφ “ d2 ξ gRδφ ñ a1 “ ´
4π 4π 48π 12
Now let’s start with a flat metric and do two changes:
ż ż
c 2 ? 2 a4 ?
δφ1 δφ2 log Z “ ´ d ξ g δφ2 ∇ δφ1 ´ dξ g δφ2 na Ba δφ1
24π 4π
ż ż
c 2 ? a
´ c a4 ¯ ?
“ d ξ g B δφ2 Ba δφ1 ` ´ dξ g δφ2 na Ba δφ1
24π 24π 4π
a4 c
Note that the second term is not symmetric under δφ1 Ø δφ2 , and so we must have the counterterm 4π “ 24π .
A variation of this argument can be used to show that c is truly a constant, independent of any worldsheet
coordinates.
L
24. Take the map 2π log z, mapping the plane to the cylinder of circumference L. We get:
ˆ ˙2 ˜ ˆ ˙2 ¸ ˆ
2πz 2 plane
˙
2π c 1 L c
T cyl 1 ´2
“ pBz q pT plane 1
´ tz , zuq “ 2
z T plane
´ 2
“ T ´
L 12 2z 2π L 24

So the zero mode of T cyl is modified. By T cyl has the expansion

ř ´2πinx
n Ln e so we see L0 gets modified
c
by ´ 24 .
Because L0 is a codimension 1 operator, it will get modified the same way, whether on the cylinder or torus.

25. Each raising operator L´n acts by raising the level by n, and so assuming each one gives a unique state not
expressible in terms of the action of the other L´k , we get that it will contribute:
1
1 ` q n ` q 2n ` ¨ ¨ ¨ “
1 ´ qn

20
 to the partition function. All together these give
1 2πiτ p∆´c{24q q ∆´c{24
ś8 n
ñ Trre s “ ś 8 n
.
n“1 p1 ´ q q n“1 p1 ´ q q

This also shows that at level n there will generically be as many states as there are partitions of the number
n.
26. Consider a nontrivial state |hy so that Ln |hy “ 0 for some n sufficiently positive. Then:
npn2 ´ 1q
0 “ xh| L´n Ln |hy “ xh| p c ´ 2nhq |hy
12
If c “ 0 we get a contradiction unless either |0y is null (and thus decouples) or otherwise h “ 0, and so we
get a vacuum state.
I think we need to add the assumption of irreducibility to have a unique ground state (ie a counterexample
would be TQFTs with multiple ground states).
27. It is enough to show that L1 and L2 acting on this state give zero, since then all other Ln can be obtained
by commutators of these two. Indeed:
3 3 3
L1 pL´2 ´ L2´1 q |1{2y “ p3L´1 ´ p2L0 L´1 ` 2L´1 L0 qq |1{2y “ p3L´1 ´ p2L´1 ` 4L´1 L0 qq |1{2y “ 0
4 4 4
3 2p22 ´ 1q 3 1 9
L2 pL´2 ´ L2´1 q |1{2y “ p4L0 ` c ´ 3pL´1 L1 ` L1 L´1 qq |1{2y “ p4L0 ` ´ L0 qq |1{2y “ 0
4 12 4 4 2
28. The null state’s field must satisfy:
« ˆ ff
3 B2
˙
3 2 ź ÿ 1{2 1 ź
pL´2 ´ L´1 q xψw ψwi y “ ´ Bi ´ xψw ψwi y (45)
4 i i
pwi ´ wq2 wi ´ w 4 B2w i

For the three-point function (holomorphic sector) this gives:

„ 
1{2 1 1{2 1 3 2 λ
2
` B1 ` 2
` B2 ´ Bw “0
pw ´ w1 q w ´ wi pw ´ w2 q w ´ wi 4 pw ´ w1 q pw1 ´ w2 q1{2 pw2 ´ wq1{2
1{2

This gives:
7λ pw1 ´ w2 q3{2
“0
16 pw ´ w1 q5{2 pw2 ´ wq5{2
which gives λ “ 0. We could have inferred this from fermion parity.
Next, for the four-point function, first note that all the ψ have the same scaling dimension, so WLOG we
can write this as: ˆ ˙
1 z12 z34
xψpz1 qψpz2 qψpz3 qψpz4 qy “ h
z12 z34 z13 z24
plugging this into (45) gives a complicated-looking differential equation, but this can be simplified substan-
tially by taking z1 “ z, z2 “ 0, z3 “ 8, z4 “ 0. Notice then that z here is indeed the cross ratio. We then
get the simpler differential equation:
2zgpzq ` 2p1 ´ z 2 qg 1 pzq ´ 3zp1 ´ zq2 g 2 pzq “ 0
This can be solved in terms of known functions (we should more specifically give boundary conditions by
specifying residues of gpzq at z “ 0, 1, 8). All in all we get:
z2 ´ z ` 1
gpzq “
1´z
Thus
1 1 1
xψpzqψpz1 qψpz2 qψpz3 qy “ ` ´
z12 z34 z14 z23 z13 z24
exactly as we would get for Wick contraction.

21
29. Assume it is not primary - then it is a descendant. By positivity of scaling dimensions, it must be a descendant
of a field of scaling dimension 0, but as we have shown two exercises ago, the only such field is the vacuum
|0y. The vacuum is translation invariant Bz 1 “ 0 and so it has no descendants of scaling dimension 1. (It
does have T as a descendant of scaling dimension 2 under Bz2 1).

30. Assume z ą w. On one hand,

ÿ
: rJ a pzq, J b pwqs :“ J a pzqJ b pwq “ a
rJm , Jnb sz ´m´1 w´n´1
m,n

On the other
Gab ifcab J c pwq ÿ
ab ´2 w
´ ¯m´1 ÿ
ab c ´m´1 ´1 w
´ ¯n
J a pzqJ b pzq “ ` ` . . . “ mG z ` ifc J m w z
pz ´ wq2 z´w m
z m,n
z
ÿ ÿ
“ mGab z ´m´1 wm´1 ` ifcab Jmc ´pm´nq´1 ´n´1
w z
m m,n
ÿ´ ¯
“ mδm`n Gab w´m´1 w´n´1 ` ifcab Jm`n
c
w´m´1 z ´n´1
m,n

so we get:
a
rJm , Jnb s “ mδm`n Gab ` ifcab Jm`n
c

31. Rewrite the first part of the action as ´ 4λ1 2 d2 ξ Trrpg ´1 Bgq2 s. Now note:
ş

δpg ´1 Bgq “ g ´1 Bδg ´ g ´1 δg g ´1 Bg

Then we can write the variation of the action as:

» ¨ ˛fi
ż ż
1 1
d2 ξTr pg ´1 Bµ δg ´ g ´1 δg g ´1 Bµ gqpg ´1 B µ gq “ 2 d2 ξTr –δg ˝Bµ pg ´1 B µ g g ´1 q ` loooooooooomoooooooooon
g ´1 Bµ gg ´1 B µ gg ´1 ‚fl
“ ‰ — ˚ ‹ffi
´
2λ2 2λ
g ´1 Bµ g B µ pg ´1 q
ż
1
d2 ξTr g ´1 δg B µ g ´1 Bµ g
“ ` ˘‰
“
2λ2

So we see that we must have g ´1 Bµ g be a conserved current if we only had the first part of the action. In
z, z̄ cooredinates we have BJ z ` B̄J z̄ “ 0. We would like both J “ J z and J¯ “ J z̄ to be separately conserved
B̄J “ B J¯ “ 0. However, this is equivalent to also having εµν Jν conserved. However Bµ Jν ´ Bν Jµ “ ´rJµ , Jν s
gives that Bµ µν Jν “ ´µν Jµ Jν ‰ 0 for nonabelian algebras.
On the other hand, the second term has variation:
ż
ik ” ı
d3 ξ εαβγ Tr pg ´1 B α δg ´ g ´1 δgg ´1 B α gqpg ´1 B β gqpg ´1 B γ gq ` perms.
8π B

this will all vanish identically as an action on B, since TrpA ^ A ^ Aq is already closed for our 1-form
A “ g ´1 dg. On the other hand, the first term in parenthesis contributes a boundary term when α is
transverse
ż ż
ik 2 ´1 β ´1 γ ik ” ˘ı
d2 ξ εβγ Tr g ´1 δg B β g ´1 B γ g
´1
`
d ξ εβγ Trpg δgg B gpg B gqq “ ´
8π BB 8π BB

Appreciate the difference between this and the factor of 3 in Di Francesco. I believe we only account for 1 of
the 3 terms, since only 1 of the 3 indices will give a transverse direction.
This gives a total equation of motion of:
1 µ ´1 ik
2
B pg Bµ gq ´ εµν B µ pg ´1 B ν gq “ 0 (46)
2λ 8π

22
 Taking the basis z, z̄, B z “ 2Bz̄ , εz z̄ “ i{2, we get:
ikλ2 “ kλ2 kλ2
ˆ ˙ ˆ ˙
´1 ´1
‰
rBz̄ pg Bz gq`Bz pg Bz̄ gqs´ iBz̄ pg ´1 Bz gqg ´1 ´ iBz pg ´1 Bz̄ gq “ 1` ´1
Bz̄ pg Bz gq` 1 ´ Bz pg ´1 Bz̄ gq
4π 4π 4π
When λ2 “ 4π{k (meaning k must be positive) the second term goes away and we get the conservation law
B̄Jz . Taking the conjugate of this equation gives the other conservation law.
¯ ´1 Bgq “ 0 Ñ ´BpB̄g g ´1 q “ 0
dpg

In particular the classical solutions factorize into the form gpz, z̄q “ f pzqf¯pz̄q. It is also quick to show that
gpz, z̄q Ñ Ωpzqgpz, z̄qΩ̄pz̄q (for Ω, Ω̄ two independent matrices valued in the same rep’n of G) keeps the action
invariant, and so we see that the G ˆ G classical invariance of the action is enhanced to a full Gpzq ˆ Gpz̄q
invariance. This is the real power of WZW models, and should be appreciated.
32. Importantly, the 3D action does not have any metric dependence. For the 2D boundary we have:
ż
1 ?
2
d2 ξ gg µν Trrg ´1 Bµ g g ´1 Bν gs
4λ
this gives a stress tensor:
ˆ ˙
π 1
Tµν “ ´ Trrg ´1 Bµ g g ´1 Bν gs ´ gµν g αβ Trrg ´1 Bα g g ´1 Bβ gs
λ2 2
we see that this is traceless. The holomorphic part is:
π k
2
TrrJ 2 s “ J a J a
´
λ 4
the constant out front can have a field strength renormalization from its classical value (because the J are
not free fields), and so we would not expect it to agree with the one given in the definition of T .
Give another reason for this discrepancy. Try to account for it.
33. This one is direct. Take z ą w
ÿ ÿ ´ w ¯n
rJ a pzq, Ri pwqs “ a ´m´1
Jm z Ri pwq “ z ´1 Tij Rj pwq
m n
z
and so we get:
a
Jm Ri pwq “ wm Tija Rj pwq
34. We have:
1 ÿ
a a
ÿ
z ´2´pn`mq : Jm Jn :“ Lm z ´2`m
2pk ` h̄q n,m k
1
Appropriately shifting, we see that Lm “ 2pk` h̄q
: Jm`n J´n : as required. The only term here that doesn’t
give zero when acting on a WZW primary is J´1 J0a which acts as J´1
a a T a and this terms appears twice, so
ij
we get that
1
|χi y “ pL´1 δij ´ T a J a q |Rj y
k ` h̄ ij ´1
is null. But we also have that:
¿
a 1 dz
xpJ´1 Rpz1 qqRpz2 q . . . RpzN qy “ J a pzqRpz1 qRpz2 q . . . RpzN q
2πi z ´ z1
z1
¿
1 ÿ dz
“´ Rpz1 qRpz2 q . . . J a pzqRpzk q . . . RpzN q
2πi i‰1 z ´ z1
zi
¿
1 ÿ dz 1
“´ Rpz1 qRpz2 q . . . Tija Rj pzk q . . . RpzN q
2πi k‰1 z ´ z1 z ´ zk
zk
ÿ a
Tij Rj pzk q
“
k‰1
z 1 ´ zk

23
 Here we chose to do this with Rpz1 q, but we could have picked arbitrary zi . This means that correlators
must satisfy: ˜ ¸
N N
1 ÿ Tia b Tja ź
Bz1 ´ x Rpzk qy “ 0
k ` h̄ j‰i zi ´ zj k“1

where Tia acts on the ith primary field in the correlator.

35. I think its instructive to do this one out in detail. First let’s take a look at just TG pzq acting on any current
J a pwq. We want the singular terms:
¿
1 1 1 dx
pJ b J b qpzqJ a pwq “ pJ b pxqJ b pzqJ a pwq ` J b pxqJ b pzqJ a pwqq
2pk ` h̄q 2pk ` h̄q 2πi x ´ z
z
Gba ifcba J c pwq
¿ „ˆ ˙ 
1 1 dx b b
“ ` J pzq ` J pxqpz Ø xq
2pk ` h̄q 2πi x ´ z px ´ wq2 x´w
z
ˆ ab b
J c pwqJ b pzq ` J b pzqJ c pwq
˙
1 G J pzq 1
“ ` fabc
k ` h̄ pz ´ wq2 2 z´w

but note that last term will have

2Gbc 2ifcbd J d pwq
J c pwqJ b pzq ` J b pzqJ c pwq “ ` ` pJ c J b qpwq ` pJ b J c qpwq
pz ´ wq2 w´z

The first term will cancel when multiplied by the anti-symmetric fabc , as will the last (regular) term. The
second term will give fabc fcbd “ ´fabc fdbc “ 2h̄δad , by the definition of dual coxeter number. On the other
hand we have Gab “ kδab so altogether we get:

1 pk ` h̄qJ a pzq J a pwq BJ a pwq

“ `
k ` h̄ pz ´ wq2 pz ´ wq2 pz ´ wq

as we wished. Note we could have run this logic in reverse, and demanded that a stress tensor must its OPE
make second term involving BJ a have coefficient 1, giving the required normalization of p2pk ` h̄qq´1 . Now
1
H
note that if we define T pzq :“ 2pk`h̄ q aPH pJ J a qpzq, then as long as we are taking the OPE with J a for
a
ř
H
a P H, we see that the singular terms are exactly the same. Indeed, we get the same factor of kδab from the
quadratic OPE term, and the sum over fabc fdbc restricts b and c to be in H by the subgroup property, so we
get h̄H . Thus pTG ´ TH qJ a “ TG{H J a is regular for a P H.
For the next step, again lets first just look at the singular terms in the TG TG OPE:
¿
1 1 dx
T pzqT pwq “ T pzqJ a pxqJ a pwq
2pk ` h̄q 2πi x ´ w
„ˆ a
BJ a pxq
¿ ˙ 
1 1 dx J pxq a a
“ ` J pwq ` J pxqpw Ø xq
2pk ` h̄q 2πi x ´ w pz ´ xq2 z´x
BJ a pxq J a pwq
¿ „ 
1 1 dx k dim G
“ ` ` pw Ø xq
2pk ` h̄q 2πi x ´ w pz ´ xq2 px ´ wq2 z´x
c{2 2T pwq BT pwq
“ 4
` 2
`
pz ´ wq pz ´ wq pz ´ wq
k dim G
here we have c “ k`h̄
as required. The same logic applies to the TH pzqTH pwq OPE, where we would get:

cH {2 2TH pwq BTH pwq k dim H

` ` , cH “
pz ´ wq4 pz ´ wq2 pz ´ wq k ` h̄H

24
 Now it remains to evaluate:
¿
1 1 dx ÿ
TG pzqTH pwq “ TG pzqJ a pxqJ a pwq
2pk ` h̄H q 2πi
x ´ w aPH
„ˆ a
BJ a pxq
¿ ˙ 
1 1 dx J pxq a a
“ ` J pwq ` J pxqpw Ø xq
2pk ` h̄H q 2πi x ´ w pz ´ xq2 z´x
BJ a pxq J a pwq
¿ „ 
1 1 dx k dim H
“ ` ` pw Ø xq
2pk ` h̄H q 2πi x ´ w pz ´ xq2 px ´ wq2 z´x
cH {2 2TH pwq BTH pwq
“ 4
` 2
`
pz ´ wq pz ´ wq pz ´ wq
so indeed TG pzqTH pwq ´ TH pzqTH pwq “ TG{H pzqTH pwq has a regular OPE. This further gives us that
TG{H pzqTG{H pwq has singular part coming from TG{H pzqTG pwq “ TG pzqTG pwq ´ TG pzqTH pwq, which gives:

pcG ´ cH q{2 2TG{H pwq BTG{H pwq

` `
pz ´ wq4 pz ´ wq2 z´w

So a G theory can be re-written as a set of “decoupled” CFTs with stress tensors TH and TG{H . Now take
G “ SUp2qm ˆ SUp2q1 . This theory have total level m ` 1. So now take the diagonal subgroup SUp2qm`1 .
We see that the G{H theory has central charge:
ˆ ˙
mˆ3 1ˆ3 pm ` 1q ˆ 3 3m 3pm ` 1q 6
cG ´ cH “ ` ´ “1` ´ “1´
m`2 1`2 m`1`2 m`2 m`3 pm ` 2qpm ` 3q
exactly coincident with the prescribed formula for the minimal models. So, we expect at m “ 1 to get the
Ising CFT.
36. We have ÿ ÿ
ψ i pzq “ ψni z ´n´1{2 ñ xψ i pzqψ j pwqy “ xψni ψm
j
y z ´n´1{2 w´m´1{2
n n,mPZ
8
ÿ
i j
“ xψm ψ´m y z ´m´1{2 wm´1{2
m“0
« ff
8 ´ ¯m
δi ÿ w 1
“? ´
zw m“0
z 2
δij z ` w
“ ?
2 zw z ´ w
the 1{2 comes from the zero-mode Clifford algebra tψ0i , ψ0j u “ δ ij .
37. We can get this directly from the Ward identity:
ˆ ˙
Bz2 Bz3 ∆ ∆ 1 ∆
xT pz1 qφpz2 qφpz3 qy “ ` ` 2
` 2 2∆
“ 2 2 2∆´2 .
z1 ´ z2 z1 ´ z3 pz1 ´ z2 q pz1 ´ z3 q pz2 ´ z3 q z12 z13 z23
Next, we can write:
w̄´2∆ ∆ ∆
xX| T pzq |Xy “ lim w̄´2∆ x0| Xp1{w̄q T pzq Xp0q |0y “ lim 2 ´2∆
“ 2.
wÑ0 wÑ0 z w̄ z
řN
Finally, let’s look at the OpN q fermion. We have that T pzq “ ´ 21 i“1 : ψ i Bψ i : so we get:
» fi

1 — ˆz ` w 1 ˙ 1 ffi N 1 N {16
xS| T |Sy “ ´ ΣN lim
— ffi
—Bw ? ´ Bw ffi “ ´ p´ 2 q “
2 i“1 zÑw – 2 zw z ´ w pz ´ wq
looooomooooon fl 2 8w w2
Normal ordering constant

as required.

25
38. This is direct:

Dθ X̂ “ pBθ ` θBz q pX ` iθψ ` iθ̄ψ̄ ` θθ̄F q “ iψ ` θBX ` θ̄F ` θθ̄B ψ̄, D̄θ̄ X̂ “ iψ̄ ` θ̄BX ` θF ` θ̄θB̄ψ

Now we only want the θθ̄ terms of pDθ X̂qpD̄θ̄ X̂q as everything else will vanish in the Berezin integral. This
gives:
ż ż ż
1 2 2 1
S“ d z dθ̄dθ θθ̄pBXBX ´ F ` iψ̄B ψ̄ ` iψ B̄ψq “ d2 zpBXBX ` iψ̄B ψ̄ ` iψ B̄ψq
2π`2s 2π`2s

we have dropped F 2 because it has no dynamics or interactions with X, ψ whatsoever.

39. Expanding
µ `iθψ µ `iθ̄ ψ̄ µ `θ θ̄F µ q
eip¨X̂ “ eipµ pX “ p1 ` iθp ¨ ψqp1 ` iθ̄p ¨ ψ̄qp1 ` θθ̄p ¨ F qeipX
Imposting EOM’s gives F “ 0 right away. Now for the rest:

Dθ X̂ µ Dθ̄ X̂ ν eipX |θθ̄ “ rpBX µ BX ν ` iψ̄ µ B ψ̄ ν ` iψ ν B̄ψ µ q ` piBX ν ψ µ qp ¨ ψ ` piBX µ ψ̄ ν qp ¨ ψ̄seipX

¯ terms. Now we get:
again using the equations of motion we get rid of the B ψ̄, dψ

rBX µ BX ν ` piBX ν ψ µ qp ¨ ψ ` piBX µ ψ̄ ν qp ¨ ψ̄seip¨X “ pBX µ ` ipp ¨ ψqψ µ qpBX ν ` ipp ¨ ψqψ ν qeip¨X

40. Following the same logic as the N “ p2, 0q case, we can now compute in the R sector:
ˆ ˙
α β 4k 1 αβ
tG0 , Ḡ0 u “ ´ δ ` 2L0 δ αβ
2 4

for this to be positive we need:

2p∆ ´ k{4q ě 0 ñ ∆ ě k{4.
In the NS sector, we have a positivity condition on

tGα´1{2 , Ḡβ1{2 u “ ´2σαβ

a
J0a ` 2δ αβ L0

The positivity condition on this operator translates to the matrix:

a
2∆1 ´ 2σαβ Ja

being positive semidefinite. But the determinant of this matrix is given by

∆2 ´ |J|2 “ ∆2 ´ j 2

So for this to be ě 0, given that ∆ ě 0, we need ∆ ´ j ě 0

41. This calculation is also direct:

´ ¯
1{2 ´ `12 pBXq2 pwq B ´ `12 pBXq2 pwq
s
T pzqT pwq “ 4
`2 s 2
`
pz ´ wq pz ´ wq z´w
`s Q 2 `s Q ´2 `2s Q2 ´6
` BXpzq ? 3 2 3
` BXpwq ? 2 3
´
2 2`s pz ´ wq 2 2`3s pz ´ wq 2 2`s pz ´ wq4
2

1{2p1 ` 3Q2 q 2T pwq BT pwq

“ 4
` 2
`
pz ´ wq pz ´ wq z´w

we thus get a central charge equal to 1 ` 3Q2 as required.

26
42. The integral over the zero mode will give no contribution from the BXB X̄ term in the action and instead
will just: ˜ ż ˜ ¸ ¸
ż
? Q ÿ
DX exp ´ d2 z g ? Rp2q ´ i pi δ 2 pz ´ zi q Xpzq
4π`s 2 i
This is a δ-functional on the pi . We have:
« ff
Q p2q
ÿ
2
δ ? R ´ i pi δ pz ´ zi q
4π`s 2 i

but this can only happen if, after integrating over z, we get:
Q ÿ ? ÿ
? χ “ i pi ñ i 2`s pi “ Qχ.
`s 2 i i

Give an interpretation of the vertex operators as “contributing curvature”.

43. Note that:
A
rL´m , J´n s “ nJ´m´n ` mpm ´ 1qδm`n
2
Take pm ` nq “ 0 and look at the central term. We see that we cannot simply identify L:m “ L´m , Jn: “ J´n
because then the commutation relation above has a central term ´mA off from the correct central term. This
can be corrected by redefining just the zero mode J0: “ J0 ` A. We see that then the hermitian conjugates
satisfy the same algebra. We see sufficiency. Is this necessary?
We now show that we cannot change the algebra in any other way and keep the commutation relations.
:
Firstly, we cannot add a (necessarily zero weight) central term to any other Jm “ J´m relation since only
J0 transforms with 0 weight. In fact we cannot form any linear combination J˜m of the Jm and expect the
commutation relations to hold, since each Jm has different eigenvalue under L0 . We can thus only rescale
the Jm - and applying rL1 , Jm s shows that to keep the commutation relation, this rescaling must be the same
for all Jm - but this would necessarily modify the central term by changing the charge A.
The same logic applies to the L:m . We cannot mix Lm for different m to define L:m since they have different
weight under L0 . There is also no consistent way to rescale all of them and keep the commutation relations
the same. The only possibility is adding a central term to the relation L:0 “ L0 , but any redefinition of this
will modify the central charge of the conjugate theory.
44. Noting that
1
bpzqBcpwq “ cpzqBbpwq “
pz ´ wq2
1
Bbpzqcpwq “ Bcpzqbpwq “ ´
pz ´ wq2
2
BbpzqBcpwq “ BcpzqBbpwq “ ´
pz ´ wq3
we can just directly compute the T T OPE:
T pzqT pwq “ p´λbpzqBcpzq ` p1 ´ λqBbpzqcpzqq p´λbpwqBcpwq ` p1 ´ λqBbpwqcpwqq
“ λ2 pbBcqpzqpbBcqpwq ` λpλ ´ 1q rpbBcqpzqpBbcqpwq ` pBbcqpzqpbBcqpwqs ` p1 ´ λq2 pBbcqpzqpBbcqpwq
λ2 ` p1 ´ λq2 ` 4λpλ ´ 1q λ2 p´bpzqBcpwq ` Bcpzqbpwqq p1 ´ λq2 pBbpzqcpwq ´ cpzqBbpwqq
“´ ` `
pz ´ wq4 pz ´ wq2 pz ´ wq2
BcpzqBbpwq ` BbpzqBcpwq bpzqcpwq ` cpzqbpwq
` λpλ ´ 1q ´ 2λpλ ´ 1q
z´w pz ´ wq3
The first term on the last line will die since we can take z Ñ w and ignore first-order terms capturing the
differences. The second term in the last line will become:
Bbpwqcpwq ` Bcpwqbpwq B 2 bpwqcpwq ` B 2 cpwqbpwq
´ 2λpλ ´ 1q ´ λpλ ´ 1q (47)
pz ´ wq2 pz ´ wq

27
 the second two terms in the first line contribute a pz ´ wq´2 term of:

λ2 p2Bcpwqbpwqq ` p1 ´ λq2 p2Bbpwqcpwqq

this will combine with the pz ´ wq´2 terms in (47) to give:

2 rλBcpwqbpwq ` p1 ´ λqBbpwqcpwqs “ 2T pwq

as required. Finally, the pz ´ wq´1 terms all collected give coefficient (dropping the w dependence, as it is
understood):

λ2 p´BbBc ` B 2 cbqq ` p1 ´ λq2 pB 2 bc ´ BcBbq ´ λpλ ´ 1qpB 2 bc ` B 2 cbq

“ ´λ2( ´ 2λBbBbc ` 1BbBc ` rλ2 ` λp1 ´ λqspB 2 cbq ` rp1 ´ λq2 ` λp1 ´ λqspB 2 bcq
((
pBbBc
((` (BcBbq
(

“ λB 2 cb ` p1 ´ λqB 2 bc ` p1 ´ 2λqBbBc “ BT
as required. So altogether we get exactly the stress tensor OPE needed to satisfy the Virasoro algebra with
central charge:

´2pλ2 ` p1 ´ λq2 ` 4λpλ ´ 1qq “ ´2p6λ2 ´ 6λ ` 1q “ 1 ´ 3Q2 , Q “ p1 ´ 2λq

45. The BRST current is:

jB pzq “ cpzqT X pzq ` pbcBcqpzq
There are several OPEs to do. Let’s start with the easier ones:
„ X 
X X c {2 2T pwq BT pwq
pcT qpcT q „ cpzqcpwq ` `
pz ´ wq4 pz ´ wq2 z´w
8
ÿ pz ´ wqn cX {2
„ 
n 2T pwq BT pwq
“´ cpwqB cpwq ` ` (48)
n“1
n! pz ´ wq4 pz ´ wq2 z´w
1 X
c cpwqBcpwq 1 12 cX cpwqB 2 cpwq 1 12 cX cpwqB 3 cpwq 2T pwqcpwqBcpwq
„´2 ´ ´ ´
pz ´ wq3 2 pz ´ wq2 6 z´w z´w
Next:
T X pzqcpwqB(¸wq cpzqBcpzqT X pwq
pcT X qpbcBcq ` pbcBcqpcT X q „ `
z´w z´w (49)
2T pwqcpwqBcpwq
„
z´w
This exactly cancels the last term in the previous expression. Now the hard one. Being careful of fermion
minus signs, I’ll underline the contractions that will give them:

pbcBcqpbcBcq “ pbcBcqpbcBcq ` pbcBcqpbcBcq ` pbcBcqpbcBcq ` pbcBcqpbcBcq

(50)
` pbcBcqpbcBcq ` pbcBcqpbcBcq `   ` pbcBcqpbcBcq
pbcBcqpbcBcq 
 

the last two terms are canceled because they contribute only pz ´ wq´1 singularities multiplying cpzqBcpwq
which is Opz ´ wq and so only contributes finite terms. The remaining terms give:
cpzqcpwq cpzqBcpwq Bcpzqcpwq BcpzqBcpwq cpzqBcpzqbpwqcpwq bpzqcpzqcpwqBcpwq
´ ` ´ ` ` ` (51)
pz ´ wq4 pz ´ wq3 pz ´ wq3 pz ´ wq2 pz ´ wq2 pz ´ wq2

The last two terms will cancel, as they contribute a pz ´ wq´1 singularity with numerator cB 2 cbc ` BcBcbc `
bBccBc ` BbccBc. All of these terms are evaluated at w, so all are zero. Now we have (all evaluated at w)

´Bcc ` cBc ´ Bcc ´ 21 B 2 cc ` BcBc ´ B 2 cc ` BcBc ´ 16 B 3 cc ` 12 B 2 cBc ´ 12 B 3 cc ` B 2 cBc

` `
pz ´ wq3 pz ´ wq2 z´w
3 2 2 3 3 2
(52)
3cpwqBcpwq cpwqB cpwq cpwqB cpwq ` 2 B cpwqBcpwq
“ 3
` 2 2
` 3
pz ´ wq pz ´ wq z´w

28
 Combining Equations (48), (49) and (52) we get:
cX
p3 ´ 12 cX qcpwqBcpwq p 32 ´ 14 cX qcpwqB 2 cpwq p 23 ´ 3
12 qcpwqB cpwq ` 23 B 2 cpwqBcpwq
jB pzqjB pwq “ ` ` (53)
pz ´ wq3 pz ´ wq2 z´w

Now for Q2B “ 0, we need to look at jB pzqjB pwq residue as z Ñ w as a function of w and ensure that this
has no residue in w. First we just need to look at the pz ´ wq´1 term and reduce it all to the integral:

2 cX 13 cX
¿ „ˆ ˙  ¿ ˆ ˙
2 1 3 3 2 1
QB “ dw ´ cpwqB cpwq ` B cpwqBcpwq “ dw ´ cpwqB 3 cpwq
2πi 3 12 2 2πi 6 12

This will vanish exactly when cX “ 26 as required.

NB in Polchinski, there is an additional cB 3 c term in the definition of jB that contributes to this OPE (which
makes Equation (53) look nicer), but the conclusion about D “ 26 is still the same.

46. This is the type of question with a two-line answer that depends on a lot of conceptual build up. It is
instructive to go through some of the details. Here I will set `2s “ 2. First lets start with the system of two
Majorana-Weyl fermions ψ 1 , ψ 2 . This has central charge c “ 1. Moreover, we can compute everything in
terms of
1 1
ψ “ ? pψ 1 ` iψ 2 q, ψ̄ “ ? pψ 1 ´ iψ 2 q.
2 2
Note both ψpzq and ψ̄pzq are in the holomorphic sector. The anti-holomorphic fields, if we considered them,
can be labeled as in polchinski by ψ̃pz̄q, ψ̄˜pz̄q. These fields give OPE:
1
ψpzqψpwq “ Opz ´ wq, ψ̄pzqψ̄pwq “ Opz ´ wq ψpzqψ̄pwq “ ` : ψ ψ̄ : pwq ` Opz ´ wq (54)
z´w

Now Jpzq “: ψ ψ̄ : pzq can be seen to have scaling dimension 1 by OPE, so it is a conserved current (and
necessarily a primary operator in a unitary theory). Indeed JJ “ pz ´ wq´2 and Jψ “ ψpz ´ wq´1 , J ψ̄ “
ψ̄pz ´ wq´1 so ψ, ψ̄ have charge ˘1 under J. From extending equation (54) to terms of order pz ´ wq the
stress energy tensor T “ ´ 12 : ψ i Bψ i :“ 12 J 2 .
Now note that this shares everything in common with the free scalar theory. The central charge c “ 1. The
up1q currents there are J “ iBφ and have the same OPE. The analogues of the fermions ψ, ψ̄ are then the
operators e˘iφpzq respectively. Indeed these have charge ˘1 under J. But it would be surprising if these
operators anti-commuted, being built out of bosons and all. In fact they do! By Baker-Campbell-Hausdorff:
1 1 1 1
eiφpzq eiφpz q “ e´rφpzq,φpz qs eiφpz q eiφpzq “ ´eiφpz q eiφpzq

since rφpzq, φpwqs “ ´ log z´w

w´z “ ´iπ The anti-commutation property comes out of the non-locality of the
vertex operators in terms of φ. We can make the exact same argument for eiφpzq e´iφpwq or any combination
thereof. So all of these fields are in fact fermionic. They have the same OPEs as the fermions above:
1
e˘iφpzq e˘iφpwq “ Opz ´ wq, e˘iφpzq e¯iφpwq „
z´w
Note also the OPE
„ ż 
iφpzq iφpwq
:e :: e : “ exp ´ dz dw logpz ´ w qδφpz 1 q δφpw1 q : eiφpzq e´iφpwq :
1 1 1 1

1
“ p1 ` iBφpwqpz ´ wq ` Opz ´ wq2 q
z´w
1
“ ` iBφpwq ` Opz ´ wq
z´w
as required.

29
We can actually perform this procedure to the bc ghosts as well, for any value of λ. The trick is to note that
we have performed it for λ “ 1{2, and now the stress-energy tensor changes to:
T λ “ T λ“1{2 ´ pλ ´ 1{2qBp: bc :q
If we still take b “ eiφ , c “ e´iφ then : bc :“ iBφ and so the stress-energy tensor looks like:
1
T λ “ ´ pBφq2 ´ ipλ ´ 1{2qB 2 φ
2
which is just the Coloumb gas model with Q “ ´ip2λ ´ 1q. The central charge is 1 ` 3Q2 “ 1 ´ 3p2λ ´ 1q2 ,
exactly as we want. The conformal weights are k 2 {2 ˘ iQk{2 Ñ 21 ˘ pλ ´ 1{2q at the lowest level, and this
is exactly λ and 1 ´ λ as desired. Note that b and c are hermitian, so we need φ to be anti-hermitian.
Equivalently we can write φ “ iρ for ρ hermitian. Then
b “ e´ρ , c “ eρ , J “ ´Bρ.
1
Note ρ has opposite OPE from φ so that Bρpzq Bρpwq „ pz´wq2
.
Now lets look at the bosonic βγ theory. Can we bosonize this too? For one, the charge is J “ ´βγ which
1
has opposite sign OPE JpzqJpwq “ ´ z´w , so we will now need ρ to have the regular-sign OPE (ie the
same as φ). We’ll just call this hermitian field φ. Let’s take β “ e´φ , c “ eφ as before and J “ ´Bφ.
Already there is an issue. If φ satisfies the standard OPE then β and γ will be anticommute. Further,
βγ “ e´φpwq eφpzq “ Opz ´ wq while by the same logic ββ „ γγ „ pz ´ wq´1 . We want ββ “ Oppz ´ wq0 q, γγ “
Oppz ´ wq0 q, βγ „ ´pz ´ wq´1 , γβ „ pz ´ wq´1 .
Another way to see that we are missing something: we can try to write a Coulomb gas model for the βγ
theory: ˆ ˙
λ λ“1{2 1 2 1 1 1 ´ 2λ 2
T “T ´ pλ ´ 1{2qBpβγq “ ´ J ´ ´ λ BJ “ ´ pBφq2 ` B φ
2 2 2 2
notice the ´ sign in front of 21 J 2 , as we want. We have a coulomb gas model with Q “ 1 ´ 2λ. This gives
a central charge 1 ` 3Q2 “ 4 ´ 6λ ` 12λ2 . On the other hand, the βγ theory should have central charge
´1 ` 3Q2 . We are off by 2.
All of this indicates that we need to add an uncoupled c “ ´2 including fermions–namely the bc fermi theory
at λ “ 1– and redefine βγ in terms of φ to incorporate this. Take η, ξ of scaling dimensions 1, 0 and charges
¯1 respectively. Then define
β “ e´φ Bξ, γeφ η.
We now have the OPE:
1 1 1
βpzqγpwq “ pz ´ wq ˆ ´ 2
“´ , γpzqβpwq “
pz ´ wq z´w z´w
This is 4.15.2. Further because ηη “ Opz ´ wq and BξBξ “ Opz ´ wq we get ββ “ Oppz ´ wq0 q and likewise
for γγ as needed. We also know how to interpolate between NS and R sectors by taking φ Ñ φ{2 etc.
The total current ´ : βγ : stays the same because we look for the constant term in the expansion:
1 1
e´φpzq eφpwq “ ´ pz ´ wq ´ Bφpwqpz ´ wq2 Ñ Bφpwq ñ J “ ´Bφpwq
` ˘
βpzqγpwq “ ´ 2 2
pz ´ wq pz ´ wq
so we identify : βγ : with Bφ, which are both ´J. This is 14.15.10. Writing out the full stress tensor now
gives:
1 1 ´ 2λ 2
´ pBφq2 ` B φ ´ ηBξ “ T λ“1{2 ` p1{2 ´ λqBpβγq
2 2
It remains to show that T λ“1{2 “ ´ 12 βBγ ` 12 Bβγ “ 21 p2Bβ γ ´ Bpβγqq. Now looking at the βγ OPE to order
z ´ w we get:
e´φpzq Bξpzqeφpwq ηpwq “ Bξpzqηpwqe´φpzq eφpwq
ˆ ˙ˆ ˙
´1 2 1 3 2 2
“ ` : Bξη : pz ´ wq ´ pz ´ wq Bφ ` pz ´ wq ppBφq ´ B φq
pz ´ wq2 2

30
 NOTE I had to assume that while ξ, η and eφ , e´φ separately anticommute with their partners, the e˘φ
fields commute with the ξ, η fields. Give an interpretation/example in condensed matter of this.
The order z ´ w term here is:
1
: Bξη : ´ ppBφq2 ´ B 2 φq
2
So this is the normal ordered product of Bβ γ. The Bpβγq “ B 2 φ term will cancel the B 2 φ term there and
we’ll get the stress tensor
1
´ηBξ ´ pBφq2 “ T λ“1{2
2
which is 4.15.8 as desired.
We can also bosonize the η, ξ theory in terms of an auxiliary bosonic field χ, but this was not necessary for
the exercise.

47. We are looking at DN boundary conditions. Let us do this directly from definitions:
? ÿ αk ´ikτ `s ÿ αk ´k
Xpτ, σq “ x ´ 2`s e sinpkσq “ x ` i ? pz ´ z̄ ´k q
k 2 k
kPZ`1{2 kPZ`1{2

`2s ÿ αk αl ´k
ñ xXpz, z̄qXpw, w̄qy “ ´ pz ´ z̄ ´k qpw´l ´ w̄´l q
2 kl
k,lPZ`1{2
2 8 „´ ¯ 
`s ÿ 1 w k`1{2 ´ w̄ ¯k`1{2 ´ w ¯k`1{2 ´ w̄ ¯k`1{2
“ ´ ´ `
2 k“0 k ` 1{2 z z z̄ z̄

Now we have
8 8 ? 2k`1
ÿ xk`1{2 ÿ p xq ? ? ?
“2 “ 2 arctanhp xq “ ´plogp1 ´ xq ´ logp1 ` xqq.
k“0
k ` 1{2 k“0
2k ` 1

Our convention on the square root branch cut is along the negative real axis. We get:

`2s ” a a a a ı
´ logp1 ´ w{zq ´ logp1 ` w{zq ´ logp1 ´ w̄{zq ` logp1 ´ w̄{zq ` c.c.
2
so the final result gives us:

`2s ” a 2
a 2
a 2
a 2
ı
´ log |1 ´ w{z| ´ log |1 ` w{z| ´ log |1 ´ w̄{z| ` log |1 ` w̄{z| .
2
We can simplify this to:
? ? ? ?
`2s
„ 
z´ w 2 z ´ w̄ 2 ? ? 2
´ log | ? ? | ´ log | ? ? | ` log | z ` w̄| .
2 z` w z ` w̄
For ND boundary conditions, the ´ between the two logs becomes a `.
Interpret this in terms of image charges

31
48. Firstly, BX B̄X requires no normal ordering constant to be added ordinarily, since it has a wick contraction
of zero. Now to go from the plane from the disk we have x “ z´i 1`x
z`i . Vice versa is z “ i 1´x . This gives
2
log |z ´ w|2 “ log |x ´ y|2 ` log | |2
p1 ´ xqp1 ´ yq
2
log |z ´ w̄|2 “ log |1 ´ xȳ|2 ` log | |2
p1 ´ xqp1 ´ ȳq
So for N N and DD boundary conditions we get:
`2s `
log |x ´ y|2 ` log |1 ´ xȳ|2 ´ 2 log |p1 ´ xqp1 ´ yq|2 ` 4 log 2
˘
xXN N px, x̄qXN N py, ȳqy “ ´
2
`2s `
log |x ´ y|2 ´ log |1 ´ xȳ|2 .
˘
xXDD px, x̄qXDD py, ȳqy “ ´
2
So NN boundary conditions correspond to putting an image charge of the same sign at 1{x˚ while DD
boundary conditions correspond to putting an image charge of opposite sign at 1{x˚ as well as a neutralizing
charge of the opposite sign at 1–corresponding to 8 in the H setting. Interpret this.
Differentiating the above with Bx B̄y shows that in either case only the logp1 ´ xȳq term contributes:
`2s 1
xBXN N pxqB̄XN N pȳqy “
2 p1 ´ xȳq2
`2 1
xBXDD pxqB̄XDD pȳqy “ ´ s .
2 p1 ´ xȳq2
This will become singular only as z approaches the boundary of the unit circle. We encounter the divergence
2
˘ `2s p1´xȳq
1
2 in the N N and DD cases respectively and so we can define

‹ ‹ `2s 1
‹ BXpzqB̄Xpw̄q‹ “ BXpzqB̄Xpw̄q ¯
2 p1 ´ z w̄q2
On the other hand for BX BX we get the normal ordering constant:
‹ ‹ `2s 1
‹ BXpzqBXpwq‹ “ BXpzqBXpwq `
2 pz ´ wq2
We have X̄p1{w̄q “ ˘Xpwq so consequently BXpwq “ ˘B̄1{w̄ Xp1{w̄q. Now its a quick check (being careful
to keep subscripts on B̄ so we know what we’re differentiating w.r.t.):

‹ ‹ `2s 1
‹ BXpzqB̄w̄ Xp1{w̄q‹ “ BXpzqB̄w̄ Xp1{w̄q ¯
2 p1 ´ z{w̄q2
`2 1
ñ ‹‹ BXpzqB̄1{w̄ Xp1{w̄q‹‹ “ BXpzqB̄1{w̄ Xp1{w̄q ¯ p´w̄´2 q s
2 p1 ´ z{w̄q2
`2 1
ñ ‹‹ BXpzqBXpwq‹‹ “ BXpzqBXpwq ` s
2 pz ´ wq2
where the extra minus sign in the Dirichlet boundary condition case removes any sign ambiguity in the last
line. Thus, we see that indeed ‹‹ BXpzqBXpwq‹‹ “ ˘ ‹‹ BXpzqB̄Xp1{w̄q‹‹ for Neumann and Dirichlet boundary
conditions respectively.

32
49. Using the doubling trick we have ψ̄pz̄q “ ψpz ˚ q. So zi “ We can compute the correlator by Wick contraction:
m 2n´m n n
ź ź ź 1 ÿ ź 1 ” 1 ı
x ψpzi q ψ̄pz̄j qy “ x ψpwi qy “ sgnpπq “ Pf
i“1 j“1 i“1
2n n! πPS w
i“1 πp2i´1q
´ wπp2iq wi ´ wj
2n

where wi “ zi for z “ 1 . . . m and wi`m “ zi˚ for z “ 1 Ñ 2n ´ m

50. I feel that this has already been done in 2.3.31. Rotating to euclidean signature, the most general solution
for X is
`2 `2 `s ÿ e´kτ
Xpτ, σq “ xµ ` s pp ` p̄qτ ` s pp ´ p̄qσ ` i ? pαk e´ikσ ` ᾱk eikσ q
2 2 2 k‰0 k

The first boundary condition X9 “ 0 at σ “ 0 gives:

αk “ ´ᾱk , p ` p̄ “ 0
while the second boundary condition X 1 “ 0 at σ “ π gives:
sinpkπq “ 0 ñ k P Z ` 1{2 p ´ p̄ “ 0
Thus we have neither momentum nor winding-number. So for the mode expansion is:
? ÿ αk `s ÿ αk
Xpτ, σq “ x ´ 2`s e´kτ sinpkσq “ x ` i ? pz ´k ´ z̄ ´k q
k 2 k
kPZ`1{2 kPZ`1{2

as desired. This gives:

`s ÿ `s ÿ
BX “ ´i ? αk z ´k´1 , B̄X “ i ? αk z̄ ´k´1
2 kPZ`1{2 2 kPZ`1{2

51. We have N scalars with BX i pzq “ Oij B̄X j pz̄q on the real axis. Because the conformal group includes the
translation group, Oij must be translationally invariant, ie it cannot depend on z. Further because X i is a
scalar B ` B̄ and B ´ B̄ both act on it in an invariant way. These are the two boundary conditions we can
set on each X i . So we see that Oij can definitely be a diagonal matrix of ˘1s. However, because all the
scalars are identical we can also transform X 1 j pz, z̄q “ Rij X i pz, z̄q, with R any orthogonal matrix (not just
special orthogonal) and still get a valid boundary condition. So O is any orbit of the matrix of ˘1s under
the conjugation action of the orthogonal group O Ñ P T OP . This can be easily appreciated as boundary
conditions for an open string along the various coordinate directions being either Neumann or Dirichlet.
Its surprising that O can’t vary on the real axis - corresponding to the D-brane changing which X i live on it.
Think about this more.
52. Everything is in the NS sector. Let’s first evaluate xψN N pzqψN N pwqy. We have
8
ÿ
´n´1 ´m´1
ÿ 1
x0| bn`1{2 bm`1{2 |0y z w “ z ´n´1 “
n,m
looooooooooomooooooooooon
n“0
z´w
δn“´m´1

For the NS sector we have the following cases:

33
 • NN: bn`1{2 ` b̄n`1{2 “ 0
• DD: bn`1{2 ´ b̄n`1{2 “ 0
• DN: bn ` b̄n “ 0

so we see that xψpzqψ̄pw̄qy will add an extra minus sign in the NN case. It will not do so in the in the DD
case. Collecting our results.
1 1
xψN N pzqψN N pwqy “ , xψN N pzqψ̄N N pw̄qy “ ´
z´w z ´ w̄
1 1
xψDD pzqψDD pwqy “ , xψDD pzqψ̄DD pw̄qy “
z´w z ´ w̄
Lastly, for the DN case, ψ now takes integer values and so:
8
ÿ ÿ 1 z`w
xψDN pzqψDN pwqy “ x0| bn bm |0y z ´n´1{2 w´m´1{2 “ z ´n´1{2 wn´1{2 ´ z ´1{2 w´1{2 “ ? .
n,m
looooomooooon
n“0
2 on
loomo 2 zwpz ´ wq
δn“´m
zero mode

Because bn “ ´b̄n we then also have

z ` w̄
xψDN pzqψ̄DN pw̄qy “ ´ ? .
2 z w̄pz ´ w̄q

53. On to the R sector.

• NN: bn ´ b̄n “ 0
• DD: bn ` b̄n “ 0
• DN: bn`1{2 ´ b̄n`1{2 “ 0

Let’s again evaluate xψN N pzqψN N pwqy. The calculation is exactly the same as the DN calculation above.
Using the above relations between the b and b̄ in the different sectors we’ll get:
z`w z ` w̄
xψN N pzqψN N pwqy “ ? , xψN N pzqψ̄N N pw̄qy “ ?
2 zwpz ´ wq 2 z w̄pz ´ w̄q
z`w z ` w̄
xψDD pzqψDD pwqy “ ? , xψDD pzqψ̄DD pw̄qy “ ´ ?
2 zwpz ´ wq 2 z w̄pz ´ w̄q
1 1
xψDN pzqψDN pwqy “ , xψDN pzqψ̄DN pw̄qy “
z´w z ´ w̄

54. There are several ways to do this. One way is directly by using the identity relating an expectation of an
exponential to the exponential of an expectation:
ˆ 2 ˙ ˆ 2 2 ˙
a a ` 1
xeiaXpzq yRP2 “ xeiaXpzq e´iaX̄pz̄q yCP1 9 exp ˆ 2 xXpzqX̄pz̄qy “ exp ´ s logp1 ` z z̄q “ .
2 2 p1 ` |z|2 qa2 `2s {2
It is not clear that we haven’t omitted a proportionality constant. Another way to compute this is to note
2
that x: Xpz, z̄qXpz, z̄q :y “ ´ `2s log |1 ` z z̄|2 and so expanding out:
8
ÿ piaqn
eiaX “ xXpz, z̄qn y .
n“0
n!

Now we do wick contractions. For each even term we need to put 2n elements in to n pairs. There are
p2n ´ 1qp2n ´ 3q . . . p3qp1q ways to do this. Simplifying we get:
8 ˆ 2 ˙n
ÿ p´1qn paq2n `s n 2
´
a2 `2s {2
¯ 2 2
n n!
´ log |1 ` z z̄| “ exp log |1 ` z z̄| “ p1 ` |z|2 qa `s {2
n“0
2 2

34
 This doesn’t look right. If instead we had:
8 ˙n
piaqn p´iaqm ÿ a2n ˆ 2
iaXpzq ´iaX̄pz̄q
ÿ
n m n! `s 1
e e “ x: Xpzq X̄pz̄q :y “ ´ logp1 ` z z̄q “
n,m“0
n! m! n 
n
!n! 2 p1 ` |z|2 qa2 `2s {2

as required.
In doing this problem, I needed to consider the eiaX e´iaX̄ correlator rather than the eiapX`X̄q correlator -
otherwise I would get an ill-defined one-point function that blows up as z Ñ 8 (ie is not a globally-defined
differential). Perhaps this comes from boundary conditions in the case of RP2 , since H1 “ Z2 and so we can
enforce anti-periodic boundary conditions that would be consistent with a negative charge vertex operator
being placed a ´1{z̄.
?
55. For the non-supersymmetric theory, we have the action (on the sphere, with ´gR2 “ 1):
ż ż ż ż
1 2 ? αβ Q 2 ? 1 2 Q
S“ d z gg Bα XBβ X ` ? p2q
d z gR X “ d zBX B̄X ` ? d2 z X
4π`2s 4π`s 2 2π` 2
s 4π`s 2
this gives a stress-energy tensor:
1 Q
T “´ 2
BX 2 ` ? B 2 X
`s `s 2
Now for N “ 1 we might expect an action of the form:
ż ż ż ż
1 2 ? αβ Q 2 ? p2q 1 2 Q
S“ d z gg Bα XBβ X ` ? d z gR X “ d zBX B̄X ` ? d2 z X
4π`2s 4π`s 2 2π`2s 4π`s 2
This gives: ?
1 Q 1 2
T “ ´ 2 BXBX ` ? B 2 X ´ ψBψ, G“i ψBX ´ iQBψ
`s `s 2 2 `s
3 Gpwq BGpwq
The T T OPE will give central charge 32 ` 3Q2 . G remains primary, so we’ll have T G “ 2 pz´wq2 ` z´w .
Finally, GG will give ? ?
2 2 
2Q2 `s QBX  QBX

1 ´`s
2T
3
` 3
`  
2
`
pz ´ wq pz ´ wq  pz ´ wq z´w
so we get ĉ “ 1 ` 2Q2 as desired.
Now for N “ 2, following the same example, we still get get same T T OPE and G˘ remains primary, so we
have the T G˘ OPE staying the same. The GG OPE will have ĉ “ 1 ` 2Q2 as before and J will have to be
modified to include B 2 X so as to remain primary under T .

56. For X a compact scalar valued in S 1 of radius R we have the solutions X “ 2πRpnσ1 ` mσ2 q, which have
vanishing Laplacian. The action of these instanton solutions is:
ż1 ż1
1 1 2 πR2
S“ dσ 1 dσ 2 |τ B1 X ´ B 2 X| “ |nτ ´ m|2
4π`2s 0 0 τ2 ` 2τ
s 2

Expanding X “ X cl ` χ, we get no cross-terms in the action. We now do the path integral over the χ with
periodic conditions around both cycles. χ separates into the zero mode χ0 ` δχ and δχ can be expanded
in terms of eigenfunctions of the laplacian on periodic functions. These are precisely e2πipm1 σ1 `m2 σ2 q with
2
eigenvalues p2πq 2
τ2 |m1 τ ´ m2 | . They form an orthonormal basis. The contribution to the action is then

1 ÿ
λm1 m2 |Am1 m2 |2
4π`2s
m1 ,m2 PZ2

The measure on the space of functions comes from the norm of δX

ÿ |Am m |2 ż 2πR
dχ0 8
ż ż ż
2 1 2 ? 1 2
ź dAm1 ,m2
||δX || “ d σ gpdχq “ ñ Dχ “ .
`s m ,m
`2s 0 `s ´8 `s
1 2 m1 ,m2 ‰t0,0u

35
Note the difference with Kiritsis. This is crucial to get the right factors of 2π in the end. This then gives:
λm1 m2 |Am1 m2 |2 d
´
∇2 ´1{2
ż ż8 4π`2
2πR ź e s 2πR ź 1 2π 2πR
Dχe´Spχq “ ˆ dAm,n “ ˆ “ ˆpdet1 q
`s p2π`s q2 `s m,n λm1 m2 `s 2π
m1 ,m2 PZ2ě0 zt0,0u ´8

Henceforth a primed sum or product means that we omit the origin 0 or t0, 0u and sum over the integers. It
remains to evaluate d ˜ ˆ ˙¸
ź 2π 1 ÿ1 2π 2
“ exp ´ log |m ` nτ |
λn,m 2 m,n τ2
Notice that this sum can be obtained by explicitly calculating the Eisenstein series
´ τ ¯s ÿ 1 1
2
Gpsq “
2π m,n |m ` nτ |2s

and evaluating 12 G1 p0q. Let’s do that. First note:

ÿ1 1 ÿ1 ÿ 1
2s
“ 2ζp2sq `
m,n
|m ` nτ | n m
|m ` nτ |2s

The derivative of 2ζp2sq at s “ 0 `yields

˘ ´2 logp2πq. On the other hand 2ζp0q is ´1, which multiplies the
τ2 s 0
order s factor in the expansion of 2π (none of the subsequent terms will have an Ops q term to multiply
this). This gives logp2π{τ2 q. Together these contribute
1
´ logp2πτ2 q
2
to 21 G1 p0q.
Note also because this is a periodic function of τ of period one, we can represent it as a Fourier series in τ
ż1 ż8
ÿ 1 ÿ
2πipnτ1 ´2πipt
ÿ 1 ÿ
2πipnτ1 1
“ e dte 2 ` n2 τ 2 qs
“ e dt 2
m
|m ` nτ | 2s
pPZ 0 mPZ
ppm ` tq 2 pPZ ´8 pt ` n2 τ22 qs
looooooooooomooooooooooon
ş1 ř
combine 0 with Z

Using a clever Gamma function manipulation (following Di Francesco here):

? ÿż 8
1 ÿ 8
ż ż8
2πippnτ1 ´tq s´1 ´xpt2 `n2 τ22 q π 2 2 2 2
dt dx e x e “ dx xs´3{2 e´xn τ2 ´π p {x`2πipnτ1 .
Γpsq p ´8 0 Γpsq p 0

Now at p “ 0 this reduces to ?

πΓps ´ 1{2q
|nτ2 |1´2s
Γpsq
?
Summing this over n gives 2 πΓps´1{2q
Γpsq ζp2s ´ 1q. We have explicit series formulae for these at s “ 0.
Extracting the first-order term (this is in fact finite at s “ 0) gives πτ32 .
Now let’s evaluate the sum over p ‰ 0. I’ll directly take s “ 3{2 here. We get a sum over an integral that is
now solvable:
? ?
πΓps ´ 1{2q ÿ ´2πipnτ1 8 ´3{2 ´xn2 τ2 ´π2 p2 {x ? ÿ π ´2πipnpτ1 `iτ2 q
ż
e x e “ πs pe ` e´2πipnpτ1 ´iτ2 q q
Γpsq pą0 0 pą0
πp

We see that the contribution to G1 p0q from this will be:

ÿ ÿ1 ÿ πτ2 πτ2
2 pq ` q̄q “ ´2 logp|1 ´ q n |2 q “ ´2 logpe 6 |ηpτ q|2 q “ ´2 logp|ηpτ q|2 q ´
p 3
loomoon p
ną0 ną0
ř 1
“ n

36
 ?
we see that the p “ 0 term cancels this last part and we are left with 12 G1 p0q “ ´ logp τ 2 2πq ´ logp|η|2 q,
and so:
R ÿ ´ πR22 |m´nτ |2
ZpR, τ q “ ? ˆ e τ2 `s .
`s τ2 |ηpτ q|2 m,n
While we’re at it, let’s simplify this even further by applying Poisson summation. We have the 1D case for
the Gaussian: ÿ 2 1 ÿ ´ π pn`i b q2
e´πan `πbn “ ? e a 2 .
n
a ñPZ
Performing this over the m variable we get
2nτ1
hkkkkikkkkj ˙2
2 ? 2 π`2
s τ2
ˆ
R2 nτ1
ÿ ´ πR
`2
pm2 ´m pnτ ` nτ̄ q `n2 |τ |2 q `s τ2 ÿ ´ πR
2 n2 |τ |2 ´ R2
m̃`i `s τ2
e s τ2 “ e `s τ2 e
m,n
R m̃,n
? 2 π`2
`s τ2 ÿ ´π R2 n2 τ2 ´
R2
s m̃2 τ ´2πim̃nτ
2 1
“ e `s
R m̃,n
? ´ ¯2 ´ ¯2
`s τ2 ÿ πpiτ1 ´τ2 q 21 `s
m̃` `R n πp´iτ1 ´τ2 q `Rs m̃´ `R n
“ e R s e s
R m̃,n
? 2 2
`s τ2 ÿ PL PR
“ q 2 q̄ 2
R m̃,n

with PL “ ?1 pm`s {R ` nR{`s q, PR “ ?1 pm`s {R ´ nR{`s q. We then get a simple form for the partition
2 2
function:
P2 P2
ÿ q 2L q̄ 2R
ZpR, τ q “ .
m̃,n
|ηpτ q|2

57. We follow Polchinski Vol 2 on advanced CFT. The following operator product arises when we calculate
correlation functions of the energy-momentum tensor:
ż
´T O “ ´Tz zpz, z̄q g d2 wφ∆,∆ pw, w̄q

We get:
„ 
∆ Bw
B̄z̄ T pz, z̄qφpw, w̄q “ B̄z̄ ` φpw, w̄q “ p´2π∆Bz δpz ´ wq ` 2πδpz ´ wqBw qφpw, w̄q
pz ´ wq2 z ´ w
Where the last line was obtained using basic delta-function identities. Integrating over w gives:

´B̄z̄ T O “ 2πgp∆ ´ 1qBφ

Thus, unless ∆ “ 1 we get that T gains an anti-holomorphic part. The exact same equation (with z Ñ z̄)
¯ zz ` BTz z̄ “ 0 gives us that
holds for T̄ . Further, the conservation equation dT

Tz z̄ “ 2πgp1 ´ ∆qφ.

There cannot be an overall constant, since this is zero when φ “ 0. Here we will define βpgq by:
ÿ1
Tii pz, z̄q “ ´2π βpgqOi pz, z̄q
i

where the sum runs over operators of dimension ď d. The trace is Taa “ 2Tz z̄ “ ´4πgp1 ´ ∆qφ so under this
deformation β “ p2 ´ 2∆qg. We now want to go to second order. The next contribution will come from:

g2
ż
1 2
´T pOOq “ ´Tzz pz, q̄ d2 wd2 w1 φpw, w̄qφpw1 , w̄1 q
2 2!

37
 Doing an OPE we get to leading order:
C
φ∆,∆ pw, w̄qφ∆,∆ pw1 , w̄1 q „ φ∆,∆ pw1 , w̄1 q
|z ´ w|2∆

where here C is the coefficient of the φ∆,∆ 3-point function. We can now preform the w, w1 integrals and
get: ż ż
2 dr
2πCg ˆ dw1 φpw1 , w̄1 q
r2∆´1
Assuming ∆ “ 1 we get a log term that must be regulated in the UV and IR. Regulation in the UV gives a
scale that breaks conformal invariance. Rescaling by 1 `  increases the log by . Equivalently we get

δg “ ´2πCg 2

This gives a second-order contribution to the beta function of Cg 2 as required. If the operator is not exactly
marginal, the second order term will still have this form, plus higher-order corrections in ∆ ´ 1 and g.

58. Generalizing the preceding analysis to a deformation by a family of marginal operators ga φa1,1 , for the
deformation to be marginal at second order in g we need the three-point function to satisfy λcab ga gb “ 0 so
that the second order term does not contribute the 1{r integral and thus does not break conformal invariance.
In this case that means that we require
λcab gaā gbb̄ “ 0.

59. Again, we work from the same chapter of Polchinski. For a general 2D QFT with a stress tensor, we can
define the quantities
F pr2 q “ z 4 xTzz pz, z̄qTzz p0, 0qy
Gpr2 q “ 4z 3 z̄ xTzz pz, z̄qTz z̄ p0, 0qy
Hpr2 q “ 16z 2 z̄ 2 xTz z̄ pz, z̄qTz z̄ p0, 0qy
From rotational invariance, these can only depend on r2 “ |z|2 . The conservation law B̄Tzz ` BTz z̄ “ 0 gives
us that:
4F9 ` G9 ´ 3G “ 0, 4G9 ´ 4G ` H9 ´ 2H “ 0
1
where F9 , G9 indicates the operator 2 rBr (ie differentiation wrt log r2 ). Note subtracting 3{4 of the second
one from the first gives:
3 3
4F9 ´ 2G9 ´ H9 “ ´ H
4 2
Define C “ 2F ´ G ´ 83 H. Note that in a CFT, where G “ H “ 0, C is exactly the central charge c.
Further, from this definition we get that in the general setting C9 “ ´ 43 H. But note that an exactly marginal
perturbation does not give the stress-energy tensor a trace, so C9 “ 0 and the central charge will remain
fixed.
This technology wasn’t developed in Kiritsis. I’m unsure how he would have wanted us to
show this.

60. Note under τ Ñ τ ` 1 the η function is invariant and we our constraint comes from:
1 2
pP ´ PR2 q P Z ñ Gij mj Gik Gkl nl “ mk nk P Z
2 L
as required. So in particular we have PL2 ´ PR2 P 2Z. We can interpret pPL , PR q as being a vector lying in an
even, Lorentzian lattice, with signature pN, N q. Note in the 1D case then get that
„ 
1 R `s R `s R `s R `s
P 1 ¨ P 2 :“ PL1 PL2 ´ PR1 PR2 “ p n ` mqp n1 ` m1 q ´ p n ´ mqp n1 ´ m1 q “ pmn1 ` nm1 q P Z
2 `s R `s R `s R `s R

38
 Going to higher dimensions and turning on G and B gives us the same result (take `s “ 1 for simplicity
here). All terms will cancel except the ones given by the relative minus sign of G on the second term
1” 1i i 1 i 1j ((1i((j (
(ı
mi n ` n mi ` (pn(n( ` n n qBij P Z
2
The last term cancels by antisymmetry. Here ni , mi P Z (note the index convention, different from Kiritsis).
Under τ Ñ ´1{τ the η function is a modular form of weight 1{2, so ηpτ qN is a modular form of weight N {2
and |ηpτ q|2N “ |τ |´N |ηp´1{τ q|2N . Let us now look at the remaining part
ÿ 1 2 1 2
Θpτ q :“ q 2 PL q̄ 2 PR
P “pPL ,PR qPΓ

is also a modular form of this weight. Let’s show this. We can use the Poisson resummation formula to
write: ÿ 1 ÿ 2πipp2 ÿ 1 ÿ ˆ
δpp ´ p1 q “ e ñ f ppq “ f pqq
p1 PΓ
VΓ p2 PΓ˚ pPΓ
VΓ qPΓ˚
2 2
here VΓ´1 is the covolume of Γ. Taking f “ eiπτ PL ´iπτ̄ PR and doing a 2N -dimensional Fourier transform, we
2 2
see that fˆpqq “ |τ1|N e´iπQL {τ `iπQR {τ̄ . We can use this to write:
„ 
ÿ 1 ÿ1 2 1 2
πipτ PL2 τ̄ PR2 q
“ ‰
Θpτ q “ exp ´ “ N exp πip´ QL ` QR q
P PΓ
|τ | VΓ QPΓ˚ τ τ̄

Now as long as Γ “ Γ˚ , that is, Γ is an even, Lorentzian, self-dual lattice. Then VΓ “ 1 and the sum over
Q P Γ˚ is the same as the sum over P P Γ. So we get

Θpτ q “ |τ |´N Θp´1{τ q

which is the exact same transformation law as the |η|2N in the denominator, and so we get that ZpRq is
indeed modular invariant.

61. We have in fact done this in the first part exercise 46.

62. Certainly this is an order 2 involution, just like R Ñ 1{R. Now we know Vm,n Ñ Vm,´n under this involution,
so
1 ÿ 1
¨rH 0 s „ C 2n,2m rV2n,2m s ` C 2n`1,2m rV2n`1,2m s “ prH 0 s ¨ rH 0 s ` rH π s ¨ rH | π|sq ` rH 0 s ¨ rH π s
n,m
2
ÿ 1
rH π1 s ¨ rH π1 s „ C 2n,2m rV2n,2m s ´ C 2n`1,2m rV2n`1,2m s “ prH 0 s ¨ rH 0 s ` rH π s ¨ rH | π|sq ` rH 0 s ¨ rH π s
n,m
2
1 1 ÿ 1
rH 0 s ¨ rH 0 π s „ C 2n,2m`1 rV2n,2m`1 s “ prH 0 s ¨ rH 0 s ´ rH π s ¨ rH π sq
n,m
2

the only consistent transformation with these OPEs is exactly:

ˆ 01 ˙ ˆ ˙ ˆ 0˙
H 1 1 1 H
“ ? π
H π1 2 1 ´1 H

63. Define the orbifold partition function as

1
ˆ ˙
` 1 ` ´ ` ´
 “ ` ` ` 
` 2 ` ` ´ ´

Note that the orbifolded theory itself has a Z2 symmetry obtained by taking all the states in the Z2 twisted
sectors to minus themselves:
˘
Ñ˘ ,
  ˘
Ñ ´˘  
` ` ´ ´

39
 I can now orbifold again by this symmetry, defining (as before):
1
ˆ ˙
˘ 1 `
“  `´ ˘` ˘´
   
` 2 ` ` ´ ´
1
ˆ ˙
˘ 1 `
“  ´´ ˘` ¯´
   
´ 2 ` ` ´ ´

Then forming the new partition function of this double orbifold theory I see that almost everything cancels:
ˆ 1 1 1 1
˙
1 `
`´ ``
    `´ “` 
2 ` ` ´ ´ `

?
64. Note first that at R{` “ 1{ 2 we get
n n
PL “ m ` , PR “ m ´
2 2
So we are summing over these lattice values in the numerator Θ of ZpRq. On the other hand, we have:

1 ÿ ˆ1 1 1 pn´1{2q2 1 pm´1{2q2
˙
2 2 2 n`m 12 n2 12 m2
p|θ2 | ` |θ3 | ` |θ4 | q “ p1 ` p´1q qq q̄ ` q 2 q̄ 2
2 n,m
2 2

This is a sum over all lattice points whose sum is an even integer union with the set of all half-lattice points,
but only half of the half-lattice points are counted in the sum. This agree exactly with the standard weighting
1
for the lattice generated by p1,?1q and 2 p1, ´1q which is exactly the original theta function numerator in the
untwisted ZpRq at R{`s “ 1{ 2.
Squaring the Ising model theta function then gives:
ˆ ˙
|θ2 θ3 | ` |θ3 θ4 | ` |θ2 θ4 | 1 1 1 1 |η| |η| |η|
` p|θ2 |2 ` |θ3 |2 ` |θ4 |2 q “ ZpRq ` ` `
4|η|2 4 |η| 2
loooooooooooooooomoooooooooooooooon 2 2 |θ2 | |θ3 | |θ4 |
1
2
ZpRq

exactly as we wanted.

