NOTES ON LIE GROUPS

1
2 NOTES ON LIE GROUPS

1. Lecture 1
1.1. Lie Groups.
Definition 1.1.1. A Lie group G is a C ∞ manifold with a group structure so that the group
operations are smooth. More precisely, the maps
m:G×G→G (multiplication)
inv : G → G (inversion)
are C ∞ maps of manifolds.
Example 1.1.2. Take G = R with m(a, b) = a + b, inv(a) = −a for all a, b ∈ R. Then G is an
abelian Lie group.
Example 1.1.3. Let V be a finite dimensional vector space over R. Then, V has a canonical
manifold structure, and is a group under vector addition. It can be shown that vector addition
and negation are smooth, so V is a Lie group.
Example 1.1.4. Let Mn (R) denote the set of all n × n matrices over R. Define
GL(n, R) = {A ∈ Mn (R) | det A 6= 0}
Then, GL(b, R) is a group unde the operations m(A, B) = AB and inv(A) = A−1 = adj A
det A where
adj A denotes the adjugate of A. As these operations are smooth on GL(n, R) considered as a
2
submanifold of Rn , GL(n, R) is a Lie group called the real general linear group. Completely
analogously, we have the Lie group
GL(n, C) = {A = Mn (C) | det A 6= 0}
the complex general linear group.
© ª
Example 1.1.5. The orthogonal group O(n) = A ∈ Mn (R) | AAT = I is a Lie group as a
subgroup and submanifold of GL(n, R).
We shall not prove the following theorem for now, but rather leave it as an advertisement of
coming attractions.
Theorem 1.1.6. (Closed Subgroup Theorem) Let G be a Lie group and H < G a closed
subgroup of G. Then, H is a Lie group in the induce topology as an embedded submanifold of
G.
As a direct consequence we have
Corollary 1.1.7. If G and G0 are Lie groups and φ : G → G0 is a continuous homomorphism,
then φ is smooth.
From the Closed Subgroup Theorem we can generate quite a few more examples of Lie
groups.
Example 1.1.8. The following groups are Lie groups:
• The real special linear group SL(n, R) = {A ∈ GL(n, R) | det A = 1}.
• The complex special linear group SL(n, C) = {A ∈ GL(n, C) | det A = 1}.
• The special orthogonal group SO(n, R) = SL(n, R) ∩ O(n).
 NOTES ON LIE GROUPS 3

• The unitary group U (n) = {A ∈ GL(n, C) | AA∗ = I}

(where A∗ denotes the Hermitian transpose of A).
• The special unitary group SU (n) = U (n) ∩ SL(n, C)
Exercise 1.1.9. Prove that each of the groups in Example 1.8 are Lie groups (assuming the
Closed Subgroup Theorem).
Example 1.1.10. We now define the Euclidean group of rigid motions, Euc(n). Let End(V, W )
denote the vector space of all linear endomorphisms from a vector space V to itself. As a set,
we have
Euc(n) = {T ∈ End(Rn ) | kT x − T yk = kx − yk∀x, y ∈ Rn }
qP
n 2
where kxk = i=1 xi . Now, one can check that if T ∈ Euc(n) and T (0) = 0, then T ∈ O(n).
Then, we can write x 7→ T x − T (0) ∈ O(n) and so T (x) = (T (x) − T (0)) + T (0). This shows
that T ∈ Rn × O(n).
We can think of Euc(n) as a slightly different set. Write
n· ¸¯ o
A v ¯ n
Euc(n) = ¯ A ∈ O(n), v ∈ R
0 1
· ¸
w
If we identify Rn with the set of all vectors of the form with w ∈ Rn , then we have
1
· ¸· ¸ · ¸
A v w Aw + v
=
0 1 1 1
Exercise 1.1.11. Is Euc(n) ∼
= Rn × O(n) as groups?
1.2. Lie Algebras.
Definition 1.2.1. A (real) Lie algebra L is a vector space over R with a bilinear map (called
the Lie bracket)
[·, ·] : L × L → L
(X, Y ) 7→ [X, Y ]
such that for all X, Y, Z ∈ L,
(1) [X, Y ] = −[Y, X]
(2) [X, [Y, Z]] = [[X, Y ], Z] + [Y, [X, Z]]
Remark 1.2.2. If we write ad(X)Y = [X, Y ], then 2) reads: ad(X) is a derivation of (L, [, ]).
Example 1.2.3. Let L = Mn (R). Then, L is a Lie algebra with the commutator ; i.e. [X, Y ] =
XY − Y X.
Exercise 1.2.4. Prove that Mn (R) is a Lie algebra with the commutator bracket.
4 NOTES ON LIE GROUPS

2. Lecture 2
2.1. Left Invariant Vector Fields.
Definition 2.1.1. Let G be a Lie group and M a smooth manifold. An action of G on M is
a smooth map
G×M →M
satisfying
(1) 1G · x = x for each x ∈ M
(2) g · (g 0 · x) = (gg 0 ) · x for each g, g 0 ∈ G, x ∈ M .
Example 2.1.2. Any Lie group G acts on itself by left multiplication. If a ∈ G is fixed, we
denote this action by
La (g) = ag ∀g ∈ G
G also acts on itself by right multiplication (we denote this by Ra ).
Remark 2.1.3. La (resp. Ra ) is a diffeomorphism for each a ∈ G since we have a smooth inverse
given by
L−1 −1
a (g) = a g = La−1 (g)
for any g ∈ G.
Remark 2.1.4. If G is a Lie group acting on a manifold M and we write gM : m 7→ g · m, then
we have a map
ρ : G → Dif f (M )
g 7→ gM
for each g ∈ G. Now, ρ(1G ) = idM and ρ(g1 g2 ) = ρ(g1 )ρ(g2 ) so that ρ is a group homomorphism
from G to the group of diffeomorphisms of M .
Example 2.1.5. Define a map L : G → Dif f (G); g 7→ Lg by Lg (g 0 ) = gg 0 . Then Lg is a
homomorphism for each fixed g ∈ G and represents the usual left action of G on itself.
Definition 2.1.6. A vector field X on a Lie group G is left invariant if
(dLg )(X(x)) = X(Lg (x)) = X(gx)
for each x, g ∈ G.
We will now use left invariant vector fields to show that the tangent space of G at the
identity, denoted T1 G, is a Lie algebra.
Proposition 2.1.7. Let G be a Lie group. Then, the vector space of all left invariant vector
fields on G is ismomorphic (as a vector space) to T1 G.
Proof. Since X is left invariant the following diagram commutes
dLa
TG / TG
O O
X X
La
G /G
 NOTES ON LIE GROUPS 5

