Version of January 9, 2009.

L-functions.
TCC course, OctDec 2008.
Kevin Buzzard
Chapter 0: Introduction.
http://tcc.maths.ox.ac.uk/syllabi/L-Functions.shtml
(or just google TCC Oxford) for books. Syllabus is mildly inaccurate (my
fault): Tate didnt give a new proof of the functional equation of the Riemann
zeta functionhe conceptually explained an older one.
Basic denitions.
If r > 0 is real and s is complex, dene r
s
:= exp(s. log(r)). Note that
[r
s
[ = r
Re(s)
.
The Riemann zeta function is a holomorphic function of a variable s, whose
denition for Re(s) > 1 is
(s) :=

n1
n
s
.
(s) :=

n1
n
s
.
Its easily checked to converge to a holomorphic function (the convergence
is absolute and locally uniform). The rst big fact is that it has a meromorphic
continuation to s C with a simple pole at s = 1 and no other poles. Well see
a proof of this in Lecture 2.
To explain the functional equation (relating (s) to (1 s) Ill need the
function
(z) :=
_

0
t
z1
e
t
dt
which converges (absolutely and locally uniformly) for Re(z) > 0 and hence
denes a holomorphic function there; well see that this also has a meromorphic
continuation to z C but I want to state the functional equation before we get
onto proofs.
The theorem (due to Riemann) is
(s) = 2
s

s1
sin(s/2)(1 s)(1 s).
This is an equality of meromorphic functions; one has to be a bit careful. For
example if s is a positive even integer then the simple zero of sin(s/2) cancels
the simple pole of (1 s) on the RHS (when we get o the introduction and
onto the details well see that has some simple poles).
Heres a nicer (more symmetric) way of writing the functional equation: this
is crucial. If we set
(s) :=
s/2

_
s
2
_
(s)
(s) :=
s/2

_
s
2
_
(s)
1
then the functional equation can be rewritten
(s) = (1 s)
Well prove this before we do anything else because its kind of important to
us. And then well look at the proof and spend the rest of the course trying to
generalise it to a conceptual proof of meromorphic continuation of a huge class
of functions.
Remarks.
1) is truly a product of local factors; one can check that (s) =

p
(1
p
s
)
1
(where the product is over all prime numbers; this is because every
positive integer is uniquely the product of primes). The factor (1 p
s
)
1
is
the local factor at p. The stu that was in but not in is the local factor
at . Well make this rigorous later. It was one of Tates many insights that
this could be formalised and massively generalised.
2) Why do we care about (s), either for Re(s) > 1, or for all s C?
Its a well-observed phenomenon that the zeta function (and its generalisations)
encode arithmetic information, especially where doesnt converge. Indeed, the
general idea is that given an arithmetic object, it could have a zeta function,
which will converge for Re(s) suciently large, and then it might be a tough
theorem (or, more likely, a profound open conjecture) that this zeta function has
a meromorphic continuation to the complex numbers, and then special values
of this function (i.e. its values at certain carefully-chosen points) might tell you
information about the original arithmetic object.
Examples of this phenomenon: (2) =
2
/6 and (4) =
4
/90 and
(12) = 691
12
/638512875
(denominator is 3
6
5
3
7
2
11.13) and (11) = 691/32760 (denominator is 2
3
3
2
.5.7.13;
numerator is prime). These numbers are related to Bernoulli numbersfor ex-
ample B
12
= 691/2730. All this was known to Euler (1700s), in some sense.
Bernoulli numbers tell us information about unramied extensions of cyclotomic
elds: so in some sense the zeta function really is telling us that the class number
of Q(
691
) is a multiple of 691.
The Riemann zeta function has a simple pole at 1 (with residue 1). Hence
there are innitely many primes! (think about the representation of (s) as
a product). Dirichlets theorem (about 150 years ago) pushed this idea a lot
further: mild generalisations of the zeta function plus their behaviour at s = 1
give his famous theorem that there are innitely many primes in an AP.
The Riemann Hypothesis is that all the zeros of the Riemann zeta function,
other than those at s = 2, 4, 6, . . ., lie on the line Re(s) = 1/2. This is a
deep open problem which, were it to be true, would have lots of applications
(it and its generalisations to other zeta functions give you all sorts of results
about the error term in the prime number theorem, or the smallest quadratic
non-residue mod p, and so on).
Generalisations of the Riemann zeta function: Ive already mentioned Dirich-
lets L-functions: Also, a number eld K has a zeta function
K
(s), with
Q
(s)
2
being the classical Riemann zeta function. The function
K
(s) has a simple pole
at s = 1 and the residue is
lim
s1
(s 1)
K
(s) =
2
r
1
(2)
r
2
h
K
R
K
w
K

_
[ D
K
[
lim
s1
(s 1)
K
(s) =
2
r
1
(2)
r
2
h
K
R
K
w
K

_
[ D
K
[
where, as usual, r
1
is the number of real embeddings K R, r
2
is half the
number of non-real embeddings K C, h
K
is the size of the class group of K,
R
K
is the regulator (this is to do with the logarithms of the fundamental units),
w
K
is the number of roots of unity in K and D
K
is the discriminant of K.
Using the functional equation (this zeta function also has a functional equa-
tion) we can recast this statement as a statement about
K
near s = 0, and it
turns out to say that
K
(s) has a zero of order r
1
+r
2
1 at s = 0 (the rank of
the class group) and the power series expansion near s = 0 looks like
(h
K
.R
K
/w
K
)s
r
1
+r
2
1
+. . . .

K
(s) = (h
K
.R
K
/w
K
)s
r
1
+r
2
1
+. . . .
So in some sense the reason (0) = 1/2 is because the rational integers are a
PID and the only units are the two roots of unity (and hence the regulator is 1).
Zeta functions hold profound arithmetic secrets. More general zeta functions
are also called L-functions. Putting Dirichlets ideas together with the general-
isations to number elds gives the correct analogue and proof of Dirichlets
theorem for the integers of a number eld.
Other things with zeta functions: automorphic forms, elliptic curves, alge-
braic varieties,. . . . Special values of L-functions and analogues of the class
number formula above give profound conjectures. For example the Birch
Swinnerton-Dyer conjecture is just the analogue of the above theorem about

K
(s) near s = 0, but for the L-function of an elliptic curve.
Hecke proved Tates theorem rst; but Tates proof was amenable to vast
generalisations and has run and run.
Chapter 1: Meromorphic continuation and functional equation of
the Riemann function.
1.1 The function.
Denition:
(z) =
_

0
t
z1
e
t
dt.
Converges for Re(z) > 0 [integrand blowing up at zero if Re(z) < 1 but not
too badly: integral converges] to a holomorphic function.
Integrate by parts: for Re(z) > 0 we have
3
(z + 1) =
_

0
t
z
e
t
dt
= [t
z
e
t
]

0
+
_

0
zt
z1
e
t
dt
= z(z).
Hence for Re(z) > 0 and n Z
1
we have
(z +n) = (z +n 1)(z +n 1)
= (z +n 1)(z +n 2) . . . (z + 1)z(z)
and hence (z) = (z + n)/[(z + n 1)(z + n 2) . . . (z + 1)(z)], and the
right hand side is meromorphic for Re(z) > n, with (at worst) simple poles at
z = 0, 1, 2, . . . , 1 n. So we can now regard as a meromorphic function on
the entire complex plane, satisfying z(z) = (z + 1).
Easy: (1) = 1 (just compute the integral), and now its an easy exercise
from z(z) = (z + 1) to check that
(n + 1) = n! for n Z
0
(z) has a simple pole at z = 0, 1, 2, . . . and no other poles. (exercise:
compute the residue at these poles).
To check the two versions of the functional equation that I gave in the rst
lecture are the same, one has to check
2
s

1/2
sin(s/2)(1 s)(s/2) = ((1 s)/2).
but I wont use this because well never use the asymmetric functional equation.
[It follows easily if you can prove
Eulers reection formula
(1 z)(z) = / sin(z)
and Legendres duplication formula
(z)
_
z +
1
2
_
= 2
12z

(2z).
]
1.2: Poisson summation.
Dene
(t) =

nZ
e
n
2
t
2
4
(a function on the positive reals); its easily checked to converge, and tends to
one (very rapidly) as t . Our goal here is to show the fundamental fact
(1/t) = t(t).
This is not at all obvious (to me)for example e
16
10
20
so its not surprising
(looking at the denition) that
(4) = 1.0000000000000000000002958 . . .
but it is surprising (to me) that
(1/4) = 4.00000000000000000000118322 . . .
(or equivalently, why, if r = e
/16
= 0.821724958 . . . then r + r
4
+ r
9
+ r
16
+
r
25
+r
36
+r
49
= 1.49999 . . ..
The fundamental fact
(1/t) = t(t)
follows from the Poisson Summation formula, which follows from the general
theory of Fourier series. Heres how a proof goes.
First let me remind you of a crucial integral:
_

e
x
2
dx = 1
because if I denotes the integral then I
2
=
_
R
2
e
(x
2
+y
2
)
dxdy which is (recall-
ing dxdy = rdrd)
_
r0
_
02
e
r
2
rdrd
which is 2[e
r
2
/2]

0
= 1. As a consequence we deduce
_

e
y
2
+2iry
dy = e
r
2
()
for r > 0 real (complete the square with x = y ir and use Cauchys theorem).
As another consequence we deduce (1/2) =

(Exercise: comes straight from
the denition after a simple substitution).
Now lets get back to . Fix t > 0 and dene (for x R) a function
f(x) = e
t
2
x
2
, and then dene
F(x) =

nZ
f(x +n)
=

nZ
e
t
2
(x+n)
2
.
Note that F(0) = (t). But note also that F is continuous and periodic with
F(x) = F(x + 1) so by the theory of Fourier series [which well do in some
generality later on, but let me just assume the classical theory now] we must
have F(x) =

mZ
a
m
e
2imx
and we can compute
a
m
=
_
1
0
F(x)e
2imx
dx.
5
a
m
=
_
1
0
F(x)e
2imx
dx
and this comes out to be

nZ
_
1
0
f(x +n)e
2imx
dx
=

nZ
_
1
0
f(x +n)e
2im(x+n)
dx
because changing x to x+n changes things by e
2imn
which is 1. And now this
is just
_

f(x)e
2imx
dx
so we have proved that
a
m
=
_

e
t
2
x
2
+2imx
dx.
a
m
=
_

e
t
2
x
2
+2imx
dx.
Now setting y = tx and r = m/t we get
a
m
= t
1
e
m
2
/t
2
from () above. So thats a
m
and now (from the denitions)
(t) = F(0)
=

mZ
a
m
= t
1
(1/t)
so were done.
Corollary: (t) 1/t for t > 0 small.
1.3: Meromorphic continuation of (s) and (s).
Dene, for Re(s) > 1,
(s) :=
_

t=0
((t) 1)t
s1
dt.
Note: this was not the denition of we saw in the rst lecture; but well
prove its the same. This function is the Mellin Transform of (t) 1, and
well now see that the relation between (t) and (1/t) translates into proof of
meromorphic continuation and functional equation for (s).
If Re(s) > 1 then this integral converges. Indeed the integral from 1 onwards
is ne, because (t) 1 is decaying exponentially, and the integral from 0 to 1
is OK because (t) is like 1/t so were just OK.
6
Now breaking up the integral at the point t = 1 we see
(s) =
_

t=1
((t) 1)t
s1
dt +
_
1
t=0
((t) 1)t
s1
dt.
The rst integral converges for all s C, and using the fact that (t) = (1/t)/t
and subbing u = 1/t we get that the second is
_

u=1
(u(u) 1)u
1s
du
and we can break this up into two pieces as
_

u=1
(u)u
s
du
_

u=1
u
1s
ds
(noting that both integrals converge in the region Re(s) > 1). The second piece
is just 1/s. Changing that back to 1 in the rst piece by adding and
subtracting
_

1
u
s
du = 1/(1 s), and putting everything together, gives us
(s) =
_

t=1
((t) 1)t
s1
dt
+
_

u=1
((u) 1)u
s
du 1/(1 s) 1/s
so
(s) =
_

t=1
((t) 1)(t
s1
+t
s
)dt
1/(1 s) 1/s
Now that integral converges for all s C to a holomorphic function which is
visibly invariant under s 1 s. We deduce that has simple poles at s = 0
and s = 1 with residues 1 and +1 respectively, and no other poles, and satises
(s) = (1 s).
So whats left is to check that (s) has got something to do with the zeta
function! And we do this by now assuming Re(s) > 1 again, and writing
(s) = 2
_

t=0

n1
e
n
2
t
2
t
s1
dt
and interchanging the sum and the integral, and observing that we can then do
the inner integral: its
_

t=0
e
n
2
t
2
t
s1
dt
_

t=0
e
n
2
t
2
t
s1
dt
and now setting u = nt we get
n
s
_

u=0
e
u
2
u
s1
du
7
and now setting v = u
2
so 2udu = dv we get
n
s
_

v=0
e
v
(v/)
s/21
(2)
1
dv
which is
n
s
2
1

s/2
(s/2)
so (s) =
s/2
(s/2)(s) for Re(s) > 1 and that was what we wanted!
Finally, lets think about poles. We saw (s) had simple poles at s = 0 and
s = 1, with residues 1 and +1 respectively, and no other poles. So
(s) = (s).
s/2
/(s/2)
(an equation which now gives us the meromorphic continuation of the Riemann
zeta function!) will have a simple pole at s = 1 with residue
1/2
/(1/2) = 1,
and will be holomorphic at s = 0 because has a simple pole at s = 0. Further-
more, the only other poles of (s) will come from zeros of the functionbut if
the function had a zero then z(z) = (z +1) implies it would have zeros with
arbitrarily large real part and hence (s) would have poles with arbitrarily large
real partbut this is impossible because (s) is holomorphic for Re(s) > 1.
What Tate did was he managed to understand the above argument to such
an extent that he could generalise it. Perhaps you cant see the wood from the
trees at the minute, but somehow the ingredients are: clever denition of ,
and two ways of evaluating it: one by brute force and one by viewing it as a
Mellin transform of a theta function, breaking up the integral into two pieces,
and using Poisson summation. This is the strategy that we shall generalise,
once we have spent at least half of the course creating the necessary machinery.
Chapter 2: Local elds.
Let k be a eld. A norm on k is [.[ : k R with
(i) [x[ 0 with equality i x = 0
(ii) [xy[ = [x[[y[
and some version of the triangle inequality, which varies from book to book.
Let me use the following variant:
(iii) Theres some constant C 1 such that [x[ 1 implies [1 +x[ C.
We say that a pair (k, [.[) consisting of a eld k and a function [.[ : k R
satisfying the above axioms is a normed eld.
(i) [x[ 0 with equality i x = 0
(ii) [xy[ = [x[[y[
(iii) Theres some constant C 1 such that [x[ 1 implies [1 +x[ C.
Theres now lots of things to do and no (for me) clear order in which to do
them. We need comments about the axioms, basic properties deducible from
the axioms, and elementary examples.
Let me rst make a comment about the axioms: Why not the triangle in-
equality? Why not [x + y[ [x[ + [y[ instead of (iii)? In fact (iii) is slightly
weaker than the triangle inequality, as can be easily seen: if [x + y[ [x[ + [y[
for all x and y then (iii) is satised with C = 2.
8
(i) [x[ 0 with equality i x = 0
(ii) [xy[ = [x[[y[
(iii) Theres some constant C 1 such that [x[ 1 implies [1 +x[ C.
The problem with the triangle equality if you take a norm on a eld which
satises the triangle inequality, and then cube it, it might not satisfy the triangle
inequality any more. On the other hand, one can easily check that [.[ is a norm
in the sense above i [.[
r
is for any r > 0 (easy exercise: replace C by C
r
).
Denition We say that two norms [.[ and [.[

on a eld k are equivalent if

theres some r > 0 such that [x[
r
= [x[

for all x k.
We only really care about norms up to equivalence.
(i) [x[ 0 with equality i x = 0
(ii) [xy[ = [x[[y[
(iii) Theres some constant C 1 such that [x[ 1 implies [1 +x[ C.
Basic properties of a normed eld: [0[ = 0, and [1[ = [1[
2
and hence [1[ = 1
so [ 1[
2
= 1 and hence [ 1[ = 1 and [ a[ = [a[.
Examples: k = R (or any subeld, for example Q), and [x[ is the usual
norm: [x[ = x for x 0 and x for x < 0.
Trivial example: the trivial norm on a eld: [0[ = 0 and [x[ = 1 for all
x ,= 0.
Back to the triangle inequality. Ive already mentioned that if a function [.[
on a eld k satises (i) and (ii) and [x + y[ [x[ + [y[ for all x and y, then it
clearly satises (iii) with C = 2.
Conversely,
Lemma. if (k, [.[) is a normed eld and if [x[ 1 implies [1 +x[ 2 (that is,
if we can take C = 2 in the denition of the norm), then [.[ satises the triangle
inequality.
Ill sketch a proof of this because the proof is slightly tricky and we use
corollaries of this result quite a bit. Let me mention some corollaries rst.
Corollary 1. Any norm is equivalent to a norm satisfying the triangle in-
equality.
Proof: C
r
2 for some appropriate r.
Corollary 2. A norm denes a topology on k: if we say that a subset U of k
is open i for all u U theres > 0 such that [u v[ < implies v U, then
the open sets satisfy the axioms for a topology.
Proof: equivalent norms dene the same open sets, and if the norm satises
the triangle inequality then d(x, y) = [x y[ is a metric and the open sets for a
metric form a topology.
OK, now onto the proof of the lemma.
Lemma. if (k, [.[) is a normed eld and if [x[ 1 implies [1 +x[ 2 (that is,
if we can take C = 2 in the denition of the norm), then [.[ satises the triangle
inequality.
Proof (sketch).
(a) The denition implies [x +y[ 2 max[x[, [y[.
(b) Hence (induction) [x
1
+x
2
+. . . +x
2
n[ 2
n
max[x
1
[, [x
2
[, . . .
(c) Hence
[x
1
+x
2
+. . . +x
N
[ 2N max[x
1
[, [x
2
[, . . . , [x
N
[
9
(choose n with 2
n1
< N 2
n
and use (b) with x
N+1
= . . . = 0).
(c)
[x
1
+x
2
+. . . +x
N
[ 2N max[x
1
[, [x
2
[, . . . , [x
N
[
(d) Hence [N[ 2N for all N Z
0
.
(e) Now use the binomial theorem, (c) and (d) to check that
[(x +y)
m
[ 4(m+ 1)([x[ +[y[)
m
for all m Z
1
.
(f) Now let m and take mth roots to get the result.
Theres a dichotomy: if we can take C = 1 in (iii) then C = 1 will also do
for any equivalent norm. But if we need C > 1 in (iii) then by replacing [.[ with
[.[
N
for some N >> 0 we can make C as large as we like (and equivalently as
small as we like, subject to it being bigger than 1, by letting N 0
+
).
Denition: a norm is non-archimedean if we can take C = 1 in (iii) above.
A norm is archimedean if its not non-archimedean. This denition is good
on equivalence classes. Note that a norm is non-archimedean i [x + y[
max[x[, [y[ for all x, y k (easy check). This is much stronger than the
triangle inequality!
The usual norm on R is archimedean. The trivial norm is non-archimedean.
Other examples: k = C and [x + iy[ =
_
x
2
+y
2
, or even [x + iy[ = x
2
+ y
2
:
these are archimedean.
A less trivial example of a norm: k = Q, choose a prime p, and dene
[p[ = p
1
(or indeed [p[ = r for any 0 < r < 1) and [q[ = 1 for any other prime
q, and extend multiplicatively (and set [0[ = 0). So we have

p
n

u
v

= p
n
for n Z and u, v integers prime to p.
I claim that this is a non-archimedean norm (its called the p-adic norm
on Q). This is such an important norm for us that Ill check the axioms.
Denition: [0[ = 0 and

p
n

u
v

= p
n
for n Z and u, v integers prime to p.
Check its a norm: (i) and (ii) are obvious. So it suces to check that for
all x and y we have [x +y[ max[x[, [y[; then (iii) will follow with C = 1.
Now [x + y[ max[x[, [y[ is clear if any of x, y or x + y is zero. In the
general case we may assume [x[ [y[, so x = p
n u
v
and y = p
ms
t
with n m,
and we see that
x +y = p
n
_
u
v
+
p
mn
s
t
_
= p
n
ut +p
mn
sv
vt
= p
n
u

10
with v

= vt and n

n, so [x +y[ = p
n

p
n
= [x[ = max[x[, [y[.
Easy exercise: check (by beeng up the proof) that in fact the p-adic norm
on Q satises [x +y[ = max[x[, [y[ if [x[ , = [y[.
More fun: check that if [.[ is any non-arch norm on any eld k then [x[ , = [y[
implies [x +y[ = max[x[, [y[. Theres a one-line proof from the axioms which
I sometimes struggle to nd.
Natural generalisation of the p-adic norm: if K is any number eld with
integers R, and if P is a non-zero prime ideal of R, then theres a P-adic norm
on K, dened by [0[ = 0 and, for 0 ,= x K, if we factor the fractional ideal
(x) as (x) = xR = P
e

i
P
e
i
i
with the product nite and only involving prime
ideals other than P, then we can dene [x[ = r
e
for any r with 0 < r < 1;
traditionally we take r = 1/N(P) where N(P) is the size of the nite eld R/P.
Again one checks that this is a norm, and indeed its non-archimedean. These
norms generalise the p-adic norm on Q.
Note that if p factors into more than one prime in R, then there is more
than one P-adic norm on K that induces a norm equivalent to the p-adic norm
on Q. For example, the (2 +i)-adic norm on Q(i) is certainly not equivalent to
the (2 i)-adic norm [because [2 +i[ = 1/5 for one of them and [2 +i[ = 1 for
the other].
There are also natural generalisation of the usual archimedean norm on Q to
a number eld: if K is a number eld then for any eld homomorphism K C
(C the complexes), the usual norm on C induces (by restriction) a norm on K.
There is a subtlety here: if : K C is a eld homomorphism then , dened
by (x) = (x), is also a eld homomorphism, and may or may not be equal
to , but and induce the same norm on K, because [z[ = [z[ on C. So
in fact were led to the following equivalence relation on eld homomorphisms
: K C dened by: and , and nothing else. The equivalence
classes are easily described: the maps K R each give one equivalence class
(the standard notation is that there are r
1
of these) and the maps K C
which dont land in R come in pairs , of equivalence classes: there are r
2
equivalence classes (and hence 2r
2
embeddings). Lets stick with this notation
throughout the course.
If K = Q() and P(X) Q[X] is the minimal polynomial of , then r
1
is the number of real roots of , and r
2
is half the number of non-real roots.
This shows that r
1
+2r
2
is the degree of P. This leads us easily to a proof that
r
1
+ 2r
2
= [K : Q].
We wont logically need the following result so I wont prove it:
Theorem. If K is a number eld, then any non-trivial norm on K is either
equivalent to a P-adic norm for a unique P (this is i its non-archimedean),
or equivalent to the valuation induced by an embedding K C, for a unique
equivalence class , of embeddings as above (this is i its archimedean).
The case K = Q is due to Ostrowski (an explicit elementary calculation),
and the general case can be deduced from this case (after a little work).
Now heres a crucial property of norms. Given a normed eld (and you
can assume C 2 if you like, because what we do here only depends on the
equivalence class of the norm) there are obvious notions of a Cauchy sequence
11
and a convergent sequence:
A sequence (a
n
)
n1
is Cauchy if for all > 0 there is M > 0 such that
m, n M implies [a
m
a
n
[ < .
A sequence (a
n
)
n1
is convergent if there exists b k such that for all > 0
theres M > 0 with n > M implies [a
n
b[ < .
Both notions only depend on the equivalence class of the norm. Every con-
vergent sequence is Cauchy (easy). We say a normed eld is complete if every
Cauchy sequence is convergent.
Examples: R with the usual norm is complete. Q with the usual norm isnt.
In fact, if were dening mathematics from the ground up we would build R
by completing Q. This is a process that works in much more generality!
Theorem. Given a normed eld (k, [.[) it has (up to unique isomorphism)
a completion (K, [[.[[), by which I mean:
(i) a complete normed eld (K, [[.[[), and
(ii) an inclusion k K which preserves the norm (so if youre thinking of
K as containing k then Im just saying that for x k we have [x[ = [[x[[),
such that
(iii) if we endow K with the topology induced by [[.[[, then the closure of k
is K.
Note that (iii) is absolutely crucial: we want R to be the completion of Q,
whereas Q C satises (i) and (ii).
Proof. Lets do existence rst. WLOG [.[ satises the triangle inequality.
Let R denote the set of all Cauchy sequences in k; its a ring with respect to
pointwise addition and multiplication; the constant sequences give a map k R
of rings (so 1 R is the sequence (1, 1, 1, . . .)).
One checks that if (a
n
)
n1
is a Cauchy sequence in k then ([a
n
[)
n1
is a
Cauchy sequence of real numbers, so its convergent. Say ((a
n
)) is its limit.
Let I denote the ideal in R of sequences which tend to zero; this is easily
checked to be an ideal. Let K denote the quotient R/I. Its easily checked that
if (a
n
) (b
n
) I then ((a
n
)) = ((b
n
)). So induces a map [[.[[ : K R.
The claim is that this works. Lets see why.
To check that K is a eld one needs to check that I is maximal; this is because
if (a
n
) is Cauchy but doesnt tend to zero then a
n
,= 0 for all n suciently large
and one can check that the sequence b
n
dened by b
n
= 1/a
n
(unless a
n
= 0 in
which case set b
n
= 59) satises (a
n
)(b
n
) (1) I; this is enough to prove that
I is maximal.
Its clear that the obvious map k K is a map of rings, and hence its an
injection because k is a eld. The rst two axioms for a norm are easily checked
to be satised by [[.[[. Were assuming that [.[ satises the triangle inequality,
and we deduce that [[.[[ does too. Hence (iii) is satised.
To check denseness of k in K we need to check that for any (a
n
) K and
> 0 we can nd a K with [[(a
n
a)[[ < , but this is easy from the denition
of Cauchy.
Finally, to check completeness of K we use the fact that Cauchy sequences
in k converge in K, and that any Cauchy sequence in K can be approximated
by a Cauchy sequence in k in a suciently sensible way to ensure that the limits
coincide.
12
So weve done existence. For uniqueness we need to check that if (K
1
, [[.[[
1
)
and (K
2
, [[.[[
2
) both work then theres a norm-preserving isomorphism of elds
K
1
= K
2
which is the identity on k. The reason for this is that the map
k K
2
can be extended to a map K
1
K
2
thus: write K
1
as the limit
of a sequence in k; this sequence converges in K
2
; send to this element. Now
check that this gives a well-dened bijectionjust follow your nose.
The only complete archimedean elds we care about in this course are the
reals and the complexes (in fact Ostrowski proved that any eld complete with
respect to an archimedean norm was equivalent to (R, [.[) or (C, [.[), with [.[
denoting the usual norm, but we wont need this: see Chapter 3 of Cassels
Local Fields).
So now we press on with the (arguably more subtle) theory of the structure
of complete non-archimedean elds. Its easy to give examples of such things:
for example lets dene the p-adic numbers to be the completion of Q with
respect to the p-adic normthe usual notation for the p-adic numbers is Q
p
(note: we havent yet proved that Q with its p-adic norm isnt complete, or
equivalently that Q
p
,= Q. But this will come out in the wash later).
It turns out that if k is a number eld and P is a non-zero prime ideal of its
integer ring, and if P contains the rational prime number p, then the completion
of k with respect to the P-adic norm is naturally a nite extension of Q
p
(Ill
prove this later but it shouldnt surprise you because k is a nite extension of
Q), so in some sense the basic example of a complete non-archimedean eld is
the p-adic numbers, and the most general example well ever use in this course
is a nite eld extension of the p-adic numbers.
Before we start on the general structure theory, let me observe that the
norm on Q
p
or more generally k
P
is discrete, in the following sense: the P-
adic norm on a number eld k has the property that theres a real number q > 1
(the way I normalised it we have q = N(P), the norm of P), such that every
element of k had norm either equal to zero, or to an integer power of q. What
does this imply about the norm on k
P
?
Recall that in the denition of the completion of a eld, the norm of a
Cauchy sequence was the limit of the norms of the elements, and hence (easy
calculation) we see that [.[ : k
P
R is also taking values in the set 0
. . . , q
2
, q
1
, 1, q, q
2
, q
3
, . . ..
We say that a norm on a eld K is discrete if theres some > 0 such that
a K and 1 < [a[ < 1 + implies [a[ = 1. In fact, because [K

[ := [a[ :
a K

is a subgroup of R
>0
its easy to check that if a norm on a eld K
is discrete then either [K

[ = 1 (the trivial norm) or theres some q R

>1
with [K

[ = q
n
: n Z.
The usual norm on the reals or complexes is of course not discrete, but the
P-adic norm on a number eld k is, and weve just seen that even the comple-
tion k
P
of k with respect to this norm is a discretely-normed eld. Dont get
confused thoughthere are blah non-archimedean norms that arent discrete
for example if k
n
is the eld Q(p
1/2
n
) (so we keep square rooting p) and k

is
the union of the elds k
n
(note that k
n
is naturally a subeld of k
n+1
) then the
p-adic norm on Q
p
extends to a non-discrete, non-archimedean norm on k

.
13
Note that k

isnt a number eld though, its an innite extension of Q.