65. Take `s “ 1 here. The partition function will still have 1 twisted sector and a single projection. So we
need to consider 4 terms. We have Z 00 “ ZpR1 , R2 q “ ZpR1 qZpR2 q. Our vertex operators are labelled by
“‰

pm1 , n1 , m2 , n2 q, and g acts as pm1 , n1 , m2 , n2 q Ñ p´1qm2 p´m1 , ´n1 , m2 , n2 q. And so:

1 Ñ´X 1
Xhkkikkj
„  ˇ ˇ „ ¯2 
1 0 ˇηˇ 1 ÿ ´ ¯2 ´
Z “ Tr1 rg q L0 ´c{24 q̄ L̄0 ´c̄{24 s “ ˇ ˇ m iπτ m
p´1q exp 2 R2 ` nR2 ´ 2 R2 ´ nR2 iπτ̄ m
2 1 ˇ θ2 ˇ η η̄ m,n
loooooooooooooooooooooooooooooooooooooomoooooooooooooooooooooooooooooooooooooon
X 2 ÑX 2 `πR2
„  ˇ ˇ „ ¯2 
1 1 L0 ´c{24 L̄0 ´c̄{24
ˇηˇ 1 ÿ iπτ
´
m 1
¯2
iπτ̄
´
m 1
Z “ Trg rg q q̄ s“ˇ ˇ
ˇ ˇ exp 2 R2 ` pn ` 2 qR2 ´ 2 R2 ´ pn ` 2 qR2
2 0 θ4 η η̄ m,n
„  ˇ ˇ „ ¯2 
1 1 L0 ´c{24 L̄0 ´c̄{24
ˇηˇ 1 ÿ m iπτ
´
m 1
¯2
iπτ̄
´
m 1
Z “ Trg rg q q̄ s“ˇ ˇ
ˇ ˇ p´1q exp 2 R2 ` pn ` 2 qR2 ´ 2 R2 ´ pn ` 2 qR2
2 1 θ3 η η̄ m,n

it is clear that the sum of all these is modular invariant. I am unsure if I should try to simplify this further.
Certainly (unlike the freely-acting orbifold case) this doesn’t look trivial. This is the CFT of fields valued in
the Klein bottle.

40
66. Take `s “ 1 here. The symmetry interchanges |m1 , n1 , m2 , n2 y Ñ |m2 , n2 , m1 , n1 y We have Z 00 “ ZpRq2 .
“‰

In the g-trace, we will need m1 “ m2 , n1 “ n2 . Then, excitations around this state must have equal mode
1 2 1 2
number in m1 , m2 and n1 , n2 to contribute to the g-trace so for each factor of q 2 PL q̄ 2 PR we have
„  „ 
0 ÿ iπ 2τ ` m ˘2 iπ 2τ̄ ` m ˘2 ź 1 ź 1
Z “ pq q̄q´2{24 exp R ` nR ´ R ´ nR 1
1 m,n
2 2 n1
1 ´ q 2n
m1
1 ´ q̄ 2m1
1 ÿ ” `m ˘2 `m ˘2 ı 2 ÿ
“ exp iπτ R ` nR ´ iπτ̄ R ´ nR “ ...
|ηp2τ q|2 m,n |ηpτ q||θ 10 pτ q|
“ ‰

On the other hand, the twisted sector we have boundary conditions X 1 pσ`2πq “ X 2 pσq, X 2 pσ`2πq “ X 1 pσq.
Applying τ Ñ ´1{τ on the preceding we get:
„  „ 
1 1 ÿ iπτ ` m ˘2 iπτ̄ ` m ˘2
Z “ exp ` nR ´ ´ nR
|ηpτ q||θ 01 pτ q| m,n 4 R 4 R
“‰
0

Taking τ Ñ τ ` 1 gives
„  „ 
1 1 ÿ
mn iπτ ` m ˘2 iπτ̄ ` m ˘2
Z “ p´1q exp ` nR ´ ´ nR .
|ηpτ q||θ 00 pτ q| m,n 4 R 4 R
“‰
1

Let us check if this is modular invariant. Clearly Z 00 maps to itself under both S and T . Under T ,
“‰
“0‰ “1‰ “1‰
Z 1 maps to itself, and Z 0 and Z 1 get exchanged by the properties of theta functions. Further, under
τ Ñ ´1{τ Z 10 and Z 01 map to one another. However, Z 11 does not map to itself under S, and we are led
“‰ “‰ “‰

to conclude that this Z2 symmetry is anomalous.

67. If we orbifold the single free scalar by acting as |m, ny Ñ p´1qm`n |m, ny we have Z 00 “ ZpRq as before,
“‰

but now: „  ÿ
0 ” ˘2 iπτ̄ ` m ˘2 ı
p´1qm`n exp iπτ
`m
Z “ 2 R ` nR ´ 2 R ´ nR
1 m,n

Taking τ Ñ ´1{τ gives that both m and n shift by 1{2

« ˙2 ˙2 ff
1 1
„  ÿ ˆ ˆ
1 iπτ m´ 2 1 iπτ̄ m´ 2
Z “ exp 2 R ` pn ´ 2 qR ´ 2 R ´ pn ´ 21 qR
0 m,n

Then doing τ Ñ τ ` 1 gives:

„  ÿ „ ¯2 ¯2 
1 m`n` 21
´
m´ 12
´
m´ 12
Z “ p´1q exp iπτ
2 R ` pn ´ 1
2 qR ´ iπτ̄
2 R ´ pn ´ 1
2 qR
1 m,n

this already looks a little “weird. Out front we don’t necessarily have a ˘1. Further, doing τ Ñ τ ` 1 again
1
‰
does not get us back to Z 0 , we need τ Ñ τ ` 3.

68. In the untwisted sector we have our vacuum state |0y, with ∆ “ ∆ ¯ “ 0 as required. Now consider the
kth twisted sector. We have creation and annihilation operators αn`k{N satisfying the same commutation
relations rαr , αs s “ rδr`s . However as X is a complex boson, the αr are complex numbers and so we have
two sets of them (which we can call αr , ᾱr following previous convention). From commuting them across,
we get:
w z k{L
8 ` ˘
1 ÿ ´ w ¯r
z w
xXpzqBXpwqy “ 2 ˆ “2ˆ
w z z´w
r“minp1,k{N q

Then, differentiating with respect to z gives:

2 ´ w ¯k{N
xBXpzqBXpwqy “ ´ p1 ´ Lk p1 ´ z
w qq
pw ´ zq2 z

41
 Taking the finite part of this ´ 12 of expression as w Ñ z gives us:

kpL ´ kq
xT y “
2L2
as required.

69. We have the scalar propagator written in terms of the eigen-modes as:

`2s ÿ 1 1
xXpzqXp0qy “ ´ e2πipmσ1 `nσ2 q
2 m,n |m ` nτ |2

Rather than trying to massage this into our appropriate logarithm of theta functions, let’s appreciate what
properties we want our correlator to have. For z Ñ 0, the small-distance behavior of the correlator should
reproduce the CP1 result, so we namely need it to go as:

`2s
´ log |z|2 ` Opzq
2
Further, the only singularity on the torus is at z Ñ 0, nowhere else. Thus we should be able to write our
correlator as
`2
´ s log Gpz, z̄q
2
where G must be a doubly-periodic harmonic function with a single zero at z “ 0 on the torus and no
poles. There are no such holomorphic functions since all non-constant elliptic functions need to have an
equal number of zeros and poles (and also more than one zero, since the coefficients of all zeros must sum
to 0). In other words, instead of looking at an elliptic function we should be looking at a section of a line
bundle over the torus with a single zero.
We see that the theta functions give us exactly this- and moreover rational functions of the theta functions
generate
“1‰ all such sections. The constraint of a single zero at z “ 0 together with modular invariance singles
out θ 1 uniquely. To give it the appropriate coefficient of the zero, we must have:

ˇ θ 1 pz, τ q ˇ2
ˇ “‰ ˇ
ˇ 1
Gpzq “ ˇ ˇ ˆ p1 ` Opzqq
ˇ
“1‰
ˇ Bz θ 1 p0, τ q ˇ

The problem is that this is a quasi-periodic foundtion in z. Under z Ñ z ` τ we get that log G Ñ log G `
z2
´2π τ2
2πτ2 ` 4πImpzq. This can be remedied by adding e 2 to G.
2 {τ q
Also, under τ Ñ 1{τ , z Ñ z{τ from the ratio we pick up a factor of | exppiπz 2 {τ q|2 “ e´2πImpz . But this
2
´2π Imz
is exactly the same factor as is picked up by e 2 , so adding this term fixes modular invariance as well.
τ

Our final result is then:

ˇ θ 1 pz, τ q ˇ2
ˇ “‰ ˇ
2
ˇ 1 ˇ ´2π pImzq
Gpzq “ ˇ “1‰ ˇ e τ2 .
ˇ Bz θ 1 p0, τ q ˇ

So we now have an explicit formula for ∆pz ´ w, τ q on the torus. The Klein bottle is given by identifying
z – ´z̄ ` τ {2. Then we expect the propagator to be

∆K2 pz ´ wq “ ∆pz ´ w, 2itq ` ∆pz ` w̄ ` it, 2itq

Next, for the cylinder we have the involution z – ´1{z̄ so we have the propagator:

∆C2 pz ´ wq “ ∆pz ´ w, itq ` ∆pz ` w̄, itq

Finally, for the Möbius strip, we have two involutions and get

∆M2 pz ´ wq “ ∆pz ´ w, 2itq ` ∆pz ` w̄, 2itq ` ∆pz ´ w ´ 2πpit ` 12 q, 2itq ` ∆pz ` w̄ ` 2πp´it ` 12 q, 2itq

42
 cyl
70. We already know how to calculate TrN S{R pp˘1qF q L0 q for the free fermion. Ω acts by sending a left-moving
state to a right-moving one and vice-versa. Only states that are left-right symmetric survive. First lets do
the NS sector. There is a single vacuum and we get:
d
8
ź θ3 p2itq
TrNS rΩe´2πtpL0 `L̄0 ´c{12q s “ e2πt{24 p1 ` e´2πtˆ2pn´1{2q q “
n“1
ηp2itq
d
8
ź θ3 p2itq
TrNS rΩp´1qF e´2πtpL0 `L̄0 ´c{12q s “ e2πt{24 p1 ` e´2πtˆ2pn´1{2q q “
n“1
ηp2itq

Note that these two are the same, since only sectors with an equal number of left movers and right-movers
contribute, and this necessarily forces F to be even. Then, for the Ramond sector we have
d
? ´2πtp1{16´1{48q ź8
´2πtpL0 `L̄0 ´c{12q ´2πtˆ2n θ2 p2itq
TrR rΩe s “ 2e p1 ` e q“
n“1
ηp2itq
TrR rΩp´1qF q L0 ´c{24 q̄ L̄0 ´c̄{24 s “ 0

where the last one is zero as before, since for any state, there is a corresponding one with opposite p´1qF
eigenvalue, related by zero-modes.

43
Chapter 5: Scattering Amplitudes and Vertex Operators
0. A worthwhile exercise (that is not in the book) is to show that we have the correct Regge behavior of
the Virasoro-Shapiro amplitude at large s, fixed t. From Stirling’s approximation for large s, we have
Γpa`sq a´b , so:
Γpb`sq „ s

Γp´1 ´ `2s t{4q Γp´1 ´ `2s s{4qΓp3 ` `2s t{4 ` `2s s{4q Γp´1 ´ `2s t{4q 1``2s t{4 Γp´1 ´ `2s s{4q
S4 ps, t, uq „ „ s
Γp2 ` `2s t{4q Γp2 ` `2s s{4qΓp´2 ´ `2s t{4 ´ `2s s{4q Γp2 ` `2s t{4q Γp´2 ´ `2s t{4 ´ `2s s{4q
Γp´1 ´ `2s t{4q 1``2s t{4 Γp3 ` `2s t{4 ` `2s s{4q sinp`2s ps ` tq{4q
“ s
Γp2 ` `2s t{4q Γp2 ` `2s s{4q sinp`2s s{4q
Γp´1 ´ `2s t{4q sinp`2s u{4q 2``2s t{2
Ñ s
Γp2 ` `2s t{4q sinp`2s s{4q

Using the same argument, in the large s, t, u limit, we get the following soft behavior:
2 2 2
s´1´`s s{4 t´1´`s t{4 u´1´`s u{4 `2
´ 2s pps`3q log s`pt`3q log t`pu`3q log uq
`2
´ 2s ps log s`t log t`u log uq
S4 ps, t, uq „ 2 2 2 „ e Ñ e
s2``s s{4 t2``s t{4 u2``s u{4

1. Note that we need 3 more c ghosts than b ghosts since the difference of the zero modes must be three.
Now,
ś c has ś scaling dimension 1 and b has scaling dimension ´2 so the total scaling of the correlator
x n`3
i“1 cpz i q n
j“1 bpwj qy will be 3 ´ n. Thus, viewed in the complex plane, we expect it to be a homogenous
rational function of degree exactly 3 ´ n.
We will have n contractions of the bs and cs with 3 cs left over. This gives:
n`3 n
ź ź pzn`1 ´ zn`2 qpzn`1 ´ zn`3 qpzn`2 ´ zn`3 q
x cpzi q bpwj qy “ ˆ c.c. ` perms.
i“1 j“1
pz1 ´ w1 q . . . pzn ´ wn q

where each permutations will pick up a sign for every odd combined permutation of the zi , wj . Another way
to do it is as follows:
As stated before, the correlator when viewed in the complex plane will be a homogenous rational function of
degree exactly 3 ´ n. That way, it will be finite at infinity. We also know that this function is antisymmetric
upon swapping any of the zi , any of the wi , or any of the zi with the wi . Further, if any of the zi “ zj or
1
wi “ wj , this function will vanish. On the other hand, if zi “ wj , we expect a contribution of a pole zi ´wj
.
There is only one such homogenous rational function:
śn`3 śn
iăj pzi ´ zj q iăj pwi ´ wj q
śn`3 śn .
i“1 j“1 pzi ´ wj q

This is indeed of degree 3 ´ n, as desired.

2. It is clear from plugging things in that when z1 Ñ 0, z2 Ñ 1, z3 Ñ 8, the 4-point tachyon amplitude becomes:
ż
8πi 2 2 2 2 2 2
lim 2 gc2 δ
26 pΣpq|z3 |2 |z3 ´ 1|2 d2 z4 |z4 |`s p1 ¨p4 |1 ´ z4 |`s p2 ¨p4 |z4 ´ z3 |`s p3 ¨p4 1`s p1 ¨p2 |z3 |`s p1 ¨p3 |z3 ´ 1|`s p2 ¨p3
z3 Ñ8 `s

here δ
“ 2πδ. Note all the terms that go to infinity cancel, since `2s p3 ¨ pp1 ` p2 ` p3 q “ ´`2s p23 “ ´4 which
cancels with the two powers of two outside the integral. Next, `2s p1 ¨ p4 “ 12 pp1 ` p4 q2 ´ 12 p`2s p21 ´ `2s p24 q “
´`2s t{2 ´ 4 etc so we get: ż
8πi 2 26 2 2
2
gc δ
pΣpq d2 z4 |z4 |´`s t{2´4 |1 ´ z4 |´`s u{2´4
`s
as required.

44
3. For a conformal transformation we have |x1ij |2 “ Ωpxi qΩpxj q|xij |2 where Ωpxi q is the local scale factor
det Bx1 {Bx evaluated at xi . Then, the N -point tachyon amplitude will pick up Ωpx1 q2 Ωpx2 q2 Ωpx3 q2 from the
three terms outside of the integral. The terms inside the integral can be written as:
2
ź
p|zij |2 q`s pi ¨pj {2
iăj

so zi in this term will pick up a power of j‰i `2s pi ¨ pj {2 “ ´`2s p2j {2 “ ´2 on its scale factor. This exactly
ř

cancels for z1 , z2 , z3 . For the other zi , we note that d2 zi will pick up the factor Ωpzi q2 upon transformation.
ş
Another way to do this is directly from noting that each d2 zi Vpi pzi , z̄i q for i ą 3 is invariant under conformal
transformation, and cpzi qc̄pz̄i qVpi pzi , z̄i q has scaling dimension zero, so transforms trivially under SL2 pCq
transformations.

4. Note that the three-point tachyon amplitude is very simple and independent of momenta aside from a delta
function: Spk1 , k2 , k3 q “ 8πi g δ 26 pΣkq.
`2 c

s

Let’s now consider the limit of a nearly on-shell particle of momenta k. From elementary field theory we get:
˙2
d26 k SS 2 pk1 , k2 , kqSS2 p´k, k3 , k4 q
ż ˆ
8πi 1
Spk1 , k2 , k3 , kq4q „ i “i gc2 δ
26 pΣki q
p2πq26 ´k 2 ` 4{`2s ` i `2s s ` 4`2s ` i

This has a pole when ´pk1 ` k2 q2 “ s “ ´4{`2s . We see that (ignoring the δ
term) this gives a residue of
2
´i 64π
`4
gc2
s

On the other hand we have from 5.2.5 a residue of:

8πi 2 4 64π 2 2
g ˆ 2π ˆ ´ “ ´i g
`2s c `2s `2s c

exactly consistent with unitarity. Note we needed every constant to be as it was so that we could get such
agreement.

5. The massless state corresponds to ζµν BX µ BX ν eip¨X . We don’t have to integrate.

Let’s calculate the correlator

ź 1
´ 2¯´ ¯´ 2
¯´ ¯
xBXpz1 qB̄Xpz1 qeik1 Xpz1 q eik2 Xpz2 q eik3 Xpz3 q y “ iCSX2 δ
26 pΣpq |zij |α ki ¨kj ´ i`2s k2
z12 `
k3
z13 ´ i`2s k2
z̄12 ` k3
z̄13
iăj

45
 with the ghost correlator this gives:
´ ¯´ ¯
´`4s 26
ź 1
iCSX2 CSgh2 4 δ
pΣpq |zij |α ki ¨kj `2 k2
z12 ` k3
z13
k2
z̄12 ` k3
z̄13
iăj

Now k12 “ 0 “ k1 ¨ k2 ` k1 ¨ k3 . On the other hand ´4{`2s “ ´k22 “ k2 ¨ k3 ` k1 ¨ k2 “ ´k32 “ k2 ¨ k3 ` k1 ¨ k3 .

Solving this gives k1 ¨ k2 “ k1 ¨ k3 “ 0 while k2 ¨ k3 “ ´4{`2s . Then, taking z1 Ñ 0, z2 Ñ 1, z3 Ñ 8 gives:

`2s X gh 26
´i C 2 C 2 δ
pΣpqζµν k2µ k2ν
4 S S
Further, we have that ζµν k1µ “ ζµν pk2 ` k3 qµ “ 0 so we can rewrite this symmetrically as

`4s X gh 26 µ ν iπ`2s 1 26 µ ν
´i C 2 C δ pΣpqζµν k k
23 23 “ ´ g δ
pΣpqζµν k23 k23 .
S S2 2 c


16 looomooon
:“8πgc1 {`2s

The overall constants can be determined from unitarity. The pole of the Veneziano amplitude at s “ 0 has
residue (using that s “ 0, s ` t ` u “ ´16{`2s ) that is a delta function times:

`2
s 2
p 8 pt´uqq
hkkkkkkikkkkkkj
8πi 2 4 Γp´1 ´ `2s t{4qΓp3 ` `2s t{4q p4πq2 2 4 2 2 2 2 pt ´ uq
2
g ˆ 2π ˆ “ ´i g ˆ p2 ` ` t{4q “ ´iπ g (55)
`2s c `2s s Γp´2 ´ `2s t{4qΓp2 ` `2s t{4q `2s c `2s s s c
s

On the other hand, factorization of this into amplitudes with massless states yields a delta function times:

1 1 2
2 pu ´ tq
ÿ µ ν σ ρ
2 2 2
iC3pt ζµν ζσρ k12 k12 k34 k34 ˆ “ iC pk
3pt 12 ¨ k34 q ˆ “ iC 3pt (56)
ζ
pk1 ` k2 q2 ` i s s

where we have used that, just as the sum over intermediate photon polarizations µ ˚ν can be replaced by just
ηµν , the sum over intermediate polarizations ζµν ζρσ be replaced by 21 pηµρ ηνσ ` ηµσ ηνρ q. Comparing equations
(55) and (56) We the get C3pt “ ´πigc . Equivalently, gc1 “ 2gc {`2s .

6. This problem is so nasty - I’m pretty sure Kiritsis meant for us to just look at scattering 4 open string states
- which in and of itself is nasty enough.
We have already determined the normalization in the previous question. It is also simple to check that it
is correct to attach gc1 to each vertex operator in the 3-point and 4-point functions by considering first the
2 tachyon Ñ 2 massless state scattering in the t and u channels, which relates the 3-point scatterings of
tachyons and massless states to one another, and then use the 2 Ñ 2 tachyon to tachyon scattering to express
its normalization in terms of the 3-point tachyon amplitude. All of this equates to taking gc1 “ 2gc {`2s .
As a warm-up lets do the three-point massless amplitude. We compute the correlator

x: BX α pz1 qeipi Xpz1 q : : BX β pz2 qeipi Xpz2 q : : BX γ pz3 qeipi Xpz3 q : ˆc.c.y

In the holomorphic part, there are two types of contribution: One where each BX contracts with an exponen-
tial and one where two of the BX contract
ś with one another and the last one contracts with an exponential.
2
Further, we see that pi ¨ pj “ 0, so the iăj |zij |`s pi ¨pj is unity. The first contribution gives:
˙3 ˆ ˙3
`2s k2α kα k1α kα k1α kα `2s
ˆ ˙ˆ ˙ˆ ˙ ˆ
1
i ` 3 ` 3 ` 2 Ñi pk1 ´ k2 qγ pk2 ´ k3 qα pk3 ´ k1 qβ
2 z12 z13 z21 z23 z31 z32 2 22

The second contribution gives

ˆ 2 ˙2 « αβ ˆ γ ˜ ¸ff
k2γ η βγ k2α k3α η αγ k1β k3β
˙ ˆ ˙
`s η k1
i 2 ` ` 2 ` ` 2 `
2 z12 z31 z32 z23 z12 z13 z13 z21 z23

46
 Multiplying this by the c contribution z12 z23 z13 ˆ c.c. and setting z1 “ 0, z2 “ 1, z3 “ 8 we get the 3-point
amplitude:

γ β `2s γ α β
πigc ζ1,αᾱ ζ2,β β̄ ζ3,γγ̄ T αβγ T ᾱβ̄γ̄ , T αβγ “ η αβ k12 ` η βγ k23
α
` η αγ k31 ` k k k . (57)
8 12 23 31

Now let’s do the four-point amplitude. First, I will work with the open string (no CP indices, so U p1q gauge
symmetry) and use some tricks at the end to get the closed string amplitude. For the open string, there are
six possible orderings of the y1 , y2 , y3 , y4 . In three of these cases we can send y1 Ñ 0, y2 Ñ 1, y3 Ñ 8 and
vary y4 . In the other three, cases we switch y2 and y3 . This amounts to swapping s Ø t. HOWEVER for
Polchinski’s trick, I only need to consider one of these six. WLOG I set y4 to be between y1 , y2 in 0, 1. I’ll
also absorb `2s in the definition of s, t, u. So we have,
ź
|yij |2ki ¨kj Ñ |y|´u |1 ´ y|´t Ø |y|´u |1 ´ y|´s
iăj

We now get three types of contributions: If all the BX α contract with each other (3 terms), if two of the
BX α contract with each other (6 terms) and the remaining two contract with one of the eiki ¨X , or if they all
contract with the eiki ¨X (1 term).
In the first case we get:

η αγ η βδ η αδ η βγ
ˆ ˙ ˆ ˙
˘2 1 1 1 ` ˘2 αβ γδ
´2`2s Ñ 2`2s
`
2 y2 ` y2 y2 ` y2 y2 η η ` `
y12 34 13 24 14 23 p1 ´ yq2 y2

Integrating y from 0 to 1 gives

igo2 δ
26
ˆ ˙
2 2 Γp1 ´ tqΓp1 ´ uq Γp1 ´ tqΓp´1 ´ uq Γp´1 ´ tqΓp1 ´ uq
ˆ p2`s q ` ` (58)
4`4s Γp2 ` sq Γpsq Γpsq

kj
Now the annoying one1 . Define Ki “
ř
j‰i yij . Note:

k3α k4β k1δ kδ

K1 “ ´k2α ´ , K2 “ k1β ` , K3 Ñ ´p1 ` yqk1γ ´ yk2γ ´ k4γ , K4 “ ` 2 .
y 1´y y y´1

igo2

δ 26
We can now write the pα1 q3 contribution as 4`4s
p2`2s q3 times:
ˆ ˙
K3 K4 αβ K1 K2 γδ K1 K4 βγ K2 K3 αδ K2 K4 αγ K1 K3 βδ
2 η ` 2 η ` 2 η ` 2 η ` 2 η ` 2 η
y12 y34 y23 y14 y13 y24
k4β
”
kδ k2δ k4α k4α k1δ k2δ
Ñ ´ pp1 ` yqk1γ ` yk2γ ` k4γ qp y1 ` y´1 qη
αβ
´ pk2α ` β
y qpk1 ` 1´y qη
γδ
´ pk2α ` y qp y ` y´1 qη
βγ

k4β k4β
ı
αδ k1δ k2δ k4α βδ
´ pk1β ` 1´y qpp1 ` yqk1γ ` yk2γ ` k4γ q ηy2 ` pk1β ` 1´y qp y ` y´1 qη
αγ
` pk2α ` y qpp1 ` yqk1γ ` yk2γ ` k4γ q p1´yq
η
2

Now we use shorthand ki`j “ ki ` kj . Looking at the first of the six terms above, we get the order pα1 q3
igo2

δ 26
to our scattering amplitude to be 4`6s
p2`2s q3 multiplying:
ż1 ´ ¯
αβ γ γ γ γ
´η dy k1δ ry ´1 k1`4 ` k1`2 s ` k2δ rpy ´ 1q´1 k1`4 ` ypy ´ 1q´1 k1`2 s |y|´u |1 ´ y|´t
0
ˆ ˙
γ Γp1 ´ tqΓp´uq γ Γp1 ´ tqΓp1 ´ uq γ Γp´tqΓp1 ´ uq γ Γp´tqΓp´uq
“ η αβ k1δ k1`4 ` k1δ k1`2 ´ k2δ k1`4 ´ k2δ k1`2 ` 5 perms.
Γp1 ` sq Γp2 ` sq Γp1 ` sq Γpsq
(59)
1
Wasted all of 1/17/20 on this. Not worth it

47
 Now for the order pα1 q4 term. This is given by contracting each BX against an exponential, yielding
K1α K2β K3γ K4δ . Again we have y in r0, 1s
˙˜ ¸
k4β
ż1
k4α ¯ ˆ kδ kδ
ˆ ´ ˙
1 4 26 2 4 α β γ γ 1
ipgo q CD2 δ
p2`s q dy k2 ` k1 ` k1`4 ` yk1`2 ` 2 y ´u p1 ´ yq´t
0 y 1 ´ y y y ´ 1

δ 26
igo2

This gives a 24 “ 16 terms. Its not terrible. The pα1 q4 term is 4`6s
p2`2s q4 times:

k2α k1β k1`4

γ
k1δ Bp´u, 1 ´ tq ` k2α k1β k1`4
γ
k2δ Bp1 ´ u, ´tq ` k2α k1β k1`2
γ
k1δ Bp1 ´ u, 1 ´ tq ` k2α k1β k1`2
γ
k2δ Bp2 ´ u, ´tq
` k2α k4β k1`4
γ
k1δ Bp´u, ´tq ` k2α k4β k1`4
γ
k2δ Bp1 ´ u, ´1 ´ tq ` k2α k4β k1`2
γ
k1δ Bp1 ´ u, ´tq ` k2α k4β k1`2
γ
k2δ Bp2 ´ u, ´1 ´ tq
` k4α k1β k1`4
γ
k1δ Bp´1 ´ u, 1 ´ tq ` k4α k1β k1`4
γ
k2δ Bp´u, ´tq ` k4α k1β k1`2
γ
k1δ Bp´u, 1 ´ tq ` k4α k1β k1`2
γ
k2δ Bp1 ´ u, ´tq
` k4α k4β k1`4
γ
k1δ Bp´1 ´ u, ´tq ` k4α k4β k1`4
γ
k2δ Bp´u, ´1 ´ tq ` k4α k4β k1`2
γ
k1δ Bp´u, ´tq ` k4α k4β k1`2
γ
k2δ Bp1 ´ u, ´1 ´ tq
(60)
The open string amplitude is then given by summing equations (58), (59) and (60) and multiplying that
2 26
δ
result by ig4`
o
αβγδ
6 . Call this Ao ps, t, u, `s , go q. Using Polchinski 6.6.23 we can write the closed string
s
amplitude as:

πigc2 `2s 4
Ac ps, t, u, `s , gc q “ ζ1,αᾱ ζ2,β β̄ ζ3,γγ̄ ζ4,δδ̄ go sinpπ`2s tqAαβγδ
o ps, t, u, `s {2, go qrAᾱo β̄γ̄ δ̄ pt, u, s, `s {2, go qs˚
go4

where ζ are our 242 closed string polarization vectors.

If we had not determined the relationship between gc1 and gc from the prior problem, we could have determined
it by using the KLT relation of the above formula from Polchsinski and specialized to relating go1 and go .
Then, we would only have needed to look at the (nice) leading order pα1 q2 term in this calculation and
observed the pole structure at s “ 0 corresponding to massless exchange. Making this agree with the square
of the 3-point amplitude would then be sufficient. We illustrate the open string case with CP factors in
exercise 11

7. There are three types of propagators to consider: bulk-bulk, bulk-boundary, and boundary-boundary. Using
shorthand Xi “ Xpzi , z̄i q, XI “ XpwI q, from 4.7.9 we have:
m n
ź ź ” ÿ 1ÿ ÿ ı
x eipi Xi eiqI XpwI q y “ δ
26 pΣp ` Σqq exp ´ pi pj xXi Xj y ´ pi qI xXi XI y ´ qI qJ xXI XJ y
i“1 I“1 iăj
2 i,I IăJ

Using the form of the propagators

`2s
xXi Xj y “ ´ plog |zi ´ zj |2 ` log |zi ´ z̄j |2 q
2
`2
xXi XI y “ ´ s plog |wI ´ zi |2 ` log |wI ´ z̄i |2 q
2
xXI XJ y “ ´`2s log |wI ´ wJ |2

we get
m
2 2 2 2 2
ź ź ź ź
δ
26 pΣp ` Σqq |zi ´ z̄i |`s pi {2 |pzi ´ zj qpzi ´ z̄j q|`s pi ¨pj |wI ´ wJ |2`s qI qJ |pwI ´ zi qpwI ´ z̄i q|`s pi ¨qI
i iăj IăJ I,i

Note an additional term which I believe Kiritsis dropped. The extension to RP2 is no more difficult. We
2
now have no boundary and the xXi Xj y propagator is ´ `2s plogpzi ´ zj q ` logp1 ` zi z̄j qq so we get:
2 2 2 2
ź ź
δ
26 pΣp ` Σqq |1 ` zi z̄i |`s pi {2 |pzi ´ zj qp1 ` zi z̄j q|`s pi {2
i iăj

48
 8. Forgetting c ghosts here, I can just integrate over all of H. The massless closed-string state of zero momentum
is given by BXpzqB̄Xpzq. Note that H “ PSL2 pRq{SOp2q, so that:
`2 gc `2s `2s π`2s
ż ż
1 1 1 dxdy VolpHq
´ s2 dz “ ´ “ ´ “ ´
2go VolpPSL2 pRqq H |z ´ z̄|2 8 VolpPSL2 pRqq H y 2 8 VolpPSL2 pRqq 2
Note that this answer is finite and invariant under conformal transformation. This gives an amplitude of
´ πi
2δ
26

p0q.
9. Let p1 be the momentum of the closed-string tachyon, and p2 , p3 the the momenta of the open string tachyons.
We get 2p2 ¨ p3 “ p21 ´ p22 ´ p23 “ 2{`2s ñ p2 ¨ p3 “ 1{`2s , 2p1 ¨ p2 “ p23 ´ p22 ´ p21 “ ´4{`2s ñ p1 ¨ p2 “ ´2{`2s

I no longer have enough freedom to fix all three points. I can send one to 8 on the real line, and fix the
position of the closed string to be i P H. The remaining open string insertion can be anywhere on the real
line, so we must integrate over this. The ghost and vertex operator correlator gives:
ż
2 2 2 2 2
pz1 ´ z̄1 qpz1 ´ w3 qpz̄1 ´ w3 q |z1 ´ z̄1 |`s p1 {2 |z1 ´ w3 |2`s p1 ¨p3 dw2 |w2 ´ w3 |2`s p2 ¨p3 |w2 ´ z1 |2`s p1 ¨p2 δ
pΣpq
R

Setting z1 “ i, w3 Ñ 8 has momentum conservation and p23

“ 1{`2s , p21 “ 4{`2s getting the w3 factors to drop
out. We are left with
? Γp´ 21 ` 2q
ż
2 2 2
2i 2`s p1 {2 dw pw2 ` 1q`s p1 ¨p2 δ
pΣpq “ 8i π δ
pΣpq “ 4πiδ
pΣpq
R Γp2q
This gives a scattering amplitude of:
4πgo2 26
´ δ
pΣpq.
`2s
10. The conformal Killing group is now SOp3q. Again, we can fix one operator to be at z “ 0, but the other one
can be at any value of |z| P r0, 1s (we have control over the phase). So we must integrate over the modulus.
We do this on the disk using the RP2 propagator. We insert one vertex operator at 0 and the other z. The
integral gives a delta function times:
ż1 ż1
2 `2s p2 {2 2 `2s p2 {2 ´`2s p2 2 2 2 2
d|z2 |cpz1 qc̄pz̄1 qcpz2 qp1 ` |z1 | q p1 ` |z2 | q |pz1 ´ z2 qp1 ` z1 z̄2 q| Ñ rdr r´`s p p1 ` r2 q`s p {2
0 0

For the closed string tachyon, we have “ p2 4{`2s .

The integral is divergent, coming from the pz ´ wq´4
singularity as the two tachyons approach one another. If we had the milder pz ´ wq´1 singularity of the
open-string tachyon, this could be fixed. REVISIT
11. To simplify this problem, as Polchinski asks in his problem 6.9, I will look at the terms that contribute to the
e1 ¨ e2 e3 ¨ e4 amplitude, which comes from contracting BX α py1 qBX β py2 q and BX β py3 qBX δ py4 q. There are six
14 ` ˘2
possible orderings for the trace in the 4-point amplitude. We get gig2o`2 δ
26 pΣpqˆ 2`2s multiplying a sum of six
o s
integrals. Using s :“ ´`2s pp1 ` p2 q2 “ ´2p1 ¨ p2 , t :“ ´`2s pp1 ` p3 q2 “ ´2p1 ¨ p3 , u :“ ´`2s pp1 ` p4 q2 “ ´2p1 ¨ p4
and the shorthand r1234s for Trpλµ1 λµ2 λµ3 λµ4 q, we get:
„ ż0 ż1 ż 8
r1234s `r1423s `r1243s p|w|´u |1 ´ w|´t qdw
´8 0 1
„ ż0 ż1 ż 8
` r1324s `r1432s `r1342s p|w|´u |1 ´ w|´s qdw
´8 0 1

49
 Note the second triplet of integrals swaps 2 with 3 so equivalently swaps s and t. We get the amplitude
igo2 26
”
e 1 ¨ e 2 e 3 ¨ e 4 δ pΣpq pr1234s ` r1432sqBp1 ´ u, ´1 ´ sq
2`2s

` pr1423s ` r1324sqBp1 ´ t, 1 ´ uq
ı
` pr1243s ` r1342sqBp1 ´ t, ´1 ´ sq

Now in the s channel, the first and third Beta functions give us poles at s “ 0 with residues ´t and ´u “ t
respectively. This gives:
igo2 26 t´u
´ δ
pΣpqe1 ¨ e2 e3 ¨ e4 pr1234s ` r2143s ´ r1243s ´ r2134sq ˆ (61)
2`2s s
On the other hand, the 3-point vertex (again just the leading order of the two terms, compare with (57)) for
massless bosons comes from the correlator
ipgo1 q3
|w12 w13 w23 | x: BX µ1 pw1 qeik1 Xpw1 :: BX µ2 pw2 qeik2 Xpw2 q :: BX µ3 pw3 qeik3 Xpw3 q :y
go2 `2s
ˆ µ3
ipgo1 q3 pµ2 3
˙
2 2 p1 2 2 2
Ñ 2 2 p´i2`s qp´2`s q 2 ` 2 ` 2 perms. |w12 |2`s p1 ¨p2 ´1 |w13 |2`s p1 ¨p3 ´1 |w23 |2`s p2 ¨p3 ´1
go `s w12 w13 w12 w23
?
2 µ1 µ2 1 µ3
“ ´igo pη p ` 2 perms.q
`s 2 12
?
using go1 “ go {p 2`s q. Adding CP factors gives:
igo
´ ? pη µ1 µ2 pµ123 ` η µ1 µ3 pµ132 ` η µ2 µ3 pµ231 q loooooooomoooooooon
pr123s ´ r321sq
2`s
f 123

We care about the e1 ¨ e2 e3 ¨ e4 term which means we only look at the p12 ¨ p34 “ t ´ u contribution in the s
channel. ż 26
d k Spk1 , k2 , kqSp´k, k3 , k4 q go2 26 t ´ u ÿ ` 125 534 ˘
i 26 2
Ñ ´i 2
δ
pΣpq ˆ f f
p2πq ´k ` i 2`s s 5

Lastly,řnote that the factors in equation (61) give Trpf 12a λa f 34b λb q, and with suitable normalization, this
gives 5 f 125 f 534 , exactly as desired.
We thus see that the amplitude indeed factorizes, respecting the structure of the U pN q gauge group.
1
12. We have p2 ` m2 “ L
`2s 0
for the open string.

50
 From 5.3.1 (and consequently 5.3.3) this gives:
transverse only
hkkkkkkikkkkkkj
dtN1 N2 ηpitq2
ż8 ż8 ż8
i V26 dt 1 ´2πtm2 i V26 dt 1 ´2πtLcyl i V26
Tr re s“ Tr re 0 s“
2 p4πq26 0 t13`1 2 p16π 2 `2s q13 0 t 13`1 2 p16π 2 `2s q13 0 t13`1 ηpitq26

All together this gives: ż8

dt 1
iN1 N2 V26
0 2t p8π `s tq13 ηpitq24
2 2

as required.

13. We already know the form of our propagators on the torus from exercise 4.69. Take

ˇ θ 1 pz ´ w, τ q ˇ2
ˇ “‰ ˇ
ˇ 1 ˇ ´2πpImzq2 {τ2
Gpz, wq “ ˇ ˇ e .
ˇ Bz θ 11 p0, τ q ˇ
“‰

This gives us
2
ź ź
x : eiki Xpzi ,z̄i q :y “ iZT 2 δ
26 pΣkq |Gpzi , zj q|`s ki ¨kj {2
i iăj

where ZT 2 which is equal to the partition function of the torus Zpτ q that we have also computed in the last
chapter. The amplitude is then:
n ż
gcn d2 τ
ż
26 1 ź ź 2
iδ
pΣkq 2 12 dz i |Gpzi , zj q|`s ki ¨kj {2
p2π`s q26 48
τ2 τ2 |η| i“1 iăj

14. We need to calculate the form of the propagators xX µ pzqX ν pwqy on the cylinder with NN boundary condi-
tions. Let’s use the image charge method. The finite cylinder can be thought of as the fundamental domain
of the quotient of the upper half plane by the action z Ñ λz for λ a real number corresponding to the
modulus of the cylinder. For X at z where 1 ă |z| ă λ we place images at each λn z in the upper half plane
as well as at λn z̄ on the lower half plane.

`2s ÿ ´ ¯
xXpzqXpwqy “ ´ log |λ´n{2 z ´ λn{2 w|2 ` log |λ´n{2 z ´ λn{2 w̄|2
2 nPZ

This gives
2
ź źź
x : eipi X :y “ δ
D pΣpq |pλ´n{2 zi ´ λn{2 zj qpλ´n{2 zi ´ λn{2 z̄j q|`s pi ¨pj
i n iăj

For open strings (operators inserted at the boundary) we must apply boundary normal ordering. We’ll get:
2
ź źź
x ‹‹ eiqi X ‹‹ y “ δ
D pΣqq |pλ´n{2 wI ´ λn{2 wJ q|2`s qI ¨qJ
i n IăJ

Lastly, for the correlations between boundary and bulk operators we’ll get:
2
źź
|λ´n{2 wi ´ λn{2 zi |2`s pi ¨qI
n i,I

Taking the product of the above three equations (with only a single momentum-conserving delta function)
gives us the X correlator on the cylinder. The CKG here is simply the compact SOp2q so it is best to ignore
ghosts, integrate the insertions over the whole cylinder and divide at the end by the volume of the SOp2q
action: λ.
There is a cleaner way to do this. From exercise 4.69 we know the cylinder propagator can be written in
terms of the torus propagator as an involution:

∆C2 pz ´ wq “ ∆pz ´ w, itq ` ∆pz ` w̄, itq

51
 2
Here ∆ “ ´ `2s log Gpz, z̄q from the problem above. This will then give us for m closed string and n open
string tachyons:
m ż n ż
gcm gon
ż8
ź
ipi Xpzi q
ź
‹ iqI XpwI q‹ D dt 1 ź ź
x :e : ‹e ‹ y “iδ

pΣp ` Σqq dzi dwI
i I
p2π`s q26 0 2t p2tq13 ηpitq24 i“1 C I“1 BC
n
`2s pi ¨pj {2 2
ź ź
ˆ rGpzi ´ zj ; τ “ itqGpzi ` z̄j ; τ “ itqs GpwI ´ wJ ; τ “ itq`s qI ¨qj
iăj IăJ
2
ź
ˆ rGpwI ´ zi ; τ “ itqGpwI ` z̄i ; τ “ itqs`s pi ¨qI {2
i,I

15. Here I assume Kiritsis meant c “ 1, since equation 3.4.3 refers specifically to closed string ground states.
The open string constraint o “ 1 comes from consistency of interactions stemming from the Jacobi identity
for Lie algebras. The one-loop contribution for the unoriented closed string comes from the cylinder + Klein
bottle + Möbius strip amplitude. As before, the only nonzero contributions come from states with an equal
number of left and right movers. All that this gives is an overall factor of c in this amplitude:
2 2
d26 p e´π`s tp
ż ż8
1 iV26 V26 c dt
ZK2 :“ TrrΩe´2πtpL0 `L̄0 ´c{12q s “ 26
c 24
ñ ΛK2 “ i
2 2 p2πq ηp2itq p2π`s q26 0 4t1`13 ηp2itq24

And working in the transverse channel t “ π{2` gives massless contribution:

ż8
26 V26
c 2 ˆ 24i ? d`
4πp2 2π`s q26 0

Similarly, the Möbius strip amplitude is given by

2 2
d26 p ζN e2πt`s p
ż ż8
1 iV26 V26 ζN dt
Z M2 “ Tro rΩe´2πtpL0 ´c{24q s “ ñ Λ M2 “ i
2 2 p2πq13 pηp2itqθ3 p2itqq12 p2π`s q26 0 2p2tq1`13 pθp2itqηp2itqq12

In the transverse channel with 2t “ π{2` now:

ż8
V
´ζN 2 1`13
ˆ 24i ? 26 d`
4πp2 2π`s q26 0

This gives a tadpole cancelation condition of:

c 226 ´ 214 ζN ` N 2 “ 0

We have N is a positive integer. Further, we have that ζ is a sign. If ζ “ ´1 then  must be negative, and
so by unitarity it is ´1, but there are no integer solutions N to 226 “ 214 N ` N 2 . Thus we need ζ “ 1 and
consequently  “ ´1, N “ 213 .

52
Chapter 6: Strings in Background Fields
Note this chapter is specific to closed oriented strings. As such, we will not consider the effects of the boundary.

0. This is not a required problem but it certainly should be 2 . Let’s calculate the β-functions of the nonlinear
sigma model. Here, I will borrow diagrams from the very nice set of TASI lecture notes of Callan and
Thorlacius
First, it is worth using a normal coordinate system for the X µ (one in which all of the Γ symbols vanish
and all higher symmetrized Γ symbols also vanish). We want to look at radiative corrections to xT`` y, since
they have integrals that are easier to handle than those for xT`´ y. From conservation this will give us the
trace anomaly for xT`´ y. We will first look at how G, B affect the trace on a flat worldsheet.
For the graviton contribution to β G , we have only only one diagram

This contributes an anomalous trace of

1
xT`´ y “ Rµν Ba X0µ B a X0ν
4

For the B contribution to β B , we have two such diagrams:

These contribute anomalous traces of:

1 1 λ
´ Hµρσ Hνρσ Ba X0µ B a X0ν , ∇ Hµνλ ab Ba X0µ Bb X0ν
16 8
respectively.
The dilaton contribution also affects the trace on the flat world sheet (even though it does not couple at
R “ 0), by affecting the stress energy tensor as it is defined by varying the action w.r.t. the metric. Kiritsis
has worked this out before and shown that the dilaton contributes pBa Bb ´ gab lqΦ to the stress energy tensor,
from which we get a dilaton contribution of lξ ΦpXpξqq to the trace. Using covariant expressions for the
D’alambertian we arrive at a contribution
1
∇µ ∇ν ΦpX0 q Ba X0µ B a X0ν ´ ∇λ ΦpX0 qHµνλ pX0 qBa X0 Bb X0 ab
2
Combining all of this together, we see that we will get the β-functions:
1 1
β G “ Rµν ´ Hµρσ Hνρσ ` 4∇µ ∇ν Φ, β B “ ´ ∇λ Hλµν ´ 2∇λ ΦHλµν .
4 2
As pointed out, these are not quite that RG beta functions (for example compare β B to the correct form
in Kiritsis), but around the fixed point, they capture the correct first order behavior. In particular their
vanishing will mean that we have no Weyl anomaly.
2
After seeing the details of this calculation, I can understand why it was omitted.

53
Now we need to account for the effects of a curved worldsheet geometry. We can account for this by looking
at a xT`´ T`´ y correlator:
δ 1
xT`´ p0qyeφ δab “ ´ xT`´ pξqT`´ p0qyδab (62)
δφpξq 4π

Again we can get this by first looking at xT`` T`` y and appealing to conservation. The Weyl anomaly comes
from this diagram:

This gives xT`´ T`´ y “ πD p2q

12 lδ pξq. Here we have a factor of D coming from each degree of freedom. This
can be used to integrate equation (62) to yield:
D D?
xT`´ y “ ´ lφ “ γR
48 24
Note that the ghosts (which are otherwise decoupled) will here contribute their factor of ´26.
We also now need to consider two-loop contributions of G, B to the T T correlator. The following diagrams
contribute:

The calculations here are very involved, but will precisely give us

α H2
p´R ` q
8 12
Finally, the dilaton both modifies the energy-momentum tensor, giving rise to a tree-level propagator con-
tribution to the two-point function:

dil T dil y “ πα1 p∇Φq2 lδ p2q pξq which will integrate to give a factor of
This contributes xT`´ α1 2 ?γR.
`´ 2 p∇Φq
Also, the dilaton gives a loop-contribution to the unmodified energy-momentum tensor:

54
 dil T dil y “ ´πα1 lΦlδ p2q pξq which will integrate to give a factor of ´ α1 lΦ?γR.
Which contributes the term xT`´ `´ 2
Altogether this gives: „ 
Φ 3 1 2 1 2
β “ D ´ 26 ` α 4p∇Φq ´ 4lΦ ´ R ` H .
2 12
as required.
δ
1. Each β-function of a coupling constant G, B, Φ as given in 6.1.5, 6.1.6, 6.1.7 is δφ of that coupling constant,
since our scaling µ “ eφ ñ log µ “ φ. Since
β Φ p2q 1
Taa “ G αβ
R ` 2 pβµν B αβ
g ` βµν ε qBα XBβ X
12 2`s
The change in effective action under an infinitesimal Weyl transformation δg αβ “ ´g αβ δφ is
ż ż „ Φ 
1 2 ? a 1 2 β ? p2q 1 ´ G ? ab B ab
¯
δ log Z “ ´δS “ d ξ gTa δφ “ d ξ gR ` 2 βµν g g ` βµν ε Ba XBb X δφ
4π 4π 12 2`s
We can integrate this to get the change after a finite conformal transformation:
ż „ ˆ ˙ 
1 2 ? Φ p2q 1 ab φ ´ G B ab
¯
d ξ gβ R φ ´ g ∇a φ∇b φ ` 2 βµν ` βµν  Ba XBb X
4π 2 2`s
this vanishes, of course, when all beta functions are zero. When β G , β B are zero we can show (exercise 3)
that β Φ is a constant, and we recover the Liouville action from before.
2. First write G explicitly in the action:
ż „ 
1 ? 1 26 ´ D
S “ 2 dD x ´ det Ge´2Φ R ` 4Gαβ ∇α Φ∇β Φ ´ Gαδ Gβ Gγζ Hαβγ Hδζ ` 2
2κ 12 3`2s
The classical equations of motion from varying the action with respect to G give
R variation 2
p∇Φq variation
hkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkikkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkj hkkkkkikkkkkj
2
0 “ Rµν ` 2∇µ ∇ν Φ ´ 
 
4∇ Φ∇ν Φ ´ 2Gµν lΦ ` 4Gµν p∇Φq `  4∇ Φ∇ Φ
  ν
µ µ
ˆ ˙
1 1 1 2
´ Hµρσ Hνρσ ´ Gµν R ` 4p∇Φq2 ´ H 2 ` 2 p26 ´ Dq
4
loooooomoooooon 2 12 3`s
looooooooooooooooooooooooooooooomooooooooooooooooooooooooooooooon
?
(63)
H 2 variation ´ det G variation
ˆ ˙
1 ρσ 1 2 1 2 26 ´ D
“ Rµν ` 2∇µ ∇ν Φ ´ Hµρσ Hν ´ Gµν R ´ 4p∇Φq ` 4lΦ ´ H ` 2
4
loooooooooooooooooomoooooooooooooooooon 2 12 3`2s
G
:“βµν

With respect to B we get:

IBP IBP
1 ´2Φ hkkikkj 2 ˆ 3 hkkikkj 1
´ e p2pδB µν pBα Bβγ ` 2 perms.qqH αβγ q Ñ e´2Φ p∇α Hαµν q Ñ ´ ∇α pe´2Φ Hαµν q “ 0
12 12 4
loooooooooomoooooooooon
B
:“βµν

Finally, with respect to Φ we get:

ˆ ˙ ˆ ˙
2 1 2 26 ´ D 2 2 1 2 26 ´ D
0 “ ´2 R ` 4p∇Φq ´ H ` 2 ´8lΦ´16p∇Φq “ ´2 R ´ 4p∇Φq ` 4lΦ ´ H ` 2
12 3`2s 12 3`2s
loooooooooooooooooooooooooooomoooooooooooooooooooooooooooon
:“´ 32 β Φ

The term in parentheses is the same as the term in parentheses the bottom line of (63). This agrees with
Polchinski 3.7.21 (with appropriate conventions adopted)
ż „ ˆ ˙ 
1 D
? ´2Φ µν G 1 2 Φ µν B 2 Φ
δS “ ´ 2 d x ´ det Ge δG βµν ´ Gµν β ` δB βµν ` 2δΦ β
2κ 2 3 3

55
 2
3. Let’s look at 3`2s
∇β Φ . We get:

1
8∇ν Φ∇µ ∇ν Φ ´ 4l∇µ Φ ´ ∇µ R ` p∇µ Hαβγ qH αβγ
6
The contracted Bianchi identity ∇µ R “ 2∇ν Rνµ together with the vanishing of βµν
G gives:

1
∇µ R “ 2∇ν Rµν “ ∇ν pHµρσ Hνρσ q ´ 4l∇µ Φ
2
which in turn gives
1 1
8∇ν Φ∇µ ∇ν Φ ´ ∇ν pHµρσ Hνρσ q ` p∇µ Hαβγ qH αβγ
2 6
The fact that H is exact gives us dH “ 0 so Brα Hβγδs “ 0. The symmetry properties of H imply that
summing over the four cyclic permutations of this gives zero. Contracting with the metric then implies a
contracted Bianchi-type identity for H, namely that ∇α Hαβγ “ 0.
Using β B “ 0 together with the Bianchi identity, we have 0 “ ∇ρ Hµνρ “ 2∇ρ Φ Hµνρ . So we have that H is
divergence-free, and ∇ρ Φ dotted with any component of H is zero. This lets us rewrite:
1 1
´ ∇ν pHµρσ Hνρσ q “ ´ H νρσ ∇ν Hµρσ
2 2
1 αβγ 1 1
∇µ pHαβγ qH “ ´ H αβγ p∇α Hβγµ ´ ∇β Hγαµ ` ∇γ Hαβµ q “ ´ H νρσ ∇ν Hµρσ
6 6 6
1 1 1 1
ñ 2
∇µ β Φ “ ∇ν Φ∇µ ∇ν Φ ´ ∇ν pHµρσ Hνρσ q “ ´ ∇ν ΦRµν ´ ∇ν Hµν
12`s 12 2 12
One last step. I am missing something.
This gives that ∇µ β Φ “ 0 as required. So β Φ “ c is a constant.

4. We get a linear dilaton giving rise to a Liouville action with Q “ 0. This is our familiar free massless boson
in 2D with 1D target space. So we get a string propagating in a single dimension.

5. Note that the only relevant parameters are `s , with units of length, and whatever length scales there are on
the manifold, all of which depend on its volume (since its compact) as V 1{D . In particular c “ β Φ depends
on `s as
c “ D ` Op`2s {V 2{D q.
I think this is correct, though it is different from Kiristis’ equation.

6. Note that a nonzero total flux of H over any closed 3-manifold is incompatible with H “ dB for a single-
valued B. We can write: i
ş i
ş
B H
e 2π`2s M “ e 2π`2s N
where M is the 2D manifold corresponding to the embedding of the world-sheet into the target space and
N is any manifold whose boundary is M . We need this to be independent of N , so for any three-cycle M3
we need: ż ż
1 1
H P 2πZ ñ 2 2 HPZ
2π`2s M3 4π `s M3

56
7. (a) We have
p2πRq2 R2
ż
H “ 2R2 sin2 ψ sin θ dψ ^ dθ ^ dφ ñ H“ “ PZ
S3 4π 2 `2s `2s
(b) The dilaton is Φ “ 0. Using Mathematica, the Ricci tensor is:

Rµν “ diagp2, 2 sin2 ψ, 2 sin2 ψ sin2 θq

Which gives a Ricci scalar of 6{R2 . From the previous part, H123 “ 2R2 sin2 ψ sin θ. From the metric
2 :“ H ρσ
being diagonal, we get that Hµν µρσ Hν is diagonal. We have

2 1 2
Hµν “ diagp8, 8 sin2 ψ, 8 sin2 ψ sin2 θq ñ β G “ Rµν ´ Hµν “0
4
as desired. Next, βµνB “ ´ 1 ∇α pH
2 µνα q. To take a contravariant divergence we divide by the volume
?
element and differentiate, but the volume element is sin2 ψ sin θ which will give H{ g is a constant, so
βµν will vanish.
1
Lastly, H 2 “ p2R2 q2 {R6 “ 2{R6 so that ´R ` 12 H 2 “ ´ R42 . Ignoring ghosts, this gives a central charge
of:
`2 6
D ´ 6 s2 ` Op`4s q “ D ´ ` Op`4s q
R k
as desired.
(c) Without using coordinates, the isometry of S 3 is G “ SOp4q “ rSUp2q ˆ SUp2qs{Z2 . TO see that
equivalence, think of of S 3 as the unit quaternions, and take SUp2q ˆ SUp2q act as unit quaternions on
the left and right. We get a right G-action by: x Ñ a´1 xb. Note the kernel is the set of pa, bq P G
ax “ xb for all x. In particular, for x “ 1 we get a “ b so the kernel lies in the diagonal subgroup.
To act trivially on all quaternions, a must be in the center, and for the unit quaternions this is exactly
˘1. So this is an injection ϕ : rSUp2q ˆ SUp2qs{Z2 Ñ SOp4q. Since SOp4q is compact and connected,
it is generated by the image of exponentiating sop4q, and so surjectivity of ϕ at the level of the Lie
algebras (which is true by dimension-counting) implies surjectivity and hence equivalence at the level
of Lie groups.
So we see that sop4q acting on S 3 is just a simultaneous left and right copt of sup2q acting on SUp2q.
Thus, we view this as the CFT of a nonlinear sigma model with target space G “ SUp2q and the left,
right copies of the sup2q action correspond to currents J “ g ´1 Bg and J “ B̄g g ´1
3k
We indeed get the central charge c “ k`2 which has the large k expansion 3 ´ 6{k ` Op1{k 2 q. Since k
in a non-negative integer in WZW models, except for the case k “ 0 corresponding to the trivial CFT,
we must have k ě 1, where we get R ě `s .

8. Here the metric has three degrees of freedom and Bµν , Φ both have only one degree of freedom (which can
be spatially varying). H, being a 3-index antisymmetric tensor, must vanish in 1 ` 1D, and so we will always
have β B “ 0. The other two constraints become:

G 1 3 “
0 “ β Φ “ ´24 ` `2s 4p∇Φq2 ´ 4lΦ ´ R
‰
0 “ βµν “ Rgµν ` 2∇µ ∇ν Φ,
2 2
Translational isometry implies that R, g depend on only the time variable t. The x variable can therefore
parameterize either S 1 or R endowed with constant metric.
Now taking the trace of the first equation implies Rptq “ ´2lΦpx, tq. Then the second equation will give:
16
“ 4p∇Φpx, tqq2 ´ 2plΦqptq
`2s

The only way for this to work is for R “ lΦ “ 0 so that ∇Φ can be a constant. We then have Φ “ αx ` βt
so that α2 ` β 2 “ 4{`2s , and g is Ricci flat everywhere (so we can pick it to be constant). In the case of either
α, β “ 0, we can also safely take x, t respectively to be periodic without having Φ be multi-valued.

57
 9. We still have β B “ 0, but β G “ Rµν ´ ∇µ ∇ν Φ while β Φ “ D ´ 26 ` 23 `2s p4p∇Φq2 ´ 4lΦ ´ Rq
This can be recast in terms of a new 4D Ricci flat metric ds2 “ F pφqdφ2 ` φR2 dΩ23 .
Using Mathematica again to take the trace of this gives Rij for i “ j ě 1 proportional to R2 φF 1 pφq `
8φF pφq2 ´ R2 F pφq. Solving this differential equation for F gives

R2 φ
F pφq “
4φ2 ` R2 c1

Setting c1 “ 0, F pφq “ R2 {4φ will also make R00 vanish. Then we can take the dilaton to be zero Φpφq “ 0.

10. As stated in the problem, upon gauging the adapted compact U p1q : θ Ñ θ ` , which has radius 2π, we
modify our derivative operator to act as Bα θ Ñ Bα θ ` Aα , where Aα gives our connection on the U p1q
principal bundle associated with gauging the Killing symmetry. The action gets modified:

R2 R2
ż ż
2
SĚ |Bθ| Ñ |Bθ ` A|2
4π`s 4π`2s

This is a new theory, but we can return to the old one by enforcing that A be pure gauge as follows: introduce
an auxiliary field φ and add to S the term
ż ż
i i
φ αβ Bα Aβ “ ´ dφ ^ A.
2π 2π

Integrating out φ gives exactly a δ-function enforcing αβ Bα Aβ “ 0. This gives that
ş A is closed, but it need
not be exact if our manifold has nontrivial topology. Going around any cycle, A can pick up a factor of
2πn.
For a closed, genus g Riemann surface, there are 2g cycles labeled by ai , bi , 1 ď i ď g coming from viewing
it as a 2g-gon. we have Riemann’s bilinear identity, namely for two closed 1-forms ω1 , ω2 ,
ż g ˆż
ÿ ż ż ż ˙
ω1 ^ ω2 “ ω1 ω2 ´ ω2 ω1 (64)
Σ i“1 ai bi ai bi

1
ş
Now take ω1 “ A, ω2 “ dφ. Now (64) gives us that 2π dφ ^ A will not be zero in general, but in the path
integral, it suffices to have it be an integral multiple of 2π, since then the nontrivial holonomies will have no
contribution to the action. We have that A can have winding 2πZ, so the only solution is to have φ have
winding 2πZ. This will exactly leave over a factor of 2πZ. So we return to our original action by introducing
the field φ of period 2π. (NB if I had kept the fields dimensionful, then φ would have period 2π{R when θ
has period 2πR)
In this new, equivalent action, we can gauge-fix θ “ 0 (do I need ghosts? No because this is abelian U p1q)
and integrate out A. We get:
`4s {R2
ż
d2 ξ pBφq2
4π`2s
so we have obtained the same action but now on a circle of radius `2s {R instead of R.
a
In doing this path integral we get a determinant
a factor of 4π 2 `2s {R2 “ 2π`s {R for each mode. Using zeta
function regularization this is equal to R{2π`s which we can understand as adding a ´ 21 logpR{2π`s q term
to the action that will couple to the curvature R (Show why), this shifting the dilaton as required.

11. We can simplify things by using the conventions of the next problem to do this one. Here, we have a single
compact coordinate θ. In our convention:
ˆ ˙
G00 G00 Aj 1
Ĝµν “ , Bµν “ Bj dθ ^ dxi ` Ai Bj bij dxi ^ dxj , φ “ Φ ´ log det G00
G00 Ai gij ` G00 Ai Aj 4

58
 From formula F.3 specialized to this case, we get that the metric and dilaton terms become
ż b ı ż „ 
D ´2Φ
”
2 D´1
a ´2φ 2 1 µ 00 1 A 2
d x ´ det Ĝµν e R̂ ` 4pBµ Φq “ d x ´ det g e R ` 4pBµ φq ` Bµ G00 B G ´ G00 pFµν q
4 4
(65)
where Fµν “ Bµ Aν ´ Bν Aµ and R̂ corresponds to the original Ĝµν while R corresponds to gij . Further G00 “
From F.6-F.9, the antisymmetric tensor changes as:
ż b ż „ 
1 D ´2Φ ijk D´1
a ´2φ 1 ijk 1 ij0
´ d ´ det Ĝ e Ĥijk Ĥ “ ´ d x ´ det g e Hijk H ` Ĥij0 Ĥ (66)
12 12 4

Here where Hij0 “ Ĥij0 and Hijk “ Ĥijk ´ pAi H0jk ` 3 perms.q. Here Hijk is defined so that it is invariant
under T -duality (TYSM Kiritsis for pre-organizing these terms for me) . Further, under T-duality
00
G00 Ñ G´1
00 “ G ñ Bµ G00 B µ G00 invariant
gij Ñ gij ñ R invariant
Ai Ñ Bi (67)
Bi Ñ Ai
1
ΦÑΦ´ log G00 ñ φ Ñ φ ñ pBµ φq invariant.
2
? A Ñ B B ´
We see that the ´ det g e´2φ as well as first three terms of equation (65). We have that Fµν µ ν
B B
Bν Bµ “: Fµν and Fij “ Hij0 . The last term of (65) will therefore become swap with the last term of (66)
and we are done.

12. This one is quick. We have

ds2 “ G00 dθ2 ` 2G00 Ai dxi dx0 ` Gij dxi dxj , B “ Bj dθ ^ dxj ` pbij ` Ai Bj qdxi ^ dxj
Certainly we have G̃00 “ 1{G00 , B̃i “ G00 Ai {G00 . Then Ãi “ Bi is consistent both for the i, 0 components
of the line element and the dxi ^ dxj components of the B-field as long as we keep b̃ij “ bij and g̃ij “ gij .
Finally, the dilaton must be shifted by Φ “ Φ ´ 12 log G00 .
13. The N commuting isometries correspond to a fibration by N -dimensional tori over each point in the base
space. As we have seen before (for strings valued in a N -dimensional torus target space), we have that modes
are described by two momenta pL , pR that Lie on an integral lattice. Naively, we can rotate pL , pR by any
GLpN q transformation, but the integrality condition restricts us to GLpN, Zq. Now GLpN q acts separately
on the left and the right momenta, but we are allowed to exchange between these two by applying T -duality,
which still preserves our Lorentzian norm, so the T -duality group gets enhanced to OpN, N, Zq.
14. This is clear, since orientation reversal acts trivially on g ab Gµν Ba X µ Bb X ν while it acts with a minus sign on
ab Bµν Ba X µ Bb X ν . The corresponding vertex operators are:
: BX µ B̄Xµ eikX :, : Gµν BX µ B̄X ν :, R : eikX :
If we assume the tachyon : eikX : is negative under parity then so are the dilaton and graviton.
This is incompatible with 6.1.10, as then parity will flip the sign of the dilaton in the exponential, substan-
tially changing the action of the theory.

59
Chapter 7: Superstrings and Supersymmetry
1. We already know that T T will have the desired OPE, since the bosons and fermions are uncoupled and we
already have shown their own respective stress tensor OPEs. Next
2
GpzqGpwq “ ´ ψµ pzqBX µ pzqψν pwqBX ν pwq
`4s
ˆ ˙ˆ 2 ˙
2 2 ηµν `s ηµν
“ ´ 4 `s ` pz ´ wq : Bψµ ψν pwq : ´ ` : BXµ BXν pwq :
`s z´w 2 pz ´ wq2
D ´ `22 BXµ BX µ pwq ´ `12 ψ µ Bψµ pwq
s s
“ `
pz ´ wq3 z´w
ĉ 2T pwq
“ 3
`
pz ´ wq z´w
Finally
ˆ ˙ ?
1 1 µ 2
µ
T pzqGpwq “ ´ 2 : BXµ BX pzq : ` ψ Bψµ pzq i 2 ψν BX ν pwq
`s 2 `s
? ˆ 2
`s ψµ BX pwq ` ψµ B X pwqpz ´ wq `2s ψµ BX µ pwq
µ 2 µ
`2s Bµ ψBX µ pwq
˙
2
“ ´i 4 ´ ´ ` p´q
`s 2 pz ´ wq2 2 pz ´ wq2 2 pz ´ wq
3 Gpwq BGpwq
“ `
2 pz ´ wq2 z´w

2. We will take the OPE of jB pzqjB pwq, but just look at the pz ´ wq´1 term as a function of w, as this, when
integrated around the origin in w will give Q2B . This is an extension of exercise 4.45, and there is nothing
conceptually further, except for some βγ manipulation. There are altogether 16 terms to consider, and we
will get c “ 15. The algebra is heavy, so I will skip this. An alternative is to do this as in Polchinski 4.3.
To do it this way, note the following OPEs:
ˆ ˙ ˆ ˙
Tmatter pzq 1 3 1 1 3
jB pzqbpwq „ ´ bcpzq ` βγpzq ` ´bBcpzq ` Bβγpzq ´ βBγpzq
z´w pz ´ wq2 4 z´w 4 4
„ 
1 1 3
“ ¨¨¨ ` Tmatter pzq ´ Bb cpwq ´ 2bBcpwq ´ Bβγpwq ´ βBγpwq
z´w 2 2
Tmatter pwq ` Tgh pwq
“ ¨¨¨ ` ñ tQB , bn u “ Ln
z´w
Similarly
Gmatter pwq ` Ggh pwq
jB pzqβpwq “ ¨ ¨ ¨ ` ñ rQB , βn s “ Gn
z´w
Now note that the Jacobi identity on QB reads:
pm´nqbm`n Ln
hkkkikkkj hkkkikkkj
trQB , Lm s, bn u ´ t rLm , bn s , QB u ´ rtbn , QB u, Lm s “ 0 ñ trQB , Lm s, bn u “ pm ´ nqLm`n ´ rLm , Ln s

So if the total central charge is zero we’ll get trQB , Lm s, bn u “ 0, implying that rQb , Lm s is independent of
the c ghost. But on the other hand this operator has ghost number 1, so it must therefore vanish. Further,
the Jacobi identity also yields

rtQB , QB u, bn s “ ´2rtbn , QB u, QB s “ 2rQB , Ln s

since we just showed that this last term vanishes, we must have QB , QB is also independent of c, but again
since Q2B has positive ghost number, we get that it is in fact zero. We can do the same argument with β and
G and get that the superstring BRST operator is zero, as long as the total central charge vanishes. This was
much cleaner than the OPE way.