so that X(a) = (dLa )1 (X(1)) for all a ∈ G. Denote by Γ(T G)G the set of all left invariant
vector fields on G. Define a map φ : Γ(T G)G → T1 G by φ : X 7→ X(1). Then, φ is linear and
injective since if X, Y ∈ Γ(T G)G and φ(X) = φ(Y )
X(g) = dLg (X(1)) = dLg (Y (1)) = Y (g)
for each g ∈ G.
Now, φ is also surjective. For v ∈ T1 G, define Xv ∈ Γ(T G)G by Xv (a) = (dLa )1 (v) for
a ∈ G. We claim that Xv is a left invariant vector field. Now, Xv : G → T G is a C ∞ map of
manifolds since if f ∈ C ∞ G, then for a ∈ G
(Xv (f ))(a) = (dLa (v))f
= v(f ◦ La )
Now, if x ∈ G we have
(f ◦ La )(x) = (f ◦ m)(a, x)
which is a smooth map of a, x (here, m is the multiplication map on G). Thus, v(f ◦ La ) is
smooth and hence so is Xv .
We now show Xv is left invariant. For a, g ∈ G, we have
(dLg )(Xv (a)) = dLg ((dLa )1 (v))
= d(Lg ◦ La )(v)
= d(Lga )(v)
= Xv (ga)
= Xv (Lg (a))
so that Xv is left invariant. Therefore φ is onto and Γ(T G)G ∼ = T1 G. ¤
We now give T1 G a Lie algebra structure by idetifying it with Γ(T G)G with the Lie bracket
of vector fields. But, we need to show that [, ] is in fact a binary operation on Γ(T G)G . Recall if
f : M → N is a smooth map of manifolds and X, Y are vector fields on M and N respectively,
we say that X, Y are f -related if df (X(x)) = Y (f (x)) for every x ∈ M . It is a fact from
manifold theory that is X, Y and X 0 , Y 0 are f -related, then so are [X, Y ] and [X 0 , Y 0 ]. But, left
invariant vector fields are La related for all a ∈ G by definition. This justifies
Proposition 2.1.8. The Lie bracket of two left vector fields is a left invariant vector field.
Thus, we can regard T1 G as a Lie algebra and make the following definition.
Definition 2.1.9. Let G be a Lie group. The Lie algebra g of G is T1 G with the Lie bracket
induced by its identification with Γ(T G)G .
Example 2.1.10. Let G = (Rn , +). What is g? Notice that for this group, La (x) = a + x, so
that (dLa )0 = idT0 Rn . So, (dLa )0 (v) = v for all v ∈ T0 Rn and thus g = T0 Rn ∼
= Rn . So the Lie
algebra contains all constant vector fields, and the Lie bracket is identically 0.
Example 2.1.11. Consider the Lie group GL(n, R). We have TI GL(n, R) = Mn (R), the set
of all n × n real matrices. For any A, B ∈ Mn (R), the Lie bracket is the commutator; that is,
[A, B] = AB − BA
6 NOTES ON LIE GROUPS

To prove this, we compute XA , the left invariant vector field associated with the matrix A ∈
T1 GL(n, R). Now, on Mn (R), we have global coordinate maps given by xij (A) = Aij , the
ijth entry of the matrix B. So, for g ∈ GL(n, R), (XA (xij ))(g) = XA (I)(xij ◦ Lg ). Also, if
h ∈ GL(n, R),
(xij ◦ Lg )(h) = xij (gh)
X
= gik hkj
k
X
= gik xkj (h)
k
P
which implies that xij ◦ Lg = k gik xkj . ¯
d¯
Now, if f ∈ C ∞ (GL(n, R)), XA (I)f = dt t=0 f (I
+ tA) so that
¯
d ¯¯
XA (I)xij = xij (I + tA)
dt ¯t=0
= Aij
P P
Putting these remarks together, we see that XA (xij ◦ Lg ) = k gik Akj = k xik (g)Akj .
We are now in a position to calculate the Lie bracket of the left invariant vector fields
associated with elements of Mn (R):

([XA , XB ](I))ij = [XA , XB ](I)xij

= XA XB (xij ) − XB XA (xij )
³ ³X ´ ³X ´´
= XA Bkj xik − XB Akj xik (I)
k k
³X ´
= Bkj xil Alk − Akj xil Blk (I)
k,l
X
= Bkj δil Alk − Akj δil Blk
k,l
X X
= Aik Bkj − Bik Akj
k k
= (AB − BA)ij
So, [A, B] = AB − BA.
 NOTES ON LIE GROUPS 7

3. Lecture 3
3.1. Lie Group Homomorphisms.

Definition 3.1.1. Let G and H be Lie groups. A map ρ : G → H is a Lie group homomorphism
if
(1) ρ is a C ∞ map of manifolds and
(2) ρ is a group homomorphism.
Furthermore, we say ρ is a Lie group isomorphism if it is a group isomorphism and a diffeo-
morphism.
If g and h are Lie algebras, a Lie algebra homomorphism τ : g → h is a map such that
(1) τ is linear and
(2) τ ([X, Y ]) = [τ (X), τ (Y )] for all X, Y ∈ g.

Now, suppose V is an n-dimensional vector space over R. We define

GL(V ) = {A : V → V | A is a linear isomorphism}

Since V ∼
= Rn , GL(V ) ∼
= GL(n, R). Note that GL(V ) ⊆ Hom(V, V ) as an open subset.

Definition 3.1.2. A (real) representation of a Lie group G is a Lie group homomorphism

ρ : G → GL(V ).

We may similarly define GL(W ) for a complex vector space W and thus the notion of a
complex representation.
There are two basic problems in Lie Theory:
(1) classify all Lie groups (and Lie algebras),
(2) classify all representations of Lie groups.
One step in this direction is the association between Lie group homomorphisms and homo-
morphisms of Lie algebras.

Theorem 3.1.3. Suppose ρ : G → H is a Lie group homomorphism. Write dρ1 = δρ. Then,
δρ : T1 G → T1 H is a Lie algebra homomorphism.