Ok so lets let K now denote an arbitrary eld equipped with a non-archimedean
norm [.[ (so [x+y[ max[x[, [y[ and so at least the triangle inequality holds).
Lets set
R = x K : [x[ 1
and
I = x R : [x[ < 1.
Its easy to check that R is a ring: the surprising part is that if x, y R then
x + y R, and this is because [x + y[ max[x[, [y[ 1. We say R is
the integers of K. Note that in the archimedean case this part already fails:
the closed unit disc is not a subring of the complex numbers. The fact that
R is a ring in the non-archimedean case is, perhaps initially at least, a little
psychologically disturbing: for example it implies that the integers are bounded
within K.
Once we have re-adjusted, its easy to check from the axioms that I is an
ideal of R, and in fact I is the unique maximal ideal of R because if r R and
r , I then [r[ = 1 so r ,= 0 and s := 1/r K has [s[ = 1/[r[ = 1 so s R, and
we see that r is a unit (exercise: this is enough).
We say that the eld R/I is the residue eld of K.
Example: K = Q with the p-adic norm. Then (writing a general rational as
a/b in lowest terms)
R = a/b : a, b Z, p b
and
I = a/b R : p [ a = pR.
R = a/b : a, b Z, p b
and
I = a/b R : p [ a = pR.
I claim that R/I = Z/pZ, and to check this all I have to do is to check that
0, 1, 2, . . . , p 1 meets every coset r + I exactly once, which follows easily
from the statement that given a, b Z with p b theres a unique blah blah
t 0, 1, 2, . . . , p 1 with a bt mod p. So the residue eld of Q with its
p-adic norm is Z/pZ.
My aim now is to basically prove a structure theorem for characteristic zero
blah non-archimedean elds K which are complete with respect to a discrete
valuation; this will easily give us enough to show that if K is the completion of
a number eld at a non-zero prime ideal then K and K

are locally compact

abelian groups, which is the buzz-word well need to do the abstract Fourier
analysis well need for Tates thesis later on.
Structure of complete discrete non-arch elds.
The rst (and main) goal of this lecture is to explain what a complete
discretely-valued non-arch eld looks likewell end up with some kind of
structure theorem, analogous to the theorem that every real has an essentially
unique decimal expansion, but with 0.1 replaced by a small number in the eld
for example the role of 0.1 is played by p in Q
p
. This structure theorem (plus
14
the associated exercises on the example sheet) will hopefully greatly clarify what
elds like the p-adic numbers (and their nite extensions) look like. Hopefully,
by the end of the lecture, well begin to have a concrete feeling about how to
compute with Q
p
, just as we have a concrete feeling about how to compute with
real numbers.
Let K be a eld complete with respect to a non-arch norm (we dont need
discreteness for the next few slides). Im going to do some abstract analysis
nowthings which will be familiar from basic analysis classes, but which will
work just as well in K because they work for any complete eld, not just the
reals or complexes.
Denition. If x
1
, x
2
, x
3
, . . . K then we say that

n1
x
n
converges if the
partial sums tend to a limit : we write

n1
x
n
= . Because were assuming
K is complete, the sum converges i the partial sums s
m
=

m
i=1
x
i
are Cauchy,
and the standard argument shows that if the sum converges then x
n
= s
n
s
n1
had better tend to zero (Cauchyness implies s
n
s
n1
gets arbitrarily small).
[

x
n
converges implies x
n
0].
The weird thing is that, in the non-arch world, the converse is true. Let
x
1
, x
2
, x
3
, . . . be a sequence in a complete non-arch eld K.
Lemma. If x
n
0 as n then

n1
x
n
converges! Furthermore, if B
is real and [x
n
[ B for all n then

x
n
= s with [s[ B too.
Proof. By an easy induction on n, using the denition of a non-archimedean
norm, we see that if [x
i
[ B for 1 i n then [

n
i=1
x
i
[ B (note that this
is a nite sum). Its easy (but crucial) to deduce from this that a sequence (a
n
)
is Cauchy if and only if a
n
a
n1
tends to zero as n . Now apply this with
a
n
=

n
i=1
x
i
to deduce that x
n
0 implies that the the a
n
form a Cauchy
sequence, and hence converge. One way of doing the second part is to prove
that if a
n
as n then [a
n
[ [[ in Rthis is true in any normed eld
(hint: WLOG triangle inequality holds; now use it judiciously).
Before we go any further, let me explain why the residue eld of a non-
archimedean normed eld is the same as the residue eld of the completion. This
is easy. Let k be a non-archimedean normed eld with completion

k. Let R, I
be the integers and maximal ideal for k, and let

R,

I denote the corresponding
things for

k. Theres a natural map k

k sending R to

R and I to

I, and
hence sending = R/I to =

R/

I. Well see a bit later that

k can be much
bigger than k (analogous to R being much bigger than Q). But. . .
Lemma. The map is an isomorphism of elds.
Proof. Injectivity is clear (k

k is norm-preserving, so R

I = I). To get
surjectivity, for r

R simply choose r k with [r r[ < 1 (this is possible by
denseness) and observe that this implies [r[ 1 and hence r R. Moreover
r r

I, so

R =

I +R which shows that is surjective.
Corollary. The residue eld of Q
p
is Z/pZ. For this is the residue eld of
Q with the p-adic norm.
Exercise: let k be a number eld and let P denote a non-zero prime ideal
of its integer ring A. Show that the residue eld of k equipped with the P-adic
norm is canonically isomorphic to A/P (hint: if R and I are the usual things
then construct a natural surjective ring homomorphism R A/P with kernel
15
equal to I). Deduce that the residue eld of k
P
is A/P.
Clarication: A is the integers of k (so, for example, Z is the integers of Q)
but we also referred to R = x k : [x[ 1 as the integers of kthis notion
of course depends on the choice of a norm on k. Sorry. If k = Q above then
A = Z and if P = (p) then R = a/b : p b and Im saying that theres a
natural map R Z/pZ.
As a consequence, we see that if k
P
is the completion of a number eld at
a non-zero prime ideal then the residue eld of k
P
is nitesuch elds have a
much more arithmetic avour than general complete normed elds (for example
you can do analysis in any complete normed eld, but if the norm is discrete and
the residue eld is nite then you can do local class eld theory (i.e., arithmetic)
too).
Exercise: consider the ring C[[T]] of power series

n0
a
n
T
n
with complex
coecients (and no convergence conditionsjust abstract power series) (NB this
exercise would work if you replaced C by any eld at all). Let k := C((T)) de-
note its eld of fractions. Check that a general element of C((T)) is

nM
a
n
T
n
with M a possibly negative integer. Dene a norm on C((t)) by [0[ = 0 and, for
f =

nM
a
n
T
n
with a
M
,= 0, set [f[ = e
M
(where e could really be replaced
by any real number greater than 1). Check that this is a non-arch norm on k,
that the integers R are C[[T]], that the maximal ideal I is TC[[T]] and that the
residue eld is C again. So we can do analysis in k but not arithmetic.
Before we go on to prove the structure theorem, lets play about a bit with
Cauchy sequences in Q with the p-adic norm, and see if any of them converge.
Example 1: Consider the sequence
3, 33, 333, 3333, . . .
in Q with the 5-adic norm.
(a) Its Cauchy! Because if a
n
is n threes then for n m we have 10
n
[
(a
m
a
n
) so [a
m
a
n
[ 5
n
.
(b) In fact its even convergent! Because 3a
n
+ 1 = 10
n+1
which tends to
zero in the 5-adic norm, so a
n

1
3
.
Example 2: Lets put the 3-adic norm on Q. Set a
1
= 1 and a
2
= 4 and
note that 3
2
[ (a
2
2
7). Lets try and nd an integer a
3
with 3
3
[ (a
2
3
7).
Lets try a
3
= 4 + 9n; then a
2
3
= 16 + 72n mod 27 so a
2
3
7 0 mod 27 i
1 + 8n 0 mod 3 so lets set n = 1 and a
3
= 13; this works.
Can we pull this trick o in general? Say m 1 and a
m
1 mod 3 and
a
2
m
7 mod 3
m
.
Can we nd a
m+1
with a
2
m+1
7 mod 3
m+1
? Lets try setting a
m+1
= a
m
+3
m
n
for some n to be determined. Then we see that
a
2
m+1
a
2
m
+ 2n.3
m
(mod 3
m+1
)
7 +t
m
.3
m
+ 2n.3
m
(mod 3
m+1
)
and we can solve 2n + t
m
0 mod 3 for n, so we can indeed nd a
m+1
whose
square is 3-adically close to 7, and by letting m go to innity we can get as close
as we want.
16
Upshot: we have a sequence a
1
, a
2
, a
3
, . . . of elements of Z, with a
n+1
a
n
a multiple of 3
n
, and a
2
n
tending to 7 in Q with the 3-adic norm.
Because a
m+1
a
m
is a multiple of 3
m
we see that the a
m
are a Cauchy
sequence, and their limit in Q
3
is visibly going to satisfy
2
= 7. Hence Q is
not complete with respect to the 3-adic norm!
Exercise: check that 7 is a square in Q
2
and 1 p < 0 is a square in Q
p
for any p > 2. Hence Q is not complete with respect to the p-adic norm, for
any p.
Remark: Ive collected up these exercises and put them on an example sheet.
See the course web page.
Now lets assume that F is a eld with a non-trivial non-archimedean discrete
norm. In this case we have seen that [F

[ := [a[ : a F

is q
n
: n Z for
some q > 1; set = 1/q < 1 and lets choose F with [[ = . We call a
uniformiser in F. As an example, if F = Q or Q
p
with the p-adic norm then
we can set = 1/p and = p, and more generally if F is a number eld k or a
completion k
P
at a prime ideal then = 1/N(P) and, even though P may not
be principal, we can nd x k an algebraic integer with (x) = PJ and P J
(for example, by uniqueness of factorization we have P
2
,= P and any x P
with x , P
2
will do), and then [x[ = 1/N(P) = so x is a uniformiser for both
k and k
P
with their P-adic norms.
Now let R be the integers of K, and let I denote the maximal ideal of R.
If is a uniformiser, then y I implies [y[ < 1 and hence [y[ [[, so y = z
with [z[ 1 and we have proved that I = () is a principal ideal.
Now here we go with the structure theorem. Let K be complete with respect
to a non-trivial non-arch norm. Let R be the integers, I the maximal ideal of
R, let be a uniformiser (so I = ()) and let denote the residue eld R/I.
Let S denote a subset of R, containing 0, such that the reduction map S R/I
is a bijection (so S is a set of representatives for R/I).
Theorem.
(a) If a
0
, a
1
, a
2
,. . . is an arbitrary innite sequence of elements of S, then
the innite sum

n0
a
n

n
converges in R, and furthermore for every element
r of R its possible to write r =

n0
a
n

n
with the a
n
as above, in a unique
way.
(b) If 0 ,= r R then r =

n0
a
n

n
with at least one a
n
,= 0 and in fact
[r[ = [[
m
, where m 0 is the smallest non-negative integer such that a
m
,= 0.
(c) A general non-zero element of K can be written uniquely as =

nM
a
n

n
with a
M
,= 0, a
n
S for all n, and we have [[ = [[
M
.
Before we go on, lets observe the consequences for Q
p
. Let Z
p
denote
x Q
p
: [x[ 1.
Corollary. A general element of Z
p
can be written uniquely as

n0
a
n
p
n
with each a
n
0, 1, 2, . . . , p 1. A general non-zero element of Q
p
can be
written

nM
a
n
p
n
with M Z, 0 a
n
p 1 and a
M
,= 0.
Note that we now see why is called a blah uniformiserits playing some
kind of analogue to the role of a local uniformiser in the theory of complex ana-
lytic functions of one variable, with the theorem giving a power series expansion
17
near a point.
Corollary. Q ,= Q
p
. Indeed we see that Q
p
is uncountable.
Exercise: if Q

and we write =

nM
a
n
p
n
with 0 a
n
< p then
check that the sequence a
n
is ultimately periodic. Hence a number like

n1
p
n!
is an explicit example of an element in Q
p
but not Q.
Lets state the theorem again because its a new lecture.
Let K be complete with respect to a non-trivial non-arch norm. Let R be
the integers, I the maximal ideal of R, and let be a uniformiser (so [[ =
with 0 < < 1 and [K

[ =
Z
, and I = ()). Let denote the residue eld
R/I. Let S denote a subset of R, containing 0, such that the reduction map
S R/I = is a bijection (so S is a set of representatives for ).
Theorem.
(a) If a
0
, a
1
, a
2
,. . . is an arbitrary innite sequence of elements of S, then
the innite sum

n0
a
n

n
converges in R, and furthermore for every element
r of R its possible to write r =

n0
a
n

n
with the a
n
as above, in a unique
way.
(b) If 0 ,= r R then r =

n0
a
n

n
with at least one a
n
,= 0 and in fact
[r[ = [[
m
, where m 0 is the smallest non-negative integer such that a
m
,= 0.
(c) A general non-zero element of K can be written uniquely as =

nM
a
n

n
with a
M
,= 0, a
n
S for all n, and we have [[ = [[
M
.
Proof of theorem.
If a
0
, a
1
, a
2
, . . . are arbitrary elements of R then [a
n
[ 1 so [a
n

n
[
n
0,
where 0 < < 1 is the real number which generates the norm group [K

[ as
above. So the sequence (a
n

n
) tends to zero, so the sum

n0
a
n

n
converges,
and furthermore [a
n

n
[ 1 for all n 0 and hence the sum converges in R.
Thats done the rst part of (a), because S R by denition.
Next note that again by denition a S implies that either a = 0 or [a[ = 1.
So now if r =

n0
a
n

n
with a
n
S and not all of the a
n
equal to zero, and
if m 0 is the smallest non-negative integer with a
m
,= 0, then
r =

n0
a
n

n
= a
m

m
+

nm+1
a
n

n
and [a
m

m
[ =
m
whereas each term in blah

nm+1
a
n

n
has norm at most

m+1
<
m
,
so the sum converges to something with norm at most
m+1
, so [a
m

m
[ >
[

nm+1
a
n

n
[ and we see [r[ = [a
m

m
[ =
m
. This does (b).
Now the uniqueness in (a) is easy: if r =

n0
a
n

n
=

n0
b
n

n
with the
a
i
and b
i
in S then 0 =

n0
(a
n
b
n
)
n
. But its easily checked that for a, b
S, either a b = 0 or [a b[ = 1. So the argument above shows that if a
n
,= b
n
for some n then [

(a
n
b
n
)
n
[ > 0, contradicting

n0
a
n

n
=

n0
b
n

n
.
Hence -adic expansions are unique, if they exist.
To nish (a) we need a construction proof: given r R we need to nd
a
n
S with r =

n0
a
n

n
. Theres a natural way to do this. Given r R we
18
consider the image r of r in , the residue eld. Choose a
0
S whose reduction
is r. Now r a
0
reduces to zero in , so [r a
0
[ < 1 and hence [r a
0
[ .
Hence r
1
:= (r a
0
)/ satises [r
1
[ 1 and we can apply the same trick to
nd a
1
with [r
1
a
1
[ . Hence [r
1
a
1
[
2
and we deduce
[r a
0
a
1
[
2
.
Set r
2
= (r a
0
a
1
)/pi
2
and continue in this way. At the Nth step we nd
[r
N

i=0
a
i

i
[
N+1
so, by denition,

n0
a
n

n
= r.
Thats done (a) and (b). For (c) we just observe that any K with ,= 0
we have [[ =
M
for some integer M, and hence
M
R (with norm 1). So

M
=

n0
b
n

n
with b
n
S and b
0
equal to a lift of the reduction of
M
in , so b
0
,= 0. So
=

nM
a
n

n
with a
n
= b
nM
.
Were done with our structure theorem; now go and do some exercises on
the example sheet.
A corollary whose importance will become clear later is:
Corollary. If K is complete with respect to a non-trivial non-arch discrete
norm, and K has integer ring R with maximal ideal I and residue eld , then
R (with the topology induced from the metric d(x, y) = [x y[) is compact i
is nite.
Proof. Because R is a metric space, compactness is equivalent to sequential
compactness, which Ill remind you means that given a sequence (r
m
)
m1
with
r
m
R we can always nd a convergent subsequence, that is m
0
< m
1
< m
2
<
m
3
< . . . such that (r
m
j
)
j0
converges. Lets rstly assume is nite and prove
that R is sequentially compact.
By the structure theorem we can write
r
m
=

n0
a
m,n

n
with a
m,n
S (a set of coset representatives for ).
Now is nite so S is nite, so we can apply the usual trick: Konigs Lemma
(which according to Wikipedia is due to Konig). Explicitly, we know that a
m,0
assumes at least one value in S innitely often; call it a
0
, and let m
0
be any m
such that a
m,0
= a
0
. Now, amongst the innitely many m > m
0
with a
m,0
= a
0
,
we know that a
m,1
takes on a value innitely oftencall it a
1
. Let m
1
be one of
19
the innitely many m with m > m
0
, a
m,0
= a
0
and a
m,1
= a
1
. Continue in this
way and we see easily that

n0
a
n

n
is the limit of the innite subsequence
r
m
0
, r
m
1
, r
m
2
, . . ..
Conversely, if is innite, then heres an innite open cover of R with no
nite subcover: for any s S the open disc centre s and radius 1 is everything
of the form s + with [[ < 1, so its s + I. Because S is a set of coset
representatives for we see that R is the disjoint union of the open sets s + I
for s S, and this is an innite disjoint cover of R by open sets, which visibly
has no nite subcover.
Corollary. If k is a number eld equipped with a P-adic norm, and if R is
the integers of k
P
, then R is compact.
Indeed, the residue eld of k
P
is A/P, where A is the integers of k in the
sense of algebraic number theory.
That corollary is very important for Tates thesis, as well see later on. I
want to nish local elds today, so, rather than developing the theory in some
kind of logical way (for example Hensels Lemma would be a natural thing to
do next) I am just going to prove the other main thing well need, which is that
if k
P
is the completion of a number eld at a prime ideal then k
P
is naturally a
nite extension of Q
p
, and Ill say a little about the structure of such extensions.
Let k be a number eld, and P a non-zero prime ideal of its integer ring. We
can think of k as a nite-dimensional vector space over Q. Now lets say that P
contains the rational prime p. The restriction of the P-adic norm [.[
P
on k, to
Q, is easily checked to satisfy [[
P
= 1 for a prime with ,= p, and [p[
P
= p
m
for some positive integer m, so [.[
P
on k induces a norm equivalent to the p-adic
norm on Q (in fact its just the mth power of the p-adic norm, where m is easily
checked to be ef, where the size of k
P
is p
f
and where (p) = P
e
.J with J and
ideal coprime to P.
Now there are inclusions of elds Q k k
P
, and k
P
is complete. Of
course k
P
might not be the completion of Q, because theres no reason to
expect that Q is dense in k
P
[the archimedean analogue of whats going on is
that C is an archimedean completion of Q(i) with i
2
= 1 but the resulting
map Q C doesnt have dense image].
[Q k k
P
]
But we can certainly take the closure of Q in k
P
. A closed subspace of a
complete metric space is complete, and its easy to check that the closure of Q in
k
P
is a eld (limit of sum is sum of limits, limit of product is product of limits,
limit of reciprocals is reciprocal of limits when this makes sense), and hence
a normed eld (the norm is induced from k
P
hence the axioms are satised).
Hence this closure must be the completion of Q with respect to the mth power
of the p-adic norm (because its a completion, we showed that completions are
unique up to unique isomorphism).
We deduce that k
P
contains a copy of Q
p
(although, as already mentioned,
the norm on k
P
restricts on Q
p
to a norm which is in general a non-trivial power
of the usual p-adic norm). Now k/Q was a nite extension,
so it wont surprise you to learn that k
P
/Q
p
will also be a nite extension.
Perhaps what will surprise you is that the degree of k
P
/Q
p
might be smaller
than that of k/Q. In fact let me prove something stronger, which will clarify
20
whats going on.
Let me start with some abstract algebra.
Let L be a eld and let K be a subeld of L. Then L is naturally a vector
space over K. The dimension of L as a K-vector space might be nite (for
example C has dimension 2 over R, Q(i) has dimension 2 over Q, Q(2
1/57
)
has dimension 57 over Q) or innite (for example C has innite dimension as
a Q-vector space). Recall that by denition a number eld is a eld k that
contains a copy of Q and such that the Q-dimension of k is nite.
Lets go back to the general case K L and lets assume that the dimension
of L as K-vector space is a nite number n. We say L is nite over K, and
L has degree n over K or even L/K has degree n. Note that L/K isnt a
quotient, its just notation.
Now for L, multiplication by is a map L L which is L-linear and
hence K-linear, so we can regard it as a linear map on an n-dimensional vector
space, and as such it has a trace and a determinant.
We dene the trace of , written Tr() or sometimes Tr
L/K
(), to be the
trace of this linear map, and we dene the norm of to be the determinant of
that linear map and write N() or N
L/K
().
Note that the trace and the norm of an element of L is an element of K.
Moreover Tr( +) = Tr() + Tr() and N() = N()N().
Example: Multiplication by x + iy C is, when you think of C as R
2
with basis 1, i, represented by the matrix
_
x y
y x
_
and hence has trace 2x and
determinant x
2
+y
2
. So Tr
C/R
(x +iy) = 2x and N
C/R
(x +iy) = x
2
+y
2
.
Example: if K L and L is nite over K of degree n, and K then
Tr
L/K
() = n, and N
L/K
() =
n
(proof: the matrix representing multipli-
cation by is scalar).
Because the norm N is multiplicative, it should be no surprise that it can
be used to extend norms (i.e. maps of the form [.[).
Lemma. Let K L with L nite over K, of degree n. Assume furthermore
that K is equipped with a non-archimedean norm [.[ that makes K complete.
Then there is a unique norm [[.[[ on L which restricts to [.[ on K. Its non-
archimedean, it makes L into a complete normed eld, and it is given by the
formula
[[[[ = [N
L/K
()[
1/n
Proof. Omitted. On example sheet. Elementary but a little long.
Note that the uniqueness statement needs K to be complete. For example
Q(i) is nite over Q but if A = Z[i] then in A we have (5) = (2+i)(2i) = PQ
and the P-adic norm and the Q-adic norm on Q(i) both extend the 5-adic
norm on Q. So in fact the lemma gives another proof that Q isnt complete
with respect to the 5-adic norm (and its not much trouble to deduce that its
not complete with respect to any p-adic norm this way).
Using this lemma lets deduce its analogue in the incomplete case (al-
though Im really only interested in the case of number elds). So now say
L/K is a nite extension of elds of characteristic zero (or more generally, a
nite separable extension of elds, if you know what that means). As weve just
21
seen, it is now no longer true that a non-arch norm on K extends uniquely to a
non-arch norm on L. Indeed, if L and K are number elds and if P is a prime
ideal of the algebraic integers of K, and P factors in L as Q
e
1
1
Q
e
2
2
. . . Q
e
r
r
in L,
then we have at least r norms on L extending the P-adic norm on K (namely
the appropriate powers of the Q
i
-adic norms for each i). But it turns out to
be true that in the general case there are only nitely many norms on L that
extend a given non-arch norm on K.
Lets x a norm [.[ on K. Were asking how to extend it to L. The key
construction is the following. Let

K denote the completion of K with respect to
[.[. Then L and

K both contain copies of K, so we can form the tensor product
L
K

K. I can write down what this is explicitly:
We know that L can be written as K(), for some L (that is, L is the
smallest eld containing K and ). Hence we can write L = K[X]/(P(X))
where P(X) is the minimal polynomial of , that is the monic polynomial of
smallest positive degree with coecients in K and having as a root. For
example C = R(i) = R[X]/(X
2
+ 1). Now if youre not completely certain
about the tensor product, you can simply dene L
K

K to be the ring

K[X]/(P(X)).
Now, considered as a polynomial with coecients in K, P(X) was irreducible,
and hence (P(X)) was a maximal ideal of K[X] (and thus L was a eld!).
However, P(X) might not be irreducible in

K[X]. One thing is for sure though,
and thats that P(X) has no repeated roots (because if it did then it would have a
factor in common with its derivative, contradicting the fact that its irreducible).
So, in

K[X], if P(X) factors, it will factor as Q
1
(X)Q
2
(X) . . . Q
r
(X) with the
Q
i
(X)

K[X] irreducible and pairwise coprime.
So by the Chinese Remainder Theorem we see that
L
K

K =

K[X]/(P(X))
=

K[X]/(
r

i=1
Q
i
(X))
=
r
i=1

K[X]/(Q
i
(X))
=
r
i=1

L
i
(this last line is a denition) where

L
i
=

K[X]/(Q
i
(X)) is a eld with a name
that is currently only suggestive of what is to come rather than being any
kind of completion of L. Note that theres a completely canonical natural map
L L
K