60
3. First a lemma: An abelian p-form field A has D´2
` ˘
p on shell DOF. To prove this, note that we have a gauge
`D˘
symmetry of A Ñ A ` BΛ which has p´1 parameters. Next, the Euler-Lagrange equations give us that the
components A0i1 ...ip´1 are non-propagating. We thus get D´1
` ˘
p massless propagating off-shell d.o.f. which
`D´2˘
have p´1 gauge symmetries left over. These can be used to enforce Coulomb gauge conditions which allow
for there to be no polarizations along one of the spatial directions. We thus get D´1
˘ `D´2˘
´ p´1 “ D´2
` ` ˘
p p
massless on-shell degrees of freedom. For Aµ this is D ´ 2 and for Bµν this is pD ´ 2qpD ´ 3q{2.
The metric has 21 DpD ´ 3q on-shell degrees of freedom. There are two ways to see this, first, that the
dynamically allowed variation δg may on-shell be described by a symmetric traceless tensor in dimension
D ´ 2 which gives
pD ´ 1qpD ´ 2q 1
´ 1 “ DpD ´ 3q
2 2
or by noting that since we are gauging translation symmetry locally, each translation makes 2 polarizations
unphysical and so we get:
DpD ` 1q 1
´ 2D “ DpD ´ 3q
2 2
as required.
We now consider the R-R, R-NS, NS-R, NS-NS sectors together. For NS-NS we have the scalar “ 1 both on-
shell and off-shell, the antisymmetric two-form, which has only transverse degrees of freedom “ 8 ˚ 7{2 “ 28
and the gravity, “ 10 ˚ 7{2 “ 35 altogether we get 64 on-shell degrees of freedom.
In both the R-NS and NS-R sector, we have a Weyl representation of dimension 25´1 “ 16. There are
µ
however only 8 on-shell degrees of freedom. Similarly, we only consider the on-shell ψ´1{2 acting on the NS
part of the vacuum which gives another factor of 8. This gives 64 fermionic variables in each sector for a
grand total of 128.
In R-R for IIA we have a 0, 2, and self-dual 4-form. This gives:
ˆ ˙ ˆ ˙
8 1 8
1` ` “ 64
2 2 4
For IIB we have a 1 and 3-form. This gives
ˆ ˙ ˆ ˙
8 8
` “ 64
1 3
so in either case we have 64 on-shell degrees of freedom here. This is consistent with each |Sy state having 8
on-shell degrees of freedom giving 8 ˆ 8 “ 64. All together, we have the same number of on-shell fermionic
and bosonic degrees of freedom.
Now for the massive case. In the NS sector you might expect the next excitations come from the bosons α´1 ,
i i j k
but this gets projected out by GSO, so in fact the next states come from ψ´3{2 , Cijk ψ´1{2 ψ´1{2 ψ´1{2 and
i j
Cij ψ´1{2 α´1 . These have dimensions 8 ` 56 ` 64 “ 128, which decomposes as the traceless symmetric 44
and three-index antisymmetric 84 representation of SOp9q. In the R sector, we must look at α´1 i |S y and
α
i
ψ´1 |Cα y for Sα , Cα suitably chosen so that the state satisfies G0 “ 0. This constraint gives a factor of two
reduction for the dimension of the space of candidate Sα . Consequently, we get 8v b 8s ‘ 8v b 8s1 which has
dimension 128. This indeed turns out to be a spinor representation of SOp9q, and it comes from looking at
the tensor product of the fundamental spinor representation with the vector representation 16s b 9v . This
turns must decompose as a sum of two spinor representations 16s ‘ 128s . One is again the fundamental,
while the other is the required 128.
For the massive states in the type IIA and type IIB, we must tensor we wish to look at the lowest-level
masses. Note we must match massive states with massive states. In this case, we match 2{α on both sides
to get massive states of mass 4{α. Since the particles already organize into representations of SOp9q on each
side, the closed string massive spectrum will again clearly organize intro representations of SOp9q. Also since
fermionic and bosonic degrees of freedom already were equal on each side, they will be equal in the closed
string as well. We will have 2 ˆ 1282 “ 32768 bosonic and fermionic degrees of freedom.

61
4. In terms of theta functions: ˜ ¸
4 4
1 ź θ3 pνi q ź θ4 pνi q
χO “ ´
2 i“1
η i“1
η
˜ ¸
4 4
1 ź θ3 pνi q ź θ4 pνi q
χV “ `
2 i“1
η i“1
η
˜ ¸
4 4
1 ź θ2 pνi q ź θ1 pνi q
χS “ ´
2 i“1
η i“1
η
˜ ¸
4 4
1 ź θ2 pνi q ź θ1 pνi q
χC “ `
2 i“1
η i“1
η
We’ll take νi “ 0 here (I assume this is what I’m supposed to do) and so θ1 “ 0 ñ χS “ χC .
For IIB we look at
1 4 a
“ ‰ 1 4 ā
“ ‰
|χV ´ χC |2 1 1 ÿ a`b θ b 1 ÿ ā`b̄ θ̄ b̄
? “ ? p´1q ˆ p´1q
p τ 2 η η̄q8 p τ 2 η η̄q8 2 a,b“0 η4 2 η̄ 4
ā,b̄“0

Under modular transformations τ Ñ τ ` 1 θ 1 Ø θ4 00 , θ4 10 Ñ ´θ4 10 while η 12 Ñ ´η 12 . In the

4 0
“‰
“‰ “‰ “‰

holomorphic and anti-holomorphic parts separately, each term in the sum picks up a minus sign that is
cancelled by the minus sign in the η 4 .
Under τ Ñ ´1{τ , the p?τ21ηη̄q8 out front is invariant. On the other hand, the θ functions transform as
θ4 00 Ñ p´iτ q2 θ4 00 , θ4 01 Ñ p´iτ q2 θ4 10 , θ4 10 Ñ p´iτ q2 θ4 01 . These are exactly compensated by the η
“‰ “‰ “‰ “‰ “‰ “‰

transformations in the denominator, and no overall sign is picked up

For IIA we have similarly
1 4 a
“ ‰ 1 4 ā
“ ‰
pχV ´ χC qpχ̄V ´ χ̄S q 1 1 ÿ a`b θ b 1 ÿ ā`b̄`āb̄ θ̄ b̄
? “ ? p´1q ˆ p´1q
p τ 2 η η̄q8 p τ 2 η η̄q8 2 a,b“0 η4 2 η̄ 4
ā,b̄“0

Again, the holomorphic part transforms as before and as we have set the νi to zero, we have the same
partition function. Using D.18, we see that each of the four above sums are zero since they are equal to a
product of θ1 “ 0.

5. Again, these are identical if I set the νi “ 0 (am I not supposed to be doing this? What do the νi represent
physically?). They are equal to
1
? p|θ4
|2 ` |θ24 |2 ` |θ34 |3 ` |θ44 |2 q

p τ2 η η̄q8 4η 4 η̄ 4 1

We have θ3 and θ4 swapping under τ Ñ τ ` 1, generating no signs in this case, while the denominator looks
like |η|24 and also doesn’t generate a sign. Then, under τ Ñ ´1{τ we have θ2 and θ4 swapping generating a
|τ |4 , identical to what is generated by the pη η̄q4 .

6. The partition function is

“h‰2
4 a 8 γ
“ ‰ “ ‰
1 ÿ Z̄E8 g 1 ÿ
„ 
het θ h 1ÿ θ̄
ZSOp16qˆSOp16q “ ? p´1qa`b`ab`ag`bh`gh 4b , Z̄E8 “ p´1qγg`δh 8δ
2 h,g p τ2 η η̄q8 2 a,b η g 2 γ,δ η̄

First look at Z̄E8 . Under modular transformations τ Ñ ´1{τ we get Z̄E8 hg Ñ Z̄E8 hg . Under τ Ñ τ ` 1,
“ ‰ “ ‰

we get Z̄E8 hg Ñ p´1qh´2{3 Z̄E8 g`h

“ ‰ “ h ‰ het
. With this, we can look at ZSOp16qˆSOp16q under τ Ñ ´1{τ
“ g ‰2 4 b
“ ‰
1 ÿ Z̄E8 h 1 ÿ a`b`ab`ag`bh`gh θ a
? p´1q
2 h,g p τ2 η η̄q8 2 a,b η4

62
 Under relabeling of a Ø b, g Ø h, this is the same. Next, under τ Ñ τ ` 1:

´4{3 Z̄
“ h ‰2 “ a ‰
1 ÿ p´1q E8 g`h 1 ÿ a`b`ab`ag`bh`gh
p´1qa θ4 a`b´1
? p´1q
2 h,g p´1q4{3 p τ2 η η̄q8 2 a,b p´1q1{3 η 4
“ h ‰2 “ a ‰
1 ÿ Z̄E8 g`h 1 ÿ b`ab`ag`bh`gh
θ4 a`b´1
“ ´ ? p´1q
2 h,g p τ2 η η̄q8 2 a,b η4
“ h ‰2 “ a ‰
4
1 ÿ Z̄E8 g1 1 ÿ 1 1 θ
“ ? p´1q1`b`ab`ag `pa`bqh`g h`h a`b´1
2 h,g1 p τ2 η η̄q8 2 a,b η4
“ h ‰2
4 a
“ ‰
1 ÿ Z̄E8 g1 1 ÿ 1`pb 1 `a`1q`pab1 `a´aq`ag 1 `pb1 h`hq`g 1 h`h θ
b1
“ ? p´1q






2 h,g1 p τ2 η η̄q8 2 a,b η4
“ h ‰2
4 a
“ ‰
1 ÿ Z̄E8 g1 1 ÿ 1 1 1 1 1 θ 1
“ ? 8
p´1qa`b `ab `ag `b h`g h 4b
2 h,g1 p τ2 η η̄q 2 a,b η

Keep in mind that x2 “ x mod 2.

Before we do the next part, let’s elaborate on why ZE8 “ 21 a,b θ8 ab is the partition function of the E8
ř “ ‰

lattice. From
“ ‰ the8sixteen fermion picture, this is just the p´1qF “ 1 in“ the
‰ NS sector (corresponding to the
χO “ 2 pθ 0 ` θ 1 q character) together with the R sector χS “ 21 θ8 10 giving the spinor representation.
1 8 0
“0‰

Indeed, the roots of E8 consist of the roots of Op16q as well as the spinor weights of Op16q. Note that the
spinor representation comes from the half-integral points, corresponding to θ 10 in the sum, while the adjoint
“‰

representation comes from θ 01 and θ 00 . Consequently the action of Si ř

“‰ “‰
that fixes the
“ ‰adjoint vectors but
flips the sign of the spinor acts on our partition function as Si ZE8 “ 21 a,b p´1qa θ8 ab . It of course also
gives rise to a twisted sector, so altogether we get the four twisted blocks Z̄E8 hg as required.
“ ‰

Since we have projected out the spinor representation, the current algebra only contains the NS currents J¯ij
corresponding to the adjoint of SOp16q, and we have two copies of this for each group of 16 fermions.
?
From the factor of p τ 2 η η̄q´8 we see that we have 8 on-shell noncompact massless bosonic excitations as
well as all of their descendants (on both left and right moving sides). We also see on the left-moving side we
get a theta-function corresponding to N “ 8 fermions transforming under a spacetime SOp8q, forming the
superpartners of the bosons. On the right side instead of the superpartner fermions, we have the 16 internal
fermions that transform in the adjoint representations.
Let’s see what massless states we can build. In the NS sector of the left-movers, we have L0 “ 1{2, L̄0 “ 1
i j
and so we get ψ´1{2 α´1 |py which gives us our usual graviton, two-form field, and dilaton. We also have
i ¯a
ψ´1{2 J´1 |py for the Op16q ˆ Op16q currents. This gives us vectors corresponding to gauge bosons valued in
the adjoint of Op16q ˆ Op16q as required.
In the R sector we have G0 “ 0, L̄0 “ 1 we’ll get a gravitino, fermion, and gaugino as before, but again this
time valued in Op16q ˆ Op16q.

7. Because we have seen that T-duality flips the antichiral U p1q B̄X Ñ ´B̄X, and we want to preserve the
(1,1) supersymmetry G in the type II string (and so must keep it as a periodic variable Why is this
absolutely necessary. Can we not work with double covers in some clever way when defining
supercurrents? ), we must consequently flip ψ̄. This corresponds to inserting p´1qFR . For the right-moving
R sector, this changes the chirality of the R spinor, taking Sα Ñ Γ9 Γ11 Sα (there can be no phase, by reality
conditions of Γ). We thus flip IIA to IIB and vice versa.
From this we get that

Fαβ “ Sα pΓ0 qβγ S̃γ Ñ Sα pΓ0 Γ9 Γ11 qβγ S̃γ “ ´ξ Sα pΓ9 Γ0 qβγ S̃γ “ ´ξF Γ9

63
 Expanding in terms of the Fµ1 ...µk gives the action:
10
ÿ p´1qk
Fαβ Ñ ´ξ Fµ1 ...µk Γµ1 ...µk Γ9
k“0
k!

This gives that

F̃µ1 ...µk ,9 “ ´ξFµ1 ...µk , F̃µ1 ...µk “ Fµ1 ...µk ,9
Then
Bµ1 C̃µ2 ...µk 9 “ ´ξBµ1 Cµ2 ...µk , Bµ1 C̃µ2 ...µk “ Bµ1 C̃µ2 ...µk 9
so that (up to a closed term)
ppq pp`1q
C̃µ1 ...µp´1 9 “ ´ξCµp´1
1 ...µp´1
, C̃µppq
1 ...µp
“ Cµ1 ...µp 9

Get rid of the ξ factor

8. We have that Ω |Sα S̃β y “ εR |Sβ S̃α y. Further, it acts trivially on Γ0 (you sure?). Now, in the operator
language we will have ΩSα Ω´1 “ 1 S̃α and ΩS̃β Ω´1 “ 2 Sβ . In any case, we must have for the bi-spinor
that ΩSα S̃β Ω´1 “ R Sβ S̃α , which gives that 1 2 “ ´R Thus, we have:

ΩFαβ Ω´1 “ ΩSα Γ0βγ S̃γ Ω´1 “ ´R Γ0βγ Sγ S̃α “ ´R Γ0βγ Fγδ Γ0δα “ ´R pΓ0 F Γ0 qβα “ ´R pΓ0 F T Γ0 qβα

I think 7.3.3 of Kiritsis has the derivation wrong. Ask Nathan/Xi.

9. When we take R “ ´1 the scalar and four-index self-dual tensor survive. In this case, we will not have
consistent interactions. Since the graviton survives, there must be an equal number of massless bosonic
and fermionic excitations. The fermions come just from the NS-R sector (there is no R-NS now), giving 64
on-shell fermionic excitations. From the NS-NS sector, the dilaton and gravity will give 1 ` 35 “ 36 on-shell
bosonic degrees of freedom. We are missing 28 bosonic degrees of freedom.
The scalar and four-index self dual tensor contribute 1 ` 21 8ˆ7ˆ6ˆ5
4! “ 36 on-shell bosonic degrees of freedom.
This is too much. The two-form, on the other hand, contributes the requisite 8 ˆ 7{2 “ 28. Consistency of
interaction thus demands we keep only the 2-form and drop the 0 and self-dual 4-form. This necessitates
R “ 1.

10. We are just looking at the open superstrings here. Any open string that consistently couples to type I or
type II string theory must have a GSO projection as well. We have already seen how the oriented open
strings look like in exercise 7.3. In the NS sector we have at ´p2 “ m2 “ 2{`2s
i
ψ´3{2 λab |p; abyN S
i j k
Cijk ψ´1{2 ψ´1{2 ψ´1{2 λab |p; abyN S (68)
i j
Cij ψ´1{2 α´1 λab |p; abyN S

In the R sector we have (for Sα suitably chosen so that the state satisfies G0 “ 0):
i
α´1 λab |Sα ; abyR
i
(69)
ψ´1 λab |Cα ; abyR

I will assume NN boundary conditions. In this case

Ωα´1 Ω´1 “ ´α´1

Ωψ´1 Ω´1 “ ´ψ´1
Ωψ´ 1 Ω´1 “ ´iψ´ 1
2 2

Ωψ´ 3 Ω´1 “ iψ´ 3

2 2

64
 So all of the terms in (68) are terms of the form Aλab |p; abyN S with the operator A transforming as A Ñ iA
´1
under parity. Doing parity twice therefore will generate a ´2N S Apγγ T qii1 |p; a1 b1 y pγ T γ ´1 qj 1 j . This is
exactly the same as in 7.3.10. Demanding that Ω act on the state with eigenvalue `1 will make it so
that λ “ iN S γλT γ ´1 . We already have N S “ ´i so λ “ γλT γ ´1 here. Imposing the tadpole cancelation
condition ζ “ 1 and we get gauge group SOp32q. So we get that states at this level will transform in the the
traceless symmetric tensor + singlet representation of SOp32q.
´1
All of the terms in (69) will transform under parity twice as as 2R Apγγ T qii1 |Sα ; a1 b1 y pγ T γ ´1 qj 1 j . We will
have the same γ matrix as in the NS sector, as required for consistency of interactions. Here, though, we
will get R “ ´1 ñ 2R “ 1 and we will get λ “ ´γλT γ ´1 (this is what we got from the massless sector with
an extra minus sign since ψ´1 , α´1 now transform with minus signs). Again we will have that these states
will transform in the symmetric representation of SOp32q.
Again we get 128 bosonic states that will transform as the 44 ‘ 84 representation of SOp9q. We will also get
fermions transforming in the 128 spinor representation as in exercise 3. All of these states will transform in
the traceless symmetric representation of SOp32q. Confirm

11. Certainly in the untwisted sector, the theory we get corresponds to tracing over the projection operator
1 µ
2 p1 ` gq where g is orientation-reversal. Now in the twisted sector, we still have X satisfies the Laplace
equation B` B´ X “ 0 so we can write
µ µ µ µ
ˆ ˙
µ 2 p ` p̄ 2 p ´ p̄ i`s ÿ αn ´inpτ `σq α̃n ´inpτ `σq
Xpσ, τ q “ x ` τ `s ` σ`s `? e ` e
2 2 2 n n n

The condition that Xpσ ` 2πq “ Xp2π ´ σq give that pµ “ p̄µ and the σ term vanishes. We must have n
is a half integer. For integer modding we have e´inpτ ˘σq Ñ e´inpτ ¯σq . For half-integer modding we have
e´inpτ ˘σq “ p´1qn e´inpτ ¯σq . We should thus have αn “ α̃n for n integral and αn “ ´α̃n We thus get
? ÿ αn ? ÿ αn
Xpσ, τ q “ xµ ` 2`2s pµ τ ` σi 2`s cospnσqe´inτ ´ 2`s sinpnσqe´inτ
n 1
n
nPZzt0u nPZ` 2

This is the twisted sector. The last sum picks up a minus sign under orientation reversal, and so will be
projected out. We are left with the equations of motion for the open string.

12. In NS we have (up to an overall irrelevant factor of i´1{2 )

ÿ ÿ
ψpσ, τ q “ ψn`1{2 epn`1{2qpτ `iσq , ψ̄pσ, τ q “ ψ̄n`1{2 epn`1{2qpτ ´iσq
nPZ nPZ

In the closed string case have that Ωψn`1{2 Ω´1 “ ψ̄n`1{2 . Given that Ωψpσ, τ qΩ´1 “ ψ̄pπ ´ σ, τ q, we directly
get Ωψn`1{2 Ω´1 “ ip´1qn ψ̄n`1{2 . For DD boundary conditions we get an extra minus sign to this, since
there Ωψpσ, τ qΩ´1 “ ´ψ̄pπ ´ σ, τ q.
In the R sector we have ÿ ÿ
ψpσ, τ q “ bn enpτ `iσq , ψ̄pσ, τ q “ b̄n enpτ ´iσq
nPZ nPZ

Following the same logic we get that Ωψn Ω´1 “ p´1qn ψn for NN and Ωψn Ω´1 “ ´p´1qn ψn for DD.
All of these cases can be summarized by

NN: Ωψr Ω´1 “ p´1qr ψr

DD: Ωψr Ω´1 “ ´p´1qr ψr .

13. Let’s clarify a bit of terminology before we begin. We are looking at just the fermions of the left moving and
right moving sides of the heterotic string theory. On the left-hand (supersymmetric) side, in the light-cone
gauge these form an Op8q
z current algebra at level 1. On the right-hand side the form a Op32q
{ current algebra
at level 1 again (why must we always have level 1? Ask Xi.).

65
The characters of OpN
{q for N even correspond to the integrable representations labeled by O, V, S, C
1
corresponding to the trivial, vector, spinor, and conjugate spinor. For our purposes (ie the heterotic string),
we do not need to distinguish between S and C, which will have the same character. The characters can be
written in terms of θ functions as
«ˆ ˙ ˆ ˙N {2 ff «ˆ ˙ ˆ ˙N {2 ff
θ3 N {2 θ3 N {2 1 θ2 N {2
ˆ ˙
1 θ4 1 θ4
χO “ ` , χV “ ´ , χS “
2 η η 2 η η 2 η

(a) Now let us first look at Op32q. The Op8q

z characters transform under τ Ñ τ ` 1 as
1

χ8O Ñ p´1q´1{6 χO , χ8V Ñ ´p´1q´1{6 χV , χ8S Ñ ´p´1q´1{6 χS

And under τ Ñ ´1{τ they transform as

1 1 1
χ8O Ñ pχ8O ` χ8V q ` χ8S , χ8V Ñ pχ8O ` χ8V q ´ χ8S , χ8S Ñ pχ8O ´ χ8V q
2 2 2

The Op32q
{ characters depending on q̄ transform the same way under τ Ñ ´1{τ but under τ Ñ τ ` 1
1
transform as
χ32
O Ñ p´1q
2{3 32
χO , χ32 V Ñ ´p´1q
2{3 32
χV , χ32
S Ñ p´1q
2{3 32
χS
Our partition functions in question can be constructed from a linear combination of products of exactly
one Op8q
z and one Op32q { character. This gives 9 possible terms χ8 χ32 ˚ . I label these in the table
1 1 i j
below. I cancel all terms that are not invariant under τ Ñ τ ` 1.
O
 O
 OV O
S
VO VV VS
 

SO S
 V SS

But we are not done. It is easy to see that while that χ8V , χ8S blocks have Taylor series Opq 1{3 q, the χO
block contains a singular term going as q ´1{6 . Similarly, χ32
O contains a singular term going as Opq
´2{3 q
32
while χV “ Opq̄ ´1{6 32 4{3
q and χS “ Opq̄ q. The tachyon can come exactly (and only!) from combining
8 32 ˚ 1{6
χO χV to get 1{|q| that will be singular and satisfy level-matching. Thus we must drop OV above
as well. We are left with four possible terms that can work.
Modular invariance under τ Ñ 1{τ further constrains this to take a form proportional to

pχ8V ´ χ8S qpχ32 32

1 ` χS q

The normalization of the identity to 1 fixes this entirely. Note that we get spin statistics for free, as the
only character combinations appearing with a minus sign are precisely those containing χ8S , associated
with the spacetime fermions.
(b) Having Op32q
{ out of the way, let’s move on to Op16q ˆ E8 . E
1
x81 has only one integrable representation
E
and thus one corresponding character, χ . As pointed out in the text, it is related to the characters of
8

{ by χE8 “ χ16 ` χ16 . Thus we have trilinear combinations χ8 pqqχ16 pq̄qχE8 pq̄q. Upon noting that
Op16q 1 O S i j
the characters of Op16q1 multiplied by χE8 transform the same way under modular transformations as
{
{ and the same combination χ8 χ16 χE8 uniquely gives the tachyon, we see that again the argument
Op32q 1 1 V
goes as before and the only viable character we can have is

pχ8V ´ χ8S qpχ32 32 E8

1 ` χS qχ “ pχ8V ´ χ8S qχE8 ˆE8 .

This is exactly the heterotic E string theory.

(c) Finally we get to the hard one: Op16q ˆ Op16q. Here we have 27 trilinear terms that can contribute.
I will write them out, and again cross out the ones that are not invariant under τ Ñ τ ` 1 as well as
double crossing out the tachyons. Here, though, the notation OO, OV, SV etc will represent just the
right-moving characters χ16 16 16 16 16 16
O pq̄qχO pq̄q, χO pq̄qχV pq̄q, χS pq̄qχV pq̄q respectively.

66
 O
 O X
 OX
 V


X O
 S OO O
V OS
 OO O
V OS

χ8O VX
ˆ 
X O 


X VV
 VS , χ8V VO VV 
ˆ  
 V
S , χ8S VO VV 
ˆ  
 V
S
S
 O SV S
 S SO SV SS
 SO SV SS


It may look that 12 independent terms remain. The fact that the characters are symmetric under
exchange of the last two labels mean that there are in fact only 12.
˚ 16 ˚
Let us look at two cases. First, assume χ8O χ16V χS does not contribute (ie its coefficient vanishes).
Then the first 9 terms are all zero. The remaining constraint of modular invariance under τ Ñ τ ` 1
constrains the partition function to take the form

pχV ´ χS q χ16 16 16 16 16 16 16 16
“ ‰
O χO ` 2αχO χS ` p1 ´ αqχV χV ` p2 ´ αqχS χS

for any value of α. Spin statistics requires all these contributions to come in with positive coefficient, so
0 ď α ď 1. Moreover, if α is non-integral we will have coefficients that are not integers in the character
expansion, which would lack a Hilbert space interpretation Think more about the integrality
condition. Thus we can have only α “ 0 and α “ 1 corresponding exactly to the Op32q and E8 ˆ E8
superstrings.
˚ 16 ˚
So our remaining possibility is that χ8O χ16V χS does not have vanishing coefficient. WLOG set this
coefficient to 1. Invariance under τ Ñ ´1{τ constrains us to:

2χ8O χ16 16 8
“ 16 16 16 16 16 16 16 16
‰
V χS ` χV αχO χO ` 2βχO χS ` p´1 ` α ´ βqχV χV ` p´1 ` 2α ´ βqχS χS
` χ8S p1 ´ αqχ16 16 16 16 16 16 16 16
“ ‰
O χO ´ 2p1 ` βqαχO χS ` p´α ` βqχV χV ` p1 ´ 2α ` βqχS χS

Again, spin-statistics requires the coefficient of all the characters involving χV to have positive sign and
all the characters involving χS to have negative sign. This makes 1 ď α, 0 ď β ď α ´ 1. Integrality then
forces α “ 1, β “ 0. More general solution? We need to impose that χ8i χ16 16
O χO has coefficient
1 or 0

Of all these theories, the first two theories have vanishing partition function - an indicator of spacetime
supersymmetry, but not necessarily an identifier. Of course, we can identify them as the heterotic string
theories, which indeed have space time SUSY. The last theory has nonvanishing partition function and thus
cannot have spacetime SUSY as the fermions and bosons do not cancel at one loop.

14. I think this problem is backwards. For 32 fermions all with the same boundary conditions, its immediate to
see that they will reproduce the partition function for the Spinp32q{Z2 string:
„ 
1 ÿ 16 a
θ
2 a,b b

Just by considering the OpN q fermion at N “ 32. On the other hand, if we split the fermions into 16 ` 16,
and consider separately boundary conditions for each of those, then our partition function is the square of
the 16-fermion system. We then get the E8 ˆ E8 lattice theta-function, as required
» fi2
1 ÿ „a 
– θ8 fl
2 a,b b

15. Note this was a Lorentzian lattice of signature pn, nq. The norm was thus PL2 ´ PR2 “ 2mn P 2Z. It is also
self dual, since it is already integral, and there is no integral sublattice.

16. We have
3 1 3
γGghost “ ´cγBβ ´ Bcγβ ´ 2γ 2 b, cTghost “ 2bcBc ´ cγBβ ´ cBγβ
2 2 2

67
 Here Kitisis’ conventions are different than Polchinski. Recall upon bosonization βpzq “ e´φpzq Bξpzq, γ “
eφpzq ηpzq. Although we can solve this problem very quickly since we already know what the stress tensor looks
like in the bosonized variables, I think it’s way more instructive to explicitly compute OPEs to Opz ´ wq.
First let’s look at the η, ξ theory, which is a fermoinic bc theory of weights 1, 0. We get
1
ξpzqηpwq “ ` : ξη : pwq ` Opz ´ wq
z´w
We can bosonize this theory in terms of hermitian χ field so that η “ e´χ , ξ “ e´χ . Using these coordinates
„ 
1 1
ξpzqηpwq “ eχpzq e´χpwq “ 1 ` pz ´ wqBχ ` pz ´ wq2 pB 2 χ ` pBχq2 q ` . . .
z´w 2
1 1
ñ Bξpzqηpwq “ ´ 2
` pB 2 χ ` pBχq2 q
pz ´ wq 2
Using this we can write

βpzqγpwq “ e´φpzq Bξpzq eφpwq ηpwq

„ „ 
1 2 2 2 1 1 2 2
“ pz ´ wq 1 ´ pz ´ wqBφpwq ` pz ´ wq ppBφq ´ B φq ´ ` pB χ ` pBχq q
2 pz ´ wq2 2

The constant term gives :βγ:“ Bφ ñ: Bpβγq :“ B 2 φ. The pz ´ wq term gives exactly the stress tensor of the
βγ theory at λ “ 0, which makes sense since this is exactly Bβγ
1 1 1 1
: Bβγ : “ ´ pBφq2 ` B 2 φ ` pBχq2 ` B 2 χ
2 2 2 ˆ2 ˙
1 2 1 1 1
ñ Tβγ “ Bβγ ´ λBpβγq “ ´ pBφq ` ´ λ B 2 φ ` pBχq2 ` B 2 χ.
2 2 2 2

In our case we have λ “ 3{2.

ˆ ˙
1 1 2 1 1 2 3
γGghost “ ´c ´ pBφq ` B φ ` pBχq ` B χ ´ Bφ Bc ´ 2γ 2 b
2 2
2 2 2 2 2
ˆ ˙
1 1 1
cTghost “ 2bcBc ` c ´ pBφq2 ´ B 2 φ ` pBχq2 ` B 2 χ
2 2 2
Altogether this gives a BRST current:
1
jB “ cTX ` γGX ` pcTgh ` γGgh q
2
3 3
“ cTX ` γGX ` bcBc ´ BφBc ´ cB 2 φ ´ γ 2 b
4 4

17. We are looking at rQB , ξe´φ{2 Sα eipX s. Therefore we should look at the 1{pz ´ wq pole in the OPE of jB
with ξe´φ{2 Sα eipX . The terms that contribute to this pole must involve pairing ξ with its conjugate η. η
appears in jB wherever γ “ eφ η appears. From the previous exercise, we see that we need only look at the
terms γGX and ´γ 2 b.
These two terms contribute poles:
« ff
: eφ GX :: e´φ{2 Sα eipX : : e3φ{2 ηbSα eipX :
´ ´
z´w z´w

The overall minus sign comes from commuting across an odd number of fermions for the Wick-contraction.
We will need to recall two things:
˜ ¸
µ `s µ β 1 ?
ψ pzq ¨ Sα pwq „ ? ? Γαβ S pwq ` 2 D Γαβ Sβ ψν ψ pz ´ wq , eφpzq e´φpwq{2 „ z ´ weφpwq{2
ν µ
2 z´w `s p 2 ´ 1q

68
 The subleading term in the first expansion is taken from Blumenhagen 13.81. That means that first term is:
?
φpzq 2
e i 2 ψ µ pzqBXµ pzq ¨ e´φpwq{2 Sα pwqeipXpzq
`s
? ˜ µ
β BX eip¨X 2 p eip¨X
¸
2? ` s Γαβ S µ ´i`s µ 1 ?
„ i 2 z ´ w eφpwq{2 ? ? ` Γν S β ψν ψ µ z ´ w
`s 2 z´w 2pz ´ wq 4`2s αβ
eφ{2 ´ µ β ¯
“´ Γαβ S BXµ ´ iΓναβ S β ψν p ¨ ψ eip¨X
`s

Note this OPE has no singularity, so we exactly got the normal ordered term we required: : eφ{2 GX Sα eip¨X :.
Altogether this gives us:
« ff
p1{2q α eφ{2 µ β µ i eφ{2 ν β 3φ{2
Vfermion pu, pq “ u ppq Γ S BX ´ Γ S ψν p ¨ ψ ` e ηbSα eipX .
`s αβ 8 `s αβ

I believe this is right, and moreover that the inclusion of the `´1
s factor is necessary for the dimensional
analysis to make sense.

18. Here, I followed the discussion of Polchinski 12.5. The picture changing operator is:

Xpzq :“ QB ¨ ξpzq

Over the sphere, the βγ path integral is equivalent to the φ, η, ξ path integral plus an additional insertion of
ξ to make up for the fact that it pics up a zero mode due to to the vacuum degeneracy it produces. Because
the expectation value is just proportional to the zero-mode of ξ, which depends on global information rather
than the specific local insertion point, xχpzqy is independent of position and we can normalize ξ so that this
is 1.
Say we have a null state. This means it is BRST exact. This means that we can rewrite its pointlike insertion
as a local operator surrounded by a BRST contour (direct, from the definition of exact). For that null state
to decouple, we need to be able to contract the BRST contour off the sphere (i.e. by pulling it off to the
north pole). The fact that ξ is inserted will seem to obstruct this. What happens now as we pull the BRST
charge to infinity is that it will circle ξ, creating the PCO Xpzq. However, when the ξ insertion is replaced
by X, the path integral will vanish since there is now no ξ insertion to avoid the zero-mode.
Now consider a path integral with a PCO insertion as well as additional BRST-invariant operators (mean-
ing. the contour integral around them of jB is zero). Then we can write Xpz1 qξpz2 q “ QB ξpz1 qξpz2 q “
p´q2 ξpz1 qQB ξpz2 q “ ξpz1 qXpz2 q where I have pulled the QB contour around the sphere (there two minus
signs, one from commuting QB across a fermionic variable and one from reversing the orientation of the
contour.)
This is interesting: although X is null, it does not vanish in the path integral, since pulling QB off of it will
make QB encircle ξpz2 q but leave behind Xpz1 q’s ξpz1 q, so the ξ zero-mode will remain saturated and we
won’t get zero.
The X can be brought near any of the local BRST closed operators to change their picture (the OPE
is nonsingular). Ie note that the main term we look at is γGX “ eφ ηGX in jB so that XOp´1q pzq “
zGX pzqOp0q Ñ G´1{2 Op0q. We can move X to any other point on the sphere - since the exact position of X
does not matter any more than the position of ξ.

19. It is enough to look at the 1{pz ´ wq term in the OPE

p´1{2q
: e´φpzq{2 Sα pzq : Vfermion pwq “ e´φpzq{2 Sα pzquβ ppqe´φpwq{2 Sβ pwqeip¨Xpwq

We will use the fact of 4.12.42:

Cαβ Γµαβ ψµ pwq
Sα pzqSβ pwq “ `?
pz ´ wqN {8 2`s pz ´ wqN {8´1{2

69
 where Cαβ is the charge conjugation matrix and here N “ 10. We also have e´φ{2 e´φ{2 “ pz ´ wq´1{4 e´φ .
This leaves the pz ´ wq´1 term to be the requisite

´φ β
Γµαβ ip¨X p´1q Γµαβ uβ
e u ppq ? ψµ e “ Vboson p “ ? , p, zq
2`s 2`s
For the second example, we will look at the pz ´ wq´1 term in the OPE
ˆ ˙
i
e ´φpzq{2 µ
Sα pzqµ BX ´ pµ ψ ψ eip¨X .
µ ν
2
The first term in parentheses will not contribute to the singular term. Also the e´φ{2 and eip¨X contract with
nothing. Here, we use 4.12.41 to evaluate
`2s pΓµν qβα Sβ pwq
ψoµmo
Sα pzq ¨ lo ψoνnpwq „ ´
2pz ´ wq
´iJ µν

The ´ sign comes from the fact that the fermion current is coming from the right this time so z and w are
swapped. This gives a variation
`2s ν p´1{2q ipµ ν `2s pΓµν qβα
e´φ{2 ipµ ν  pΓµν qβα Sβ eip¨X “ Vfermion puβ “ , p, zq
4 4
20. We are in type I. We have
1
x: cV p´1q pw1 q : : cV ´1 pw2 q : : cV 0 pw3 q :y ` 1 Ø 2, x1 ą x2 ą x3
`4s go2
That constant out front is not obvious from Kiritsis, c.f. the discussion in Polchinski 12.4 and
allow for another factor of `2s since the fermions are dimensionful

The relevant expectation values are

´1
xcpw1 qcpw2 qcpw3 qy “ |w12 w13 w23 |, xe´φpw1 q e´φpw2 q y “ w12 ,
xψ µ pw1 qψ ν pw2 qy “ `2s η µν w12
´1
, xX9 µ pw1 qeik1 X pw2 qy “ ´2i`2 k µ eik1 X w´1
s 1 12

In the matter CFT we get (here ki ¨ kj “ 0 so the pure eikX terms contract to 1):

xψ µ pw1 qeik1 ¨Xpw1 q ψ ν pw2 qeik2 ¨Xpw2 q piX9 ρ ` 2k3 ¨ ψψ ρ qeik3 ¨Xpw3 q y
ˆ µν ρ
η µν k2ρ η µρ k3ν ´ η νρ k3µ
˙
4 10 η k1
“ 2`s δ
pΣkq ` `
w12 w13 w12 w23 w13 w23
So altogether we get an amplitude of (taking x1 Ñ 0, x2 Ñ 1, x3 Ñ 8)
ˆ µν ρ
2i`4s gopen
3
η k1 x23 η µν k2ρ x13
˙
i 10 µρ ν νρ µ
2
ˆ ? δ
pΣkq ` ` η k3 ´ η k3 pr123s ´ r132sq
gopen `4s 2`s x12 x12
igopen 10 ρ µ
“ ? δ
pΣkq pη µν k12 ` η µρ k31
ν
` η νρ k23 q pr123s ´ r132sq
2`s
Note unlike the Bosonic string this is exactly the same as the ordinary Yang-Mills amplitude, there is no k 3
correction term (what would correspond to at TrF 3 term in the Lagrangian).

70
21. This is also in type I. We should put the gaugini in the ´1{2 picture and the boson in the ´1 picture.

We have
1
x: cV p´1q pw1 q : : cV ´1{2 pw2 q : : cV ´1{2 pw3 q :y ` 1 Ø 2, x1 ą x2 ą x3
`4s go2
The relevant expectation values are
´1{4 ´1{2 ´1{2
xcpw1 qcpw2 qcpw3 qy “ |w12 w13 w23 |, xe´φpw1 q{2 e´φpw2 q{2 e´φpw3 q y “ w12 w13 w23 ,
`2
ñ xSα pw1 qSβ pw2 qψ µ pw3 qy “ ?s pCΓqµαβ w12 w13 w23
´5{4 ´3{4 ´1{2 ´1{2
xSα pw1 qSβ pw2 qy “ Cαβ w12
2
So altogether this gives
i a `2s 10 igopen 10
ˆ pgopen `s q 2
gopen ? δ
pΣkqCΓµαβ ˆ pr123s ´ r132sq “ ? δ
pΣkqCΓµαβ pr123s ´ r132sq
`4s gopen
2
2 2`s
This is k-independent so is an even simpler amplitude that the last in some sense.
22. We are now in type II. Gravitons are NS-NS states. We take two of them in the p´1, ´1q picture and the
remaining one in the p0, 0q picture. Again now, the constant demanded from unitarity now gets modified to
8pii
g 2 `6
We look at
c s

8πi 2gc3 D“ ´φ´φ̄ µ σ ik1 X ‰ “ ´φ´φ̄ ν ω ik X ‰ “ ρ i ρ λ i λ ik3 X

‰ E
cc̃e ψ ψ̃ e pz 1 q cc̃e ψ ψ̃ e 2
pz 2 q cc̃pBX ´ k ¨ ψψ qpBX ´ k ¨ ψψ qe pz 3 q
gc2 `6s `2s 2 2
Let’s just look at the holomorphic part of the matter CFT, and the calculation goes almost exactly as in the
last problem
1
xψ µ pz1 qeik1 ¨Xpz1 q ψ ν pz2 qeik2 ¨Xpz2 q piX9 ρ ` k3 ¨ ψψ ρ qeik3 ¨Xpz3 q y
ˆ µν ρ 2
µν ρ µρ ν νρ µ˙
1 η k1 η k2 η k3 ´ η k3
“ `4s ` `
2 z12 z13 z12 z23 z13 z23
` 4
ρ µ
Ñ s loooooooooooooooomoooooooooooooooon
pη µν k12 ` η µρ k31ν
` η νρ k23 q
4
“:V µνρ
So the total amplitude becomes
πigc δ
10 pΣkqV µνρ V σωλ
consistent with Polchinski.
23. We can put all our gaugini in the ´1{2 picture thankfully.

71
 ?
Our vertex operators are gopen `s λα e´φ{2 Sα eikX . The relevant two-point correlator is
`s pCΓµ qαβ ψµ
Sα pzqSβ pwq „ ?
2pz ´ wq
From considerations of the singularity structure, we get that the four-point correlator is:
`2s pCΓµ qαβ pCΓµ qγδ `2 pCΓµ qαγ pCΓµ qαδ `2 pCΓµ qαδ pCΓµ qβγ
` s ` s
2z12 z34 z23 z34 2z13 z24 z32 z42 2z14 z23 z42 z43
Take z1 “ 0, z2 “ w, z3 “ 1, z4 “ 8. In order for the term going as 1{z to cancel so that the integral over
the line is well-defined, we need the (physical on-shell condition):
Γµαβ Γµγδ ` Γµαγ Γµβδ ` Γµαδ Γµβγ “ 0

and defining ´pk1 ` k2 q2 “ s etc. gives

`2s 1 ´`2s s´1
ż
i a 2
4 10
ˆ pgopen `s q δ
pΣkq ˆ x p1 ´ xq´`s u´1 pΓµαβ Γµγδ ` xΓµαγ Γµβδ qr1234s
`4s gopen
2 2 0
2 `2s 10 Γp´`2s sqΓp´`2s uq
ˆ ˙
igopen µ µ µ µ
“´ δ
pΣkq 2 ˆ puΓαβ Γγδ ´ sΓαδ Γβγ qr1234s ` 2 perms.
2 Γp1 ´ `2s s ´ `2s uq
The minus sign comes from pulling an s or u out of the Γ functions. The factor of 2 comes from summing
over both orientations. Altogether we can write this as
Γp´`2s sqΓp´`2s uq
ˆ ˙
2 2 10
´ 8igopen `s δ
pΣkqKpu1 , u2 , u3 , u4 q r1234s ` 2 perms.
Γp1 ´ `2s s ´ `2s uq
1
Kpu1 , u2 , u3 , u4 q “ pu ū1 Γµ u2 ū3 Γµ u4 ´ s ū1 Γµ u4 ū3 Γµ u2 q
8
24. The bosonic action in 11D is:
ż b „ 
1 11 1 2 1 M1 ...M11
d x ´ det Ĝ R ´ G `  GM1 ...M4 GM4 ...M8 ĈM9 M10 M11
2κ2 2 ¨ 4! 4 p144q2
where G4 is the field strength of the 3-form Ĉ. From Appendix F, we have that the dilaton Φ “ 0 in 11D.
So the field σ will just be σ “ ´2φ “ 12 log G11 11 , and A here is as it is in appendix F. Directly using the
bosonic equation F.3 gives the terms
ż „ 
1 8 ? σ 1 2σ 2
d x ´ge R ´ e F2
2κ2 4
(Here the 41 Bµ G11 11 B µ pG11 11 q´1 ) will exactly cancel the 4Bµ φdµ φ.
Now let’s look at the 3-form potential contribution. Because F is antisymmetric in all four indices, and we
only are compactifying along one dimension, only the first two terms of F.28 can contribute. They give
ż „ 
1 10 ? σ 1 2 1 ´2σ 2
“ 2 d x ´ge ´ F ´ e H3
2κ 2 ¨ 4! 4 2 ¨ 3!
and pH3 qµνρ “ Brµ pB2 qνρs “ BCµν 11 so that H32 “ Gµσ Gνλ Gρκ Hµνρ Hσλκ and Cµνρ “ Ĉµνρ ´ pĈνρ 11 Aµ `
2 perms.q consistent with F.30.
Finally, let’s look at that last M1 ...M 11 term. At first it looks quite scary. Note we can write this last term
as 61 dĈ3 ^ dĈ3 ^ Ĉ3 . I have 11 indices to pick to be index 11. If I pick any of the indices of the last Ĉ I get
the term
1
dC3 ^ dC3 ^ B2
12κ2
If I pick either of the dC terms, then after an integration by parts I get the same term. So the same term
contributes three times Revisit this logic. We thus get the requisite action contribution:
ż
1
d10 xB2 ^ dC3 ^ dC3
4κ2

72
25. Under A1 Ñ A1 ` d and C3 Ñ C3 ` H3 we see that obviously R, F2 , B2 , and H3 will stay the same. Now
dC3 Ñ dC3 ` d ^ H3 while A ^ H3 Ñ A ^ H3 ´ d ^ H3 . Thus, F4 will stay the same.
It remains to look at the variation of B2 ^ dC3 ^ dC3 . This is

B2 ^ d ^ H3 ^ dC3 ` B2 ^ dC3 ^ d ^ H3 .

These two terms cancel by antisymmetry of the indices.

26. Defining C31 “ C3 ` A ^ B2 give the above transformation as A1 Ñ A1 ` d, C31 Ñ C31 ` H3 ` d ^ B2 so
that dC31 Ñ dC31 ` d ^ H3 ´ d ^ H3 “ dC31 .
Now G4 “ dC31 ´dA^B2 (Kiritsis wrote a small A, which I believe is a typo). Under the same transformation
we get that G4 is invariant as required.
Further, the transformation of C3 Ñ C3 ` dΛ2 implies that C31 Ñ C 1 ` dΛ2 ñ dC31 Ñ dC31 ñ G4 Ñ G4 as
required. So C31 now transforms trivially under the A transformation.

27. Take S “ S0 ` ie´φ . The SL2 pRq transformation acts on IIB supergravity as:
ˆ ˙ ˆ ˙ˆ ˙
aS ` b B2 d ´c B2
SÑ , Ñ
cS ` d C2 ´b a C2

The latter is also how H3 , F3 will transform together. Now think of S as the modular parameter τ , e´φ “
S2 “ τ2 is the imaginary part. Think of H3 , F3 as periods ω1 , ω2 The IIB action in the Einstein frame can
be written as
1 |G3 |2
ż „  ż
1 10 ? 1 BS B S̄ 1 2 1 G3 ^ Ḡ3
SIIB “ 2 d x ´g R ´ 2 ´ ´ F5 ` 2
C4 ^
2κ 2 S2 2 ¨ 3! S2 4 ¨ 5! 8iκ S2
BS B S̄
Now R, C4 , and F5 do not change. The term S2
transforms under SLp2, Rq exactly like the invariant
measure dττ dτ̄
2 .
2

Finally, any term consisting of a pair G3 , Ḡ3 in the numerator (either wedged or wedged with a hodge star)
divided by S2 will also remain modular invariant, as a quick Mathematica check confirms for us:

28. Again for some reason Kiritsis writes a small a, which again I think is a typo. We need to find which gauge
transformations need to be modified for C0 , B2 , C2 , C4 . There is a Chern-Simons term only in the definition
of F5 “ dC4 ´ C2 ^ H3 so we see that the C0 Ñ C0 ` c (for c a constant, the only closed 0-form) keeps the
action invariant.
Taking B2 Ñ B2 ` dΛ1 will keep H3 and therefore F5 invariant, so this transform is legitimate.
Also, taking C4 Ñ C4 ` dΛ3 will keep F5 invariant as well and will modify the Chern-Simons term in the
full action C4 ^ H4 ^ F3 by closed form, which will give no contribution.

73
 Finally, taking C2 Ñ C2 ` dΛ1 will change F5 Ñ F5 ´ dΛ1 ^ H3 . This must be compensated by changing
C4 Ñ C4 ` 21 Λ1 ^ H3 ` 12 dΛ1 ^ B2 . Then F5 will be invariant. Moreover, the Chern Simons term in the
action C4 ^ H3 ^ F3 have a variation
1 1
Λ1 ^ H3 ^ H3 ^ F3 ` dΛ1 ^ B2 ^ H3 ^ F3
2 2
After integration by parts this variation will contribute nothing, as required.

29. Clearly dim Op32q “ 32 ˆ 31{2 “ 496, which is necessary. For N “ 32 we also get from 7.9.29 that
TrpF 6 q “ 15 trpF 4 qtrpF 2 q where tr is the trace of the curvature form in (an associated bundle for) the
fundamental representation Also using 7.9.30 we get TrpF 4 q “ 24trpF 4 q`3ptrpF 2 qq2 and TrpF 2 q “ 30trpF 2 q.
Then, both sides of equation 7.9.26 become
15 15
15trpF 4 qtrpF 2 q “ ´ trpF 2 q3 ` trpF 2 q3 ` 15trpF 4 qtrpF 2 q
8 8
and we have agreement!

30. For an individual E8 , we have exactly

1 1 1 1
TrpF 6 q “ TrF 2 “ TrF 2 ¨ pTrF 2 q2 ´ pTrF 2 q3
7200 48 100 14400
Now for E8 ˆ E8 we have the property that TrpF n q “ Tr1 pF n q ` Tr2 pF n q where Tri means tracing over the
direct summands in the associated bundle that are acted on by the ith E8 . Then we will have the relations
1 1
TrpF 6 q “ pTr1 pF 2 q ` Tr2 pF 2 qq3 , TrpF 4 q “ pTr1 pF 2 q ` Tr2 pF 2 qq2
7200 100
So replacing TrpF 2 q with Tr1 pF 2 q ` Tr2 pF 2 q in the prior derivation gives us again exact matching and thus
anomaly cancelation.
On the other hand U p1q has no Casimirs so TrpF m q “ 0 for all m. In particular this allows us to take
E8 ˆ U p1q248 or U p1q496 as a gauge group and remain anomaly-free. Check with Nick. Reconcile this
with

31. Now let us turn to the SOp16q ˆ SOp16q theory. To check anomalies, we look at the chiral terms. In this
case we have massless content consisting of spin 1{2 Majorana-Weyl fermions transforming in the p16, 16q
(positive chirality), and p1, 128q ‘ p128, 1q (negative chirality) representations. We also have massive fermion
fields in the p128, 128q representation that do not contribute to the anomaly. Note that we do not have a
gravitino, as there is no spacetime SUSY in this theory.
The positive chirality p16, 16q MW fermions have field strength F` “ F1 b 1 ` 1 b F2 valued in the vector
representation.
The negative chirality p1, 128q ‘ p128, 1q MW fermions have field strength F̂´ “ F̂1 ‘ F̂2 valued in the spinor
representation.
Our anomaly polynomial is thus
1
pI pR, F` q ´ I1{2 pR, F´ qq
2 1{2
n
Both representations have dimension 256, so the 64 p. . . q term in 7.9.22 cancels (note we did not need to
use that the dimension of the gauge group was 496 here!). We are left with (using tr for the trace in the
fundamental representation and trs for the spinor rep’n)

trrF`6 s trrF`4 sTrrR2 s trrF`2 s TrrR4 s pTrrR2 sq2

ˆ ˙
´ ` ´ `
720 24 ¨ 48 256 45 36
˜
6 4 2 2 ˙¸ (70)
trS rF̂´ s trS rF̂´ sTrrR s trS rF̂´ s TrrR s pTrrR2 sq2
4
ˆ
´ ´ ` ´ `
720 24 ¨ 48 256 45 36

74
 From explicitly expanding out pF1 b 1 ` 1 b F2 q2,4,6 we get:

trF`2 “ 16ptrF12 ` trF22 q

trF`4 “ 16ptrF14 ` trF24 q ` 6trF12 trF22
trF`6 “ 16ptrF16 ` trF26 q ` 15trF12 trF24 ` 15trF14 trF22

Together with the results 7.4E, 7.5E relating trS to tr we get

trS F´2 “ 16ptrF12 ` trF22 q

trS F´4 “ ´8ptrF14 ` trF24 q ` 6ptrF12 ` trF22 q2
15 `
trS F´6 “ 16ptrF16 ` trF26 q ´ 15ptrF12 trF14 ` trF22 trF24 q ` ptrF12 q3 ` ptrF22 q3
˘
4
Altogether we get

We thus get a Green-Schwarz term

ż „ 
10 4 4 1 2 2 2 2 2 2
9 d xB Tr1 pF q ` Tr2 pF q ´ pTr1 pF q ` Tr2 pF q ´ TrpF1 qTrpF2 q
4

We have exhausted the set of supersymmetric chiral anomaly-free theories, so the question remains whether
there are any non-supersymmetric theories that are chiral and anomaly-free in 10D. We will have only MW
fermions and perhaps self-dual 5-form fields contributing. It does not seem possible to cancel the IA pRq with
just the I1{2 pR, F q, so I expect any such 10 D non-SUSY theory will in fact contain only MW fermions.
They must come in pairs of opposite parities with equal particle number to cancel the gravitational anomaly.
Exhaustively showing this seems really difficult. Xi didn’t know the full answer
I know for a fact there is at least one other anomaly free theory in 10D, namely the U Spp32q open string
Sugimoto theory (c.f. question 7.36).

32. We are orbifolding by a Z2 . In the sector of the left-moving worldsheet fermions, only p´1qF acts nontrivially.
The twisted blocks are „  1 4 a
“ ‰
h 1 ÿ a`b`ab`ag`bh`gh θ b
Zf ermions “ p´1q
g 2 a,b“0 η4

On the E8 ˆ E8 s the untwisted block is just p 21 ab θ̄8 ab {η̄ 8 q2 . Performing the projection g requires that a, b
ř “ ‰

match for both factors, giving:

1 “a‰ “0‰ “0‰
1 ÿ p´1qb θ̄8 b p2τ q 1 θ̄8 0 ` θ̄8 1
„ 
0
ZE82 “ “
η̄ 8 p2τ q 4 η̄ 4 pτ qθ̄4 10 pτ q
“‰
1 4 a,b“0

Taking τ Ñ ´1{τ gives “0‰ “1‰

1 θ̄8 0 ` θ̄8 0
„ 
1
ZE82 “
4 η̄ 4 pτ qθ̄4 01 pτ q
“‰
0

75
 Finally taking τ Ñ τ ` 1 gives “0‰ “1‰
1 θ̄8 1 ` θ̄8 0
„ 
1
ZE82 “
4 η̄ 4 pτ qθ̄4 00 pτ q
“‰
1
The full partition function is thus
1 „  „ 
1 ÿ h h
Z“ ZE82 Zf ermions
2 h,g“0 g g

We see that this is modular invariant, as individually both Zf ermions and ZE82 are invariant under τ Ñ ´1{τ .
Their anomalous changes under τ Ñ τ ` 1 from the η 4 powers in the denominator are cancelled in pairs.
The gauge group corresponds to the invariant (diagonal) E8 sublattice of E8 ˆ E8 (Confirm). At the
massless level, we still have the gravity supermultiplet (G, B, Φ), as well as gauge bosons with gauge group
E8 from the untwisted sector. What about the twisted sector?
The gravitino has been projected out, so this theory no longer has spacetime supersymmetry. The theory is
still chiral, and since the partition function is modular invariant, we are also guaranteed that it is anomaly
free. However, it has a tachyon.
Unfinished

33. Let’s assume we do not have self-dual 2-form gauge fields that give self-dual 3-form field strengths and we
do not consider an IA contribution. Recall we can write
D{2
ź xi {2
I1{2 “
i“1
sinhpxi {2q

where xi are the off-diagonal entries in the 2 ˆ 2 block decomposition of R0 “ dω. All of this is easy to do
in Mathematica.

For n spin 1/2 fermions and a gravitino we thus get the forms

TrpR4 q pTrR2 q2 TrF 2 TrF 4

ˆ ˙
n
I1{2 pR, F q “ ` ´ TrR2 `
576 10 8 96 24
49 43
I3{2 pRq “ TrpR4 q ´ TrpR2 q2
576 ˆ 2 576 ˆ 8
The Anomalies.nb also has the 10D cancelation if anyone is interested.

76
34. In the absence of a linear dilaton background, the RR fields simply satisfy the equations of motion d‹Fp`2 “ 0,
as well as the Bianchi identities dFp`2 “ 0. We want to show that, the tree level effective action of type
II SUGRA in the string frame will have no coupling at tree level between the RR field strengths and the
dilaton.
In a linear dilaton background Φ “ ?Q X 9 , the supercurrent G will be modified to
2`s
?
2 ? 1
G“i ψ ¨ BX ´ i 2 Φ,µ Bψ µ ñ G0 9 ? ψ0 ppµ ` iΦ,µ q
`2s 2
I’m not sure about a possible constant factor multiplying the second term in the definition of G, but it is as
in Polchsinki 12.1.18. Acting on the RR ground states, ψ0 gives an additional Γ matrix, but the Φ µ term
will modify the Bianchi and free massless equations as:

pBµ ´ Φ,µ q ^ F “ pBµ ´ Φ,µ q ^ ‹F “ 0 ñ eΦ de´Φ F “ eΦ d ‹ pe´Φ F q “ 0.

This implies that we should view F̂ “ e´Φ F as the field strength, and so the RR states correspond to eΦ F
(ie they already incorporate a factor of eΦ ). The RR charges are surface integrals of F̂ “ dC. Thus the
dilaton coupling to the RR field strength F̂ 2m is e2mΦ e2pk´1qΦ F 2m . In particular, at tree level the F 2 term
does not couple to the dilaton.

35. The (minimal) supergravity multiplet contains the left-handed 3{2 gravitino as well as a right-handed self-dual
3-form field. The tensor multiplet contains a left-handed anti-self-dual 3-form field and the right-handed 1{2
dilatino. Combining one of the NT tensor multiplets with the gravity multiplet gives an anomaly contribution
of:
I3{2 ´ I1{2
The vector multiplet contains the left-handed gaugino. The hypermultiplet (which BTW Kiritsis has not
yet defined this) apparently contains a right-handed hyperino (wow fancy).
So far this gives
I3{2 ` pNV ´ NH ´ 1qI1{2 pRq
But we have NT ´ 1 addition tensor multiplets which will then contribute

I3{2 ` pNV ´ NH ´ NT qI1{2 pRq ` pNT ´ 1qIA pRq

A quick calculation for an anti-self-dual tensor gives

7TrR4 pTrR2 q2
ˆ ˙
IASD “ ´ ´
1440 144 ˆ 4
the minus sign out front is from being anti -self dual.
As before, in order to have factorization of the anomaly polynomial for the GS mechanism to work, we need
the TrR4 terms to cancel. This gives our desired constraint
49 pNV ´ NH ´ NT q 7
` ´ pNT ´ 1q “ 0 ñ NH ´ NV ` 29NT “ 273
144 ˆ 8 144 ˆ 40 144 ˆ 10
I think Kiritsis has a typo in this equation and it should be `29NT rather than ´29. This is consistent with
BBS exercise 5.9.

36. Another 10D nonsupersymmetric string theory without tachyon! This one is open+closed. The Op16qˆOp16q
is the only closed non-SUSY string theory in 10D without tachyon. Is this the only open one? The
relevant reference is arXiv:hep-th/9905159
This is a theory of strings stretching D9 ´ D9̄ branes.
We have λ, λ̃ are positive chirality spinors belonging to the adjoint of Sppnq (equivalent to the symmetric
representation and traceless antisymmetric representation of Sppmq respectively, while ψ, ψ̄ are

77
negative chirality spinors belonging to the bi-fundamental representation of SOpnq ˆ SOpmq. We take n “
0, m “ 32.
As in the SOp16q ˆ SOp16q example, the lack of spacetime SUSY means there is no gravitino contribution,
and we look only at the massless fermion content:

Iλ ` Iλ̄ ´ 2Iψ

The gravitational anomaly cancels for free since we have the same number of left and right chirality fermions.
λ̄ does not contribute to the gauge or mixed anomaly, since it transforms trivially under USpp32q.
Finish

78
Chapter 8: D-Branes
1. First, a simple magnetic monopole for a 1-form gauge field in D spacetime dimensions has a radial magnetic
field Br “ Ω Q̃r1D´2 where ΩD´2 “ 2π d{2 Γpd{2q is the volume of a unit D ´ 2 sphere. This way, the flux of
D´2
the solution over any D ´ 2 sphere surrounding the (point) monopole will be Q̃1 .
Upon taking the Hodge star we get the solution is F “ Q̃1 sin θdθ ^ dφ. We can write this as A “
Q̃1 pc ´ cos θqdφ. Taking c “ 1 we get A vanishes at θ “ 0 (which we need since the φ coordinate degnerates
there) while taking c “ ´1 we get A vanishes at θ “ π, which we also need.
We cannot have both solutions, and so we realize we are dealing with two As, corresponding to local sections
of a line bundle over S 2 on different hemispheres. Let A` be well-defined on all points on S 2 except θ “ π.
Then A` is a section of a line bundle on the punctured sphere. The punctured sphere is contractible so any
fiber bundle over it is trivial, so A` is just a function on the punctured sphere S 2 ztθ “ πu. So let’s define
A` “ Q̃1 p1 ´ cos θqdφ. Similarly, we define A´ to be the nonsingular A on the sphere with θ “ 0 removed,
namely A´ “ Q̃1 p´1 ´ cos θqdφ.
On the overlap, A` ´ A´ differ by an integer, which labels the degree of “twisting” of this line bundle over
S2.
For a p form, our monopole will now be spatially extended in p ´ 1 directions. Label these (locally),
by x1 . . . xp´1 . Time is x0 . Locally transverse to these coordinates will be r, ϕ1 . . . , ϕD´1´p , where ϕi
parameterize a D ´ 1 ´ p sphere enclosing the monopole. The field strength looks like:

F “ Q̃p ΩD´p´1

where Ω is the canonical D ´ p ´ 1-sphere area form:

Ω “ sinD´p´2 pϕ1 q sinD´p´3 pϕ2 q . . . sinpϕD´p´2 q dϕ1 ^ . . . dϕD´p´1

This can be written (unfortunately unavoidably) in terms of a hypergeometric function:

sinD´p´1 pϕ1 q
ˆ ˙
1 D´p´1 D´p`1 2
A “ 2 F1 , , , sin pϕ1 q dϕ2 ^ ¨ ¨ ¨ ^ dϕD´p´1
2 2 2 D´p´1

there is no need for an overall constant, as the function above vanishes at both ϕ1 “ 0 and π, however this
is compensated by the hypergeometric function having a branch cut at ϕ1 “ π{2. Across this cut, it will
have a discontinuity set by an integer depending on the convention of the arcsin function, and again we will
have A` ´ A´ differing by an integer. The same quantization condition follows.
Again A` will be defined on the S D´p´1 sphere minus the south-pole (this is homeomorphic to the D ´ p ´ 1
ball, and hence contractible, so again the line bundle trivializes and A` is a bona-fide function for any D, p)
and A´ is similarly defined on the sphere with the excision of the north pole.

2. Our simply charged point particle with a Wilson line A9 “ χ{2πR turned on will have an action

m2
ż ˆ ˙
1 M 9
S “ dτ x9 x9 M ´ ` qA9 x9
2 2
looooooooooooooooomooooooooooooooooon
L

qχ
The canonical momentum will be pi “ x9 µ for µ “ 0 . . . 8 and p9 “ x9 9 ` 2πR . Consequently, our hamiltonian
is
qχ 2 m2
„ 
M 1 µ qχ 1 qχ qχ
H “ pM x9 ´ L “ pµ p ` p9 pp9 ´ q ´ pp9 ´ q ´ ` pp9 ´ q
2 2πR 2 2πR 2 2πR 2πR
1´ qχ 2 ¯
“ pµ pµ ` pp9 ´ q ` m2
2˜ 2πR ¸
2πn ´ qχ 2
ˆ ˙
1 µ 2
“ pµ p ` `m
2 2πR

79
3. For a string satisfying Dirichlet boundary conditions, the total momentum is not conserved (along the
directions associated with the D boundary conditions). This is easily interpretable as momentum transfer
to the brane that it is attached to.
χ ´χ
4. For an open string of state |ijy, A9 will act as i2πi j . Since this is an open string with no winding, we can
only have momentum contribution, and so we will get a mass formula
1 ˆ ˙2
N̂ ´ n χi ´ χj
m2ij “ 2
` ´
`2s R 2πR

In particular at the lowest (massless) level for a string without momentum we will get the desired spectrum
ˆ ˙2
χi ´ χj
m2ij “
2πR

5. For completeness, we will do both the gauge boson and scalar scattering. These come from the NS sector,
and are given by:
a,µ gp
V´1 “ gp λa ψ µ e´φ ceipX , V0a,µ “ ? λa piX9 ` 2p ¨ ψ ψ µ qceipX
2`s
We can explicitly scatter four such gauge bosons - two in the ´1 picture and two in the 0 picture.

iCD2 δ
10 pΣkq ˆ gp4 xcV´1 py1 qcV´1 py2 qcV0 py3 qcV0 py4 qy ` 5 perms.
iδ
p`1 gp4
“ ˆ xrψ µ1 eik1 X sy1 rψ µ2 eik2 X sy2 rpiX9 µ3 ` 2k3 ¨ ψψ µ3 qeik3 X sy3 rpiX9 µ4 ` 2k4 ¨ ψψ µ4 qeik4 3X sy4 y
gp2 `4s 2`2s
ˆ xcpy1 qcpy2 qcpy3 qy xe´φpy1 q e´φpy2 q y

Take y1 “ 0, y2 “ 1, y3 “ 8 and integrate y4 “ y from 0 to 1 (then we’ll have 5 more terms coming from
permuatations). There are five contributions.

• Contracting the ψ µ1 p0qψ µ2 together and allowing the remaining 4 terms at y3 , y4 to contract either
amongst themselves or with various vertex operators.
• Not contracting the first two ψ and Contracting the iXpy9 3 q with any of the vertex operators while
contracting the last ψ with the first two
• Not contracting the first two ψ and contracting the iXpy9 4 q with any of the vertex operators while
contracting the third ψ with the first two
9 and contracting the ψ at y1 with the various ψ at y3 (consequently the ψ at y2 with
• Forgetting the iX,
the ψ at y4 )
• Swapping 3 with 4 in the above (this gives an overall minus sign by fermionic statistics)

Integrating this will give two types of terms: η ab η cd and η ab k c k d . Our shorthand replaces the superscript µi

80
by just i. Below, I underline the terms that contribute to the first type:
2 2
#
igp2 δ
p`1 1 y 2`s k1 ¨k4 y 2`s k2 ¨k4 ´ k4 k24 ¯
ż „ 
2 12 2 34 2 2 34 3 4 2 2 3 3 1
dy ´ ` s η 2` s η ` p2` s q p´η k 3 ¨ k4 ` k k
4 3 q ` p2` s q pk 2 ` k4 q `
`4s 0 2`2s y y´1
` `2s p2`2s q2 pk23 ` yk43 qp´k41 η 24 ` k42 η 14 q
“ ‰
”´ k 4 k4 ¯ ı
` `2s p2`2s q2 1
` 2 p´k31 η 23 ` k32 η 13 q
y y´1
2 2
p2` q “ 13 24
` `2s s η η k3 ¨ k4 ` η 34 k31 k42 ´ η 13 k42 k34 ´ η 24 k43 k31
‰
y´1
+
p2` 2 q2 “
´ `2s s η 14 η 23 k3 ¨ k4 ` η 34 k32 k41 ´ η 13 k42 k34 ´ η 24 k43 k31
‰
y
(71)
Using s “ ´2`2s k1 ¨ k2 etc we see the underlined terms contribute

igp2 δ
p`1 Γp1 ´ uqΓp1 ´ tq

„ 
12 34 Γp1 ´ uqΓp´tq 14 23 Γp´uqΓp1 ´ tq 14 23
p´p1 ` sqη η q ` η η s` η η s r1423s
`2s Γp2 ` sq Γp1 ` sq Γp1 ` sq
igp2 δ
p`1 Γp´uqΓp´tq
“ p´tuη 12 η 34 ´ suη 13 η 24 ´ stη 14 η 23 qr1423s
`2s Γp1 ` sq

The non-underlined terms are more involved but end up contributing twelve terms that yield:

igp2 δ
p`1 Γp´uqΓp1 ´ tq 2 12 3 4 igp2 δ
p`1 2 Γp´uqΓp´tq “ 12 3 4

„ 
‰
2
2` s η k k
2 1 ` . . . r1423s “ 2
2`s tη k2 k1 ` 11 perms. r1423s
`s Γp1 ` sq `s Γp1 ` sq

Here my Mandelstam variables are dimensionless. The result with dimensionful Mandelstam variables is:
2 2
ˆ ˙
2 p`1 2 Γp´`s sqΓp´`s uq
igp δ
`s K4 pki , ei qpr1234s ` r4321sq ` 2 perms. (72)
Γp1 ` `2s tq

with K4 pki , ei q “ ´tue1 ¨ e2 e3 ¨ e4 ` 2spe1 ¨ e3 e2 ¨ k4 e4 ¨ k2 ` 3 perms.q ` 2 perms.