Proof. It is enough to show that any two left invariant vector fields on G and H are ρ-related.
So, let X ∈ Γ(T G)G and X̃ ∈ Γ(T H)H . Then, for each a, g ∈ G, we have

(ρ ◦ La )(g) = ρ(ag)
= ρ(a)ρ(g)
= (Lρ(a) ◦ ρ)(g)
8 NOTES ON LIE GROUPS

so that ρ ◦ La = Lρ(a) ◦ ρ. Now,

dρa (X(a)) = dρa (La (X(1)))

= d(ρ ◦ La )(X(1))
= d(Lρ(a) ◦ ρ)(X(1))
= dLρ(a) (dρ(X(1)))
= dLρ(a) (δρ(X(1)))
= dLρ(a) (X̃(1))
= X̃(ρ(a)) (since X̃ is left invariant)

and thus, X and X̃ are ρ-related. ¤

3.2. Lie Subgroups and Lie Subalgebras.

Definition 3.2.1. Let G be a Lie group. A subset H of G is a Lie subgroup if

(1) H is an abstract subgroup of G,
(2) H is a Lie group and
(3) the inclusion ι : H ,→ G is an immersion.
A linear subspace h of a Lie algebra g is a Lie subalgebra if h is closed under the Lie bracket
in g.

Example 3.2.2. Let g be a Lie algebra, and v ∈ g a nonzero vector. Then Rv, the span of v,
is a Lie subalgebra of g.

Example 3.2.3. Let H ,→ G be a Lie subgroup. By Theorem 3.2, T1 H ,→ T1 G is a Lie

subalgebra.

Example 3.2.4. Consider the Lie group Gl(2, R). The subgroup
n· ¸ o
cos t sin t
SO(2) = |t∈R
− sin t cos t

is a Lie subgroup. We will compute the Lie algebra of SO(2). By Example 4.2, it will be
sufficient to find a nonzero vector in T1 SO(2) since this Lie algebra is of codimension 1 in
T1 GL(n, R). Now,
¯ · ¸ · ¸¯
d ¯¯ cos t sin t − sin t cos t ¯
= ¯
dt ¯t=0 − sin t cos t − cos t − sin t t=0
· ¸
0 1
=
−1 0

Thus, the Lie algebra so(2) of SO(2) is

n· ¸ o
0 x
so(2) = |x∈R
−x 0
 NOTES ON LIE GROUPS 9

Example 3.2.5. Consider the Lie subgroup O(n) of GL(n, R). We compute o(n), the Lie
algebra of O(n). Let A(t) be a path in O(n) with A(0) = I. Then, since BB T = I for all
B ∈ O(n), we have A(t)A(t)T = I for every t. Thus,
¯ ¯
d ¯¯ d ¯¯
0= ¯ I = A(t)A(t)T
dt t=0 dt ¯t=0
³ d ¯¯ ´ ³ d ¯¯ ´
= ¯ A(t) A(0)T
+ A(0) ¯ A(t)T
dt ¯ t=0 dt ¯ t=0
0 0 T
= A (0) + A (0)
Thus, we have o(n) ⊆ S where
© ª
S = X ∈ Mn (R) | X + X T = 0
2
To prove equality, we proceed by dimension count. Now, dim S = n 2−n (the easiest way to
see this is to write down the form of a general element of S and determine where to place 1’s).
View I ∈ Sym2 (Rn ), the set of all symmetric n × n real matrices. Then, I is a regular value of
the map
A 7→ AAT
where A ∈ GL(nR). Thus,
dim o(n) = dim GL(n, R) − dim Sym2 (Rn )
n2 − n
= n2 − ( + n)
2
n2 − n
=
2
We can conclude, therefore, that o(n) = S.
We conclude this section by discussing the induced maps δm and δinv on the Lie algebra of
a Lie group G.
Proposition 3.2.6. Let G be a Lie group. Then, for all X, Y ∈ g,
(1) δm(X, Y ) = X + Y .
(2) δinv(X) = −X.
Proof. 1. First note that since δm = (dm)1 is linear, it is enough to prove that (dm)1 (X, 0) =
d
X. So, let γ : (a, b) → G be a curve with γ(0) = 1 and dt γ(t) = X. Then,
δm(X, 0) = (dm)1 (X, 0)
¯
d ¯¯
= m(γ(t), 1)
dt ¯t=0
¯
d ¯¯
= γ(t)
dt ¯t=0
= X
2. Now, m(γ(t), inv(γ(t))) = 1 for each t ∈ (a, b). Consider F : G → G defined by
F (g) = gg −1 . Denote by ∆ the diagonal map ∆(g) = (g, g) for each G ∈ G. Then, F =
10 NOTES ON LIE GROUPS

m ◦ (idG × inv) ◦ ∆. Thus,

0 = (dF )1 (X) = ((dm)1 ◦ (d idG )1 × (d inv)1 ◦ (d∆)1 )(X)
= X + (d inv)1 (X)
and so δinv(X) = −X. ¤
 NOTES ON LIE GROUPS 11

4. Lecture 4
4.1. Topological Groups.
Definition 4.1.1. A topological group G is a topological space which is a group and has the
properties that the group operations are continuous.
Lemma 4.1.2. Let G be a connected topological group. Suppose H is an abstract open subgroup
of G. Then H = G.
Proof. For any a ∈ G, La : G → G given by g 7→ ag is a homeomorphism. Thus, for each
a ∈ G, aH ⊆ G is open. Since the cosets partition G, and G is connected, we must have
|G/H| = 1. ¤
Lemma 4.1.3. Let G be a connected topological group, U ⊆ G a neigborhood of 1. Then U
generates G.
© ª
Proof. For a subset W ⊆ G, write W −1 = g −1 ∈ G | g ∈ W . Also, if k is a positive integer,
we set W k = {a1 . . . ak | ai ∈ W }. Let U be as above, and V = U ∩ U −1 . Then, V is open and
v ∈ V implies that v −1 ∈ V . Let H = ∪∞ n
n=1 V . Then, H is a subgroup and we claim that H is
open. Notice that H is precisely the subgroup generated by U , so if we prove that H is open,
then H = G and the Lemma is proved.
If V k is open, then V k+1 = ∪a∈V (aV k ) is open since left multiplication is a homeomorhism.
By induction, V n is open for every n. Thus H is open. ¤
We will use these results to prove that Lie subalgebras correspond to connected Lie sub-
groups. But, first, we’ll need to develop some more terminology and recall some results from
Differential Geometry.
Definition 4.1.4. A d-dimensional distribution D on a manifold M is a subbundle of T M of
rank d.
Question: Given a distribution D ⊆ T M , does there exist for each x ∈ M an immersed
submanifold L(x) of M such that Ty L(x) = Dy for every y ∈ L(x)? A necessary condition for
this question to be answered in the affirmative is: If X, Y ∈ Γ(D), then [X, Y ] ∈ Γ(D).
Definition 4.1.5. A distribution D on a manifold M is integrable (or involutive) if for every
X, Y ∈ Γ(D), [X, Y ] ∈ Γ(D). An immersed submanifold L ⊆ M is an integral manifold of D if
Tx L = Dx for every x ∈ L.
We’ll get some mileage out of the following theorem and proposition for which we omit the
proofs.
Theorem 4.1.6. (Frobenius) Let D be a d-dimensional integrable distribution on a manifold
M . Then, for all x ∈ M , there exists a unique maximal, connected,immersed integral subman-
ifold L(x) of D passing through x.
Proposition 4.1.7. Suppose D ⊆ T M is an integrable distribution and L ⊆ M an immersed
submanifold such that Ty L = Dy for every y ∈ L. Suppose f : N → M is a smooth map of
manifolds and F (N ) ⊆ L. Then, f : N → L is C ∞ .
Assuming both of these results, we’ll prove
12 NOTES ON LIE GROUPS