K, sending to X, and hence a map L
r
i=1

L
i
so, by projection,
maps L

L
i
for each i.
Theorem. L/K nite as above, and [.[ a norm on K. Then there are only
r extensions [[.[[
i
(1 i r) of [.[ to L, and if L
i
denotes L equipped with
the ith extension then the completion of L
i
is (after re-ordering if necessary)
isomorphic to

L
i
.
Proof. Let [[.[[ be any norm of L extending [.[ on K. Let

L denote the
completion of L with respect to this norm. Then the closure of K in

L is
22
isomorphic to

K (we saw this argument once today already). Now consider the
subeld

K() of

L. Clearly

K() is a nite extension of

K. Moreover

K()
inherits a norm from

L. So by the lemma-to-be-proved-on-the-example-sheet,

K() is complete! In particular its a closed subspace of

L that contains L and
hence it is

L. Let Q(X) denote the minimal polynomial of over

K. Then
Q(X) divides P(X) (because P() = 0) and hence Q(X) is one of the Q
i
(X)
above and

L =

K[X]/(Q(X))
=

K[X]/(Q
i
(X))
=

L
i
for some i.
Conversely, each

L
i
is visibly a nite extension of

K and hence inherits a
unique norm extending that on

K, and the inclusion L

L
i
induces a norm
on L.
All that remains is to show that distinct is induce non-equivalent norms on
L. But this is clearif the norms corresponding to two distinct is were equiv-
alent, then the completions would be isomorphic as L-algebras, but Q
i
() = 0
in

L
i
whereas Q
j
() ,= 0 in

L
i
if i ,= j.
Ill remind you that I stated earlier in the course, without proof, a theorem
saying that the only non-arch norms on a number eld k were the P-adic norms;
its not hard to use the above argument to reduce this statement to the case
of k = Q, which can be checked directly using a brute force argument due to
Ostrowski.
As a nal remark, we can now deduce that k
P
is a nite extension of Q
p
if
p P. For weve just shown that k
P
is a direct summand of the ring k
Q
Q
p
and hence the dimension of k
P
/Q
p
is at most the dimension of k/Q.
[Remark: people who want to see more of the theory of elds with norms have
two excellent choices for booksCassels Local elds, which does everything
I did here but which is also completely stued with beautiful applications of the
theory to number elds and Diophantine equations and lots of other things, and
Serres Local Fields which is more highbrow in nature and which goes much
further than Cassels, going as far as proofs of the main theorems of local Class
Field Theory.]
Chapter 3: Haar measure and abstract Fourier theory.
3.1: Introduction.
If f is a continuous function R C such that
_

[f(x)[dx converges, then

f has a Fourier transform

f : R C, dened by

f(y) =
_

f(x)e
iyx
dx.

f(y) =
_

f(x)e
iyx
dx.
In general

f bears little resemblance to f. Lets do an example to stress this:
23
lets say f(x) = 1/(1 +x
2
). Then

f(y) =
_

e
iyx
1 +x
2
dx.
We do this integral via closing up the contour and getting back from + to
via a big arc [z[ = R. We have a choice of two arcsupper half plane and
lower half planeand which one we choose turns out to depend on the sign of y.
Lets imagine closing up via the upper half plane. So y is always real, but
now were thinking of x as a complex number with big positive imaginary part.
If we want the integral along the big arc to be small then wed better make
sure that the integrand is small. So closing up along the top will work if y < 0
(because then were integrating something whose value is at most c/R
2
along
an arc whose length is O(R)).
And so, for y < 0,

f(y) =
_

e
iyx
1 +x
2
dx
= lim
D
_
D
e
iyx
1 +x
2
dx
where D is a contour that looks like a D lying on its back, and is getting bigger
and bigger. Now this integral is just going to be 2i times the sum of the
residues at the poles of e
iyx
/(1 + x
2
) for x in the upper half plane. The only
pole is at x = i, the residue is e
y
/(2i) and we deduce

f(y) = e
y
for y < 0.
A similar argument shows

f(y) = e
y
if y > 0 (now using the lower half
plane). Finally

f(0) = [tan
1
(x)]

= so we conclude

f(y) = e
|y|
.
The purpose of this was just to show that

f is of an entirely dierent nature to
f.
Summary: if f(x) = 1/(1 + x
2
) then

f(u) = e
|y|
. So in this case

f
is rapidly decreasing (this means

f(y).P(y) tends to zero as [y[ , for
any polynomial P C[X]) but not dierentiable, whereas f was innitely
dierentiable but decreasing not particularly quickly.
Two very elementary exercises about Fourier transform:
(1) If g(x) = f(x +r) (r real) then g(y) =

f(y)e
iry
.
(2) If g(x) = f(x)e
ix
then g(y) =

f(y ).
[Proof: change of variables]
This also indicates that

f is very much not like f: its transforming in a
dierent way.
But heres a nice thing: sometimes

f also has a Fourier transform (for ex-
ample the

f we just saw is certainly continuous and integrable). So we can take
24
the Fourier transform again! And (1) and (2) together imply that

f behaves in
a similar way to f (for example if g(x) = f(x +r) then

g(z) =

f(z r)).
Now lets try our toy example f(x) = 1/(1 +x
2
), so

f(y) = e
|y|
. Then

f(z) =
_

e
|y|
e
izy
dy
and this integral can be done easily because the integrand has an indenite
integral. Split the integral into
_
0

+
_

0
; the rst integral is

_
0

e
yizy
dy
= [e
y(1iz)
/(1 iz)]
0

= /(1 iz)
and the second one is /(1 + iz) so the sum is 2/(1 + z
2
) and

f is looking
remarkably similar to f. In fact, for this f, we have

f(x) = 2f(x) = 2f(x)

because f was even.
But the general theorem is that if f is now an arbitrary function which
is, say, innitely dierentiable and rapidly decreasing (much weaker conditions
will do, but these will suce for us), then

f is also innitely dierentiable and
rapidly decreasing, so

f makes sense and

Theorem (Fourier Inversion Theorem)

f(x) = 2f(x).
Now not only will I freely confess that I have no idea (yet) how to prove the
above statement, but also, more importantly, before I had read Tates thesis, I
would never have believed that there would or could be some abstract version
of this theorem which would, say, work over the p-adic numbers (what would
play the role of 2, for example??
So now I know better. In fact the Fourier transform should be thought of as
some sort of duality sending functions on one R to functions on a dual R,
and the Fourier inversion theorem is some form of the statement that the dual
of the dual is the function you started with (up to some fudge factors).
A good analogy is with nite abelian groups G. Say G is nite abelian,
and let

G be the set of (1-dimensional) characters of G. Then

G is a group
non-canonically isomorphic to G. Now for f : G C, dene

f :

G C by

f() =
1
[G[

gG
f(g)(g).
Exercise: if

G is identied with G in the following STRANGE way: let g G

dene a group homomorphism

G C

by sending to (g
1
) [NOT (g)],
then

f(g) = (1/[G[).f(g).
25
[

f(g) = (1/[G[).f(g).]
Note the minus sign in the Fourier inversion theorem corresponds to the
strange identication of G with its double dual, and the fudge factor 2 corre-
sponds to the fudge factor 1/[G[. The analogy in fact is more than an analogy
our goal in this chapter is to formulate and prove an abstract Fourier inversion
theorem and both the above things will be special cases. We need to start by
coming up with an integration theory that works in much more generality than
blah Riemann/Lesbesgue integration. Before we do that, I need to introduce the
objects well be integrating on: locally compact Hausdor topological groups.
3.2: Locally compact Hausdor topological groups.
So on the real numbers we have the Riemann Integral. Im going to explain
in this lecture and the next a far more general integration theory that will work
on an arbitrary locally compact Hausdor topological group. So I have to start
by explaining what a locally compact Hausdor topological group is.
A topological group is a group G equipped with a topology on G such that
m : G G G and i : G G dened by m(x, y) = xy and i(x) = x
1
, are
continuous (where G G is equipped with the product topology). Examples:
any group, with the discrete topology. The real numbers with its usual topology.
The non-zero real numbers with its usual topology. If K is any normed eld
then K with the topology coming from the norm.
[One might ask whether continuity of m implies continuity of i. It doesnt:
for example if G is the integers with the order topology (so the open sets are the
empty set, the whole thing, and all sets of the form n, n + 1, n + 2, n + 3, . . .
then multiplication is continuous but inverse isnt).]
Heres a slightly more subtle example: if K is a normed eld, then K

,
with the topology induced from K, is a topological group. The reason one
has to be careful here is that one has to check that inverse is continuous
but it is (exercise), because the topology is coming from a metric. Ill perhaps
make the cryptic remark that if R is an arbitrary topological ring (so + and
and are continuous) then its unit group, with the induced topology, is not
always a topological group, because inverse really might not be continuous in this
generality; this can however be xed by embedding R

into R
2
via u (u, u
1
),
and giving it the subspace topologythen R

really is a topological group.

Pedantic exercise: if K is a normed eld, then check that the two topologies
Ive just put on K

(the subspace topology coming from K, and the one coming

from K
2
) coincide.
Back to examples: If K is a normed eld and G is an algebraic group over
K (for example GL
n
or Sp
n
or something) then G(K) is a topological group (so
for example GL
n
(R) and GL
n
(Q
p
) are topological groups, or E(Q
p
) for E/Q
p
an elliptic curve, and so on). I wont prove these things because we dont need
them, but theyre not hard.
If g G then the map G G G sending h G to (g, h) is continuous
(think about the denition of the product topology, or the universal property)
and hence if we x g G then left multiplication by g, the function G
G sending h to gh, is continuous, and similarly right multiplication by g is
26
continuousand even homeomorphisms, because left/right multiplication by
g
1
is a continuous inverse. In particular G is blah homogeneousif a, b G
then left multiplication by ba
1
is a homeomorphism G G sending a to b. In
words, G looks the same at a and at b; the group of homeomorphisms of G
acts transitively on G, if you prefer.
A remark on non-Hausdor groups: It turns out that if G is a topological
group which isnt Hausdor, and if e is the identity element, then e isnt
a closed set, and its closure H is a normal subgroup of G such that G/H is
naturally a Hausdor topological group. Using this argument, questions about
topological groups can frequently be reduced to the Hausdor case, and well
only be concerned with Hausdor groups in practice anywayfor example, all
the examples we saw above were Hausdor; moreover topologies will usually
come from metrics and hence will automatically be Hausdor.
So lets get on. Let G be a hausdor topological group. We want to integrate
a class of continuous functions G C. Which ones? Well, probably not all of
themfor example if G = R then f(x) = 1 for all x wont be integrable. So
lets restrict, at least for the time being, to continuous functions which vanish
outside a compact setthis is a good niteness condition. Unfortunately, in this
generality, there might not be any such things! For example if G = Q with its
subspace topology coming from R, a continuous function G C which vanishes
outside a compact set must be identically zero (exercise). This is unsurprising
who would do integration on Q?? Heres a nice condition which will at least
ensure the existence of lots of functions which vanish outside a compact set:
Denition. If X is a topological space and x X, we say that U X is
an open neighbourhood of x X if U is open and x U. We say S X is a
neighbourhood of X if x is in the interior of S. We say that X is locally compact
if every x X has a compact neighbourhood.
For the rest of this chapter, we are only interested in locally compact Haus-
dor topological groups. Lets call them LCHTGs.
Note that to check that a topological group is locally compact, it suces to
nd a compact neighbourhood of the identity (by homogeneity).
Good examples of LCHTGS: if K is either the real numbers, or the complex
numbers, or k
P
for k a number eld, then K (considered as a group under
addition) is a LCHTG; for the reals and the complexes a compact neighbourhood
of the identity is the closed unit ball, and for k
P
the ring of integers will work,
once we remember that the residue eld is nite in this setting. Moreover, I
claim that K

(with the subspace topology induced from that of K) is also a

LCHTGif K = k
P
then this follows because 1+R is a compact neighbourhood
of 1, and in the archimedean case consider the closed ball centre 1 radius 1/2.
Now if X is a Hausdor topological space then for f : X C continuous,
its easily checked that the following are equivalent:
(1) f vanishes outside a compact set (that is, theres some compact K X
such that f(x) = 0 if x , K), and
(2) the support of f is compact
[recall that the support of a function f : X C is just the closure of the
set x X : f(x) ,= 0].
So for a LCHTG G, lets dene /(G) to be the continuous functions G R
27
with compact support. The point is that this space is very rich (and in particular
non-zero!). Well see this in a second, using Urysohns Lemma. First let me
remark that wed really like to integrate more functions than those in /(G), but
/(G) is a good start and turns out to be where the work is; well expand our
horizons later by taking limits. First Ill show /(G) is big. We need a lemma
in order to prove Urysohns Lemma.
First let me remind you that if X is compact Hausdor and C U X
with C compact and U open, then we can nd V open and D compact with
C V D U. For C and K := XU are compact and disjoint, and by
Hausdorness its easy (and fun!) to nd C V and K W with V and W
open and disjoint.
Denition, for convenience: S T X are subsets of a topological space,
we say that T is a neighbourhood of S if T is a neighbourhood of s for all s S.
Lemma. If X is locally compact and Hausdor and C X is compact,
then every neighbourhood of C contains a compact neighbourhood of C.
Proof. It suces to check that if C U X with C compact and U open,
then there exists V open and D compact with C V D U. Heres the
idea; each c C has a compact neighbourhood N
c
, with interior M
c
. Now C
is covered by nitely many of the M
c
, and the union X

of the corresponding
N
c
is compact and a neighbourhood of C. Replacing X with X

and U with
U

= X

U reduces us to the case where X is compact, which we just did.

Lemma (Urysohns Lemma): if X is a locally compact Hausdor topo-
logical space, and C U X with C compact and U open, then theres a
continuous function f : X R such that
(1) f(x) = 1 for x C
(2) f(x) = 0 for x , U
(3) 0 f(x) 1 for all x X
(4) Supp(f) is contained in U and is compact.
Proof.
The proof is sort of constructive (but does involve innitely many choices).
We apply the previous lemma innitely often, basically. Set C(1) = C and dene
C(0) to be any compact neighbourhood of C in U.
[C = C(1); C(0) a compact neighbourhood of C(1)]
Now by the previous lemma (applied to C and the interior of C(0)) we get
a compact C(1/2) such that C(0) is a neighbourhood of C(1/2) and C(1/2)
is a neighbourhood of C(1). Applying this trick again we get C(1/4) between
C(1/2) and C(0), and C(3/4) between C(1) and C(1/2). Continuing in this way,
we construct compacts C(i/2
n
) for all n 1 and positive odd i with 0 < i < 2
n
,
such that C() is in the interior of C() for < .
Now heres the magic: for 0 r 1 an arbitrary real, dene C(r) =

sr,s=i/2
nC(s). Now we have a decreasing sequence of compact sets; lets
nish the job by dening C(r) = for r > 1 and C(r) = X for r < 0.
We dene f : X R by letting f(x) be the supremum of the with
x C(). This sup visibly exists and is at most 1, the support of f is within
C(0), and indeed all properties we require of f are easy to check, the one with
the most work being continuity, which goes like this: f(x) < if and only if
28
x ,
<
C() and the intersection is closed so the complement is open, and
f(x) > if and only if x
>
Int(C()) (with Int meaning interior), which is
also open, and so the x with < f(x) < are an open set, and thats enough.
Corollary If G is a LCHTG then /(G) separates points and in particular
is non-zero.
3.3: Haar Measure/integral on a LCHTG.
As people probably realise, Im preparing these lectures on the y, but I
might actually have to number lemmas in this section because the results are
elementary but sometimes slightly tricky. Ill sometimes be sketchy with the
easier proofs however, for the following reason: the only groups for which well
actually need Haar integrals/measures are: (i) the reals and complexes (where
the Haar integral is just the Lesbesgue integral), (ii) the p-adics and nite
extensions thereof (where you can dene the measure by hand), and (iii) the
adeles (well get to these) (where you can dene Haar measure as just a product
of things in (i) and (ii)). My main motivation for going through this stu really
is that its the kind of thing I wish I had been taught as a graduate student.
Let G be a LCHTG. Weve seen that /(G) is non-zero, and moreover if we
dene /
+
(G) to be the f : G R in /(G) such that f(x) 0 for all x and
f(x) > 0 for some x (that is, f isnt identically zero), then weve also see that
/
+
(G) isnt zero either. The reason were sticking with the reals rather than
the complexes is that were after some kind of integral
_
G
which will take
an element of /(G) to a real number and is guaranteed to take an element of
/
+
(G) to a non-negative real number; if we worked with the complexes then
we couldnt enforce this sort of positivity condition so easily.
Its easy to dene what a Haar integral isthe hard part is existence and
uniqueness. If f : G C and x G then dene f
x
: G C by
f
x
(g) = f(gx
1
).
So its just f composed with right multiplication by x
1
(dont read anything
signicant into the x
1
; its easier to T
E
X f
x
than
x
f). Note that f /(G)
implies f
x
/(G) and similarly for /
+
(G).
Denition. A Haar Integral on G is a non-zero R-linear map
: /(G) R
such that
(1) (f) 0 for f /
+
(G)
(2) (f) = (f
x
)
Idea: think (f) =
_
G
f.
Remark: as Ive already mentioned, ideally one would like to integrate more
functions than just those with compact support, but these will come later on
without too much trouble. The hard work is all in
Theorem. If G is a LCHTG then a Haar integral exists on G, and further-
more if
1
and
2
are two Haar integrals, then theres some c > 0 such that
c
1
(f) =
2
(f) for all f /(G).
As I mentioned before, we will principally be interested in the case G = K
or K

for K a completion of a number eld; in the archimedean case we have

29
K isomorphic to R or C and we can use Riemann integration or Lesbesgue
integration to produce a Haar measure (dont forget that its dx/x in the K

case to make it invariant under multiplication), and in the non-arch case one can
check that /(G) is generated by step functions so (by linearity and translation-
invariance) we only need to dene the measure of the characteristic function of

n
R, which can be q
n
, and that does existence in all the cases we need. So
in some sense this lecture and the next are not really logically necessary. So I
might race through the elementary-undergraduate-exercise parts of some proofs
a bit, although I will stress all of the key ideas.
In fact the proof contains several ideas. If you think about how the Riemann
integral works, we rst integrate rectangular functions (like the characteristic
function of an interval) and then bound more general functions above and below
by rectangular functions. The problem at this level is that such a rectangular
function might not be continuous. Moreover, G is only a group, not a eld, so
we cant yet say things like this open set is twice as big as this one. We x this
by, instead of using rectangular functions, setting up an approximate theory
using an arbitrary element of /
+
(G)indeed our rst goal is, for F /
+
(G),
to dene an approximate integral
F
.
Notation: for f, g : G R we say f g if f(x) g(x) for all x G, and
we say f < g if f g and f ,= g. So, for example, /
+
(G) is the f /(G) with
f > 0.
Lemma 1. Say f, F /
+
(G). Then there exist real numbers
1
, . . . ,
n
0
and x
1
, x
2
, . . . x
n
G such that

1in

i
F
x
i
f.
Hence if is a Haar integral, we have (f) (

i
)(F).
[think: what does this lemma mean?]
Proof. We know F(t) = r > 0 for some t G, so by continuity theres some
open neighbourhood U of e G such that F(ut) > r/2 for all u U. Now
the support of f is covered by translates Uh of U and hence by nitely many
translates; this gives the x
i
(if Uh is in the cover then one of the x
i
will be
t
1
h). Finally note that f /(G) so f is bounded, say by M; now we can
just let all the
i
be M/(r/2). The nal statement follows immediately from
positivity, linearity and translation-invariance.

[(f) (

i
)(F)]
As a consequence, which will guide us later, we deduce that for any F
/
+
(G) and for any Haar integral we have (F) > 0 [or else taking F with
(F) = 0 shows that is identically zero on /
+
(G); but for f /(G) we
have 2f = ([f[ + f) ([f[ f) and both bracketed terms on the right are in
/
+
(G) 0, so is identically zero, contradiction].
If we pretend that F is one of those rectangle functions then this motivates
the following denition: for f, F /
+
(G) we set (f : F) to be the inf of the

i
over all the possibilities for
i
in the lemma, that is, the inf over all the
possible ways of choosing
i
and x
i
with f

i
F
x
i
.
30
Exercise: prove (f : h) (f : g)(g : h) by observing that if f

i

i
g
x
i
and g

j
h
y
j
then f

i,j

i

j
h
y
j
x
i
.
Now (f : F) looks like a good candidate for the integral of f, at least if the
support of F is small, but in fact as well as rounding errors caused by F not
being ne enough, theres a normalisation issue: if we replace F by 2F, say,
we see (f : 2F) = 2(f : F). So if we want to dene the integral of f as some
kind of limit of the (f : F) as, say, the support of F tends to zero, we need
to scale things.
So heres a crucial remark that shows that scaling is possible. Let f, F be as
above (both in /
+
(G)). We dened (f : F) to be the inf of the

i
such that

i
F
x
i
f. Now any function in /
+
(G) has a positive supremum, which
it attains. Furthermore

i

i
F
x
i
f implies that (

i
) sup(F) sup(f)
which gives us a lower bound

i

i
(sup(f)/ sup(F)). Hence blah blah blah
blah (f : F) sup(f)/ sup(F) > 0 and weve shown that (f : F) > 0 for all
f, F /
+
(G).
So heres the next good idea:
FIX ONCE AND FOR ALL A FUNCTION /
+
(G) (it doesnt matter
what it is). We know that if a Haar integral exists, it will take to something
positive. We want to dene approximate Haar integrals and tease the ex-
istence of a Haar integral from these approximate ones. For the approximate
Haar integrals to be compatible we will simply force each of them to integrate
to 1.
Denition. For f, F /
+
(G) dene

F
(f) := (f : F)/( : F).
The idea:
F
might not be a Haar integral but its a good rst approximation.
Well see that in fact as the support of F gets smaller the
F
become better
and better approximations to a Haar integral.
Exercise: Use the previous exercise to check that
F
(f) (f : ).
Now this denition of
F
is great because its impervious to linear changes
of F. In fact it almost does the job already, at least for functions in /
+
(G):
F
is positive on /
+
(G), its translation-invariant, and satises
F
(f) =
F
(f)
for > 0. Its normalised in the sense that
F
() = 1. Unfortunately its not
additive; its trivial to check that
F
(f
1
+f
2
)
F
(f
1
) +
F
(f
2
) (easy exercise)
but theres no reason why equality should hold (and it wont, in general).
[
F
(f) := (f : F)/( : F) and
F
(f
1
+f
2
)
F
(f
1
) +
F
(f
2
)]
We need more than subadditivity, we need an approximate additivity,
which is given by
Lemma 2. Let G be a LCHTG. Say f
1
, f
2
/
+
(G). Say > 0. Then
theres a symmetric open neighbourhood V (that is, V = v
1
: v V ) of the
identity in G (depending on f
1
and f
2
and ) such that, for any F /
+
(G)
with support in V , we have

F
(f
1
+f
2
)
F
(f
1
) +
F
(f
2
) .
Proof. Let C be the union of the supports of f
1
and f
2
. Choose (Urysohn)
q /
+
(G) with q(x) = 1 for x C. Choose some tiny > 0 (well say how tiny
31
laterwe could take a vote on this issue if the audience is suciently oended
by this idea though); Ill tell you now that < 1 though.
Set p = f
1
+f
2
+q, so p(x) for x C. The key construction is to dene
(for i = 1, 2) the functions
h
i
(x) = f
i
(x)/p(x) (if x C)
= 0 (if x , C)
[before you askthere were overow vbox issues]
[p = f
1
+f
2
+q
h
i
(x) = f
i
(x)/p(x) (if x C)
= 0 (if x , C)
]
We need to ensure that the h
i
are continuous! The sum of h
1
and h
2
is
approximately the characteristic function of C. One checks easily that h
i

/
+
(G) (the support of h
i
is closed in C) and 0 h
1
+h
2
1. Now continuous
with compact support implies uniformly continuous, by the usual argument, so
we can let W be a suciently small open neighbourhood of the identity such
that [h
i
(x) h
i
(y)[ < /2 whenever xy
1
or x
1
y W. By replacing W with
W w
1
: w W we may even ensure that W = W
1
. Note that V will be
our W when weve decided upon a .
So now choose any F /
+
(G) with support in W. By Lemma 1 we can
nd
j
and x
j
with p

j

j
F
x
j
.
[p

j

j
F
x
j
]
Weve just bounded p above by translates of F, and now we can bound the
f
i
above by translates of F too. First note that F
x
j
(t) = 0 if t , Wx
j
, and for
t Wx
j
we have [h
i
(x
j
) h
i
(t)[ < /2 for 1 i 2. So in either case we have
h
i
(t)F
x
j
(t) (h
i
(x
j
) +/2)F
x
j
(t).
Hence
f
i
= ph
i

j
h
i
F
x
j

j
(h
i
(x
j
) +/2)F
x
j
.
Hence, by denition,
(f
i
: F)

j
(h
i
(x
j
) +/2).
And, because h
1
(x
j
) +h
2
(x
j
) 1, we deduce
(f
1
: F) + (f
2
: F)

j
(1 +).
32
(f
1
: F) + (f
2
: F)

j
(1 +).
Now the last thing we chose were the
j
and x
j
with p

j

j
F
x
j
, so by
letting these vary we deduce from the previous equation
(f
1
: F) + (f
2
: F) (1 +)(p : F)
and hence (dividing by ( : F))

F
(f
1
) +
F
(f
2
) (1 +)
F
(p)
for any F /
+
(G) with support in W. You can now presumably see that were
on the right track, because p is approximately f
1
+f
2
.
In fact p = f
1
+f
2
+q, and we deduce (from subadditivity of
F
) that

F
(f
1
) +
F
(f
2
) (1 +)(
F
(f
1
+f
2
) +
F
(q))

F
(f
1
+f
2
) +.R,
with R =
F
(f
1
+f
2
) + 2
F
(q).
[
F
(f
1
) +
F
(f
2
)
F
(f
1
+f
2
) +.R, with R =
F
(f
1
+f
2
) + 2
F
(q).]
Now unfortunately R depends on F which depends on W which depends on
, but fortunately you all did the exercise earlier which showed
F
(f) (f : ),
and hence R (f
1
+ f
2
: ) + 2(q : ), which were all chosen before . So now
choose such that ((f
1
+f
2
: ) +2(q : )) < , let V denote the corresponding
open set W, and were home.