Now for the four transverse scalar amplitude, our vertex operators look like:
a,µ gp
V´1 “ gp λa ψ µ e´φ ceipX , V0a,µ “ ? λa pX 1 ` 2p ¨ ψ ψ µ qceipX
2`s
here X 1 “ Bσ X “ 2iBX. Moreover, we have Dirichlet boundary conditions on both the Xs and the fermions
ψ. The ψ are still in the NS sector since we’re looking at the (bosonic) scalar field scattering.
Crucially, the correlators for the ψ field are the same for DD boundary conditions. We had xX9 µ pzqX9 ν pwqy “
2 2
´ `2s η µν pz ´ wq´2 while the correlators for X 1 pick up a minus sign xX 1 i pzqX 1 j pwqy “ `2s η ij pz ´ wq´2 . This
ends up giving the exact same result however, since the vertex operators contain X 1 pzq while the prior ones
contain iXpzq.9
Finally, contracting X 1 i with any of the eikX will give zero, since the open strings only have momenta parallel
to the Dp brane while the X 1 i is transverse. This gives a simpler amplitude than (72):

Γp´`2s sqΓp´`2s uq
ˆ ˙
igp2 δ
p`1 `2s K41 pr1234s ` r4321sq ` 2 perms. (73)
Γp1 ` `2s tq

with K41 “ ´ptuδ12 δ34 ` suδ13 δ24 ` stδ14 δ23 q. In the case where there are no CP indices we expand the ΓΓ{Γ
functions: ˆ ˙ ˆ ˙
2 p`1 2 1 2 2 2 2 p`1 2 1 2ps ` t ` uq
igp δ
`s K4 ˆ 4 ` 4 ` 4 “ igp δ
`s K4 ˆ “0
`s su `s st `s tu `4s stu
So to leading order in the string length this is zero. This is consistent with the U p1q DBI action, as the
scalars do not directly interact with the U p1q gauge field Aµ (in general a real scalar cannot be charged

81
 under a U p1q gauge field). That is, at leading order the action is free in the X fields. Taking ξ µ “ X µ for
µ “ 0 . . . p and X i independent functions, we get:

δ αβ δij
ż b ż b ż
dp`1 ξ det GM N Bα X M Bβ X N “ dp`1 ξ det δαβ ` δij Bα X i Bβ X j Ñ dp`1 ξ Bα X i Bβ X j
2
This is just a free theory. Its also quick to see that the 3-point function of the transverse scalars vanishes at
tree level in string perturbation theory.

6. I have done the previous problem in full generality, including CP indices. So now let’s again look at the s
channel. As s Ñ 0 so that t “ ´u we get from the δ12 δ34 term a pole in s going as:

gp2 t
ˆ ˙
1 1
2 p`1 2
´igp δ
`s tuˆ 4 pr1234s ` r4321sq ` 4 pr1243s ` r3421sq “ ´iδ
p`1 2 pr1234s`r4321s´r1243s´r3421sq
`s su `s st `s s

We can rewrite this as:

gp2 t g2
p`1 p t
´iδ
p`1 pTrp12r34sq ´ Trpr34s21q “ ´iδ Trpr12sr34sq
`2s s `2s s

The pole at s “ 0 corresponds to an exchange of a gluon from the 12 pDµ X I q2 term in 8.6.1.
We also have a further term that does not involve a pole in s. Let’s still take 1 and 2 equal. Expanding to
this order we find:
gp2 ´ ¯
´iδ
p`1 2 Trpr12sr34sq ` Trpr13sr24sq ` Trpr14sr23sq
`s
This comes from exactly the potential term 14 rX I , X J s2 in the effective action 8.6.1. Come back to that
last term

7. Momentum conservation will imply pk “ 0 for the NSNS states. Our vertex operator will take the form
˚
ζµν cc̃e´φ e´φ̃ ψ µ ψ̃ ν eikK X . We can use the doubling trick to get ζµν cpzqcpz ˚ qe´φpzq e´φpz q ψ µ ψ̃ ν eikK X and we
are automatically in the ´2 picture.
µν
The states from in the p ` 1 parallel directions give just the correlator xψ µ piqψ ν p´iqy “ ´ η2i (importantly
NN fermions in NSNS have 2-point function ´1{pz ´ w̄q c.f. 4.16.22). We also get a δ
p`1 from momentum
conservation.
δ ij
The states in the transverse (Dirichlet) directions give 2i correlator. Defining the diagonal matrix Dµν “
pη αβ , δ ij q we get a correlator proportional to

gc p`1 µν p2π`s q2 Tp
´ δ pkk qD “ ´ Vp`1 Dµν
2`2s gp2


2

Check with Victor. Confirm the tension relation. This diagonal tensor Dµν allows for a nonvanishing
dilaton and graviton tadpole, but will not couple to the antisymmetric Kalb-Ramond B-field.

8. Our RR fields have picture pr, sq for r, s half-integers. In order to have total picture ´2 on the disk, we need
to pick this to be the (asymmetric) p´3{2, ´1{2q picture. The construction of this operator is complicated.
I expect that the p´1{2, ´1{2q operator that we are familiar with is basically e´φ G0 times the p´3{2, ´1{2q
operator. This means that the p´3{2, ´1{2q will be one less power of momentum and one less gamma
matrix than the p´1{2, ´1{2q operator. Picking out the p ` 2 form part of the p´1{2, ´1{2q operator (in
Blumenhagen’s convention 16.21) gives

1 `s αβ 1 ` Fµ1 ...µp`2 µ1 ...µp`2 αβ

? F S α pzqS̃β pz̄qe´φ{2´φ̄{2 Ñ ?s pΓ q S α S̃β e´φ{2´φ̄{2
4 2 4 2 pp ` 2q!

Here S α “ Sα: Γ0 as is standard in a spinor product. BRST will require that F and ‹F be closed.

82
 Changing picture means removing one power of momentum and one gamma matrix. This anti-differentiates
F , which must be proportional to the potential C since prµ1 Cµ2 ...µp`2 s “ Fµ1 ...µp `2 . It is then reasonable to
expect the corresponding p´3{2, ´1{2q operator to be proportional to

Cµ1 ...µp µ0 ...µp αβ

e´3φ{2´φ̄{2 pΓ q S α S̃β
pp ` 1q!

Note both e´3φ{3 Sα and e´φ̃{2 Sβ remain primary operators, having dimensions 3{8 ` 5{8, so this is indeed
a reasonable guess. Now, for the D-brane boundary conditions we have
9
ź
K
pδ S̃qβ pz̄q “ Sβ pz q, ˚ K
δ “ δi, δ i “ Γi Γ11
i“p`1

This reflects the Sβ spinor along all the Dirichlet directions and keeps it the same across the Neumann
directions.
From 5.12.42 of Kiritsis I expect the leading-order of the Sα S̃β correlator to be Cαβ {pz ´ z̄q10{8 and the
e´3φ{2 ẽ´φ{2 will contribute pz ´ z̄q´3{4 to make this a primary correlator transforming as Cαβ pz ´ z̄q´2 . For
Neumann boundary conditions, Cαβ is the charge conjugation matrix. For the D-brane, it will be δ K C.
Only in the IIB case will Cαβ will be nonzero between S α and S̃β since Sα and S̃β transform in the same
representations. Each β i changes the chirality. So the amplitude in IIB will vanish if we have an even number
of β i , equivalently 9 ´ p is odd, so we will have only odd dimensional branes in IIB as required and even
dimensional branes in IIA as required.
We thus get an amplitude proportional to:
µ0 ...µp
gc δ
p`1 Cµ0 ...µp gc δ
p`1 Cµ0 ...µp pp`1q
A“i 2 2 TrpΓµ1 ...µp Γp`1 . . . Γ9 Γ11 q “ i 2 2
gp `s pp ` 1q! go `s pp ` 1q!

Comparing with the 8.4.4, which should factorize as Appk q2 G9´p ppK qδ p`1 ppk q we see that the normalization
of our on-shell amplitude is in fact:
µ ...µ
0 p
? Cµ0 ...µp pp`1q
3´p
A “ iVp`1 2πp2π`s q
pp ` 1q!

This is consistent with other results c.f. Di Veccia, Liccardo Gauge Theories from D-Branes, arXiv:0307104
but I think they’re not incorporating 1{αp “ 2κ210 in the propagator. Taking this factor into account and
dividing by it followed by taking a square root gives us an on-shell amplitude of:
0 p µ ...µ
1 Cµ0 ...µp pp`1q
A “ iVp`1 “ iVp`1 Tp Cp`1 ^ pp`1q .
p2π`s qp `s gs pp ` 1q!
ş
This is exactly what would come from a minimal coupling term of the form iTp Cp`1 .

9. We take one vertex operator to be in the p´1, ´1q picture and gauge fix it to lie at z “ i, and take the
other in the p0, 0q picture lying at iy, fixing y to range from 0 to 1. In doing the doubling trick, we take
X̃ µ “ Dνµ X ν , ψ̃ µ “ Dνµ ψ ν and φ̃ “ φ, c̃ “ c.
We wish to calculate the correlator:
gc2 2 1 1
´ 2 4 2
xrψ µ ψ̃ µ̄ eik1 X s0 rpiBX µ ` k2 ¨ ψψ ν qpiB̄X ν̄ ` k2 ¨ ψ̄ ψ̄ ν̄ qeik2 X sy y
go `s `s 2 2

The simplest way to do this problem is to recognize that after the doubling trick has been applied, we are
calculating the a correlator of four fields at collinear insertions on the Riemann sphere.
1 1
ε1µµ̄ ε2ν n̄u Dµµ̄1 Dνν̄1 xV´1
µ µ
pp1 q, V´1 pDp1 q, V0ν pp2 q, V0ν pDp2 qy

83
 We can map these four points to 0, 1, y, 8 and take the integral to be from y “ 0 to y “ 1, with appropriate
jacobian.
This is nothing more than the 4-gluon amplitude calculated previously, which gave
go2 p`1 2 Γp´`2s sqΓp´`2s uqΓp´`2s sq
2i δ
`s Ñ K4 pki , ei q
gp2 Γp1 ` 2`2s tq

Here we do not have three terms for y in different regions - we only have one. Our momenta are p1 , p2 , Dp1 , Dp2
giving: The only caveat is that in that case, we had boundary normal ordering. We can view this as
`here
s “ `there
s {2. This is reflected by halving the momenta of the open strings.

pthere
1 Ñ phere
1 {2, pthere
2 Ñ phere
2 {2
pthere
3 Ñ D ¨ phere
1 {2, pthere
4 Ñ D ¨ phere
2 {2

This gives us (here we do not have three terms for y in different regions - we only have one)

go2 p`1 2 Γp´`2s sqΓp´`2s uq go2 p`1 2 Γp`2s p1 ¨ p2 {2qΓp`2s p1 ¨ Dp1 {2q
2i δ ` s Ñ K pk ,
4 i i e q Ñ 2i δ
`s
gp2 Γp1 ` 2`2s tq gp2 Γp1 ` `2s p1 ¨ p2 {2 ` `2s p1 ¨ Dp1 {2q

Following Myers, we can write t “ ´2p1 ¨ p2 as the momentum transfer to the brane, and q 2 “ p1 ¨ Dp1 {2 as
the momentum flowing parallel to the brane. The Gamma ratio then looks like:
Γp´`2s t{4qΓp`2s q 2 q
Γp1 ` `2s t{4 ` `2s q 2 q

If we held t fixed and took the large q limit we would get a series of open string poles. This corresponds to
a closed string splitting in two, with its ends on the D-brane as an intermediate state.
Now let’s hold q 2 fixed and take the large t limit. This is the limit of large energy transfer - the Regge limit.
From the ratio of ΓΓ{Γ we see that there are closed string poles. This can be interpreted as the closed strings
interacting with the long-range background fields generated by the presence of the Dp brane.

Comment about pole structure

10. In the D9 brane case, we have seen that the open string boundary only preserves the sum of Q ` Q̃. If
we T-dualize in the 9 direction, we act on the right-moving sector by spacetime parity, so that necessarily
B̄X 9 Ñ ´B̄X 9 , ψ̃ 9 Ñ ´ψ̃ 9 , S̃α Ñ δ 9 S̃ şα (up to a phase in that last one). Here δ 9 “ Γ9 Γ11 . Our spacetime
1
supersymmetry generator Q̃α “ 2πi dz̄ e´φ{2 Sα therefore will be mapped to δ 9 Q̃. Thus, in the T-dual
picture we preserve the supercharge Q1 ` δ 9 Q̃1 .
Iterating this procedure in other directions we get that in general we preserve Q ` δ K Q̃, with δ K “ i δ i ,
ś
where i runs perpendicular to the brane. Note that T-dualities along different directions do not commute!
They commute up to a p´1qFR , and so the order that we do them matters. In this case the δ i act by
left-action.

84
11. (As in Polchinski section 13.4) From the previous problem, we see that the first D-brane preserves the
1 ´1 1
supercharges Q ` δ K Q̃ while the second preserves the supercharges Q ` δ K Q̃ “ Q ` δ K pδ K δ K Q̃q so the
supersymmetries that will be preserved must be of both forms. This is in one-to-one correspondence with
´1 1
spinors invariant under δ K δ K . This operator is a reflection in the direction of the ND boundary conditions
(the directions orthogonal to the Dp1 brane in the Dp brane). Since in either IIA or IIB p and p1 must differ
by an even integer, the number of mixed boundary conditions–call it ν–must be even. Then we can write
´1 1 ´1 1
δ K δ K as a product of rotations by π along each of the ν{2 planes δ K δ K “ eiπpJ1 `¨¨¨`Jν{2 q . Each j acts
´1 1
in a spinor representation, so that eiπJi has eigenvalues ˘i. If ν{2 is odd, this makes δ K δ K “ ´1 so this
will not preserve supersymmetry. We thus need ν{2 even, or ν “ 0 mod 4.
From this I posit that the static force between two branes vanishes precisely when ν “ 0 mod 4.

12. Now let’s confirm this guess with an amplitude calculation. Take p1 ď p. We work in lightcone gauge. We
do a trace over an open string with p1 NN boundary conditions, p ´ p1 “ ν DN boundary conditions, and
8 ´ p DD boundary conditions.
We begin from the open string point of view in calculating the cylinder amplitude. Chapter 4 has done the
NN, DD, and DN boson amplitudes for us. The difficulty lies almost entirely in the fermions. Recall the
following:
8
ź
η “ q 1{24 p1 ´ q n q
n“1
d d d
8 8 8
θr00 s ź θr10 s ? ź θr01 s ź
“ q ´1{48 p1 ` q n`1{2 q “ 2 q 1{24 p1 ` q n q “ q ´1{48 p1 ´ q n`1{2 q
η n“0
η n“0
η n“0

Further from 4.16.2 recall that for modes bn`1{2 , bn corresponding to NS and R sectors the NN and DD
boundary conditions give:

• NN: b̄n`1{2 “ ´bn`1{2 , b̄n “ bn

• DD: b̄n`1{2 “ bn`1{2 , b̄n “ ´bn

For DN we have the same result as for DD but now the R sector is half-integrally modded and the NS
sector in integrally modded. Now lets compute partition functions. Our final answer will be a sum over spin
structures NS+, NS-, R+, R-. Taking q “ e´2πt we see Trrq L0 ´c{24 s “

• NS+:
a
– NN: q ´1{48 n p1 ` q n`1{2 q “ θr00 s{η
ś
a
– DD: q ´1{48 n p1 ` q n`1{2 q “ θr00 s{η
ś
? a ?
– DN: 2q ´1{48 q 1{16 n p1 ` q n q “ θr10 s{η ( 2 when raised to a power counts ground state degen-
ś
eracy)
• NS-:
a
– NN: q ´1{48 n p1 ´ q n`1{2 q “ θr01 s{η
ś
a
– DD: q ´1{48 n p1 ´ q n`1{2 q “ θr01 s{η
ś

– DN: 0
• R+
? a
– NN: 2q 1{24 n p1 ` q n q “ θr10 s{η
ś
? a
– DD: 2q 1{24 n p1 ` q n q “ θr10 s{η
ś
a
– DN: q ´1{48 n p1 ` q n`1{2 q “ θr00 s{η
ś

• R-
– NN: 0
– DD: 0
a
– DN: q ´1{48 n p1 ´ q n`1{2 q “ θr01 s{η
ś

85
Notice NN vs DD boundary conditions have no effect on fermion contribution to partition function. This
is because, although the left moving and right-moving modes are identified differently, the mode excitations
look exactly the same.
a
On the other hand for NN and DD the bosons will contribute 1{η and will contribute η{θr01 s for DN.
Thus we have the following contributions to the partition function (here N is the number of NN boundary
conditions): ´ ¯2
∆x
´2πt 2π`
˜ “ ‰ ¸p8´νq{2 ˜ “ ‰ ¸ν{2
θ 00 θ 10
ż
VN dt e s
N S` “ ?
p2π`s qN 2t p 2tqN η 8´ν θr0 s{η ν{2 η η
` ˘
1
´ ¯2 ˜ “ ‰ ¸8
∆x
´2πt 2π`
θ 01
ż
VN dt e s
N S´ “ ´ ? δν“0
p2π`s qN 2t p 2tq η 8
N η
´ ¯2 ˜ “ ‰ ¸p8´νq{2 ˜ “ ‰ ¸ν{2
∆x
´2πt 2π`
θ 10 θ 00
ż
VN dt e s
R` “ ´ ?
p2π`s qN
` 0 ˘ν{2
2t p 2tqN η 8´ν θr s{η η η
1
´ ¯2
∆x
ż ´2πt 2π`
VN dt e s
R´ “ ? δν“8
p2π`s qN 2t p 2tqN
All theta and eta functions are evaluated at it. The circumference of the cylinder is 2πt. The relative
signs in front of the different contributions come from a combination of defining the NS vacuum to have
negative fermion and modular invariance (equivalently spacetime spin-statistics). Note when ν “ 4 we only
get contributions from NS+ and R+, which exactly cancel. Similarly when ν “ 4 or 8, by the abstruse
identity of Jacobi we will get cancelation again.
We can interpret our result as a one-loop free energy. Differentiating this w.r.t. ∆x would then give us
our force. For ν “ 0, 4, 8 we do not get a force, consistent with the D-brane configuration preserving
supersymmetry.
For the sake of completeness, and to clear my own confusion once and for all, I will also do this from the
POV of the boundary state formalism (not developed in Kiritsis). For a good reference see the last chapter
of Blumenhagen’s text on conformal field theory.
For a single free boson, after the flip pσ, τ qopen Ñ pτ, σqclosed the boundary states |N y , |Dy must satisfy
pαn ` α̃´n q |N y “ 0, pαn ´ α̃´n q |Dx y “ 0,
This gives boundary states:
1 ź 1 ÿ
|N y “ ? e´ n α´n α̃´n |0, 0; 0y “ |m,
~ Θm;
~ 0y
p2π`s 2q1{2 n m“tm
~ iu
? ż
dk ipx ź ´ 1 α´n α̃´n
|Dx y “ p2π`s { 2q1{2 e e n |0, 0; ky
2π n

The overall normalization came from comparing with cylinder amplitudes. Θ here is CPT reversal. Similarly
for a fermion
pψn ` ψ̃´n q |N y “ 0, pψn ´ ψ̃´n q |Dx y “ 0,
So with GSO projection we get:
ź ź
|N, NSNSy “ PL PR eψ´r ψ̃´r |0y , |N, RRy “ PL PR eψ´n ψ̃´n |0y
r n
ź ź
´ψ´r ψ̃´r
|D, NSNSy “ PL PR e |0y , |D, RRy “ PL PR e´ψ´n ψ̃´n |0y
r n

Here r runs over half-integers in the NSNS sector and n runs over integers in the RR sector. PL “ 21 p1 `
p´1qF q, PR “ 12 p1 ` p´1qF̃ q are our GSO projections, defined to project out the tachyon in the NS sector
and project out one of the spinors in the R sector.

86
For the boson, it is quick to see that (` “ 1{t)

V V
xN | e´π`pL0 `L̃0 ´c{12q |N y “ ? “ ?
2π`s 2ηpi`q p2π`s q 2tηpitq
´ ¯2
∆x
ż ´2πt 2π`
2π`s dk ik∆x ´π`2s p2 {2t e s
xD| e´π`pL0 `L̃0 ´c{12q |Dy “ ? e e “
2tηpitq 2π ηpitq
d d
1 1 ηpi`q ηpitq
xD| e´π`pL0 `L̃0 ´c{12q |N y “ ? ś 2n
“ “1‰ “ “0‰
2 n p1 ` q q θ 0 pi`q θ 1 pitq

These are exactly what we’ve already gotten many times before from our trace over the open string bosonic
states. The states |N y , |Dy must be a sum of both the RR and NSNS sector fermion states. We do not know
the relative coefficients.
Let’s look at the NSNS contributions. For the NN boundary conditions, the NSNS sector with projection
consists of two terms:
˜ “‰ ¸#N N {2 ˜ “ ‰ ¸#N N {2
´π`pL0 `L̃0 ´c{12q θ 00 pi`q θ 00 pitq
xN, NSNSunproj | e |N, NSNSunproj y “ “
ηpi`q ηpitq
˜ “‰ ¸#N N {2 ˜ “ ‰ ¸#N N {2
FL “FR ´π`pL0 `L̃0 ´c{12q θ 01 pi`q θ 10 pitq
xN, NSNSunproj | p´1q e |N, NSNSunproj y “ “
ηpi`q ηpitq

Replacing N with D would give the exact same factor in both cases WHY? (explain: bc we need to match
on both sides and so both minuses cancel in the exponent). For DN boundary conditions the NSNS sector
give the two terms:
˜ “‰ ¸ν{2 ˜ “ ‰ ¸ν{2
0 1
θ pi`q θ pitq
xD, NSNSunproj | e´π`pL0 `L̃0 ´c{12q |N, NSNSunproj y “ 1
“ 0
ηpi`q ηpitq
˜ “‰ ¸ν{2 ˜ “ ‰ ¸ν{2
FL “FR ´π`pL0 `L̃0 ´c{12q θ 00 pi`q θ 00 pitq
xD, NSNSunproj | p´1q e |N, NSNSunproj y “ “
ηpi`q ηpitq

Now let’s look at the RR sector. For NN boundary conditions, it contributes:

˜ “‰ ¸#N N {2 ˜ “ ‰ ¸#N N {2
1 0
θ pi`q θ pitq
xN, RRunproj | e´π`pL0 `L̃0 ´c{12q |N, RRunproj y “ 0
“ 1
ηpi`q ηpitq

xN, RRunproj | p´1qFL “FR e´π`pL0 `L̃0 ´c{12q |N, RRunproj y “ 0

By the argument before, we get the same for DD boundary conditions. Finally, with DN boundary conditions
we get

xD, RRunproj | e´π`pL0 `L̃0 ´c{12q |N, RRunproj y “ 0

˜ “‰ ¸ν{2 ˜ “ ‰ ¸ν{2
1 0
θ pi`q θ pitq
xD, RRunproj | p´1qFL “FR e´π`pL0 `L̃0 ´c{12q |N, RRunproj y “ 0
“ 1
ηpi`q ηpitq

Together this is exactly consistent with what we get from tracing over the open string. We can work back
to get relative normalizations.

87
 This shows that the massless RR and NSNS fields mediate the force. Moreover the NSNS fields without and
with projection correspond respectively to the unprojected NS and R open string states while the RR fields
without and with projection correspond to the projected NS and R open string states.
χ9
13. First recall that for a constant vector potential A9 “ 2πR corresponds to a T -dual picture of a D-brane
2
at position ´χR̃ “ ´2π`s A9 . Now consider a magnetic flux F12 we can write a (nonconstant now) vector
potential that gives this flux as A2 “ F12 X 1 . We T -dualize along X 2 and get X 2 “ ´2π`2s F12 X 1 . Then
tan θ “ ´2π`2s F12 .

Although we were working with D1 and D2 branes, we could have done the exact same calculation for F01
on a D1 brane and recovered a D0 brane tilted in the X 0 ´ X 1 plane (ie boosted). Such a D0 brane has the
usual point-particle action: ż b
SD0 “ ´T0 dX 0 1 ` pB0 X 1 1 q2

Because the D0 brane and the D1 brane describe the same physics, this action should be identical to the D1
action. Note that B0 X 1 1 is infinitesimally exactly tan θ calculated above. We get the action
ż a
SD1 “ ´T1 dX 1 dX 2 1 ` p2π`2s F12 q2

Of course, because the branes couple to strings, the only gauge invariant combination under transformations
of the Kalb-Ramond B field is F “ B ` 2π`2s F . We thus get:
ż a
SD1 “ ´T1 dX 1 dX 2 ´ detpG ` Fq

We can tilt this brane and T -dualize to pick up EM field strengths in arbitrary dimension up to 9.

14. Let’s T -dualize. This describes two D4 branes that are tilted only along the x1 -x5 plane, and are otherwise
parallel in the x2 , x3 , x4 directions. T-dualizing x2 , x3 , x4 makes these into D1 branes tilted in the x1 ´ x5
plane. See the solution (74) to the next problem. Now setting ν2,3,4 “ 0 will give poles from the theta

88
 function in the denominator. This is to be expected, from the NN boundary conditions that always come
with a volume divergence factor in that direction. We regulate this divergence by replacing:
„ 
1 L
θ piνt, itq´1 Ñ i ?
1 ´3
ηpitq 2π`s 2t
Thus we get a potential: ´ ¯2
∆x
´2πt 2π`
θ 11 piνt{2, itq4
“‰
L3
ż8
dt e s
´i
p2π`s q4 t p2tq4{2 ηpitq9 θ 11 piνt, itq
“‰
0

Let’s take the distance to be large. The small t contributions are then most important. The θ-function ratio
contributes a factor of e´3π{4t t´3{2 while the η ´9 contributes e3π{4t t9{2 Finish the details here

3 4
L `s sin pθ{2q
We get a potential that decays as ´ p∆xq 5 sinpθq giving another attractive force going as 1{p∆xq6 .

15. Following Polchinski, we define variables Z i “ X i ` iX i`4 , i “ 1, . . . , 4. Let the σ “ 0 endpoint be on the
untilted string. Then at σ “ 0 we have B1 <Z a “ =Z a “ 0 and at σ “ π on the tilted string we have
B1 <peiθa Z a q “ =peiθa Z a q.
We see that the field that satisfies this is given by Z a pw, w̄q “ Z a pwq ` Z̄ a p´w̄q for w “ σ 1 ` iσ 2 . Using
the doubling trick we have Z a pw ` 2πq “ e2iθa Z a pwq (and similarly for the conjugate). This gives a mode
expansion with νa “ θa {π
`s ÿ αra irw
Z a pwq “ i ? e .
2 rPZ`ν r a

The a:modes are then linearly independent. Taking q “ e´2πt as usual for open string partition functions,
and restricting 0 ă φa ă π we get:
1 1 1 2 2
q 24 ´ 2 pνa ´ 2 q q ´νa {2 ηpitq
ś8 “ ´i
θ 11 piνa t, itq
m`νa qp1 ´ q m`1´νa q
“‰
m“0 p1 ´ q

Where we have used

„  8 8
1 1
ν{2
ź
m m`ν m´ν 1 1
` ν2
ź
θ piνt, itq “ iq pq ´q
8
´ν{2
q p1´q qp1´q qp1´q q “ iηpitqq 8
´ 24
p1´q m`ν qp1´q m`1´ν q
1 m“1 m“0

So the angles act like chemical potentials to make the theta functions nonzero. Now its time for the fermions
(oh boy!). In each NS and R sector (projected and unprojected) the boundary conditions shift by νa . We
thus get e.g. for NS unprojected:
„  8 “0‰
0 1
`νa2 {2 νa2 {2 θ 0 piνa t, itq
ź
´ 24 m`1{2`νa m`1{2´νa
Z “q p1 ´ q qp1 ´ q q“q
0 m“1
ηpitq

In total, then we will see that the fermion part gives us

ś ν 2 {2 « ź 4 4 4 4
ff
4 2
eνa {2 θ 11 piνa1 t, itq
„  „  „  „  “‰
aq 0 1 0 1
a ź ź ź ź
θ piνa t, itq ´ θ piνa t, itq ´ θ piνa t, itq ´ θ piνa t, itq “
2ηpitq4 a“1 0 a“1
0 a“1
1 a“1
1 a“1
ηpitq

89
 This last equality follows from the full abstruse identity of Jacobi, where φ11 “ 21 pφ1 ` φ2 ` φ3 ` φ4 q,
φ12 “ 12 pφ1 ` φ2 ´ φ3 ´ φ4 q, φ13 “ 12 pφ1 ´ φ2 ` φ3 ´ φ4 q, φ14 “ 12 pφ1 ´ φ2 ´ φ3 ` φ4 q and the νa1 are defined
identically. Inserting the DD conditions along the 9 direction that denotes separation, we get the full potential
as a function of the separation ∆x
´ ¯2
∆x
´2πt 4 “1‰
piνa1 t, itq
ż8
dt e 2π`s ź θ
V “ ´2 ˆ ? “11‰ (74)
0 2t 2π`s 2t a“1 θ 1 piνa t, itq
The initial overall factor of two comes from two orientations of the open string. At long enough distances,
the exponential factor forces small t to contribute primarily. The complicated ratio of θ-functions becomes
a ratio of sines. Then we get
4 2
sinpπνa1 q 8
ż
ź dt ´ p∆xq t
? e 2π`2s
a“1
sinpπνa q 0 2π`s 2tt
|∆x| ś4 sinpπνa1 q
Taking the integral and analytically continuing, we get a potential that looks like ´ 2π`2 a“1 sinpπνa q , giving
s
1 ś4 sinpπνa1 q
an attractive, constant force, of ´ 2π`2 a“1 sinpπνa q .
s

16. Let the first brane at σ “ 0 have no electric field and put an electric field F01 on the second brane at σ “ π.
The endpoints of the string are charged. We have the following action (take e, the charge of the string
endpoint to be 1) ż ż
1 2 2
´ dσdτ rpBσ Xq ` pBτ Xq s ` dτ Aµ Bτ X µ
4π`2s σ“π
Upon variation, we get a boundary term:
ż ż
1
´ dτ Bσ X δXµ ` dτ δpAν Bτ X ν q
µ
2π`2s
ż ż
1 µ
“´ dτ Bσ X δX µ ` dτ Bµ Aν δX µ Bτ X ν ´ Bτ Aν
2π`2s
ż ż
1
“´ dτ Bσ Xµ δX ` dτ Fµν δX µ Bτ X ν
µ
2π`2s
ñ Bσ Xµ ´ 2π`2s Fµν Bτ X ν “ 0

This gives mixed boundary conditions on the X 0 and X 1 which can be written as
ˆ ˙ ˆ ˙ˆ ˙
Bσ X 0 2 0 1 Bτ X 0
“ 2π`s E
Bσ X 1 1 0 Bτ X 1

with E “ F10 . Note that we have been careful in raising the µ index. I will define Z ˘ “ X 0 ˘ X 1 and have
Bσ Z ` “ Bσ Z ´ “ 0 at σ “ 0 and pBσ ´ 2π`2s EBτ qZ ` “ pBσ ` 2π`2s EBτ qZ ´ “ 0 at σ “ π. The modes thus
satisfy Neumann-Mixed boundary conditions. Following a modification of exercise 2.14 and solving these
boundary conditions we get that the modes must be labeled by ν “ ´iu{π ` Z. Here u “ atanhv is the
rapidity. Now let’s compute a cylinder diagram. Let’s assume for now that we are scattering D1 branes (the
problem does not explicitly give p, p1 ). It will look like the wick-rotation of the integral considered in the
previous two questions. We have an amplitude
2
´t p∆xq2
θ 11 put{2π, itq4
ż8 “‰
dt e 2π`s
´iVp ˆ 2 ˆ
2t p2π`s qp p2tqp{2 ηpitq9 θ 11 put{π, itq
“‰
0

Note however that since the first argument of the θ functions is real, we have poles at t “ πn{u, ν “ u{π for
1
n an integer. Upon deforming the integration contour, we can use the identity x´i “ πδpxq ` Ppxq to pick
up poles at t “ n{ν, at odd integers n (so that the four-order zero in the numerator doesn’t cancel them),
giving:
2
´n p∆xq
θ 11 p n2 , itq4
“‰
ÿ e 2π`2s ν
πVp (75)
odd
p2π`s qp p2n{νqpp`2q{2 2ηpin{νq12
nPZ

90
 The imaginary part of the amplitude can be interpreted (after T -duality) as resonances (ie bound states) of
the D-branes.
17. From the last problem we can write: In terms of B` , B´ , we have:
ˆ ˙ ˜ 1`E 2 2E
¸ˆ ˙
B` X 0 1´E 2 1´E 2 B´ X 0
“ 1`E 2 (76)
B` X 1 2E
2 2
B´ X 1
1´E 1´E

Here E “ 2π`2s E. Taking E close to zero recovers NN boundary conditions on X 0 , X 1 .

(It’s worth noting that that the speed of light here will translate to a maximum electric field |E| ă T on the
brane. This provides one motivation for the necessity of a nonlinear electrodynamics, namely DBI). taking
E “ 0 give N boundary conditions on X 0 , X 1 . T-dualizing X 1 gives D boundary conditions on X̃ 1 . Now,
boosting the brane along X 0 and
B` X 0 B´ X 0
ˆ ˙ ˆ ˙ˆ ˙ˆ ˙ˆ ˙
γ vγ 1 0 γ ´vγ
“
B` X̃ 1 vγ γ 0 ´1 ´vγ γ B´ X̃ 1
˙ ˜ 1`v2 2v
¸ˆ
B´ X 0 B´ X 0
ˆ ˙ ˆ ˙ˆ ˙ˆ ˙ˆ ˙ˆ ˙
B` X 0 γ vγ 1 0 γ ´vγ 1 0 1´v 2 1´v 2
ñ “ “ 1`v 2
B` X 1 vγ γ 0 ´1 ´vγ γ 0 ´1 B´ X̃ 1 2v B´ X̃ 1
1´v 2 1´v 2

This exactly what we had before, with v “ 2π`s E.

This argument is quite simple from the abstract picture: taking A1 “ EX 0 , T -dualizing in the direction of
X 1 gives a D-brane lying at X1 “ 2π`2s EX 0 giving a velocity 2π`2s E. This can also be obtained by analytic
continuation of question 8.13.
18. Using the fact that this scattering problem is exactly T-dual to the electric field problem mentioned before,
we return to (75) and consider b “ ∆x to be small. In this case the large t regime dominates (corresponding
to a loop of light open strings). First we perform a modular transformation to get
θ 11 piν{2, i{tq4
“‰
tb2
ż8
Vp dt p6´pq{2 ´ 2π` 2
A“ t e s
ηpi{tq9 θ 11 piν, i{tq
“‰
p8π 2 `2s qp{2 0 t

Now weşfollow Polchinski and rewrite A in terms of an integral over the particle’s path rpτ q2 “ b2 ` v 2 τ 2 ,
A “ ´i dτ V prpτ q, vq. Then we get V from reversing a Gaussian integral to be
“1‰ 4
tr 2
ż8
2Vp p5´pq{2 ´ 2π`2s tanh πν θ 1 piν{2, i{tq
V pr, vq “ i dtt e “1
‰
p8π 2 `2s qpp`1q{2 0 ηpi{tq9 θ 1 piν, i{tq
The large-t limit is now direct:
tr 2
tanh u sin4 put{2q
ż8
2Vp`1 dt ´
2π`2
V pr, vq “ ´ e s
p8π `2s qpp`1q{2
2
0 t p1`pq{2 sinputq
Using steepest descent at zeroth order, the t that dominates is of order 2π`2s {r2 so that ut « 2π`2s v{r2 is
?
the leading contribution. Justify why ut is Op1q. This then gives that r „ `s v. If we go at very small
velocities we can probe below the string scale.

91
 On the other hand, a slower velocity means that the time it takes to probe this distance is longer δt “ r{v “
?
`s { v. This implies that δxδt ě `2s . This looks like a type of noncommutative geometry with α1 playing the
role of Planck’s constant now.
Combining this with the usual position-momentum uncertainty relation
g`s
1 ď δx mδv “ g`s δxδv ñ ∆x “ ,
δv
?
we can minimize simultaneously `s v, g`s {v by having v „ g 2{3 giving δx “ `s g 1{3 . At weak coupling this is
smaller than the string scale.
2 9i
19. The action is a factor of two off from Polchinski’s. The momenta are pi “ gs `s pX ` rAt , X i sq. Then we get:
ż ”g `
s s 1 ı
H“ dtTr pi pi ` rX i
, X j 2
s
4 2gs `s p2π`2s q2
1{3
Defining g 1{3 `s Y i “ X i , pi “ pYi {gs `s , this sets the length scale, which coincides with what we got in the
previous question by less rigorous arguments. Now we have Y i is dimensionless and get a hamiltonian
1{3 ż
gs ”1 1 i j 2
ı
H“ dtTr pYi pYi ` rY , Y s
`s 4 2p2πq2

So the only dimensionful content of this hamiltonian comes from g, `s appearing in the overall normalization.
1{3
This gives an energy scale of gs {`s . For strong coupling gs ą 1 this probes deeper than the string scale.

20. This is pretty direct. Since the metric Gµν does not depend on X i for i “ p ` 1 . . . 9, we have that each
X i is killing, in particular the metric takes a block-diagonal form where only the first pp ` 1q2 Gµν entries
have nontrivial coordinate dependence and the remaining metric is just the identity matrix δij along the Xi
directions (we didn’t even have to do this since 8.5.1 has ηµν the flat metric. Is my logic here even right?).
Take the ansatz A Ñ pAµ , Φi q. We thus get Fµν in the first pp ` 1q2 entries and Bµ Φi in the off-block-diagonal
piece. We can rewrite this as a determinant of just the pp ` 1q piece Justify this step
b
´ detpGµν ` 2πFµν ` Bµ Φi Bν Φi q

21. The bosonic part of this is immediate. Write the fields Ai řin the dimensionally reduced dimensions as X I
and we immediately get TrF10 Ñ TrrFd`1 ` 2rDµ , X s ` I,J rX I , X J s2 s. The fermionic part will reduce
2 2 I 2

to:
pTrχ̄Γµ Dµ χq10D Ñ Trrχ̄Γµ Dµ χ ` λ̄a Γi rXi , λb ss
Where now the χi are fermions that break the 16 representation of SOp9, 1q into a representation of SOpd ´
1, 1q ˆ SOp10 ´ dq. For d “ 3 we get N “ 4 SYM and this is p2, 4q ` p2̄, 4̄q, corresponding to four Weyl
spinors.

92
22. At the minimum of the potential, all X I lie in a cartan and mutually commute. The Aij correspond to
open strings moving between the D-branes at positions XI . The ground states of these open strings have a
mass squared of pXI ´ XJ q2 {2π`2s , so indeed the mass is linear in the separation. Confirm. Understand
Lie-Theoretic perspective.
ş
23. The worldvolume coupling to the RR 2-form looks like iT2 C2 . For the brane tilted in the x1 , x2 plane we
can write this explicitly as: ż
i dx0 dx1 pC01 ` C02 tanpθqq
ş
Now T-dualizing in the x2 direction changes the C01 form to the RRş 3-form C3 , giving the standard i C3
term. On the other hand, the second term gets reduced to ´2π`2s dx0 dx1 dx2 C0 F12 , where I have used
exercise 8.13 to write tan θ “ ´2π`2s F12 . So we get a leading coupling to the three-form and a sub-leading
coupling (in `2s ) to the one-form. This is a hint of the Meyers effect.
4
24. For the D2 brane the CP odd terms are C3 , C1 ^ F and ´C1 ^ p2π` sq
48 rp1 pT q ´ p1 pN qs. In the frame described
by exercise 7.26, C3 transforms trivially under A transformations, and under its own gauge transformations
it only adds an exact term which does not modify the CS action.
F transforms trivially under A transformations and under B transformations only modifies the action by a
closed term again.
Equation I.14 is not in any standard frame. The Dilaton plays no role here. The
I feel I am missing something.

25. Gauge transformations of the axion C0 are just shifts C0 Ñ C0 ` a. C0 couples to F2 through the Chern-
Simons term: ż
C0 ^ TreF ^ G

Because of the Bianchi identity, dF “ 0, and the same holds for any trace of any polynomial of F . Similarly
G is also a closed form. Therefore shifting C0 gives an integrand term TreF ^ G which is closed.

26. For trivial flat-space background ηµν , we have gab “ Ba Xµ Bb X µ . Take Mab “ Ba Xµ Bb X µ ` 2π`2s Fab and
M “ det Mab . Taking the DBI variation w.r.t. Xµa and Aa respectively gives:

T a ´? ¯
B ´M Mab´1
Bb X µ “ 0
2
2π`2s T a ´? ´1
¯
B ´M Mab “0
2
Its rather nasty to evaluate that inverse matrix. On the other hand, taking X 9 to be the only nontrivial
function of the ξ, and depending only on the radial distance r from a central point, and setting all Ai “ 0
with A0 a function of r alone, we get M “ 1 ` δa“r,b“r pBr X 9 q2 ` 2π`2s pδa“r,b“0 ´ δa“0,b“r qE. We take
E “ Br X 9 . The determinant is then p∇X 9 q2 p1 ` 2π`2s q.
Note that if the second equation of motion holds, the first equation of motion implies that we would want for
Cp
B r Br X 9 “ 0, namely that X 9 is a harmonic function of r. On a p brane this is X 9 “ rp´2 . In this case, the
determinant, as well as M ´1 will vanish when covariantly differentiated by B r , giving us our desired second
equation of motion.
This solution is known as a BI-on (BI for Born-Infeld). It represents an infinitely long open string ending
on our p-brane.

27. Let’s have G, B, Φ, C2 trivial. We get

ż ż
1 2
a 1
S“ d ξ 1 ´ p2π`2 F q2 `
s 01 d2 ξ C0 p2π`2s qF01
2π`2s g 2π`2s

93
 Pick the gauge A0 “ 0. Our variable is then A1 . From this, we get a canonical momentum conjugate to A1
equal to:
1 p2π`2s q2 F01
´ a ` C0 F01
2π`2s g 1 ´ p2π`2s F01 q2
Consider putting the D1 brane in a circle. Now since C0 acts as a θ term, consider putting an integer m for
C0 . The momentum is quantized, and in particular there is a gap between the zero momentum ground state
and the next state up. We get:

2π`2s F01 gm
a “ gm ñ 2π`2s F01 “ a
2
1 ´ p2π`s F01 q2 1 ` m2 g 2

We have a Hamiltonian
1 1
H“ a
2π`2s g 1 ´ p2π`2s F01 q2
And from the quantization condition on the electric field we obtain from the Hamiltonian a set of quantized
tensions
1 a
T “ 1 ` m2 g 2 .
2π`2s g

28. The D3 -D´1 system has #ND“ 4 and so preserves 1{4 supersymmetry (1{2 the SUSY of the D3 brane
itself). Similarly, the instanton configurations satisfying ‹F2 “ ˘F2 . The supersymmetric variation of the
gaugino is δλ9Fµν Γµν . The Γµν are generators of SOp4q “ SUp2q ˆ SUp2q, and the (A)SD conditions on
F2 will ensure that only half the generators (the first or second SUp2q) will appear in the variation. Thus
instanton configurations are also 1/2 BPS on the worldvolume.
CS term contains 12 p2π`2s q2 T3 C0 F2 ^
ş
To confirm that these instantons really are D´1 branes,ş note that the
F2 . For a nontrivial instanton configuration we get F2 ^ F2 “ 8π 2 . Thus the instanton coupling to C0 is
p2π`s q4 T3 “ T´1 , exactly the charge of the D´1 brane. Does this exclude the possibility of objects
with the same charge and BPS properties as D´1 branes, but that don’t have interpretations
as endpoints of open strings?
I expect the moduli space to have dimension 4n, corresponding to the space (technically Hilbert scheme) of
n points on R4 .

29. This configuration is invariant under x1 translations as well as under time x0 . The exact same BPS properties
discussed in the previous question apply here. The state is half-BPS on the worldvolume both from the POV
of string theory and from the POV of the low energy SYM theory having half the gaugino variations vanish.
The same instanton action argument in the previous question gives us that 12 p2π`2s q2 T4 C1 F2 ^ F2 yields a
ş

coupling p2π`s q4 T4 “ T0 to the C1 form.

For N D5 branes the low-energy effective theory is SUpN q SYM, and we obtain the moduli space of SU pN q
instantons. The dimension now becomes 4N k justify . I expect that the moduli spaces of D1-D5 bound
states are identical to the moduli space in the previous problem.

94
30. First, the pair of 5-branes in the 12345 and 12367 dimensions are parallel in the 123 directions and 900
rotated in two directions. This gives 2 sets of N D and 2 sets of DN boundary conditions, on the strings
´1 1
which gives us ν “ 4. In this case, following Polchinski the spinor β “ β K β K has an equal number of
eigenvalues ´1 and 1. So half of the original 16 spinors preserved by the first D-brane will be preserved by
the combination of both.
Now take a third D-brane in the 12389 direction, perpendicular to both the first two. The same argument
shows that we brane another half of the supersymmetry, giving 4 supersymmetries left in this configuration.
In other words it is 81 BPS. Confirm

31. Adding a D1 string gives 4 ND boundary conditions with each of the other D-branes. This breaks the
1
supersymmetry in half again, preserving 2 supersymmetries now. It is 16 BPS.
If I were to add it along direction 4 it would have 5 ND boundary conditions with the second two branes,
which preserves no SUSY, so the latter configuration has nothing preserved.

32. Note this is Op2q and not SOp2q, so instead of getting one D-brane at θR̃ and the other at ´θR̃, we get one
“half” D8 brane at x9 “ 0 (the location of one orientifold plane) and its image at π R̃ (the location of the
other).

33. We work with the compact real form USpp2N q “ Spp2N q X Up2N q. In this case any symplectic matrix can
again be diagonalized to be of the form peiθ1 , e´iθ1 , . . . q. Again we interpret this as D-branes on both sides
˘θi R̃ of the orientifold plane. The generic gauge group is U p1q2N . If m branes lie at either orientifold plane
we get an enhancement Spp2mq. When all N branes and their images lie on one of the orientifold planes, we
recover the full symmetry.

34. Due to the negative tension, an excitation on it has even lower energy, corresponding to a negative norma
state which is forbidden in a unitary theory by positivity. What more can I say?

35. There is a mistake in Kiritsis’ equation G.8. We should have A “ ´H ´1 pρq not ´Hpρq. The way to see
ρ
that is by noting that ´H ´1 pρq “ ´ ρ`Q “ ´ r´Q Q
r “ 1 ` r . The constant 1 is gauge and hence irrelevant,
Q
while the second term is the proper electric potential that will give rise to a Ftr “ r2
.
Also, this problem asks us to work in N “ 2, D “ 4 SUGRA, so the appropriate equations should be that
the variation of each dilatino by a Killing spinor is zero. In this SUGRA, there are two Majorana gravitinos
ψµ,A , A “ 1, 2 with four components each, for a total of 8 SUSYs. Consequently, the variations involve two
Majorana spinor parameters A , A “ 1, 2. We will use lower indices to indicate chiral and upper indices to
indicate anti-chiral fermions. The gravitino variation is then (c.f. Freedman Supergravity Section 22.4)
1
δψµ,A “ ∇µ A ´ Fνρ γ νρ γµ εAB B (77)
4
Here we have ∇µ “ Bµ ` 41 ωµab Γab with ω spin connection 3

Because of the chirality we can replace F with F ´ in the above equation. Now, if we use spatial coordinates
~x, |x| “ ρ, the metric takes the form
ds2 “ ´H ´2 dt ` H 2 d~x2
Take e2A “ H ´2 , then we have the frame fields e0̂ “ eA dt, eî “ e´A dxi . We will use hats to denote frame
indices a, b. Our spin connection is then:

ω0̂î “ ´e2A Bi A dt, ωîĵ “ ´Bj Adxi ` Bi Adxj

First let’s look at the µ “ 0 constraint of equation (77)

1 1
Bt A ` ω0 ab Γab A ´ Fνρ γ νρ γ0 εAB B
4 4
3
?
This corresponds to setting κ “ 2 in Friedman’s Supergravity 22.69.

95
 Now because the solution is static, Bt is killing and we expect that Bt A “ 0. Further, the only contribution
to ω0 ab is ω0 0̂î since only this has a dt (NB the double sum gives a factor of 2). Similarly for the second
term, since there is only an electric field, we only care about ν, ρ P t0, iu (NB the double gives a factor of 2).
Finally, we have only an electric field F0i “ ´Bi At (At is the vector potential, not to be confused with A).
This yields:
1 1
´ e2A Bi A γ 0̂ γ î A ´ p´Bi At qγ 0 γ i γ0 εAB B “ 0
2 2
1 A A î 0̂ 1
ñ ´ e Bi e γ γ A ` p´Bi At qγ i εAB B “ 0
2 2
ñ eA Bi eA γ î γ 0̂ A ´ Bi At eA γ î εAB B “ 0
ñ Bi eA γ 0̂ A “ Bi A0 εAB B
Here, the hatted γ-matrices are the familiar ones from flat space. We thus need (up to an irrelevant gauge
constant) A0 “ ˘H ´1 and
A “ ¯γ 0̂ εAB B (78)
Since we require ´H ´1 to match the electromagnetic potential A, and so that it is asymptotically unity, we
have H “ p1 ´ Qr q´1 “ ρ`Q Q
ρ “ 1 ` ρ . This verifies the extremal RN solution.
We have not yet derived the spatial dependence of . Taking µ “ i we get
1 1
Bi A ` ωi ab Γab A ´ Fνρ γ νρ γi εAB B
4 4
Now we must use that ωi ĵ k̂ “ ´Bn Apδij δkn ´ δik δjn q. Using the µ “ 0 constraint we get:

1 1
Bi A ´ Bk A γi k̂ A ´ F0i γ 0
γ i
γi εAB B “ 0
2 2
1 1
Bi A ` Bk A e´A γ i k̂ A ¯ Bi eA Hγ 0̂ εAB B “ 0
2 2
1 î k̂ 1 A ´A
Bi A ` Bk A γ A ´ Bi e e A “ 0
2 2

Now the γ îk̂ is nothing more than a generator of rotations acting on A . Since we are assuming spherical
symmetry, γ îk̂ A “ 0 and we are left with the differential equation:
1
Bi A “ Bi AA ñ A “ e1{2A 0
2
where 0 is a constant spinor satisfying (78).
Because the constraint A “ ¯γ 0̂ εAB B applies to half the space of spinors at any given point, we have that
the extremal RN solution is half-BPS.

36. Again take coordinates xi so that

dt2
ds2 “ ´ ` H 2 pρqpdx2i q
H 2 pρq
Upon the choice A “ ´ ¯ γ 0̂ εAB B , we had the relation Bi H ´1 “ Bi A0 “ Fi0 . The field equation B ‹ F “ 0
?
Bi ´gg 00 g ii Fi0 “ Bi H 2 Bi H ´1 “ Bi2 Hpxi q

Thus we have that H is a harmonic function of the flat Laplacian in transverse space.
Q
We see that H from the previous problem takes the simple form 1 ` |x| , which is obviously harmonic in 3+1
dimensions.
A more general solution
ř allowing for for multiple charged extremal black holes amounts to nothing more than
Qi
replacing H with 1 ` i |x´x i|
, which remains harmonic, and thus preserves half supersymmetry. This looks
like a bunch of extremal black holes whose pairwise electric repulsions cancel their gravitational attractions.

96
37. Again we are working in N “ 2 SUGRA. The metric takes the form
2 2
„ 
2 2 ´dt ` dρ 2
ds “ Q ` dΩ2
ρ2

This corresponds to 2D AdS times a sphere of constant radius. Both spaces have radius Q. The spinor
equation is as before, now with Tµν “ ´ L1 gµν . Consider just AdS2 . Define the operator

1
D̂µ A “ ∇µ A ´ γµ εAB B
2Q
Consider now
1 1
rD̂µ , D̂ν sA “ p Rµνab γ ab ` q
4 2Q2
And since AdS2 is a maximally symmetric space Rµνab “ ´peaµ ebν ´ eaν ebµ q{L2 and so the commutator
vanishes identically. This is the integrability condition we need. At each point, the spinor bundle is dimension
N ˆ 2rD{2s - for N “ 2 AdS2 this is 4. We see that any spinor can be transported by the connection D̂µ to
define a spinor field on all of AdS2 , and thus we get that the space is maximally supersymmetric.
The exact same arguments (with Q Ñ ´Q) apply for the positively curved 2-sphere of the same radius.
The product of two maximally supersymmetric spaces is maximally supersymmetric Confirm. We get
8 Killing spinors. We now see that the Bertotti-Robertson universe preserves full supersymmetry, and
thus the extremal RN black hole plays the role of a half-BPS soliton that interpolates between two fully
supersymmetric backgrounds (flat space and AdS2 ˆ S 2 ).
1 2
38. The variation of 2pp`2q! Fp`2 “ Fp`2 ^ ‹Fp`2 directly gives ‹Fp`2 “ 0.
Varying the dilaton gives

0 “ ´2e´2Φ rR ` 4p∇Φq2 s ´ ∇pe´2Φ 8∇Φq “ ´2e´2Φ rR ` 4p∇Φq2 s ` 16e´2Φ p∇Φq2 ´ 8e´2Φ lΦ

ñ R “ 4p∇Φq2 ´ 4lΦ

Finally, writing the metric explicitly in the action

? 1 ?
´ge´2Φ rg µν Rµν ` 4g µν Bµ ΦBν Φs ´ 2
´gFp`2
2pp ` 2q!

Let’s look how each term changes when we vary ?1 δ

.
´g δg µν
?
• ´ge´2Φ R

1
Ñ pRµν ` gµν l ´ ∇µ ∇ν qe´2Φ ´ gµν e´2Φ R
2
´ 1 ¯
“ e´2Φ Rµν ´ gµν R ` gµν p´2lΦ ` 4pBΦq2 q ´ p´2∇µ ∇ν Φ ` 4Bµ ΦBν Φq
2
?
• ´ge´2Φ 4g µν Bµ ΦBν Φ Ñ 4e´2Φ Bµ ΦBν Φ ´ 2e´2Φ pBΦq2
1 ? 1 1
• ´ 2pp`2q! ´gg µ1 ν1 . . . g µp`2 νp`2 Fµ1 ...µp`2 Fν1 ...νp`2 Ñ ´ 2pp`1q! 2 `
Fµν 2
4pp`2q! gµν F .
2 “F
Here Fµν µ... F
ν...

Combining these all together and using the dilaton equations of motion gives

1 2 1 e2Φ ´ 2 gµν ¯
e´2Φ Rµν ` 2∇µ ∇ν Φ ´ Fµν ` gµν F 2 “ 0 ñ Rµν ` 2∇µ ∇ν Φ “ Fµν ´ F2
2pp ` 1q! 4pp ` 2q! 2pp ` 1q! 2p ` 2

exactly as required.

97
39. This problem was very time-consuming to do out explicitly. The only resource that was of any help for
cross-checking was Ort́in’s “Gravity and Strings”.
First consider the possibility of Φ “ 0, F “ 0. In this case we have no stress tensor and are left with static
vacuum Einstein equations, spherically symmetric in 10 ´ p dimensions and translationally invariant in p
dimensions. In that case we know that our solution is nothing more than the Schwarzschild solution in 10 ´ p
dimensions times a transverse p-dimensional space:
1
ds2 “ ´f prqdt2 ` dx2i ` dr2 ` r2 dΩ28´p
f prq

Reproducing the arguments from black holes in 4D, we see that

d 8´p d d
pr f prq logpf qq “ pr8´p f 1 prqq “ 0
dr dr dr
ie f prq must be harmonic in the transverse dimension. After rescaling coordinates to have appropriate
asymptotic behavior, we get:
r07´p
f prq “ 1 ´ 7´p
r
for some constant r0 related to the ADM mass of the solution. So we see that Hprq “ 1 when the dilaton
and p ` 2-form field strength are turned off. The curvature is R “ 0. Now let’s turn on Φ. We expect small
Φ will correspond to small H.
To do this, let’s look at the simplest case, p “ 0. Take the Ansatz (which you can convince yourself to be
completely general given the symmetry of the problem)

dr2
ds2 “ ´λprqdt2 ` ` Rprq2 dΩ2d´2
µprq
?
We will later set d “ 10, λ “ µ “ f prq{ H. Let’s explicitly calculate the Christoffel symbols. There are
three categories: Γs involving just r, t Γs involving mixed r, Ω, and Γs involving just the Ω variables. I will
use a, b, c to index the angular variables ψa , whose metric is R2 dΩ2ab “ qa δab , and I use 1 to denote ordinary
partial differentiation w.r.t. r.

Γrtt “ 12 µλ1 Γttr “ 12 λ´1 λ1 Γrrr “ ´ 12 µ´1 µ1

1
Γarb “ δba RR Γrab “ δab µpR2 q1 Rqa2
Γabc “ θbąc δba cot ψb ` δcąb θac cot ψc ´ θbąa δbc cot ψa qqab

That last Christoffel symbol looks particularly nasty. Thankfully, by using the fact that the sphere is a
symmetric space, we will not need to use it explicitly.
Now let’s directly compute the Ricci tensor.

Rµν “ Bρ Γρµν ´ Bµ Γρρν ` Γρρσ Γσµν ´ Γσµρ Γρσν

?
In what follows, it is useful to recall the identity Γµµν “ Bν log ´g. The nonzero terms will be Rtt , Rrr , Rab .
Respectively they are:
ρ σ ρ
Rtt “ Br Γrtt ´ 
Bt Γ
 ρ r
ρt ` Γρr Γtt ´ Γtρ Γσt
1 ? 1 λ1 1 1
“ pµλ1 q1 ` Br log g pµλ1 q ´ 2 µλ
2 2λ2
1 λ ´? λ 1 ¯ 1
“ ? Br ´gµ “ λ∇2 log λ
2 g λ 2

It’s important for this next one to note that qa {R2 is independent of r. It’s equally important to appreciate
that the final combination of Γ symbols is the only thing that would appear in the absence of r dependence

98
 d´3
in R. In this case, because the d ´ 2 sphere is a symmetric space, we’d have Rab “ g .
R2 ab
Indeed, this is
exactly what the final term gives. We thus get
ρ σ ρ
Rab “ Br Γrab ´  Γ
Ba ρ r
ρb ` Γρr Γab ´ Γaρ Γσb
1 ´
2 1 qa
¯ ? qa d´3
“ ´ δab Br µpR q 2 ` Br log ´g δab µpR2 q1 2 ` qa δab
2 R R R2
d´3
“ gab p´∇2 log R ` q
R2
The next onea
is a bit different. Less cancelation. The last term will sum over pσ, ρq “ pr, rq, pt, tq, pa, aq. Also
?
note ´g “ λ{µRd´2 .

Rrr “ Br Γrrr ´ Br Γρρr ` Γρρσ Γσrr ´ Γσrρ Γρσr

1 ´1 a 1 a 1 pλ1 q2 1 2 1 2 pR1 q2
q ´ Br2 logp λ{ µRd´2 q ´ Br logp λ{µRd´2 q pµ´1 µ1 q ´
µ 11
“´ pµ ´ µ pµ q ´ pd ´ 2q
2 2 4 λ2 4 R2
1 d´2 2 1 R 1 a
“ ´ Br2 logpλq ´ R ´ pd ´ 2q Br log λ{µ
2 R d ˆ c R˙
2
1
1 ´1 2 d´2 λ 1 µ
“ ´ µ ∇ log λ ´ R
2 R µ λ

Altogether we get a Ricci scalar:

ˆ c ˙1
2 d´2 pd ´ 2qpd ´ 3q d ´ 2 a µ
R “ ´∇ logpλR q` 2
´ λµ R1
R R λ
? ?
Now let’s take λ “ µ “ f prq{ H and R “ H 1{4 , ´g “ H pd´2q{4 .

d´4 pd ´ 2qpd ´ 3q f prq

´∇2 logpf prqrd´2 H 4 q` ´ pd ´ 2q prH 1{4 q2
r2 H 1{2 rH 3{4
Br rH pd´4q{4 r8 f prqBr s
The Laplacian takes the form rd´2 H pd´2q{2
which simplifies the above to:

Br2 pf prqrd´2 H pd´4q{4 q pd ´ 2qpd ´ 3q

´ `
rd´2 H pd´2q{2 rd´2 H 1{2
The dilaton equation of motion is
R “ 4p∇Φq2 ´ 4∇2 Φ
Since a nonzero Φ is what gives an H away from 1, we might hypothesize a relationship log H9Φ, meaning
we should replace Φ with logpH α q in the dilaton equation. Let’s also take f prq to be not different from the
rd´3
Schwarzschild solution: f prq “ 1 ´ r0d´3 . So far we will not be so bold as to assume anything about H.
We also at this point need to specialize to d “ 10, otherwise no nice simplification occurs. Straightforward
algebra then gives:

99
To get rid of the term quadratic in H 1 prq we need 3 ` 8p1 ´ 2αqα “ 0 ñ α “ 3{4.
After that, these will only be equal if

L7
8H 1 ´ rH 2 “ 0 ñ H “ 1 ´
r7
The above solution is the most general given that H Ñ 1 as r Ñ 8.
Now let us generalize this to higher dimensions. We add p xi in the parallel dimensions of the solution.

dr2
ds2 “ ´λprqdt2 ` νprqd~x2i ` ` Rprq2 dΩ2d´2
µprq

This gives two new Christoffels. Here is a complete list

Γrtt “ 21 µλ1 Γttr “ 12 λ´1 λ1 Γrrr “ ´ 12 µ´1 µ1

1
Γarb “ δba RR Γrab “ δab µpR2 q1 Rqa2
Γabc “ θbąc δba cot ψb ` δcąb θac cot ψc ´ θbąa δbc cot ψa qqab
Γrij “ ´ 12 δij µν 1 Γirj “ 12 δji ν ´1 ν 1

Our nonzero Ricci components are now Rtt , Rrr , Rab , Rij . The primary way that the new dimensions will
?
contribute is by modifying ´g We get:

p10´pq 1
Rtt “ Rtt ´ pµλ1 plog νq1
4
p10´pq 1
Rab “ Rab ´ pgab µplog νq1 plog Rq1
2
1
Rrr p10´pq
“ Rrr ` ppµνq´1{2 ppµνq1{2 plog νq1 q1
2
1
Rij “ δij ν∇2 ν
2
This gives a Ricci curvature of:
1 1
R “ Rp10´pq ` pp´∇210´p log ν ´ ν ´1 ∇2 ν ` µpplog νq1 q2 q
2 2
Making the necessary replacements we get

Br2 pf prqrd´2 H pd´4q{4 q pd ´ 2qpd ´ 3q

´ `
˜r H
d´2 pd´2q{2 rd´2 H 1{2
8´p p6´pq{4
? ¸
1 Br rf r H Br log H ´1{2 s H 8´p p6´2pq{4 ´1{2 1 f ´1{2 2
` p ´ ´ 8´p p8´2pq{4 Br rf r H Br H s` pBr log H q
2 r8´p H p8´pq{4 r H 2 H 1{2

r7´p 7´p
It makes sense to take the ansatz f prq “ 1 ´ r07´p and H “ 1 ` Lr7´p . Further, the relationship between H
and Φ can be guessed from reasoning in the p “ 0 case to go as Φ9H p3´pq{4 , or alternatively we can establish
this from first principles by algebra

100
This immediately gives that the dilaton term will equal the scalar curvature only when α “ p3 ´ pq{4.
We have thus proved the form of f, H, Φ. Let’s finally look at the RR field.
For now let us ignore the issues with self-duality at p “ 3. Take the the p ` 1 form has flux in the radial
but not angular directions in transverse space. The only nonzero component of Fp`2 is given by Fr0...p . The
equation of motion gives: `?
´gg µ0 ν0 g ... Fµ0 ... “ 0
˘
Br
?
Now we already have g “ r8´p H p4´pq{2 , while we will have raising for each index 0 . . . p as well as r, giving
a factor of f prqH pp`1q{2 H ´1{2 {f prq “ H p{2 . Altogether the differential equation becomes:

Br r8´p H 2 H r
κ{r8´p
Immediately we must have F “ H2
. This means that F is proportional to H 1 prq{Hprq2 .
I don’t know how to easily get this constant of proportionality without knowing the decay properties of Rµν
as r Ñ 8 Return to this. I know it must scale roughly as a positive power of L. It turns out to be:
d
r7´p Hp1 prq
Fr0...p “ ´ 1 ´ 07´p 2
L Hp prq

101
 Can I do this all by somehow “boosting” Schwarzschild?

40. From the expression 8.8.9 of the electric field in terms of H we have (assuming p ă 7)

r6`p
b b
Er “ p7 ´ pqL p7´pq{2
L7´p ` r07´p Ñ p7 ´ pqLp7´pq{2 L7´p ` r07´p rp´8
pr7 ` Lrp q2

Integrating this over the 8 ´ p sphere will give

b
Ω8´p p7 ´ pq p7´pq{2
Tp N “ L L7´p ` r07´p .
2κ210

41. Using the standard ADM formula (cf, eg Carrol)

ż
1
M“ 2 g µν pgµα,ν ´ gµν,α qnα dS
2κ10 S 8 ´p
cµν
Then for a metric that looks like gµν “ ηµν ` hµν with hµν “ r7´p
.