Theorem 4.1.8. Let G be a Lie group with Lie algebra g and h ⊆ g a Lie subalgebra. Then,
there exists a unique connected Lie subgroup H of G with T1 H = h.
Proof. Consider D ⊆ T G given by Da = dLa (h) for a ∈ G. Then, D is a distribution. We claim
D is integrable. To prove this, let v1 , . . . , vk be a basis of h. Let V1 , . . . , Vk be the corresponding
left invariant vector fields on G. Then, {V1 (g), . . . , Vk (g)} is a basis of Dg . Also, we have
[Vi (g), Vj (g)] = dLg ([Vi , Vj ](g))
since the bracket of left invariant vector fields is left invariant.
Now, for arbitrary sections X, Y of D, write them (locally) as
X
X= xi Vi
i
X
Y = yj Vj
j

where xi , yj ∈ C ∞ (G) for each i, j. So,

X X X
[X, Y ] = xi Vi (yj )Vj + i, jxi yj [Vi , Vj ] − Vj (xi )yj Vi
i,j i,j

each term of which is in Γ(D), and hence [X, Y ] ∈ Γ(D).

If we now apply the Frobenius Theorem, we get an immersed, connected, maximal subman-
ifold H of G such that 1 ∈ H and T1 H = h. The claim is that H is a Lie subgroup of G. To
show H is a subgroup, fix some x ∈ H. Consider x−1 H = Lx−1 (H). Then, 1 = xx−1 ∈ x−1 H
and for all a ∈ G, we have
Tx−1 a (x−1 H) = dLx−1 (Ta H)
= dLx−1 (dLa h)
= dLx−1 a h
= Dx−1 a
So, x−1 H is tangent to D. Since H is connected, x−1 H is connected, and by maximality and
uniqueness of H, we have x−1 H ⊂ H. Therefore, H is a subgroup of G.
Finally, we need to show that m|H×H and inv |H are C ∞ . But, m : H × H → G is C ∞
and m(H × H) ⊆ H. By Proposition 5.7, multiplication is a smooth binary operation on H.
Similarly, inv is smooth on H and thus H is a Lie subgroup. ¤
 NOTES ON LIE GROUPS 13

5. Lecture 5
5.1. Simply Connected Lie Groups. Recall now Theorem 3.2, which states that if ρ : G →
H is a Lie group morphism, then δρ : g → h is a map of Lie algebras. Is the conerse true? That
is, if G, H are Lie groups with Lie algebras g and h respectively and τ : g → h is a map of Lie
algebras, does there exist a Lie group morphism ρ : G → H with δρ = τ ? Unfortunately, the
answer is “not always”. We can answer affirmitively when G is connected and simply connected
however. Let’s recall a couple of definitions from basic topology.
Definition 5.1.1. A connected topological space T is simply connected if T is arcwise con-
nected and every pointed map f : (S 1 , 1) → (T, ∗) is homotopically trivial.
Definition 5.1.2. A continuous map p : X → Y` is a covering map if for each y ∈ Y , there
exists a neighborhood U of y such that p−1 U = Uα where Uα ⊂ X is open for each α, and
p|Uα is a homeomorhism.
Lemma 5.1.3. Let φ : A → B be a map of Lie groups with (dφ)1 : a → b an isomorphism.
Then,
a) φ is a local diffeomorphism and
b) If B is connected, φ is onto.
Proof. Consider the following commutative diagram
φ
A /B

La Lφ(a)
² ²
A /B
φ

which can also be viewed element-wise as

φ
1_A Â / 1B
_
La Lφ(a)
² ²
a / φ(a)
φ

From this we can conclude that

(dφ)1 = (dLφ(a) )−1
φ(a) ◦ (dφ)a ◦ (dLa )1

Now, since (dφ)1 is an isomorphism, (dφ)a is an isomorphism for every a ∈ A. Invoking the
Inverse Function Theorem, we see then that φ is a local diffeomorphism. In particular, φ is
an open map, so φ(A) is an open subgroup of B. Now if B is connected, Lemma 5.2 yields
φ(A) = B and thus φ is onto. ¤
Lemma 5.1.4. Let φ : A → B be a surjective Lie group map that is a local diffeomorphism.
Then, φ is a covering map.
Proof. Let Λ = ker φ. Since φ is a local diffeomorphism, there exists an open neighborhood U
of 1A such that φ|U is injective and so U ∩ Λ = 1A . Since A is a Lie group, the multiplication
map m : A × A → A is continuous and so there exists an open neighborhood V of 1A such that
14 NOTES ON LIE GROUPS

m(V × V ) ⊆ U . That is, V V ⊆ U . Let W = V ∩ V −1 , then W W −1 ⊆ U . We claim that for

every λ, λ0 ∈ Λ, λW ∩ λ0 W = ∅ if and only if λ 6= λ0 .
To prove this claim, suppose λW ∩λ0 W = ∅ for some λ, λ0 ∈ Λ. Then, there exists w, w0 ∈ W
so that λw = λ0 w0 . But then, (λ0 )−1 λ = w0 (w)−1 and so (λ0 )−1 λ ∈ Λ ∩ U and thus (λ0 )−1 λ = 1.
Now, what we’ve just proved is that ker φ = Λ is discrete, so
a
φ−1 (φ(W )) = ΛW = λW
λ∈Λ

homeomorphism φ|λW : λw 7→ φ(w). Thus, for each b ∈ B and a ∈ φ−1 (b),

and we have a `
φ−1 (bφ(W )) = λ∈Λ aλW . Therefore, the fibers of φ are discrete and φ : A → B is a covering
map. ¤