Corollary 3. Given f
1
, f
2
, . . . , f
n
/
+
(G) and > 0 there exists a symmet-
ric open neighbourhood V of the identity in G such that whenever F /
+
(G)
has support in V , we have

F
(

i
f
i
)

F
(f
i
) .
Proof. Induction.
In the last lecture we proved some lemmas and in this lecture we need one
or two more, but we also need to actually prove existence and uniqueness of the
Haar integral. There are several ways to do this; Ill use a method that I found in
P. J. Higgins book An introduction to topological groups because in my view
the uniqueness proof is the least painful out of all the references Ive seen (its
still pretty painful though :-( ). Were going to deduce existence and uniqueness
of Haar integrals from some Zorns Lemma argument applied within the real
vector space /(G) of continuous functions with compact support. Heres the
abstract linear algebra well need to pull this o.
Let V denote a real vector space. A non-empty subset E of V is called
convex if v, w E and 0 1 implies v + (1 )w E. A subset C of
V is called a cone if c C and > 0 implies c C. One checks that for E
non-empty, E is a convex cone i
(i) v E and > 0 implies v E
33
(ii) v, w E implies v +w E.
(examples: (r, 0) R
2
with r > 0 or r 0). Finally, we say that a convex
cone E is open if w E and f V implies that theres some > 0 such that
w + f E for all with [[ < . So the (r, 0) examples above arent open,
but (r, s) with r, s > 0 would be. Note that the complement of a cone is a cone,
but the complement of a convex cone might not be convex.
A subspace H of V is called a hyperplane if V/H is 1-dimensional.
[convex cone: E = E for > 0, and E +E E]
Note that a Haar integral is a non-zero R-linear map /(G) R and is
hence, up a non-zero constant, determined by its kernel (the functions whose
integral is zero). The kernel will be a hyperplane in /(G) and our existence and
uniqueness proofs of Haar integrals will be done via existence and uniqueness
of hyperplanes with certain properties.
Proposition. (Haar integral machine) Say V is a real vector space, E is
an open convex cone in V , and W is a subspace of V that doesnt meet E.
(i) Theres a hyperplane H W such that H E is empty.
(ii) If furthermore V E (the complement of E) is convex, then H is unique.
Proof. This is elementary, unsurprisingly.
(i) By Zorns Lemma one can choose a maximal subspace H containing W
and missing E and the claim is that its a hyperplane. Set D = H + E. Its
easy to check that D is an open convex cone (H and E are convex cones, and
E is open). By assumption H E is empty, so H D is also empty.
First I claim that V is the disjoint union of H, D and D. Disjointness is
trivial (if D D was non-empty then use convexity to show 0 D which is
false). The fact that V is the union of H, D and D follows from maximality:
if v V with v , H then H+Rv intersects E and hence Rv meets H+E = D,
but 0 , D so v D for some choice of sign.
Now I claim that H is a hyperplane. Note that H ,= V because E is non-
empty, so V/H has dimension at least 1. Say v, w V generate a 2-dimensional
subspace of V/H and lets get a contradiction. Well, w , H so (after changing
sign if necessary) we may assume w D. Similarly (after changing sign of v
if necessary) we may assume v D.
Now consider the line joining v to w; think about it as the image of [0, 1].
Because D is an open convex cone, one checks easily that the intersection of
D with this line is an open interval containing 0 but not 1. Similarly the
intersection of the line with D is a open interval containing 1 but not 0. But
D and D are disjoint, and two disjoint open intervals cant cover a line, so we
have v +(1)w H for some 0 < < 1 and theres a linear relation in V/H
between v and w, the contradiction we seek.
(ii) Let E

denote the complement of E in V ; then E

is assumed convex
and is hence a convex cone. Now 0 , E so E (E) is empty; let X be the
complement of E (E). Now X = (E

) (E

) so its a convex cone, and

X = X, so X is a vector subspace of V . The argument from (i) (with X
replacing H and E replacing D) shows that X is a hyperplane; moreover any
subspace of V disjoint from E will be contained in X, so were home.
The application is of course the following. Let G be a LCHTG, set V =
34
/(G), let W be the subspace of V spanned by all functions of the form f f
x
for f /(G) (so everything in L should integrate to zero), set E = /
+
(G) +W
(think of E as everything for which the axioms imply that the integral should
be positive). I claim that the hypotheses of the Haar integral machine are
satised. Lets check these in a second, but lets rst observe that if they are,
then (i) gives us a Haar integral, and (ii) (if it applies, that is, if E

is convex)
gives us uniqueness up to a positive scalar. For (i) gives us a hyperplane H; let
denote any R-linear isomorphism V/H R. Then clearly is linear and
translation-invariant; furthermore if f, g /
+
(G) then the line from f to g lies
within /
+
(G) so doesnt meet H, and hence (f) and (g) have the same sign,
so either or is a Haar measure. Conversely any kernel of a Haar integral
will contain W and be disjoint from E, so if (ii) applies then theres only one
possibility for the kernel.
[W spanned by f f
x
; E = /
+
(G) +W]
So what is left to do? For existence of a Haar integral, we just need to check
the hypotheses of the proposition (that is, that W E is empty and that E is
an open convex cone; weve done the work to prove these easily though). For
uniqueness up to positive scalar we need to check that the complement of E is
convex (we need another lemma to do this). Lets do existence rst because it
actually helps with uniqueness.
Existence of Haar integral.
To check W E is empty we just have to check W /
+
(G) is empty. This
isnt a surprising result, because everything in W should integrate to zero, and
nothing in /
+
(G) should. But lets give the proof. Now W is generated by
things of the form f f
x
; furthermore using the 2f = ([f[ +f) ([f[ f) trick
we can check that W is generated by things of the form f f
x
for f /
+
(G).
So if W /
+
(G) is nonempty then we can nd f, f
i
/
+
(G) and x
i
G with
f =

i
(f
i
f
x
i
i
).
We rewrite as
f +

i
f
x
i
i
=

i
f
i
and for any > 0 we use Corollary 3 to nd an open neighbourhood V of the
identity such that for any F /
+
(G) with support in V , we have

F
(f +

i
f
x
i
i
)
F
(f) +

F
(f
x
i
i
) .
Now we easily get a contradiction, for choosing F as above (Urysohn), we
35
see

F
(f
i
)
F
(

i
f
i
)
=
F
(f +

i
f
x
i
i
)

F
(f) +

F
(f
x
i
i
)
( : f)
1
+

F
(f
i
)
so if we had chosen with 0 < < ( : f)
1
then we get a contradiction.
All thats left for existence is the check that /
+
(G) +W is an open convex
cone. Its clearly a convex cone; the issue is openness. If f = p + q is an
arbitrary element of /
+
(G) + W, and k /(G) is arbitrary, we need to show
f k /
+
(G) +W for 0 < small. Heres how. By Lemma 1 we can bound
[k[ above by

i

i
p
x
i
and WLOG not all of the
i
are zero. So
f k p +q

i
p
x
i
= p +q

i
p

i
(p
x
i
p)
= p(1

i
) +q

with q

W, so f k /
+
(G) +W if 0 < < (

i
)
1
, and we have proved
the existence of the Haar integral.
For uniqueness we need to show that (with the above notation) E

is convex.
The reason we dont yet have enough is that we have only approximated a
function from abovewe now really need to approximate a function in /(G)
uniformly across G. To do this we need the a standard application of the bump
functions that Urysohns lemma gives us.
Lemma 4. If G is a LCHTG and f /
+
(G) and W is any neighbourhood
of the identity in G, then one can nd x
1
, x
2
, . . . , x
n
all in the support of f and
f
1
, f
2
, . . . , f
n
/
+
(G) with the support of f
i
in Wx
i
, and

i
f
i
= f.
Proof. This is easy. First choose a compact neighbourhood N of the identity
in W, with interior U. Now the support of f is compact so its covered by nitely
many Ux
i
, 1 i n, with x
i
all in the support of f. By Urysohn, there exists
h
i
/
+
(G) which is identically 1 on Nx
i
and whose support is contained within
Wx
i
. Now set h =

i
h
i
and f
i
(x) = f(x)h
i
(x)/h(x) for x in the support of f,
and f
i
(x) = 0 otherwise. Its an easy check that this works.
Say F : G C is symmetric if F(x) = F(x
1
) for all x. The following
lemma is the last piece of the puzzle.
Lemma 5. (Uniform approximation) If f /
+
(G) and > 0, then there
exists some neighbourhood V of the identity in G such that for every symmetric
F /
+
(G) with support contained in V there are real numbers
1
,
2
, . . . ,
n

36
0 and x
1
, x
2
, . . . , x
n
G such that
[f(x)

i
F
x
i
(x)[ <
for all x G.
Proof. By uniform continuity we can choose a neighbourhood V of the
identity such that [f(x) f(y)[ < /2 whenever y V x and this V is going
to work. Say F /
+
(G) is symmetric with support in V . Then of course the
support of F
x
is within V x, and one deduces easily that [f(x) f(y)[F
x
(y)

2
F
x
(y) for all x, y G, so by denition we have (as functions of y)
[f(x)F
x
f.F
x
[

2
F
x
.
[f(x)F
x
f.F
x
[

2
F
x
(1)
(as functions of y). Now for any > 0, by uniform continuity of F, we can nd
some neighbourhood W of the identity such that
[F(y) F(z)[ <
for all y Wz, and hence, for any x G, [F
x
(y) F
x
(z)[ < for all y Wz.
Now by (bump function) Lemma 4, applied to f and W, we write f =

i
f
i
with f
i
/
+
(G) and the support of f
i
in Wx
i
. The same trick as above gives
us
f
i
(y)[F
x
(y) F
x
(x
i
)[ f
i
(y)
for all x, y G (check separately for y Wx
i
and y , Wx
i
).
A labour-saving observation now is that F is symmetric so F
x
(x
i
) = F
x
i
(x),
and summing the last equation over i we get
[f(y)F
x
(y)

i
f
i
(y)F
x
i
(x)[ f(y).
This latter equation is true for all x, y G, and hence (by denition)
[f.F
x

i
F
x
i
(x)f
i
[ f. (2)
Recall now equation 1:
[f(x)F
x
f.F
x
[

2
F
x
(1)
and we get
[f(x)F
x

i
F
x
i
(x)f
i
[ /2F
x
+f (3)
(an inequality of functions of y) for all x. What we did here was used uniform
continuity of f and uniform continuity of F
x
to get two good approximations
for f.F
x
, and we reaped the consequences.
[ [f(x)F
x

i
F
x
i
(x)f
i
[ /2F
x
+f (3) ]
A painless way to nish now is to assume the existence of a Haar integral! We
have already proved this so its OK. Apply a Haar integral to this last equation
37
(observing that if /(G) then [[ and [[, so [()[ ([[)) and
deduce
[f(x)(F)

i
F
x
i
(x)(f
i
)[ /2(F) +(f).
Its an easy check that (f) (f : F)(F) [look at the denition of (f : F) and
apply ] and we deduce
[f(x)(F)

i
F
x
i
(x)(f
i
)[ (/2 +(f : F))(F).
Now divide by (F), set
i
= (f
i
)/(F), let be /(3(f : F)) and we get
[f(x)

i
F
x
i
(x)[ <
and we have won.
Corollary 6. Set E = /
+
(G) + W as before. Then, for any f /(G),
there exists some h /
+
(G) such that for every > 0, either f + h E or
f h E.
Proof. Let C be the support of f; let D be a compact neighbourhood of C.
By Urysohn theres h /
+
(G) with h(x) > 2 for x D. This h will work. For
we can write f = f
1
f
2
with f
i
/
+
(G), and by the previous Lemma (uniform
approximation) both f
1
and f
2
can be uniformly approximated by scalings of
translates of any symmetric function with support in some V , which is WLOG
symmetric and satises V C D.
The trick now is if F
0
/
+
(G) has support in V then F(x) = F
0
(x) +
F
0
(x
1
) is symmetric with support in V , and we can uniformly approximate f
1
and f
2
using F, and hence we can nd
1
, . . . ,
n
Rwith [f(x)

i
F
x
i
(x)[ <
2 for all x G.
We have rigged it so that f and F
x
i
all have support in D, so we can conclude
that
[f

i
F
x
i
[ < h.
Now if =

i
then k := F

i
F
x
i
W and
f h < F k < f +h.
But this implies what we want: if 0 then weve shown f +h > k W so
f + h /
+
(G) + W, and if 0 then weve shown f h < k and hence
f h E.
Uniqueness of Haar integrals.
As usual W is generated by f f
x
, E = /
+
(G) +W and all we need to do
is to prove that the complement of E in V = /(G) is convex. The complement
is certainly a cone, so we need to show its a convex cone, so we need to check
that if f
1
, f
2
E

(the complement of E) and f

1
+ f
2
E then we have a
contradiction. This is now easy. Write f = f
1
f
2
and apply the previous
corollary to deduce that theres some h /
+
(G) such that for all > 0 either
f + h E or f h E. But E is open and f
1
+ f
2
E, so theres some
38
> 0 such that f
1
+f
2
h E. If f +h E then 2f
1
E, and if f h E
then 2f
2
E, and either one is a contradiction.
We now consider consequences of the existence and uniqueness of Haar inte-
grals, and extend our range of denition somewhat. We established the existence
of a Haar integral which, by denition, was impervious to right translations.
These are sometimes called right Haar integrals. We could instead demand
that the integral of f /(G) was equal to the integral of the function g f(xg)
[invariance under left translation]. Such a gadget would then be called a left
Haar integral. But these exist and are unique too:
Theorem. Left Haar integrals exist and are unique up to a positive con-
stant.
Proof. If f /(G) then the function

f : x f(x
1
) is also in /(G) and
if we dene
L
(f) = (

f) then
L
is a left Haar integral i is a right Haar
integral.

Note that the left Haar integral might not be (a positive constant times)
the right Haar integral! The moral reason for this is that its not hard to nd
a LCHTG G with a subgroup H and g G with gHg
1
a proper subset of
H. If there were a left and right invariant Haar integral on G then a good
approximation to the characteristic function of H would have the same measure
as a good approximation to the characteristic function of gHg
1
which cant
happen because gHg
1
is strictly smaller than H.
Exercise: Let G be the matrices
_
a b
0 1
_
in GL
2
(R). Conjugating by g :=
_
2 0
0 1
_
sends
_
a b
0 1
_
to
_
a 2b
0 1
_
so if f is any continuous function on R with compact
support and which is increasing on (, 0), has f(0) > 0, and is decreasing
on (0, ), then the function F on G dened by F(
_
e
t
b
0 1
_
) = f(t)f(b) (and
F = 0 if a < 0) is continuous with compact support and satises x F(x)
F(gxg
1
) /
+
(G), which is enough to show that no bi-invariant measure can
exist ((F) > (gFg
1
) for any right Haar measure).
If is a (right) Haar integral on G then lets write
_
G
f(x)d(x)
for (f); its a more suggestive notation. Then we have the following result:
Fubinis Theorem. If G and H are LCHTGs with Haar integrals and ,
and f /(GH) then
_
G
__
H
f(x, y)d(y)
_
d(x)
and
_
H
__
G
f(x, y)d(x)
_
d(y)
exist, are equal, and are both right Haar integrals on GH.
Proof. Existence is easy. First, f has compact support so has support within
a compact set of the form C D (projection of a compact set is compact), so
certainly the inner integral
_
H
f(x, y)d(y) exists and, as a function of x, has
compact support.
39
We need to check
_
H
f(x, y)d(y) is continuous as a function of x but this
isnt hard (big hint: if f is supported within C D and we choose k /
+
(H)
which equals 1 on D then by uniform continuity we have that x close to x

implies [f(x, y) f(x

, y)[ k(y) so changing x to x

changes the integral by

at most (y) which can be made arbitrarily small). It follows easily that both
integrals are Haar integrals. To check that theyre the same, it suces to check
that they agree on one positive function! So check it for f(x, y) = p(x)q(y) with
p /
+
(G) and q /
+
(H), where both integrals are just (p)(q) ,= 0.

Finally let me talk about extending our range of integrable functions. This
is rather formal, really.
Let G be a LCHTG and let U denote all the functions f : G R +
which are pointwise limits of increasing sequences f
1
f
2
f
3
. . . with
f
n
/(G). If f U then one can check that (f) := lim
n
(f
n
) R +
is well-dened and independent of the choice of f
n
. Set U = f : f U,
dene on U by (f) = (f) R .
Denition. A function f : G R is summable if there exists
g U and h U with g f h and, crucially,
sup
gf,gU
(g) = inf
hf,hU
(h).
The common value is dened to be (f) R (note: it cant be innite).
Note that a summable function certainly doesnt have to be continuous.
Exercise: if G = R then check that the characteristic function of [0, 1] is
summable, and has integral equal to 1 (if the Haar measure is normalised in
the usual way). Similarly check that the characteristic function of a point is
summable and has integral equal to zero.
If L
1
(G) denotes all the summable functions, then theres a natural norm
on L
1
(G), dened by [[f[[ = ([f[) (one can check [f[ L
1
(G)). Unfortu-
nately there are plenty of functions in L
1
(G) with [[f[[ = 0 (for example the
characteristic function of a point, if G = R). Say a function f is null if [[f[[ = 0.
Denition. L
1
(G) is dened to be L
1
(G) modulo the null functions.
One can check that L
1
(G) is in fact a real Banach space. In fact more
generally, if 1 p < one can dene L
p
(G) to be the functions f : G
R such that [f[
p
is summable, one can dene a norm on L
p
(G) by
[[f[[
p
= ([f[
p
)
1/p
and then let L
p
(G) be the quotient of L
p
(G) by the subspace
of f with [[f[[
p
= 0. It turns out that these are all Banach spaces (this needs
a little proof), which are absolutely fundamental to the further development of
the theory. Note also that L
2
(G) is a real Hilbert space, because one can make
sense of
f, g) =
_
G
f(x)g(x)d(x)
for f, g L
2
(G). One can also tensor all these spaces with the complexes
to get complex Banach spaces and a complex Hilbert space in the usual way.
All these spaces are independent of the explicit choice of Haar measure, but the
40
inner product on L
2
(G) does depend on the choice (it aects things by a scaling
factor).
Convolution (denition below) denes a product on L
1
(G); this is not hard
to check. To make this work we have to x a choice of Haar measure . Now if
f, g L
1
(G) then we dene f g L
1
(G) by
(f g)(z) =
_
G
f(zy
1
)g(y)d(y).
One checks that this is dened on L
1
(G), descends to L
1
(G), and is associative
and norm-non-increasing ([[f g[[
1
[[f[[
1
[[g[[
1
).
One can dene a measure on G associated to the integral ; one says that a
subset A G is measurable if its characteristic function
A
is summable, and
one denes (A) = (
A
).
3.4: Overview of Pontrjagin duality and Fourier inversion.
Ive decided/realised that one simply needs to assume too much measure
theory/spectral theory to give a reasonable presentation of this stu :-( and,
given that I do actually want to spend some lectures talking about Tates the-
sis, Ive decided that its impossible to give full proofs here (it would probably
take 6 or so lectures to go through the details) and hence I may as well just give
an overview of results. The original paper by Cartan and Godement (Theorie
de la dualite et analyse harmonique dans les groupes abeliens localement com-
pacts) is a good reference, and it seems to me that the 40-page Chapter 3 of
RamakrishnanValenza is, to a large extent, an English translation of this paper
(and chapter 2 of Ramakrishnan-Valenza is 30 pages of spectral theory and so
on which one needs as prerequisites).
If you want to see a complete presentation of this stu then, these 70 pages
are perhaps one place to look. It is possible to read this stu, but it would be
helpful if you knew e.g. what a Radon measure was and knew some of the basic
spectral theory of Banach algebras, and a fair bit of functional analysis too (the
BanachAlaoglu theorem, the KreinMilman theorem and so on). I dont know
if there is a simpler way to get to the results in the cases that were interested
in. I do know a low-level proof of the Fourier Inversion theorem for p-adic elds
but we also need this result in an adelic setting.
I will prove some basic results, and then give precise statements of deeper
theorems.
Let G be a topological group (later on it will be locally compact and Haus-
dor, of course, but we dont need that yet). The big new assumption now that
we do need, is that G must be abelian. The non-abelian story is much more
subtle (it occupied much of Harish-Chandras mathematical life and there are
still plenty of questions left unanswered). Even the abelian case needs some
work (c.f. those 70 pages I just mentioned).
So let G be an abelian topological group. Dene

G, the dual of G, to be the
group of continuous group homomorphisms
: G S
1
with S
1
= z C : [z[ = 1 the circle group (remark that if G isnt abelian then

G should probably contain some higher-dimensional unitary representations so

the non-abelian theory diverges at this point).
41
[G an abelian topological group;

G = : G S
1
]
One checks easily that

G is a group (the product of two continuous group
homomorphisms is continuous, and if is continuous then so is g (g)
1
).
Our rst job is to make

G into a topological group. There are subtleties here. In
the analogous linear theory (topological vector spaces) there is more than one
way to topologise the dual space of a topological vector space (look up Weak
topology on Wikipedia, for example).
Heres how were going to topologise

G. At the minute all we need to assume
is that G is an abelian topological group. If K is a compact subset of G and V
is a neighbourhood of the identity in S
1
then dene
W(K, V ) :=

G : (K) V .
Note that 1, the identity character (the one sending all g G to 1 S
1
) is in
all W(K, V ). We dene a topology on

G by letting the W(K, V ) be a base of
neighbourhoods of 1. Explicitly, a subset U of

G is dened to be open i for all
U there is K and V (possibly depending on ) such that W(K, V ) U.
Lemma. Let G be an abelian topological group. Then the construction
above does dene a topology on

G, and moreover

G becomes a topological
group with respect to this topology.
[W(K, V ) :=

G : (K) V .]
To check that weve dened a topology on

G we rst need to check rstly that
the empty set and the entire space are open (which just boils down to checking
that at least one set of the form W(K, V ) exists; for example K = e for e G
the identity will do). Next we need to check that an arbitrary union of open sets
is open, which is obvious. Finally we need to check that the intersection of two
open sets is open, which boils down to checking that W(K
1
, V
1
) W(K
2
, V
2
)
contains some W(K, V ); but this is true because one can set K = K
1
K
2
and
V = V
1
V
2
.
To check that multiplication on

G is continuous with respect to this topology,
we need to ensure that if = then, for all W(K, V ) there is W(K
1
, V
1
) and
W(K
2
, V
2
) with
W(K
1
, V
1
)W(K
2
, V
2
) W(K, V ).
This immediately simplies to
W(K
1
, V
1
)W(K
2
, V
2
) W(K, V )
because G is abelian.
Given K and V , we need K
i
, V
i
with
W(K
1
, V
1
)W(K
2
, V
2
) W(K, V ).
If we set K
1
= K
2
= K then all thats left is to check that for all neighbourhoods
V of the identity in S
1
there exists V
1
and V
2
with V
1
V
2
V , which is clear
because S
1
is itself a topological group!
Of course we could have done this more explicitly: if V contains e
i
with
< < then we could let V
1
= V
2
= e
i
: [[ < /2. While were at it,
42
Notation. If 0 r < then dene N(r) := e
i
: [[ < r.
Exercise. Let G = R be the real numbers, under addition. Prove that the
continuous maps G S
1
are precisely those of the form r e
irt
with t R
(hint: consider a neighbourhood N() in S
1
and its pre-image in G; this gives
a map (, ) (, ) which is additive whenever this makes sense; draw
some conclusions). Check that that this identication of

R with R induces an
isomorphism of topological groups

R = R.
Proposition. Say G is an abelian topological group and

G is its dual,
topologised as above.
(i) If G is discrete (that is, if all subsets are open) then

G is compact.
(ii) If G is compact then

G is discrete.
Proof.
(i) Consider

G as a subset of Hom(G, S
1
), the (arbitrary set-theoretic) maps
from G to S
1
. This latter space is just

gG
S
1
; give it the product topology
(reminder: a basis for the product topology, when considering an innite prod-
uct, are the subsets which are products of open sets such that all but nitely
many of the prodands are the entire space).
First I claim that

G is a closed subset of Hom(G, S
1
); this is because its
complement is clearly open, as if (ab) ,= (a)(b) then one can choose neigh-
bourhoods V
x
of (x) for x a, b, ab with V
ab
V
a
V
b
= .
Next I claim that the subspace topology on

G is the compact-open topology.
The compact subsets of G are just the nite subsets, and with this in mind
its easy to check that a set is open in one topology i its open in the other
(both topologies give

G the structure of a topological group and this reduces the
question to one about neighbourhoods of the identity, which just follows from
the denitions).
Finally I claim that this does it, and this is because Tychanovs theorem
says that a product of compact spaces is compact, so

g
S
1
is compact, and a
closed subspace of a compact space is compact.
(ii) If G is compact then I claim that subset 1 containing only the trivial
character is an open subset of

G, and because

G is a topological group this
will suce to prove that

G is discrete. To check that its compact we need to
show that W(K, V ) = 1 for some K and V ; take K = G and V = N() for
any < /3. If (G) N(1) then (G) is a subgroup of N(), but the only
subgroup of S
1
contained in N() is 1; hence W(G, N()) = 1 and were
done.
I will now start stating things without proof.
Theorem. If G is locally compact Hausdor, then so is

G.
I have seen a low-level proof of this, and a more abstract one. The low-level
proof (which is long, but completely elementaryit could be a long example
sheet question, with hints) goes as follows.
[Theorem. If G is locally compact Hausdor, then so is

G.]
One checks rstly that if K is a compact neighbourhood of the identity in
G then blah W(K, N(/6)) is a compact neighbourhood of the identity in

G (a
43
dull check), and secondly that this suces (which boils down to checking that
as K shrinks, W(K, N(/6)) gives a basis of neighbourhoods of 1).
The higher-level proof (which one needs later on in the theory anyway)
proceeds by rst introducing the Fourier transform of f L
1
(G). Notational
note: from now on, L
p
(G) will always denote the complex L
p
-functions, rather
than the real-valued ones, so its the thing I originally called L
p
(G), tensored
over R with C.
Fix a Haar measure on G. If f L
1
(G) then dene

f :

G C by

f() =
_
G
f(y)(y)dy.
Note that f L
1
(G), and [f(y)(y)[ = [f(y)[, and its easy to check that
the integrand is also in L
1
(G) (do it!). In particular the integral makes sense.
We call

f the Fourier transform of f.
Example: if G = R and f L
1
(G), and if we identify

G with R by asso-
ciating the real number r with the character x e
ixr
, and if we use the usual
Lesbesgue measure as our Haar measure, then we see that

f(r) =
_
R
f(x)e
irx
dx
which is the denition of Fourier transform that I learned as an undergraduate.
[

f() =
_
G
f(y)(y)dy.]
If however one identies

G with R by identifying r with the character x
e
2ixr
then one gets the denition of the Fourier transform which is used at the
top of the Wikipedia page about Fourier transforms. Finally, if one sticks to
x e
ixr
but uses the Haar measure which is

2 times Lesbesgue measure,
then one gets a third way of normalising things, which according to Wikipedia is
another popular choice. Which choice you prefer depends on why youre taking
Fourier transforms, but the point of this discussion is that all three choices are
covered by our denition.
Back to G locally compact and abelian. For every f L
1
(G) we get a
function

f :

G C (pedantic remark that even though f isnt a function
because two functions which dier on a null set are the same element of L
1
,

f
really is a function).
Dene the transform topology on

G to be the weakest topology that makes
every

f continuous. A computation which is basically elementary (if you know
that the continuous functions with compact support are dense in L
1
(G) and
that L
1
(G) is a Banach space, something I didnt prove) but long shows that
the transform topology coincides with the compact-open topology (note that
local compactness here is essential for this strategy even to make sense, as we
used a Haar measure). So we get another proof of G-locally-compact-implies-