Ω8´p
M “ T00 Vp “ pp7 ´ pqc00 ´ η00 η ab cab q
2κ210
Ω8´p Vp 1 1
“ 2 pp7 ´ pqpr07´p ` L7´p q ` r07´p ` Lp7´pq{4 q
2κ10 2 2
Ω8´p Vp
“ 2 pp8 ´ pqr07´p ` p7 ´ pqL7´p q
2κ10

Revisit- something seems off

42. Note that

L7´p 1 1
f´ pρq “ 1 ´ 7´p 7´p
“ 7´p 7´p
“
r `L 1`L r Hprq
Similarly
r07´p ` L7´p r7´p ` r07´p f prq
f` pρq “ 1 ´ 7´p 7´p
“ 7´p 7´p
“
r `L r `L Hprq

102
 ´1{2
This confirms that the
? dt and d~x terms are indeed consistent, and that the f´ pρq in front of the transverse
part is our desired H. Next note that:
1{p7´pq
ρ7´p ´ L7´p r2
„
1 r 1´ 5´p
f´ pρq 7´p “ “ ñ ρ2 f´ pρq 7´p “ ρ2 2 “ r2
ρ7´p ρ ρ

So the angular part is consistent. Lastly, f` {f´ “ f prq which is the required coefficient for dr2 . It remains
to cancel the jacobian:

ρ6´p ´ 6´p 2 ´ 12´2p 2

5´p
´1´ 7´p
dr “ dρ “ f´
7´p
dρ ñ dr “ f´
7´p
dρ “ f´ dρ2
r6´p
So a
Hprq 2 5´p
´1{2 f´ ´1´ 7´p ´ 1 ´ 5´p dρ2
dr “ f´ f´ dρ2 “ f´ 2 7´p
f prq f` f` pρq
43. Before doing any supersymmetric manipulations, we should know the spin connection.
Take the extremal p-brane metric to be of the form

ds2 “ e2Aprq dxµ dxν ηµν ` e2Bprq dxi dxj δij

1
In this case we have A “ ´B “ 4 log Hprq. Take the frame fields

eµ̂ “ eA dxµ , eî “ eB dxi

Then
ÿ
deµ̂ “ Br AeA dr ^ dxµ “ Bi AeA dxi ^ dxµ “ eî ^ ω µ̂î ñ ωµ̂ν̂ “ 0, ωµ̂î “ p´qµ“0 Bi A eA´B dxµ
i
ÿ
î B i
de “ Br Be dr ^ dx “ Bi BeB dxj ^ dxi “ eĵ ^ ω îĵ ñ ωîĵ “ Bj Bdxi ´ Bi Bdxj
j

Using our extremal form of the solution, I can further write

H1
eΦ “ gs2 H p3´pq{4 “ gs2 epp´3qA , Fr01...p “ ¯ “ ˘4A1 e4A
H2
The ˘ corresponds to brane/anti-brane solutions.
` 1˘
In 10D N “ 2 SUGRA coupled to matter, represent the Killing spinor as  “ 2 . We have the gravitino
and dilatino variations:
1 eΦ
0 “ δψµ,A “ pBµ ` ωµab Γab q ` F{ Γµ Pp`2 
4 8
eΦ
0 “ δλ “ B{ Φ ` p´1qp p3 ´ pqF{ Pp`2 
4
Here we are took what was written the democratic formulation of Kiritsis Appendix I.4, setting all fields
equal to zero except for the dilaton and relevant RR p ` 2 field strength. The extra factor of two in the last
terms on both lines comes from counting F̂n ¨ F̂n and F̂10´n ¨ F̂10´n on equal footing.
1`n
As written there, for IIA we have Pn “ pΓ11 qn{2 σ 1 and for IIB we have Pn “ σ 1 for 2 even and iσ 2 for
1`n
2 odd.
Let’s first look at the dilatino variation. We get 4

pp ´ 3qA1 Γr  ˘ eAp3´pq p´1qp pp ´ 3qA1 e4A Γr0...p Pp`2  “ 0

ñ p1 ˘ p´1qp eAp1`pq Γ01...p Pp`2 q “ 0
ñ p1 ˘ p´1qp Γ0̂1̂...p̂ Pp`2 q “ 0
4
I have set gs “ 1 for all of this. I don’t understand how any of this could work without being modified for arbitrary gs .

103
 Here the Γ matrices with hatted (vielbein) indices are the familiar 10D Dirac matrices, as in Freedman and
Van Proyen Supergravity. We need our constant of proportionality κ “ 2 in order for the above combination
of matrices to have a nontrivial null space. Note we could inversely have taken this as a way to take the
profile of Φ “ epp´3qA and get the profile of F to be ˘4A1 e4A .
Locally, then, this is a linear algebraic constraint on the space of spinors at a given point, which half of the
spinors will satisfy.
Now let’s look at longitudinal the gravitino variation. Similar to the case of the RN black hole, we expect
Bµ  “ 0 since Bµ in the longitudinal direction is Killing.

1 epp´3qA 1 4A r0...p
Bµ  ` ωµab Γab  ¯ 4A e Γ Γµ Pp`2  “ 0
4 8
1 1
ñ ´ eA´B A1 Γr̂µ̂  ¯ epp`1qA A1 Γr0...p Γµ Pp`2  “ 0
2 2
1 1 1 1 r̂0̂...p̂
ñ ´ A Γr̂µ̂  ¯ A Γ Γµ̂ Pp`2  “ 0
2 2
ñΓµ̂  ˘ Γ0̂...p̂ Γµ̂ Pp`2  “ 0
ñp1 ˘ p´1qp Γ0̂...p̂ Pp`2 q “ 0

This is exactly the same constraint as the one that the dilatino gave us. This also directly confirms our
assumption: Bµ  “ 0 longitudinally, since we can subtract the dilatino variation from the above gravitino
one.
Meanwhile in the transverse directions we no longer expect Bi  “ 0. It is important to note that the
components of the spin connection ωiab Γab will only be nonvanishing for a, b “ tj, ku being transverse
coordinates, in which case Γjk is proportional to the infinitesimal rotation generator. By assumption of
spherical symmetry (just as in the RN case of problem 35), this must vanish Γjk  “ 0.

1 ab eApp´3q 1 4A r0...p

0 “ Br  ` ω Γab  ˘ 4A e Γ Γr Pp`2 
4 r
 8
1
“ Br  ˘ A1 Γr̂0̂...p̂ Γr Pp`2 
2
1
“ Br  ˘ p´1qp`1 A1 Γ0̂...p̂ Pp`2 
2
1 1
“ Br  ´ A  ñ  “ eA{2 0
2
where 0 is a constant spinor satisfying the linear algebraic constraints previously given.
We thus have that indeed our configuration is half-BPS.

44. We write again the spin connection found in the last problem:

eµ̂ “ eA dxµ , eî “ eB dxi

Hatted indices always denote the vielbein indices.

Then
ÿ
deµ̂ “ Br AeA dr ^ dxµ “ Bi AeA dxi ^ dxµ “ eî ^ ω µ̂î ñ ωµ̂ν̂ “ 0, ωµ̂î “ p´qµ“0 Bi A eA´B dxµ
i
ÿ
î B i
de “ Br Be dr ^ dx “ Bi BeB dxj ^ dxi “ eĵ ^ ω îĵ ñ ωîĵ “ Bj Bdxi ´ Bi Bdxj
j

From this, we can get the Riemann curvature using Rα̂β̂ “ dω αβ ` ω αγ ^ ω γβ . First Rµ̂ν̂ is the easiest:

î 2pA´Bq
dω
Rµ̂ν̂ “  µ̂ν̂ ` ωµ̂î ^ ω ν̂ “ e
 pBAq2 dxµ ^ dxν “ Rµν

104
Note that last expression is unhatted.
Next is Rµ̂î

Rµ̂î “ dωµ̂î `ωµ̂ĵ ^ ω ĵν̂

“ rBj Bi AeA´B ` Bi ApBj A ´ Bj Bqsdxj ^ dxµ ´Bi AeA´B Bj Bdxµ ^ dxi ` Bj AeA´B Bi Bdxµ ^ dxj

“ eA´B rpBi Bj A ` Bi ABj A ´ Bi BBj A ´ Bj BBi Aqsdxj ^ dxµ ´ eA´B Bj ABj Bdxµ ^ dxi
We can get Rµ;i (note unhatted) by multiplying this by eA´B and Riµ by multiplying this by ´eB´A .
Finally Rîĵ :

ω µ̂ `ωîk̂ ^ ω;k̂ĵ

Rîĵ “ dωîĵ ` ωîµ̂

 ^ ĵ
“ Bk Bj Bdxk ^ dxi ´ Bk Bi Bdxk ^ dxj `Bk BBj Bdxi ^ dxk ´ Bk BBi Bdxj ^ dxk ´ pBBq2 dxi ^ dxj

“ ´pBBq2 dxi ^ dxj ` pBk Bi B ´ Bk BBi Bqdxj ^ dxk ´ pBk Bj B ´ Bk BBj Bqdxi ^ dxk “ R;j
i

To evaluate Rµνρσ Rµνρσ amounts to summing the squares of all the entries in the curvature two form when
expressed in only vielbein indices. We can do this in Mathematica:
I can’t get c` , c´ exactly right. The best attempt is in ”exact p brane solutions.nb”. The
general L and r dependence in both cases matches though, and I’m not getting any p ´ 3
factors, so I can believe this result.
To get the Ricci tensor, we must to the appropriate contractions. Importantly, if a longitudinal index must
be summed over this gives an extra factor of p ` 1 while if a transverse index must be summed over this
gives an extra factor of 9 ´ p.
´ 8´p 1 ¯
Rµν “ Rµρνρ ` Rµiν i “ ´ηµν e2pA´Bq pp ` 1qpA1 q2 ` A2 ` A ` A1 B 1 p9 ´ p ´ 2q
r
Here A1 “ Br A is differentiation with respect to the radial coordinate. The Ricci tensor has no components
mixing transverse and longitudinal directions:
ρ j
Rµi “ 
Rµρi Rµji
 `   “0

Finally the annoying one, for which I looked at Stelle’s Lectures on p-Branes 9701088 :
´ ¯
Rij “ Riµj µ ` Rikj k “´δij B 2 ` pp ` 1qA1 B 1 ` p7 ´ pqpB 1 q2 ` 2p7´pq`1
r B 1 ` d A1
r
xi xj
` r2
pp7´pqB 2 ´ 7´p
r
B 1 `pp`1qA2 ´ p`1
r
A1 ´2pp`1qA1 B 1 `pp`1qpA1 q2 ´p7´pqpB 1 q2 q

xi
In this last part I rewrote Bi “ r Br .
We can evaluate this directly in Mathematica. For Rµν Rµν and R we get:

105
 The last line is in agreement with the expression for R in Kiritsis 8.8.31
45. Exercise 7.7 shows that, upon T -dualizing along the x9 direction we get
pp`1q pp`1q
C̃µ1 ...µp 9 “ Cµppq
1 ...µp
, C̃µppq
1 ...µp
“ C̃µ1 ...µp 9
In transverse space, our pp ` 1q-form C has components only along the longitudinal directions. Upon T-
dualizing, we pick up the 9 index in the C form, and thus get that our brane has a pp ` 2q form charge. We
thus expect this to be a p ` 1 brane wrapping that additional x9 direction. I’m unsure if this wants us to
explicitly give the form of that solution, since doing it in a compact space seems a bit harder.
46. Let’s assume p ă 7. When λ " 1 the perturbative stringy description is no longer valid. For an extremal
p-brane, we know from problem 40 that:
9´p
2κ210 Tp N L 7´p
ˆ ˙
7´p gs N Γp 2 q
L “ ñ “
p7 ´ pqΩ8´p 2π`s 7 ´ p 2π 9´p
2

So λ “ 2πgs N " 1 gives that L " `s , meaning that the throat size is macroscopic. We can thus probe it
without having to see distances smaller than the string scale.
2p7´pq
When p ą 3 we see from our calculation of R in problem 44 that R blows up as rLpp´3q{2 as r Ñ 0. This will
become order `´2
s at
˙2{pp´3q
`2s
ˆ
r«
Lp7´pq{2
When p ă 3 the formula for R indeed is seen to go to zero. On the other hand the string coupling grows as
˙p3´pq{4
L7´p
ˆ
Φ p3´pq{4
e “ gs H “ gs 1 ` 7´p
r
So if gs is the string coupling “at infinity” which we can take to initially be small, then it will become
appreciable at
r “ Lp´1 ` gs´4{p3´pq q1{pp´7q

106
 so for gs sufficiently small, the quantity in parentheses will be quite large and be raised to a negative power,
so that this is a small fraction of the throat size.

47. This is only a slight variant of exercises 8.38-9, and in fact is a bit easier. Our action is
ż „ 
1 10 ? ´2Φ 2 1 2
S“ 2 d x ´ge R ` 4p∇Φq ´ pdBq
2κ10 2 ¨ 3!

Let’s first vary Φ. We get

„ 
´2Φ 1
2 2
0 “ ´2e R ` 4p∇Φq ´ pdBq ´ ∇pe´Φ 8∇Φq
2 ¨ 3!
e´2Φ
“ ´2e´2Φ R ´ 8e´2Φ p∇Φq2 ` pdBq2 ´ 8e´2Φ lΦ ` 16p∇Φq2 e´2Φ
3!
pdBq2
ñR “ 4p∇Φq2 ´ 4lΦ `
2 ¨ 3!
Next for the B-field we will just have
d ‹ e´2Φ dB “ 0
Finally, varying g is the hardest, but we’ve done most of the work already in the other problem:
?
• ´ge´2Φ R

1
Ñ pRµν ` gµν l ´ ∇µ ∇ν qe´2Φ ´ gµν e´2Φ R
2
´ 1 ¯
“e ´2Φ
Rµν ´ gµν R ` gµν p´2lΦ ` 4pBΦq2 q ´ p´2∇µ ∇ν Φ ` 4Bµ ΦBν Φq
2
?
• ´ge´2Φ 4g µν Bµ ΦBν Φ Ñ 4e´2Φ Bµ ΦBν Φ ´ 2e´2Φ pBΦq2
e´2Φ ? e´2Φ e´2Φ
• ´ 2pp`2q! ´gg µ1 ν1 . . . g µp`2 νp`2 Fµ1 ...µp`2 Fν1 ...νp`2 Ñ ´ 2pp`1q! 2 `
Fµν 2
4pp`2q! gµν F .
2 “F
Here Fµν µ... F
ν...

Combining these all together, dividing through by e´2Φ and using the dilaton equations of motion gives
1
Rµν ` 2∇µ ∇ν Φ “ H2
2 ¨ 2! µν
Here Hµν “ Hµρσ Hνρσ as we’ve had before (i.e. in chapter 6).
Ok next let’s take the ansatz as in Kiritsis:
dr2
ˆ ˙
2 2
ds “ ´f pRqdt ` dx2i ` Hprq 2 2
` r dΩ3 .
f prq

With dilaton
e2Φ “ gs2 Hprq
The field strength written is wrong (as you can see by noting that as r Ñ 8 the magnetic flux integral
goes to zero). We can find the correct expression by noting that d ‹ dB “ 0 trivially since ‹dB has a dr
component. The only nontrivial equation is the Bianchi identity, giving (by spherical symmetry)

dB “ 0 ñ B “ cω

for ω “ dψ ^ sin ψdθ ^ sin ψ sin θdφ the unit volume form on the sphere. Let’s see what this constant c
should be from the dilaton equations. We get

107
 a
So when c “ ´2L 1 ` r02 {L2 we get our dilaton.
a
By Hodge-dualizing, this also gives credibility for the 1 ´ r02 {L2 constant in the p-brane solu-
tion, which would have required the more complicated Rµν equation.
Finally the least trivial equation of motion is also straightforward:

48. Let’s review the extremal near horizon limit first. There, when r ! L we can just write

dr2
ds2 “ ´dt2 ` d~x ¨ d~x ` L2 ` L2 dΩ23
r2
? r?
Defining γ “ N `s log g gives dγ 2 “ L2 {r2 giving
s `s N

ds2 “ ´dt2 ` d~x ¨ d~x ` dγ 2 ` N `2s dΩ23

This
? looks like flat space times a constant-radius sphere with a linear dilaton background going as Φ “
γ{ N `s .

108
 Next let’s look at the near-extremal case. We take r “ r0 cosh σ so that f prq “ 1 ´ r02 {r2 “ tanh2 σ.
Meanwhile
Hprq 2 L2 r2 sinh2 σdσ 2
dr “ p1 ` 2 q 0 “ L2 ` r02 cosh σ 2 “ Hprqr2
f prq r0 cosh σ tanh2 σ
So we get a metric
´ tanh2 σdt2 ` d~x ¨ d~x ` pN `2s ` r02 cosh2 σqpdσ 2 ` dΩ23 q
At large N , rescaling t this looks like

´ tanh2 σN `2s dt2 ` d~x ¨ d~x ` N `2s pdσ 2 ` dΩ23 q,

which looks like a 2D black hole solution in σ, t space, after rescaling

49. Let’s write the spin connection. Take e2A “ Hprq so that φ ´ φ0 “ A. Our frame fields look like:

eµ̂ “ dxµ , eAprq dxi

for µ parallel and i transverse. It looks like ωµν “ ωµi “ 0 while

ωîĵ “ ´Bj Adxi ` Bi Adxj

similar to what we had before.

We again write the gravitino and dilatino variation in 10D type II SUGRA, neglecting this time the RR
forms but incorporating the N 2-form contribution:
1 1
0 “ δψµ,A “ pBµ ` ωµab Γab q ` H{ P
4 4 µ
1
0 “ δλ “ B{ Φ ` HP
{
2
Here P “ Γ11 b 12 in type IIA and ´132 b σ 3 in type IIB.
The dilatino variation gives
1
Br φΓr  ˘ p´2L2 q sin2 ψ sin θΓψθφ P
2
1
H r L2
“ Γ  ¯ 3{2 3 Γψ̂θ̂φ̂ P
2H H r
H 1 r̂ L2
“ Γ  ¯ Γψ̂θ̂φ̂ P
2H 3{2 H 3{2 r3
ñ ´L2 p1 ˘ Γr̂ψ̂θ̂φ̂ Pq “ 0

109
 This is an algebraic constraint that is satisfied by half the space of spinors at any given point. This makes
the solution half-BPS, so long as the profile of ε can be chosen so that the gravitino vanishes.
The δψµ variation longitudinal to the solution is trivial. The transverse variation is

1 1
pBi ` ωijk Γjk qi ` Hijk Γjk P
4 4
Crucially, though, Γjk is the generator of rotations. By rotational symmetry we thus reduce this to Bi  “ 0,
implying that prq “ 0 is a constant spinor.
It is also worth noting that transverse to the NS5 brane is precisely the extremal BH solution in 5D, which
preserves half SUSY by the same arguments as before. Parallel to it is flat space (which preserves all SUSY).
The product spacetime therefore preserves half.

50. We have the same equations as when we were solving the for the NS5 brane. This time, the de´2Φ ‹ dB
constraint is nontrivial, and we must have a field strength. Because the field is electrically charged under
the field, I expect
B „ H ´1 prq
6
For H “ 1 ` Lr6 the relevant harmonic form in transverse space. I don’t have much justification for this other
than the fact that - in every problem I’ve seen this seems to hold true. Now let’s take the ansatz that the
metric and dilaton look like

ds2 “ H α p´dt2 ` dx21 q ` H β dK x2 , eΦ “ H γ

?
Then ´g “ H α`4β and we get

e´2Φ ‹ dB “ H ´α`3β´2γ´2 r7 H 1 prq

We want de´2Φ ‹ dB “ 0 so we must have

´α ` 3β ´ 2γ ´ 2 “ 0

The simplest guess would be α “ ´1, γ “ ´1{2. This turns out to work. First look at the dilaton EOM:

Next, look at the metric’s EOM

110
 Perhaps the easier thing to do was look for a BPS solution. In either case we are done. My (reasonable)
guess for the non-extremal version of this would be to keep the dilaton and NS field the same and modify
the metric to be
dr2
H ´1 prqp´f prqdt2 ` dx21 q ` ` r2 dΩ27
f prq
r06
where f prq “ 1 ´ r6
. I have not checked this, but it seems right based on experience at this point. s
51. Let’s start with IIB. The untwisted sector will contain closed string states that are invariant under the
I4 p´1qFL combination. The twisted sector will localize to the 5-plane left invariant by the inversion. Let’s
say that this is labeled by x0 . . . x5 and x6 . . . x9 are the coordinates reflected under the orbifold. The
supersymmetries QL “ Q, QR “ Q̃ both transform in the 8s representation. On the 5-plane, this decomposes
under SOp4qk ˆ SOp4qK as
8s Ñ r2k ˆ 2K s ` r2̄k ˆ 2̄K s
Here I4 acts with ´ on the 2 b 2̄ vector representation of SOp4qK , leaving the 2 b 2̄ SOp4qk alone. We take
I4 to flip the sign of only the 2K spinor. p´1qFL acts with a ´ sign on only Q. Together, this leaves

Q P 2k ˆ 2K , Q̃ P 2̄k ˆ 2̄K

invariant. These preserved generators give p1, 1q6 supersymmetry. The exact same argument would give that
the IIA twisted sector has p2, 0q6 SUSY. These rigid supersymmetries have a unique massless representation,
namely the vector and tensor multiplets respectively, so this is what we would expect to get.
Let’s check this explicitly for the twisted sector of the IIB orbifold. The parallel αµ do not get twisted
boundary conditions, but the transverse αi get acted on by a ´ from the I, so will get half-integrally
modded. For the fermions, the ψ µ are affected by the p´1qFL and so will become integrally modded in the
NS sector and half-integrally modded in the R sector. The ψ i are additionally affected the I and so remain
half-integrally modded in the NS sector and integrally modded in the R sector.
In both R and NS sectors, we have the same number of periodic and anti-periodic bosons and fermions, so
the ground state energy vanishes in both sectors. Massless excitations are described purely in terms of the
ground states of the system. The bosonic ground state is unambiguous. In the NS sector there are four ψ0i
transforming in the SOp4qK vector representation while in the R sector there are four ψ0µ transforming in the
SOp4qk vector representation. These lead to ground states transforming as 2 ` 2̄.
The effect of the p´1qFL is to change the left-moving GSO projection in the twisted sector (c.f. exercise
11.29). Thus in both NS-NS and R-R we get GSO projections:
1
p1 ´ p´qFL qp1 ` p´qFR q
4

111
For NS-NS this means:
p2L L R R L R
K ` 2̄K q b p2K ` 2̄K q Ñ 2̄K b 2K

This transforms in the vector representation of SOp4qK and can thus be interpreted as 4 scalars with SOp4q
R-symmetry.
For R-R we get the same:
p2L L R R L R
k ` 2̄k q b p2k ` 2̄k q Ñ 2̄k b 2k

This is a vector with little group SOp4q.

So the NS-NS states give 4 scalars and the RR states give the vector. These are consistent with the spectrum
of a single NS5 brane in type IIB, and the same logic holds for IIA. Understand how this connects with
Sen’s articles on non-BPS particles

112
Chapter 9: Compactification and Supersymmetry Breaking
In collaboration with Alek Bedroya

1. We compactify the heterotic string along just one dimension, making it a compact circle of radius R with
all 16 Wilson lines turned on.
Each noncompact boson contributes
1
?
τ 2 η η̄
The fermions on the supersymmetric side contribute
1 a 4
“ ‰
a`b`ab θ b
ÿ
p´1q
a,b“0
η4

The pp, pq compact bosons and 16 complex right-moving fermions that can be written as the pair ψ I pz̄q, ψ̄ I pz̄q
have the action as in E.1 (setting `s “ 1)
ż ż ż
1 2
a ab α β 1 2 ab α β 1 a ÿ
¯ ` Y I B̄X α sψ̄ I
d σ det gg Gαβ Ba X Bb X ` d σ Bαβ Ba X Bb X ` d2 σ ´ det g ψ I r∇ α
4π 4π 4π I

Here α, β are the toral coordinates for the compact spacetime and YαI is the Wilson line along torus cycle α.
To evaluate the path integral, as we did in the purely bosonic case, we have a factor of
?
det G
p{2
τ2 pη η̄qp

coming from evaluating the determinant pdet ∇2 q´1{2 of the bosons. This multiplies a sum over instanton
contributions labelled by mα , nα taking values in a pp, pq-signature lattice with classical action
ÿ π α β
e´ τ pG`Bqαβ pm`τ nq pm`τ̄ nq ˆ fermions.
mα ,nα

The fermion contribution depends via the Wilson lines on the configuration of the X α . In each such instanton
sector, the fermion path integral with a constant background Wilson line is equivalent to a free fermion with
twisted boundary conditions. For simplicity, let’s compactify just on S 1 , and denote θI “ Y I n, φI “ ´Y I m.
We get boundary conditions:
I
ψ I pσ ` 1, σ2 q “ ´p´1qa e2πiθ
I
ψ I pσ, σ2 ` 1q “ ´p´1qb e´2πiφ
where a, b “ 0, 1 denotes anti-periodic/periodic boundary conditions respectively. We know that (in the
absence of Wilson lines) the determinant of B acting on complex fermions is:

θ ab
“ ‰
deta,b B “
η

Let us now investigate the twisted boundary conditions. For simplicity its enough to take a “ b “ 0 (all
antiperiodic). We have two different ways to write the partition function. As a product over modes, we have
ψm , ψ̄m modes, with respective weights m ´ 12 ´ θ, m ´ 12 ` θ Check against Polch 16.1.16 and respective
fermion numbers ˘1 relative to the ground state. The fermion number of the ground state has no canonical
value (as far as I can see). On the other hand, the ground state energy is given by the standard mneumonic
1
to be ´ 24 ` 12 θ2 . This gives:
0
8 “‰
θ2 1
θ2 {2 θ 0 pφ ` θτ |τ q
ź
2iπφF H ´ 24 m´1{2`θ 2πiφ m´1{2´θ ´2πiφ
Trθ re q s“q 2 p1 ` q e qp1 ` q e q“q
m“1
η

113
For other boundary conditions, we can apply the same logic to get
“a‰
θ2 {2 θ b pφ ` θτ |τ q
q
η

The overall phase is still a mystery. Writing θ ab φθ as a new theta function, we can fix the phase by requiring
“ ‰“ ‰

modular invariance
« ff« ff « ff« ff « ff« ff « ff« ff
0 θ 0 θ 0 θ 0 θ
θ pτ ` 1q “ θ pτ q θ pτ ` 1q “ θ pτ q
0 φ 0 φ`θ 1 φ 0 φ`θ
« ff« ff « ff« ff « ff« ff « ff« ff (79)
1 θ iπ{4 1 θ 1 θ iπ{4 1 θ
θ pτ ` 1q “ e θ pτ q θ pτ ` 1q “ e θ pτ q
0 φ 1 φ`θ 1 φ 0 φ`θ

Even from the first of these conditions, we see that we need a term going as eiθφ out front. After adding
this in, all other transformations will hold automatically. The τ Ñ ´1{τ transformation will thus hold
automatically. Interpret this as an anomaly? Yes, Narain, Witten do this in Section 3 of their
paper. It seems careful anomaly analysis is not enough and one must indeed impose modular
invariance by hand.
Altogether then the 16 complex antiholomorphic fermions contribute in each instanton sector:
1 ź 16 “a‰
ř I I I 1 ÿ θ̄ b pφ ` τ̄ θ|τ̄ q
e´iπ I θ pφ `τ̄ θ q
2 a,b“0 i“1 η̄

Giving a total partition function as in the second (unnumbered) equation of Appendix E:

» fi
2 1 ź 16 „  1 4 a
“ ‰
R ÿ ´ πR
|m`nτ | 2 ř I I I 1 ÿ a 1 1 ÿ θ
–? e τ2 e´iπ I nY pm`nτ̄ qY Y θ̄ pY I pm ` τ̄ nq|τ̄ qfl ˆ 7{2 b
τ2 η η̄ 17 m,n 2 a,b“0 i“1 b τ η 7 η̄ 7 2 a,b“0 η
4
2

From the properties of the theta functions in Equation (79), the underlined fermionic sum has the exact
same transformation properties as a sum of θ16 terms and thus makes the full partition function modular
invariant.
Each theta function can be written in sum form as:
„ „ 
a θ 2
ÿ 1 a 2 a b ÿ 1 a 2 1 a a
θ “ eπiθφ q θ {2 q 2 pn´ 2 q e2πipn´ 2 qpφ`τ θ´ 2 q “ q 2 pn`θ´ 2 q e2πiφpn` 2 θ´ 2 q´πibpn´ 2 q
b φ nPZ nPZ

Then we get the following expression for the underlined fermionic term:
1 16
1 ÿ ź ÿ 1 pk`nY I ´ a q2 ´2πimY I pk` 1 nY I ´ a q`πibpk´ a q
q̄ 2 2 e 2 2 2
2 a,b“0 I“1 kPZ
1
1 ÿ ÿ 1 pqI `nY I ´ a q2 ´2πimY I pqI `nY I ´ a q`πibpk´ a q
“ q̄ 2 2 e 2 2
2 a,b“0 I 16
q PZ
1 ÿ ” 1 pqI `nY I q2 ´2πimY I pqI ` 1 nY I q ř I 1 I I 1 2 I I 1 I 1
ř I 1 ı
“ q̄ 2 e 2 p1 ` p´1q I q q ` q̄ 2 pq `nY ´ 2 q e´2πimY pq ` 2 nY ´ 2 q p1 ` p´1q I pq ´ 2 q q
2 I 16
q PZ
I I 2 I I 1 I
ÿ
“ q pq `nY q e´2πimY pq ` 2 nY q
q I PΛ16

We note that the second-to last line is indeed the sum over the roots of Op32q augmented with one of the
spinor weight lattices. Altogether the compact dimensions contribute:
„ 2 
R ÿ πR I I 2 I 1 I
? exp pm ` nτ qpm ` nτ̄ q ` πiτ pq ` nY q ´ 2πimY pk ` Y q
τ2 η η̄ 17 I 16
τ2 2
mPZ,nPZ,q PΛ

114
 To put this whole thing into Hamiltonian form, we proceed as in the bosonic case and perform a Poisson
summation over m. The terms that contribute are:
2 2 nR2 τ1
´ πR n2 τ12 ´n2 πR2 τ2 ´ πR m2 ´2πimY I pq I ` 21 nY I q´i
ÿ
e τ
2 e τ2 τ2

m
2 ? ÿ πτ R2 τ
´ πR τ2
n2 τ12 ´n2 πR2 τ2 ´ 2 pm`Y I pq I ` 12 nY I q´in τ 1 q2
“e 2 τ e R2 2
R m
2 2 2 ? ÿ πτ τ2
´ πR n τ1 ´n2 πR2 τ2 τ2 ´ 22 pm`Y I pq I ` 12 nY I qq2 `πR2 τ1 n2 `2πipm`q I ` 12 nY I qnτ1
“e τ 2 e R 2
R m
? ÿ πτ
´n2 πR2 τ2 τ2 2 I I 1 I 2 I 1 I
“e e´ R2 pm`Y pq ` 2 nY qq `2πipm`q ` 2 nY qnτ1
R m

Together with the other terms this gives us

1 ÿ 1 I I 2 2 2 πτ2 I I 1 I 2 I 1 I

17
q 2 pq `nY q e´n πR τ2 e´ R2 pm`Y pq ` 2 nY qq `2πipm`q ` 2 nY qnτ1
η η̄ In,m,q
1 ÿ 1 I I 2 1 1 I I 1 I 2 1 1 I I 1 I 2
“ 17 q 2 pq `nY q q 2 p R pm´Y pq ` 2 nY q`nRq q̄ 2 p R pm´Y pq ` 2 nY q´nRq
η η̄ I n,m,q

where I’ve flipped m Ñ ´m at the end there. We get momenta

1 1 m 1 I I
kL “ pm ´ q I Y I ´ nY I Y I q ` nR “ ` npR ´ Y Y q ´ qI Y I
R 2 R 2
1 1 m 1 I I
kR “ pm ´ q Y ´ nY I Y I q ´ nR “
I I
´ npR ` Y Y q ´ qI Y I
R 2 R 2
I
kR “ q I ` nY I

consistent with Polchinski with m Ð nm , n Ð wn , Y I Ð RAI and α1 “ 0 (might be off by a factor of

2 for kRI rel. to Polchinski but I think I’m consistent with Ginsparg). We only care about the

SOp1, 1, Zq T-duality group coming from the compact x9 . This does not act on the Y I as far as I can see
CHECK
The SOp16, Zq on the other hand acts on the Y I as in the standard vector representation.

2. I am going to re-do the computations of appendix F Hatted indices denote the 10D terms. Greek indices
from the start of the alphabet denote compact 10-D-dimensional indices while greek indices from the middle
of the alphabet denote noncompact D-dimensional indices.
The 10D action is b ż
1 1
d10 x ´Ĝ10 e´2Φ̂ rR̂ ` 4p∇Φ̂q2 ´ Ĥ 2 ´ TrF̂ 2 s ` Op`2s q
12 4
I “ B ÂI ´ B ÂI and Ĥ 1 ř I I
with F̂µν µ ν ν µ µνρ “ Bµ B̂νρ ´ 2 I µ F̂νρ ` 2 perms.. Here I is the internal 16-dimensional
index for the heterotic string.
We take the 10-bein (r, a denote D and 10 ´ D 10-bein indices, hatted indices r̂, µ̂ should not be confused
for 10-bein indices!!) ˜ ¸
r Aβ E a
ˆ µ ˙
r̂ eµ µ β µ̂ er ´eνr Aαν
eµ̂ “ er̂ “
0 Eαa 0 Eaα
This gives us the metric: ˜ ¸
Gµν ´ Aαµ Gαβ Aβν Gαβ Aβµ
Gµ̂,ν̂ “
Gαβ Aβν Gαβ
As we’ve done before in chapter 7, we then define
1 A
φ“Φ´ log det Gαβ , Fµν “ Bµ Aν ´ Bν Aµ
4

115
With this, the compactification of R ` 4p∇φq2 is clear:
ż
? 1 1 Aα Aβ
dD ge´2φ rR ` 4Bµ φB µ φ ` Bµ Gαβ B µ Gαβ ´ Gαβ Fµν Fµν s
4 4
The first and second terms are clear. The third term makes up for the redefinition of Φ in terms of φ while
the last term is the standard KK mechanism generating a gauge field strength from the compact dimensions.
Next, let’s look Ĥ. Because we have no sources for the H field, Ĥ is on the compact cycles. We can define
the D-dimensional fields using the 10-bein as:

Hµαβ “ erµ eµ̂r Ĥµ̂αβ “ Ĥµαβ (80)

Hµνα “ erµ esν eµ̂r eν̂s Hµ̂ν̂α “ Ĥµνα ´ Aβµ Ĥναβ ` Aβν Ĥµαβ (81)
Hµνρ “ erµ esν etρ eµ̂r eν̂s eρ̂t Ĥµ̂ν̂ ρ̂ “ Ĥµ̂ν̂ ρ̂ ` r´Aαµ Ĥανρ ` Aαµ Aβν Ĥαβρ ` 2 perms.s (82)

The point of defining these coordinates in terms of the 10-bein coordinate is that now, we can just directly
separate the Ĥµ̂ν̂ ρ̂ Ĥ µ̂ν̂ ρ̂ sum into terms without worrying about the metric, and yield directly:
ż
? 1 3 3
dD ´ge´2φ r´ Hµνρ H µνρ ´ Hµνα H µνα ´ Hµαβ H µαβ s
12 12 12

The method is the same for the F tensor. We define new Wilson lines and field strengths:

YαI “ AIα , AIµ “ erµ eµ̂r ÂIµ̂ “ ÂIµ ´ YαI Aαµ

I “ B AI ´ B AI , F̃ I “ B Y I . This gives me F̂ I “ F I ` B pY I Aα q ´
I can define F in the standard Fµν µ ν ν µ µα µ α µν µν µ α ν
I α
Bν pYα Aν q. By redefining
I I
F̃µν “ Fµν ` YαI Fµν
A,α

I . For the compact coordinates its more simple and I take F̃

we can equate this with F̂µν I
µα “ Bµ Yα . Again
F̃αβ vanishes since we cannot have internal sources. This yields directly
ż 16
? 1 ÿ I I,µν 2 I I,µα
dD x ´ge´2φ r´ F̃ F̃ ´ F̃µα F̃ s
4 I µν 4

Its not good enough for us to write everything in terms of an abstract H 3-form. We want to relate H to B
and Y . From our relationship in 10D we can directly write:
1ÿ I
Hµαβ “ Bµ Bαβ ` pY Bµ YβI ´ YβI Bµ YαI q
2 I α

1
YαI YβI we get
ř
Taking Cαβ “ B̂αβ ´ 2 I
ÿ
Hµαβ “ Bµ Cαβ ` YαI Bµ YβI
I

Next
1ÿ I
Hµνα “ Bµ Bνα ´ Bν Bµα ` pÂν Bµ YαI ´ ÂIµ Bν YαI ´ YαI Fµν
I
q
2 I
We define the B field using not just the vielbein but also the gauge connection:
1ÿ I I
Bµα :“ B̂µα ` Bαβ Aβµ ` Y A , B
Fµν “ Bµ Bν ´ Bν Bµ
2 I α µ

Then using (81) we get

B A β ÿ
Hµνα “ Fαµν ´ Cαβ Fµν ´ YαI Fµν
I

116
 Finally, using both vielbein and connection
1 ÿ
Bµν “ B̂µν ` rAαµ Bνα ` AIµ Aαν YαI ´ pν Ø µqs ´ Aαµ Aβν Bαβ
2 I

And this gives us

1
Hµνρ “ Bµ Bνρ ´ Lij Aiµ Fνρ
j
` 2 perms.
2
where Lij is the p10 ´ D, 26 ´ Dq-invariant metric and we have combined Aαµ , Bαµ , AIµ into a length 36 ´ 2D
vector.
Now the full action is:
ż
? 1
dD ge´2φ rR ` 4Bµ φB µ φ ´ Hµνρ H µνρ
12
1 αβ 1 A α A µνβ 1 I I,µν
´ G Hµνα H µνβ ´ Gαβ Fµν F ´ F̃µν F̃
4 4 4
1 1 1 I I,µα
´ Hµαβ H µαβ ` Bµ Gαβ B µ Gαβ ´ F̃µα F̃ s
4 4 2
A , the middle line can be combined into
Using our expressions for Hµνα and F̃µν

G ` C T G´1 C ` Y T Y ´C T G´1 C T G´1 Y T ` Y T

¨ ˛
1˝ i
´ ´G´1 C G´1 ´G´1 Y T ‚ Fµν F µν j
4
Y G´1 C ` Y ´Y G´1 1 ` Y G´1 Y T ij

α
here F i “ pF A , F B α , F I q. Call the matrix M ´1 and notice that LM L “ M ´1 , and indeed we get M
transforms in the adjoint of SOp26 ´ D, 10 ´ Dq.
Similar arguments would give that the last line becomes 18 TrBµ M B µ M ´1 (Too much algebra).
From this, its immediate that any SOp10 ´ D, 26 ´ Dq transformation on the scalar matrix (adjoint rep) and
array of vector bosons (vector rep) will preserve both of these last two terms. It will also preserve H since
j
it depends on the invariant Bνρ and SO-invariant combination Lij Aiµ Fνρ .

3. The action for IIA in the string frame is

ż b „  ż
1 10 ´2Φ̂ 2 1 µ̂ν̂ ρ̂ 1 2 1 2 1
d x ´Ĝ e rR̂ ` 4p∇Φ̂q ´ Ĥµ̂ν̂ ρ̂ Ĥ s ´ F2 ´ F ` 2 B2 ^ dC3 ^ dC3
2κ210 12 4 2 ¨ 4! 4 4κ
1
Doing the same reduction as before, the R̂ ` 4p∇Φ̂q2 ´ 12 H 2 term becomes:
ż
? ” 1 A α A µν 1 1 1 1 ı
d4 ´ge´2φ R ` 4Bµ φB µ φ ´ Fµν F α ` Bµ Gαβ B µ Gαβ ´ Hµνρ H µνρ ´ Hµαβ H µαβ ´ Gαβ Hµνα H µνα
4 4 12 4 4
ż
? ” 1 1 1 ı
“ d4 ´ge´2φ R ` 4Bµ φB µ φ ´ Hµνρ H µνρ ´ Mij´1 Fµν i
F µν j ` TrrBµ M B µ M ´1 s
12 4 8

Here we used H as in the last problem and the matrix M consisting of the 21 Gαβ and 15 Bαβ . The F i are
the field strengths of the 6 ` 6 U p1q vectors coming from G and B compactification.
ˆ ˙
1 i j ´1 G ` B T G´1 B ´B T G´1
Hµνρ “ Bµ Bµρ ´ Lij Aµ Fνρ ` 2 perms. M “
2 ´G´1 BG´1 G

The Hµνρ can be dualized to provide a sixteenth scalar coming from the B field. By analogy to 9.1.13, in
the string frame I would expect to write:

e´2φ Hµνρ “ Eµνρσ ∇σ a

117
The Bµν equations ∇µ pe´2φ Hµνρ q are now automatically satisfied. The axion EOMs come from the Bianchi
identity:
1 1
E µνρσ Bµ Hνρσ “ ´ Lij E µνρσ Fρσ
i j
Fµν i
“ ´Lij F̃µν F j µν , i
F̃µν “ E µνρσ Fρσ
2 2
Here we have defined the dual 2-form as required. This can now be recast as the equation of motion for the
axion (contracting the Es gives a 4):
1 i
∇µ pe2φ ∇µ aq “ ´ Lij Fµν F̃ j µν
4
With this, we can dualize the action in terms of the axion to yield:
ż
? ” 1 1 1 1 ı
d4 ´ge´2φ R ` 4Bµ φB µ φ ´ e4φ pBaq2 ` e2φ aLij Fµν i
F̃ j µν ´ Mij´1 Fµν
i
F µν j ` TrrBµ M B µ M ´1 s
2 4 4 8
We could also do this in the Einstein frame and get exactly the same action as in 9.1.15 with the M matrix
as we have it (no sum over heterotic internals).
The only thing left is the RR fields. We follow Kiritis’ treatment of the 4-form field strength. We use the
10-bein to get:
Cαβγ “ Ĉαβγ
Cµαβ “ Ĉµαβ ´ Cαβγ Aγµ
Cµνα “ Ĉµνα ` Ĉµαβ Aβν ´ Ĉναβ Aβµ ` Cαβγ Aβµ Aαν
Cµνρ “ Ĉµνρ ´ pAαµ Ĉνρα ` Aαµ Aβν Cαβρ ` 2 perms.q ´ Cαβγ Aαµ Aβν Aγρ
Let’s now define the field strengths. Now we must have Fαβγδ “ 0 since the internal dimensions do not
contain sources for the field. What remains is
Fµαβγ “ Bµ Cαβγ
γ
Fµναβ “ Bµ Cναβ ´ Bν Cµαβ ` Cαβγ Fµν
β
Fµνρα “ Bµ Cνρα ` Cµαβ Fνρ ` 2 perms.
α
Fµνρσ “ pBµ Cαβγ ` 3 perms.q ` pCσρα Fµν ` 5 perms.q
α are taken to mean F A ):
Then this gives the contribution (here all two-lower one-upper index Fµν
ż
p4q 1 ? a
SRR “ ´ d4 ´g det Gαβ rFµνρσ F µνρσ ` 4Fµνρα F µνρα ` 6Fµναβ F µναβ ` 4Fµαβγ F µαβγ s
2 ¨ 4!
It is important to realize that in 4-D the 4-form field strength coming from the 3-form has no dynamical
degrees of freedom. It plays the role of a cosmological constant Check w/ Alek.
The two-spacetime-index term can be directly dualized. It corresponds to 6 ˆ 5{3 “ 15 vectors. The three-
spacetime-index term can be dualized to become the kinetic term for 6 scalar axions aα with no interaction
term.
p4q
The Fµαβγ correspond to kinetic terms of the 6 ˆ 5 ˆ 4{3! “ 20 scalars Cαβγ .
Let’s do a similar thing for the 2-form field strength. There, we get Cα “ Ĉα , Cµ “ Ĉµ ´ Cα Aαµ . The
corresponding field strength is Fαβ “ 0, Fµα “ Bµ Cα and Fµν “ Bµ Cν ´ Bν Cµ ` Cα Fµν α . We then get

contribution ż
p2q 1 ? a
SRR “ ´ d4 ´g det Gαβ rFµν F µν ` 2Fµα F µα s
4
p2q
Again Fµν can be written in terms of dual fields F̃µν “ Eµνρσ F p2q ρσ . This is one gauge fields and six further
scalars.
Return and think about the effect of the CS terms. I bet they make the RR field equations
non-free.

118
4. First note that using the OPE

δ IJ
ΣI pzqΣ̄J pwq “ ` pz ´ wq1{4 J IJ pwq
pz ´ wq3{4

the xJ II ΣJ Σ̄J y correlator can be evaluated as

1{4
z23
xJ II pz1 qΣJ pz2 qΣ̄J pz3 qy “ pδ IJ ´ 14 q
z12 z13
1
pδ IJ ´ 4 q ´3{4
Taking z1 Ñ z2 we see a singularity going as z12 z23 . Meanwhile taking the JΣ OPE gives

xΣpz2 qΣ̄pz3 qy q ´3{4

q “ z
z12 z12 23

So we see that under J I the charge of ΣJ is 3{4 if I “ J and ´1{4 otherwise. We have 4 J II , and notice
that the total charge under all four of each ΣI is always zero. Consider the following combination of charges,
which provides a basis for the ΣI charge space

J˜1 “ J 11 ` J 22 ´ J 33 ´ J 44
J˜2 “ J 11 ´ J 22 ` J 33 ´ J 44
J˜3 “ J 11 ´ J 22 ´ J 33 ` J 44

Under each of J˜i we have the following charges

Σ1 Ñ p 12 , 21 , 12 q, Σ2 Ñ p 21 , ´ 12 , ´ 12 q, Σ3 Ñ p´ 12 , 21 , ´ 12 q, Σ4 Ñ p´ 12 , ´ 12 , 12 q
Σ̄1 Ñ p´ 12 , ´ 21 , ´ 12 q, Σ̄2 Ñ p´ 12 , 12 , 12 q, Σ̄3 Ñ p 12 , ´ 12 , 12 q, Σ̄4 Ñ p 12 , 12 , ´ 12 q
ř 1 2
These are exactly all combinations, and we can define the three bosonic fields φi with T “ i 2 pBφi q so
that
Σ1 “ exp ip 12 φ1 ` 21 φ2 ` 12 φ3 q , Σ2 “ exp ip 12 φ1 ´ 12 φ2 ´ 12 φ3 q , etc.
“ ‰ “ ‰

Each of these ΣI , Σ̄I has dimension 3{8 as required.

Let’s look at the supercurrent Gint . It can be written in terms of an eigenbasis of the commuting J˜i . In
particular look at J˜1 . ÿ
Gint “ eiqφ1 T pqq
q

Now consider the OPEs Gint ¨ Σ1 and Gint ¨ Σ̄1 . As observed in the chapter, both of these have only the
singular term going as pz ´ wq´1{2 . Together both of these require that q in G can only be ˘1. We can
repeat this argument for J˜2 , J˜3 to see that Gint must be a sum of 6 terms:

eiq1 φ1 Z1 ` e´iq1 φ1 Z̄1 ` eiq2 φ2 Z2 ` e´iq2 φ2 Z̄2 ` eiq3 φ3 Z3 ` e´iq3 φ3 Z̄3

Each Zi , Z̄i must be dimension one operators, so they are themselves bosonic fields iBX˘ i . We thus have
int
ř3 ˘ i
that G “ i“1,˘ ψi BX˘ . This is exactly the supercurrent for six free boson-fermion systems and will
give (under anticommutator) the stress tensor of a six free boson-fermion systems. This is exactly a toroidal
CFT.

5. The relevant partition function is not difficult to compute, as we can follow 9.4’s example but not do the
twist on the internal p0, 16q part. Firstly the fermions on the left-moving (SUSY) side have orbifold blocks
under the shifts as before:
2 a θ a`h θ a´h
1
“ ‰ “ ‰ “ ‰
θ
„ 
h 1 ÿ b b`g b´g
Zψ “ p´1qa`b`ab 4
g 2 a,b“0 η

119
Similarly we’ve already constructed the bosonic blocks before. They are given by 4.12.10 as:

η 2 η̄ 2
„  „ 
0 Γ4,4 h
Z4,4 “ 4 4 , Z4,4 “ 24 “1´h‰ “1´h‰
0 η η̄ g θ2 1´g θ̄2 1´g

Γ2,2
Then the p2, 2q part is untouched, yielding η 2 η̄ 2
as is the p0, 16q part. We get the partition function

“h‰ ´ ř
1 1 “a‰8 ¯2
θ2 ab
“ ‰ “a`h‰ “a´h‰
a,b“0 θ̄
1 Z4,4 g 1 θ b`g θ b´g
Γ2,2 1 ÿ 1 ÿ 2 b
Z het “ 2 2
ˆ 2 2
ˆ p´1q a`b`ab
4
ˆ 16
loηomoη̄ on 2 h,g“0 τ2 η η̄ 2 a,b“0 η η̄
loooooooooomoooooooooon
loooooooomoooooooon looooooooooooooooooooooomooooooooooooooooooooooon
1 4
2 3

Let’s see how each term transforms under τ Ñ ´1{τ . 1 stays invariant. 2 have Z4,4 hg Ñ Z4,4 hg with τ2 η 2 η̄ 2
“ ‰ “ ‰

invariant. 3 is the only nontrivial one. We will do it explicitly in the next step. 4 will remain invariant.
“1‰ “ ‰
iπ{4 while θ ´1 picks up e´3iπ{4 . The other two
Under τ Ñ τ ` 1, we must be careful, as θ 0 picks up an e 0
nonzero theta functions simply do θ ab Ñ θ a`b´1
“ ‰ “ a ‰

1, 2, remain invariant, with 2 making us change variables g 1 , h1 “ g, h ` g ´ 1. The η functions in the

denominators of 3 and 4 leave over an 1{η̄ 12 which contributes a ´ sign.
Let’s look at 3. First when h “ 0, g “ 0 we have p´1qa`b`ab θ4 ab and τ ` τ ` 1 will send this to ´ itself as
“ ‰

required to cancel the η̄ 12 ´ sign.

The other terms looks like (after canceling θ 11 )
“‰

« ff4 « ff4 « ff4 « ff4

0 1 0 1
h “ 0, h “ 0 : θ ´θ ´θ ´θ “0
0 0 1 1
« ff2 « ff « ff « ff2 « ff « ff « ff2 « ff « ff « ff2 « ff 
«ff « ff2 « ff2 « ff2 « ff2
0 1 ´1 1 2 0 0 1´1   1 2 0 0 1 1 0
h “ 1, g “ 0 : θ θ θ ´θ θ θ ´ θ  θ θ ´θ θ θ “θ θ ´θ θ “0
0 0 0 0 0 0 1 1 1 1  1
 1 0 0 0 0

« ff2 « ff « ff « ff2 « ff « ff « ff2 « ff « ff « ff2 « ff 
«ff « ff2 « ff « ff « ff2 « ff2
0 0 0 1 11  0 0 0 1 1 1 0 0 0 0 0
h “ 0, g “ 1 : θ θ θ ´ θ  θ θ ´θ θ θ ´θ θ θ “θ θ θ ´θ θ “0
0 1 ´1 0 1 ´1 1 2 0 1
  2 0 0 1 1 1 0

« ff2 « ff « ff « ff2 « ff « ff « ff2 « ff « ff « ff2 « ff 
« ff « ff2 « ff2 « ff2 « ff2
0 1´1   1 2 0 0 1 ´1 1 2 0 1 0 0 1
h “ 1, g “ 1 : θ  θ  θ ´θ θ θ ´θ θ θ ´θ θ 2 θ 0 “ ´θ 0 θ 1 ` θ 1 θ 0 “ 0

0 1 ´1 0 1 ´1 1 2 0 1
 
Ok, so in fact this partition function is zero. This should not be surprising, since naively we are just
breaking supersymmetry in half, and so we should still expect fermions and bosons to run in loops such that
the vacuum energy vanishes. Naively, then we would again say “zero is modular invariant” and be done with
it- but not so fast. There are still phases we can pick up, say from τ Ñ τ ` 1 that would not be visible given
the vanishing of the partition function, but would nonetheless spoil modular invariance.
One way around this is to turn on the chemical potential νi in the“ theta functions to prevent vanishing.
Effectively, then, we ignore the Jacobi identity and don’t just set θ 11 “ 0. Then, let’s look at how each
‰

term transforms under τ Ñ τ ` 1. Again, the terms not involving θ 11 will cancel independently of νi “ 0 or
“‰

not, and after simplifying things ,we have

« ff4 « ff4 « ff4 « ff4 « ff4 « ff4 « ff4 « ff4
0 1 0 1 0 1 0 1
p0, 0q : θ ´θ ´θ ´θ Ñθ `θ ´θ `θ ð ´p0, 0q
0 0 1 0 1 0 0 1
« ff2 « ff2 « ff2 « ff2
0 1 0 1
p1, 0q : ´ 2θ θ Ñ ´2i θ θ ð i ˆ p1, 1q
1 1 0 1
« ff2 « ff2 « ff2 « ff2
1 1 1 1
p0, 1q : 2θ θ Ñ ´2θ θ ð ´p0, 1q
0 1 0 1
« ff2 « ff2 « ff2 « ff2
0 1 0 1
p1, 1q : ´ 2θ θ Ñ ´2iθ θ ð i ˆ p1, 0q
0 1 1 1

120
 So we see p0, 1q (ie the projected part of the untwisted sector) goes to its negative as required. On the other
hand, the twisted sector has p1, 0q and p1, 1q swap, but with a factor of i instead of ´1. This is not good
enough for modular invariance.
Under τ Ñ ´1{τ the sectors appropriately get sent to one another except for the twisted projected sector
“ ‰2
which picks up a factor of ´1 from the θ 11 , so this too is not modular invariant.
It is worth adding that Polchinski remarks in 16.1 that for abelian orbifolds (of the type T n {H with H and
abelian group), the only obstruction to modular invariance is τ Ñ τ ` 1
Indeed,
ř4 weřsee that this twist violates 16.1.28 of Polchinski, where we hae r2 “ 0, r3 “ r4 “ 1 and so
16
i“2 ri ´ k“1 sk 2 ‰ 0 mod 2N when N “ 2.

6. Now the partition function is given by

“h‰ “a‰8
θ2 ab θ a`h
1 1
“ ‰ “ ‰ “a´h‰ 1 6 γ
“ ‰ “γ`h‰ “γ´h‰ 1 ř1
het Γ2,2 1 ÿ Z4,4 g 1 ÿ a`b`ab b`g θ b´g 1 ÿ θ̄ δ θ̄ δ`g θ̄ δ´g 2 a,b“0 θ̄ b
ZN “2 “ 2 2 ˆ ˆ p´1q ˆ ˆ
η̄ on 2 h,g“0 τ2 η 2 η̄ 2 2 a,b“0
loηomo η4 2 γ,δ“0 η̄ 8 η̄ 8
looooooomooooooon
loooooooomoooooooon looooooooooooooooooooooomooooooooooooooooooooooon loooooooooooooooomoooooooooooooooon
1 5
2 3 4

Things will still remain invariant under τ Ñ ´1{τ for the reasons “given
‰ above, now applied to both 3 and 4.
The only important subtlety is now in the p1, 1q sector the E8 θ̄6 11 will contribute a ´1 sign, as necessary
to cancel the twisted projected left-moving fermion sector.
Next, under τ Ñ τ ` 1, the exact same arguments apply to 3 and 4, namely the untwisted sector of the
left-handed fermions picks up ´1 phase as required to cancel with the η̄. The twisted sectors look like:
« ff8 « ff8 « ff8 « ff8 « ff4 « ff4 « ff4 « ff4
0 1 0 1 0 1 0 1
p0, 0q : θ̄ ` θ̄ ` θ̄ ` θ̄ Ñ θ̄ ` θ̄ ` θ̄ ` θ̄ ð p0, 0q
0 0 1 0 1 0 0 1
« ff6 « ff2 « ff6 « ff2 « ff6 « ff2 « ff6 « ff2 « ff6 « ff2 « ff6 « ff2 « ff6 « ff2 « ff6 « ff2
0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0
p1, 0q : θ̄ θ̄ ` θ̄ θ̄ ` θ̄ θ̄ ` θ̄ θ̄ Ñ ´iθ̄ θ̄ ´ iθ̄ θ̄ ´ iθ̄ θ̄ ´ iθ̄ θ̄ ð i ˆ p1, 1q
0 0 0 0 1 1 1 1 1 0 0 1 0 1 1 0
« ff6 « ff2 « ff6 « ff2 « ff6 « ff2 « ff6 « ff2 « ff6 « ff2 « ff6 « ff2 « ff6 « ff2 « ff6 « ff2
0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1
p0, 1q : θ̄ θ̄ ´ θ̄ θ̄ ` θ̄ θ̄ ´ θ̄ θ̄ Ñ θ̄ θ̄ ´ θ̄ θ̄ ` θ̄ θ̄ ´ θ̄ θ̄ ð p0, 1q
0 1 0 1 1 0 1 0 1 0 0 1 0 1 1 0
« ff6 « ff2 « ff6 « ff2 « ff6 « ff2 « ff6 « ff2 « ff6 « ff2 « ff6 « ff2 « ff6 « ff2 « ff6 « ff2
0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0
p1, 1q : ´ θ̄ θ̄ ´ θ̄ θ̄ ´ θ̄ θ̄ ´ θ̄ θ̄ Ñ iθ̄ θ̄ ` iθ̄ θ̄ ` iθ̄ θ̄ ` iθ̄ θ̄ ð i ˆ p0, 1q
0 1 0 1 1 0 1 0 1 1 0 1 0 1 1 0

So we get that the untwisted sector remains the same, while each of the two twisted sector components
change by a factor of i. This combines with what we know about the left-moving fermions to make every
combined contribution change with a ´ phase which exactly cancels the η-functions. The result is modular
invariant.
To verify the spectrum, as remarked in the text when we act by orbifold on the E8 ˆ E8 we break down
r120s ‘ r128s of Op16q. We get: r120s Ñ r3, 1, 1s ‘ r1, 3, 1s ‘ r1, 1, 66s ‘ r2, 1, 12s ‘ r1, 2, 12s and 128 Ñ
r2, 1, 32s ‘ r1, 2̄, 32s in SUp2q ˆ SUp2q ˆ Op12q.
The Z2 action takes the spinors of the two SUp2q subgroups to minus themselves, keeping the conjugate
spinors in variant. Projecting by this keeps r3, 1, 1s ‘ r1, 3, 1s ‘ r1, 1, 66s, r1, 2̄, 32s. This organizes into
r3, 1s ‘ r1, 133s ‘ r2, 56s P SUp2q ˆ E7 . Here 56 is the fundamental representation and 133 is the adjoint
representation of E7 .
Now let’s organize our coordinates into µ “ 2, 3 indicating the spatial coordinates in lightcone gauge, and
pair the remaining 6 coordinates into Z i “ ?12 pX 2i ˘ iX 2i`1 q, i “ t2, 3, 4u. Let’s organize the different sector
contributions based on how they transform under the Z2 twist:
• Untwisted Sector
– Left-handed side:
∗ NS - The zero-point energy is ´1{2 and we thus have massless states coming from single fermion
excitations.
µ 4,5
` : ψ´1{2 , ψ´1{2
6,7,8,9
´ : ψ´1{2

121
 ∗ R - The zero-point energy is 0 from equal number of bosons and fermions and our massless
excitation comes from the ground state. Under the rotation e2πips2 φ2 ´s3 φ3 q the ground states
organize as follows:

`: | 12 , 12 , 21 , 12 y , |´ 12 , ´ 12 , 12 , 12 y , | 12 , 21 , ´ 12 , ´ 12 y , |´ 12 , ´ 12 , ´ 12 , ´ 12 y
´: | 12 , ´ 12 , 12 , ´ 12 y | 12 , ´ 12 , ´ 12 , 12 y |´ 12 , 12 , 12 , ´ 12 y |´ 12 , 12 , ´ 12 , 12 y

Note we only have an even number of ` signs in any of the ground states by GSO projection.
These won’t matter for the massless bosonic spectrum.
– Right-handed side
The zero-point energy is ´1, so we either have a bosonic excitation:
µ 4,5
`: α̃´1 , α´1
6,7,8,9
´: α̃´1

Or a weight 1 excitation from the current algebra:

`: |a` y P r3, 1, 1s ‘ r1, 133, 1s ‘ r1, 1, 128s

´: |a´ y P r2, 56, 1s

So, the untwisted bosonic massless states must be the Z2 -invariant combinations of left (NS) and right
movers. We get
µ ν : G , B , Φ.
– ψ´1{2 α̃´1 µν µν
µ µ 4,5
– ψ´1{2 |a` y - vector boson in the adjoint of SUp2q ˆ E7 ˆ E8 . This combines together with ψ´1{2 α̃´1
4,5 µ
and ψ´1{2 α̃´1 to produce an extra U p1q4 .
4,5 4,5
– ψ´1{2 |ay Y ψ´1{2 α̃14,5 - complex scalar transforming in the adjoint of U p1q4 ˆ SUp2q ˆ E7 ˆ E8
6,7,8,9 6,7,8,9
– ψ´1{2 α̃´1 - 16 neutral real scalars.
6,7,8,9 ´
– ψ´1{2 |a y 4 real scalars transforming in the r2, 56, 1s representation of SUp2q ˆ E7 ˆ E8
Here Kiritsis does not mention the presence of the dilaton with the other 16 real scalars. I assume this
is an accidental omission.
• Twisted Sector
For the transformation g, we have 4 points on each T 2 that are equivalence classes with the transformed
point gx. This means that we have 4 ˆ 4 equivalence classes that we must include in the spectrum for
the twisted sector. This will be the same as looking at the spectrum for 1 class of twist and taking it
16-fold.
Equivalently, because fixed points correspond to the equivalence classes in this case, note that our
transformation has fixed points given by p0, 12 , τ22 , 12 ` τ22 q ˆ p0, 12 , τ23 , 21 ` τ23 q on the respective T 2 s. The
products give 16 fixed points. So we will have 16 copies of the spectrum at the fixed point p0, 0q on our
T 4 Appreciate this. Are you sure its not 32?
– Left side The bosonic oscillators will be shifted by 1{2
The fermionic oscillators will also be shifted by 1{2.
∗ NS - The zero-point energy is now ´ 14 ` 14 “ 0 and so we get only one ground state - the
vacuum.
∗ R - The zero-point energy remains zero. The zero modes that give the vacuum are now obtained
from ψ 2,3,4,5 . We thus get 2 ground states after GSO projection, which will end up giving us
the two requisite gravitinos
– Right side:
This is the hardest part. We use complex fermion language for the current algebra. We separate it
into two parts λ˘,1...8 , λ˘,9...16 . We get massless states from the pR, N Sq and pN S, N Sq states.

122
 ∗ (NS,NS) Here the ground state energy is ´1{2. We thus get the following states contributing:
6,7,8,9
α´1{2 , λ˘3...8
´1{2

The first one will get GSO projected out (as will anything with an even number of fermions).
The second one will transform as the r12s of SOp12q. In line with this, we can also construct
three other copies of r12s (or r12s):
λ˘3...8 ˘1 ˘2
´1{2 λ0 λ0
ISNT THIS 5?
The other state we can build that does not get GSO projected out is:
6,7,8,9 ˘,1,2
α´1{2 λ0

This gives 4 ˆ 2 copies of the r2s of SUp2q.

∗ (R, NS) Here the ground state energy is 0. We have zero modes coming from the 12 fermions
λ˘,3...8 giving 26 ground states giving the 32 and 32 spinors of SOp12q, one of which will get
projected out by GSO.
α0 alone will get GSO projected out, so does not contribute to the spectrum.
Together the two copies of r32s ` r12s ` r12s of SOp12q combine together to form the two copies of
the r56s of E7 and we get 8 copies of the 2 of SUp2q.
Altogether our gauge multiplets lie in 2 ˆ r1, 56, 1s and 8 ˆ r2, 1, 1s.
Thus we get the twisted bosonic states coming from |0yN S |ay giving us 32 scalars in the r1, 56, 1s and
128 scalars in the r2, 1, 1s.
The zero-point energy calculations are here:

7. Under τ Ñ τ ` 1 its quick to see that compactifying on any pd, d ` 16q Lorentzian lattice and orbifolding by
a Zn shift symmetry of {N will give a transformation
„  „ 
iπh2 2
N h 4πi{3 N h
τ Ñτ `1:Z “e e N Z
2
g h`g
where the first exponential factor comes from the η̄ ´16 and the second factor comes from shifting p2L ´ p2R
2
which is otherwise even by h{N which gives Nh 2 p2L ´ 2R q “ h2 2 {N 2 .
The τ Ñ ´1{τ phase „  „ 
h
N ´ 2πihg
2
N g
τ Ñ ´1{τ : Z Ñe N Z
h ´h

123
 can similarly be proven from straightforward Poisson resummation.
This problem specializes to N “ 2.
For 2 {2 “ 1 mod 4 the twisted sector picks up a phase under τ Ñ τ ` 1 and one can see that this phase is
`i, just as in the last problem. This is what was necessary to combine with the left-moving fermions to give
a modular invariance. Note this happens only when 2 {2 “ 1 mod 4.
Under τ Ñ ´1{τ the twisted sector’s projected part picks up a factor of ´1, exactly what we need to cancel
the ´1 on the left-moving side.
8. The partition function for our general heterotic N “ 2 compactification takes the form:
1
“h‰ “h‰ 1 θ 2 a θ a`h θ a´h
“ ‰ “ ‰ “ ‰
het 1 ÿ Γ2,18 g Γ 4,4 g 1 ÿ b b`g b´g
ZN “2 “ 8 24 4
2 h,g“0 τ2 η η̄ 2 a,b“0 η

We seek to compute τ2 B2 where B2 “ Trrp´1q2λ λ2 s over our string’s Hilbert space. To do this, consider the
following helicity generating partition function:
1
“h‰ “h‰ 1 θ a pνqθ a θ a`h θ a´h
“ ‰ “ ‰ “ ‰ “ ‰
1 ÿ Γ2,18 g Γ 4,4 g 1 ÿ b b b`g b´g
Zpν, ν̄q “ Trrq L0 q̄ L̄0 e2πiνλL ´2πiν̄λR s “ ¯
ξpνqξpν̄q (83)
2 h,g“0 8
τ2 η η̄ 24 2 a,b“0 η 4

Here
8
ź p1 ´ q n q2 sin πν θ11 pνq
ξpνq “ “
n“1
p1 ´ q n e2πinν qp1 ´ q n e´2πinν q π θ1 pνq
plays the role of exchanging the traces over the bosons in the non-compact spatial (3,4) directions with traces
that involve the helicity.
I apply formula D.21 in Kiritsis to simplify the theta functions to:
“a‰ “a‰ “a`h‰ “a´h‰
θ2 11 p ν2 qθ 1´h
1
“‰ “ ‰ ν “1`h‰ ν
1 ÿ θ b pνqθ b θ b`g θ b´g 1´g p 2 qθ 1`g p 2 q
4
“ (84)
2 a,b“0 η η4

This vanishes at least as fast as ν 2

We must now take (83) this and apply
´ 1 1 ¯2
Bν ´ B̄ν̄ Zpν, ν̄q.
2πi 2πi
Because our generating function (83) vanishes as ν 2 thanks to (84), we only need to look at Bν2 .
To obtain a nonzero result we thus need to act with Bν2 . On these terms for each h, g. First note that for
h“ 4 2
2
“1‰g “ 0 (84) vanishes as ν so will not contribute. For ph, gq ‰ p0, 0q, the terms vanish as ν due to the
θ 1 , exactly cancellable by taking two derivatives on that term. Thus, we need only worry about the zeroth
order behavior of everything else: ξ „ 1 and θ 1´h
“ ‰ “1`h‰ “1´h‰ “1`h‰
1´g θ 1`g pν{2q „ θ 1´g θ 1`g p0q. We are left with
“h‰
π 2 1 1 ÿ Γ2,18 g 16η 2 η̄ 2
„  „  „ 
1´h 1`h ` 1 ˘2
´ θ θ Bν θ
4 p2πq2 2 τ2 η 8 η̄ 20 θ2 1´h 2 1´h
“ ‰ “ ‰
1´g θ̄ 1´g
1´g 1`g 1 ν“0
h,g‰p0,0q
« “1‰ “1‰ “0‰ ff
4η 6 Γ2,18 1 Γ2,18 0 Γ2,18 1
“´ “0‰ ` “0‰ ´
2τ2 η 6 η̄ 18 θ̄2 10
“‰
2
θ̄ 0 2
θ̄ 1
“1‰ “1‰
Where we have used θ 2 “ ´θ 0 , as well as θ11 |ν“0 “ 2η 3 We now use the identity

θ̄2 θ̄3 θ̄4 “ 2η̄ 3

and recover
Γ2,18 11 θ̄2 θ̄3 Γ2,18 10 θ̄2 θ̄3 Γ2,18 01 θ̄3 θ̄4
“‰ “‰ “‰
τ2 B2 “ ´ ´ ` .
η̄ 24 η̄ 24 η̄ 24

124
 9. The gravitini can only come from the untwisted left-moving R sector (spinor spacetime index) tensored
2,3
with an α̃´1 on the right (vector spacetime index). The zero-point energy of the left-moving R sector
is 0 from equal numbers of bosons and fermions. Because our group acts on the (bosonized) fermions
the same way it acts on the bosons, we get that Z22 gives the three nontrivial elements given by rotations
e2πips1 φ1 ´s2 φ2 q , e2πips1 φ1 ´s3 φ2 q , e2πips2 φ1 ´s3 φ2 q , with φ0 corresponding to the spacetime fermions not appearing.
We see that the only spinors which are invariant under these three transformations take the form

|˘ 21 , 12 , 12 , 12 y , |˘ 21 , ´ 12 , ´ 12 , ´ 12 y

And we must have an even number of signs by GSO projection, so we in fact get two supersymmetries
2,3 1 1 1 1 2,3
preserved: α̃´1 | 2 , 2 , 2 , 2 y , α̃´1 |´ 12 , ´ 12 , ´ 21 , ´ 21 y, providing the ˘3{2 states only one gravitino.