We also have the following fact from topology, stated without proof:
Lemma 5.1.5. Let φ : A → B be a covering map of topological spaces with B simply connected.
Then φ is a homeomorphism.
We are now in a position to answer the question posed in the beginning of this section.
Theorem 5.1.6. Let G be a connected and simply connected Lie group with lie algebra g and
H a Lie group with Lie algebra h. Given a Lie algebra morphism τ : g → h, there exists a
unique Lie group morphism ρ : G → H such that δρ = τ .
Proof. First note that
graph(τ ) = {(X, τ (X)) ∈ g × h | X ∈ g}
is a subalgebra of g × h since
[(X1 , τ (X1 )), (X2 , τ (X2 ))] = ([X1 , X2 ], [τ (X1 ), τ (X2 )])
= ([X1 , X2 ], τ [X1 , X2 ])
So, by Theorem 5.8, there exists a connected Lie subgroup Γ of G × H so that T1 Γ = graph(τ ).
The claim is that Γ is the graph of the Lie group morphism ρ we are trying to construct, and
hence it is sufficient to show that Γ is in fact a graph. Formally, if Γ is a graph, then we have
G ×O 5H
­ 55
­­­ 55
­ Â 55
π1 ­­ ? 55π2
­­ Γ GG 55
­ ww w GG 5
­­­ www GG 55
GG 5
w π2 |Γ GG# 5½
¥­­{www π1 |Γ
G H
and can simply define ρ = π2 ◦ (π1 |Γ )−1 .
Now, (dπ1 |Γ )(1,1) : graph(τ ) → g is an isomorphism, so π1 |Γ is a local diffeomorphism by
Lemma 6.3, and evidently π1 |Γ : Γ → G is a surjective group homomorphism. By Lemma 6.4,
π1 |Γ is a covering map. Since G is simply connected, π1 |Γ is a homeomorphism.
Finally, define ρ : G → H by ρ = π2 ◦ (π1 |Γ )−1 . Since Γ is a subgroup, ρ is a homomorphism
and graph(ρ) = Γ. This gives us the Lie group morphism we want.
 NOTES ON LIE GROUPS 15

We now have to establish the uniqueness of such a Lie group homomorphism. Suppose
ρ̃ : G → H is another such Lie group morphism, then
T(1,1) (graph(ρ̃)) = graph(τ ) = T(1,1) (graph(ρ))
Since graph(ρ̃) and graph(ρ) are connected subgroups of G × H with the same Lie algebra,
they must be identical. Therefore, ρ̃ = ρ. ¤
16 NOTES ON LIE GROUPS

6. Lecture 6
6.1. The Exponential Map. Given a Lie group G and its Lie algebra g, we would like to
construct an exponential map from g → G, which will help to give some information about the
structure of g.
Proposition 6.1.1. Let G be a Lie group with Lie algebra g. Then, for each X ∈ g, there
exists a map γX : R → G satisfying
(1) γX ¯(0) = 1G ,
d ¯¯
(2) γX (t) = X and
dt ¯t=0
(3) γX (s + t) = γX (s)γX (t).
Proof. Consider the Lie algeba map τ : R → g defined by τ : t 7→ tX for all X ∈ g. Now, R
is connected and simply connected, so by Theorem 6.6. there exists a unique Lie group map
γX : R → G such that (dγX )0 = τ ; which is to say
¯
d ¯¯
γX (t) = X
dt ¯ t=0
¤
This motivates the following definition:
Definition 6.1.2. Let G be a Lie group with Lie algebra g. Define the exponential map
exp : g → G by exp(X) = γX (1).
Lemma 6.1.3. Let G be a Lie group with Lie algebra g and X ∈ g. Write X̃ for the left
invariant vector field on g with X̃(1) = X. Then,
φt (a) = aγX (t)
is the flow of X̃. In particular, X̃ is complete; i.e. the flow exists for all t ∈ R.
Proof. For a ∈ G, we have
¯ ³ d ¯¯ ´
d ¯¯ ¯
aγ (t) = (dL ) γ (t)
dt ¯t=s dt ¯t=s
X a γX (s) X

³ d¯ ¯ ´
= (dLa )γX (s) ¯ γX (t + s)
dt ¯t=0
³ d ¯¯ ´
= (dLa )γX (s) ¯ γX (s)γX (t)
¯
dt t=0
³ d ¯¯ ´
= (dLa )γX (s) ¯ Lγ (s) (γX (t))
dt ¯t=0 X
³ d ¯¯ ´
= (dLaγX (s) )1 ¯ γX (t)
dt ¯t=0
= (dLaγX (s) )1 (X)
= X̃(aγX (s)) (since X̃ is left-invariant)
So, aγX (t) is the flow of X̃ and exists for all t. ¤
 NOTES ON LIE GROUPS 17

Lemma 6.1.4. The exponential map is C ∞ .

Proof. Consider the vector field V on G × g given by
V (a, X) = (dLa (X), 0)
Then, V ∈ C ∞ (G, g) and the claim is that the flow of V is given by ψt (g, X) = (gγX (t), X).
To prove this claim, consider the following:
¯
d ¯¯
(gγX (t), X) = (dLgγX (s) (X), 0)
dt ¯
t=s
= V (gγX (s), X)
from which we can conclude that γX depends smoothly on X
Now, we note that the map φ : R × G × g defined by φ(t, a, X) = (aγX (t), X) is smooth.
Thus, if π1 : G × g → G is projection on the first factor, (π1 ◦)(1G , X) = γX (1) = exp(X) is
C ∞. ¤
Lemma 6.1.5. For all X ∈ g and for all t ∈ R, γtX (1) = γX (t).
Proof. The intent is to prove that for all s ∈ R, γtX(s) = γX (ts). Now, s 7→ γtX (s) is the
integral curve of the left invariant vector field tX˜ through 1G . But, tX˜ = tX̃, so if we prove
that γX (ts) is an integral curve of tX̃ through 1G , by uniqueness the Lemma will be established.
To prove this, first let σ(s) = γX (ts). Then, σ(0) = γX (0) = 1G . We also have
d d
σ(s) = γX (ts)
ds ds ¯
d ¯¯
= d γX (u)
du ¯u=ts
= tX̃(γX (ts))
= tX̃(σ(s))
So, σ(s) is also an integral curve of tX̃ through 1G . Thus, γtX (s) = γX (ts) and, in particular,
when s = 1 we have γtX (1) = γX (t). ¤
We’ll now use this to prove a rather nice fact about the exponential map.
Proposition 6.1.6. Let G be a Lie group and g its Lie algebra. Identify both T0 g and T1 G
with g. Then, (d exp)0 : T0 g → T1 G is the identity map.
Proof. Using the result established in Lemma 7.5, we have
¯
d ¯¯
(d exp)0 (X) = exp(0 + tX)
dt ¯t=0
¯
d ¯¯
= γtX (1)
dt ¯t=0
¯
d ¯¯
= γX (t)
dt ¯t=0
= X
¤
18 NOTES ON LIE GROUPS

Corollary 6.1.7. a) For all t1 , t2 ∈ R, exp((t1 + t2 )X) = exp t1 X + exp t2 X.

b) exp(−tX) = (exp(tX))−1 .
 NOTES ON LIE GROUPS 19

7. Lecture 7
7.1. Naturality of exp. In this section, we reveal a property that will be used liberally in
discussions to come and provides an important relationship between morphisms of Lie groups
and morphisms of Lie algebras.
Theorem 7.1.1. Let φ : H → G be a morphism of Lie groups. Then, the following diagram
commutes:
δφ
h /g

exp exp
² ²
H /G
φ

That is to say, exp is natural.