G-locally-compact if we check that the transform topology is locally compact,

which follows from Gelfands theory of commutative Banach algebras, applied
to L
1
(G).
This latter approach (via the Fourier transform) might seem heavy-handed,
but in fact all of these techniques, and more, seem to be needed later on anyway.
44
The next step in the theory, at least in the development Ive seen, is to
consider an arbitrary Radon measure on

G with the property that (

G) 1
(note: Haar measure may well not have this property! We are not demanding
that is invariant under right translations). For such a measure we dene its
Fourier transform T

to be the function G C such that
T

(y) =
_
b
G
(y)d ()
and in some sense the crucial result seems to me to be an intrinsic characterisa-
tion of these functions on G; the functions T

that arise in this way are precisely
the functions which are are essentially bounded by 1 and are of positive type
(see below). The argument needs some graduate-level functional analysis and
measure theory,
in the sense that it needs results which seem to be standard but which
I didnt see in my undergraduate courses on functional analysis and measure
theory).
Once one has all this, one can prove the rst form of the Fourier inversion
formula. Here G is an abelian LCHTG. First a denition. Say that : G C,
continuous and bounded, is of positive type if for any f /(G) we have
_
G
_
G
(s
1
t)f(s)dsf(t)dt 0.
Fourier inversion formula (rst form).
There exists a Haar integral on

G with the following property: If f
L
1
(G) with Fourier transform

f :

G C, and if f is furthermore a C-linear
combination of functions of positive type, then
f(y) =
_
b
G

f()(y)d ().
f(y) =
_
b
G

f()(y)d ().
This is hard work. If one could interchange the integrals on the right hand side
then it might perhaps be easier, but the problem is that
_
G
(y)(t)dy probably
wont converge. I would almost certainly make a fool of myself were I to try
and summarise the 16-page proof in Ramakrishnan-Valenza.
As a consequence of the Fourier inversion formula, and I know of no simple
proof of this statement, we get
Theorem. If G is an abelian LCHTG and z G is not the identity charac-
ter, then there exists

G with (z) ,= 1.
Proof. If no such exists, then for every f L
1
(G) we would have

f =

f
z
.
Hence for every f for which the Fourier inversion formula applies, we would have
f = f
z
. But by Hausdorness we can nd a neighbourhood U of the identity
with U Uz empty.
We next nd a neighbourhood V with V
2
U and V symmetric; nally we
observe that if is real-valued supported in V and (1) = 1 then f :=

45
(with

(g) = (g
1
)) is of positive type but has support disjoint from that of
f
z
, a contradiction.
Corollary. If G is an abelian LCHTG then the obvious map G

G is
injective.
It will of course turn out that if G is an abelian LCHTG then G

G is
an isomorphism. But we have used analysis (rather than topology) to prove
injectivity, and in particular we used that G was locally compact. If G is an
arbitrary abelian topological group then one can still make sense of

G but I
dont know, and very much doubt, if G

G is bijective (or even injective) in

this generality; consider the case of double-duality of a vector space for some
analogue of thisV = V

i V is nite-dimensional.]
So from now on lets say G is an abelian LCHTC. Ill explain how the theory
can be developed.
Next one checks that G

G has the property that the induced map from

G to its image (with the subspace topology) is a homeomorphism onto a closed
subspace (this argument is elementary).
Now one checks that for f, g L
1
(G) we have

f g =

f g. This is just
an unravelling of things (once one has realised that maps L
1
(G) L
1
(G) to
L
1
(G), which can be proved using Fubini: in fact [[f g[[
1
[[f[[
1
[[g[[
1
.
If f L
2
(G) then set

f(x) = f(x
1
), so

f L
2
(G), and dene h = f

f
(so h is integrable and of positive type: this is some analogue of the fact that
if A is a real matrix then A
t
A is positive semidenite). Now unravelling the
denitions we see that if and are Haar measures normalised so that the rst
form of Fourier inversion holds, then
_
G
[f(x)[
2
d(x) = h(1)
=
_
b
G

h()d ()
=
_
b
G
[

f()[
2
d ()
(the second = is Fourier inversion, the other two are elementary). So the inte-
grals of [f[
2
and [

f[
2
coincide. This is the rst form of the Plancherel theorem.
But in fact, by a density argument one can now conclude
Plancherel Theorem.
For G an abelian LCHTG one can extend the Fourier transform uniquely to
an isometric isomorphism

: L
2
(G) L
2
(

G).
One has to be careful here: I am not asserting that if f L
2
(G) then the
original denition of

f that I gave makes sense. All Im saying is that the map,
which we dened using an integral, extends to give a map on all of L
2
(G) in
some way.
From this one gets, without too much trouble,
46
Pontrjagin duality.
If G is an abelian LCHTG then the obvious map G

G is a group-theoretic
isomorphism and a topological homeomorphism.
And then nally this leads us to a cleaner version of the Fourier Inversion
theorem:
Fourier Inversion Theorem (nal form).
Fix Haar measures on G and

G. Then there exists a positive real constant
c > 0 such that if f L
1
(G) is continuous, and

f L
1
(

G), and if we identify

G with

G, then

f(x) = cf(x
1
)
for all x G. Furthermore, for any choice of Haar measure on G theres a unique
Haar measure on

G which ensures c = 1.
We havent given a complete proof of this. I do know complete proofs in
certain explicit cases. I currently dont know whether the proofs I know suce
to cover the instances needed for Tates thesis, but Ill probably nd out within
a few weeks.
Case studies.
1) G = R. Here its not so hard to give a proof. The trick is to introduce
the following rapidly-decreasing functions: for t > 0 and x R xed, consider
(y) = e
iyxt
2
y
2
. One explicitly computes the Fourier transform of this (good
clean fun) and now, instead of integrating

f(y)e
iyx
with respect to y, one in-
tegrates

f(y)(y). The trick is that this is easily shown to be the integral of
f(r)

(r). Now one lets t tend to zero from above, and uses the dominated con-
vergence theorem (and the fact that the Fourier transform is continuous, which
also needs to be checked, and which also follows from the dominated convergence
theorem).
2) G = S
1
with its usual topology, so

G = Z with the discrete topology. In
this case, the Fourier inversion theorem simply says that for a periodic function f
on R (that is, a function on S
1
), the Fourier series of f converges to f. This
just boils down to the statement that if z is the inclusion S
1
C then the
functions z
n
: n Z is an orthonormal basis for the Hilbert space L
2
(S
1
).
Orthonormality is easy, and checking that the functions give a basis is just
a standard application of the Stone-Weierstrass theorem (polynomials in z and
1/z separate points, and z = 1/z on S
1
). Orthonormality also gives Plancherels
theorem (which is called Parsevals theorem in this context; Parseval was 1799
and thinking about Fourier series, Plancherel was 1910).
3) G = Q
p
or k
P
: well come back to these. In some sense theyre much easier
(in the sense that you dont have to remember what the dominated convergence
theorem or the Stone-Weierstrass theorem are!). Ill treat these cases carefully
in the next chapter.
Chapter 4. Local zeta functions.
The reference for this is chapter 2 of Tates thesis. Throughout this section
K will be a eld which is either the reals, an algebraic closure of the reals (you
can think the complexes but perhaps a more pedantic way of thinking about
it is the complexes except that there is no way of distinguishing between the
47
two square roots of 1), or a nite extension of Q
p
for some p. In the global
applications, K will be the completion of a number eld k with respect to a
norm. There is another case where everything in this section applies, and that
is the equicharacteristic case, that is, K = F
q
((t)), where F
q
is a nite eld
with q elements and F
q
((t)) is the eld obtained by adjoining 1/t to the integral
domain F
q
[[t]] of power series. Personal preferences mean that I will stick to the
number eld case, rather than the function eld case, later on, but one might
want to bear in mind that there are no obstructions to making this sort of thing
work in the function eld case.
In all cases (K = R, K

= C, K = k
P
nite) we have got a canonical equiva-
lence class of norms on K, and we have seen that K is a locally compact abelian
group under addition (in the p-adic case the crucial observation was that the
residue class eld was compact). Now heres a completely wacky construction:
we are going to single out a canonical norm in each equivalence class. Heres
the idea. Choose a random Haar integral on K. For K

consider the
function

: /
+
(K) R dened by

(f) = (x f(x)).
In words,

is Haar measure, stretched by multiplication by . One checks

easily that

is well-dened and is also a Haar measure, and hence c

= ,
where c = c() is a positive real number. One checks easily that c() does not
depend on the choice of it is truly intrinsic. The reason we didnt see this
structure before is that were not just thinking of K as an additive group, were
using its ring structure.
Lets write [[ := c(), and dene [0[ = 0.
[c()

= ]
This choice is a canonical choice of norm on K. One does need to check
its a normbut this is easy by a brute force calculation, which Ill now do: in
each case we see that were reconstructing the norm Ive already put on these
eldsbut now we see that the norm I put on them is the natural norm.
1) If K = R then (think about a good approximation to the characteristic
function of [0, 1]) [[ is just the usual absolute value of .
2) If K = C then (think about the characteristic function of a square) we
see [x + iy[ = x
2
+ y
2
, so our canonical norm is the square of the usual norm
(and hence doesnt satisfy the triangle inequality, which is the unique reason
that I didnt make the triangle inequality an axiom earlier).
3) If K = Q
p
then lets compute [p[. Well, if is the characteristic function
of Z
p
, the integers of Q
p
, then really is continuous with compact support.
Now Z
p
is the disjoint union of a + pZ
p
for a = 0, 1, 2, . . . , p 1, so by nite
additivity and translation-invariance of Haar measure we see that if is the
characteristic function of pZ
p
then p() = (), so
p
() = () = p()
and hence that [p[ = p
1
. So in fact the canonical norm on Q
p
is just the usual
p-adic norm, normalised the way I normalised it.
4) More generally (easy check) if k
P
is a uniformiser and q is the size
of the residue eld A/P (A the global integers of the number eld k), then we
showed that the residue eld of k
P
has size q (residue elds dont change under
completion) and one checks easily that the canonical norm sends to 1/q.
48
5) [optional extra] K = F
q
((t)). Then again we see that the index of tF
q
[[t]]
in F
q
[[t]] is q, so multiplication by t is making things q times smaller, so [t[ =
q
1
again the norm of a uniformiser is the reciprocal of the size of the residue
eld.
4.1: the dual of (K, +).
The next thing well do is to compute

G, where G = (K, +). As ever in this
section, K is either the reals, the complexes, a completion of a number eld at
a prime ideal, or, if were feeling adventurous, a the eld of fractions of a power
series ring over a nite eld. Lets call these things local elds for simplicity,
and lets always endow them with their canonical norms.
Theorem. If G = (K, +) is a local eld, considered as a group under
addition, then

G is isomorphic to (K, +) (not in a particularly canonical way,
mind).
Remark. The case K = R was an exercise earlier.
Theorem. If G = (K, +) is a local eld, considered as a group under
addition, then

G is isomorphic to (K, +).
Remark. I am going to be lazy and give Tates proof, which appears to me to
assume (a consequence of) Pontrjagin duality, but which works for an arbitrary
locally compact complete normed eld (so, for example, it works for a power
series eld over a nite eld). On the example sheet Ill give an explicit proof
when K/Q
p
is nite.
Proof. First let me assume that

K ,= 0 (of course here K is considered as a
group under addition). Well check this later in a case-by-case way, although if
you believe Pontrjagin duality then its obvious because

K ,= 0 implies K = 0.
Anyway, lets x once and for all a non-zero element of

K, and later on Ill
write one down explicitly just to prove one exists.
Now consider the map i : K

K (which depends on ), dened by letting
i() be the character x (x) (so again were crucially using both the additive
and multiplicative structure of K). Its easily checked that i()

K and that
the induced map i : K

K is a group homomorphism. Injectivity is also easy:
if (i())(x) = 1 for all x K then is trivial on K which is impossible if
K = K, because is non-trivial, so had better not have an inverse.
Its slightly more delicate to nish the job. Well follow Tate and again
assume Pontrjagin duality. A consequence of this duality is that one can show
that the annihilator construction, sending a subgroup X of an abelian LCHTG
G to the subgroup of

G consisting of characters which vanish on X, induces an
order-reversing bijection between the closed subgroups of G and of

G.
Apply this to the closure of i(K)

K and we observe that the corresponding
closed subgroup X of K must be be contained in the set x K : (i())(x) =
1 K but this set is easily checked to be 1. Hence the image of K is
dense in

K.
Next I claim that the map i : K

K is a homeomorphism onto its image.
To check this we need to remember the denition of the topology on

K: a general
neighbourhood of the origin was given by W(L, V ) with L K compact and V
a neighbourhood of the identity in S
1
. So what we have to do is to rst check
49
that each W(L, V ) contains an i(B(0, )) (the open ball centre zero radius
in K), and conversely that each i(B(0, )) contains i(K) W(L, V ) for some
L, V . Both of these are easy; Ill do the slightly harder of the two, which is the
latter one. Given > 0 we need to come up with with L and V such that if
K and (L) V then [[ < , and we do this thus. Choose k K with
(k) ,= 1. Let L be a huge closed disc centre 0 radius M (these are compact,
as is easily checked), with the property that [k[ < M, and let V be any open
neighbourhood of the identity in S
1
such that (k) , V . Then for K, if
[[ then k L so i() , W(L, V ) which is what we want.
Finally lets show that i(K) is a closed subspace of

K; this will do us because
we already know that its dense. Because any compact set in K is bounded, its
easy to check that the identity in

K has a countable basis of neighbourhoods
(this isnt logically necessary, I dont think, but its psychologically satisfying
for what follows). For example if C
M
denotes the closed disc centre zero radius
M and V
M
is e
i
: [[ < 1/M then N
M
:= W(C
M
, V
M
) will do. So now
choose an arbitrary

K. For each integer M 1 choose x
M
K with
i(x
M
) N
M
(we can do this by density of the image of i). Its easily checked
(its an argument similar to the one showing i was a homeo onto its image, but
it seems to me to be not quite a formal consequence of what we already have)
that the x
M
form a Cauchy sequence, so x
M
x K, and we have i(x
M
)
and i(x
M
) i(x), so by Hausdorness we have = i(x) and were home.
Actually thats not really the end of the proof because I still need to exhibit
a non-zero

K. Lets do this on a case-by-case basis.
1) K = R. Then dene (x) = e
2ix
.
2) K = Q
p
. Then, by the structure theorem for elements of Q
p
we can
write any k K as k =

n=N
a
n
p
n
with a
n
0, 1, 2, . . . , p 1. Set
q(k) =

1
n=N
a
n
p
n
Q (so q(k) = 0 i k Z
p
). Its an easy check that q is a
group homomorphism Q
p
/Z
p
Q/Z and hence that (k) = e
2iq(k)
will work.
3) K a nite extension of K
0
:= Q
p
or R; then the trace map T
K/K
0
is
an additive map K K
0
and its surjective (its multiplication by [K : K
0
]
on K
0
), so we take this map and then just compose it with the relevant map
coming from (1) or (2).
Note that if we write (y) = e
2i(y)
with : K R/Z then our map i is
just (i(x))(y) = e
2i(xy)
.
4) K = F
q
((t)). Then coecient of t
1
is a surjection K F
q
, and F
q
is
just a nite-dimensional vector space over F
p
, so choose a non-zero linear map
F
q
F
p
(if we want to x one then we should use the trace mapbut if you
dont know about separable extensions it might not be immediately clear to you
that this is non-zero), and nally x e
2i x/p
gets you from F
p
to S
1
, where
x x is a lifting Z/pZ Z.
Its convenient, but not essential, to x a non-zero character of K once and
for all; in cases (1)(3) above Ive written down precisely one character, so lets
always use this one. Note that this is not a canonical choice however; it seems
to me that K and

K are not canonically isomorphic.
While were here, lets x a choice of Haar integral on K. If K = R then
lets choose the obvious onethe one such that the integral of the characteristic
50
function of [0, 1] is 1. If K = C then lets choose twice the obvious
onethe one such that the integral of the characteristic function of [0, 1]
[0, 1] is 2 (unsurprisingly, this 2 is related to the fact that our norm on C is the
square of the usual norm).
If K is a nite extension of Q
p
then we do something perhaps a bit more
surprising. If K has degree n over Q
p
then the integers R of K are isomorphic
to (Z
p
)
n
as a Z
p
-module (this is elementary if you know that nitely-generated
torsion-free modules over a PID are free) and one can dene the discriminant
in the usual way: choose a Z
p
-basis e
1
, e
2
, . . . , e
n
for R, and dene an n n
matrix A
ij
whose (i, j)th entry is the trace of e
i
e
j
. The determinant of this
matrix is in Z
p
(easy) and generates an ideal called the discriminant ideal of
K/Q
p
; it is a non-zero ideal (this is a standard fact from eld theory, coming
from separability) and is well-dened independent of all choices (easy). Say the
discriminant ideal is p
m
Z
p
. Lets dene our Haar integral on K by letting the
integral of the characteristic function of R be p
m/2
R. [One can check that
if k/Q is a number eld then the discriminant of k
P
/Q
p
will be Z
p
if P is
unramied in k, and in particular if you regard k as xed then all but nitely
many of its P-adic completions will have the property that their discriminant
ideals will be Z
p
.]
Why are we labouring over these choices? Well, we have xed a choice
of Haar measure on K, and we have xed an isomorphism K

K, so we
get an induced Haar measure on

K, and were now in a position to apply
Fourier transforms twice. Recall that if f is continuous and f L
1
(K) with

f L
1
(

K) = L
1
(K) then we know that

f(x) = cf(x) where c is some constant

depending only on our choices of Haar measure.
Proposition. With the choices we made, c = 1.
Non-proof. It suces to check for just one function, and well have to com-
pute loads of Fourier transforms quite soon, so Ill postpone this. We dont need
this result at all, its just psychologically satisfying.
4.2: The dual of (K

, ).
Let K be as usual. Actually not really interested in the Pontrjagin dual
of K

, were much more interested in the continuous group homomorphisms

K

, that is we are dropping the unitary assumption on our characters

in this section. Let me call a continuous map G C

a quasi-character of
G. The main results we need here are rather easy to prove. Let U denote
x K

: [x[ = 1the units of K. Say that a quasi-character is unramied if

its trivial on U.
Lemma. The unramied quasi-characters c : K

are all of the form

[[
s
for s complex. If K = R or C then s is uniquely determined, but if
K is p-adic or F
q
((t)) then s is only determined modulo
2i
log(q)
Z with q the size
of the residue eld. In either case, however, the unramied quasi-characters are
naturally a 1-dimensional complex manifold.
Proof. Recall [[
s
means e
s. log(||)
. Everything is immediate once we observe
that K

/U can be explicitly computed as the image [K

[ of the norm map in

R
>0
. If K = R or C then [.[ gives an isomorphism K

/U R
>0
, and if
K is non-arch then [.[ : K

/U q
Z
is an isomorphism. Assuming that you
51
know that the continuous group homomorphisms R
>0
C

are all of the form

x x
s
then were home.
For simplicity we now x a continuous splitting of the map [.[ : K

[K

[,
that is, we choose a subgroup V of K

with the property that every element of

K

is uniquely of the form uv with u U and v V . Again its impossible, in

general, to do this naturally.
If K = R or C then we can just let V be the positive reals. If K is non-arch
then lets choose a uniformiser K and let V be
Z
; this is easily checked to
work. Now K

= U V so Hom(K

, C

) = Hom(U, C

) Hom(V, C

).
Notation: for K

, write = v for U and v V . This clearly

depends on our choice of V but well only use this notation temporarily.
Corollary. The quasi-characters c : K

are all of the form

( ).[[
s
with s C and a character of U.
Proof. This is just an explicit rephrasing of the statement Hom(K

, C

) =
Hom(U, C

) Hom(V, C

).
Note that that corollary gives the group of all quasicharacters the structure
of a 1-dimensional complex manifold (again given by the s variable; U is compact
and we regard Hom(U, C

) = Hom(U, S
1
) =

U as discrete). Pedants might like
to check that this
complex structure is independent of the choice of V . In fact is well-dened
independent of the choice of V (its just c[U) and s is well-dened up to the
2i/ log(q) ambiguity mentioned earlier.
Explicitly, what is happening is that if

U is the set of characters of U, then
the quasi-characters of K

are just one copy of either C or C/

2i
log(q)
Z for each
element of

U, the dictionary being that if is any character of U and c is any
quasi-character of K

which restricts to , then the connected component of c

in the manifold of quasi-characters is just c.[.[
s
for s C.
Lets do examples to see what these manifolds look like.
[ c : K

are all of the form ( ).[[

s
with s C and a
character of U.]
The easiest example to think about is K = R; then K

= R
>0
, U = 1,
and a quasicharacter of K

is just a sign (where we send 1) and a complex

number (where we send the positive reals), and the complex structure is given
by the complex number; the space is just two copies of the complexes. If K = C
then K

= S
1
R
>0
, U = S
1
, and the characters of S
1
are just Z, so here
the quasicharacters are countably innitely many copies of the complex plane,
indexed by the integers. If K is an algebraic closure of R then we get the same
thing, but where Z is replaced by an innite cyclic group.
If K = Q
p
then we get a cylinder C/(
2i
log(p)
Z) for each primitive Dirichlet
character (Z/p
n
Z)

and in the general non-arch case the picture is a

generalisation of this.
Let me make two denitions before we go any further: using the corollary
above one sees that if we have a quasi-character of K

as in the corollary, then

[c[ := [.[ c : K

R
>0
(note that the target C

always has the usual norm

[x + iy[ =
_
x
2
+y
2
) must just be the map [[

with = Re(s) (note

that this is well-dened even in the non-arch case). This real number is called
52
the exponent of c, and quasi-character of exponent at least is going to be
our analogue of complex number with real part at least later on. Also,
lets say that two quasi-characters are equivalent if their ratio is unramied
this is just the same as demanding that they are in the same component of the
quasi-character manifold.
We need now to x a Haar integral on K

. If f /(K

) then the function

x f(x)/[x[ is a function on K0 and (by compactness of support) its
extension to K (send 0 to 0) is continuous with compact support, so we can
integrate it using our xed choice of Haar measure on K; the resulting functional
is easily checked to be a Haar integral
1
on K

. If K is archimedean then this

will be our xed choice of Haar integral on K

. If however K is p-adic then

were going to do something else.
Lets compute
1
(
U
) (the characteristic function of the units) in the p-adic
case. Because [.[ is trivial on U, we see
1
(
U
) = (
U
) and we normalised on
K so that the integers R had measure p
m/2
, where p
m
Z
p
is the discriminant
ideal of K. So, because U = RR, we see (
U
) = (1 1/q)(
R
) = (1
1/q)p
m/2
. It will be convenient to choose a Haar measure

on K

such that
U has measure 1 for almost all P, as P runs through the primes of a number
eld k and K = k
P
, so we dene

on K

by

=
q
q1

1
in the P-adic case.
Then

(
U
) = p
m/2
if K has discriminant p
m
, and in particular if K = k
P
then

(
U
) = 1 for all but nitely many P (a number eld has a discriminant
and if P is coprime to this discriminant then the discriminant of k
P
is just Z
p
).
4.3: Local analytic continuation and functional equations.
In some sense weve done nothing much in this chapter so farapart from
the check that K is isomorphic to

K, all weve done is made explicit choices of
things. Heres the rst hint that something magic is happening though. Fix K
as usual, and say f : K C is continuous, with f L
1
(K), and such that

f is
also continuous and in L
1
(

K) = L
1
(K) (recall we have xed an identication of
K with

K). Assume furthermore that x f(x)[x[

and x

f(x)[x[

are both
in L
1
(K

) for any > 0. One might ask whether any non-zero such functions
exist, but well see plenty of examples later on, and in fact its an easy exercise
to come up with some examples. Let Z = Z(K) be the set of such functions.
In words these conditions imply that f is racing to zero more quickly than any
polynomial in [x[ as [x[ gets big (thats the L
1
(K

) condition), and that f is

bounded near zero (thats the L
1
(K) condition) and furthermore that

f has the
same properties.
Easy exercise: why would asking f L
1
(K

) be asking a bit too much?

[Hint: remember Haar measure on K

isnt the same as that on K; consider

whats happening near the origin].
Let Q denote the set of quasi-characters of K

. If c Q then lets write

Re(c) for the exponent of c. Given f as above, dene a function (f, ) on
c Q : Re(c) > 0 by
(f, c) =
_
K

f(t)c(t)d

(t).
So were using the multiplicative Haar measure on K

dened above.
53
Lemma. The function (f, ) (converges and) is holomorphic on the com-
plex manifold c Q : Re(c) > 0.
Proof. This rather fancy-sounding lemma is actually elementary to prove.
Convergence is not an issue because our assumptions on f imply that the in-
tegrand dening this local zeta function is in L
1
(K

)we have [f(x)c(x)[ =

[f(x)[[x[
Re(c)
which is in L
1
(K

) by denition.
The complex structure near c Q is given by c.[.[
s
for s C small, so all we
have to do is to check that (f, c.[.[
s
) is holomorphic in s, for s small enough.
So we have to dierentiate (f, c.[.[
s
) with respect to s, and if you write out the
denition of dierentiation, and remember all the boundedness assumptions
weve made on f, you see that you can dierentiate under the integral sign! This
reduces us to checking that
_
K

f(t)c(t)[t[
s
log([t[)d

(t) converges, but it does

because for [t[ big, [t[ beats log([t[) and the integral converges for all suciently
large exponents, and for [t[ small, [t[

beats [ log([t[)[ (where [t[ is chosen so

small that < Re(s) works, which makes the integral converge by assumption).