10.

11. As before, the twist acts the same way on the bosons and (left moving) fermions. Already at this level, we
see that the only invariant states |s1 , s2 , s3 , s4 y must satisfy s2 “ s3 “ s4 so we will have the (GSO projected)
possibilities:
| 12 , 12 , 12 , 12 y , |´ 12 , ´ 12 , ´ 12 , ´ 21 y
providing again the ˘3{2 states of a single gravitino.
To avoid anomaly from ground state energy mismatch, we need the condition of Polchinski 16.1.28
4
ÿ 16
ÿ
ri2 ´ s2I “ 0 mod 2N
i“2 I“1

Here N “ 6. Note that our ri “ p1, 1, ´2q already sums to 6, so we must have the same for our si that
determines the Γ16 action.
I am confused why Kiritsis is saying there is only one such action of Z3 on Γ16 . As long as s2i “ 0 mod 6
ř
we should get a consistent theory, as shown in Table 16.1 of Polchinski.
The simplest such twist (aside from the trivial one that leaves the E8 ˆ E8 untouched) would be to act on
the first 3 complex fermions of forming the first E8 group in the same way as we act on the complexified
bosons and left-moving fermions, namely by

λ̃˘,1,2,3 Ñ e˘2πiβ1,2,3 λ̃˘,1,2,3 , β1 “ β2 “ 31 , β3 “ ´ 23

while the remaining λ˘,4...16 are left untouched. Let’s now get the massless spectrum under Z3 “ t1, r, r2 u

• Untwisted
– Left-moving The bosons are labeled by
• Twisted by r
• Twisted by r2

12. The Nijenhuis tensor is defined in terms of the almost-complex structure p1, 1q tensor Jji as

Nijk “ Jil pBl Jjk ´ Bj Jlk q ´ Jjl pBl Jik ´ Bi Jlk q “ Jil pBrl Jjs
k
q ´ Jjl pBrl Jisk q “ Jil p∇rl Jjs
k
q ´ Jjl p∇rl Jisk q

We are able to replace partial derivatives with covariant derivatives and vice versa because the non-tensoriality
can only enter through the Christoffel symbols Γ, as follows

´ Γqrljs q  
Jil pΓkrrl Jjs
r k l k r k k r l k r l 2 r k r k
Γkrl(pJ(jr(
Jil(
´(Jjl Jir q “ 0
(
((
 Jq q´Jj pΓrrl Jis ´ Γrlis
Jq q “ Γrrl Jjs Ji ´Γrrl Jis Jj “ p´q pδ
i Γ
rj ´ δj Γri q`(
 
  
So we see because everything is antisymmetrized that the Nijenhuis tensor is indeed a tensor.

125
13. It is easiest to directly construct coordinate patches on CPN . We define N such patches to consist of N
complex coordinates z1 “ Z1 {Zi . . . zi´1 “ Zi´1 {Zi , zi`1 “ Zi`1 {Zi , . . . zN “ ZN {Zi that are valid for all
parts of CPN where Zi ‰ 0. It is clear that these coordinates cover the whole manifold, since one Zi is
always not equal to zero, so any given point is always in a coordinate patch.
Moreover, the transition functions between different patches U, U 1 are simply fractional linear transformations
of the zi , zi1 , so are holomorphic. This is enough to give a globally defined complex structure (vanishing Nijk )
the the manifold.

14. We begin with the existence of a Killing spinor

r∇m , ∇n sξ “ Rrs,mn γ rs ξ “ 0

Now, multiplying by γ n , we get

0 “ γ n γ rs Rrs,mn ξ “ pγ nrs ` g nr γ s ´ g ns γ r qRrs,mn ξ

The first term vanishes by the Bianchi identity. The other two terms give:

2Rns γ s ξ “ 0
¯ r to get
Now we can multiply by ξγ
¯ r γ s ξ “ Rns J r
0 “ Rns ξγ s

Since the complex structure is invertible, this gives Rns “ 0, so indeed our space is Ricci flat.

15. To verify the masslessness of the graviton, it is enough to look at linearized gravity and confirm that the
perturbations satisfy the massless spin 2 condition.
To

16.

17.

18. The NSNS fields give a graviton, an antisymmetric tensor, and 81 scalars. From the RR sector we get
another scalar from the axion C2 , another 2-index antisymmetric tensor and 22 scalars from C2 . Finally the
self-dual 4-form C4 gives 19 anti-self-dual and 3 self-dual two-index antisymmetric tensors.
The supergravity multiplet contains two left-handed Weyl gravitini. The tensor multiplet contains two Weyl
fermions of opposite chirality from the gravitini.
In order to cancel anomalies, we must be able to apply the Green-Schwartz mechanism
Thus we get NT “ 21. Now IIB has a self-dual 5-form. For each of 20 2-cycles wrapped we get

126
Chapter 10: Loop Corrections to String Effective Couplings
It is not likely that this will be relevant to my research. I will skip it indefinitely for now.

1.

127
Chapter 11: Duality Connections and Nonperturbative Effects
1. Taking the expression for a toroidal heterotic compactification from exercise 9.1
» fi
2 1 ź 16 „  1 4 a
“ ‰
R ÿ ´ πR
|m`nτ | 2
´iπ
ř
nY I pm`nτ̄ qY I Y I 1
ÿ a I 1 1 ÿ θ b
–? e τ2 e I θ̄ pY pm ` τ̄ nq|τ̄ qfl ˆ 7{2
τ2 η η̄ 17 m,n 2 a,b“0 i“1 b 7 7 2
τ2 η η̄ a,b“0 η 4

Using θ function identitiess as in the second equation in appendix E, we get

1
a ´ 2nY I
2
„ 
R ÿ ´ πR |m`nτ |2 1
ÿ
iπmY I Y I n´iπbnY I
Γ1,17 pR, Y q “ ? e τ2 e θ̄
τ2 m,n 2 a,b“0 b ´ 2mY I

Now take Y I “ 0 for I “ 1 . . . 8 and Y I “ 1{2 for I “ 1 . . . 16. Then

I I I I 2 I
ź ř ř
eiπmY Y n´iπbnY “ eiπm I pY q ´iπb I Y “ 1
I

and we can ignore “this

‰ term. “ Similarly because we are taking a product over 16 θ̄, no phases will interfere
with us replacing θ uv with θ ´u
‰
´v for integer u, v. This gives us the desired first step

1 „ 8 „
a`n 8
2

´ πR |m`nτ |2 1 a
ÿ ÿ
Γ1,17 pR, Y q “ R e τ
2 θ̄ θ̄
m,n
2 a,b“0 b b`m
“ a`n ‰
Now again because we have enough θ b`m that phases do not interfere, we see that we only care about n, m
modulo 2 in the fermion term. We know how to divide the partition function of the compact boson into
parity odd and even blocks by doing the Z2 stratification corresponding to the πR translation orbifold of the
circle. This gives our desired answer:
„  ÿ „ 8 „
a`h 8

1ÿ h 1 a
Γ1,1 p2Rq θ̄ θ̄
2 h,g g 2 a,b b b`g

with
´πR2
ÿ „ 
Γ1,1 p2Rq “ 2R exp |2m ` g ` p2n ` hqτ |2
m,n
τ2

2. As before, take the ansatz

ds2 “ e2Aprq ηµν dxµ dxν ` e2Bprq dxi ¨ dxi , A012 “ ˘eCprq ñ Gr012 “ ˘C 1 prq eCprq

The BPS states in 11D require only the gravitino variation to vanish:
1 PQ 1 8
δψM “ BM  ` ωM ΓP Q  ` GP QRS ΓP QRS ΓM  ´ GM QRS ΓQRS 
4 2 ¨ 3! ¨ 4! 2 ¨ 3! ¨ 4!
We have worked out ω in 8.43.

ωµ̂ν̂ “ 0, ωµ̂î “ p´qµ“0 Bi A eA´B dxµ , ωîĵ “ Bj Bdxi ´ Bi Bdxj

Let’s look first at M “ µ parallel. Since  is Killing we expect no longitudinal variation and we get
1 1 1 1
Bµ ` A1 eA´B Γµ̂r̂  ˘
0 “ C 1 prqeC
Γr012
Γ
µ ¯ C prqeC Γµ Γr012 
2 2 ¨ 3! 3!
1 1
“ A1 eA´B Γµ̂r̂  ¯ C 1 eC´B´2A Γµ̂r̂0̂1̂2̂ 
2 3!
1
ñ 0 “ A1  ¯ C 1 eC´3A Γ0̂1̂2̂ 
3

128
If we would like these two terms to be proportional, then we should take C “ 3A, and we get the following
condition for 
p1 ¯ Γ0̂1̂2̂ q “ 0
So half the dimension of the space of spinors satisfies this at any given point. We thus get
For M “ i transverse, we recall Γij generates rotations, so assuming rotational invariance in the transverse
space, we’ll cancel this. We get
1 jk  1 
r012

1
Br  ` ω r
Γjk  ` Gr012
 Γ Γ r  ¯ Gr012 Γ012  “ 0
4
  2
 ¨ 3!
 3!
1
ñBr  ¯ Gr012 Γ012  “ 0
3!
e´3A 1 C 0̂1̂2̂
ñBr  ¯ Ce Γ 
3!
Solving this gives us that
prq “ eCprq{6 0
for 0 some constant spinor. We still do not have a relationship between C and B. This can be obtained by
not assuming rotational invariance but rather imposing cancelation of the second and third terms above as
follows:
1 1
Bj B Γîĵ  ˘ Bj C eC Γj012 Γi 
2 2 ¨ 3!
1 1
“ Bj B Γîĵ  ˘ Bj C eC´3A Γîĵ 0̂1̂2̂ 
2 2 ¨ 3!
1
ñ Bj B ` Bj C “ 0
3!
where we have used the condition on  already obtained. Thus C “ 3A “ ´6B. Finally Let’s look at G’s
equation of motion:
1 3
dG “ 0, d‹G` M N OP QRST GM N OP GQRST “ 0
3! p144q2
By assumption, the term quadratic in G vanishes. What remains gives us:

0 “ Br pe3A`8B e´6A´2B C 1 prqeC q “ Br pe´3A`6B`C C 1 q “ Br pC 1 e´C q ñ Br2 e´C “ 0

So we have that e´C “ Hprq as required, where

L6
Hprq “ 1 `
r6
I’m happy with this. I could use Mathematica to show that the other EOM:
ˆ ˙
1 2 2 1 P QR 1 2
RM N ´ gM N R “ κ TM N , κ TM N “ 4GM P QR GN ´ gM N G
2 2 ¨ 4! 2
is satisfied - but this is barely different from what I’ve done several times before for the D-branes and
fundamental string solutions in chapter 8.
As before, this generalizes straightforwardly to multi-membrane configurations.
32π 2 N `6s ‹G
The charge of the M2 brane with H “ 1 ` r6
is given by integrating 2κ211
on a seven-sphere at infinity.
Here 2κ211 “ p2πq8 `911 Asymptotically we will get the field strength going as
32 ˆ 6π 2 N `611
r6
π4
Altogether, using Ω7 “ 3 this gives a total charge of

π 4 32 ˆ 6π 2 N `611 N
9 “
3 8
p2πq `11 p2πq2 `211

129
 This is exactly consistent with 11.4.10-13, with µ “ N “ 1 corresponding to a single M2 brane.
Calculating the Ricci scalar curvature in fact gives a constant as r Ñ 0 so we do not encounter a divergence.
This signifies that this is just a coordinate singularity and we can extend past.

Finally, we can take the near-horizon limit and get

r4 L2 i
ds2 “ η µν dx µ
dx ν
` dx ¨ dxi
L4 r2
r4 L2
“ 4 ηµν dxµ dxν ` 2 dr2 ` L2 dΩ27
L r
?
Take now r “ L{ z to get the first term to look like 1{z 2 while not affecting the second term much:
1
pηµν dxµ dxν ` 4L2 dz 2 q ` L2 dΩ27
z2
We can rescale z, xµ and see that this geometry is AdS4 ˆ S 7

3. The M5 brane is now magnetically charged under C3 . Now the equations of motion d ‹ dC “ 0 are trivially
satisfied but the Bianchi identity is nontrivial, giving

L3
Br2 H “ 0 ñ H “ 1 `
r3
The metric form can be fixed by analyzing the gravitino variation similar to before. Longitudinally:
1 1
0 “ A1 eA´B Γµ̂r̂ ` C 1 eC`A´4B Γθ̂1 θ̂2 θ̂3 θ̂4 µ̂
2 2 ¨ 3!
1
ñ A1  ` C 1 eC´3B Γr̂θ̂1 θ̂2 θ̂3 θ̂4 
3!
We see that we must take C “ 3B and A “ ´C{6, and we get the half-BPS condition:
ˆ ˆ
p1 ´ Γ7̂8̂9̂1011 q “ 0

The transverse components will give the profile for .

1
Br  ` C 1 eC´3B Γθ̂1 θ̂2 θ̂3 θ̂4 r̂ 
2 ¨ 3!
and this gives a profile
 “ e´C{12 0
The membrane charge is given by integrating G on a 4-sphere whose area is given by 8π 2 {3, so we get

8π 2 3πN `311 N
9 “
8
3 p2π q`11 p2π`11 q5 `11

Again we get that the Ricci scalar tends to a constant as r Ñ 0, giving regularity at the horizon. Again,
this signifies that this is just a coordinate singularity and we can extend past.

130
 Taking the near-horizon limit we arrive at
r L2 r L2
ds2 “ ηµν dxµ dxν ` 2 dxi ¨ dxi “ ηµν dxµ dxν ` 2 dr2 ` L2 dΩ24
L r L r
Now take r “ L{z 2 yielding
1
pηµν dxµ dxν ` 4L2 dr2 q ` L2 dΩ24
z2
so again after rescaling the same was as before we get AdS7 ˆ S 4 .
As before, a solution can consist of an arbitrary number of M 5 branes at different places, in which case we
get
ÿ Li
Hprq “ 1 `
i
|r ´ ri |3
This remains half-BPS.
4. First look at the field strengths. The general M5 brane solution For a uniform distribution of M5 charges,
we know that in the transverse (3D) space the potential must now decay as
ż
L 2L
H “ 1 ` dx11 11 2
“1` 2
|~r ´ x ê11 | r10D
where L depends on the density of the distribution. Then the 3-form field strength in 10D will just be

pdBqabc “ abce Be H

Given this source in 10D, we have already worked out Einstein’s equations in Chapter 8. Another way to
see this is that we remain half-BPS after adding even an infinite number of parallel branes.
We have that e4Φ{3 “ G11,11 so that eΦ “ H 1{2 consistent with the NS5 solution.
Using the perscription of dimensional reduction in appendix I.2, we take eσ “ e2Φ{3 “ H 1{3 . Using gµν “
S , we see that multiplying by H 1{3 takes us to the string frame NS5 metric solution.
e´σ gµν

ds2 “ ηµν dxµ dxν ` Hprqdxi ¨ dxi

131
 This is exactly the NS5 metric in string frame.
S “ eΦ{2 g E and multiply the string frame by e´Φ{2 “ H ´1{4 to get us to the Einstein
We can further take gµν µν
frame.

5. Recall the BPS D6 brane in 10D is described by

L
H ´1{2 ηµν dxµ dxν ` H 1{2 d~x ¨ d~x, H “1` , L “ gs `s N {2, F “ L dΩ2 , eΦ “ gs2 H ´3{4
|x|

This means that e´2Φ{3 “ H 1{2 . Multiplying ds2string by this factor, we the 10D part of 11D metric

ηab dγ a dγ b ` V d~x ¨ d~x

Here we’ve picked notation consistent with the problem so that γ 0...6 “ x0...6 , Hprq “ V prq, and xi is the
same.
Note also that ż
1 L4π
F “ “ nTp ñ 2L “ `s ngs
2κ210 S2 p2πq7 `8s gs2
This should be supplemented by the metric component in the internal 11th dimension, given by e4Φ{3 pdγ `
Aµ ¨ d~xq2 “ V ´1 pdγ ` Aµ ¨ d~xq2 where Aµ is the 10D gauge field generated by the monopole solution.
Now A cannot be globally defined because of the monopole. Given L “ 2N , it takes the same form as Aµ
does in 3D about a monopole of charge n “ N {`s .
We could have taken a more “active” approach, demonstrating that this metric ansatz does indeed solve
Einstein’s equations, and shown that for the field strength to satisfy the Bianchi identity in this geometry it
needed to indeed be a harmonic function of the transverse coordinates taken with flat metric.

6. The DBI action for a two-brane in flat space with vanishing B-field and constant dilaton is given in euclidean
signature as ż ż
a
´T2 d x detpδab ` Ba X Bb X ` 2π`s Fab q ` i C p3q ^ TrreF s ^ G,
3 µ ν 2

where the second integral consists of Chern-Simons terms that we will ignore in this argument. We can work
with the field variable F rather than A by imposing the Bianchi identity “by hand”, namely writing the
(non-CS) part of the action as
ż „ 
3
a
µ ν 2
i abc
´T2 d x detpδab ` Ba X Bb X ` 2π`s Fab q ` λ Ba Fbc
2

132
 This last term can just as well be integrated by parts to give abc Ba λFbc .
We now introduce an auxiliary V variable to rewrite the action as
ż „ 
3 1 µ ν 2 11 i abc
´ T2 d x V detpδab ` Ba X Bb X ` 2π`s Fab q ` `  Ba λFbc
2 2V 2
ż „ 
3 1 1 2 2 2 11 i abc
“ ´T2 d x V p1 ` p2π`s q Fab ` . . . q ` `  Ba λFbc
2 2 2V 2
here . . . involves terms depending on the Ba X µ . The equations of motion for F then give

abc Ba λ
Fab “ ´i
p2π`2s q2 V
Substituting this back in gives
ż „ 
3 1 1 2 ´2 2 11
´T2 d x V p1 ` p´ ` 1qp2π`s q pBλq ` . . . q `
2 2 2V
Integrating out V gives us the square root action again, but now with F replaced by Bλ, a new coordinate
ż a
´T2 d3 x detpδab ` Ba X µ Bb X ν ` p2π`2s q´2 Ba λBb λq

Taking X “ λ{2π`2s gives our desired result

I have only shown classical equivalence. How to I prove this is quantum-mechanically true as
well?

7. We are looking at the transformation τ Ñ ´1{τ . We see that

´1 ´C0 ` ie´Φ
C0 ` ie´Φ Ñ “
C0 ` ie´Φ C02 ` e´2Φ
C0 e´Φ
So we see C0 Ñ ´ C 2 `e ´Φ Ñ
´2φ and e C02 `e´2Φ
. On the other hand, C0 will not affect the C2 , B2 transfor-
0
mations. Nor will it affect C4 , which remains invariant
In the Einstein frame the metric is invariant. That means that e´Φ{2 gstring is invariant, which means gstring
transforms as e´Φ{2 times the Einstein frame metric. Consequently, in the string frame gstring
1 “ e´Φ gstring
(I think Kiritsis is wrong here, and Polchinski agrees with this)
Am I missing anything with that last one?

8. There’s effectively nothing to derive. Translating the Einstein frame means multiplying all lengths by e´Φ{4 .
´1{4 ?
At fixed dilaton this is gs . Given `2s in the denominator will then contribute a factor gs overall, that’s
exactly what was done here.

9. We have that C4 is invariant. That means that objects charged under C4 remain charged under C4 , with the
same charge. These are precisely the D3/anti-D3 branes. Now recall the DBI action has coupling constant
1
gY2 M “ “ 2πgs
p2π`2s q2 T3
note that this is dimensionless, as it should be for a gauge theory in 4D. At low energies, the closed strings
decouple we can reliably trust the DBI action, considering the D-brane gauge theory on its own. In the
absence of axion, the SLp2, Zq of IIB takes gs Ñ 1{gs . This corresponds to

4π 2
gY2 M Ñ
gY2 M

So this is the Weak-Strong Montonen-Olive duality of N “ 4 SYM.

133
 The only subtlety is that one must take care to include the Chern-Simons terms in the DBI action in order
to get the full duality, specifically ż
C0 TrrF ^ F s.

At fixed C0 “ θ{2π this produces the instanton number. The duality C0 Ñ C0 ` 1 is a bona-fide duality of
the N “ 4 theory, a consequence of the fact that instanton charge is quantized.
Is there anything else that I can say that constitutes any form of “showing” that this fact is
true? The only thing is I think I’m assuming that the D3 brane is the only object charged
under C3 at leading order in `s . Can I safely assume this?

10. I’ll start from the F1 string rather than the D1, not that it matters. Let us look at the macroscopic solution
in the Einstein frame, so we multiply the string frame solution obtained in the chapter 8 exercises by H 1{4 .
We get:
L6
ds2E “ H ´3{4 p´dt2 ` pdx1 q2 q ` H 1{4 d~x ¨ d~x, H “1` 6
r
2κ2 T
Here L6 “ 6Ω 10 F 1
7
“ 32`6s gs2 π 2 Note this is the same metric as the D1 solution, and indeed the metric will
stay the same for all pp, qq strings.
The C0 field has been set to zero. For F1 the dilaton and B-field have the profile

eΦ “ gs H ´1{2 , B01 “ H ´1

and indeed the dilaton has the inverse of this for the D1 while B and C exchange. Indeed, consider the
SL2 pZq action ˆ ˙
a b
Λ“ .
c d
Here, we have ad ´ bc “ 1, implying c, d are relatively prime. This will correspond to the fact that pp, qq
bound states only exist for p, q relatively prime, since otherwise there is a decay process of marginal instability
allowing the pp, qq system to separate into two or more sub-systems. Further S “ C0 ` ie´φ and C2 , B2
transform as ˆ ˙ ˆ ˙ ˆ ˙ˆ ˙
aS ` b B2 T ´1 B2 d ´c B2
SÑ , Ñ pΛ q ¨ “
cS ` d C 2 C 2 ´b a C2
There is a subtlety in the problem, which resolved the ambiguity in our choice of Λ. I learned of it from
reading arXiv:hep-th/9508143. The subtlety is as follows: We need to fix the dilaton’s asymptotic value as
r Ñ 8 so as to define the vacuum of our string theory. First, consider φ, C0 “ 0 asymptotically, i.e. S Ñ i.
We then stay within the SOp2q Ă SL2 pRq that fixes S “ i. We want to take p1, 0q to the string p, q. This is
now uniquely determined: ˆ ˙
1 p ´q
Λ“ a
p2 ` q 2 q p
Applying this to B2 , C2 given that we start with only NS charge p1, 0q gives
ˆ ˙ ˆ ˙
B2 H ´1 p
“a
C2 p `q q
2 2

Upon doing this, the B2 , C2 fluxes will have coefficients that get modified from just p, q by a factor of
? 1 , so will no longer be integers satisfying the quantization condition. We can fix this by modifying
p2 `q 2
a
T Ñ Tp,q “ p2 ` q 2 T . Since this only serves to modify L, which was an arbitrary parameter of the classical
solution, this still remains a valid solution.
L6 6 2κ2 Tp,q
a 2
2 ` q 2 2κ T1,0 “
a
This means: Hp,q “ 1 ` rp,q 6 , Lp,q “ 6Ω7 “ q 1 2 6Ω7 q12 ` q22 L6 .
Our solution is now:
ˆ ˙ ´1 ˆ ˙ 1{2
B2 Hp,q p ipHp,q ´ q
ds2E “ Hp,q
´3{4
p´dt2 ` pdx1 q2 q ` Hp,q
1{4
d~x ¨ d~x “a S “ χ0 ` ie´φ “
C2 p `q q
2 2 1{2
iqHp,q ` p

134
 Note that as r Ñ 8, S Ñ i as we expect. Now, let us generalize this for different asymptotic values of the
dilaton and axion. After applying Λ, we can further apply
ˆ ´φ {2 ˙
1 e 0 χ0 eφ0 {2
Λ “
0 eφ0 {2
S now asymptotes to
e´φ0 {2 i ` χ0 eφ0 {2
“ χ0 ` ie´φ0
0 ` eφ0 {2
exactly as we want. To get the right final field strengths, take Λ initially arbitrary:
ˆ ˙
T ´1 cos θ ´ sin θ
pΛ q “
sin θ cos θ
Again, applying this will break our quantization condition. Now, the electric charges transform contragradi-
ently from the fields strengths, which means that
ˆ ˙ ˆ ´φ ˙ ˆ ˙
QB e 0 cos θ ` χ0 sin θ 1 p
“ eφ0 {2 “: a
QC sin θ ∆p,q q
We can solve this to get
eφ0 {2 ´φ0 eφ0 {2 eφ0 {2
sin θ “ a e q ñ cos θ “ a pp ´ χ0 qq ñ eiθ “ a pp ´ Sqq
∆p,q ∆p,q ∆p,q
1{2
The asymptotic value of the charges of B2 , C2 is thus given by pp, qq{∆p,q . Unimodularity gives:
1 “ eiθ e´iθ ñ ∆p,q “ eφ0 |p ´ qS|2 “ eφ0 pp ´ qχ0 q2 ` e´φ0 q 2
This coincides with the invariant
ˆ 2 ˙ˆ ˙
|S| S1 p
“ eφ0 pp ´ qχ0 q2 ` e´φ0 q 2
` ˘
p q S2´2
S1 1 q
So in full generality we get the tension:
b
Tp,q “ eφ0 pp ´ qχ0 q2 ` e´φ0 q 2 TF 1
1
Where TF 1 “ 2π`2s
is the tension in the string frame.
Because (aside from redefining L) the metric is unchanged, the singularity structure of pp, qq strings is no
different from p1, 0q or p0, 1q strings. Neither of these has a regular horizon. Confirm
11. First, by naive reasoning - there is no reason to write the full effective action to see what β should look
like. From the perspective of IIA in the string frame, we have coupling gA “ R̃11 {`s “ R11 τ2 {`s “ R11 {`s gB .
Recognizing RB “ `2s {R1 1 we can write this as RB`sgB . Because the translation of metrics between the 11-D
´ ¯2{3
2{3 `s
frame and the standard string frame in IIA involves the factor gA we get β “ RB gB . What about
conversion to the Einstein frame?

135
Now let’s do it the long way. The 11-D SUGRA Lagrangian is
„ 
1 1 2
LD“11 “ 2 R ´ |G4 | ` G4 ^ G4 ^ Ĉ3
2κ11 2

In this problem we’ll ignore the Chern-Simons terms.

Let’s take M-theory to 9 dimensions. The metric takes the form
˜ ¸
gµν ` Gαβ Aαµ Aβν Gαβ Aβµ
Gµ̂ν̂ “
Gαβ Aβν Gαβ

Here
eσ
ˆ ˙
1 τ1
Gαβ “
τ2 τ1 |τ |2
?
with eσ “ det G the Kähler parameter (area) of the torus. The metric’s R term becomes:
1 σ” µ 1 µ αβ 1 A α A µν β
ı
e R ` B σBµ σ ` Bµ G αβ B G ´ G αβ Fµν F .
2κ211 4 4
A “ B Aα ´ B Aµ α. We now have two U p1q field strengths. Using the fact that we have the explicit
Here Fµν µ ν ν
form of the torus metric, we can further write this as:

p2πR11 q2 τ2 σ 1 pBτ1 q2 1 pBτ2 q2 1 eσ ` A,1

„ ˆ ˙ ˆ A,1 ˙
1 2 A,2
˘ 1 τ1 F
e R ` pBσq ´ ´ ´ F F 2 (85)
8 9
p2πq `11 2 2 τ2 2 2 τ2 2 2 ¨ 2! τ2 τ1 |τ | F A,2

The kinetic 3-form potential yields a four-form, two three-form, and a two-form field strength:
2 τ
R11 ” 1 1 1
2 σ p3q µνρ p2q µν
6 9 e ´ p4q
Fµνρσ F p4q µνρσ ´ Gαβ Fµνρα
p3q
F β ´ Gαβ Gγδ Fµναγ Fβδ s
p2πq `11 2 ¨ 4! 2 ¨ 3! 2 ¨ 2!

Again, using the explicit form of the metric we can write this as
2 τ
ˆ ˙ ˆ p3q ˙
R11 2 σ
” 1
2 1 e´σ ` p3q ˘ 1 τ1 F 1 e´2σ p2q p2q µν
6 9 e ´ |F4 | ´ F H p3q
2 p3q ´ 2 p|τ |2 ´ τ12 qFµν12 F 12 s
p2πq `11 2 2 ¨ 3! τ 2 τ1 |τ | H 2 ¨ 2! τ2
2 τ ” eσ
ˆ ˙ ˆ p3q ˙
R11 2 2 1 1 ` p3q p3q
˘ 1 τ1 F e´σ 2
ı
“ ´ |F 4 | ´ F H 2 ´ |F2 |
p2πq6 `911 2 2 ¨ 3! τ2 τ1 |τ | H p3q
2
p2q p2q
Here the last 2-form field strength is defined as Fµν :“ Fµν12 .
This action not in any standard frame. Let’s take it to the Einstein frame g11 “ e´2{7σ gE :
2 τ ” 1 pBτ1 q2 1 pBτ2 q2 1 e9σ{7 ` A,1
ˆ ˙ ˆ A,1 ˙
R11 2 3 2 A,2
˘ 1 τ1 F
R ´ pBσq ´ ´ ´ F F
p2πq6 `911 7 2 τ22 2 τ22 2 ¨ 2! τ2 τ1 |τ |2 F A,2
e6σ{7
ˆ ˙ ˆ p3q ˙
e´12σ{7 2 1 e´3σ{7 ` p3q p3q
˘ 1 τ1 F 2
ı
´ |F2 | ´ F H ´ |F 4 |
2 2 ¨ 3! τ2 τ1 |τ |2 H p3q 2

The IIB SUGRA Lagrangian in the string frame is

„ „  
1 ´2Φ 2 1 2 1 2 1 2 1 2
e R ` 4p∇Φq ´ |H3 | ´ |F1 | ´ |F3 | ´ |F5 |
2κ210 2 2 2 4

supplemented by ‹F5 “ F5 . Taking this to 9 dimensions, the NSNS terms become

2πRB ´2φ
”
2 2 e2ρ A 2 1 2 e´2ρ 2
ı
2 e R ` 4p∇φq ´ pBρq ´ |F | ´ |H3 | ´ |H2 |
p2πq7 `8s gB 2 2 2

136
with G10,10 “ e2ρ , φ “ Φ ´ 21 ρ. The RR forms give

RB ρ
” 1
2 1 2 e´2ρ 2 1 2 e´2ρ 2 ı
e ´ F ´ F ´ F ´ F ´ F .
2
p2πq6 `8s gB 2 1 2 ¨ 3! 3 4 2 looomooon
4 ¨ 5! 5 4 ¨ 5! 4
dualize

Here F2 comes from F3 and F4 from F5 . We can dualize the 9D F5 to give the canonical normalization to
the F4 term.
RB ” eρ eρ e´ρ e´ρ ı
2 2 2 2
2 ´ |F1 | ´ |F3 | ´ |F2 | ´ |F4 | .
p2πq6 `8s gB 2 2 2 2
It is important to T-dualize this to get to IIA. This takes φ Ñ φ, ρ Ñ ´ρ, Gp9q Ñ Gp9q and also swaps H2
and F A . Lastly, we have gB 2 {R “ g 2 {R . We then get
B A A
«
RA ´2φ
”
2 2 e2ρ A 2 1 2 e´2ρ 2
ı
LIIA “ 2 e R ` 4p∇φq ´ pBρq ´ |F | ´ |H 3 | ´ |H2 |
p2πq6 `8s gA 2 2 2
ff
e´ρ 2 e´ρ 2 eρ 2 eρ 2
´ |F1 | ´ |F3 | ´ |F2 | ´ |F4 |
2 2 2 2

Now let’s take this to the Einstein frame gS “ e4{7φ gE :

«
E RA 4 2 2 e2ρ´4φ{7 A 2 e´8φ{7 2 e´2ρ´4φ{7
LIIA “ 2 R ´ p∇φq ´ pBρq ´ |F | ´ |H3 | ´ |H2 |2
p2πq6 `8s gA 7 2 2 2
ff
e´ρ`2φ eρ`10φ{7 e´ρ`6φ{7 eρ`2φ{7
´ |F1 |2 ´ |F2 |2 ´ |F3 |2 ´ |F4 |2
2 2 2 2

Comparing |τ1 | with |F1 |2 since these are the only two scalars that aren’t minimally coupled, we get ´ρA `
2φ “ ´2 logpτ2 q. T-dualizing to get back to IIB gives ρB ` 2φ “ 2ΦB “ ´2 log τ2 implying that τ1 “ C0 and
τ2 “ e´Φ in IIB as required.
Comparing the F4 coefficient gives ρA ` 2φ{7 “ 6σ{7. This gives σ “ 43 ρA ` 13 ΦB “ ´ 43 ρB ` 13 ΦB . This gives
A3{2 g ´1{2 „ RB´2
, close to what is desired. Expressing the relevant quantities in terms of the fundamental
units of their respective frames, this gives our desired relationship

`2s A3{2 1 R113

2 “ ñ 2 “
RB p2π`11 q3 g 1{2 RB `5s g 5{2

Off by a factor of g 1{2

Note that in the IIA action, F A and F2 have coefficients that differ by ´ρA ` 2φ “ 2ΦB . We should thus
identify them with e9σ{7˘ΦB of the M theory action. This implies that 9σ{7 “ 23 ρA ` 3φ{7, exactly what
we got from the F4 coefficient. The same argument for the F3 , H3 terms in both theories gives the same
difference between them, and their average gives the same relationship. Finally, the lone H2 term in IIA
compared to the F2 gives the same dependence as well, giving three nontrivial checks that what we’ve done
is correct.
Finally let’s get the conversion factor. To go from 11D to the string frame we must do e4{7φ e2{7σ . We now
understand σ “ ´ 43 ρB ` 13 ΦB and φ “ Φ ´ ρ2B we get the relationship
ˆ ˙2{3
2 4 2 `s
σ ` φ “ ´ pρB ´ ΦB q ñ β “
7 7 3 RB gs

as required. The dilaton dependence is flipped, fix!

137
12. There is a subtlety in this problem involving the form of the metric. Recall that the Einstein frame metric
gE gets mapped to itself under S-duality gE “ gE 1 . This implies that the string frame metric g “ eΦ{2 g is
S S
related to its S-dual by:
1
gS “ e´Φ gS1
We can verify this at the level of the solutions to the string equations of motion:

D5 : ds2E “ H ´1{4 dxk ` H 3{4 dxK , ds2S “ H ´1{2 dxk ` H 1{2 dxK , eΦ “ gs H ´1{2
NS5 : ds2E “ H ´1{4 dxk ` H 3{4 dxK , ds2S “ dxk ` HdxK , eΦ “ gs H 1{2

We see that the string frame metric are related in this way except for the issue of rescaling by gs . This means
we should redefine length so that ds2S asymptotes to gs ηµν for the NS5 metric why don’t we modify D5
instead?.

First, let’s calculate the energy of the F1 string stretched between two D5 branes. Directly from the Nambu-
Goto action, noting that the parallel X µ will be along the τ direction while the transverse X i will be along
the σ direction we can write
d
Bτ X µ Bτ Xµ Bτ X µ Bσ Xi pGµi ` Bµi q
ż ż
a 1
SN G “ ´TF 1 d2 ξ detpGab ` Bab q “ ´ d2
ξ µ
2π`2s Bτ X Bσ Xi pGµi ` Bµi q Bσ X i Bσ Xi
ż
1 a a
´1{2 B X µ B X µ H 1{2 B X i B X i
“´ dσdτ H τ τ σ σ
2π`2s
żπ ż
1
“´ 2
dσ|Bσ X i | dτ |Bτ X µ |
2π` s 0
loooooooooomoooooooooon
mS

This gives the string and Einstein frame mass:

żπ
1 g 1{4
mS “ |Bσ X 1 |dσ ñ mE “ ∆x1
2π`2s 0 2π`2s

For the D1 stretching the two NS5s, we apply the same logic to the DBI action:
ż ż ż
2
a 1 ´Φpxi q
a a
SDBI “ ´ d ξ TD1 ´ detpGab ` Bab q “ ´ 2
dσe Bσ X Bσ Xi dτ Bτ X µ Bτ Xµ
i
2π`s
? ż ż
g a
“´ 2g
dσ|Bσ X i | dτ Bτ X µ Bτ Xµ
2π` s
looooooooooooomooooooooooooon
mS

138
 Again we get string and Einstein frame mass:
ż
1 1
mS “ 2 ? dσ|Bσ X 1 | ñ mE “ ∆x1
2π`s g 2π`s g 1{4
2

The masses agree under S duality: g Ñ 1{g.

13. The argument will go very similar to how it did for the string-like objects. Again, call pp, qq “ p1, 0q the NS5
brane (magnetically charged under B2 ) with pp, qq “ p0, 1q the D5 brane (magnetically charged under C2 ).
Again, first take the axio-dilaton S to asymptote to i. The NS5 solution in the Einstein frame is:
L2
ds2E “ H ´1{4 ηµν dxµ dxν ` H 3{4 d~x ¨ d~x, H “1`
r2
2κ210 TN S5
Here L2 “ Q 2Ω3 “ Q`2s . We also have

eΦ “ gs H 1{2 , pdBqθφψ “ ´Br H

This time, the magnetic charges transform in the same way as the field strengths (since they are associated
with the Bianchi identity, not the EOMs), giving
ˆ ˙ ˆ ˙ ˆ ˙
QB eφ0 {2 cos θ 1 p
“ φ {2 ´φ {2 “: a
QC χ0 e 0 cos θ ` e 0 χ0 sin θ ∆p,q q

Solving this gives

e´φ0 {2 eφ0 {2 eφ0 {2

cos θ “ a p ñ sin θ “ a pq ` pχ0 q ñ eiθ “ i a pq ` Spq
∆p,q ∆p,q ∆p,q

Unimodularity gives
∆p,q “ eφ0 |q ` pS|2 “ eφ0 pq ` pχ0 q2 ` e´φ0 p2

We thus get that the 5-brane tension in the Einstein frame satisfies a similar relation to the case of 1-branes:
b
Tp,q “ e´φ0 p2 ` eφ0 pq ` pχ0 q2 T
1
with T “ the appropriate dimensionful constant.
p2πq5 `6s
a a
For the general pp, qq-brane solution, we get Lp,q “ ∆p,q L “ ∆p,q `2s

14. We’re going to work in the Einstein frame. After compactifying on T 2 we will get scalars not just from the
φ and C0 term but also from C2 , B2 , and the 3 metric components Gαβ .
The torus moduli in 14 Bµ Gαβ B µ Gαβ will take the same form as in Equation (85), namely

pBT q2 1 µ αβ 1 pBτ12 q 1 pBτ12 q 1 pBT q2

` Bµ G αβ B G “ ´ ´ `
T2 4 2 τ22 2 τ22 2 T2
a
Here T “ det Gαβ is the Kähler modulus and does not belong to the SLp2, Rq{U p1q coset. We have that
2
the axio-dilaton is ´ 12 |BS|
S2
. The scalars coming from B2 , C2 give kinetic terms:
2

1 G11 G22 ´ pG12 q2 1 S2 1 pBµ C12 q2

´ pS2 Bµ B12 q2 “ ´ pBµ B12 q2 , ´
2 S2 2 T2 2 T 2 S2
Altogether the scalars have appear as:
1 pBτ12 q 1 pBτ12 q 1 pBT q2 1 |BS|2 1 S2 1 pBµ C12 ` C0 H3 q2 ı
ż
? ”
d8 x ´g T R ´ ´ ` ´ ´ pB µ B 12 q2
´
2 τ22 2 τ22
looooooooooomooooooooooon 2 T2 2 S22 2 T2 2 T 2 S2
loooooooooooooooooooooooooooooooooooooomoooooooooooooooooooooooooooooooooooooon
SLp2,Rq{U p1q SLp3,Rq{SOp3,Rq

139
 1 pBT q2 2
Taking things to the new Einstein frame will get rid of the T out front, and modify 2 T2 Ñ ´ 23 pBT
T2
q

It remains to find the metric for SLp3, Rq{SOp3q. Because SOp3q is maximally compact, we can write the
metric on this space in a set of global coordinates known as Borel gauge. This is given by taking the Einbein
on T 3 symmetric space to be the exponentiation of the SLp3, Zq Borel sub-algebra: L “ exprχi Ei s exprφi Hi s.
From this, the T 3 metric is M “ LLT , and the kinetic terms are then TrrBµ M B µ M´1 s. By choosing the χi
and φi judiciously we see

here et “ T, eΦ “ S2´1 . It is worth stressing that this exactly recovers our kinetic terms. Everything matches
perfectly.

15. The the field strengths coming from the two-forms yield the following terms in the Einstein frame Lagrangian
ż
? ” 1 |F ` C H |2 1 1 |F3Ð5 |2 ı
3 0 3
d8 x ´g T 2{3 ´ ´ S2 |H3 |2 ´
2 S2 2 2 T2
The two-form field strengths F3 , H3 are unaffected by dualities of the torus. F5 can be dualized to an F3
as well, and we also get a further F3Ð5 by wrapping the D3 around the torus, which will combine with the
2
F3Ð5 to give a single (canonically normalized) field strength invariant under symmetries of the torus. Thus,
the F3 , F3Ð5 , H3 are invariant under the SLp2, Zq part of the U-duality group involving τ1 , τ2 .
We can indeed write these terms in a manifestly SLp3, Rq-invariant form, namely as
¨ ˛
H3
1 ` 1
˘
´ H3 F3 F3Ð5 M ˝ F3 ‚
2 1
F3Ð5

Here though, we should take care that it is really F3Ð5 ` B12 F3 ` C12 H3 that forms the kinetic term of the
action. Understand this, as well as the C0 H3 in the Einstein frame generally.

16. The metric will contribute 6 scalars while the 3-form C3 will contribute a seventh. We understand how to
generally build T 3 metrics from the last problem. Indeed, L there is the einbein not on the symmetric space
itself but on the torus T 3 . Given a Borel subgroup of SLp3, Rq, the einbein for the unit torus is specified by
three twist “axion” parameters χ1 , χ2 , χ3 and two dilaton parameters φ1 , φ2 as:
¨ ˛ ¨ φ1 {3´φ2 {2 ˛
1 χ1 χ2 e 0 0
L “ exprχi Ei s exprφi Hi s “ ˝0 1 χ3 ‚˝ 0 eφ1 {3`φ2 {2 0 ‚
0 0 1 0 0 e 2 {2
´φ

140
 We see directly that the parameters of this three-torus coincide exactly with the scalars C0 , C12 , B12 , T, Φ in
IIB compactified on T 2 from the prior problem.
The 3-torus volume parameter, which we will call T (not to be confused with T ) in the prior problem,
together will have kinetic terms

pBT q2 1 pBT q2 1 |C0Ð3 |2 2 pBT q2 1 |C0Ð3 |2

ż ż
8? 8?
d ´g T rR ` ´ ´ s “ d ´g T rR ` ´ s
T2 3 T2 2 T2 3 T2 2 T2
Taking this to the Einstein frame:

1 pBT q2 1 |C0 |2
ż
?
d8 ´grR ´ ´ s
2 T2 2 T2

This is exactly the SLp2, Zq-invariant action, which came from the perturbative T -duality in the earlier
problem. We see they are neutral under SLp3, Zq, while the other 5 belonging to the SLp3, Rq{SOp3q coset
are neutral under this SLp2, Zq. This re-derives the results for scalars of Section 11.6.
From the M-theory perspective, the three distinct 2-form potentials come from wrapping the C3 around
different T 3 cycles from 11D.

17. From M-theory, the 3-form C p3q descends directly down to a 3-form in the 8D picture. This has a field
strength G4 with kinetic term ż „ 
8 ? 1 2
d x g T ´ |G4 |
2
which is the same in both the original and the Einstein frames. We could have started with ‹G4 in 11D,
giving the 8D action: ż „ 
8 ? ´1 1 2
d x gT ´ |G4 |
2
The Chern-Simons term further contributes a further topological piece:
ż
?
d8 x g C0 T G4 ^ G4

Summing these all together gives the standard SLp2, Rq invariant bilinear form. Thus, SLp2, Rq acts by
electric magnetic duality, transforming the tuple pG, ‹Gq in the 2 representation.
Slightly incomplete, understand the origin of the action better. Look at 9506011

18. Taking IIB down to 5D and looking at conserved vectors (coupling to point-like objects). First, note we can
wind any of the pp, qq strings around any of the 5 cycles of T 5 , giving 10 vector currents. We also get 5
KK currents from the dimensional reduction that are T -dual
` to
˘ the string modes and together forms a 10 of
SOp5, 5q. We also have the D3 brane winding around any 53 “ 10 cycles. Finally, the D5-brane and NS5
can wrap the torus giving an additional 2 charges. The NS5 is a singlet of SOp5, 5q. The D-branes are all
T-dual and give a 16-dimensional representation, which is either the spinor or conjugate spinor depending
on whether we start from IIA or IIB.
Altogether we get 1+10+16. This is exactly the 27 representation of E6 under U -duality. This gives a
total of 27 point-like charges, which are the 27 different electric charges than can be carried by black holes
in 5D.

19. Note that rescaling the string length by eγ will correspond rescaling the metric by e´2γ . So relationships
between the string lengths are inverse-square-root proportional to the relationships between the metrics.
het
At the level of the supergravity theory, we have g I “ e´Φ g het is a symmetry of the theory. Then the string
length scales must obey
M het
b
`Is “ `het g het ñ M I “ a s
s s s
gshet

141
20. The effective action will look like
e´2σ
ż
V ? 1
d4 x g e6σ e´2φ rR ` ¨ ¨ ¨ ´ |H2 |2 ´ TrrF 2 ss
p2πq7 `8s gs2 2 4

Taking this to the proper string frame requires g Ñ e´6σ g. This gives

e4σ e6σ
ż
V 4 ? ´2φ 2
d x ge rR ` ¨ ¨ ¨ ´ |H2 | ´ TrrF 2 ss
p2πq7 `8s gs2 2 4

Because σ is such that V e6σ is strictly larger than the order of `s , then both of the gauge fields will have
coupling constants that go as Op`8s gs2 {V e4σ q or Op`8s gs2 {V e6σ q. This is only going to be Op1q if gs " 1. In
this case, we can (after a possible T -duality that doesn’t change the coupling substantially, esp. if one of
the dimensions is reasonably close to the order of the string length already) apply the type I - heterotic O
duality to get a weakly coupled type I description.

21. Since the B-field strength H het gets mapped directly to the RR field strength H I , we expect that the objects
electrically charged between them should get mapped to one another. This means the heterotic fundamental
string gets mapped to the D1 brane in type I. Their magnetic cousins should also be swapped, which will
interchange the heterotic NS5 with the type I D5 brane. At the classical level this is easy to see, since the
two branes have the same supergravity solution. Clearly this is not enough, eg in IIA vs IIB the worldvolume
theories of the NS5 are radically different.
To understand the quantum mechanical equivalence, we need to understand the origin of the Spp2q on the
D5 in type I and the NS5 in the heterotic picture. This question is answered (using nontrivial arguments
involving ADHM) first in Witten “Small Instantons in String Theory”. Return and understand this
when you know more N “ 2 SUSY.

22. Certainly we see that heterotic-type I together with T -duality will relate both heterotic strings together,
and connect this with type I which, after T-dualizing and moving the orientifold plane appropriately, will
connect with the other type II string theories.
It remains to look at the self-duality of type IIB. For this, we took a leaf from Sen’s paper. Let’s look at IIB
on a Z2 orientifold T 2 {p´1qFL ¨ Ω ¨ I where I is the inversion z Ñ ´z on the torus and Ω is worldsheet parity
inversion. This manifold has 4 singular points that each carry ´4 RR charge why. Since it is compact, we
must cancel this by placing 4 D6 branes at each of the 4 points for a total of 16. In this case, the geometry
of the tetrahedron is flat everywhere except for the 4 deficit angles of π at each vertex. The singularities at
the verticies are of D4 “ SOp8q type, so this theory has an unbroken SOp8q4 gauge symmetry. The torus
has moduli T, τ together with axiodilaton S. There is no B field in the orientifold.
Now, let’s T-dualize both cycles of the torus. This keeps us in IIB, but takes us to pT 2 q1 {Ω, undoing the
effects of p´1qFL I. Type IIB on this space is just Type I on pT 2 q1 , but with SOp32q broken down to SOp8q4 .
Now it is time to dualize to heterotic O theory. We see that we have heterotic string theory on pT 2 q1 with
gauge group broken down to SOp8q4 .
Let’s match the moduli:

• The τ modulus is the same in IIB and the heterotic theory.

• The torus in the heterotic picture now has a B89 scalar that gets mapped to the axion C0 in IIA. B89
can combine with the heterotic torus volume to provide another modular parameter ρ “ B ` iVhet .
• The standard parameter in compactification on a torus is Ψhet “ Φhet ´ 41 log det Ghet 1
αβ “ Φhet ´ 2 log Vhet .
This will be mapped to ´ 21 ΦIIB ` log VIIB where VIIB is the original T 2 radius.

We know that heterotic on T 2 has T-duality Op18, 2; Zq. This has a subgroup SOp2, 2q „ SLp2, Zq ˆ SLp2, Zq1
that does not affect the Wilson lines but acts only on the torus parameters. Both τ and ρ transform under
fractional linear transformations of the two SLp2, Rq separately, while Ψhet remains unaffected.
Now, taking VIIB Ñ 8, the two SLp2, Zq symmetries remain unbroken. One of these can be identified with
large diffeomorphisms of the torus, and so combines with spacetime diffeomorphisms in the large V limit.

142
 The remaining SLp2, Zq then becomes the S-duality group. The SOp8q gauge theory living at each of the
vertices is not seen, since the singularities and accompanying D7 branes have “flown off” to infinity.
That Ψhet remains unaffected means that GIIB e´ΦIIB {2 is an invariant under SLp2, Zq. So the volume as
measured in the frame of that modified metric is an invariant. This is exactly the Einstein frame metric.
We have also seen in the chapter that the M theory - heterotic E duality can be obtained through a chain of
dualities involving heterotic O - type I together with the M theory - type IIA. We are only asked to reproduce
dualities between string theories in this question however.

23. The D9 brane is orthogonally projected, as we know from tadpole conditions on it from chapter 7, and the
same argument with the cylinder gives a ?12 reduction of tension relative to type II.
For a D1 brane interacting with itself, the gravitational contribution in the cylinder amplitude also has a
extra 12 factor due to the orientation-projection. Thus, the total tension of the D1 brane is lowered by a
factor of ?12 relative to type II as required.
Naively we could apply the same argument to D5 branes, which would then violate the D1-D5 Dirac quan-
tization by a factor of 2.
However, from an analysis of the cylinder amplitude for 59 and 95 strings with orientation projection, we
get the constraint 259 ζ5 ζ9 “ 1. By consistency of interactions of 59 strings with 55 and 99 strings, we get
259 “ 255 “ 299 “ ´1. Consequently, the D5 brane will have opposite orientation projection than the D9
brane, namely the symplectic one. Taking the determinant of γ “ ζγ T however gives ζ N “ 1, so ζ “ ´1 will
only work for N even. Another way to say this is: “symplectically projected branes must move in pairs”.
Thus, the “fundamental” D5 brane should be thought of as a D5 with Spp2q index a “ 1, 2. Repeating the
T II ?
cylinder amplitude calculation gives a factor of 22 , which translates to a tension of 2 ˆ ?52 “ 2T5II .

24. The crucial component of this is to note that at Dp worldvolume theory contains a CP-odd term coupling
to the lower-dimensional forms going as:
ż ż
p`1 F 2 2
iTp d xC ^ Trre s ^ G Ą iTp p2π`s q dp`1 x Cp´3 TrrF ^ F s

in the absence of an NS-NS background.

Consider the 9-brane with an instanton background in the 5678 directions with instanton number obtained
from integrating over x5,6,7,8 ż
TrrF ^ F s
d4 x “k
p2πq2
For the case of k “ 1, the CP-odd term simplifies to
ż ż
2 2 2
iT9 p2πq p2π`s q d x C6 “ iT5 d6 x C6
6

This is exactly the CP-odd term for a D5 brane. In the limit of vanishing instanton size, this sources RR
fields in the same way with the exact same RR charge. It is also a BPS state, so has the same mass as a D5
brane. This satisfies all the criteria to qualify as a D5 brane.
We can extend this to k localized D-branes and see the exact same coupling
k
ÿ ż
iT5 d6 x C6
i“1 x5...8 “x i

as k distinct D5 branes. For nonvanishing instanton size, this describes D5 branes “dissolved” in the D9.
This argument can be carried over for an arbitrary pair pp, p ´ 4q.

143
25. We have already seen by general arguments that we need the number of Newman-Dirichlet conditions to be
a multiple of 4 so that the NS and R sectors have a chance of having degeneracy. I will repeat the argument
here.
In the R sector, the zero-point energy is always zero because of the equal number of periodic fermions and
bosons. The excitations above this will have integer or half-integer weights.
1 1
In the NS sector, the NN and DD fermions and bosons contribute zero point energies ´ 24 and ´ 48 , so
1 1 1
´ 16 total. The ND sector bosons and fermions contribute 48 and 24 , ie the opposite. Altogether for ν ND
boundary conditions we get:
p8 ´ νq ν 1 ν
´ ` “´ `
16 16 2 8
This ground state and its excitations above it will have half-integer weight when ν “ 0 mod 4.
Since type I string theory necessitates 32 D9 branes to cancel out the O9 tension, we are only allowed
ν “ 8, 4, 0 giving D1, D5, and D9 brane configurations preserving supersymmetry in the theory. In the
text, we have seen that D1, D5, D9 all lead to consistent worldvolume excitations that respect GSO and
Ω-projection

26. Let’s review the logic so far. For supersymmetric open strings in the NS sector, we are principally interested
in the ψr states. Orientation projection acts as on the NN string as Ωψr “ i2r ψr (in both NS and R sectors)
and on the DD string as Ωψr “ ´i2r ψr . For the R sector ground states, supersymmetry requires that for all
directions NN (D9 brane) R “ ´1: that is, Ω |Ry “ ´ |Ry.
When we add indices, writing the NS state as |p, ijy, for NN strings the massless levels are given by
µ
ψ´1{2 λij |p, ijy. We get the constraint λ “ ´iN S γλT γ. WLOG we can either have γ “ 1 for SOpN q
with ζ “ 1 or γ “ iω for SppN q with N even, ζ “ ´1. In either case the Jacobi identity require N S “ ´i.
This gives that λ “ ´γ T λγ ´1 for the massless level. In both cases this corresponds to the adjoint represen-
tation. In the DD case, we get an extra minus sign, giving λ “ γ T λγ ´1 . This corresponds to the symmetric
traceless representation plus a singlet.
For the D1 brane, the above discussion already shows us that in the 1-1 NS sector, we get the 8 DD scalars
transforming the symmetric traceless plus single representation of SOpN q together with the 2 NN scalars
transforming in the adjoint.
For the 1-1 R sector, before orientation projection we have the 16` ground state from GSO. The orientation
projection acts as
Ω |Sα , i, jy “ ´eiπps1 `s2 `s3 `s4 q γ |Sα , i, jy γ ´1
What Kiritsis writes can’t be the adjoint for N “ 1. We need it to have dim 1 in that case,
but it would have dim 0. I believe that the correct thing is that we have 8 fermions forming the 8´ (ie
left-moving) and in the symmetric representation of SOpN q while we have 8 forming the 8` (ie right-moving)
but in the adjoint of SOpN q (these disappear for N “ 1).
In the 1-9 sector, we have 2 NN and 8 DN boundary conditions. The NS ground state energy is positive, so
this will not contribute. The massless states come from the R ground state in the DN part combined with
the Op1, 1q spinor from the R sector of the NN part. The fermions are right-moving (chirality +) as before.
We get 32 indices from the D9 brane and N from the D1 brane. The orthogonal projection guarantees that
these transform in the pN, 32q bi-fundamental representation. Orientation projection disallows for the second
copy of this spectrum (ie the 1-9 string with the orientation reversed).

27. To get to the D5-brane from the D9-brane we T-dualize four times. In this problem we focus only on the
5-5 strings. Again, the R-sector contains the GSO-projected 16` spinor before orientation projection. We
must decompose under SOp5, 1qk ˆ SOp4qK . The projection condition in the R sector reads:

Ωλij |Sα , ijy “ ´eiπps1 `s2 q γλij γ ´1 |Sα , ijy

Here γ “ iω, since we have the symplectic Spp2q projection for the D5. Then, for s1 ` s2 odd, the 6D fermion
is negative chirality, and we require λ “ ´γλT γ ´1 . This gives a negative-chirality fermion in the adjoint
representation of Spp2q, completing the vector multiplet.

144
 For s1 ` s2 even, the 6D fermion is positive chirality and we require λ “ γλT γ ´1 which will leave the skew-
traceless antisymmetric representation plus a singlet. For Spp2q this is just the singlet, so we get a single
positive chirality fermion, completing the hypermultiplet.
I think I’m off by a sign?

28. From the D5-D5 analysis of the previous problem, we immediately see the generalization to general Spp2N q.
µ
The R sector yields fermions in the Spp2q adjoint combining with the vectors ψ´1{2 in the adjoint, yielding
the vector multiplet. The DD boundary conditions reverse the projection sign for ψ´1{2 i λij |p, ijy yielding
a sum of the skew-traceless antisymmetric representation plus a singlet. I assume this is the same as the
two-index symmetric rep, by analogy to SOpN q, where a similar thing happens. We also know that the R
sector also provides (positive chirality) fermions to combine with this to form the hypermultiplet.

Finally, we must look at the D5-D9 spectrum. We have 4 ND boundary conditions and 6 NN ones. For 4
ND boundary conditions, the NS sector ground states also contribute to the massless spectrum. The ND
conditions these consists of ground states transforming in the 4 of SOp4q, combining with the singlet NS
ground state of the 6 NN coordinates. This yields 4 scalars.
In the R sector, the massless states come from the bosonic ND ground state combining with one of the 4 NN
R sector states giving an SOp5, 1q spinor. After GSO projection, this gives a chirality ` fermion, completing
the hypermultiplet. This part is a bit shifty, thing about it
Each of these states has 32 labels from the D9 brane, and 2N labels from the D5 brane. Therefore, we get
that this hypermultiplet in fact transforms in the p2N, 32q bi-fundamental. Again, orientation projection
simply restricts us to not have a second copy of this spectrum from 5-9 strings of opposite orientation.
Say we pull apart m D5 branes. Because the D5 branes move in pairs in type I, we must have m an even
integer. The 5-9 strings now all have positive zero-point energy and will not contribute to the massless
spectrum. The 5-5 strings remain the same, but transforming in Spp2N ´ mq instead of Spp2N q.

29. We can focus on the purely chiral left-moving CFT, since this is the only part that the orbifold acts on
nontrivially. Immediately, we see that the untwisted sector corresponds to the NS states, which are the same
between IIA and IIB.
In the twisted sector, we again have NS and R fermions. Because the NS fermions are taken to minus
themselves, they are now integrally modded while the R fermions become half-integral. Again, the R fermions
will be projected out by the p´1qFL . The 8 NS fermions will give two (unprojected) ground states 8 ` 8̄ of
fermion numbers 1, ´1 respectively. In Polchinski’s convention, the original |0y NS ground state has fermion
number ´1, so the only the C operator on top of this will give something that is unprojected. In Kiritsis’
convention, the NS ground state has fermion number 1 but we we take p´1qF “ ´1 for GSO. In either case,
we can only keep the C operator. In the original IIA we kept the S on the left and the C on the right. Now
we keep C on both sides giving IIB (we could have done the same with p´1qFR , and C, C or S, S both yield
IIB, since they are related by parity).
Orbifolding IIB by this symmetry is the same as orbifolding twice. This necessarily must return us back to
IIA.

145
 The M theory parity orbifold differs from this p´1qFL orbifold primarily in that it includes fixed points, on
which the twisted sectors localize.
30. Start with the heterotic E theory and compactify on a circle. n units of KK momentum on this circle will be
T-dualized to n units NS flux in the O(32) theory, ie a string wrapping the circle n times. Upon S-duality,
this will correspond to a D1 brane wrapping the circle in type I n times. We T-dualize again to get a D0
brane in the type I’ theory carrying n units of charge. In the strong coupling limit, this is understood as n
units of momentum in the eleventh direction.
31. The bosonic part of the vector multiplet on a single boundary is given by
ż
1 ?
´ 2 d10 x ´g10 TrrF 2 s
4λ
At first glance, λ would appear arbitrary. Anomaly cancelation will yield an exact value for it in terms of
the eleven-dimensional gravitational coupling. Kiritsis writes explicitly λ2 “ 2πp4πκ211 q2{3 , which gives that
the dimensionless ratio λ6 {κ411 “ p2πq3 p4πq2 “ 128π 5 . This is as in Horava and Witten, but it is not obvious
that this is how λ is determined from κ11 from first principles.
Let’s recall anomaly cancelation in 10D. Recall that for the type I supergravity theories, it was crucial to
have enough 10D vector multiplets to cancel the TrrR6 s terms, giving n “ 496
We can view the M-theory orbifold R11 {Z2 as giving rise to two twisted sectors (as in string theory). In 11D
we have
Γ1 . . . Γ11 “ 1
The supersymmetries preserved by the Z2 action are those that satisfy (WLOG) Γ11  “ . This means that
in the 10D perspective, this gives rise to chiral fermions in the 16` . Although in the smooth part of the
bulk, there cannot be a gravitational anomaly, the incorporation of a boundary (or more) can. A general
diffeomorphism in the bulk will not lead to any anomalous variation δΓ of the effective action. WLOG, take
a diffeomorphism on R10 and pull it back to the orbifold by making it constant along the interval S 1 {Z2 .
The anomaly is the standard one in 10D. The boundaries must therefore contribute massless multiplets. The
only such candidate is a vector multiplet. By symmetry, each must contribute the same number of vector
multiplets. The prior paragraph then shows that each must contribute 248.
In order to apply Green-Schwarz, we crucially need a two-form B, that we are guaranteed in all 10D string
theories. The ˇanswer here comes from the pullback of the 3-form A3 to the boundaries H1 , H ş 2 giving
B2 , B21 “ Aµν11 ˇH1 ,H2 respectively. The anomaly polynomial in 10D of the form I4 ^ X8 leads to B ^ X8 ,
but one can see that X8 for E8 ˆ E8 involves no cross terms from either E8 , and can in fact be written as
I12 pR, F1 , F2 q “ Iˆ12 pR, F1 q ` Iˆ12 pR, F2 q (86)
Iˆ12 pR, Fi q “ Iˆ4 pR, Fi qIˆ8 pR, Fi q (87)
1 1
Iˆ4 pR, F q “ trR2 ´ trF 2 (88)
2 2
1 ` 1 1
Iˆ8 pR, F q “ ´ Iˆ4 pR, F q2 ` ´ trR4 ` ptrR2 q2
˘
(89)
2 8 32
1
Here tr “ 30 Tradj for E8 . This gives Chern-Simons terms of the form:
ż ż
B2 ^ I8 pR, F1 q ` B21 ^ I8 pR, F2 q
H1 H2

This does not quite go far enough, in that we would not be able to recover λ6 {κ211 from this.
This next part I got from Horava and Witten 9603142 First note that we would like our boundary
theory to be locally supersymmetric. As it stands, it is not. The standard way in SUGRA (although I did
not know this because I don’t know enough SUGRA at this point) is to add the gravitino interaction with
the supercurrent to the Lagrangian:
ż ż
1 10 ? 1 ? a
´ 2 d x ´g ψ̄SY M “ ´ 2 d10 x ´g ψ̄A ΓA F{ χa
4λ 4λ

146
After some work, we see that the only term with uncanceled supersymmetric gauge variation is
ż
1 10 ?
d x ´g ψ̄A ΓABCDE FBC
a a
FDE 
16λ2
The only way to cancel this is to modify the 11D Bianchi identity. The reason is that, in checking invariance
under local supersymmetry for the 11D Lagrangian, there is an integration by parts that involves the Bianchi
identity dG “ 0. If one instead modifies it to
? κ2
dG11 ABCD “ ´3 2 2 δpx1 1qtrpFrAB FCDs q
λ
We can then locally write

κ2
G11,ABC “ B11 CABC ` ? 2 δpx11 qΩCS
ABC (90)
2λ
2
ΩCS
ABC “ trpAA FBC ` AA rAB , AC s ` perms.q (91)
3
CS
ñ dΩ “ 6trpFrAB FCDs q (92)
CS
δ Ω “ dtrrF s (93)
κ2
ñ δ C11,AB “ ´ ? 2 δpx11 qtrpF q (94)
6 2λ

Note that the anomalous variation in equation (94) is similar to the δB „ trrF s for the string theory 2-form.
This gives that the 11D Chern-Simons 11D interaction has variation

1 κ2
ż ż
1
C ^G^GÑ ? trrF s ^ G ^ G
3! 2! 6 2λ2 H

The value of GABCD on one of the hyperplanes is quickly seen to be

3κ2
GABCD |H “ ´ ? 2 trrFrAB FCDs s
2λ
Altogether the anomalous variation looks like

κ4
ż
´ trrF strrF 2 s2
128λ6
On the other hand, the variation from the 10D chiral fermions in the vector multiplet gives
ż
1 1
TrrF 5 s
2 p4πq5 5!

where the trace is taken in the adjoint.

2 2
TrTrF 5
It is only for E8 that we have the nice identity TrX 6 “ pTrX7200
q
and similarly TrrF 5 s “ 7200 . After
identifying tr “ Tr{30, we can write TrX 6 “ 15 2 3
4 ptrX q etc. We get
ż
15
TrrF 5 s
8p4πq5 5!

This will cancel exactly when κ and λ are related as required.

Thus, we see that the Green-Schwarz term already present in 11D SUGRA plays a crucial role in canceling
the gauge anomaly. We needed the GS terms to be bulk objects, as if they were simply δ-function supported
on the boundary this would a) not seem very much like quantum gravity, and b) give gauge variations of
boundary interactions proportional to δp0q.

147
 Note that the CS term for the gauge field in the supergravity action was classical (as opposed to the 10D
superstrings, where they are 1-loop effects). Consequently, a classical theory with the SUGRA multiplet in
the bulk and the gauge multiplet on each boundary is classically inconsistent.
Given our understanding of 10D anomalies, we then expect (correctly) that the full gauge and gravity
variation will modify the Bianchi identity as:
? κ2 ´ 1 ¯
dG11 ABCD “ ´3 2 2 δpx1 1q trpFrAB FCDs q ´ trpRrAB RCDs q
λ 2
These trR2 terms are not required from classical 11D SUGRA, so they must arise as quantum effects of
M-theory.
The anomaly cancelation term usually takes the form:
ż
C ^ I8

Note the suggestive way I8 is written in equation (89). Because we have seen that G9I4 on the boundary,
the Chern-Simons term C ^ G ^ G on the boundary reduces to C ^ I42 . This is a part of I8 . We thus expect
the remaining part to take the form:
? ż ˆ ˙
2 1 4 1 2 2
C ^ ´ R ` ptrR q
p4πq3 p4πκ2 q1{3 8 32
Again, this is a purely quantum effect of M-theory.

32. Here M is a 20 ˆ 20 matrix. It is quick to see that M LM “ L for the 20 ˆ 20 matrix

¨ ˛
0 14 0
L “ ˝14 0 0 ‚.
0 0 116

This means that M is an element of Op4, 20q. We can act on it as a bi-fundamental representation (on left
and right). This is more subtle, because not all Op4, 20q matrices have the form of M . Showing
that M keeps the same form would take too much time. This ensures that the last term is invariant.
The 4 ` 4 ` 16 “ 24 gauge fields from the compactification can be directly seen to transform in the contra-
gradient representation of Op4, 20q. This ensures that the second-to-last term is invariant. All other terms
are invariant.