Proof. Fix X ∈ g. Consider the curves
σ(t) = φ(exp(tX))
τ (t) = exp(δφ(tX))
Now, σ, τ : R → G are Lie group homomorphisms with σ(0) = τ (0) = 1. So,
¯ µ ¯ ¶
d ¯¯ d ¯¯
σ(t) = (dφ) exp(tX)
dt ¯t=0 dt ¯t=0
1

= (δφ)(X)
¯
d ¯¯
= τ (t).
dt ¯ t=0
So, σ(t) = τ (t) for all t. ¤
Corollary 7.1.2. Let H ⊆ G be a Lie subgroup of a Lie group G. Then, for all X ∈ h,
expG (X) = expH (X).
In particular, X ∈ h if and only if exp(tX) ∈ h for all t.
Theorem 7.1.3. Every connected Lie group G is a quotient G̃/N where G̃ is a simply connected
Lie group of the same dimension as G and N is a central discrete normal subgroup of G̃. Both
G̃ and N are unique up to isomorphism.
Proof. Recall that the universal covering space of a topological space is the unique (up to deck
isomorphism) simply connected covering space. We will use, but not prove, the fact that every
connected Lie group has a universal covering space.
Let G̃ be the universal covering space of G, and denote by p the covering map. Let 1̃ = p−1 (1).
Denote by m̃ the lift of the multiplication map m : G × G → G to G̃ uniquely determined by
m̃(1̃, 1̃) = 1̃. Similarly, inv : G → G lifts to G̃ as well. Thus, G̃ is a group. Also, p is a Lie
group homomorphism by definition of m̃:
p(m̃(a, b)) = m(p(a), p(b)).
Now, kernels of covering maps are discrete, and evidently, G ∼
= G̃/ ker p.
20 NOTES ON LIE GROUPS

It remains to prove that N = ker p is central, that is for all g ∈ G̃ and n ∈ N , gng −1 = n.
Fix n ∈ N . Define φ : G̃ → G̃ by φ(g) = gng −1 . Since N is normal φ(G) ⊆ N . Now, G̃ is
connected, so φ(G) is connected since φ is continuous. But N is discrete, so φ(G) is a single
point. We have φ(1) = n and hence φ(G) = n. Therefore, N is central. ¤
Corollary 7.1.4. If G is a connected topological group, then the fundamental group π1 (G) is
abelian.
Our last result for this section deals with subgroups of Lie groups.
Proposition 7.1.5. Lie groups have no small subgroups, i.e. if G is a Lie group, then there
exists a neighborhood U of the identity so that for all g ∈ U there exists a positive integer N
(depending on g) having the property that g N 6∈ U .
Proof. recall first that (d exp)0 : g → g is the identity. By the Inverse Function Theorem, there
exists neighborhoods V 0 of 0 in g and U 0 of 1 ∈ G so that exp : V 0 → U 0 is a diffeomorphism.
Let U = exp( 21 V 0 ). We claim that U is the desired neighborhood.
If g ∈ U , then g = exp( 12 v) for some v ∈ V 0 . Then,
µ ¶ µ ¶
n 1 1
g = exp v . . . exp v (n times)
2 2
N
for any positive
¡ integer
¢ n. Now, given v, pick N so that 2v ∈ V 0 \ 12 V 0 . Then, g N ∈
exp(V 0 ) \ exp 21 V 0 = U 0 \ U . ¤
 NOTES ON LIE GROUPS 21

8. Lecture 8
8.1. Ad, ad and exp.

Definition 8.1.1. Let G be a Lie group and V a vector space. A representation of a Lie group
is a map ρ : G → GL(V ) of Lie groups.

For a Lie group G, consider the action of G on itself by conjugation: for each g ∈ G we have
a diffeomorphism cg : G → G given by cg (a) = gag −1 . Notice that cg (1) = 1, and we have
an invertible linear map (dcg )1 : g → g. Now, cg1 g2 = cg1 ◦ cg2 for all g1 , g2 ∈ G, and hence
(dcg1 )1 ◦ (dcg2 )1 = (dcg1 g2 )1 .

Definition 8.1.2. The Adjoint Representation of a Lie group G is the representation Ad :

G → GL(g) defined by

Ad(g) = (dcg )1

The adjoint representation of a Lie algebra g is the representation ad : g → gl(g) = Hom(g, g)

defined by

ad(X) = (d Ad)1 (X)

Proposition 8.1.3. Suppose G is a Lie group. Then, for all t ∈ R, g ∈ G and X ∈ g we have
(1) g exp(tX)g −1 = exp(t Ad(g)(X)) and
(2) Ad(exp(tX)) = exp(t ad(X)).

Proof. For the first statement, apply naturality of exp to the diagram

Ad(g)
g /g
exp exp
² cg ²
G /G

Similarly, to prove 2, apply the naturality of exp to the diagram

ad / gl(g)
g
exp exp
² ²
Ad /
G GL(g)

Example 8.1.4. We compute what Ad and ad are as maps when G = GL(n, R). Recall that
for any A, g ∈ G we have the conjugation map cg (A) = gAg −1 . Note that congugation is linear.
22 NOTES ON LIE GROUPS

Thus, for X ∈ g we have

Ad(g)(X) = (dcg )I (X)
¯
d ¯¯
= cg (exp(tX))
dt ¯t=0
¯
d ¯¯
= g exp(tX)g −1
dt ¯t=0
µ ¯ ¶
d ¯¯
= g exp(tX) g −1
dt ¯t=0
= gXg −1 .
Also, for X, Y ∈ g
¯
d ¯¯
ad(X)Y = Ad(exp(tX))Y
dt ¯t=0
¯
d ¯¯
= exp(tX)Y exp(−tX)
dt ¯t=0
µ ¯ ¶ µ ¯ ¶
d ¯¯ d ¯¯
= exp(tX) Y exp(−0X) + exp(0X)Y exp(−tX)
dt ¯t=0 dt ¯t=0
= XY + Y (−X)
= [X, Y ]
the commutator of the matrices X, Y .
The second result of the previous example generalizes nicely to any Lie group.
Theorem 8.1.5. Let G be a Lie group. Then, for any X, Y ∈ g,
ad(X)Y = [X, Y ]
Proof. First note that
¯
d ¯¯
ad(X)Y = Ad(exp tx)Y
dt ¯t=0
¯
d ¯¯
= d(Cexp tx )1 (Y ).
dt ¯t=0
Also, recall that we have shown:
(1) cg (a) = gag −1 = (Rg−1 ◦ Lg )(a),
(2) (dLg )1 (Y ) = Ỹ (g), where Ỹ is the left invariant vector field with Ỹ (1) = Y ,
(3) the flow φỸt of Ỹ is given by