But thats not the big local insight; the big insight is that (f, ) has a
meromorphic continuation to all of Q! The global zeta functions coming later
will be products of local zeta functions for Re(c) suciently large. It is however
important to note that this local analytic continuation result certainly does not
give the meromorphic continuation of the global zeta functions that are coming
lateran innite product of meromorphic things might not be meromorphic (it
might not even converge). Let me illustrate this by remarking that well shortly
see that an example of this local meromorphic continuation statement is the
statement that the function

i0
p
is
, which converges for Re(s) > 0, can be
rewritten as 1/(1 p
s
), which is meromorphic for all s C. This observation
is clearly not enough to meromorphically continue blah

p
(1 p
s
)
1
= (s).
So lets prove this local meromorphic continuation, and even a local func-
tional equation. We need an analogue of s 1 s for the functional equation;
its c c, where c is dened by c(x) = [x[/c(x)). Note that this hat has noth-
ing to do with Fourier transforms, its just an elementary denition. Note also
that Re( c) = 1 Re(c). But also note that in general c wont be equivalent to
cthey could well lie on dierent components of Q.
Note that we havent used any of our boundedness assumptions on

f so far,
weve only used the L
1
ness of f. Well use L
1
ness of

f now though.
Lemma. If 0 < Re(c) < 1 and f, g Z then
(f, c)( g, c) = (g, c)(

f, c).
[note that all integrals obviously converge.]
[(f, c)( g, c) = (g, c)(

f, c).]
Perhaps I should remark that I have very little true understanding of this
equation. I can prove it though, in fact its dead easy to prove, it just follows
from unravelling and Fubini. Lets see the proof.
First note that what we have to do, to prove the lemma, is to prove that if
left hand side is L(f, g), then L(f, g) = L(g, f).
54
If we substitute in the denition of (, ) twice on the left hand side, we
get
_
K

f() g()c(/)[[d

()d

()
and by Fubini we can integrate in whatever order we want as long as 0 < Re(c) <
1 (I only proved Fubini for /(G H) but it extends to L
1
). If I think about
doing the integral over rst, for xed , then I can use invariance of

()
under multiplication to change to = /, and making this substitution
shows that the integral equals
_
K

f() g()c(
1
)[[d

()d

().
_
K

f() g()c(
1
)[[d

()d

().
Now lets give a name to the integral
H
f,g
() :=
_
K

f() g()[[d

;
then the integral were trying to prove something about is
_
K

H
f,g
()c(
1
)[[d

().
Now if H
f,g
() = H
g,f
() then visibly this latter integral (which has no other
fs and gs in) will also be unchanged if we switch f and g, which is exactly what
we wanted to prove. So were now reduced to showing H
f,g
= H
g,f
. But recall
that

was, up to a constant which depended only on K, just (x)/[x[, so

H
f,g
() =
_
K
f() g()d()
where note now the integral is over K, and now by denition of g we see
H
f,g
() =
_
KK
f()g()e
2i()
d()d()
recalling that our identication of K with

K sent x to the character y
e
2i(xy)
. But now of course were home, because switching f and g is just
the same as changing notation (, ) (, ).
Reminder: weve just proved that if 0 < Re(c) < 1 and f, g Z then
(f, c)( g, c) = (g, c)(

f, c).
Corollary. If, for each component C of Q, we can nd one explicit function
f = f
C
Z such that (

f, c) doesnt vanish identically on c C : 0 < Re(c) <
1 and such that (c) := (f, c)/(

f, c) has a meromorphic continuation to all
c C, then for any g Z, the function (g, c) has meromorphic continuation
to all of Q, and satises the functional equation
(g, c) = (c)( g, c)
55
for c C.
Proof. Trivial. The left hand side is holomorphic for Re(c) > 0, our assump-
tions on show that the right hand side is meromorphic for Re(c) < 1, and the
lemma we just proved shows that the two sides agree on the overlap.
4.4: Tidying up.
Here Ill give explicit examples of f
C
as promised above, check that the
corresponding (c) has meromorphic continuation, and also check the assertions
about our choice of Haar measure being self-dual, which will of course come
out in the wash.
I have to start somewhere so lets start with K = Q
p
, the simplest case where
we havent done any explicit integrals yet. Recall that if Z
p
is k K : [k[ 1,
the integers of K, then Z
p
is a ring, and the invertible elements Z

p
of this ring
are easily seen to be just k K : [k[ = 1 = U.
What we have to do to perform the local meromorphic continuation is, for
each character : Z

p
S
1
, we need to nd a function f = f

Z with
(

f, c) not identically zero on the region 0 < Re(c) < 1 of the component of
Q corresponding to , and such that we can meromorphically continue (c) :=
(f, c)/(

f, c) to all of this component.
A reminder of normalisations: additive Haar measure on Q
p
is normalised
so that the characteristic function
Z
p
of Z
p
has integral 1. Note that
Z
p

/
+
(Q
p
) and by additivity of Haar measure we have (
a+p
n
Z
p
) = p
n
for
all n Z. In particular locally constant functions with compact support are
easy to integratebut these things are easily checked to uniformly approximate
anything in K(Q
p
), so integration on Q
p
is in fact easy. One last reminder:
multiplicative Haar measure

on Q

p
is
p
p1
times dz/[z[, the usual trick to
turn additive Haar measure into multiplicative Haar measure.
First lets do the component of Q corresponding to the trivial character of
Z

p
. Let f be the characteristic function of Z
p
. Now lets get it straight in our
heads what we have to do.
First we need to check f Z, which involves checking some boundedness
conditions on f, computing

f and checking the boundedness conditions on this
function too.
Next we need to compute (f, c) and (

f, c) for c in the component of Q =
Hom(Q

p
, C

) corresponding to the trivial character of U (that is, for c : Q

p

C

such that c[Z

p
is trivial).
Finally we need to check that (

f, c) is not identically zero for such c, and
that (f, c)/(

f, c) has a meromorphic continuation to the entire component.
To do all of this we just need to unwind the denitions and then gure out
the integrals.
To check f Z we rst need that f is continuous (easy), that f is integrable
(its even in /
+
(Q
p
) so its certainly integrable), and that f(x)[x[

is in L
1
(Q

p
)
for > 0. This needs checking, not least because f(x) is not in /
+
(Q

p
)f is
non-zero arbitrarily close to zero, and (think about the automorphism x 1/x
of Q

p
) this means f doesnt have compact support on Q

p
. But f is visibly
a pointwise increasing limit of functions with compact support (consider the
characteristic functions f
n
of Z
p
p
n
Z
p
for n large) so we had better compute
56
the integrals of f
n
(x)[x[

on Q

p
and check they converge; if they do then we
will have proved f(x)[x[

is summable and hence in L

1
(Q

p
).
Well
_
Q

p
f
n
(x)[x[

(x)
=
_
Z
p
\p
n
Z
p
[x[

(x)
=
n1

i=0
_
p
i
Z

p
[x[

p
p 1
[x[
1
d(x)
=
p
p 1
n1

i=0
(p
i
Z

p
)p
i(1)
=
p
p 1
n1

i=0
p 1
p
p
i
=
n1

i=0
p
i

i=0
p
i
=
1
1 p

as n , if > 0. So the boundedness properties of f are satsied. Note: I

am being lazy and writing (X) for (
X
), the characteristic function of X.
Now we need to compute

f. Well, by denition

f(x) =
_
Q
p
f(y)e
2iq(xy)
d(y)
where q is that function Q
p
/Z
p
Q/Z dened by take the fractional part.
This is just
_
Z
p
e
2iq(xy)
d(y)
so lets do this integral. We always have y Z
p
so if x Z
p
then q(xy) = 0
and we just get (Z
p
) = 1.
On the other hand, if x , Z
p
and e
2iq(x)
= ,= 1 and [x[ = p
n
, n 1 (so
is a primitive p
n
th root of unity), then e
2iq(xy)
only depends on y mod p
n
and
_
Z
p
e
2iq(xy)
d(y) =

p
n
1
i=0

i
= 0. Hence

f = f.
But this is greatrstly we have proved that c = 1 in the Fourier Inversion
theorem, as claimed earlier (

f = f(x) = f(x)), and secondly we have proved

f Z, because

f = f has all the same boundedness properties as f.
Next we need to compute (f, c) and (

f, c) on the region 0 < Re(c) < 1 of
the component corresponding to the trivial character of U, the general element
of which is x [x[
s
for s C = C/
2i
log(p)
Z. We stick to 0 < Re(s) < 1. Now
57
(being much lazier about the dierence between L
1
and /(Q

p
) this time)
(f, c) =
_
Q

p
f(x)[x[
s
d

(x)
=
p
p 1
_
Z
p
\{0}
[x[
s1
d(x)
=
p
p 1

j=0
_
p
j
Z

p
p
j(s1)
d(x)
=

j=0
p
js
= (1 p
s
)
1
(which looks surprisingly familiar!). Note the last line is OK because Re(s) >
0 so [p
s
[ < 1.
Similarly (

f, c) =
_
Q

p
f(x)[x[
1s
d

(x) which (recalling that were assum-

ing 0 < Re(s) < 1), by exactly the same calculation but with 1 s replacing
s, comes out to be (1 p
(1s)
)
1
. Hence (

f, c) is not identically zero on
the component were interested in, and (c) = (f, c)/(

f, c) =
1p
s1
1p
s
, which
looks less familiar, but later on youll see why youre not expected to recog-
nise this functionthese (c) will only show up explicitly at the bad primes.
The crucial observation that we need, however, is that (c) has a meromorphic
continuation to all s C, which is clear, and so our job is done.
Let me sketch the ramied case in this setting, because there (c) is quite
dierent. Let denote a Dirichlet character of conductor p
n
, n 1, that
is, a map : (Z/p
n
Z)

(note that the image lands in S

1
) that doesnt
factor through (Z/p
n1
Z)

. The natural map Z

p
Z/p
n
Z induces a map U
(Z/p
n
Z)

and hence a character of U, so we get a component of Q parameterised

by s C/
2i
log(p)
Z whose typical element sends x = p
n
u to (u)p
ns
= ( x)[x[
s
[were setting V = p
Z
with notation as above].
We need an f = f
C
for this component. Lets set f(x) = 0 if [x[ > p
n
,
and f(x) = e
2iq(x)
for [x[ p
n
(note [x[ p
n
implies q(x) = a/p
n
for some
integer a). So again f is locally constant (indeed its constant on cosets of Z
p
in
p
n
Z
p
) and takes values in the p
n
th roots of unity. Its clear that f L
1
(Q
p
),
and we can write f =
Z
p
+ f

with f

K(Q

p
) C, and we already showed

Z
p
.[.[

L
1
(Q

p
) for > 0, which implies f.[.[

L
1
(Q

p
) for > 0.
[f(x) = e
2iq(x)
for [x[ p
n
.] Next we need to compute the Fourier trans-
form of f. We can do this by brute force or by a trick. Heres the brute force
method:

f(x) =
_
Q
p
f(y)e
2iq(xy)
d(y)
=
_
p
n
Z
p
e
2iq(yxy)
d(y)
=
_
p
n
Z
p
e
2iq(y(1x))
d(y).
58
The same cancelling phenomenon (adding roots of unity) hence shows

f(x) =
0 if [1 x[ > p
n
(were adding roots of unity), but if [1 x[ p
n
then
[y(1 x)[ 1 for y p
n
Z
p
and the integrand is just 1, showing that the
integral is just (p
n
Z
p
) = p
n
times the characteristic function of 1 + p
n
Z
p
.
Note that this integral is bounded away from zero (as n 1) so

f /
+
(Q
p
)
and /
+
(Q

p
), and hence f Z.
[The trick way to do this last computation is to observe that the f here is just

Z
p
rescaled and multiplied by a character, and so we can compute the Fourier
transform of our f from the Fourier transform of
Z
p
using basic properties of
Fourier transforms.]
Next we need to compute (f, c) and (

f, c) for c of the form p
n
u (u)p
ns
for 0 < Re(s) < 1. These calculations are very similar to the ones we have
already done (although perhaps slightly tougher, because in some cases its
trickier to check that certain sums of roots of unity are zero). Ill stick them on
the example sheet and just tell you the answers: if I got it right then
(f, ( x)[x[
s
) =
p
ns+1n
p 1
p
n
1

j=1
(j)e
2ij/p
n
(the inner sum is called a Gauss sum) and (

f, ( x)
1
[x[
1s
) just turns out to
be p/(p 1), a constant!
So the ratio (f, c)/(

f/ c) is of the form A.B
s
with A a constant involving a
Gauss sum, and B a positive real constant (a power of p in fact), which means
that the ratio has meromorphic continuation to the component C and again
were done.
More precisely, the ratio is p
n(s1)

p
n
1
j=1
(j)
j
with = e
2i/p
n
. The
following observation is now surely worth remarking on. The Dirichlet character
we were just consideringwe were doing local calculations with it, but we can
also consider the global function (or L-function, as its more commonly known)
attached to this character, which is (for Re(s) > 1)

m1
(m)/m
s
=

(1 ()
s
)
1
,
the latter product being over all primes . This L-function has a meromorphic
continuation to all of C, which turns out to be holomorphic in this case, because
we assumed the conductor was p
n
for n 1.
We have (1) = 1 for some choice of sign. If (1) = 1 and if we
multiply this L-function by the usual fudge factor
s/2
(s/2), then we get
a new function (, s) satisfying
(, s) =
_
_
p
ns
p
n
1

j=1
(j)
j
_
_
(, 1 s).
A similar sort of thing is true if (1) = 1 but then the fudge factors and the
functional equation are slightly dierent. The moral is that this time the local
59
integrals arent showing up as components of the L-function, but the ratio (c)
is showing up in the functional equation.
I am not going to plough through all the other cases. The computations
are a little long but completely elementary and prime example sheet fodder.
The crib is Tates thesis, end of chapter 2. Heres the answers. If K is a
nite extension of Q
p
then the only extra subtlety is that we used the trace
map to dene K

K, and when doing the calculations one needs to compute
K : Tr
K/Q
p
(v) Z
p
v R where R is the integers of K. Clearly
this set contains R, and is not all of K (because it doesnt contain p
n
for
n large),so its a fractional ideal of K, but what you may not know is that
if we write it as
r
R then the norm to Q
p
of
r
generates the discriminant
ideal, which simplies some constants a bit. The answers are the same as
in the K = Q
p
case: for the unramied quasi-characters one lets f be the
characteristic function of K : Tr
K/Q
p
(v) Z
p
v R and one checks
(f, [.[
s
) = p
m(s1/2)
/(1q
s
) and (

f, [.[
1s
) = (1q
s1
), and in the ramied
case one makes a sensible choice of f and turns out to be of the form A.B
s
with A involving a Gauss sum.
If K = R then there are two components: on the component x [x[
s
use
f(x) = e
x
2
, and on the component x sgn(x)[x[
s
, with sgn(x) the sign
of x, use f(x) = xe
x
2
, and now use your 1337 Fourier Transform sk1llz to
check that in the rst case, when c(x) = [x[
s
we have (f, c) =
s/2
(s/2) and
(

f, c) =
(1s)/2
((1s)/2), so the ratio is meromorphic and furthermore we
have seen the ratio before! The ratio shows up when writing (1 s)/(s) as a
product of simpler functions (i.e. the fudge factors in the functional equation).
So now youre beginning to see some of the insights herethe fudge factors in
the functional equation may have local explanationsfor example the factor
is coming from the archimedean valuation on Q. If you look at the functional
equation for the zeta function for a number eld, you will see several factors,
coming from the real and complex norms on the eld, and one can check that
they are the same factors that come up in these calculations.
The answer on the sgn(x)[x[
s
component is similar, but one ends up with

s+1
2
(
s+1
2
), which is precisely the fudge factor that one has to use in the
functional equation for the Dirichlet L-function when (1) = 1.
If K = C then the components are parametrized by the integers. Lets
say the nth component is the quasicharacters whose restriction to S
1
C is
z z
n
. For n 0 Tate chooses the function f
n
(x+iy) = (xiy)
n
e
2(x
2
+y
2
)
,
and for n 0 he chooses f
n
(x + iy) = (x + iy)
n
e
2(x
2
+y
2
)
. It turns out
that

f
n
= c
n
f
n
where c
n
is an explicit root of unity (proof by basic integrals
and induction on n) and the local zeta values are again just powers of and
functions, for example if n 0 then (f
n
, re
i
r
s
e
in
) = (2)
1s+
n
2
(s +
n
2
)
and the other answers are similar. For an explicit list of the answers, look at
the end of chapter 2 of Tates thesis or the new example sheet.
Summary of what I just breezed through:
All local zeta functions have meromorphic continuations. The local zeta
functions attached to our favourite functions (the fs we used) looked like (1
p
s
) on the unramied non-arch components and involved the function and s
60
in the real and complex cases. These local factors are precisely what one multi-
plies together to get the function (s) (the Riemann zeta function multiplied by
the fudge factors at innity). The local zeta functions on the ramied com-
ponents in the non-arch case are messier, but the ratio (f, c)/(

f, c) involves
Gauss sums.
And let me stress once more that these local calculations do not even come
close to analytically continuing the usual zeta function; we need more to do this.
Chapter 5. The adeles and ideles.
The Pontrjagin dual of Z (with the discrete topology) is R/Z. But the
Pontrjagin dual of Q (with the discrete topology) turns out to be an absolutely
huge uncountable compact topological group, rather surprisingly! The dual
turns out to be related to some kind of innite product of all the completions
of Q at once, as we will see later on. But we have to be careful here: if I have
innitely many non-empty locally compact topological spaces X
i
, their product
turns out not to be locally compact in general (because the denition of the
product topology has, as basic open sets, products of open sets U
i
, but all but
nitely many of the U
i
have to be equal to X
i
and this makes it hard to nd
a compact neighbourhood of such a product). So we have to be carefulthe
product over all p of Q
p
isnt locally compact and hence we cant do Haar
integration on it.
Heres a partial x:
Lemma. If we have a collection X
i
of locally compact Hausdor topological
groups, and furthermore if all but nitely many of them are compact, then the
product of the X
i
is a locally compact Hausdor topological group.
Proof. Given a basic open neighbourhood

i
U
i
of a point (x
i
) in the prod-
uct, all but nitely many of the U
i
are equal to X
i
by denition, and are hence
compact, so we leave them alone, and the rest of the U
i
we can shrink to V
i
,
a compact neighbourhood of x
i
, and the product of the V
i
is a compact neigh-
bourhood of (x
i
) in

i
U
i
. So the product (with its product topology) is locally
compact, and the rest is easy (checking hausdorness, and that multiplication
and inverse are continuous).
The problem we now face is that the completions of Q with the p-adic and
real norms are all locally compact, but none of them are compact. Here is the
abstract construction that gets around this.
5.1: The restricted direct product.
Heres the set-up. We have a set I (typically innite), a locally compact
Haudsor topological group G
i
for all i I, and, for all but nitely many i, a
given xed subgroup H
i
of G
i
which is both open and compact. Say S
0
is the
nite subset of I for which no H
i
is given. Say S is any nite set containing S
0
.
Then we can form G
S
:=

iS
G
i

iS
H
i
; this is locally compact (with the
product topology) and, as a set, sits naturally inside

i
G
i
. But no one nite
set S is better than any other, so we now take the union (within

i
G
i
), as S
gets bigger, of the G
S
. Call this union G.
Then G is clearly a group (its a directed union of groups; G
S
G
T
S
ST
).
To make it a topological group we just have to give a basis for the topology near
the identity, and we can do this by choosing any S S
0
and saying that a basis
61
of neighbourhoods of the identity in G is just a basis of neighbourhoods of the
identity in the subgroup G
S
. Its an elementary exercise to check that this
independent of S (check that a basis of neighbourhoods is given by

i
N
i
with
1 N
i
G
i
and N
i
= H
i
for all but nitely many i) and that this construction
makes G into a locally compact topological group.
If all the G
i
are furthermore abelian then we have
0

iS
0
H
i
G (
iS
0
G
i
/H
i
) (
iS
0
G
i
)
so G is a sort of mixture of a direct product with a direct sum.
Note that each G
i
is naturally a subgroup of G. An element of G can be
thought of as an element (g
i
) of

i
G
i
with the property that g
i
H
i
for all
but nitely many i.
Notation:
G =

G
i
.
Not very good notation, because it doesnt say what the H
i
are. Rotten luck.
From now on, assume that all the G
i
are abelian.
The following things are all elementary to check and I will only hint at proofs.
[Reminder: G

i
G
i
is the (g
i
) with g
i
H
i
for all but nitely many i]
1) If c : G C

is continuous, then c
i
:= c[G
i
is trivial on H
i
for all
but nitely many i, and hence one can make sense of the character

i
c
i
on G
(because its a nite product) and one can check that

i
c
i
= c. [Proof: because
c is continuous, if V is a small neighbourhood of 1 then c
1
(V ) is open in G
and hence contains a subgroup of the form

iS
H
i
; but c(

iS
H
i
) is now a
subgroup of V and for V small enough the only subgroup is 1].
2) If H
i
is a compact open subgroup of G
i
(note that open implies closed,
because H
i
is the complement of the open set
gH
i
gH
i
) and if we dene H

i
to be the annihiliator of H
i
in

G
i
, then H

i
is also compact and open. [Proof:
the dual of H
i
is

G
i
/H

i
so compactness of H
i
implies discreteness of

G
i
/H

i
implies openness of H

i
etc].
3) The Pontrjagin dual of G =

i
G
i
(restricted product with respect to the
H
i
) is

G =

G
i
(restricted product with respect to the H

i
). [Proof: weve
seen that a unitary character c of G is a product of its components, and that
conversely given a bunch of c
i
all but nitely many of which are trivial on H
i
we can multiply them together to get a c, and now one just unravels this.]
4) Say S S
0
, so G
S
=

iS
G
i

iS
H
i
makes sense and is a LCHTG.
If we choose a Haar integral
i
on each G
i
(i S) and on each H
i
(i , S) and
we normalise the H
i
ones such that (1) = 1 (where 1 is the constant function
on H
i
, which is in /(H
i
)), then theres a unique natural Haar measure

i

i
on G
S
, with the property that if N
i
is a subset of G
i
for all i with the property
that N
i
= H
i
for all but nitely many i and that

i
N
i
G
S
, and if
N
i
is
summable for all i, then (
N
) =

i

i
(
N
i
) [this is only a nite product of
course].
5) Hence if we x Haar measures
i
on each G
i
with the property that

i
(
H
i
) = 1 for all but nitely many i, we get a natural Haar integral on G,
62
given by
(f) = lim
S
(f[
G
S
)
for any f /(G), the limit being taken over all S S
0
, and this limit will exist
(because the support of f will be contained within one of the G
S
, so in fact the
sequence is ultimately constant).
6) (extension of 5 to summable functions). Say, for each i, we have a
summable function f
i
on G
i
with the property that f
i
[H
i
= 1 for all but nitely
many i. Dene a function f on G by f((g
i
)) =

i
f
i
(g
i
) (a nite product!).
Then the integral of f[G
S
is just

iS

i
(f
i
) and if the innite product con-
verges absolutely (for example if f =
H
i
for all but nitely many i), we will
have (f) =

i
(f
i
).
7) If for each i we have a continuous summable f
i
: G
i
Cwith the property
that

f
i
:

G
i
C is also continuous and summable, and if furthermore f
i
=
H
i
for all but nitely many i, then f =

i
f
i
makes sense (and is a nite product
wherever it is evaluated), its continuous and summable, and

f :

G C is just

f
i
, which is also continuous and summable.
8) Finally, if we x Haar integrals on G
i
and

G
i
for all i with the property
that the integrals are self-dual (that is

f(x) = f(x), so the positive constant

that may be involved is in fact 1 for each i) and if
i
(H
i
) = 1 = (H

i
) for all
but nitely many i, then the product Haar integrals are also self-dual.
Recall last time: we had a collection G
i
(i I) of locally compact Hausdor
topological groups, a nite set S
0
I, and, for all i , S
0
(so, for all but nitely
many i), we had a compact open subgroup H
i
of G
i
.
Given this data we can form G :=

i
G
i
, the restricted product of the G
i
with respect to the H
i
. As a group its the elements (g
i
)

i
G
i
such that g
i

H
i
for all but nitely many i (where this nite set is allowed to vary). The easiest
way to think about the topology is to realise that G
S
0
:=

iS
0
H
i

iS
0
G
i
is an open subgroup, with the usual product topology on it. It turns out that
G is also locally compact and Hausdor, its Haar measure can be thought of as
the product of the Haar measures on G
i
as long as these are normalised such
that (H
i
) = 1 for all but nitely many H
i
, and the Pontrjagin dual of G is
just the restricted product of the

G
i
with respect to H

i
, the annihiliator of H
i
in

G
i
.
In fact we only need two examples for Tates thesis and in both cases the G
i
(and hence G) will be abelian.
5.2: The adeles and ideles.
Let k be a number eld, so a nite extension of Q. [The theory works just
as well for function eldsthat is, nite extensions of F
p
(T), but Id like to
emphasize the number eld case, especially as I was too lazy to nish the proof
of the meromorphic continuation of the local zeta functions in the function eld
case!]
Let I be the following set: theres an element of I for each non-zero prime
ideal P of R, the algebraic integers in k, and theres also an element of I for
each equivalence class of eld homomorphisms : k C, with .
63
Recall from ages ago that each element of I gives us an equivalence class of
norms on k; the prime ideals P give us P-adic norms, and the maps k C give
us norms induced from the standard norm on C.
The elements of I are called places of k, and a typical element of I is tradi-
tionally denoted v (for valuation, I guess, which is another word for norm). For
each v I let G
v
denote the completion k
v
of k with respect to the norm in-
duced by v. Let S
0
denote the norms coming from k Cthese are called the
innite places [this set is empty in the function eld case, and nite but non-
empty in the number eld case]. For v , S
0
(a nite place) the completion
k
v
= k
P
of k has a ring of integers R
v
; let this be H
v
.
Dene the adeles of k, written A
k
, to be the restricted product of the k
v
with respect to the R
v
.
Lets write this out explicitly in the case k = Q: we have A
Q
is the subgroup
(in fact its easily checked to be a subring) of
Q
2
Q
3
Q
5
. . . R
consisting of (g
2
, g
3
, g
5
, . . . , g

) with the property that g

p
Q
p
for all p, and
g

R, and, crucially, that g

p
Z
p
for all but nitely many p.
Indeed in the general case one can easily check that the topological group A
k
has a natural ring structure induced by componentwise multiplication (because
H
i
is a subring of G
i
for all nite places i).
Thats the rst construction we will use. As you can see, the nite and the
innite places are behaving quite dierently (the innite places have no H
v
) and
its common to write
A
k
= A
f
k
k

with A
f
k
the nite adeles, namely

P
k
P
, the restricted product over all
the nite places, and k

the innite adeles, namely the nite product

[]
k

,
with [] = , the equivalence class of , and where k

= R if : k R and
k

= C if [] = , with : k C with image not landing in R.
An absolutely crucial observation is that the diagonal map k

v
k
v
sending to (, , , . . .) has image landing in A
k
; this is because any k
can be written = a/b with a, b R, the integers of k, and b ,= 0, and the
factorization of (b) into prime ideals only involves nitely many prime ideals of
R, and if S is S
0
union this nite set then H
v
= R
v
for all v , S.
Thats the rst construction we will use. The second is the ideles of k, which
Ill denote A

k
, and which is the restricted product of the k

v
with respect to
the R

v
. This is a topological group. As the name indicates,
Lemma. The ideles are the units in the ring of adeles.
Remark. Note that this is an algebraic statement; it says nothing about the
toplogies of the adeles or ideles.
Proof. If (g
v
) A
k
has an inverse, then certainly all of the g
v
are non-
zero and the inverse is (g
1
v
). For both (g
v
) and (g
1
v
) to be in A
k
we need
g
v
R
v
for almost all v (n.b. almost all means for all but nitely many)
and g
1
v
R
v
for almost all v. This means g
v
R

v
for almost all v, which is
precisely the assertion that (g
v
) is an idele.
64
Historical note: ideles were invented/discovered before adeles. Ideles were
introduced by Chevalley, and he actually called them ideal elements, which
he abbreviated id.ele. which became idele. It was later realised that they
were the units of a ring, which Tate calls the ring of valuation vectors in his
thesis.
It was Weil that introduced the terminology adele, for additive idele. If
you look at Serres CV (for example at the beginning of Vol. 1 of his collected
works) youll see that his mothers name was Adele, but Serre once told me
that he had nothing to do with the introduction of the terminology, and merely
found it ironic that his mothers name ended up being used in mathematics.
Pedantic/irrelevant remark (which we wont use later). The inclusion A

k

A
k
is continuous (because a basic open neighbourhood of the element (1, 1, 1, 1, . . .)
in A
k
is

v
N
v
with N
v
= R
v
for all but nitely many v, and hence its pullback
to A

k
will contain

v
M
v
with M
v
= R

v
for all but nitely many v). How-
ever the inclusion is not a homeomorphism onto its image; the problem is that

v<
R

v|
K

v
is open in the ideles but not in the subspace topology
(because any neighbourhood of 1 in the adeles will contain elements of the form
(1, 1, 1, 1, . . . , 1, , 1, . . . , 1)
(with in the vth place and a uniformiser in k
v
), for all but nitely many
v. The way to x this up turns out to be the trick I mentioned earlier: give
A
k
A
k
the product topology and embed A

k
into this product by sending u
to (u, 1/u); now the restricted product topology on A

k
is indeed the subspace
topology.
An absolutely crucial function on the ideles of a number eld is the norm
function. For any completion k
v
of a number eld we have written down a
canonical norm (the one where the norm of is how much an additive Haar
measure is stretched under multiplication by ). Lets call this norm [.[
v
now.
Note that for v nite and u
v
R

v
we have [u
v
[
v
= 1. Hence there is a function
[.[ : A

k
R
>0
dened by
[(g
v
)[ =

v
[g
v
[
v
with the usual remark that, for any given v, this is a nite product. Ill
refer to this function as the global norm but note that its a continuous group
homomorphism rather than a norm on a eld in the sense we talked about
earlier.
Unsurprisingly, given that a Haar integral on A
k
can be thought of as a
product of local Haar integrals, it turns out that this norm on A

k
is just the
factor by which multipliaction by an idele is stretching the additive Haar integral
on the adeles.
Our goal, of course, is to develop some machinery to work with the following
sort of idea. Let me just stick to the case k = Q. Lets dene a function on the
ideles A