33. The heterotic action

e´2Φ e´Φ ´1 i j µν 1
ż
?
d6 x ´GrR ´ B µ ΦBµ Φ´ |H|2 ´ Mij Fµν F ` TrrBµ M B µ M ´1 ss
2 4 8
1
Hµνρ “ Bµ Bνρ ´ Lij Aiµ Fνρ
j
` 2 perms.
2
and the IIA action:
e´2Φ eΦ ´1 i j µν 1
ż ż
? 1
d6 x ´GrR ´ B µ ΦBµ Φ ´ |H|2 ´ Mij Fµν F ` TrrBµ M B µ M ´1 ss ` d6 x Lij B ^ F i ^ F j
2 4 8 2
Hµνρ “ Bµ Bνρ ` 2 perms.

I will take the shorthand 12 Mij´1 Fµν

i F j µν “ |F |2

The EOMs for G give respectively

1 1 1 1
Rµν ´ gµν R ´ 2p∇µ Φ∇ν Φ ´ gµν p∇Φq2 q ´ 2
Rgµν ´ e´2Φ pHµν ´ gµν |H|2 q ´ e´Φ pFµν ´ gµν |F |2 q ` pM termsq
2 2 2 2
1 1 1 1
Rµν ´ gµν R ´ 2p∇µ Φ∇ν Φ ´ gµν p∇Φq2 q ´ 2
Rgµν ´ e´2Φ pHµν ´ gµν |H|2 q ´ eΦ pFµν ´ gµν |F |2 q ` psame M termsq
2 2 2 2

148
 2 “ 1H
here Hµν ν 2
4 µρσ Hρσ etc. All terms are invariant under Φ Ñ ´Φ, including the terms involving H , since
we will have
1 2
e´2Φ H 1 “ e2Φ pe´2Φ ‹ Hq2 “ e´2Φ |H|2
and the same for Hµν .
The EOMs for Φ give respectively:

e´Φ 2
Het E: ∇2 Φ ` e´2Φ |H|2 ` |F | “ 0
2
eΦ 2
IIA: ∇2 Φ ` e´2Φ |H|2 ´ |F | “ 0
2

The EOMs for the Aµ give respectively:

Het E: e´2Φ p‹Hq ^ F ´ dpe´Φ Mij ‹ F j q “ 0

IIA: ´dpeΦ Mij ‹ F j q ` H ^ F “ 0

This is equivalent under Φ Ñ ´Φ, e´2Φ H “ ‹H 1 .

The EOMs and Bianchi identity for the Bµν give respectively

Het E: ´dpe´2Φ ‹ Hq “ 0, dH ´ F ^ F “ 0
´2Φ
IIA: ´dpe ‹ Hq ` F ^ F “ 0, dH “ 0

The fact that the H duality exchanges the Bianchi identity and the EOMs speaks to the fact that it is an
electric-magnetic duality of strings.
The EOMs for the M terms are Φ, A, B independent. Consequently, the M matrices can be directly identified
between the two theories.

34. Consider just the 3D space of the xi . Note that V is harmonic, and consequently F :“ ´ ‹ dV is a closed
2-form on that space. We can view F as a curvature 2-form on a principal U p1q bundle, and can thus write
(upon picking a trivialization of the U p1q) a potential A giving dA “ F . Call the the U p1q bundle X. We
will write the connection as A “ A ` dγ.
For each of the three xi there is a symplectic form on the 4D U p1q bundle given by:

ωi :“ A ^ dxi ` V ‹ dxi ñ dω “ ´ ‹ dV ^ dxi ` dV ^ ‹dxi “ 0

Here we have used that all the dxi forms are (canonically) pulled back from R3 and in R3 , ‹α ^ β “ α ^ ‹β.
Now, define a different basis of symplectic forms on X by

Ω1 “ ω2 ` iω3 “ A ^ pdx2 ` idx3 q ` iV dx1 ^ pdx2 ` idx3 q

defining z 1 “ x2 ` ix3 this gives:

Ωi “ A ^ dz i ` iV dxi ^ dz i “ V pV ´1
A ` idx1 q ^ dz i .
looooooomooooooon
α1

The kernel of this form on the U p1q-bundle is 2D. For Ω1 it is spanned by

B̃x2 ` iB̃x3 , V Bγ ` iB̃x1

Here each Bxi is lifted to the U p1q tangent space by using the connection A to identify the appropriate
horizontal subspace. We identify this as a holomorphic tangent space. Similarly Ω̄1 would complete the
basis of Tp M and give the anti-holomorphic tangent space. Thus, each Ωi gives a distinct stratification into
holomorphic and anti-holomorphic tangent spaces. The closedness of Ω guarantees integrability. Defining
3 separate complex structures Ij to act as `i on the jth holomorphic tangent space and as ´i on the jth

149
 anti-holomorphic tangent space, we can easily check that pointwise they reproduce the quaternion algebra.
This makes the manifold hyper-Kähler, with metric given by:

ds2 “ V p<pα1 q2 ` =pα1 q2 q ` V p<pdz1 q2 ` =pdz1 q2 q

“ V ´1 pAq2 ` V |d~x|2
“ V ´1 pAdx ` dγq2 ` V |d~x|2

In particular, V can take the form of the multi-center potential in the problem.
Could we not have just exhibited a 3 Killing spinors? Are there such? In any case, this was
more instructive
Lastly, to see the asymptotic limit, we can take the xi to collide. At a distance, V will look like Nr . This
corresponds to an F with N units of flux asymptotically. The circle bundle over the R3 will asymptotically
looks like an S 1 fibration over S 2 . For N “ 1, this is simply the Hopf fibration. For higher N , the connection
is N times larger, which makes the U p1q circle N times smaller, and corresponds to a fiberwise quotient of
S 3 Ñ S 2 by ZN .
Did Kiritsis mean to write N ?

35. We have seen that to apply the GS mechanism, the heterotic B-field must have a modification of its Bianchi
identity from

κ2 GR ” 2i ı
H “ dB ´ ΩY3 M pAq ` Ω pωq, Ω3 pAq “ Tr A ^ dA ´ A ^ A ^ A
g2 3 3
κ2
I have absorbed the factor of g2
into these fields. This will cancel the anomalous change in B

κ2
δB “ TrrΛF0 ´ ΘR0 s
g2
ş
required to have a GS term B ^ X8 cancel the full anomaly. The modified Bianchi identity is

dH “ ´trrF ^ F s ` trrR ^ Rs

Here tr is taken always in the fundamental. The second addition important for the cancelation of gravitational
anomalies. Following the logic in question 33, by duality, a modification of the Bianchi identity in heterotic
string requires a modification in the B equations of motion on the heterotic side corresponding to the addition
to the action of a term: ż
´ B^R^R

The CS terms are thus: ż

B ^ ptrpF ^ F q ´ trpR ^ Rqq

Further, Vafa and Witten confirm this term from a 1-loop calculation using the elliptic genus on the IIA
side in 9505053. This gives a nontrivial 1-loop test of the 6D string-string duality. This problem only asks
me to assume the duality in performing the match.

36. In our case, wrapping 3-branes around 2-cycles give rise to two-forms. As one 2-cycle shrinks B to zero size,
we get a tensionless string, of tension approximately |VolpBq|{gs . For each isolated singularity of K3 (type
ADE) there is such a tensionless string theory. Note that this is not yet the (2,0) SCFT, since we have not
taken any sort of IR limit that would lead us to expect that the theory is conformal. We still have mass
scales. This is an interacting QFT of light strings.
Upon compactifying on an S 1 , we can T-dualize to type IIA, where now we have the familiar appearance
of massless states associated to a 3-cycle shrinking in K3. The IIA theory sees massless particles emerge at
this transition, corresponding to the tensionless strings of IIB wrapping the S 1 .

150
37. Here, our cycle is C “ şni B i . Take a euclidean D2 brane wrapping this cycle. The total volume (counting
orientation) will be |ni B i Ω| “ |ni Z i |.
Because of the BPS property of the 3-cycle, we will still have M “ Tp |Z|, giving us
1
Sinst “ |ni Z i |.
p2πq2 `3s g
It is worth remarking that we get contributions from all winding numbers of D instantons in this case, while
in the IIB case, it looks like only the singly-wrapped D3 brane is stable.
Is there anything else I should say? Reproduce Vafa+Ooguri’s calculation?

38. In IIB, we have seen that as a three-cycle shrinks, a (BPS) D-brane wrapping this cycle contributes a
hypermultiplet that becomes massless as the volume goes to zero. At the conifold point, we get a new
massless multiplet. Resolving this singularity by expanding the 2-cycle corresponds to giving an expectation
value to the massless hypermultiplet from the D-brane. In general, these hypermultiplets will have a potential.
See the discussion on page 378.
From this POV, condensation of D-branes has the interpretation of topology change! For IIA the (instantonic)
D2 branes instead serve to smooth out the singularity, which corresponds to the hypermultiplet moduli space
receiving quantum corrections.
This does not answer the question, though - which was about the resolution of the two-cycles. However,
using the tool of mirror symmetry, we can posit a guess. A two-cycle shrinking in IIA causes a singularity in
the vector multiplet, and maps to the familiar three-cycle shrinking in IIB. In IIA, then, we expect a wrapped
D2 brane to contribute to a massless hypermultiplet. On the other hand, we expect quantum effects in IIB
to smooth out this singularity.
Check against literature.

39. To simplify this problem, I will reduce the heterotic theory directly from 10D and keep in mind that reduction
on tori commutes. In the string frame I get
ż „ 
4 ? ´2φ 2 1 2 1 ´1 i j µν 1 ij µ
d x ´ge R ` 4pBφq ´ |H3 | ´ M ij Fµν F ` TrrBµ M B Mij s
2 4 8

Upon taking g Ñ e2φ g we get the Einstein frame:

„ 
e´4φ e´2φ ´1 i j µν 1
ż
4 ? 2 2 ij µ
d x ´g R ´ 2pBφq ´ |H3 | ´ M ij Fµν F ` TrrBµ M B Mij s
2 4 8

Here the Mij scalar matrix lives in the SOp6, 22q coset, and there are 28 fields F i . We can dualize the H3 to
a scalar axion through the relation (as in 9.1.12)

e´4φ H “ ‹BC0 ñ e´4φ |H|2 “ e4φ pBC0 q2

In performing this dualization, the B equations of motion are automatically satisfied.

∇µ pe´4φ Hµνρ q “ ∇µ pεµνρ σ Bσ C0 q “ 0

The Bianchi identity for H must now be imposed by hand

εµνρσ Bµ Hνρσ “ ´Lij Fµν

i
F̃ j µν

This corresponds to adding the term

1 i
C0 Lij Fµν F̃ j µν
4
to the Lagrangian. We then combine the axion with φ to give:
ż „ 
4 ? 1 BSB S̄ 1 i 1 1
d x ´g R ´ 2 ´ S2 M ´1 ij Fµν F j µν ` S1 Lij Fµν
i
F̃ j µν ij µ
` TrrBµ M B Mij s
2 S2 4 4 8

151
 where here S “ C0 ` ie´2φ (note the 2φ, by contrast with 10D).
For the IIA side we get:
ż
? ´ 1
d x ´Ge R ` |Bσ|2 ` Bµ Gαβ B µ Gαβ ´ |BΦ|2 ´
4 σ
4
e´2Φ e´2Φ αβ 1
|H3 |2 ´ G Hα µν Hβµν ´ e´2Φ´2σ pBµ Bαβ q2
2 4 2
eΦ ´1 i j µν eΦ ´1 αβ i j µβ 1 ¯
´ M ij Fµν F ´ M ij G Fµα F ` TrrBM BM ´1 s
2 2 8
We take this to the Einstein frame g Ñ e´σ g giving
ż
? ´ 1 1
d4 x ´g R ´ |Bσ|2 ` Bµ Gαβ B µ Gαβ ´ |BΦ|2 ´
2 4
e´2Φ`2σ e´2Φ`σ αβ 1
|H3 |2 ´ G Hα µν Hβµν ´ e´2Φ´2σ pBµ Bαβ q2
2 4 2
eΦ`σ ´1 i j µν eΦ ´1 αβ i j µβ 1 ¯
´ M ij Fµν F ´ M ij G Fµα F ` TrrBM BM ´1 s
2 2 8

In this case, the dilaton and axion fields enter the mass matrix M , as does the torus complex modulus,
whose kinetic term is the third term in the action above. The complexified Kähler modulus parameters of
the torus remain Why? This part needs elaboration. Altogether, including the CS term, this reduces
to
eσ ´1 i j µν 1
ż
? ´ 1 1 1 ¯
d4 x ´g R ´ |Bσ|2 ´ e´2σ pBµ Bq2 ´ M ij Fµν F ` TrrBµ M ij B µ Mij s ` BLij Fµν
i
F̃ j µν
2 2 4 8 4
Defining the parameter T “ B ` ieσ we get
ż
? ´ 1 BT B T̄ 1 1 1 ¯
d4 x ´G R ´ 2
i
´ T2 M ´1 ij Fµν F j µν ` T1 Lij Fµν
i
F̃ j µν
` TrrBµ M ij B µ Mij s
2 T2 4 4 8

Comparing the above action with the heterotic one we identify S with T , giving (unprimed indicates heterotic,
primed indicates IIA)
1
C0 “ Bαβ (external SL2 ), C01 “ Bαβ
1
(internal)
1 1
e´2φ “ eσ (external SL2 ), e´2φ “ eσ (internal)
This makes us also identify g “ g 1 and M “ M 1 as well as all the gauge fields descending from 6D Aiµ “ pAiµ q1
and the two descending from the T 2 metric Aαµ “ pAαµ q1 .
Electric-magnetic duality acts least trivially on Hµνρ in 6D. Consequently, it will act nontrivially on its axion
and scalar descendants B, C0 (as we have seen). The 2-form field strength F B remains. From 6D we have
e´2φ H “ ‹H 1 . This descends to
1 1 µνρσ εαβ B 1
e´2φ Gαβ pFβ,µν
B
´ Lij Yβi Fµν
j A,γ
´ Cβγ Fµν q ´ a ‹ FA “ ? pFβ,ρσ q
2 2 g

where we see on the heterotic side we have additional terms due to how Ĥµνρ is defined in that case account
for additional axion term there.
Relatively unfinished.

40. I will just demonstrate this on the Bosonic sector. p´1qFL sends the RR fields to minus themselves (ie
C0 , C2 , C4 ), while S swaps B2 , C2 and flips the axion part of the axio-dilaton τ Ñ ´τ̄ . Conjugating p´1qFL
by S will flip the sign of C4 and B2 and also take τ Ñ ´τ̄ . The untwisted sector will thus be without
B2 , C0 , C4 leaving only G, φ, C2 . This is the closed-string sector of type I.
On the other hand, we have shown that orbifolding IIB by just p´1qFL yields just IIA. At the level of bosonic
fields, we already see that these operations do not commute.

152
41. It is worth appreciating that this duality was known before the Horava-Witten construction.

First note that the moduli space of the heterotic string on T 3 is given by the coset space

R` ˆ SOp19, 3; ZqzSOp19, 3q{pSOp19q ˆ SOp3qq

with R` parameterizing the dilaton. Now, in string compactifications K3 has a moduli space coming from
cosets of SOp4, 20q{SOp4qˆSOp20q. This includes the complexified Kähler modulus, which takes into account
the NSNS B-field. M theory lacks this parameter, and consequently the Kähler component of moduli space
involves only real moduli (ie metrics). This gives SOp3, 19q{SOp3qˆSOp19q. The volume gives another factor
of R.
The low-energy effective actions also match. Wrapping the M-theory A3 on K3 gives one 3-form C3 and 22
1-forms Ai1 . We get an action:
ż ż a
1 11
a 1 2 7 4σ 2
ÿ1 1
2 d G 11 pR ` |dA3 | q Ñ d G 7 re pR ` pBσq ´ |dAi1 |2 ` moduliq ´ |dC3 |2 s
2κ11 2 i
2 2

Upon rescaling g Ñ e´4σ g and taking φ “ 3σ we arrive at:

ż a ÿ1 1
d7 G7 re´2φ pR ` pBσq2 ´ |dAi1 |2 ` moduli ´ |dC3 |2 qs
i
2 2

This exactly matches with the heterotic theory. We go to strong heterotic coupling by taking σ Ñ 8, ie
taking the volume of the K3 to be large.

42. Because A3 is odd under the Z2 transformation, we must ` ˘wrap it on either a 1-cycle or a 3-cycle to have
things survive. There are 5 1-cycles giving 5 vectors and 52 “ 10 2-cycles giving 10 0-forms in 6D. Further,
the internal metric has 5 ˆ 6{2 “ 15 even terms that survive. Altogether we get 5 2-forms, 25 scalars, and
10 vectors.
This is N “ p2, 0q (chiral) supergravity consisting of the supergravity multiplet and five tensor multiplets,
each of which contains an anti-self-dual two-form field (the 5 self-dual parts are part of the SUGRA multiplet).
Cancelation of anomalies (prove and understand compared to the N “ p1, 0q case) require NT “ 21.
We are missing sixteen tensor multiplets.
The orbifold has 25 “ 32 fixed points which we expect will lead to twisted sectors. We shouldn’t be too
sure of how things go, though, because we don’t know how to deal with twisted sectors of M-theory. It
initially looks paradoxical that we need 16 extra tensor multiplets but have 32 fixed points. This is resolved
in Witten 9512219. The solution is to recognize the fixed points as 32 magnetic sources of charge ´1{2 for
the G4 field. The constraint that the total charges should vanish is satisfied when 16 of these sources have
a five-brane on top of them of `1 magnetic charge. Each fivebrane can be seen to support a single tensor
multiplet, giving our desired 16. We interpret the five scalar in each multiplet as describing the position of
the fivebrane inside T 5 {Z2 .
This now gives the massless spectrum of type IIB on K3.
The question is how to arrange the fivebranes in such a way that we can see the duality to IIB on K3.
The equivalence implies that when any circle shrinks show more rigorously, we would expect to recover
weakly coupled IIB on K3. The naive guess I have is to arrange them in an alternating “checkerboard”

153
 pattern. (Witten confirms this.) Now, in the limit where any circle shrinks to zero size, the opposite
charge sources cancel, giving zero 4-form field strength in the 6D spacetime, consistent with the fact that
the 3-form has been projected out on the M-theory side and doesn’t exist on the IIB side.
As a second check, we can further compactify on S 1 . We get IIA on T 5 {Z2 vs IIB on S 1 ˆK 3 . T-dualizing the
latter along S 1 (the only 1-cycle!) gives IIA on S 1 ˆ K 3 which is equivalent to heterotic on S 5 . S-dualizing
this gives type I on S 5 which is T-dual to IIA on T 5 {Z2 as an orientifold (why do we need to act with
Ω too?)

43. I will work with covariant derivatives and take the axiodilaton fields in terms of the S, S̄ basis. I will write
the equations of motion for the axiodilaton as:
ˆ ˙
¯ BS BS B̄ S̄ B B̄S BS B̄ S̄ BS B̄ S̄ ´ BS B̄S BS B̄S
∇ 2
´2 3
“ 2
´2 3
`2 3
ñ B B̄S ` 2 “0
pS ´ S̄q pS ´ S̄q pS ´ S̄q pS ´ S̄q pS ´ S̄q S̄ ´ S

Where it is important to note that we can write B B̄S for the laplacian in complex 2D coordinates instead of
∇∇S. We thus get our desired EOM.

44. Recall that as a holomorphic function of z, S should have positive imaginary part, and have its image
restricted to the fundamental domain F. This mapping should be finite energy density. From the effective
action we compute the energy density by pulling back as:
ż ż
i 2 BS B̄ S̄ i
E “´ 2 d z 2
“ 2 d2 S B B̄ logpS ´ S̄q
κ10 pS ´ S̄q κ10 F

At this point we apply Stokes’ theorem to get a boundary integral:

ż ż
i i dS
dSB logpS ´ S̄q “
κ210 BF κ210 BF S ´ S̄

The vertical lines of the fundamental domain have the same values but are traversed in opposite orientation
picture. Therefore, only the semicircle counts. This integral is readily evaulated:
ż π{3
dθ ieiθ iπ
“´
2π{3 eiθ´ e´iθ 6
π
This gives our desired final answer of 6κ210
: the angular defect due to a D7 brane.

154
Chapter 12: Compactifications with Fluxes
It is not likely that this will be relevant to my research. I will skip it indefinitely for now.

1.

155
Chapter 13: Black Holes and Entropy in String Theory
1. We begin with
dr2
ds2 “ ´F prqCprqdt2 ` ` Hprqr2 dΩ22
Cprq
and Cprq vanishes at the horizon r “ r0 while all other functions are positive for r ě r0 and everything
asymptotes to 1 as r Ñ 8. Now, let’s do a wick rotation t Ñ iτ with τ Euclidean time. We get
dr2
F prqCprqdt2 ` ` Hprqr2 dΩ22
Cprq
Now at r “ r0 `  we see that the geometry takes the form
dr2
F pr0 qC 1 pr0 qpr ´ r0 qdt2 ` ` r02 dΩ22
C 1 pr0 qpr ´ r0 q
The last term is simply the expected metric on a 2-sphere of fixed radius r0 . The other two terms give a
metric
d2
ds2 “ F pr0 qC 1 pr0 qdt2 ` 1
C pr0 q
?
Take u “ ?2 1  then du “ ?d
giving us
C pr0 q C 1 pr0 q

C 1 pr0 q2 2 2
ds2 “ F pr0 q u dt ` du2
4
This describe a conical deficit geometry in polar coordinates. In order to obtain a smooth geometry, we need
the requirement that
2
τ ` 2π ˆ a “τ
1
C pr0 q F pr0 q
Giving an inverse temperature of
1 4π
“β“ a
T 1
C pr0 q F pr0 q
This formula generalizes directly to higher-dimensional black holes.
2. In what follows, recall the area formula for a general KN Black hole of mass charge and spin pM, Q, Jq is:
a
2
A “ 4πpr` ` a2 q, r` “ M ` M 2 ´ a2 ´ Q2 , a “ J{M

In particular an extremal Kerr black hole has area 8πM 2 .

(a) The areas of the individual Schwarzschild black holes are

4πp2Mi q2 “ 16πMi2

each. The area of their composite must then be ě 16πpM12 ` M22 q. Because they start as almost
stationary, the total angular momentum in the center of mass frame is zero, so the final black hole will
be (essentially) Schwarzschild. So we get

Mf2 ě M12 ` M22

? ?
initial masses were equal, we’d get Mf ě 2M so that E “ 2M ´ 2M and E{pM1 ` M2 q “
If the ?
1 ´ 1{ 2. Let’s write WLOG M2 “ γM1 with γ ď 1 then
a
Mf2 ě p1 ` γ 2 qM12 ñ E “ p1 ` γqM1 ´ 1 ` γ 2 M1
a a
E p1 ` γqM1 ´ 1 ` γ 2 M1 1 ` γ2 1
ñ “ “1´ ě1ď1´ ?
M 1 ` M2 p1 ` γqM1 p1 ` γq 2
as required.

156
 (b) For two extremal RN black holes we have r` “ M so each has area 4πM 2 . They will collide to form a
neutral (perhaps rotating) black hole. The area law gives us
b
4πppMf ` Mf2 ´ a2 q2 ` a2 q ě 2 ˆ 4πM 2 .

This bound is sharpest if we take the final state to be extremal Kerr a “ M , giving
8πMf2 ě 8πM 2 ñ Mf ě M
We get
E 1
E ď 2M ´ M ñ ď .
2M 2
(c) Such a decay would look like
? ?
2Mf2 ě M 2 ñ 2Mf ě M ñ M ´ 2Mf ď p 2 ´ 2qMF ă 0.
This is a contradiction.
a
3. We have that n “ ?g1rr Br “ f prqBr so that
a
1 1 f prq ´ f 1 prq ¯
Kµν “ ? Br Gµν “ diag f 1 prq, ´ , 2r, 2r sin2 θ
2 grr 2 f prq

Contracting with the 3 ˆ 3 boundary inverse metric hµν “ diagpf prq´1 , r´2 , r´2 sin´2 θq which has no r
component gives ? ˆ 1 ˙
f f 4 a rf 1 ` 4f ˇˇ
K“ ` “ f ˇ
2 f r 2rf r“r0

where r0 is large and formally infinite. We can then evaluate

?
4πr2 β f ˇˇ
ż ?
1 r 1
ˇ
hK “ Kˇ “β prf ` 4f qˇ
ˇ
8πG BM 8πG r“r0 4G r“r0
2
Directly evaluating this for f prq “ 1 ´ 2GM
r ` Qr2 gives
ˆ 2 ˙
β Q
` 2r0 ´ 3GM
2G r02
This is the gravitational boundary term contribution to the classical action. The gravitational bulk term is
zero since the Ricci scalar vanishes for the RN solution. The electromagnetic contribution is
żβ ż ż r0
Q2 4πQ2 1
ż ˆ ˙ ˆ ˙
1 ? µν 1 2 1 β 2 1 1
gFµν F “ dτ dΩ2 r dr 4 “ β ´ “ Q ´
16πG M 8πG 0 r` r 8πG r` r0 2G r` r0
All together, as r0 Ñ 8 we get action:
Q2
ˆ ˙
β
SRN “´ 2r0 ´ 3GM `
2G r`
Note that there is one divergent term, namely the one linear in r0 in the boundary action, but this is
insensitive to the properties of the RN black hole and is also present in flat space. It is then sensible to
define a regularized (renormalized) action by subtracting this term off. In doing this subtraction, there is
an ambiguity of how we should define the inverse ?temperature of the reference flat space subtraction. The
appropriately redshifted temperature Justify is β f , giving reference action:
β a β
Sf lat “ ´ r0 f pr0 q “ ´ p2r0 ´ 2GM ` Op1{r0 qq
G 2G
The renormalized Euclidean action is thus
β Q2 β
IRN “ SRN ´ Sf lat “ pM ´ q “ pM ´ µQq “ βF
2 Gr` 2

157
4. The specific heat C is given by the coefficient in

dM “ M CdT

For Schwarzschild, T “ p8πGM q´1 so this is

dM
dM “ ´M C ñ C “ ´8πGM
8πGM 2
Which is negative. This should not be so surprising, given that by increasing the energy (ie mass) of the
Schwarzschild black hole we make a larger one which thus have lower temperature. It is worth noting that,
including units, this is proportional to ~1 .

5. First off, at a “ 0 Kerr-Newman reproduces the RN black hole, which we already know is a solution of the
Einstein-Maxwell system.
Further, it is quick to check using Mathematica that at Q “ 0 the Kerr metric is itself Ricci-Flat: Rµν “ 0
so is indeed a solution of the vacuum Einstein equations (away from r “ 0).

When Q ‰ 0 we get a nonzero Ricci tensor (the Ricci scalar still vanishes since classical electrodynamics is
conformal).

The Ricci tensor must correspond to an electromagnetic stress-energy tensor. It comes from an electric po-
arQ sin2 θ
tential of the form Aµ “ p rQ
Σ , 0, 0, ´ Σ q

158
For r very large we get an electric field going as qr2 {Σ2 „ q{r2 corresponding to the electric field for a charge
q, and we also get a magnetic field dying şoff as a cos θ{r3 corresponding to the field from a spinning charged
1 1
ş
source. Said another way, we see that 4π ‹F “ q and 4π F “ 0 asymptotically, so we have just an electric
charge q.
We can verify mass and angular momentum using the killing vectors Bt and Bφ respectively using the formulas
in Wald 12.3.8-9 ż ż
1 c d 1
´ abcd ∇ pBt q “ M abcd ∇c pBφ qd “ aM
8πG 16πG

For the KN black hole metric, the only singularities can come from Σ “ 0 or ∆ “ 0. Σ is only zero for
a ą 0 when r “ 0, θ “ π{2. This corresponds to the curvature singularity of the black hole (in fact despite
deceptive coordinate choice, this takes the form of a ring S1 ˆ R as is revealed in Kerr-Schild coordinates).
The horizons come from grr becoming singular, namely ∆ “ 0 which occurs at
a
r2 ´ 2GM r ` a2 ` Q2 “ 0 ñ r˘ “ M ˘ M 2 ´ a2 ´ Q2 .
These give the outer and inner horizons.
The horizon area is given by
ż π ż 2π ˇ żπ b
? 2 ` a2 q2 ´ ∆a2 sin2 θ
dθ dφ gθθ gφφ ˇ “ 2π dθ sin θ
ˇ
pr`
0 0 r“r` 0
But ∆ “ 0 at the horizon so this trivializes to
a
2
4πpr` ` a2 q “ 4πppm ` m2 ´ a2 ´ Q2 q2 ` a2 q
The entropy of the black hole is then
A 2
S“ “ πpr` ` a2 q
4
Taking care to write things in terms of J and not a now, by holding J, Q fixed, let’s vary M and get

159
 The Hawking temperature is thus a
1 M 2 ´ a2 ´ Q2
TH “ 2 ` a2
2π r`
Now let’s fix S and Q. We get

Which gives us that ˆ ˙ ˆ ˙´1

´ BM ¯ dS dS a
Ω“ “´ “ 2
BJ Q,S dJ Q,M dM Q,J r` ` a2
Finally let’s hold S, J fixed and do the same procedure, giving

ˆ ˙ ˆ ˙´1
´ BM ¯ dS dS Qr`
µ“ “´ “ 2 ` a2
BQ J,S dQ J,M dM Q,J r`

The full form of the first law is then

dM “ T dS ` ΩdJ ` µdQ

We obtain an extremal black hole when M “ a2 ` Q2 , as this is the minimum value of M where r` is a
well-defined radius. At this value, r` “ r´ .
Thermodynamic stability comes from minimizing the Gibbs free energy:

G “ M ´ T S ´ ΩJ ´ µQ

Note that for flat space, G “ 0, so if G ą 0 for any of these black holes, thermal fluctuations will eventually
drive their decay to flat space.
Plugging in what we have gives

Notice that if J ą 0 then this will always be greater than zero, by virtue of the fact that M ą Q always.
If we take J “ 0, we get that this is still thermodynamically unstable unless Q “ M and the black hole is
extremally charged.

6. The Hawking evaporation rate gets modified as

σabs pωq d3 k
ΓH “
~ ´ qΦqq ¯ 1 p2πq3
exppβp~ω ´ ~s ¨ Ω

~ is the angular momentum product (orientation of ~s relative to Ω

where ~s ¨ Ω ~ matters). The ¯ is for bosons
and fermions respectively.
Return to understand how this generalizes to systems more broadly

160
7. This is direct - we take M theory on T 6 . Wrap Q1 M2 branes along x9 ´ x10 , Q2 M2 branes along x7 ´ x8 ,
and Q3 M2 branes along x5 ´ x6 .
Take the M-theory S 1 to be along x10 . Now, taking this to be microscopic first, we get Q1 strings wrapping
x9 together with Q2 , Q3 D2 branes in IIA.
T -dualize along x5 , x6 to get Q3 D0 branes, Q2 D4 branes wrapping x5´8 , and Q2 D2 branes wrapping x5 , x6 .
The F1 around x9 is unaffected.
Finally, T -dualize along x9 , taking IIA to IIB and giving Q3 D1 branes and Q2 D5 branes, while replacing
the F1 (ie B-flux) with KK momentum of the system along the x9 . This is exactly the D1-D5 system.

8. The D5 and D5 branes are both BPS. We know that, upon toroidal compactification,
In D “ 10 have the D1 stretch x0 “ t, x5 “ γ and the D5 stretch x0 , . . . , x5 , where we write γ a , a “ 1 . . . 4
to be the new D5 directions. These will form the direction of the T 4 .
Upon compactifying on T 4 ˆ S 1 , the logic we used to for the 10D solution will still carry over to 5D. We will
still write the extremal metric in terms of functions H1,5 that must be harmonic w.r.t. the flat metric of the
4D transverse space.
Then the D1 brane solution gives

´dt2 ` dγ 2 a a r16
ds2D1 “ ? ` H1 dγ a ¨ dγ a ` H1 dxi ¨ dxi , H1 “ 1 ` , e´2Φ “ H1´1 , F05i “ Bi pH1´1 q
H1 r6

While the D5 brane gives

´dt2 ` dγ 2 dγ a ¨ dγ a a r52
ds2D5 “ ? ` ? ` H5 dxi ¨ dxi , H5 “ 1 ` , e´2Φ “ H5 , Fijk “ ´ijk Br H
H5 H5 r2

I think this problem has a typo and Kiritsis means D1 , D5 not N1 , N5 . Further, Kiritsis (likely borrowing
from Maldacena’s thesis) writes F05i “ ´ 12 Bi pH ´1 ´ 1q. This factor of 1{2 is different from what I’m used
to seeing both in Kirtsis and Blumenhagen. This stems from a different choice of normalization for the
Kalb-Ramond and RR forms in Maldacena’s thesis. I am unsure why this different normalization exists, but
at any rate I will ignore the factor of 1{2. Finally, the overall sign in F disagrees also with Kiritsis and
Blumenhagen, and I think the total D-brane charge in Maldacena counts anti-D-branes in our scheme.
When superimposing a D1 and D5 solution, the dilaton and field strength contributions add while the metric
contributions get multiplied. One way to see this is, because the solution remains BPS, we only need to
solve the first-order BPS equations.
For a p-Brane, as we have seen, the Killing spinors have spatial profile prq “ H 1{8 0 regardless of p. The
linear equations for spinors coincide with the D-brane equations L “ ˘Γ0 . . . Γp R . We know that for the
1-5 system these can be simultaneously solved, giving a 1/4 BPS state.
I could do this in more detail... but I’ve computed enough Killing spinors by this point.
The combined 10D solution thus gives:
c
´dt2 ` dγ 2 H1 a a H5
? ` dγ ¨ dγ a ` H1 H5 dxi ¨ dxi , Fr05 “ Br H ´1 , Fijk “ ´H 1 prq, , e´2Φ “
H1 H5 H5 H1

The next step is compactification. Upon wrapping D5 and D1 around a T 5 , dimensional reduction freezes
out γ, γ a dependence of the metric and fields. The T 5 is parallel to the D5, so the D5 solution will look
identical to how it looked before. The D1 also wraps a cycle of the T 5 . Compactifying the other 4 directions
will look like a periodic arrangement of D1 branes, which effectively serves to remove γ a dependence from
the D1 contribution to the solution. Think about this. Is it really true that the metric warps the
same regardless of where on T 5 I am? More likely that they are taking T 5 small and neglecting
it, or we’re thinking about a uniform distribution of D1s on T 5 .
Finally, the D1 solution can be given momentum.

161
 9. Ignoring the ´1{2 discussed before, we can verify the charge from direct integration along a S 3 . First, the
electric charge
2r12 4π 2 p2π`s q4 V 2
ż ż
1 1 Q1
2 ‹F “ 2 3
“ ´ 7 8 2
r1 “ “ Q1 T1
2κ10 S 3 ˆT 4 2κ10 S 3 ˆT 4 r p2πq `s gs 2π`2s g
for r12 “ `2s gs {V , as required.
For the magnetic charge, Fθφψ “ θφψr Br H5 “ ´H51 so we get
2r52 4π 2
ż ż
1 1 Q5
2 F “ 2 dΩ 3 3
“ 7 8 2
r52 “ “ Q5 T5
2κ10 S 3 2κ10 S 3 r p2πq `s gs p2πq5 `6s gs
for r52 “ `2s gs , as required.
We can also derive cp from the KK solution Do this.
In the non-extremal case, this generalizes quite directly.
p2π`s q4 V gs 2r12 V r02 sinh2 a1 coth a1 r02 sinh 2a1 r02 sinh 2a1
ż ż
Q1 1
“ ‹F “ coth a1 “ “ ñ Q1 “
2π`2s gs 2κ210 S 3 ˆT 4 p2πq7 `8s gs2 S 3 r3 2πgs2 `4s 4πgs `2s c1 2c1
Similarly:
2r52 r02 sinh2 a5 coth a5 r02 sinh 2a5 r02 sinh 2a5
ż ż
Q5 1 1
5 6
“ 2 ‹F “ coth a5 “ “ ñ Q5 “
p2πq `s gs 2κ10 S3 p2πq7 `8s gs2 S3 r3 p2πq5 gs2 `8s 2p2πq5 `6s gs c5 2c5
For the KK momentum, I assume it can be read off from the dtdγ term justify (Maldacena writes this too,
r2 sinh 2a
in his thesis below 2.34), which goes as r02 sinh ap cosh ap “ 0 2 p “ cp Qp , giving KK momentum
r02 sinh 2ap
Qp “
2cp
10. First, by analogy to 5D we expect an extremal metric of the form
4
ź ri
´λ´1{2 dt2 ` λ1{2 pdr2 ` r2 dΩ22 q, λ“ p1 ` q
i“1
r
?
This will have a nonzero area 4π r1 r2 r3 r4 only when all the ri ‰ 0. On the other hand the total mass is
M “ 4i“1 Mi with Mi “ ri {4G. The question is what the charges correspond to at the level of a brane
ř
construction.
Towards this end, let’s take IIA and compactify on T 6 . We consider a D6 brane wrapping x1 . . . x6 together
with a D2 wrapping x1 , x6 . 6 ´ 2 “ 4 is good, makes the state 1/4 BPS. We can also add KK momentum
along the 1 direction.
The crucial principle (Maldacena 2.5) is that if a scalar diverges at the horizon, the d-dimensional character
of the solution is lost. For a single p-brane p ‰ 3 the dilaton goes either to 8 or 0. In the case of the D1-D5
system, we needed branes symmetric about p “ 3 and differing by 4 in order to give a BPS state with the
dilaton tending to a constant 41 log H1 {H5 Ñ 12 log r1 {r5 .
We see now that this does not work with a D6 and D2. For a p-brane e´2Φ “ H pp´3q{2 giving that D6-D2
3{2 ´1{2
gas a dilaton going as e´2Φ “ H6 H2 . There’s no (even dimensional) D-brane we could add in type 1
that would save us, and adding fundamental strings would only give e´2φ “ Hf , which would not help.
But there is another extended object with the correct dilaton dependence as e´2Φ “ H ´1 . This is the NS5
brane! But will adding it break supersymmetry completely? On the contrary, the SUSY constraints from
the D6 and D2 and KK momentum are:
L “ Γ0123456 R , L “ Γ016 R , L “ Γ01 L , R “ ´Γ01 R
The NS5 brane wrapping 1 . . . 5 would give L “ Γ012345 L , R “ ´Γ012345 R . This can be rewritten as
L,R “ ˘Γ6 Γ0123456 R,L “ Γ6 L,R
But L “ ˘Γ6 L already follows from the prior supersymmetry constraints, so adding NS5 breaks nothing!

162
11. Directly applying the formula derived in problem 1 with F “ f ´1{3 , C “ f ´1{3 h gives
2
2π a “ 2πr0 coshpa1 q coshpa5 q coshpap q
F pr0 qC 1 pr0 q

12. This is direct by writing the differentials in terms of variables r0 , a1 , a5 , ap :

The variations do not involve arbitrary changes in the N˘i . This is not obvious from the form of the first
law as far as I can tell, but N˘i do need to be discrete in the brane interpretation.
13. I have done this problem for Andy’s class on quantum black holes. I will copy the full answer below: BTZ
as a Quotient of AdS3 The objective of this problem is to describe the precise way in which the BTZ
black hole arises as a quotient of AdS3 . Take the embedding space to be R2,2 , with metric:

η “ diagp´1, ´1, 1, 1q

Denote the coordinates of the embedding space by xµ “ px0 , x1 , x2 , x3 q. AdS3 is given by xµ xµ “ ´`2 . The
Killing vectors generating isometries are given by Jµν “ xν Bµ ´ xµ Bν . The most general Killing vector is
then ω µν Jµν .
Define the identification subgroup of AdS3 by

P „ etξ P, t P 2πn, nPZ

For this identification to make physical sense, it should not give rise to closed timelike curves. Unfortunately,
in some regions, the ξ used in this construction do give rise to CTCs. Luckily, however, they are bounded
by a region where ξ ¨ ξ “ 0. The part of the spacetime where ξ ¨ ξ “ 0 is then interpreted as a singularity in
the causal structure, and the region where ξ ¨ ξ ă 0 is cut out of the spacetime. Let
r` r´
ξ“ J12 ´ J03
l l
(a) Write down ωµν for this Killing vector, and the corresponding Casimir invariants for the
SOp2, 2q isometry group, which are given by
1
I1 “ ωµν ω µν , I2 “ µνρσ ωµν ωρσ
2
We have
r` r´
ω12 “ ´ω21 “ , ω03 “ ´ω30 “ ´
l l
Giving Casimirs:
2 ` r2
r` ´ r` r´
I1 “ ´2 , I2 “ 4
l2 l2

163
(b) Find the allowed region ξ ¨ ξ ą 0 in terms of x1 , x2 .
We can write:
¨ r´ ˛ ¨ r´ 3 ˛
´ l x3 ´ l x
r` r´ ˚ r` x2 ‹ ˚ r` x2 ‹
ξ“ px2 B1 ´ x1 B2 q ´ px3 B0 ´ x0 B3 q “ ˚ l ‹ ˚ l
˝´ r` x1 ‚ “ ˝ r` x1 ‚
‹
l l l l
r´
l x0 ´ rl´ x0
This vector has norm:
1 2 2 1 2
ξ2 “2
rr` px1 ´ x22 q ´ r´
2
px23 ´ x20 qs “ 2 pr` 2
´ r´ qpx21 ´ x22 q ` r´
2
l `
This is ě 0 when (assuming r` ą r´ )
2 l2
r´
x21 ´ x22 ě ´ 2 ´ r2
r` ´
2 q, pr 2 , r 2 q, pr 2 , 8qu and identify whether the boundaries
(c) Find the regions for ξ ¨ ξ P tp0, r´ ´ ` `
between them are timelike, spacelikeor null:
Now let’s look at
2
r` ď ξ 2 ñ pr`
2 2
´ r´ qpx21 ´ x22 q ě pr´
2 2 2
` r` ql ñ l2 ď x21 ´ x22
Next
2
r´ ď ξ 2 ď r`
2
ñ 0 ď x21 ´ x22 ď l2
Finally
2 l2
r´
0 ď ξ 2 ď r´
2
ñ´ 2 2
2 ´ r 2 ď x1 ´ x2 ď 0
r` ´
All boundaries between these regions are null. The boundary x21 ´ x22 “ l2 implies x20 ´ x23 “ 0 and so
is a null surface (cone). Similarly, the boundary x21 ´ x22 “ 0 implies x` “ ˘x´ which is again null
surface.
a a
(d) For region I use the coordinate transform given by x0 “ Bprq sinh t̃, x1 “ Aprq cosh φ̃, x2 “
r2 ´r2
Aprq sinh φ̃, x3 “ Aprq cosh t̃ where A, B “ l2 r2 ´r¯2 and t̃, φ̃ “ 1l p˘ trl˘ ¯ r¯ φq to write the
a a
` ´
metric in a form
´NK2 dt2 ` NK´2 dr2 ` r2 pNφ dt ` dφq2
Note we get x21 ´ x22 “ Aprq, x23 ´ x24 “ Bprq, and Bprq “ Aprq ` l2 so that B ´ A “ l2 as required
2 ´r 2 ql2
pr` ´
in AdS. At r “ r` , A “ ´ 2 ´r 2
r`
exactly saturating the boundary of region 1. Simple differential
´
manipulations

164
 gives us b
2 qpr 2 ´ r 2 q
pr2 ´ r´ ` r` r´
NK “ , Nφ “
lr lr2
For region II, taking r´ ă r ă r` makes B negative, so we will keep x1 , x2 as before and instead define

x0 “ ´p´Bprqq1{2 cosh t̃, x3 “ ´p´Bprqq1{2 sinh t̃

Here we have flipped the sign of B together with exchanging sinh and cosh (so as to remain in coordinates
satisfying the AdS constraint). This is exactly as in Kiritsis 13.7.8, referred to as the “standard
BTZ form”
We keep t̃, φ̃ the same. This gives the same value for NK and Nφ . Lastly for region III, A also becomes
negative, and we redefine x1 , x2 while keeping x0 , x3 from region II:

x1 “ p´Aprqq1{2 sinh φ̃, x3 “ p´Aprqq1{2 cosh φ̃

The here r ranges from 0 to 8 while t, φ are unrestricted and range from ´8 to 8
(e) Compute the Killing vector ξ in the pt, r, φq coordinates and perform the identification. You
should recognize the metric found in the previous part as the BTZ geometry. Identify
M, J in terms of r˘ and write the casimir invariants from part a) in terms of M, J
0 1 2
,x ,x ,x q 3
By computing the Jacobian Bpx Bpt,r,φq and judiciously guessing what vector I should push forward
(way easier than trying to compute inverse Jacobians to pull back ξ), I see that ξ “ Bφ in our new basis:

Indeed, it is easy to see that Bφ is killing from directly applying the Killing equation, and now we see
it comes directly from a combination of the manifest Jµν symmetries in the embedding space. We can
thus identify φ as a periodic variable and retain the space as a solution to Einstein’s equations.
Looking at the NK2 and Nφ2 contributions to g00 we get:

2 ` r2
r´ ` r2
´g00 “ ´ `
l2 l2
r2 `r2
This is a black hole in AdS with mass M “ ´ l2 ` . Similarly, from the dt dφ component, we see that
2r2 Nφ corresponds exactly to the angular momentum. We thus get
2 ` r2
r` ´ 2r` r´
M“ , J“
l2 l
Note for global AdS when r´ Ñ 0, r` Ñ ´l2 , we get mass ´l2 . Kiritsis adjusts the definition of
mass by `1 making it 0 in global AdS, so that it counts only the mass of the black hole.
This gives
I1 “ ´2M, I2 “ 2J{l

165
14. The KK reduction is not too bad:
ż a ´ ” 1 1 ı
d5 x det g e´2φ R ` 4Bµ φB µ φ ` Bµ Gαβ B µ Gαβ ´ Gαβ Fµν
α µν β
F
4 4
1? 1 ? 1? ¯
´ GGαβ Gγδ Hµαγ H µγδ ´ GGαβ Hµνα Hβµν ´ GHµνρ H µνρ
4 4 12
with φ “ Φ ´ 14 log det Gαβ . Here H comes from the two form not from the B field.I’m almost certain
that the formula in Kiritsis is wrong
We can rewrite this as
ż a ´ ” 1 ? ˘ 1 ı
d5 x det g e´2φ R ` 4Bµ φB µ φ ´ Gαβ Gγδ BGαγ BGβδ ´ e2φ GBCαγ BCβδ ´ Gαβ Fµν
α µν β
`
F
4 4
1? µν 1 ? ¯
´ GGαβ Hµνα Hβ ´ GHµνρ H µνρ
4 12

Now we must take this to the Einstein frame. We perform a Weyl rescaling g Ñ e4φ{3 g. This rescales fields
(and changes the kinetic φ term) to give us the requisite action

1
ż
? ” 4 1 ?
S5 “ 2 ´g R ´ pBφq2 ´ Gαβ Gγδ pBGαγ BGβδ ` e2φ GBCαγ BCβδ q
2κ5 3 4
e´4φ{3 e2φ{3 ? e´2φ{3 ? ı
´ α µν β
Gαβ Fµν F ´ GGαβ Hµνα Hβµν ´ GHµνρ H µνρ
4 4 12

Each of the field strengths will obey:

α
d ‹ re´4φ{3 Gαβ Fµν s“0
2φ{3
? αβ
d ‹ re GG Hα µν s “ 0
?
d ‹ re´2φ{3 G Hµνρ s “ 0
?
∇µ rGαβ Gγδ e2φ GBµ Cβδ s “ 0

The dilaton will obey:

8 1 ? 4 α β µν 1 2φ{3
? 1 ?
l φ´ e2φ G Gαβ Gγδ BCαγ BCβδ ` e´4φ 3Gαβ Fµν F ´ e GGαβ Hµνα H µνβ ` e´2φ{3 GHµνρ H µνρ
3 2 3 6 18
Finally, the metric will obey:
1 4 1 ?
Rµν ´ gµν R ´ Bµ φBν φ ´ Gαβ Gγδ pBµ Gαγ Bν Gβδ ` e2φ GBq
2 3 4

The solution in question has nonzero (magnetic) Hµνρ and (electric) Hµν , which are functions of r alone.
Finish
show solution
b
15. Write Gαβ “ H H2 pδαβ ` hαβ q
1

Finish
5 to be nontrivial.
16. Here we only take the two-form field strengths H5µν Fµν
By redefining φ̃ “ φ ` 21 ν5 , λ “ ´ 12 φ ` 43 ν5 , we get

4 4
´pB φ̃q2 ´ pBλq2 “ ´ pBφq2 ´ pBν5 q2
3 3

166
 The φ term matches, and the ν5 kinetic term together with the four pBνq2 terms exactly reproduces the
expected ´ 14 Bµ Gαβ B µ Gαβ . Let’s look at the field strengths
? 2 2 4
e´2φ{3 GHµνρ “ e´2φ{3`ν5 `4ν Hµνρ “ e 3 λ`4ν Hµνρ
2
? 4
e2φ{3 GG55 H5νρ
2
“ e2φ{3´ν5 `4ν H5νρ
2
“ e´ 3 φ`4ν H5νρ
2

8
5 2 5 2
e´4φ{3 G55 pFµν q “ e´4φ{3`2ν5 pFµν q “ e 3 λ pFµν
5 2
q

These all exactly match. Note that these scalars have a potential- they are not minimally coupled. Conse-
quently, at the horizon, where the field strengths diverge, we expect that the values of these scalars will be
fixed by the equations of motion.

17. This is direct:

Importantly, the derivative of the inner R profile goes as Opr02 q, which makes it subleading in determining
B. Thus, B is Opr02 q. On the other hand, this derivative’s r02 dependence is important for determining the
incoming flux: Imphr3 R˚ Br Rq. For completeness I will add this part of the notebook as well:

18. Redefining K to be unitless, we have a relationship

e´iωx
eix cos θ “ K Y000 ` higher moments
x3{2
Let’s integrate both over S 3 , giving
żπ żπ ż 2π
J1 pxq eiωx ?
dθ dφ dψ sin2 θ sin φeix cos θ “ 2π 2 “ K 3{2 2π 2
0 0 0 x x
The asymptotic form of J1 is c
2
J1 pxq „ cospz ´ 3π{4q
πx
So then, up to a phase ? ? ?
2π 2π “ K 2π 2 ñ K “ 4π
as required.

167
19. In spacetime dimension d ` 1, one can write a black hole metric as:

r0d´1
„ 2 
2 2 1{pd´1q dr ?
ds “ ´f ´1`1{pd´1q
hdt ` f ` r dΩd´1 ñ ´g “ f 1{pd´1q rd´1 ,
2 2
h“1´
h rd´1

I will ignore the details of f for now, since it depends a lot on the dimension and charges. h is simply what
reproduces the Schwarzschild solution for a totally uncharged black hole. What does matter is that the
pd´1q
horizon is at r “ r0 , giving a horizon area A “ Ωd´1 r0 f pr0 q1{2 , meaning that to leading order as r Ñ r0
we have
2pd´1q
RH d´1 A
f prq « 2pd´1q , RH “
r Ωd´1
0
This is all we need for the near-horizon data.

What matters is that the throat size that is determined by f is much larger than the extremality parameter
r0 . For a minimal scalar lφ “ 0 in d ` 1 dimensions we get
„ 
h ´
d´1
¯
2
Br hr Br ` ω f Rω prq “ 0
rd´1

For r small and close to r0 define the coordinate σ by

dr
dσ “ ñ Bσ “ hprqrd´1 Br
hprqrd´1

This give us that the wave equation becomes

rBσ2 ` ω 2 rd´1 f prqsRω prq “ 0

Now, in the r Ñ r0 limit this simplifies to

2pd´1q d´1
d´1
rBσ2 ` ω 2 RH sRω prq “ 0 ñ R “ Ãe´iωσRH « Ãp1 ´ iωσRH q

where we have picked the sign in the exponent so that the wave in the near region is purely incoming. In the
final approximation, we’re looking at the extreme near-horizon limit. Further, keeping rω small but looking
´d`2
at large r compared to r0 we see that the behavior of σ is given by σ « ´ rd´2 , yielding

d´1
RH
Rprq « Ãp1 ´ iω q (95)
pd ´ 2qrd´2

Now for the far region:

168
 d´1
Redefining ψ “ r R and introducing the tortoise coordinate dr˚ “ dr{h so that Br˚ “ hBr we get
2
˜ ¸˜ ¸
” d2 pd ´ 1qpd ´ 3q r0d´2 d ´ 1 r0d´2 2
ı
´ 2 ` 1 ´ 1 ` ´ ω f prq ψprq “ 0
dr˚ 4r2 rd´2 d ´ 3 rd´2
looooooooooooooooooooooooooooooooooooomooooooooooooooooooooooooooooooooooooon
V pr˚ q

This reproduces 13.8.4 when d “ 4. We don’t expect this to be solvable, and so we will work at it by
matching. Again, r0 , rp ! rm ! r1 , r5 .
For r large, r “ r˚ and we can divide through by ω giving ρ “ rω. The equation then reduces to
ˆ 2 ˙
d pd ´ 1qpd ´ 3q
` 1 ´ ψ
dρ2 4ρ2
This has a solution in terms of Bessel functions:
c
πρ
ψ“ rAJ´1`d{2 pρq ` BY´1`d{2 pρqs
2
This implies that asymptotically:
1 π π
R « pd´1q{2 reiωr e´iπ{4 e´ipd´2q{4 pA ´ Beipd´2q 2 q ` e´iωr eiπ{4 eipd´2q{4 pA ´ Be´ipd´2q 2 qs
r
The absorption probability is then
ˇ 1 ` B ei 12 pd´2q ˇ2
ˇ ˇ
A
Γ“1´ˇ ˇ .
ˇ ˇ
ˇ 1 ` B e´i 12 pd´2q ˇ
A
We need an odd number of spatial dimensions for this to work, reflecting the fact that the Bessel functions
degenerate in these cases. There is probably a cleaner way here.
Taking ρ “ rω ! 1 at r “ rm for matching, the Bessel functions will become at leading order:
c ˜ ¸
1 πωr 21´d{2 ´1`d{2 2pd´2q{2
R« A prωq `B
pωrqpd´1q{2 2 Γpd{2q Γp2 ´ d{2qprωqd{2´1
? p1´dq{2 ? pd´3q{2 (96)
π2 π2
“A `B
Γpd{2q Γp2 ´ d{2qpωrqd´1
Now let us match (96) onto (95). This is direct, and gives:
A Γpd{2q B Γp2 ´ d{2qpωRH qd´1 2p3´dq{2 B 22´d Γp2 ´ d{2qpωRH qd´1
“ p1´dq{2 ? , “i ? ñ “i
à 2 π à πp2 ´ dq A p2 ´ dqΓpd{2q
We will then get in the ω Ñ 0 limit:
ˇ 23´d Γp2 ´ d{2qpωR qd´1 ˇ
H
Γ“ˇ
ˇ ˇ
ˇ
pd ´ 2qΓpd{2q
Now, following the discussion of 13.18, we must account for the conversion factor K from partial waves to
plane waves. In d spatial dimensions we get this to be
d
p2πqd´1
K“
ω d´1 Ωd´1
This gives
ˇ p2πqd´1 Γp d q 23´d Γp2 ´ d{2qpωR qd´1 ˇ
H
σabs “ Γ|K|2 “ ˇ d´1 d{22
ˇ ˇ
ˇ
ω 2π pd ´ 2qΓpd{2q
ˇ 2π d{2´1 Rd´1 Γp2 ´ d{2q ˇ ˇ d´1
2π d{2´1 RH π ˇ
ˇ H ˇ ˇ ˇ
“ˇ ˇ“ˇ ˇ
pd ´ 2q pd ´ 2qΓpd{2 ´ 1q sinpd{2 ´ 1q
d´1
2π d{2 RH
“ “ AH .
Γpd{2q
as required. Literally after all that I’m off only by a factor of 2.

169
20. We have seen that the Ricci scalar for a Dp brane solution takes the exact form:

L2p7´pq pp ` 1qpp ´ 3qpp ´ 7q2

R“ p´3
4r 2 pr7´p ` L7´p q5{2

For p “ 3, this vanishes identically. On the other hand, the dilaton EOM yields

R “ 4p∇Φq2 ´ 4l Φ

since the solution Φ is a constant, at leading order about a classical solution, this equation reads

lΦ “ 0.

Thus, the dilaton is indeed minimal. We do not expect such nice simplification for other Dp branes.
Now, let us calculate cross section per
aunit D3 branea
volume. We consider s-wave scattering. This wave
?
equation in the D3 background grr “ Hprq, g “ r5 Hprq translates to

L4
ˆ ˙ ˆ ˙
´ 1
rr ? 2 tt
¯ 1 5 2
? 1 5 2
0 “ ? Br g gBr ` ω g Rprq “ ? Br r Br ` ω H R ñ 5 Br r Br R ` ω 1 ` 4
g r5 H r r

As before, lets redefine ψ “ r5{2 R. We get the equation

L4 L4
„ ˆ ˙  ˆ ˙
2 2 15 15 2
0 “ ψ prq ` ω 1 ` 4 ´ 2 ψprq ñ Vef f “ 2 ´ ω 1 ` 4
r 4r 4r r

Again, this does not look like it has an analytic solution (actually apparently it does and its a Matthieu
function, but we don’t really know that). Taking ρ “ ωr and looking at ρ " 1, we drop the L4 {ρ4 term and
obtain solutions c
πρ
ψ“ rAJ2 pρq ` BY2 pρqs
2
For large ρ these asymptote to
1
R« reirω pA ´ iBqe3iπ{4 ` e´irω pA ` iBqe´3iπ{4 s
2r5{2
For the small r limit, on the other hand, we get

pLωq2 pLωq2
c „ 
πρ
ψ“ ÃJ2 p q ` B̃Y2 p q
2pLωq2 ρ ρ

For this to be an incoming wave in this region, we require the combination

pLωq4 pLωq2 pLωq2

„ 
R “ Ã 2 J2 p q ` iY2 p q (97)
ρ ρ ρ

We want to match this at an intermediate value of r. Take ρ ! 1, and look at low frequencies. This will
allow us to write the first solution as
c c
1 πρ 1 πω 5 2
R “ 5{2 pAJ2 pρq ` BJ2 pρqq « A´ a B
r 2 8 2 r ω 3 π{2
4

We see that for small r, the second term blows up, and we expect that for matching to hold, we must take
B “ 0.
Meanwhile, equation (97) for small r gives in the ω Ñ 0 limit an expansion:

4iÃ
R«´
π

170
For these to match we must have:
c c
1 πω 5 4ià à 1 ω5π3
A“´ ñ “ i
8 2 π A 32 2
The conserved flux is
1 “ 5 ˚ ‰
F“ r R Br R ´ c.c.
2i
For the incoming wave this is Fin “ ´ω|A{2|2 . For the absorbed one, a quick Mathematica computation
8 4
gives Fabs “ ´ 2Lπω |Ã|2

This gives
Fabs |Ã|2 8L8 ω 3 π 2 pLωq8
Rabs “ “ “
Fin |A|2 π p16q2
Now, following the discussion of 13.18, we must account for the conversion factor K from partial waves to
plane waves. In D spatial dimensions we get this to be
d
p2πqD´1
K“ Ñ 32π 2
ω D´1 ΩD´1

in our case of D “ 6. Altogether we get

π 4 L8 ω 3
σabs “ K 2 Rabs “
8
as required.
The fact that the B coefficient did not come into play confirms once again that the near-horizon regime is
all that matters in the calculation of σabs
For higher partial waves, the Bessel functions involved take the form J``2 . The outer region will continue
to have B “ 0 enforced, and look like
A
J2`` pρq
ρ2
while the inner region will look like

pLωq2 ” pLωq2 pLωq2 ı

à J2`` p q ` iY2`` p q
ρ2 ρ ρ

This will give a match like à „ pωLq´2` A. This makes it so that their ratio squared goes as pωLq4` . The flux
calculations remain the same. Altogether we expect σabs to scale as L8 ω 3 pLωq2` for higher partial waves.
Finally, the Hawking emission rate remains zero, since it involves a factor of e´βω and β “ 1{T “ 8 for an
extremal p-brane. There are likely corrections to this beyond the semiclassical level of analysis.

171
21. Let’s expand:
8
ź p1 ` q n q8
n q8
“ 1 ` 16q ` 144q 2 ` 960q 3 ` 5264q 4 ` . . .
n“1
p1 ´ q

For N “ 0 pT 4 qN {SN is just a point which has trivial cohomology ring with dimension 1.
For N “ 1 we recover T 4 which has 4 ˆ 4 cocycles generated as an alternating algebra by the elements
dxi , i “ 1 . . . 4, giving dimension 16. Note that we should view dxi as fermionc elements corresponding to
the odd cohomology, and even elements such as 1, dxi ^ dxj , dx1 ^ dx2 ^ dx3 ^ dx3 as bosonic.
For N “ 2 we get T 8 {S2 identifying points of two separate T4 s. Each individual T4 has all of its cycles
remaining intact, giving 2 ˆ 24 “ 32 cycles untouched. The remaining 28 ´ 24 cycles are half-killed, giving
2 ˆ 24 ` p28 ´ 2 ˆ 24 q{2 “ 144. Although this gives the right answer, I see that its not the most generalizable
way to look at things. There will always be an untwisted sector of this orbifold, as well as twisted sectors in
1-1 correspondence with conjugacy classes of SN . The untwisted sector simply considers N particle states
on T 4 . There are 8 fermionic elements and 8 bosonic elements in the cohomology. The types of 2-particle
states are thus:
8ˆ9 8ˆ7
` ` lo8omo
ˆ o8n “ 128
2 on
loomo 2 on
loomo
bose-fermi
bose-bose fermi-fermi

The twisted sector here is a single copy of T 4 , and any cycle is allowed. We thus get an additional 16 terms,
giving 144 as desired.
Now let’s look at N “ 3, the first case where SN becomes nonabelian. We expect an untwisted sector,
corresponding to the system of 3 point particles on T 4 . This gives
8 ˆ 9 ˆ 10 8 ˆ 7 ˆ 6 8 ˆ 9 8ˆ7
` ` ˆ8` ˆ 8 “ 688
3! 3! 2! 2!
as well as two twisted sectors, in 1-1 correspondence with the conjugacy classes p123q and p12qp3q of S3 .
The former gives a single T 4 , whose cohomology is 16. The other gives two (independent!) T 4 s, whose
cohomology then is 16 ˆ 16. Altogether we get:

688 ` 162 ` 16 “ 960

as required.
Let’s finally do N “ 4.

The generating function, for a manifold M with f odd cycles and b even cycles, consists of taking the
a , a “ 1 . . . dim H ˚ pM q. For each S twisted sector of n copies of M , we
generators of H ˚ pM q to be α´1 n
introduce “twisted modes” α´na .

Then, the generating function consists of taking products over all n so that for a given SN , the full
ř orbifold
is built up from taking all the ways one can partition N in terms of subsectors twisted by ni , ni “ N .
Thus, we look for the q N coefficient in:
8
ź p1 ` q n qf
.
i“1
p1 ´ q n qb

22. A derivation of Cardy’s Formula:

Take the CFT to have continuous spectrum, which we can write in terms of a δ-function based ρp∆q as
ż8
Zpτ q “ d∆ρp∆qe2πiτ ∆
0

172
 We can invert this using a Bromwich integral:
ż
ρp∆q “ dτ Zpτ qe´2πiτ ∆
C

where C is the contour running parallel but slightly above the real axis, enclosing the upper half plane. In
the q-disk this would run close to the boundary of the disk.
Now, we would like an expression for Zpτ q as = τ Ñ 0, namely the high-temperature limit. We know that
Zpτ Ñ 8q “ dim H0 , the space of ground states, which we take to consist of only a unique |0y, so we take
this to be 1. Further, we know that
2πic
q ´c{24 Zpτ q “ e´2πiτ c{24 Zpτ q “ Zp´1{τ qe 24τ

is modular invariant. This implies that

c c
Zpτ Ñ 0q « Zp8qe2πi 24τ “ e2πi 24τ

We now can approximate the integral:

ż8
c
ρp∆q « dτ e´2πipτ ∆´ 24τ q
´8

This gives a stationary value at c

c c
∆` 2
“0ñτ “i
24τ 24∆
Plugging this back and interpreting ρ as just an expected number of states at a given level ΩpN q gives
c
c?
b
2π c∆
ΩpN q “ ρp∆q « e 6 ñ S “ log ΩpN q « 2π N
6
I did this just for the left-movers, but taking left and right movers together gives the desired result:
c
c a a
S “ log ΩpNL , NR q « 2π p NL ` NR q
6

23. As we’ve seen before, a single free boson the partition function is Trpe´βL0 q “ η ´1 while for a two free
fermions, bosonization give the identical result. For a single free fermion then, at leading order (which
means just retaining the same central charge) this gives η ´1{2 . For nf copies of this system we simply
exponentiate to obtain the leading piece:
˜ ¸ 3 nf
8 2
ź 1
(98)
n“1
1 ´ e´βn{R

This can be directly written using q “ e´β{R ie τ “ iβ{2πR

˜ ¸ 3 nf
2
q 1{24
ηpτ q

We need the high 1{24 is subleading and can be ignored. Now the η function
a temperature limit, for which q
satisfies ηpτ q “ i{τ ηp´1{τ q d
2R
2πR ´ p2πq
ηpβn{R Ñ 0q “ e 24β
βn
The square-root term is also subleading and we obtain to leading order
3 p2πq2 R
log Z “ nf
2 24β

173
which is exactly what we want, once we remove units from R, R Ñ `s R.
In general, we also have fermionic contributions modifying the numerator in equation (98). Here, again
τ “ iβ{2πR. For a single periodic or antiperiodic fermion we will have traces that gives partition functions
of the form: d
8
ź β q 1{24 θ2 pτ q
TrP re´βL0 {2πR s “ p1 ` q 2πR n q “
n“0
η
d
8
´βL0 {2πR
ź β q 1{24 θ3 pτ q
TrA re s“ p1 ` q 2πR pn`1{2q q “
n“0
η
Taking the infinite temperature limit sets τ Ñ 0, q Ñ 8 giving respectively
d
θ4 p´1{τ q 2πi p2πq2 R
« e 48τ “ e 48β
ηp´1{τ q
d
θ3 p´1{τ q 2πi p2πq2 R
« e 48τ “ e 48β
ηp´1{τ q

So in both cases we retain the same contribution as the η ´1{2 divergence, with sub-leading terms being
different.

174
Chapter 14: The Bulk/Boundary (Holographic) Correspondence
1. The two diagrams are as follows:

2. Let us focus on vacuum diagrams. We will rescale the fundamental field (call it φ) so that the action has a
leading factor going as N . Schematically this is:
ˆ ˙
1 2 2
N TrF ` trpDφq
λ
By the same argument as for fermions in QED, any fermion worldline will give a closed curve, which we
interpret at giving a boundary of the Riemann surface. Moreover, each fundamental field loop will contain
exactly as many propagators as vertices (except for the trivial disconnected loop).
Each fundamental vertex will contribute N while each propagator will contribute N1 . The connected fun-
damental loops could be viewed as not contributing N because each fundamental line will join with gauge
boson lines which are already traced over. The fact that the fundamental loops do not contribute a face is
consistent with their interpretation as enclosing boundaries of a Riemann surface. Thus, the introduction of
each fundamental loop is equivalent to including a cycle with no interior, so we will not count the face, and
the vertices and edges don’t contribute either. The counting is then unmodified
ˆ ˙E ˆ ˙V
λ N
N F “ N χ λE´V
N λ

3. I think that the normalization of this operator (which is the same as in MAGOO) is just wrong. I think it
should be Φa “ Trr i Xini s McGreevy confirms this.
ś

Our single trace operator is a product over the distinct Xi fields

ź n
Φa “ Tr Xi i
i
ř
Each such insertion has i ni external legs.