φỸt (a) = a(exp tX) = Rexp tX (a),

d¯
¯
(4) [X̃, Ỹ ](a) = dt t=0
d(φX̃ X̃
−t )(Ỹ (φt (a))),
(5) (exp tX)−1 = exp(−tX).
 NOTES ON LIE GROUPS 23

We now put 1-5 together:

¯
d ¯¯
ad(X)Y = dR (dLexp tX Y ) (by 1,5)
dt ¯t=0 exp(−tX)
¯
d ¯¯
= dR (Ỹ (exp tX) (by 2)
dt ¯t=0 exp(−tX)
¯
d ¯¯
= d(φX̃ X̃
t )(Ỹ (φt (1)) (by 3)
dt ¯
t=0
= [X̃, Ỹ ](1) (by 4)
¤
24 NOTES ON LIE GROUPS

9. Lecture 9
9.1. Normal Lie Subgroups. Problem: Given a Lie group G and N ⊂ G a normal Lie
subgroup, what properties does the Lie algebra n of N have?
Since N is normal in G, for all g ∈ G and for all n ∈ N , we have

gng −1 ∈ N =⇒ cg (N ) ⊆ N
=⇒ d(cg )1 (n) ⊆ n
=⇒ Ad(g)n ⊆ n
Thus, for any X ∈ n, any Y ∈ g and any t ∈ R, we have
¯
d ¯¯
Ad(exp tY )X ∈ n =⇒ Ad(exp tY )X ∈ n
dt ¯t=0
d¯
¯
But, dt t=0
Ad(exp tY )X = ad(Y )X = [Y, X] and hence [Y, X] ∈ n.
Definition 9.1.1. A Lie subalgebra n of a Lie algebra g is an ideal if [g, n] ⊆ n.
So, the above discussion proves
Proposition 9.1.2. If N is a normal Lie subgroup of a Lie group G, then its Lie algebra n is
an ideal in g.
It’s natural to wonder whether the converse of this proposition is true. That is, if G is
a Lie group and n is an ideal in g, is it necessarily true that the unique subgroup N of G
corresponding to n is normal? We need an extra hypothesis on G:
Proposition 9.1.3. Suppose G is a connected Lie group with Lie algebra g, n ⊆ g an ideal
and N the unique connected Lie subgroup with n ⊆ T1 N . Then N is a normal subgroup of G.
Proof. For all X ∈ n, Y ∈ g we have [Y, X] ∈ n. So, [Y, [Y, X]] = (ad(Y ) ◦ ad(Y ))(X) ∈ n. It
follows that ad(Y )n (X) ∈ n. But then,

X 1
Ad y n (X) = exp(t ad(Y ))(X)
n!
= Ad(exp tY )X ∈ n
for all t ∈ R. Since G is connected, exp g generates G and so Ad(g)X ∈ n for all g ∈ G, X ∈ n.
This implies that cg (exp X) = exp(Ad(g)X) ∈ N . But N is connected, and so exp n generates
N . Hence, gag −1 ∈ N for any g ∈ G, a ∈ N . ¤

9.2. The Closed Subgroup Theorem I. We now start a proof of The Closed Subgroup
Theorem:
Theorem 9.2.1. If G is a Lie group, H ⊆ G a closed subgroup, then H is a Lie subgroup of
G.
What we will actually prove is that H (as above) is an embedded submanifold of G and the
theorem will follow from a couple of results we will now establish.
 NOTES ON LIE GROUPS 25

Lemma 9.2.2. Let G be a Lie group. Suppose H ⊆ G is an abstract subgroup and an embedded
submanifold. Then, H is a Lie subgroup.
Proof. It’s enough to show that the multiplication map m : H × H → H is smooth. Fix
(x, y) ∈ H. Since m : H × H → G is continuous, there exists a neighborhood U of xy = m(x, y)
in G and a neighborhood V of (x, y) ∈ H so that m(V ) ⊆ U . Since H is embedded in G, there
exists a neighborhood U 0 ⊆ U of xy in G and a coordinate map φ : U 0 → Rn (n = dim G) such
that

φ(U 0 ∩ H) = φ(U 0 ) ∩ (Rk × {0})

where k = dim H. Now, φ ◦ m : m−1 (U 0 ) ∩ V → Rn is C ∞ and the image of this composition
lies in Rk × {0}. Letting p : Rn → Rk be the standard projection, we see that p ◦ φ ◦ m :
m−1 (U 0 ) ∩ V → Rk is C ∞ . Hence, m is C ∞ . ¤
Theorem 9.2.3. An abstract subgroup H of a Lie group G is an embedded submanifold if and
only if H is closed in G.
Proof. We first prove that if H is an embedded submanifold, it is closed in G. Since H is
embeddced, there exists a neighborhood U of 1 ∈ G such that U ∩ H is closed in U . Let H
be the closure of H and suppose y ∈ H. Then, yU −1 ⊆ G is a neighborhood of y in G. We
conclude that yU −1 ∩ H 6= ∅. Pick a point x ∈ yU −1 ∩ H. Then, x = yu−1 for some u ∈ U .
Thus, x−1 y = u ∈ U .
Now, Lx−1 : G → G is a homemorphism so that Lx−1 (H) = H. By continuity, Lx−1 (H) = H.
Thus, x−1 y ∈ H. We now argue that x−1 y ∈ H ∩U . Since x−1 y ∈ H ∩U , there exists a sequence
{hn } ⊂ H ∩ U so that hn → x−1 y. But recall that H ∩ U is closed in U and so x−1 y ∈ H ∩ U .
Since this implies y ∈ xH = H, we must have H ⊆ H. Thus, H is closed in G.
¤
26 NOTES ON LIE GROUPS