Q
thus: for p a prime number, dene f
p
on Q
p
to be the characteristic
function of Z
p
. Dene f

on R to be e
x
2
. Dene f : A

Q
C by f((g
v
)) =
65

v
(f
v
(g
v
)) (a nite sum). Now consider the function
s
_
A

Q
f(x)[x[
s
d

(x) (1)
where

denotes the Haar measure on the ideles which is the product of the
local Haar measures

on Q

p
and R

. This integral will not converge for a

general s C; the integrand isnt L
1
. A sucient condition for the integrand to
be L
1
is that all the local integrands are L
1
and furthermore that the product of
the local integrals is absolutely convergent. But we already worked these local
integrals out, at least at the nite places: at the nite places we have
_
Q

p
f
p
(x)[x[
s
p
d

(x)
and when we were meromorphically continuing local zeta functions we checked
that this was L
1
for Re(s) > 0 and that its value was

j0
p
js
= (1 p
s
)
1
.
At the innite place, I skipped the calculation so lets do it now: we need to
compute
_
R

e
x
2
[x[
s
(dx/[x[)
= 2
_

0
e
x
2
x
s1
dx
and setting y = x
2
this is

1
_

0
e
y
(y/)
s2
2
dy
=
s/2
(s/2)
by denition of the function, if Re(s) > 0 (and the integral doesnt converge
absolutely at zero if Re(s) 0). Hence a necessary and sucient condition for
the adelic integral (1) to converge is that

p
(1 p
s
)
1
converges absolutely,
and for Re(s) > 1 this will be the case because the product is just

n1
n
s
=
(s). So for Re(s) > 1 the adelic integral (1) will converge, and it will converge
to
(s) :=
s/2
(s/2)(s).
I proved in the second lecture that (s) had a meromorphic continuation to
s C and satised (s) = (1 s). We now have an adelic interpretation of
the statement.
If we can also give an adelic proof of (s) = (1 s), by interpreting our
original proof adelically, one might hope that the idea will generalise to all
number elds. Indeed, our main theorem will be the meromorphic continuation
of a wide class of integrals on idele groups, and we will recover a theorem of
Hecke whose original proof was a real tour de force.
We have chosen Haar measures on k
v
and isomorphisms k
v
=

k
v
in such
a way that the Fourier inversion theorem on k
v
is true on the nose (the fudge
66
factor constant is 1). For each v our map k
v


k
v
was of the form x (y
e
2i
v
(xy)
) where
v
, which we called at the time, was some explicitly given
map k
v
R/Z. For Q
p
it was Q
p
Q
p
/Z
p
Q/Z R/Z, the middle map
being called q. For k
P
/Q
p
nite it was the trace map k
P
Q
p
followed by
the above map. For the reals it was x x sending R to R/Z and for the
complexes it sent (x +iy) to 2x.
Note that for all nite v, we see that R
v
is in the kernel of
v
. So, by the
usual trick, we get a map
: A
k
R/Z
dened by
((g
v
)) =

v
(g
v
)
which is, as usual, a nite sum. I could now cheat and say that there was an
induced map
A
k


A
k
sending x to y e
2i(xy)
, which was a restricted product of isomorphisms, and
is hence an isomorphism. But let me make a very pedantic remark: this last
statement is true, but not completely formal: something needs to be checked.
The problem is that

A
k
is the restricted product of the

k
v
with respect to the
R

v
, the annihiliators of R
v
. [Reminder: for G an abelian LCHTG and H a
closed subgroup, H

G is the characters of G which are trivial on H.]
Hence to make sure that we really do get a continuous map A
k


A
k
this
way, it would suce to check that our xed local isomorphisms k
v
=

k
v
sent
R
v
to R

v
for all v. But they dont! By denition, R

v
is the characters of k
v
that vanish on R
v
, whereas our local isomorphism sends r R
v
to the function
y e
2i
v
(ry)
, so maps R
v
to the functions which vanish on x k
v
:
v
(xy) =
0y R
v
and this is the inverse dierent of k
v
, which is not always equal
to R
v
. However, an explicit calculation shows that if v is unramied in the
extension k/Q then this inverse dierent is R
v
again (this calculation would
take me too far aeld at this point, unfortunately), and hence R
v
becomes
identied with R

v
for all but nitely many v, which is good enough to ensure
that we get an isomorphism A
k


A
k
this way.
Hence for f L
1
(A
k
) we can (using our xed choice of normalisations of
Haar integrals and our xed map A
k


A
k
) consider its Fourier transform as
a function on A
k
again. Explicitly

f(x) =
_
f(y)e
2i(xy)
d(y).
And because our local Fourier transforms satised Fourier inversion on the nose,
we check (by using a non-zero test function which is a product of L
1
functions
on the factors) that

f(x) = f(x)
for f L
1
(A
k
).
67
Let me nish this chapter with some comments on the relationship between
the adeles of a number eld and the adeles of a nite extension of this eld. Ill
stick to the case of k/Q but what I say is true for general extensions.
We showed, when analysing extensions of norms to nite eld extensions,
that a given norm [.[ on the bottom extends in at least one, but at most nitely
many ways to a norm on the top. We showed something more precise, in fact
we showed that if L/K was a nite extension of elds of characteristic zero (or
more generally a nite separable extension), and [.[ was a norm on K, and

K
was its completion (note: this hat has nothing to do with Pontrjagin duality),
then L
K

K was a nite sum of elds, and these elds were precisely the
completions of L at the norms on L which extend [.[.
Applying this to the extension k/Q, we nd that k
Q
Q
p
will be isomorphic
to the direct sum of all the completions of k at all the norms extending the p-
adic norm on Q, and one can re-interpret the classical result
i
e
i
f
i
= [k : Q]
(with (p) =

i
P
e
i
) as simply saying that these extensions must just be the
P-adic norms for p P.
[Alternatively one can prove this directly, as is done in Cassels book, and
then derive this formula

i
e
i
f
i
= [k : Q] from it; there is a little work to be
done here though, which I wont do]. The upshot is that
k
Q
Q
p
=
pP
k
P
and the analogous result at innity is that
k
Q
R =
[]
k

.
Now the closure of R, the integers of k, in
pP
k
P
, is just its completion in
each component, which is

P
R
P
, and from this it follows that
A
k
= A
Q

Z
R = A
Q

Q
k.
More generally one checks that for L/k a nite extension of number elds, the
same proof gives that A
L
= A
k

k
L.
One can also deduce from these decompositions that traces and norms can
be computed locally. For example
Tr
k/Q
() = Tr
k
Q
Q
p
/Q
p
() =

pP
Tr
k
P
/Q
p
()
and similar results for norms, and similar results at the innite places too.
Chapter 6: The main theorem.
As you have surely realised by now, our strategy is as follows. Were going
to dene global zeta integrals as integrals of f(x).[x[
s
on the ideles, for f
carefully-chosen functions. We are going to use things weve proved in the
course to meromorphically continue these functions to all s C. In the local
case these meromorphic continuation proofs were of the form check it for one f
and deduce it for all f by some trick involving Fubinis theorem. In the global
setting the result is deeper and we will obtain our meromorphic continuation
from some adelic version of Poisson summation.
68
Recall that the crucial fact in the proof of the meromorphic continuation of
the Riemann zeta function was that (1/t) = t(t) for some function , and the
proof of that latter fact came from some concrete form of the Fourier inversion
theorem, which was just the statement that the Fourier series of a periodic
function F(x) did in fact converge to F(x).
Tates insight, which has run and run, is that in this adelic setting, the
correct analogue of the set R/Z is the set A
k
/k. Let me run o a few things we
know about the inclusion Z R. Firstly, Z is discrete, R is locally compact,

R
(the Pontrjagin dual) is isomorphic to R again, and if we use the isomorphism
x (y e
2ixy
) to identify R with

R then we see that the annihilator Z

of
Z (that is, the elements r R such that e
2irn
= 1 for all integers n) is just
Z again.
Hence the Pontrjagin dual of the discrete group Z is the compact group
R/Z, and the dual of the exact sequence
0 Z R R/Z 0
is itself. Finally the action of Z on R admits a natural fundamental domain
(that is, a subset D of R, namely [0, 1), with the property that the induced map
D R/Z is a bijection), and the measure of D, with respect to the standard
Haar measure on R, is 1.
Were going to prove analogues of all of these things today, with Z replaced
by a number eld k, and R replaced by A
k
. For example well soon see that
k embeds into A
k
as a discrete subgroup. So what will be the analogue of our
proof of the functional equation of the theta function?
When thinking about the function, we obtained our function F(x) orig-
inally as F(x) =

nZ
f(x + n), with f(x) = e
t
2
x
2
a function on R. The
analogue of f in this setting will be a carefully-chosen function on A
k
which
is suciently rapidly decreasing, and such that for all adeles x, the sum

k
f(x + ) converges absolutely. We then apply Fourier inversion to get
some fact, and show that this fact is precisely what is needed to give us the
meromorphic continuation and functional equation of the Riemann zeta func-
tion and a gazillion other functions too, all of which come out in the wash.
Historical interlude (non-examinable).
The theory of automorphic forms was really getting o the ground in the
1950s, when Tates thesis was written, but the classical theory tended to revolve
around considering functions on groups like GL
n
(R) which were invariant, or
transformed in some simple way, under the subgroup GL
n
(Z). In the 1950s
there was a move away from this setting to the adelic setting of functions on
GL
n
(A
k
) which were invariant under the discrete subgroup GL
n
(k), and this
insight enabled one to reformulate various notions such as Hecke operators in a
purely local form. Indeed, Hecke operators could now be interpreted as operators
in a purely local setting coming from the representation theory of GL
n
(k
v
),
giving a huge new impetus to the representation theory of p-adic groups.
There were practical consequences too in that the theory of Hecke operators
for Hilbert modular forms was very dicult to set up globally, if the integers of
69
the base eld were not a PID, because no natural analogue at P of the matrix
_
p 0
0 1
_
GL
2
(Q) existed if P was a non-principal prime. The adelic reformula-
tion of the theory removes all of these problems because even though P is not
a principal ideal, the element (1, 1, 1, 1, . . . , 1, 1, . . .) (with a uniformiser at
P) is still a perfectly good idele (indeed this was one of Chevalleys motivations
for introducing these things).
6.1. The additive theory, and the adelic Poisson summation for-
mula.
Lets prove that Z is to R as k is to A
k
. Heres the rst big reason for
believing this:
Proposition. The subspace topology on k coming from the embedding
k A
k
is the discrete topology (all sets are open). And the quotient A
k
/k is
a compact topological space.
Well prove this soon; rst well construct a fundamental domain for k in
A
k
, analogous to [0, 1) in R. Lets do this by trying to understand how k ts
into A
k
at the nite places, and then thinking about the innite places.
Consider the group that I called G
S
0
when setting up the general theory of
restricted products: this is just

v<
R
v

v|
k
v
.
The intersection of k (embedded diagonally) with this group is the elements
of k which are integers at all nite places. If 0 ,= k and we write the
fractional ideal () as

i
P
e
i
i
, and if one of the e
i
is negative, then we have
, R
P
i
, by denition. Hence the intersection
k
_
_

v<
R
v

v|
k
v
_
_
is just the elements of k which generate integral ideals, which is just another
way of saying the (global) integers R of k.
Now lets think about whats going on in

[]
k

, the innite adeles. Note

that this space just looks like RR. . . RCC. . . C, where there
are, say, r copies of R, and s copies of C, and we also note that the number
of eld homomorphisms k C is just r + 2s (recalling that we only get one
completion for each pair of complex conjugate maps k C).
Now if e
1
, e
2
, . . . , e
n
is a Z-basis for the ring of integers R in k, then the de-
nition of the discriminant of k is just (up to sign) the square of the determinant
of the square matrix (
i
(e
j
))
i,j
, where
i
runs through the eld maps k C.
Let us write [d[ for the absolute value of the discriminant of k/Q. For later use
it will be helpful to know
Lemma. The image of R in k

[]
k

(embedded diagonally) is a lattice,

and, with respect to our choices of Haar integrals on the k

, the measure of a
fundamental domain for this lattice is just
_
[d[, with d the discriminant of k.
Remark. A fundamental domain for a lattice R
n
is just a connected
set S with non-empty interior such that every element of R
n
can uniquely be
70
written +s with and s S. One way of constructing such a thing is to
write down a basis e
1
, e
2
, . . . , e
n
for and let S be

i
e
i
with 0
i
< 1
for all ia fundamental parallelogram for .
Proof of lemma. If k is totally real (that is, all k C land in R) then the
result is immediate: the volume of the fundamental domain of a lattice in R
n
is
just the absolute value of the matrix whose entries form a basis for the lattice.
But if k has complex places then we have to be a little careful.
The problem is that if is a map k C whose image does not land in R,
and if (e
j
) = x +iy, then in the usual discriminant calculation (which uses all
embeddings, both and ) we will see a contribution from x + iy and x iy.
But in the innite adele computation we only see , taking values in something
we can thinking of as R
2
, giving us coordinates of x and y. Now we have
_
x +iy
x iy
_
=
_
1 i
1 i
__
x
y
_
and the absolute value of the determinant of
_
1 i
1 i
_
is 2.
So with respect to the naive measure on the innite adeles, (which is dxdy
at the complex places) the volume of a fundamental domain for the lattice R
will be
_
[d[.2
s
, because we lose a factor of 2 at each complex place. However
the normalisation of Haar measure that we chose for the complex innite places
was not the naive onewe inserted a factor of 2! Hence with our xed choice
of Haar measure the volume is again
_
[d[.
Now its convenient to make the following denition. Dene D

to be the
following fundamental parallelogram for the lattice R in k

v|
k
v
: choose
a Z-basis (e
j
)
1jn
for R and consider the e
j
as elements of k

; they form a
lattice (because the discriminant of a number eld is non-zero!). Dene D

to
be the box whose typical element is

n
j=1

j
e
j
with 0
j
< 1. Note that
the closure D

of D

is obtained by letting the

j
range through [0, 1], and
the interior D
o

is obtained by restricting to
j
in (0, 1). In particular D

has
compact closure and non-empty interior.
Now lets dene D A
k
to be the product D
f
D

, with D
f
A
k,f
simply being

v
R
v
. Note that D
f
is an open subgroup, and hence a closed
subgroup, of A
k,f
, and hence the closure of D in A
k
is simply D
f
D

, which
is compact, and the interior is D
f
D
o

. I now claim that D is a fundamental

domain for the action of k on A
k
. More precisely,
Lemma. Any element of A
k
can be written uniquely as d + with d D
and k.
Proof. Given an adele (g
v
) of k, it is in R
P
at all but nitely many nite
places P, by denition. Choose some 0 ,= b R whose prime factorization
contains a suciently high power of each of the P for which g
P
isnt integral
to ensure bg
P
R
P
for all P. Now for each prime ideal P dividing (b), say P
e
exactly divides b and consider the equation a bg
P
modulo P
e
. By the Chinese
Remainder Theorem these equations can all be solved at once within R, and we
set
0
= a/b k. Now g
P

0
R
P
for all P[(b) and hence for all P, because
g
P
and a/b are integral at all other nite places.
71
We have now rigged it so that (g
v
)
0
has nite part in D
f
, but its innite
part might not be in D

; however this can be xed by subtracting an appro-

priate
1
R (because D

is visibly a fundamental domain for R acting on

k

).
We see that we have written A
k
= D+k now, and all that is left is to prove
that this decomposition is unique. But this is easy: if d
1
+
1
= d
2
+
2
then
t := d
1
d
2
=
2

1
(D D) k, and looking at the nite places we see
t k is in D
f
D
f
= D
f
is integral at all nite places, so t R, and looking
at the innite places we see t = 0 because 0 is the only element of R =

i
Ze
i
in D

n
i=1

i
e
i
: 1 <
i
< 1.
We can now prove something promised earlier:
Proposition. k A
k
is discrete and the quotient is compact.
Proof. Discreteness follows because D has a non-empty interior. More pre-
cisely, if d is any adele in the (non-empty) interior D
o
of D then D
o
d is an
open set in A
k
containing 0, and conversely if k is in D
o
d then we have
d

d = for d, d

D and hence d

= d + , so = 0. Hence D
o
d is an
open set in A
k
whose intersection with k is just 0 and hence for any k,
D
o
d + is an open set in A
k
whose intersection with k is .
Compactness follows because A
k
/k is a continuous image of D; the lemma
implies that the map is surjective.
Now lets prove A
k
/k-analogues of the other R/Z-results we mentioned ear-
lier.
Proposition. The measure of (the characteristic function of) D (with re-
spect to our xed choice of Haar measuse on A
k
) is 1.
Proof. D = D
f
D

. We computed the measure of D

as
_
[d[. The way
we normalised our local Haar measures at the P-adic places was such that if R
P
is the integers of k
P
then (R
P
) = p
m/2
, where p
m
was the (absolute value
of the) discriminant of k
P
/Q
p
. But the global discriminant of k/Q is just the
product of the local discriminants, and hence the measure of D
f
with respect to
our choices is [d[
1/2
. Hence the measure of D is the product of the measures
of D
f
and D

, which is 1!
Proposition. Our xed isomorphism A
k


A
k
(dened by x (y
e
2i(xy)
)) sends the closed subgroup k isomorphically onto the closed subgroup
k

of characters of

A
k
which are trivial on k.
Reminder. Our xed map R

R sends x to y e
2ixy
, so sends Z to
the characters y e
2iny
for n Z. The R/Z analogue of this proposition
is the statement that the intersection of the kernels of all of these characters is
precisely Z again.
Proof of proposition. We need to check that the set of characters y
e
2i(ry)
, for r k, is precisely the set of characters that vanish on k. So we
need to check
(i) If k then () = 0
(ii) If y A
k
and (y) = 0 for all k then y k.
Recall ((g
v
)) =

v

v
(g
v
), a nite sum, and the
v
are trace down to
Q
p
or R, and then use q : Q
p
/Z
p
Q/Z or x x : R R/Z.
72
(i) is true because is a sum of local traces, and if k then Tr
k/Q
() Q,
and this reduces (i) to the case k = Q. Its clearly true that (n) = 0 for n Z
(because all the
p
are zero), so by additivity it suces to check that (1/p
e
)
vanishes for all p prime and e 1. Now nally I realise why Tate inserted the
minus sign in his denition of his local for the reals: we have
q
(1/p
e
) = 0
for all q ,= p, we have
p
(1/p
e
) = 1/p
e
and we have

(1/p
e
) = 1/p
e
, and
the sum in R/Z is zero. So (i) is proved.
For (ii) we use a trick. We have proved A
k
/k is compact, so if k

denotes the
annihilator of k in

A
k
, and if we identify

A
k
with A
k
via our xed isomorphism,
then we know k k

from (i). Now k

is the annihilator of k and hence the

Pontrjagin dual of A
k
/k which weve seen is compact. Hence k

is discrete, and
a closed subgroup of A
k
. So k

/k is discrete in A
k
/k, and closed too, so its
compact, so its nite.
So for k

theres some positive integer n such that n = k. But

/n k and /n is now a torsion element of A
k
, which contains no torsion
other than zero. So = /n k.
So now we really see that the inclusion k A
k
is very formally similar to
the inclusion Z R. In particular we see that the Pontrjagin dual of A
k
/k
is k

= k (with the discrete topology), and hence the Pontrjagin dual of k as

A
k
/k (analogous to the Pontrjagin dual of Z being R/Z). But in some sense
the advantage of the adeles over the reals is that the adeles can be broken
into local factors, and the arithmetic of k is easier than the arithmetic of its
integers.
Now we prove the analogue of the transformation property of the function.
Instead of working with (the analogue of) f(x) = e
t
2
x
2
we set things up, for
the time being at least, in more generality: well use a general function f for
which well just assume everything converges.
First we observe that we have a natural Haar measure on the compact group
A
k
/k: a function on A
k
/k can be thought of as a periodic function on A
k
(that is, one satisfying f(x+) = f(x) for k) and, for a continuous function
of this type, one checks easily that dening (f) =
_
D
f(x)d(x) where is our
xed Haar measure on the adeles but the integral is only over our fundamental
domain D, gives us a Haar measure on A
k
/k.
On the other side, if we endow k with the discrete topology then a natural
Haar measure is just counting measure: a continuous function with compact
support is just a function f : k C which vanishes away from a nite set, and
we can dene (f) =

k
f(). With these choices of Haar measure on k and
A
k
/k, what is the constant in the Fourier inversion theorem? In other words,
if we invert F : k C twice, well get x cF(x). What is c?
Lemma. c = 1.
Proof. We just need to check this for one non-zero function. So lets let
F be the characteristic function of 0. Then

F is the function on

k = A
k
/k
which sends a character : k S
1
to

k
F()() = (0) = 1. Hence

F is
the constant function on A
k
/k, sending everything to 1. Now we dont have to
evaluate

F everywhere, we only need to evaluate it at 0, regarded as the trivial

73
character of A
k
/k. By the choice of our Haar measure on A
k
/k, we see

F(0) =
_
D
1d(x)
and we computed the integral of D as being 1, so

F(0) = F(0), and hence c = 1.

(new lecture) Summary of where we are:
1) k a number eld. We have xed an identication of A
k
with its Pontrja-
gin dual

A
k
. We checked that this isomorphism sends the closed (and discrete)
subgroup k of A
k
isomorphically onto its annihiliator k

(recall that the anni-

hilator of k is just by denition the elements of

A
k
which vanish on k). We
deduce from this that the Pontrjagin dual of A
k
/k is isomorphic to k (because
its canonically isomorphic to k

). We checked that k was a discrete subgroup of

A
k
and that the quotient A
k
/k was compact. I remarked (and indeed stressed)
that this was very much analogous to Z R being discrete and R/Z being
compact, the identication of R with its dual sending Z to its own annihilator,
and Z hence being the dual of R/Z.
2) We xed a choice of Haar measure on A
k
/k, namely
_
D
, where D is
our fundamental domain for the action of k on A
k
(analogous to [0, 1) in R).
This choice has the nice property that the integral of the constant function is 1
(because (D) = 1). We xed a choice of Haar measure on k with the discrete
topology, namely the counting measure. We computed enough of the Fourier
transform of the Fourier transform of the characteristic function of 0 on k to
deduce that the Fourier inversion theorem holds in this situation with constant
term equal to 1. I remarked that, much to my annoyance, I did not know the
full proof of the Fourier inversion theorem in this generality (but that it was
certainly true, because it was true for all locally compact abelian groups, its
just that the proof involves too much analysis for me.)
Now a reminder of something from long ago. The Fourier inversion the-
orem on R/Z, when unravelled, just tells us the classical fact that if F is
a continuous function on R/Z, viewed as a periodic function on R, and if
a
m
=
_
D
F(x)e
2imx
dx is its mth Fourier coecient, where m Z and
D = [0, 1), and if

m
[a
m
[ converges, then F(x) =

mZ
a
m
e
2imx
. We ap-
plied this very early on to a function F(x) of the form F(x) =

nZ
f(x + n)
where f was a function which was rapidly decreasing, and we deduced

nZ
f(n) = F(0) =

mZ
a
m
.
And we computed a
m
using this trick:
a
m
=
_
1
0

n
f(x +n)e
2imx
dx
=
_
1
0

n
f(x +n)e
2im(x+n)
dx
=
_
R
f(x)e
2imx
dx =

f(m)
74
and so

nZ
f(n) =

mZ

f(m) as long as everything convergesthis is the

classical Poisson summation fomula.
Lets now do exactly the same thing, but on A
k
/k instead of R/Z.
If F L
1
(A
k
/k) (that is, F is a function on the adeles and F(x+) = F(x)
for k, and furthermore if
_
D
F(x)d(x) < ), then lets dene

F : k C
by

F() =
_
D
F(x)e
2i(x)
d(x).
Lemma. With notation as above, if

k
[

F()[ converges, then

F(x) =

F()e
2i(x)
.
Proof. This is just the Fourier inversion theorem spelt out, together with
the fact that c = 1, which we proved last time.
Corollary. F(0) =

F().
Remark. As Ive mentioned already, Im slightly bothered by the fact that
Ive not actually proved the Fourier inversion theorem. However the proof for
R/Z is not hard, and the proof for Q
p
can be done by hand, and it looks to me
like A
k
/k is built up from things that look like this, and so I wonder whether one
would be able to give a hands-on proof, avoiding all the functional analysis
which I had to assume.
One last explicit denition: if f L
1
(A
k
) then, surprise surprise, dene

f : A
k
C by

f(y) =
_
A
k
f(x)e
2i(xy)
d(x), the usual Fourier transform,
once we have identied A
k
with its dual.
Theorem (Poisson summation, revisited.) If f L
1
(A
k
) is continu-
ous, if

k
f(x + ) converges absolutely and uniformly for x A
k
, and if

k
[

f()[ also converges, then

k
f() =

f().
Proof. (c.f. section 1.2.) Dene F : A
k
C by F(x) =

k
f(x + ).
Now by assumption the sum converges uniformly on A
k
, so F is continuous
and periodic. Hence F, considered as a function on A
k
/k, is continuous with
compact support and is hence L
1
. Moreover, for k we have (c.f. formula for
a
m
in 1.2)
75

F() =
_
D
F(x)e
2i(x)
d(x)
=
_
D

k
f(x +)e
2i(x)
d(x)
=

k
_
D
f(x +)e
2i(x)
d(x)
=

k
_
D
f(x +)e
2i((x+))
d(x)
=
_
A
k
f(x)e
2i(x)
d(x)
=

f()
[where the interchange of sum and integral is OK because the sum converges
uniformly on D, which has nite measure, and Ive also used the fact (proved
earlier) that k ker(), which I proved when showing k = k

.] Hence

k
f() = F(0)
=

F()
=

f()

6.2 The multiplicative theory.

We just showed that k A
k
was discrete, with compact quotient. Well now
show that k

k
is discrete, but perhaps one doesnt expect the quotient to
be compact, because R

/Z

= R
>0
isnt compact.
In fact heres a proof that A

k
/k

isnt compact. Recall that we have a norm

function [.[ : A

k
R
>0
, dened as a product of local norms.
Lemma. If k

then [[ = 1.
Proof. Lazy proof: A
k
= A
Q

Q
k and [.[ factors through the norm map
A
k
A
Q
(if you believe that the P-adic norms are the only norms on k
extending the p-adic norm on Q, which is true and not hard and in Cassels,
but I didnt prove it). This reduces us to the case k = Q. In this case, by
multiplicativity of the norm, we need only check the cases = 1 and = p
prime.
Now = 1 is a global unit so has local norm equal to 1 everywhere, and
= p also has global norm 1 because [[
q
= 1 for all q ,= p, [[
p
= p
1
and
[[

= p, so the product of the local norms is 1.