175
 For Φa distinct, we consider:
m
1 ÿ
S “ N TrrdX i dX j ` cijk X i X j X k ` . . . s ` N ga pxqΦa pxq
λ a“1
Here the ga are taken to be functions of x.
We then compute:
m
ź 1 δ δ
x Φa pxa qy “ ... log Zrtga us
a“1
N m δg1 px1 q δgm pxm q
We can compute vacuum bubbles for this modified action as before. We now get a new vertex type involving
the single-trace operator Φa . Because it still appears with a coefficient N in the action, we can still apply
the same counting logic to the calculation of log Z. The spherical contribution dominates.
Thus, again the leading contributions to the free energy goes as N 2 , and we get that the correlator expression
behaves as N 2´m . In particular, the three-point function vanishes as 1{N , as said in the text.
1
4. Again, let’s take the operator without the N2
out front. To add in a double-trace operator requires two
factors of N out front.
We thus add to the action the term:
N
ÿ
Snew “ N 2 ga pxqΨa pxq
a“1
We get m-point correlators by differentiating m times by N 2 ga . With our choice of Snew , log Z still satisfies
the same Euler characteristic rules as before, and has a dominant contribution going as N 2 . We see that the
two-point function then goes as N ´2 . Normalizing the two-point function two 1 requires that we look at the
fields Ψ̃a “ N Ψa . The three point function then goes as N ´1 .
Understand the Silverstein paper about the connecting S 2 s contributing for double twist.
5. We add to the action the following (schematic) terms:
N TrrpDqq2 ` pDq̄q2 ` q q̄ ` q q̄Φ ` . . . s
Again the propagators will give 1{N and the vertices will give N , so we can apply the same analysis to get
that planar diagrams dominate. For the two point correlator of q q̄, we must introduce a quark loop into our
worldsheet. The lowest-genus such surface is the disk, with two q q̄s inserted on the boundary. This is genus
1. Differentiating with respect to N twice?I get a two-point function going as N ´1 . To get this to be unity
I must rescale my mesons operator to be N q q̄. Now the m-point correlation function goes as N m{2 N ? 1´m .

We thus get scaling behavior behavior N 1´m{2 for mesons. In particular the 3-point function goes as 1{ N .
6. It is important that in this case, for both SOpN q and Spp2N q, the fundamental representation F is real.
Consequently, the adjoint can be written (up to 1{N corrections) as the antisymmetrized (resp symmetrized)
part of F b F. In double-line notation we can understand the gluons as being labeled by two “fundamental
lines”. Because there is no difference between F and F, there is no inherent orientation to the strips, and
we can twist to form unoriented surfaces. Thus, the string theory that these would correspond to must
necessarily be non-oriented.
The difference between the orthogonal and symplectic projection will be in the relative sign of a propogator
with intermediate twist between Op2N q and Spp2N q. For Op2N q we have the same sign contribution between
the propagator and the propagator-with-crosscap. For Spp2N q, we have the opposite sign.
Draw this
We can talk about a large N expansion of diagrams identically. The only additional ingredient is incorporat-
ing points where edges swap. These play roles identical to cross-caps. The diagrams we can draw will have V
vertices with N {λ coefficient, E edges with λ{N coefficient, F faces, with N coefficient, and C “cross-caps”
with N ´1 coefficient. Altogether these gives
N V ´E`F ´C λE´V “ N χ λE´V
generalizing the prior discussion.

176
7. Take L “ 1. The relationships 14.4.7 become:

`s “ λ´1{4 “ p4πgs N q´1{4

and
p2πq7 `8s 2 p2πq7 `8s 1 π4
GN “ gs “ “ .
16π 16π p4π`4s N q2 4N 2

8. Starting with IIB the gravitational constant is 16πGN “ p2πq7 `8s gs2 . The volume of S 5 is π 3 L3 . We get:

8π 3 `8s 2
G5 “ g
L5 s
Recall in AdS/CFT, the coupling constant λ is the 4th power of L in string units:

L4
“ 4πgs N
`4s

Substituting this gives:

8π 3 L3 πL3
G5 “ “
p4πN q2 2N 2
This is a nice relationship, independent of the string length, and only dependent on the size of AdS and the
number of D-branes.

9. Massless and massive vector fields in AdS

(a) Massless Case Take the gauge Au “ 0. Let Aµ pu, ~xq “ u∆´1 Aµ p~xq. Note that this way Aµ dxµ has
scaling dimension u∆ . The equations of motion for the Maxwell theory give:
ˆ ˙
? MN ? MA NB 1 4 ∆´1
0 “ BM p ´gF q “ BM p ´gg g BrA ABs q “ Bu u Bu pu Aµ p~xqq ` . . .
up`2

where . . . are terms with higher powers of z. Altogether this is:

p∆ ´ pqp∆ ´ 1qpAµ p~xq “ 0.

This implies that either ∆ “ 1 or ∆ “ p.

Consequently this gives a scaling dimension of d´1 to J µ , exactly what we want for a conserved current.
(b) Massive case Now again we have:

? ? m2
0 “ BM p ´gF M N q ´ ´g 2 AN
L
? ? m2
“ BM p ´gg M A g N B BrA ABs q ´ ´g 2 AM
ˆ ˙L
1
“ Lp´2 Bz u4 Bu pu∆´1 AM p~xqq ´ Lp m2 z ∆´p´1 AM ` . . .
up`2

Then this gives a new quadratic equation for ∆:

c
2 2p`1 pp ´ 1q2
0 “ p∆ ´ 1qp∆ ´ pq ´ L m ñ ∆ “ ˘ ` m2 L2
2 4
We see that the massive field picks up an Az component which cannot be gauged away, as the mass
term is not gauge invariant. Because gauge invariance is lost, we see that the corresponding operator
in the CFT does not have dimension d ´ 1 anymore and no longer gives rise to a conserved current.

177
10. Taking Aµ Ñ Aµ ` Bµ  gives:
ş ş
dd x J µ Aµ ´ dd x Bµ J µ
W rAµ s Ñ W rAµ ` Bµ s “ xe y “ W rAµ s
here Aµ is the source for the boundary J µ current in the CFT. So the partition function is invariant under
any local gauge transformation of Aµ . This gives us that a global U p1q in the CFT corresponds to a gauged
U p1q on the boundary.
11. We couple our CFT stress tensor T µν to an external field hµν . Note that under any shift hµν ` Bµ ν ` Bν µ
we get: ş d ş d
µν µν
W rhµν s Ñ W rhµν ` Bµ ν ` Bν µ s “ xe d x T hµν ´2 d x ν Bµ T y “ W rhµν s
Thus, the translation invariance given by Bµ T µν “ 0 of the CFTd has given diffeomorphism invariance in the
bulk. This is gravity.
12. In global coordinates the metric is
L2 `
ds2 “ ´dτ 2 ` dθ2 ` sin2 θdΩ23
˘
2
cos θ
Let’s find u by integrating:
ż θ1 ż u1
dθ 2du
“ ñ u “ tan 2θ
0 cos θ 0 p1 ´ u2 q
Consequently,
4u2 2 sin2 θ
“ tan θ “
p1 ´ u2 q2 cos2 θ
so we get agreement for the du, dΩ3 terms. Finally
p1 ` u2 q2 1
2 2
“
p1 ´ u q cos2 θ
and we get agreement for the dτ term, without having to rescale τ . Since 0 ă θ ă π{2 we have 0 ă u ă 1 as
required.
13. Take dτ “ 0. First, take the points to be on opposite sides of the AdS cylinder. We get a distance equal to
ż 1` ˆ ˙ ˆ ˙
2du 2´ 2
2
“ 2 log « 2 log
1´ 1 ´ u  
Given that AdS is a homogenous space, we can use symmetry arguments from group theory to get the
general formula, replacing 2 with |x1 ´ x2 |. For finding the geodesic distance through direct methods, a
better coordinate system would be global coordinates. In AdS3 this looks like:
X´1 “ L cosh ρ cos T
X0 “ L cosh ρ sin T
X1 “ L sinh ρ cos θ
X2 “ L sinh ρ sin θ
this generalizes directly to AdSp`2 . The argument will be the same there. Note
ds2 “ L2 p´ cosh2 ρdT 2 ` dρ2 ` sinh2 ρdθ2 q.
Now take dT “ 0. The Lagrangian takes the form:
b
L “ L ρ9 2 ` sinh2 ρ θ92
Take τ “ ρ so that the EOM for θ quickly gives:
c
θ1 “ a
sinh ρ sinh2 ρ ´ c2
Take c “ sinh ρ0 . In AdS this corresponds to the minimum distance ρ that the geodesic will approach, since
there we have θ1 “ 8 ñ dρ{dθ “ 0.

178
 To get ∆θ, we integrate this, giving
?
cosh ρ 2 sinh ρ0
∆θ “ 2 arctan ? ` c0
cosh 2ρ ´ cosh 2ρ0
The factor of 2 comes from the fact that we need to do the ρ integration twice to get the full geodesic curve.
Now we must take ρ Ñ 8 to approach the boundary of AdS. In this case, the equation for ∆θ simplifies to:
∆θ ´ c
tan “ sinh ρ0
2
Taking θ Ñ 0 (corresponding to ρ0 Ñ 8) shows that c “ ´π{2. This gives

∆θ 1
tan “
2 sinh ρ0
The total length of the trajectory is:
¨ d ˛
sinh2 ρf ´ sinh2 ρ0 ‚
ż ż
sinh ρ ˝ cosh ρf `
b
2 1 2
L dρ 1 ` sinh ρ pθ pρqq “ L dρ a “ 2L log
sinh2 ρ ´ sinh2 ρ0 cosh ρ0 cosh2 ρ0

Take ρf Ñ 8. The leading behavior of this goes as

2eρf
ˆ ˙
ρf θ
2L log “ 2L log 2e sin
cosh ρ0 2

Now note that for x, y coordinates in Rd`2 lying on the unit S d`1 , we have
θ θ
|x ´ y|2 “ pcos2 θ ´ 1q2 ` sin2 θ “ 4 sin2 ñ |x ´ y| “ 2 sin .
2 2
θf
Finally, sinh ρf “ tan θf . This gives θf “ π{2 ´ , with  “ e´ρf . Then ũ “ tan 2 « 1 ´ . So indeed we get
the final entropy formula:
A “ 2L logp|x1 ´ x2 |{q

14. The volume element goes as

ż 1´
u3 16L4 2 16π 2 L4
16L4 2 4
dudΩ3 « 2π “
0 p1 ´ u q 63 33

The 3 scaling valid in the small  limit is exactly area scaling.

15. Because of a horizon, particles approaching this horizon will be arbitrarily redshifted. This implies that
the frequencies reaching the boundary can be shifted to arbitrarily low values, giving a continuum of states
above the vacuum state with no mass gap.

179
16. Yes (Maldacena already does this in his seminal paper). Take the branes to be separated by a distance r.
The supergravity solution will look like

1 ? 4πgN `4s
ds2 “ ? dx2k ` Hdx2K , H “1`
H r4

and take `s , r Ñ 0 while holding U “ r{`2s fixed (we’ve done this type of near-horizon analysis in chapter
11). This keeps the masses of the stretched strings fixed even as we bring the branes together. If all the
branes are coincident, the coordinate r in the supergravity solution gives an equivalent coordinate U “ r{`2s ,
giving the metric
U2 dU 2 a
„ 
a
ds2 “ `2s ? dx2k ` 4πgN 2 ` 4πgN dΩ25
4πgN U
Now, pulling M of the N D3 branes off by a distance W gives a supergravity solution:
» d fi
U 2 a dU 2 MU 4 a
ds2 “ `2s – ? b dx2k ` 4πg 2 N ´ M ` 4
` 4πgN dΩ25 fl
MU4
4πg N ´ M ` |U ´W |4 U |U ´ W |

As long as U " W we still effectively see AdS5 ˆ S5 . For smaller values of U , this splits into two separated
AdS backgrounds, with two different near-horizon limits. We can trust these limits when both M and N are
large, but splitting off single branes gives geometries with singular curvatures that we cannot trust.
But since finite U already means that we have taken the near-horizon limit, the entire moduli space R6N {SN
is visible at that level.

17. The eigenvalues of the Laplacian on a 5-sphere of radius L are in correspondence with the quadratic casimir
of SOp6q for the p0, k, 0q representations, giving:
kpk ` 5q
L2
For a proof of this, consider a homogenous harmonic function of degree k. Any such homogenous harmonic
function takes the form f “ |x|k mi Ymk i pγq. Look then at the Laplacian
ř

f “ ´∇2 p|x|´k f q “ kpk ` 6 ´ 2q|x|´pk`2q f ` x´s

∇2
f

Thus the spherical harmonics on a unit S 5 have eigenvalue kpk `4q. Rescaling the sphere will contravariantly
rescale these eigenvalues by L´2 as required.

18. The wave equation for a massive scalar field is quickly seen to be:

u2 ” 2 p ı
∇2 φ “ Bu ´ B u ´ B 2
t ` B ¨ B φ “ m2 φ
L2 u
Upon Fourier transforming

dp`1 q
ż
φpu, xq “ φpu, qqeiq¨x , q ¨ x “ ´q 0 t ` ~q ¨ ~x
p2πqp`1
we get
m2 L2
„ 
p
Bu2 2
´ Bu ´ q ` B ¨ B ´ φpu, qq “ 0
u u2
solving this in Mathematica directly yields two solutions
1`p 1`p 1a
φ˘ pu, qq “ Au 2 Jν piquq ` Bu 2 Yν piquq, ν“ pp ` 1q2 ` 4m2 L2
2
We can rewrite this in terms of modified Bessel functions as:
1`p 1`p
Au 2 Iν pquq ` Bu 2 Kν piquq

180
 These two solutions have the desired scaling dimensions of ∆˘ respectively (K will have to be defined
differently from how it is defined in mathematica and wikipedia).
Now WLOG take x1 “ 0. Rotating to Euclidean space, it is a quick check to see that the bulk-to-boundary
propagator as written in L.51 does indeed satisfy the massive Laplace equation precisely when ∆p∆´p´1q “
m2 L2 .

Next, we see that

u∆ ζ p dζ
ż ż ż ż8
1
d x f pu, x; 0q “ dp`1 x
p`1
“ up`1´∆ p`1
d y “ up`1´∆ Ωp
pu2 ` x2 q∆ p1 ` ζ 2 q∆ 0 p1 ` ζ 2 q∆

This last integral can easily be evaluated using Γ-functions. The final result is then:

Γp 1`p
2 qΓp∆ ´
1`p
2 q p`1 Γp∆ ´ 1`p
2 q
up`1´∆ Ωp “ up`1´∆ π 2
2Γp∆q Γp∆q

Thus, the normalized bulk-to-boundary propagator is:

Γp∆q u∆
p`1 .
π Γp∆ ´ 1`p pu2 ` |x ´ x1 |2 q∆
2 q
2

In the above, we pick ∆ “ ∆` . By convolving a boundary configurations φ0 pxq with this propagator, we
obtain a field φ in AdS satisfying the massive Laplace equation.

u∆
ż
Γp∆q
φpu, xq “ p`1 dp`1 x1 2 φ px1 q
1 |2 q∆ 0
1`p
π 2 Γp∆ ´ 2 q pu ` |x ´ x

This is clear. It is also quick to see that as u Ñ 0 the leading behavior comes from the bulk-to-boundary
propagator going as up`1´∆ φ0 pxq “ u∆´ φ0 pxq. This is exactly proportional to the leading solution, which
asymptotes with the lower power ∆´ .
It is worth noting that there is another very clean way to obtain this propagator, as originally written in
Witten’s paper. Take the delta function source to be at u “ 8 and let’s look for a solution of the massive
wave equation. Because the source is at 8 the solution has full symmetry under translations in x, and so
can only depend on u. The equations of motion are:

Bu u´p´2 u2 Bu φpuq ´ m2 L2 φ “ 0 ñ φpuq “ u∆ , ∆p∆ ´ p ´ 1q “ m2 L2

this gives the correct ∆˘ as required. To relate this to our solution for a δ function at 0 we must do an
inversion, which consists of taking u, x Ñ u2u,x
`|x|2
yielding our desired propagator.

19. The extrinsic curvature K is defined as

1
Kµν “ p∇µ nν ` ∇ν nµ q, K “ hµν Kµν
2
where nµ is the normal vector and h is the pullback of the metric g to BM . When the coordinate system is
hypersurface-orthogonal, then this simplifies nicely to
1
Kµν “ nρ Bρ Gµν
2

181
 For the poincare patch, we have the normal vector ´ ?g1 Bu “ ´ Lu Bu . Note the minus sign, because u Ñ 0
uu
gives the boundary, which is the opposite from the outward normal orientation. The contraction gives:
1 2 p`1
Kµν “ pp´q2 hµν q ñ K “
2 L L
as required.

20. This problem is done for p “ 3, but I will solve it for general p. The subleading order equation of motion
for φ of the form u∆ pφ0 ` u2 φ2 q is

u∆ ∆p∆ ´ p ´ 1qφ0 ´ u∆ m2 L2 φ0 ` u∆`2 lφ0 ` u∆`2 p∆ ` 2qp∆ ´ p ` 1qφ2 ` u∆´2 m2 L2 φ2 “ 0

At leading u∆ order we get the quadratic constraint on ∆, giving ∆˘ as solutions. Solving for φ2 at subleading
order gives:
lφ0
φ2 “ ´
p∆ ` 2qp∆ ´ p ` 1q ` m2 L2
Plugging in for ∆ “ ∆´ yields:
lφ0
φ2 “ ´
4∆´ ` 2 ´ 2p
This is consistent with what is written when p “ 3. Taking now ∆ Ñ p ` 1 ´ ∆ gives
lφ0
A2 “
4∆´ ´ 6 ´ 2p
Again, taking p “ 3 gives the correct 4p∆´ ´ 3q denominator.

21. Let’s repeat the argument for clarity. We stick to p ` 1 “ d “ 4. We have a bulk-to-boundary propagator
given by:
1 u∆ Γp∆ ´ 2q
K∆ pu, x; x1 q “ 2 1 2 ∆
, C3 “ π 2
C3 pu ` |x ´ x | q Γp∆q
At this stage we take ∆ to be arbitrarily. We do not identify it with ∆˘ . In the small u-limit the propagator
looks like:
C3´1
ˆ ˙
4´∆ 1 2 ∆ 2
u pδpx ´ x q ` Opu qq ` u ` Opu q
|x1 ´ x12 |2
A general field φ0 pxq on the boundary sources a the bulk field φpu, xq to take the form:
ż
φpu, xq “ dx1 Kpu, x; x1 qφ0 px1 q
ˆ ż 1
˙ (99)
4´∆ 2 ∆ ´1 4 1 φ0 px q 2
“u pφ0 pxq ` u φ2 pxq ` . . . q ` u C3 d x ` u A2 pxq ` . . .
|x ´ x1 |2∆
Here we have written the additional terms that we obtained in the prior exercise. This gives for the on-shell
action at leading order:
3
Mpl
ż
? ˇ
Son-shell “ ´ d4 x gg uu φBu φˇ
ˇ
2 u“
pMpl Lq3
ż ż
Kpu, x1 ; xqBu Kpu, x2 ; xq ˇˇ
“´ d4 x1 d4 x2 φpx1 qφpx2 q d4 x
u3
ˇ
2 u“

This last integral is given by:

4C3´1
ż
4´2∆ 4 ∆ 1
p4 ´ ∆qu δ px1 ´ x2 q ` ` 2 u2∆´4 d4 x ` ...
|x1 ´ x2 | C3 |x ´ x1 |2∆ |x ´ x2 |2∆

When 1 ă ∆ ă 3 the remaining terms vanish. The third term doesn’t though, at least not for
1 ă ∆ ă 2. Different kind of counterterm needed?

182
 Now let’s take 3 ă ∆ ă 4. the third term and all of its higher-order contributions will vanish. on the other
hand, not only will the first term require a counter-term going as φ2 , but so will the u6´2∆ term. From
Equation (99) we see that this must go as
u2
u4´∆ pδpx ´ x1 q ´ lx δpx ´ x1 qq
4p3 ´ ∆q
Then we get divergent terms from:
u2 u2
ż ˆ ˙ „ ˆ ˙
4 ´3 4´∆ 4´∆
d xu u δpx ´ x1 q ´ lx δpx ´ x1 q Bu u δpx ´ x2 q ´ lx δpx ´ x2 q
4p3 ´ ∆q 4p3 ´ ∆q
u6´2∆ u6´2∆
“ p4 ´ ∆qu4´2∆ δ 4 px1 ´ x2 q ` p4 ´ ∆q lx1 δ 4 px1 ´ x2 q ` p6 ´ ∆q lx δ 4 px1 ´ x2 q
4p∆ ´ 3q 4p∆ ´ 3q 1
10 ´ 2∆
“ p4 ´ ∆qu4´2∆ δ 4 px1 ´ x2 q ` u6´2∆ lx δ 4 px1 ´ x2 q
4p∆ ´ 3q 1
This altogether contributes:
pMpl Lq3 6´2∆ 10 ´ 2∆
ż
´  dxφ0 lφ0
2 4p∆ ´ 3q
In the original action, expanding φ to quadratic order in  gives
2`4´∆ pM Lq∆ pM Lq3 p4 ´ ∆q 2p4´∆q 2 2p4´∆q`2 φ0 lφ0
φ “ 4´∆ φ0 ` lφ0 ñ ´ 2
φ “  φ0 ` pM Lq3 p4´∆q `. . .
4p∆ ´ 3q 2 2 4p∆ ´ 3q
This shows that the φ2 term contributes a φ0 lφ0 term as well.
We thus need a counterterm action given by:
ż ? ˜ 3 3
¸
M ∆´ 2 Mpl L 1
Sct “ h φ ` φlφ
2L 2 4p∆ ´ 3q
Here, now, l is taken with respect to the boundary metric h , meaning it absorbs two factors of {L. This
is correct :)
22. Our correlator is:
B ˇ
Son-shell pφ0 pxq “ λk eikx ` λq eiqx qˇ
ˇ
xφpkqφpqqy “
Bλk Bλq λk ,λq “0
We take the on-shell action and get
M3
ż
?
Son-shell pφq “ ´ d4 x gg uu φBu φ
2
M3
ż ˇ
d4 kd4 qλk λq δ
4 pk ` qqu´3 φpu, pqBu φpu, qqˇ
ˇ
“´
2 u“

Using the form L.50 of the bulk-to-boundary propagator we can solve this:

183
 Factor of 2 off from Gubser, Klebanov, Polyakov, but I don’t think this problem asks for the
coefficients, rather just the form of the correlator.
There is always a term going as logpp{2q. All other terms are positive polynomials in p and so will contribute
contact terms that we will need to supply counterterms to renormalize. There is an easy way to see this.
A term going as p0 simply contributes a δpxq in position space. Even polynomials in p therefore contribute
terms of the form ln δpxq. All of these need to be regulated and subtracted.
We thus recognize the pattern and get
ˆ 2 2˙
2 4 p  sinpπνq ´ p ¯2ν M2 Γp3 ´ ∆q 4 ´ p ¯2∆´4
xφpkqφpqqy “ ´M δ
pk ` qq log “ ´ p2∆ ´ 4q δ pp ` qq
Γpνq2 2


2 2 Γp∆ ´ 1q 2

Up to a factor of 2 this is consistent with 3.40 of MAGOO. My argument at the moment works only for
integral ν, but based off of remarks that I have read, this is what should be expected.
Integrating this is not hard if you know a trick:

2π 2 8 ip|x|
ż ż8
2ν`3 e 1
dpp log p Ñ 2 dpp2ν` eipx
p2πq4 0 p|x| 8π 0

And look at the Opq part of this expansion.

Altogether this gives:
1
xOpxqOpyqy «
|x ´ y|2∆
Where I am not sure about the constant, but am sure about the x-scaling.

23. In this problem I will freely exchange p ` 1 and d whenever suitable. The dual current is constrained to have
scaling dimension p. Upon choosing a gauge:

Au pu, xq “ 0, ∇ µ Aµ “ 0

We therefore have that A satisfies the differential equation:

ˆ ˙
M 1 4 ´1 1
0 “ ∇ FM N “ Bu p`2
u u Bu Aν ` p`2 u4 u´1 Bµ Brµ Aνs
u u

You will notice that there is an extra factor of u accompanying A in this PDE. This should be viewed as
turning µ into a Vielbein index, so that dxµ has trivial scaling properties.
The full solution Aµ pu, xq is then given by:
ż p`1
p{2 d p
u aµ ppqeipx Kp{2 ppuq, pµ aµ ppq “ 0.
p2πqp`1

It will be nicer to write the bulk-to-boundary propagator in position space. For this, I’ll follow Witten’s
argument by putting the δ function source at u “ 8. This simplifies things since the solution must be
independent of the x variables. Take the solution A “ f puqdxi . This gives the equations of motion for a free
field:
1
d ‹ dA “ Bu p`2 u4 f 1 puq “ 0 ñ f puq “ u∆ , ∆ “ d ´ 2
u
Doing the inversion again we have:

ud´2 xi ud´2 2xi ud´1

ˆ ˙
d´2 i 2 2 i
u dx Ñ 2 d “ pu ´ xi qdx ´ du
pu ` |x|2 qd´2 u2 ` |x|2 pu2 ` |x|2 qd pu2 ` |x|2 qd´2

adding the exact term: « ˆ ˙d´1 ff

1 xi u
´ d
d´1 u u2 ` |x|2

184
 d´1
and rescaling by an overall factor of d´2 yields:

ud´2
ˆ ˙
i i du
Gµi pu, x; 0q “ 2 dx ´ x (100)
pu ` |x|2 qd´1 u

This does not satisfy Lorenz gauge, but as expected can be brought to satisfy it by adding appropriate pure
gauge terms
Gµi pu, x; x1 q Ñ Gµi pu, x; x1 q ` Bµ Λpu, x; x1 q
Freedman et al have a slightly different propagator, which can be obtained from this one by gauge transform.
It takes the form:
Γpdq ud´2 Γpdq ud´2 ´ xµ xi ¯
Gµi “ d 2
I
2 d´1 µi
pxq “ d 2 2 d´1
δµi ´ 2
2π 2 Γp d2 q pu ` |x| q 2π 2 Γp d2 q pu ` |x| q x2

This propagator can be seen to naturally come from the embedding space formalism.
Upon integration by parts, the action leads to only a boundary term:
ż ż
1 µν 1
Fµν F Ñ A ^ ‹F
4g 2 2g 2
The nonvanishing components will come from the parts of ‹F that do not involve a du. Consequently, we
only need to calculate the du parts of F . Said equivalently, the on-shell action is:

dudd z
ż ż
1
Son´shell “ 2 dx1 dx2 Ji px1 qJj px2 q Gνi u4 Br0 Gνsj (101)
2g ud`1

Computing this is straightforward, using (100):

ud´3 i ud´3
Brµ Gνsi “ pd ´ 2q du ^ dx ´ dxi ^ du
pu2 ` |x|2 qd´1 pu2 ` |x|2 qd´1
ud´1 i ud´3 xi xj
´ 2pd ´ 1q 2 du ^ dx ` 2pd ´ 1q dxj ^ du
pu ` |x|2 qd pu2 ` |x|2 qd
ud´3 i ud´1 i ud´3 xi xj
“ pd ´ 1q 2 du ^ dx ´ 2pd ´ 1q du ^ dx ´ 2pd ´ 1q du ^ dxj
pu ` |x|2 qd´1 pu2 ` |x|2 qd pu2 ` |x|2 qd
(102)
For the u Ñ 0 limit, recall the scalar propagator for a field with dimension d ´ 1 would take the form:

Γpd ´ 1q ud´1 Γpdq ud´1 Γpdq 1

d 2 ` |x ´ x1 |2 qd´1
“ d 2 ` |x ´ x1 |2 qd´1
Ñ uδpx´x1 qdxi `ud´1 d 1 |2d´2
dxi
d
π Γp 2 ´ 1q
2 pu d
2π 2 Γp 2 q pu d
2π 2 Γp 2 q |x ´ x

So that Gµi pu, x; x1 q will (up to a constant) approach δpx ´ x1 qdxi in the u Ñ 0 limit. On the other hand,
(102) will approach (up to a constant):
˜ ¸
1 x i xj
Br0 Gisj “ ud´3 δij ´ 2 12 12
|x12 |2d´2 |x12 |2

The leading ud´3 cancels exactly against the metric-dependent terms in the integral (101). The final result
is proportional to: ˜ ¸
δ ij xi12 xj12
ż
1
dx1 dx2 Ji px1 qJj px2 q ´2
g2 |x12 |2d´2 |x12 |2d
Giving a two-point correlator (after rescaling):
1
xJi px1 qJj px2 qy “ Iij px12 q, Iij pxq “ δij ´ 2x̂i x̂j
|x12 |2d´2

185
24. I think this is pretty direct. If a field φ0 diverges as ∆ şin the IR, it must couple with an operator O of
scaling dimension d ´ ∆ in order for the interaction term φ0 O to be conformally invariant. This situation
is generic
What more does this question ask for?
25. In what follows, there are many variables and the story becomes rapidly confusing if one does not understand
what everything stands for. I will review this. My conventions will mostly by those of Kiritsis, although
occasionally I will adopt notation from de Haro, Skenderis, and Solodukin arXiv:000223

Symbol Definition
gpu, xq Full AdS5 Metric
Rµν Ricci Curvature of g
The coefficient of u2n in the expansion of g about u “ 0.
g p2nq Note that g p4q is undetermined by g0 . Its trace and
divergence is however determined.
hp4q The coefficient of u4 log u2
ρ “ u2 Alternative coordinate, usually easier to work with.
p0q
Rij , R Ricci curvatures of gij only
tij Undetermined integration constant in g p4q
2
γpxq “ L2 gij p, xq Induced metric on the renormalization hypersurface u “ 
xTij y Stress tensor of varying renormalized action w.r.t. g p0q
Tij rγs Stress tensor w.r.t. metric on renormalized hypersurface
TijA Stress tensor of varying anomaly A w.r.t. g p0q .

The theorem of Fefferman and Graham states that a general asymptotically-AdS metric can be written as
ds2 du2 gij i j
“ ` 2 dx dx , gij pu2 , xq “ g p0q pxq ` u2 g p2q qpxq ` . . .
L2 u2 u
Einstein’s equations in this setting are:
1 6
Rµν ´ gµν R “ 2 gµν
2 L
It will be rather annoying to derive how this looks like in terms of g and its derivatives. Thankfully,
Henningston and Skenderis already have this formula in equation 6 of their paper, and I am happy to quote
it directly. Take ρ “ u2 . Denote differentiation with respect to ρ by g 1 . Then we get
ρr2g 2 ´ 2g 1 g ´1 g 1 ` Trpg 1 qg 1 sij ` Rij ´ pd ´ 2qgij
1
´ Trpg 1 qgij “ 0 (103)

Here all traces mean TrpXq “ g ab Xab . At leading order in ρ, the first term is set to zero and we get for d “ 4
at u “ 0:
p2q 1 p2q 1
gij ` gij g ab gab “ Rij
2 2
Taking the ansatz:
p2q 1 1
gij “ αRij ` βRgij ñ g ab gab
1
“ pα ` 4βqR ñ β ` pα ` 4βq “ 0 ñ β “ ´ α
2 6
We see that to get the Ricci tensor to match we need α “ 1{2. Thus we get the solution
p2q 1 1
gij “ Rij ´ Rgij
2 12
as required. To next order, we have:
p4q p4q p2q

4g
` 6hij ´ 2pg p2q g ´1 g p2q qij ` 
Trpg p2q
 qgij
 ij
p2q p4q p4q p2q

` Rij ´ 4g 
ij ´ 2hij ´ 2Trrg p4q sgij ´ Trrhp4q sgij ´ 
Trrg
 p2q
sgij ` Trrg p2q g p2q sgij “ 0 (104)

p4q 1 p2q p0q p2q
ñ ´4phij ` gij Trhp4q q “ ´2g p2q g ´1 gij ´ 2Trrg p4q sgij ` Rij ` Trrg p2q g p2q sgij
4

186
 Note that g p4q has canceled. This is generic. In d dimensions g pdq will cancel. This is a reflection of the fact
that there are generally two solutions to the Einstein equations. We cannot determine g p4q uniquely without
an additional constraint. We can still trace over both sides and get a relation between traces. Alternatively,
this comes from the Rrr part of the Einstein equations:
1
Trrg p4q s “ Trrg p2q g p2q s.
4

The Einstein equations for Riρ give the further constraint that

0 “ ∇i pg ab gab
1
q ´ g ab ∇b gia
1
ñ ∇i Trg 1 “ ∇j gij
1

I’m missing how to actually get g p4q in order to get agreement with Kiritsis and 0002230.
Thankfully, explicit calculation of g p4q is not necessary for anomaly analysis.
This gives the final desired result:

p4q 1 p0q ” ı 1 1 p2q

gij “ gij pTrg p2q q2 ´ Trrpg p2q q2 s ` pg p2q q2ij ´ gij Trpg p2q q ` tij
8 2 4
Consistency of divergence and trace requires:
1” ı
∇i tij “ 0, tii “ ´ pTrg p2q q2 ´ Trrpg p2q q2 s
4

Now, revisiting (104), the trace conditions Trg p4q “ Trrpg p2q q2 s and Trhp4q “ 0 simplify it to:

p4q 1 1 p0q 1 p2q

hij “ pg p2q q2 ` Trrpg p2q q2 sgij ´ R
2 8 4 ij
1 p2q 2 1 p0q 1 k p2q p2q
“ pg q ` Trrpg p2q q2 sgij ` p∇ ∇i gjk ` ∇k ∇j gik ´ ∇2 g p2q ´ ∇i ∇j Trg p2q q
2 8 8
where we have used identities for the variation of the Ricci tensor from appendix C.2. This can be written
as: ˆ ˙
p4q 1 kl 1 1 2 1 1 1 2 1 2 kl p0q
h “ Rijkl R ` ∇i ∇j R ´ ∇ Rij ´ RRij ` ∇ R ` R ´ Rkl R gij
8 48 12 24 32 3 3
ˆ „ ˙
1 δ 1 kl 1 2
“ ´a ´ Rkl R ´ R
g 0 δpg p0q qij 8 3

As a bonus, for d “ 2, Equation (103) gives:

1
Rij “ Trpg ´1 g 1 qgij ñ R “ Trpg ´1 g 1 q
2
Together with ∇i gij “ ∇i gij Trg 1 we get

p2q 1 p0q
gij “ pRgij ` tij q
2
with tij divergence-free and tii “ ´R We then get:

L c
hTij i “ “´ R
16πGN 24π
This is the 2D Weyl anomaly.

26. Take:
L3
ż ˆż ˙
4 du a 1 a
d x det gpu, xq ´ p1 ´ uBu q gpu, xq|u“ (105)
2πG5  u5 u4

187
 It will be useful to recall:
a 1
´ detpg ` hq “ exp logp´ detpg ` hqq
2 „ 
a 1 ´1
“ deg g exp tr logp1 ` g hq
2
„ 
a 1 ´1 1 ´1 2
“ deg g exp trrg h ´ pg hq s
2 2
ˆ ˙
a 1 ´1 1 ´1 2 1 ´1 2
“ deg g 1 ` trpg hq ´ trrpg hq s ` trrg hs
2 4 8

This implies
b ˆ ˙
a 1 2 p2q 1 4 p4q 1 4 2   1
p4q 4 p2q 2 1 4 p2q 2
gpu, xq “ g p0q 1 ` u Trrg s ` u Trrg s ` u log
uTrrh s ` u Trrg s ´ u Trrpg q s

2 2 2
 8 4

Again, indices are raised and lowered with g p0q .

At zeroth order we get:
L3 1 L3 L3 1
´ “ ´6 ñ A0 “ ´6
8πG5 4 2πG5 4 16πG5 4
At second order we get:

L3 1 1 L3 1 1
2
Trrg p2q
s ´ p1 ´ q u2 Trrg p2q s “ 0 ñ A2 “ 0
4πG5 2  2πG5 2 2
At fourth order only the first term of (105) contributes and we get:

L3 ´1 1 1 ¯ L3 1`
log 2 Trrg p4q s` Trrg 2 s2 ´ Trrpg p2q q2 s “ ´ log 2 Trrg 2 s2 ´ Trrpg 2 q2 s
˘
´
4πG5 2 8 4 16πG5 2
1`
Trrg 2 s2 ´ Trrpg 2 q2 s “ A
˘
ñ A4 “
2

27. In order to get the correct counterterms, we need to solve for Trg p2q , Trrpg p2q q2 s in terms of the induced
metric on the renormalization surface, that is at u “ . I will call this γ (Kiritsis calls it h) as as not to be
2
confused with hp4q !. We have γ “ L2 gij . This gives:

4 ? 1 2 1 4 ´
b ˆ ¯˙
p2q p2q 2 p2q 2
g p0q “ 4 γ 1´ Trg ` ´2Trrpg q s ` Trrg s
L 2 L2 8 L4

Next, recall:
p2q 1 1 p0q
gij “ Rij ´ Rgij
2 12
where Rij is taken w.r.t. g p0q This gives:
« ff
1 L2
p0q s
ˆ „ ˙
p2q 1 p0q 1 p2q δRrg 1 ij 1 2
Trg “ Rrg s “ Rrgij s ´ gij “ Rrγs ` R rγsRij rγs ´ Rrγs
6 6 δg
p0q 6 2 2 6
ij

Finally, to leading order:

L4 1 1 1 ij 1 L4 1 2
Trrpg p2q q2 s “ 4
pRij rγs ´ Rrγs γ ij q pR rγs ´ Rrγs γ ij
q “ 4
pRij rγsRij rγs ´ Rrγs2 q
 2 6 2 6  4 9
As a check, we see (up to quadratic order in the curvature):

L4 1 1 L4 1
ˆ ˙ ˆ ˙
1 p2q 2 p2q 2 2 1 ij 2 2 ij 1 2
A4 “ rTrrg s ´Trrpg q ss “ 4 Rrγs ´ pRij rγsR rγs ´ Rrγs q “ ´ 4 Rij rγsR rγs ´ Rrγs
2  2 36 4 9  8 3

188
 All together we get

L3
ż b „ 
4 p0q
6 2
´ d x g ´ 4 ´ log  A4
16πG5 
3 6 2
ż „ 
L 4 ? p2q p2q 2 2
“ d x γ 6` Trg ` Trrg s ` log  A4 ` . . .
16πG5 2 L2
L3
ż „ ˆ ˙
1 4 ? 6 L 2 ij 1 2
“ d x γ ´ Rrγs ´ log  Rij rγsR rγs ´ Rrγs
16πG5 L 2 8 3

Where the . . . denotes “up to finite terms”. I think Kiritsis has a mistake and it should be ´ L2 R not
`. 0002230 confirms this.

28. Let’s review first. The stress tensor comes from varying the renormalized action by the initial source field
g p0q . This is given by:
2 δ L2
xTij y “ lim a S ren rγs “ Tij rγs
Ñ0 gij p, xq δg ij p, xq 2
Thus, we can look at variations with respect to the induced metric γ at u “  and take  Ñ 0 at the very
end.
There are two variations to consider. The first is the variation of the on-shell effective action. Generalizing
the argument giving 14.8.35 to a u-dependent metric, we get:

L3
ˆ ˙
sugra 1 ´1 3
Tij “´ pKij ´ Kγij q “ ´ ´B gij p, xq ` gij p, xqTrrg p, xqB gp, xqs ´ 2 gij p, xq
8πG5 8πG5 

Varying the counterterm action with respect to γ gives directly:

L3
„ ˆ ˙ 
c 1 3 L 1 2 A
Tij “ ´ γij ´ Rij ´ Rγij ´ log  Tij
8πG5 L 2 2 2

The next step is to write these in terms of the g p2nq , hp4q .

It is useful to note:
1 2
ˆ ˙
k kl 1 2 1 2 p0q
Rij rγs “ Rij rg0 s ` Rik R j ´ 2Rikjl R ´ ∇i ∇j R ` ∇ Rij ´ ∇ Rgij .
4 L2 3 6
1 2
ñ Rrγs “ Rrg0 s ´ Rij Rij
4 L2

where the curvatures on the RHS are those of the metric g p0q . Now we get:

L2 sugra
xTij y “ rT ` Tijc s
2 ij «
L3 L2 p2q p0q p2q 1 1 p0q 2
´
p4q A
¯
“´ p´g ij ` g Trgij ` R ij ´ Rg q ´ log  2h ` Tij
8πG5 2 2 4 ij
p2q 1 p0q
´ 2g p4q ´ hp4q ´ gij Trg p2q ´ gij Trg 2
2 ff
2
ˆ ˙
L 
1 1 p0q L p2q L p0q 1 p0q
´
2Rik Rkj ´ 2Rikjl Rkl ´ ∇i ∇j R ` ∇2 Rij ´ ∇2 Rgij ´ Rgij ` gij Rij Rij ´ ∇2 Rgij
8
L2 3 6 4 8 6

L2
We see that the 2
and log 2 terms vanish by 14.8.39, 14.8.43. The curvature terms combine to give yet
p4q
another two copies of ´hp4q qij , and the remaining terms cancel everything but tij from the gij to give:

L3 p4q
xTij y “ r2tij ` 3hij s
8πG5

189
29. In embedding space, it is easy, using L.10 to calculate:
2σ :“ ηM N pξ ´ ξ 1 qM pξ ´ ξ 1 qN “ ´p∆X 0 q2 ` p∆X p`1 q2 ` p∆X i q2
´ ¯2 ´ ¯2 ´ ¯2
1 L2 `x2 L2 `x1 2 u´u1 L2 `x2 L2 `x1 2 Lx1
“ ´ u´u
2 ` 2u ´ 2u1 ` 2 ´ 2u ` 2u1 ` Lx
2u ´ 2u1
˜ ¸
ˆ 2 2 2
L2 L2 L2 x2 x1
˙ ˆ ˙ ˆ ˙
L x x1
“ ´pu ´ u1 q ´ 1 ´ ´ 1 ´ 1 ` L2 ´
2u 2u u u u u u u1
pu ´ u1 q2 ` px ´ x1 q2
“ L2
uu1
Then η 2 “ 1 ` σ. Now let us calculate:
ÿ
´iGF “ x0|T φpxqφpx1 q|0y “ Θpx ´ x1 q φ∆` n` pxqφ∆` n` px1 q ` px Ø x1 q.
n,`

The modes of a scalar field are given by:

φ∆,`,n “ N∆,n,` sin` θ cos∆˘ θ 2 F1 pa, b, c; sin2 θq Y` pΩp qpΩp q,
1 1 p`1
a “ p` ` ∆˘ ´ ωLq, b “ p` ` ∆˘ ` ωLq, c “ ` ` , ωL “ ∆˘ ` ` ` 2n
2 2 2
Because of the homogeneity of AdS, pick the AdS coordinate origin at x1 . Then φ∆` ,`,n px1 q “ 0 for ` ‰ 0.
This reduces the sum to:
Γp p`1 8 8 p`1
e´i∆t cos∆ θ ÿ p 2 qn Γp∆ ` nq
ˆ ˙
2 q ´i∆` ´n ∆ ` n
ÿ
2 2
p`1 e N∆,n,` φ∆,0,n pxqφ∆,0,n p0q “ p`1 p´1 2 F1 1 ; sin θ e´2in|t|
2π 2 2π 2 L p n!Γp∆ ` n ` q 2 pp ` 1q
n“0 n“0 2

I have started using (ascending) Pochammer symbols. Writing this in terms of a Jacobi poylnomial:
8
ÿ Γp∆ ` nq p´1
,∆´ p`1
p´1 P n
2 2
pcos 2θqe´2in|t|
n“0 Γp∆ ` n ` 2 q

Using a Jacobi polynomial identity from 45.1.4 “A table of series and products” by Eldon R. Hansen we get
the full greens function to be
Γp∆qe´i∆t cos∆ θ cos2 θ
ˆ ∆ ∆`1 ˙ ˆ ∆ ∆`1 ˙
´2i|t| ´∆ 2 2
Γp∆q ´2∆ 2 2
1
p1`e q 2 F1 ; “ η 2 F1 ; 4
p`1
2π 2 Γp∆ ` 1´p qLp ∆ ` 1´p
2 cos2 t 2
p`1
∆`1 π 2 Γp∆ ` 1´p qLp ∆ ` 1´p
2 η
2 2
cos t
And η 2 “ cos θ is the geodesic distance. Up to a minus sign this is correct. I think Kiritsis means 2
∆`1

in the denominator. Either ∆ “ ∆` or ∆ “ ∆´ works.

30. Upon taking u Ñ 0, η ´4 goes to zero and the 2 F1 Ñ 1. What remains is (I’m including Kiritsis’ minus sign):
˙∆ ˙∆
u∆ Γp∆q
ˆ ˆ
Γp∆q ∆ u1 1 u1
´ p2uq “ ´
u1 2 ` x2 2p∆ ´ p`1 2
p`1 p`1
2∆`1 π 2 Γp∆ ` 1´p q 2 2 q π 2 Γp∆ ´
1`p
q u1 ` x2 2

This is the negative of what is in Klebanov and Witten.

u∆
G∆ pu, x; u1 , x1 qt Ñ K∆ px; u1 , x1 q
p ` 1 ´ 2∆
Take a source field φpu, xq in the bulk. As u Ñ 0 away from the source we have φpu, xq Ñ u∆ Apxq.
The expectation value xOpxqy is given by contracting the bulk source with a bulk-to-boundary propagator
going to x. On the other hand, contracting with a bulk-to-bulk propagator and taking u close to 0 gives
u∆ Apxq. Thus, we see:
u∆ 1
u∆ Apxq “ xOpxqy ñ Apxq “ xOpxqy
p ` 1 ´ 2∆ p ` 1 ´ 2∆

I think the bulk-to-bulk green’s function should have a minus sign from how Kiritsis defined it.

190
31. We are looking at the Opξq contribution to the 3-point correlator of φ fields. Note that at ξ “ 0 the theory
is noninteracting and the three point function vanishes. The on shell action is:
3 ż
M3 ÿ M 3ξ
ż
? ?
Son-shell “ d5 xBµ p gφi B µ φi q ´ d5 gφ1 φ2 φ3
2 i“1 2

the contribution due to the cubic term has been calculated. For that, it is enough to look at the Opξ 0 q
solution. Now let us look at the first term. For this, we need to look at the Opξq part of the solution. Using
the regulated bulk-to-bulk propagator G we have:
ż ż 4 1 1 ż
d x du
4 2 2 i
φi pu, xq “ d x Ki pu, x; x qφ0 pxq`ξ G pu, x; u , x q d4 x1 d4 x2 Kj pu1 , x1 ; x1 qKk pu1 , x1 ; x2 qφj0 px1 qφk0 px2 q
1 1
u1 5
Its contribution will look like
M 3 ÿ d4 x uu
ż ˇ
g φ φ
ˇ
i Bu i
u5
ˇ
2 i u“
» ¨ ˛ fi
M 3 ÿ d4 x – d4 x1 du1 d4 x2 ˝
ż ż
1 1 2 2 1 1 ‚ 1 1 fl
“ξ G  p, x; u , x q Bu Kpu, x; x q ` Kp, x; x qBu Gpu, x; u , x q W pu , x q
2 i p u1 5 looooooomooooooon
0 u“
ż
1 1 4 4 1 1 1 1 j k
W pu , x q “ d x1 d x2 Kj pu , x ; x1 qKk pu , x ; x2 qφ0 px1 qφ0 px2 q

Using L.66, the last term becomes:

∆´1´p du1 d4 x1
ż ż
4 4 2 2
´ d x d x Kp, x; x q K pu1 , x1 ; xqW pu1 , x1 q
Lp u1 5
Taking  Ñ 0 we can count powers to see that this scales as ∆´1´p`∆ “ 2∆´4 Ñ 0 for ∆ ą pp ` 1q{2.
What about outside that range?

32. This is a difficult problem in general. Hong Liu has a method of doing this using conformal partial waves.
Return
33. We can use a conformal transformation that maps the circle to a line. In this case, the minimum area surface
is just a plane in AdS5 . The renormalized area of this flat plane is just zero, since there is no curvature that
would give a separation of length scales against which to do a subtraction. Thus, for a straight line we have

W rLs “ 1

191
 Applying a special-conformal transformation will take a line to a circle:

xi ` bi x2 z
xi Ñ , zÑ
1 ` 2bi xi ` b2 x2 1 ` bi xi ` b2 x2
Take only b2 ‰ 0. A quick check shows that this takes the line x2 , x3 , x4 “ 0 to the circle x3 , x4 “ 0,
b2 x2 ` b22 px21 ` x22 q “ 0. This is centered at x1 “ x3 “ x4 “ 0, x2 “ 1{2b2 . It has radius a “ 1{2b2 .
It is now quick to see that the z surface is expressible as a hemisphere in the standard coordinate system. If
r is the distance from the center we get
a a
z “ a2 ´ r2 ñ r “ a2 ´ z 2

We can thus integrate over 0 ă z ă a and θ. The induced metric on the worldsheet is:

L2 ` 2 2 ˘ L2 z2 L2 a2
ˆ ˆ ˙ ˙ ˆ ˙
2 2 2 2 2 2 2 2 2 2
r dθ ` dz ` dr “ pa ´ z qdθ ` 1 ` dz “ pa ´ z qdθ ` dz
z2 z2 a2 ´ z 2 z2 a2 ´ z 2

Calculating the renormalized area from the determinant of the induced metric yields:
ża
L2 8 L2 ´ a
ż
1 dz a ¯
S“ A “ dθ “ ´ 1
2π`2s 2π`2s 0  z
2 `2s 

subtracting off the divergence from the area associated to a wilson loop that has a cylinder as boundary, we
effectively drop the ´1 term. This gives a wilson loop going as
?
2 {`2
W rCs “ e´Sren “ eL s “e 4πλ

I assume that making use of this conformal transformation is what was meant by guessing the qualitative
form using symmetries.

34. The monopole-monopole static potential is the EM dual of the quark-quark potential. It is probed by a D1
brane in the bulk. The tension of a D-string is 1{gs times that of an F-string. Going through the same
steps for calculating the rectangular Wilson line gives the same potential as we got in 14.9.11 with an extra
gs´1 “ p4πqgY´2M c
2
4π 2N p4πq
b 2
4π 2 2g 2 N 2
1 YM g YM
´ 1 4 “´ 1 4
Γp 4 q l Γp 4 q l
This is exactly S-duality sending gY2 M Ñ p4πq{gY2 M .

35. We derive the following

• Hawking Temperature:
In Euclidean signature, all we need to look at is the part of the metric going as:
a
f prq 2 Hprq
a dτ `
Hprq f prq
?
Now applying exercise 13.1 with F “ 1, Cprq “ f prq{ H we get exactly:

C 1 pr0 q r0
TH “ “ b
4π π L̃4 ` r04

• Entropy
The entropy is extensive so will be proportional to the (regularized) worldvolume V3 . The remainder
will be proportional to the area of the 5-sphere π 3 r5 around the brane. In order to get the correct

192
 ?
volume factors, we take the determinant of the full metric, not including t and r, yielding g “ r5 H 1{2 .
Altogether then we get c c
L̃4 L̃4
V3 π 3 r05 1` r04
V3 r05 1` r04
“
4GN 25 π 3 `8s gs2
as required.
• Chemical Potential
The chemical potential should be the value of the p-form field at r “ r0 , multiplied by Vp Tp so as to be
extensive and proportional to the tension. From 8.8.7
r02 a H3 prq ´ 1 V3 L2
Vp Tp H3 pr0 q “ a
L2 H3 pr0 q p2πq3 `4s gs r04 ` L4

Checking that the 1st law is obeyed is easy-peasy-lemon-squeezy

36. We will calculate the partition function of 4D N “ 4 SYM in volume V (flat space) at temperature β. For
a single (noninteracting) bosonic field with one degree of freedom, we get:
„ ż 
ź ź 1 V3 3 ´β|p|
ZB pβq “ Zp pβq “ “ exp ´ d p logp1 ´ e q
p p
1 ´ e´β|p| p2πq3

Then
ÿ e´β|p| π 2 V3
ż
V3 V3 4πΓp3q
ñ log ZB pβq “ d3 p “ ζp4q “
p2πq3 n
n p2πβq3 90 β 3
The same argument for a single fermion gives:

V3
ż ÿ p´1qn e´β|p| V3 4πΓp3q 7 π 2 V3
3
ñ log ZF pβq “ d p “ ηp4q “
p2πq3 n
n p2πβq3 8 90 β 3

where η is the Dirichlet eta function, giving the alternating zeta series. We have N 2 p6 ` 2q bosonic degrees
of freedom and 8N 2 fermionic degrees of freedom. Altogether this gives the partition function:
7 π 2 V3 π 2 2 V3
log Zpβq “ 8N 2 p1 ` q “ N 3
8 90 β 3 6 β
The expected energy is then:
B π2 2
E“´ log Z “ N V3 T 4
Bβ 2
The free energy is F “ β ´1 log Z “ E ´ β ´1 S. This gives the entropy to be
1 1 2
S “ βE ´ log Z “ ´pβBβ ` 1q log Z “ p ` qπ 2 N 2 V3 T 3 “ π 2 N 2 V3 T 3
6 2 3

193
 exactly as written.
Let us further verify 14.10.10 and 14.11.11 in their entirety. First, note that for a D-brane solution
´f prqdt2 ` d~x ¨ d~x a
ˆ 2
L̃4 r4
˙
2 dr 2 2
ds “ a ` Hprq ` r dΩ5 , Hprq “ 1 ` 4 , f prq “ 1 ´ 04 .
Hprq f prq r r
we have shown before that the parameter N is given by
b
2
L̃ L̃4 ` r04 b
4 4 2
N“ ñ L “ 4πg `
s s N “ L L̃4 ` r04
4πgs `4s
This gives the relevant AdS scale L. From previous calculations of the mass, we have
b
4 4 L̃2 L̃4 ` r 4
V3 p4L̃ ` 5r0 q 0 1 V3 r04
M“ « V 3 ` p5 ´ 2q
27 π 4 gs2 `8s 4πgs `4s p2πq3 `4s gs 27 π 4 gs2 `8s
π 2 3
« N V3 T3 ` 3V3 N 2 TH´4 “ N V3 T3 ` EY M
8 4
up to quartic corrections in the string length. Next, the chemical potential is also easy:
V3 L̃2 1 V3 r04 1
Φ“ 3 4
b “ T V
3 3 ´ “ T3 V3 ´ π 2 V3 N 2 TH4 .
p2πq gs `s L̃4 ` r4 2 L̃2 p2πq3 gs `4s 2
0

Finally, the entropy: b

V3 r03 r04 ` L̃4 1 3
S“ “ π 2 V3 TH3 N 2 “ SY M
4p2πq3 gs2 `8s 2 4
3
So remarkably, the interacting system has quantities lowered by exactly 4 from their free field values.
37. Let’s cut open the Wilson line at t “ 0. Then we have
żβ
L “ P exp i Aµ x9 µ dτ
0

We initially consider gauge transformations that are periodic in the time direction. That means that this
object transforms as L Ñ g ´1 Lg. If we consider now the gauge transformations periodic up to an element
of the center, we get L Ñ g ´1 Lgh. Closing the loop again, we get an additional insertion of h in the line.
Since h is central, it is equal to an N th root of unity ζ times the identity matrix. This is just a scalar, so
inserting it to give a twisted sector gives the transformation:
L Ñ ζL
as required.
38. The following diagrams will give the leading-order contributions to the scalar mass at one-loop:

194
 Here wiggly lines are gluons, straight lines are scalars, and straight lines with arrows are the fermions.
To extract the mass dependence, it is sufficient to allow for the external legs to go to zero momentum.
?
We get the thermal one-loop contribution to the mass going as λT
A mass for the fermion is already generated at tree-level due to the anti-periodic identification implicit in
putting the theory at finite temperature.
Finish, ask Pavel

39. There is a typo. It should be r2 dΩ23 , not r3 dΩ23 .

As we know by now, for a hypersurface orthogonal to the coordinate system,
1
Kµν “ nr Br gµν
2
Here nρ Bρ is the unit vector in the r direction, whose r component is ?1 . We get:
gr r
ˆ 1 prq
˙
1 1 1 1 gtt 6
Kµν 1
“ ? p´gtt prqdt2 ` 2r3 dΩ23 q ñ ? `
2 grr 2 grr gtt prq r

For AdS Schwarzschild the metric is

˙´1
r2 r2
ˆ ˙ ˆ
2 wM 2 wM 16G5
ds “ ´ 1 ` 2 ´ 2 dt ` 1 ` 2 ´ 2 dr2 ` r2 dΩ3 , w“
L r L r 3π 2

Now, calculating K for this we get:

c
2r4 ` 2L2 M w
ˆ ˙
1 r2 wM 6 4
1` 2 ´ 2 ` Ñ
2 L r r5 ` L2 r3 ´ L2 M rw r L
? r 3
as r Ñ 8. This is independent of M . As r Ñ 8, g on a constant-r slice approaches Lr Ñ L´1 ´4 which
? 4
is again independent of M . Defining h “ L4 yields
? ?
K h “ 4L4 h

Thus, the subleading terms go to zero as r Ñ 8, and so the difference in the GH term between AdS-
Schwarzschild and thermal AdS vanishes.

40. The inverse temperature β obtained from requiring no conical singularity is given by:

2πL2 r` L
β“ 2 2
ñz“
2r` ` L β

here z is the effective dimensionless temperature. Knowing that:

d c
1 1 4wM
r` “ L ´ ` 1`
2 2 L2

let us now look at ´b ¯

dz Lw 1 ` 4M
L 2
w
´ 2
“ b ¯¯3{2
dM ´ ´b
π 8M
L2
w
` 2 L2 1 ` 4M
L2
w
´ 1

For small M , this is negative, giving negative heat capacity (ie small black holes evaporate). For large M
this is positive, giving positive heat capacity (ie large black holes are thermodynamically stable, fed by the
reflection of their radiation off of the AdS boundary).

195
41. For a quantum mechanical system in 1D, there is a finite energy associated with tunneling from one minimum
to the other. As a result, instead of the?field localizing to one of the minima ψ “ ψL or ψ “ ψR , there are
finite energy configurations pψL ˘ ψR q{ 2 that can lower the Hamiltonian further. If instead we had an
infinite potential associated with crossing from one to the other, such ground states would be disallowed.
For arbitrary local field theories in finite volume, this argument generalizes to show that there can never be
a phase transition, since tunneling from one minimum to another only costs finite action (energy). Note,
however, that this argument does not commute with taking a large N limit N Ñ 8, which is what is
encountered in the text.
42.
43.
44.
45.
46.
47.
48. Newton’s constant in 10D is 16πG10 “ 2κ210 “ p2πq7 gs2 `8s . Then compactifying on T 4 ˆ S 3 to get to 3D, with
the volume of the T 4 being p2π`s q4 V4 and the volume of the S 3 being 2π 2 L3 we have
2κ210 p2πq3 gs2 `4s gs2 `4s
2κ23 “ ñ G 3 “ “
p2π`s q4 V L3 2π 2 16π ˆ 2π 2 ˆ V ˆ L3 4V L3
Using Brown-Henneaux, this further gives:
3 L 6V L4
c“ “ 2 4 “ 6Q1 Q5
2 G3 g s `s

49. Because this is so crucial, I will first review the argument by Brown and Henneaux.
50. Let us first express the expectation value of the stress energy tensor in the 2D CFT in terms of the asymptotic
metric of 3D gravity. We have done this for AdS5 /CFT4 in section 14.8.3. We have the equations:
`2 p0q
´1
Trrgp0q gp2q s “ ´ R, ∇k gkj “ ∇j Trgp0q
2
Thus only the trace-free part of g p2q is left undetermined. This can be parameterized by hzz , hz̄,z̄ .
Altogether we get:
1 1
Tzz “ ´
hzz , Tz̄,z̄ “ ´ hz̄ z̄
4` 4`
Focusing on a holomorphic transformation we get:
ds2 Ñ ds2 `

51. Say I had a conserved global charge for a continuous (internal) symmetry. This provides a local current J µ
in the theory. I take this current operator near the boundary of AdS - this defines an operator in the CFT.
The corresponding boundary local operator will thus generate a global symmetry on the boundary theory.
By preceding arguments (exercise 14.10), the corresponding bulk symmetry must be gauged.
Incorporate Harlow’s argument.
52. First consider two noninteracting CFTs, CFT1 and CFT2 . The stress energy tensor T decomposes into a
sum T1 ` T2 . The corresponding geometry will be a product of two separate (noninteracting) copies of AdS5 .
Thus, one graviton h1 ` h2 remains massless while the other h1 ´ h2 obtains a mass squared proportional to
1
N 2 L2
.

196
1 Chapter 15: Applications of the Holographic Correspondence
1. Taking U “ r{`2s and gY2 M “ gs p2πqp´2 `sp´3 fixed as `s Ñ 0, we have that at the scale U ,
e2Φ “ gs2 H p3´pq{2 ñ gef
2 2
f “ gY M N U
p´3
.
In the extremal case the electric field is:
H1 gs N p2π`s q7´p
Fr01...p “ ´ “
H2 Ω8´p H 2 r8´p
gs N 2π`s 7´p 6´p p7 ´ pq2 Ω8´p 6´p `2s p7 ´ pq2 p2πq2p´9 Ω8´p 6´p
ˆ ˙
Ñ r “ r “ U
Ω8´p L2 p2π`s q7´p gs N gY2 M N

2. The original near-horizon metric is:

« a ff
U p7´pq{2 g d N
Y M p
`2s a p´dt2 ` dx ¨ dxq ` pdU 2 ` U 2 dΩ28´p q
gY M dp N U p7´pq{2
The sphere factor is direct and yields:
a
`2s dp N U pp´3q{2 gY M dΩ28´p
The other factor will require our change of variables. Pulling out the same overall factor as before, we are
left with:
U 5´p dU 2
„ 
2
a pp´3q{2 2
`s dp N U gY M 2 p´dt ` dx ¨ dxq ` 2
gY M dp N U
Upon making the substitution:
˜ a ¸2
2gY M dp N dU 2 du
U 5´p “ ñ “´
p5 ´ pqu U 5´p u
we get: „ 
4 1 2 2
pdu ´ dt ` dx ¨ dxq
p5 ´ pq2 u2
Exactly AdS with radius 4{p5´pq. I’m not sure how Kiritsis is absorbing the gY M - strictly speaking
the metric in 15.1.17 is off by that factor is the dΩ5 is to be unital.
3. For an extremal brane it is straightforward to get the curvature in terms of the dilaton EOM, and indeed
we’ve done this in an exercise for chapter 8, as well as having it written explicitly in 8.8.31.
Schematically: d
`2 1 1 U 3´p
`2s R „ pp´3q{2 s p7´pq{2 „ b „ „
r L gef f gY2 M N
gs `sp´3 U p´3 N
as required.
4. Ok here the limits are subtle and worth discussing. I’m following section 13.7. There are two horizons.
Near-horizon means near the outer horizon. In order to take this limit successfully, we must take r0 ! L. In
fact, we must take r0 Ñ 0 in a controlled way. Expectedly, we must hold U0 “ r0 {`2s fixed alongside U “ r{`2s
while taking r0 , r, `2s to zero at the same rate.
U 7´p
For this reason, it is safe to replace H by L7´p {r7´p as before, and also to replace f by 1 ´ U07´p in the
nonextremal solution. We then recover exactly the near-horizon extremal solution with the dt2 and dU 2
terms modified by f :
´f prqdt2 ` dx ¨ dx a
„ 2 
2 dr 8´p
ds “ ? ` Hprq ` r dΩ8´p
Hprq f prq
« a ˙ff
gY M dp N dU 2
ˆ
2 U p7´pq{2 2 2
Ñ `s a p´f pU qdt ` dx ¨ dxq ` ` U dΩ8´p .
gY M dp N U p7´pq{2 f pU q

197
 with
U07´p
f pU q “ 1 ´ .
U 7´p
5. Let’s start with the Hawking temperature. From excercise 13.1 it is simply
p5´pq{2
C 1 pr0 q p7 ´ pqU0
TH “ “ a
4π 4πgY M dp N

The ADM mass above extremality is given by (again, I think there must be something wrong with
equation 8.8.14)
´13`3p
Vp 7´p 2´10`2p p9 ´ pqπ 2
p9 ´ pqr0 “ Vp U07´p
2κ210 gY4 M Γp 9´p
2 q
as required.
The entropy density will come from the area of the horizon at U “ U0 .
Vp a pp7´pq{4 ´p8´pqp7´pq{4 8´p Vp
pgY M dp N q4´p U0 U0 U0 “ 5 6 8 2
4G10 2 π `s gs

By straightfoward algebra, this is equal to the messy expression that Kirtisis has.

6. Area law behavior is indicative of confinement, which is what we would qualitatively expect in

198
Chapter 16: String Theory and Matrix Models
1. The Nambu-Goto action is
ż a
´T2 d3 ξr det ĝ ` Ĉαβγ α βγs, ĝαβ “ Gµν Bα X µ Bβ X ν , Cαβγ “ Cµνρ Bα X µ Bβ X n uBγ X ρ

Let’s set Cαβγ “ 0. The EOM for the scalar field is quickly seen to be lX “ 0, where l is the Laplacian
from the induced metric.
In the Polyakov action, the equations of motion for γ are the vanishing the energy-momentum tensor, giving:
1
Bα X µ Bβ Xµ ´ γαβ pγ γδ Bγ X µ Bδ Xµ ´ 1q
2
This is harder to solve than the p “ 1 case, as we can’t just take the square root of the determinant of both
sides. Taking the ansatz that γαβ “ λBα X µ Bβ Xµ we get:
1
λγαβ ´ γαβ p3λ ´ 1q
2
And we get a solution with λ “ 1. Note there is no Weyl rescaling here. Similarly, the X field must satisify
1 ?
? Bα p ´hhαβ Bβ X µ q “ 0
´h
which agrees with lX “ 0 upon the identification of the induced and auxiliary metrics.
Note importantly that the p-brane action for p ‰ 1 requires a cosmological constant term.
?
2. Take the gauge γ00 “ ´ det ĝij with γ0i “ 0 so that ´γ “ det gij . The action then becomes:
T2 ? ´ 00 9 9 ¯ T
2 9
´ γ γ X ¨X ´1 “ pX ¨ X9 ` det ĝij q
2 2
Here there is a small typo in Kiritsis. Rewriting
1
det ĝij “ B1 X µ B1 X ν B2 Xµ B2 Xν ´ B1 X µ B2 X ν B1 Xµ B2 Xν “ ´ tX µ , X ν utXµ , Xν u
2
We thus get total action: ż ˆ ˙
T2 3 9 µ 9 1 µ ν
d ξ X Xµ ´ tX , X utXµ , Xν u
2 2
Giving equations of motion
: µ “ ttX µ , X ν u, Xν u
X
Taking now lightcone gauge X ` pτ, σ1 , σ2 q “ τ
The transverse momenta are:
δL `
pi “ “ T2
9 i “ p X9 i
X
δpBτ X i q V
The Hamiltonian is thus ż
p´ X9 ´ ´ L ` d2 ξ pi X9 i

´ ¯1{4
N
3. This rescaling is very straightforward once one has the hamiltonian 16.1.14. One rescales X i Ñ V2 Xi
´ ¯´1{4
N
and t Ñ 4V 2
yielding:

X9 i X9 i X9 i X9 i
ż ż
T2 T2 N T2
d σ X9 i X9 i Ñ
2
d2 σ “ Trr s
2 4 V2 2 4 2
and
ż ż ˆ ˙
T2 2 i j T2 N 2 1 i j T2 1 i j
d σtX , X utXi , Xj u Ñ ´ d σ rX , X srXi , Xj s “ Tr ´ rX , X srXi , Xj s
4 4 V2 4 4 4

199
4. For a string, imagine a rectangular spike of cross-section  and length L. Its total energy is 2L ` , where
 does not multiply L now. Therefore, taking L large will give a large energy deviation, regardless of how
small we take . Thus, the string is stable against decaying into these small spikes.

5. Its immediate that the Cµνρ term multiplies a Nambu bracket, by antisymmetry. Now by permutation
invariance we can write:
1 µ ν ρ
tX , X , X utXµ , Xν , Xρ u “ B1 X µ B2 X ν B3 X ρ αβγ pBα Xµ Bβ Xν Bγ Xρ q
6
Its not hard to see that this reproduces the formula for a 3x3 determinant, as we have an antisymmetric
object involving one element from every row and column multiplied together, all with unital coefficients.
The bracket is not associative show

6.

200