Remark. Its not hard to give a direct computational proof for general k.
Tate also notes that theres a pure thought proof which goes as follows: [[ is
the factor by which additive Haar measure on the adeles is scaled, and because
(D) = 1 we will have [[ = (D). But D is a fundamental domain for
76
k = k and its not hard to check now that (D) must then be (D) [consider
D =
k
(D(D+)) etc to see that fundamental domains have the same
measure.]
Now its clear that [.[ : A

k
R
>0
is surjective (its even surjective when
you restrict to one innite place), so certainly one cant hope that A

k
/k

is
compact (because it has R
>0
as a homomorphic image).
Denition. Let J be the kernel of [.[, with the subspace topology coming
from A

k
. We have dropped one factor of R
>0
going from A

k
to J. But its
enough, because
Proposition. k

is a discrete subgroup of J and J/k

is compact.
Proof. We follow the same strategy for showing k is discrete in A
k
, but well
need some standard facts about class groups and unit groups of number elds,
which of course Ill assume. In fact the proposition is equivalent to the union of
the following statements: the rank of the unit group of k is r +s 1 (with r the
number of real and s the number of complex places), the regulator is non-zero
(which comes out of the standard proof of the unit group rank statement), the
number of roots of unity in k is nite, and the class number of k is nite.
Dont take the following proof too seriously: we dont really need the precise
volumes that come out. Just believe that the proof is the same as in the
additive case, but messier.
So heres how the argument goes (c.f. the construction of D). Dene

E
f
=

v<
R

v
(A
f
k
)

. (Im putting tildes on because the

E Im about to build
wont quite be a fundamental domain). Then k

E
f
is the elements of k

that are units at all nite places, and hence when written a/b have (a) = (b);
this is just the units R

of R k. Our choices of Haar measure imply that

(

E
f
) = [d[
1/2
.
At the innite places we take logs: the product of the maps log([.[) : k

R
give us a map R

R
r+s
whose image lands in the hyperplane consisting
of vectors the sum of whose entries is zero. Now its a standard result that
the image of R

is a lattice in this hyperplane, and the kernel is the roots of

unity. Let

L

be a fundamental domain for this lattice, and we let

E

be the
pre-image of

L

in ker([.[) : k

R
>0
; then

E :=

E
f


E

has measure
[d[
1/2
.2
r
(2)
s
Reg
k
.
Explanation: the discriminant factor comes from the nite places, the 2
r
and (2)
s
coming from the units at the innite places, which were killed by the
logs, and Reg
k
is, by denition, the volume of the fundamental domain of

L

,
which is by denition the regulator of the number eld and is known to be non-
zero and nite. Moreover,

E is almost a fundamental domain for k

J. The
problems are rstly that we lost track of the roots of unity (so

E is too big by a
factor of the number of roots of unity) and secondly that we cannot multiply any
nite idele by some element of k

to put us in

E
f
(the multiplicative version
of the CRT argument fails), so

E is too small by a factor of k

(A
f
k
)

/

E
f
, and
(A
f
k
)

/

E
f
=
v<
Zv is the group of fractional ideals, so its quotient by k

is
the class group of k, which is known to be nite. One now checks that

E can
be modied a nite amount to ensure that it becomes a fundamental domain
E for k

in J, with measure [d[

1/2
.2
r
(2)
s
Reg
k
h/w with h the class number
77
and w the number of roots of unity.
As I say, dont take all those delicate numbers too seriously, but do note
that E has compact closure and non-empty interior, and that J =
k
E a
disjoint union, so k

is discrete in J with compact quotient.

Heres a nice consequence of compactness. Note that just as in the local case
we consider quasicharacters of multiplicative groups, rather than just characters.
Corollary. If c : k

k
R
>0
is a continuous group homomorphism,
then c = [.[

for some real number .

Proof. c(k

J) is a compact subgroup of R
>0
and is hence 1. So c factors
through A

k
/J which, via the norm map, is R
>0
, and now taking logs were
done, because the only continuous group homomorphisms R R are x x.
6.3: Statement and proof of the main theorem.
Denitions. If c : k

k
C

is a continuous group homomorphism

then we say its a quasi-character of k

k
. Weve just seen that [c[ : k

k

R
>0
is of the form x [x[

; dene Re(c) = . We let the set of quasi-characters

of k

k
be a Riemann surface as in the local case, by letting the component
of c : k

k
C

be c.[.[
s
: s C. Note that in this case the Riemann
surface is just an innite union of copies of the complex numbers, indexed by
the group

J of characters of J. If c is a quasi-character of k

k
then let c be
the character x [x[/c(x); note that Re( c) = 1 Re(c).
Remark. I know very little about

J.
Recall that in the local setting we had a set Z consisting of functions for
which everything converged, and dened (f, c) for f Z and c a quasi-
character with positive real part, as some sort of integral. Heres the analogy of
this construction in the global setting.
Let Z denote the set of functions f : A
k
C satisfying the following
boundedness conditions:
Firstly, we demand f is continuous and in L
1
(A
k
), and also that

f : A
k
C
is continuous and in L
1
(A
k
).
Secondly (a condition that wasnt present in the local setting), we demand
that for every y A

k
, the sums

k
f(y(x + )) and

k

f(y(x + ))
converge absolutely, and moreover the convergence is locally uniform in the
sense that its uniform for (x, y) DC for D our additive fundamental domain
and C an arbitrary compact subset of A

k
.
Thirdly, we demand that f(y).[y[

: A

k
C

and

f(y).[y[

are in L
1
(A

k
)
for all > 1 (note: this was > 0 in the local setting).
What are the reasons for these conditions? The rst two mean that we
can apply Poisson summation to f and indeed to the map x f(yx) for any
y A

k
. The local uniform convergence in the second condition is so that we
can interchange a sum and an integral at a crucial moment. The third condition
means that our global multiplicative zeta integral will converge for Re(s) > 1.
Denition. If f Z and c : k

k
C

is a quasi-character with
Re(c) > 1, dene
(f, c) =
_
A

k
f(y)c(y)d

(y)
78
(the Haar measure on A

k
being, of course, the product of our xed Haar mea-
sures

v
on k

v
).
The last condition in the denition of Z ensures the integral converges. Our
main goal is:
Theorem. If f Z then the function (f, .) is holomorphic on the Riemann
surface of quasi-characters c with Re(c) > 1, and has a meromorphic continua-
tion to all quasi-characters. Assume furthermore that f(0) ,= 0 and

f(0) ,= 0.
Then (f, .) has simple poles at the quasi-characters c(x) = 1 and c(x) = [x[,
and no other poles (and $1,000,000 attached to its zeros). Finally it satises
the (very elegant!) functional equation
(f, c) = (

f, c).
Well now start the proof of this, which of course is going to be a not-
too-tough application of everything we have. But what else do we need to
do in this course? Well the only other thing to do is to check that the theo-
rem has some contentthat is, that Z contains some non-zero functions and
that, as special cases of the theorem, we are proving the meromorphic continu-
ation of Dirichlet L-functions, zeta functions of number elds, zeta functions of
Grossencharacters,. . . .
[extra

f L
1
condition.] Before we prove the theorem let me make some
denitions and prove some lemmas. We have J A

k
, the kernel of the norm
function. Just as in the local case lets split this by nding I A

k
isomorphic
to R
>0
such that A

k
= I J. We do this by just choosing an innite place [
0
]
of k and letting I be the copy of the positive reals in k

0
. We identify I with
R
>0
so that the norm map induces the identity R
>0
R
>0
, so if
0
happens
to be a complex place then, because our complex norms arent standard, what
were doing here is letting I be the positive reals in C

but letting the map

R
>0
I be t

t.
For f Z and Re(c) > 1, we rstly break o this factor of I in the denition
of the zeta integral: we write
(f, c) =
_
IJ
f(y)c(y)d

(y)
=
_

t=0
_
bJ
f(tb)c(tb)d

(b)dt/t
=
_

t=0

t
(f, c)dt/t
where our measure on J is the one such that its product with dt/t on I gives
us

on A

k
, and the last line is the denition of
t
(f, c) :=
_
J
f(tb)c(tb)d

(b).
Lets think a little about

t
(f, c) =
_
J
f(tb)c(tb)d

(b).
We know that the integral dening (f, c) converges, by assumption, for Re(c) >
1, and hence the integrals dening
t
(f, c) will converge (at least for all t away
79
from a set of measure zero). But these integrals are very docile: for b J we
have [b[ = 1 by denition, so if Re(c) = then [c(tb)[ = [tb[

= t

is constant on
J, and hence if the integral dening
t
(f, c) converges for one quasi-character c
(which it almost always does) then it converges for all of them.
[
t
(f, c) =
_
J
f(tb)c(tb)d

(b).]
The problem, of course, is not in the convergence of the individual
t
(f, c);
its that as t goes to zero then f(tb) will be approaching f(0) and if this is non-
zero, which it typically will be, then the integral of this function over the non-
compact J might be getting very big, so
_
1
t=0

t
(f, c)dt/t will probably diverge
if, say, < 0 (because then t

is also getting big). This is the problem we have

to solve.
Note also that weve written
(f, c) =
_

t=0

t
(f, c)dt/t
and that this is one of the crucial tricks. If f =

v
f
v
with f
v
on k
v
then
we could compute the global integral as a product of local integralsbut in
applications this would just tell us that our global zeta function is a product
of local zeta functions, which will not help with the meromorphic continuation.
The insight is to compute the integral in this second way. Note that Iwasawa
independently had this insight in 1952.
In case youve not realised, let me stress that (f, c) isnt a generalisation
of the zeta function, its a generalisation of (s), that is, the zeta function
multiplied by the fudge factor at innity, and the t in the integral above is
precisely the t that we had at the beginning of section 1.3 right at the beginning
of the course. The strategy is now clear: we break the integral over t up into
two parts, one of which will converge for all c, and the other of which we will
manipulate and, by making the substitution u = 1/t and applying Poisson
summation, turn into a form which also converges.
Recall that the closure of the fundamental domain E for k

in J is compact,
so the integrals below are nite (as the integrands are continuous). Using J =
k

.E we get

t
(f, c) =

_
E
f(tb)c(tb)d

(b)
=

_
E
f(tb)c(tb)d

(b)
=
_
E
_

k

f(tb)
_
c(tb)d

(b)
where the rst equality is the denition, the second uses the fact that

is a
multiplicative Haar measure on J and that c is trivial on k

, and the third is

an interchange of a sum and an integral which is justied by our rather strong
uniform convergence assumptions on f Z and the observation that the closure
of E is a compact subset of A

k
.
80
Exactly the same argument (changing f to

f Z, c to c and t to 1/t) shows
that blah
1/t
(

f, c) =
_
E
_

f(b/t)
_
c(b/t)d

(b).
[
t
(f, c) =
_
E
_
k

f(tb)
_
c(tb)d

(b)]
Now that sum over k

looks almost like a sum over k, but rstly the term

= 0 is missing (so well have to add it in) and secondly were not summing f()
but f(tb). So well have to work out what the Fourier transform of x f(txb)
is. In other words, we need to see how the additive Fourier transform scales
under multiplication. In the application of the lemma below well have = tb.
Lemma. If f : A
k
C is continuous and in L
1
(A
k
), if A

k
is xed
and if g(x) := f(x) then g(y) =
1
||

f(y/).
Proof. An elementary computation. We have
g(y) =
_
A
k
f(x)e
2i(xy)
d(x)
and setting x

= x we have d(x

) = [[d(x) and hence

g(y) =
_
A
k
f(x

)e
2i(x

y/)
d(x

)/[[
=
1
[[

f(y/)
as required.
So now lets add in the missing = 0 term to
t
(f, c), apply Poisson
summation, and see what happens. Recall we just showed that
t
(f, c) =
_
E
_
k

f(tb)
_
c(tb)d

(b) and that

1/t
(

f, c) =
_
E
_

f(b/t)
_
c(b/t)d

(b).
Key Lemma. For an arbitrary t > 0 and c we have

t
(f, c) +f(0)
_
E
c(tb)d

(b)
=
1/t
(

f, c) +

f(0)
_
E
c(b/t)d

(b).
Proof. As weve already remarked, the formulas we have just derived for

t
(f, c) and
t
(

f, c) involve sums of k

.
The LHS of the lemma is hence what you get when you add the missing
= 0 term: its
_
E
_

k
f(tb)
_
c(tb)d

(b) (1).
Similarly the RHS is
_
E
_

f(b/t)
_
c(b/t)d

(b) (2).
So we need to show (1) = (2). The internal sum over k screams out for an
application of Poisson summation, which, when applied to the function x
81
f(txb) (were allowed to apply Poisson summation because of our assumptions
on f) gives

k
f(tb) =

(x f(txb))() =

k
1
[tb[

f(/tb).
Hence formula (1) is equal to
_
E
_

f(/tb)
_
c(tb)/[tb[d

(b)
and now making the substitution b 1/b, which doesnt change Haar measure,
this becomes
_
E
_

f(b/t)
_
c(t/b)[b[/[t[d

(b)
=
_
E
_

f(b/t)
_
c(b/t)d

(b)
which is (2)! This proves the lemma.
Were nally ready to meromorphically continue our global zeta integrals.
But before we do, lets try and gure out exactly what that fudge factor was
that we had to add to
t
(f, c) to make that argument work in that last lemma:
we added f(0) times
_
E
c(tb)d

(b).
What is this? Well if c(x) = [x[
s
is trivial on J then c(tb) = t
s
is constant
for b E (indeed, for b J), so the integral is just t
s

(E) and we computed

the measure of E earlier to be 2
r
(2)
s
hR/(w
_
[d[)its some nite non-zero
number, anyway. But if c is non-trivial on J then, because its always trivial
on k

, it descends to a non-trivial character on the compact group J/k

=E
and the integral will hence be zero (distinct characters are orthogonal). So in
fact we have
Corollary. If c is non-trivial on J and f Z and t > 0 then
t
(f, c) =

1/t
(

f, c).
Were nally ready to prove the main theorem! Ill re-state it.
Theorem. If f Z then the function (f, .) is holomorphic on the Riemann
surface of quasi-characters c with Re(c) > 1, and has a meromorphic continua-
tion to all quasi-characters. Assume furthermore that f(0) ,= 0 and

f(0) ,= 0.
Then (f, .) has simple poles at the quasi-characters c(x) = 1 and c(x) = [x[,
and no other poles, (and $1,000,000 attached to its zeros). Finally it satises
the functional equation
(f, c) = (

f, c).
Proof. For Re(c) > 1 the LHS zeta integral converges (by assumption on
f) and is holomorphic in the c variable (dierentiate under the integral). By
82
denition, (f, c) =
_

t=0

t
(f, c)dt/t, which converges by assumption for Re(c) >
1, and now we break the integral up into two parts:
(f, c) =
_

t=1

t
(f, c)dt/t +
_
1
t=0

t
(f, c)dt/t.
Now just as in the argument for the classical zeta function, I claim that
the integral for t 1 converges for all c, because the ideles tb showing up in
the integral all have [tb[ = [t[[b[ = [t[ 1 so if the integral converges for e.g.
Re(c) = 2 (which it does, by assumption, as 2 > 1) then it converges for any c
with Re(c) < 2 (because the integrand is getting smaller).
That term isnt the problem. The problem term is the integral from 0 to
1, which typically only converges for Re(c) > 1. So lets use the previous
lemma, which has some content (Poisson summation) and see what happens.
The simplest case is if c(x) ,= [x[
s
for any s (that is, c is non-trivial on J). In
this case those extra fudge factors in the previous lemma disappear, and we see
_
1
t=0

t
(f, c)dt/t =
_
1
t=0

1/t
(

f, c)dt/t
=
_

u=1

u
(

f, c)du/u
and this last integral also converges for all quasi-characters k

k
C

because u 1 so convergence again gets better as Re(c) gets smaller. Moreover

our new expression for (f, c), namely
(f, c) =
_

t=1

t
(f, c)dt/t +
_

u=1

u
(

f, c)du/u
converges for all c and makes it clear that (f, c) = (

f, c) (and that its holo-
morphic for all c not in the component [.[
s
). The proof is complete in this
case!
Were not quite nished though: we need to deal with the component
c(x) = [x[
s
, where the argument is slightly messier because we pick up fac-
tors of f(0)
_
E
c(tb)d

(b) = f(0)t
s

(E) and

f(0)
_
E
c(
1
t
b)d

(b). In this case

(writing c(x) = [x[
s
now), the extra factors well see in the calculation will be
(for Re(s) > 1)
f(0)

(E)
_
1
t=0
t
s
dt/t
= f(0)

(E)[t
s
/s]
1
0
= f(0)

(E)/s
and
_
1
t=0
(

f(0)
_
E
[b/t[
1s
d

(b))dt/t
=

f(0)

(E)
_
1
t=0
t
s2
=

f(0)

(E)[t
s1
/(s 1)]
1
0
=

f(0)

(E)/(s 1).
83
These functions (cst /s and cst /(s1)) clearly have a meromorphic continuation
to s C! So we have, for c(x) = [x[
s
with Re(s) > 1,
(f, c) =
_

t=1

t
(f, c)dt/t +
_
1
t=0

t
(f, c)dt/t
=
_

t=1

t
(f, c)dt/t +
_

u=1

u
(

f, c)du/u
+

(E)(f(0)/s +

f(0)/(s 1))
and now we really have proved the theorem because this latter expression makes
sense as a meromorphic function for all s C, the integrals are all holomorphic
for all s C, and the expression is invariant under (f, c) (

f, c).
Weve even computed the residues of (f, [.[
s
) at s = 0 and s = 1; theyve
come out in the wash.
They are f(0)

(E) and

f(0)

(E) respectively. Recall that we computed

(E) = 2
r
(2)
s
Reg
k
h/w
_
[d[.
Short chapter 7: Applications!
We have left open the logical possibility that Z = 0, in which case our
theory is empty. Lets check it isnt!
Example of a non-zero f Z: lets build f : A
k
C as a product of f
v
. If
v is nite lets just let f
v
be the characteristic function of R
v
. If v is innite and
real set f
v
(x) = e
x
2
and if v is complex set f
v
(x +iy) = e
2(x
2
+y
2
)
. At the
innite places weve rigged it so

f
v
=

f. At the nite places,

f
v
is p
m/2
times
the characteristic function of the inverse dierent of f, where p
m
generates the
discriminant ideal of k
v
, so

f
v
= f
v
at the unramied places but not at the
ramied places.
We now have a problem in analysis: we need to check f Z. First lets check
f and

f are in L
1
(A
k
). Well, locally they are integrable, and at all but nitely
many places the local integral is 1, so the innite product trivially converges
and gives the global integral.
Next lets check the third condition; we need to check that f(y).[y[

is in
L
1
(A

f
) for > 1, and similarly for

f. Well the local factors are certainly
in L
1
indeed, they are in L
1
for > 0, because we checked this when we
were doing our local zeta integrals. But this isnt enough to check that the
product is L
1
: we need to check that the innite product of the local integrals
converges. We evaluated the local integrals at the nite places, when doing our
local calculations, and they were (1 p

)
1
for k = Q (I did these in class)
and more generally p
m/2
(1 q

)
1
if k is a nite extension of Q and were
doing the computation at a P-adic place,
with residue eld of size q and discriminant ideal (p
m
) (I mentioned these
on the example sheet; the proof is no more dicult). So we need to check
that

P
(1 N(P)

)
1
converges for > 1and it does; this is precisely
the statement that the zeta function of a number eld converges for Re(s) > 1,
which is proved by reducing to k = Q and then using standard estimates. This
argument applies to both f and

f, which are the same away from a nite set of
places.
84
Finally we have to check the second condition (the one that let us apply
Poisson summation and interchange a sum and an integral). Let y be a xed
idele, let x be a xed adele, and lets rst consider

k
f(y(x +)).
First I claim that this sum converges absolutely. Because look at the support of
f: at the nite places its supported only on integral ideles A
f
k

v<
R
v
,
so,
whatever y and x are, f(y(x + )) will actually equal zero if, at any place,
the denominator of y beats the denominator of xy. So this sum, ostensibly
over all of k

, is really only over a fractional ideal in k, and now convergence is

trivial because at the innite places (and there is at least one innite place) f
is exponentially decreasing, and there are only nitely many lattice points with
norm at most a given constant.
Now why is the convergence locally uniform? Its for the same reason. If
y and x vary in a compact then the fractional ideal above might move but for
compactness reasons the lattice wont get arbitrarily small (its not dicult to
write down a formal proof) and its hence easy to uniformly bound the sums
involved.
So the main theorem applies! What does it say in this case?
Well, (f, [.[
s
) and (

f, [.[
1s
) are closely related to, but not quite, the zeta
function of k. Indeed if we write S

for the innite places of k and S

f
for the
nite places which are ramied in k/Q then (f, [.[
s
) =

v
(f
v
, [.[
s
) (the right
hand integrals are local zeta integrals), which expands to

vS

(f
v
, [.[
s
)

vS
f
(Nv)
m
v
/2

P
(1 N(P)
s
)
1
and

v|

v
(f
v
, [.[
s
) is a load of gamma factorsexactly the fudge factors
which you multiply
k
(s) =

P
(1 N(P)
s
)
1
by to get (denition)
k
(s).
So (f, [.[
s
) =
k
(s)[d[
1/2
with d the discriminant of k. Now (

f, [.[
1s
) is
almost the same, except that

f ,= f at the nite ramied places: the local in-
tegral of f
v
at the nite place is easily checked to be p
ms
/(1 q
s1
), so for
Re(1 s) > 1 we have (

f, [.[
1s
) =
k
(1 s)[d[
s
and we deduce

k
(1 s) = [d[
s1/2

k
(s).
Slightly better: if we set
Z
k
(s) =
k
(s).[d[
s/2
=
k
(s).

v|
(f
v
, s).[d[
s/2
then we get
Z
k
(1 s) = Z
k
(s).
This is the functional equation for the Dedekind zeta function, that is, the
zeta function of a number eld.
85
Moreover, we know that the pole at s = 1 of (f, [.[
s
) is simple with residue

f(0)

(E) =

f(0)2
r
(2)
s
Reg
k
.h/(w
_
[d[), and

f(0) = [d[
1/2
, so the pole at
s = 1 of
k
(s) = (f, [.[
s
)[d[
1/2
has residue 2
r
(2)
s
Reg
k
.h/(w
_
[d[). Moreover
the local zeta factors at the real innite places are
s/2
(s/2) which equals 1
at s = 1, and at the complex innite places are (2)
1s
(s) which is again 1 at
s = 1, so we deduce
lim
s1
(s 1)
k
(s) = 2
r
(2)
s
Reg
k
.h/(w
_
[d[)
which is called the analytic class number formula and which is used crucially
in both analytic arguments about densities of primes and in algebraic arguments
in Iwasawa theory.
Remark. Iwasawa noted that applying the theory to the function above,
without assuming the classical facts about class groups and unit groups that we
needed when analysing J/k

, in fact showed that

_
J/k

1 < , and hence that

one could deduce the niteness of the class group and nite-generation of the
unit group of a number eld via this calculation.
Lets do one more example, if we have time: Dirichlet L-functions.
Let N 1 be an integer, and : (Z/NZ)

be a character. By
CRT we can write =

p|N

p
with
p
: (Z/p
e
Z)

, where p
e
[[N. Our
calculations for a fundamental domain of k

in J, when applied to k = Q, show

that A

Q
= Q

p
Z

p
R
>0
, with the rst factor embedded diagonally. Hence
naturally gives rise to a character of

p
Z

p
(use
p
if p[N and 1 otherwise)
and hence to a character c : Q

Q
C

(make c trivial on R
>0
). We write
c =

v
c
v
. If p N then c
p
: Q

p
C

is trivial on Z

p
and c
p
(p) = (p)
1
.
Lets now choose f so that (f, c[.[
s
) is not identically zero and lets see what
the resulting function of s is. If p N then we just let f
p
be the characteristic
function of Z
p
. If p[N then we let f
p
be the function we used on the example
sheet when computing on the component corresponding to
p
. Note that we
dont care what f
p
is!
At innity we let f

(x) = e
x
2
if (1) = 1 and f

(x) = xe
x
2
if
(1) = 1; these are the function we used in our local calculations in the R
case.
We set f =

v
f
v
. The same arguments as above show f Z. We have
(f, c.[.[
s
) = (f

, c

.[.[
s
)

p|N
(f
p
, c
p
.[.[
s
).L(
1
, s)
because if p N then one easily computes (f
p
, c
p
.[.[
s
) = (1 (p)
1
p
s
)
1
.
Similarly
(

f,

c.[.[
s
) = (

f

.[.[
s
)

p|N
(

f
p
,

c
p
.[.[
s
)L(, 1 s)
and the trick here is not to attempt to work out (f
p
, c
p
, [.[
s
) or (

f
p
,

c
p
.[.[
s
)
but to remember that these local zeta integrals both converge for 0 < Re(s) < 1
and that we worked out their ratio (c
p
.[.[
s
) when doing the local calculations!
86
The ratio was just p
e(s1)

p
e
1
j=1
(j)
j
p
e, the Gauss sum. [Note in passing
that in particular we never used the local meromorphic continuation results to
prove the global ones, we merely use the local ones to see the explicit form of
the functional equation.] If (, s) denotes L(, s) times the factor at innity,
we deduce
(
1
, s)N
s
W = (, 1 s)
where W is an explicit algebraic number that depends only on and N and is
basically a sum of roots of unity.
Finally Ill remark that there are more general quasi-characters k

k

C

than those above. The general such thing is usually called a Hecke character
or a Grossencharacter. If is such a gadget, then is unramied at all but
nitely many nite places, and dening f
v
at these unramied places to be
just the characteristic function of R
v
, and f
v
at the other places to be the f
v
we used when analysing the local
v
, the equation (f, .[.[
s
) = (

f,

.[.[
s
)
unravels to become the meromorphic continuation and functional equation for
the L-function of the Grossencharacter that Hecke discovered in his original tour
de force!
THE END
87