10. Lecture 10
10.1. The Closed Subgroup Theorem II. We have yet to complete the proof of the Closed
Subgroup Theorem. What remains is to show that an abstract closed subgroup H of a Lie
group G is an embedded submanifold. By Lemma 9.5, H is then a Lie subgroup of G.
Proof. Fix a norm k · k on g. Choose neighborhoods W 0 of 0 ∈ g and W of 1 ∈ G so that
exp : W 0 → W is a diffeomorphism. Let V 0 = W 0 ∩ (−W 0 ). Take V = exp(V 0 ) and note that
a ∈ V =⇒ a−1 ∈ V . Define

log := (exp |V )−1 : V → V 0

and let
½ ¾
∃ sequences {hn } ⊂ H ∩ V, {tn } ⊂ R≥0 with
h = X ∈ g|
(1) limn→∞ hn = 1 and (2) limn→∞ tn log hn = X
Claim 1: There exists a neighborhood U 0 of 0 ∈ h such that exp(U 0 ) ⊆ H.
To prove this claim, first take U 0 = V 0 ∩ h. Then, for any X ∈ U 0 there are sequences {hn }
and {tn } so that (1) and (2) hold. Since limn→∞ hn = 1, limn→∞ log hn = 0. Denote by btn c
the largest integer less than or equal to t. We then have

lim (tn − btn c) log hn = 0

n→∞
so that X = limn→∞ tn log hn = limn→∞ btn c log hn . In light of this, we see that

exp(X) = lim exp(btn c log hn )

n→∞
= lim (exp(log hn ))btn c
n→∞
= lim hbt
n
nc
∈H
n→∞
bt c
since hn n ∈ H for all n and H is closed. But this proves the claim that exp(U 0 ) ⊆ H.
Claim 2: h is a linear subspace of g.
Piick X ∈ h. As before, we then have sequences {hn } and {tn } satisfying the conditions (1)
and (2) above. Now, {h−1 n } ⊂ V ∩ H and

lim (h−1
n ) = ( lim hn )
−1
n→∞ n→∞
−1
= 1 =1
while

lim tn log h−1

n = − lim tn log hn
n→∞ n→inf
= −X ∈ h.
Also, for all t ≥ 0, we have limn→∞ (t(tn ) log hn ) → tX. Hence, tX ∈ h for all t ∈ R.
 NOTES ON LIE GROUPS 27

Now, if X, Y ∈ h, then for t sufficiently small, tX, tY ∈ U 0 and so exp tX exp tY ∈ H. In

addition, since limt→0 exp tX exp tY = 1, exp tX exp tY ∈ V . Hence for t sufficiently small,
exp tX exp tY ∈ V ∩ H. But,
¯
d ¯¯ 1
¯ log(exp tX exp tY ) = lim log(exp tX exp tY )
dt t=0 t→0 t
= X +Y
Let tn = 1/n and hn = exp tn X exp tn Y . Then hn ∈ H and limn→∞ hn = 1. It follows that
limn→∞ tn log hn = X + Y ∈ h.
Claim 3: For any neighborhoos U 0 of 0 ∈ h, exp(U 0 ) is a neighborhood of 1 ∈ H.
Suppose the claim doesn’t hold. Then, there exists a neighborhood U 0 of 0 ∈ h and a sequence
{hn } ⊂ H \ exp(U 0 ) such that hn → 1. Choose a linear subspace k in g so that g = h ⊕ k.
Consider φ : h ⊕ k → G defined by φ(X ⊕ Y ) = exp X exp Y . Then (dφ)0 = X + Y and so φ is a
diffeomorphism in a neighborhood of 0 ∈ h ⊕ k. Thus, there are sexuences {Xn } ⊂ h, {Yn } ⊂ k
so that hn = exp Xn exp Yn for n sufficiently large. Note that Yn 6= 0 since hn 6∈ exp(U 0 ).
Now, hn ∈ H implies that exp(Xn ) ∈ H and so exp(Xn )−1 hn ∈ H. On the other hand,
1/kYn kYn is bounded. By passing to a subsequence, we may assume that 1/kYn kYn → Y ∈ k
with kY k = 1. Let hn = exp(Yn ) so that for n large, hn ∈ V ∩ U and Yn = log(hn ). Since
Yn → 0, hn → 1 and 1/kYn k log hn → Y ∈ h, a contradiction.

From the three claims above, we can conclude that there exists a linear subspace h of g, a
neighborhood V 0 of 0 ∈ g and a neighborhood V of 1 ∈ G such that
(1) exp : V 0 → V is a diffeomorphism,
(2) exp(V 0 ∩ h) is a neighborhood of 1 in H.
So, log : V → V 0 are coordinates on G adapted to H. This in turn implies that for all x ∈ U ,
log ◦Lx−1 : Lx (V ) → g are coordinates adapted to H. Hence H is an embedded submanifold
of G. ¤
28 NOTES ON LIE GROUPS

11. Lecture 11
11.1. Applications of the Closed Subgroup Theorem. We begin with a remark about
the Lie algebra of a closed subgroup of a Lie group. If H is a closed subgroup of a Lie group
G, then the Lie algebra h of H is

{X ∈ g | exp tX ∈ H, ∀t ∈ R}
Since exp is natural, exp tX ∈ H for all t. Conversely, if exp tX ∈ H for all t,
¯
d ¯¯
X = ¯ exp tX ∈ T1 H = h
dt t=0
Our first application of the Closed Subgroup Theorem is a rather surprising result about
continuous group homomorphisms.
Theorem 11.1.1. Suppose H and G are Lie groups and f : H → G is a continuous group
homomorphism. Then, f is smooth.
For the proof of Theorem 11.1, we’ll need Sard’s Theorem.
Definition 11.1.2. Let f : M → N be a C ∞ map of manifolds. A point y ∈ N is a regular
value of f if for any x ∈ f −1 (y), (df )x : Tx M → Ty N is onto.
Theorem 11.1.3 (Sard’s Theorem). Let f : M → N be a smooth map of manifolds. Then,
the set of regular values of f is dense in N .
Remark 11.1.4. If f −1 (y) = ∅, y is still a regular value. Hence, if f (M ) is a single point in N ,
the complement N \ f (M ) is still dense and consists of regular values.
Proposition 11.1.5. Suppose f : A → B is a Lie group homomorphism. Then,
(i) if f is onto, (df )a : Ta A → Tf (a) B is onto for all a ∈ A and
(ii) if f is 1-1, (df )a : Ta A → Tf (a) B is 1-1 for all a ∈ A.
Proof. Recall first for any a ∈ A, Lf (a) ◦ f = f ◦ La . So, (dLf (a) )1 ◦ (df )1 = (df )a ◦ (dLa )1 .
Consequently, using the fact that Lg is always a diffeomorphism,

dim ker(df )a = dim ker(df )1 and

dim im(df )a = dim im(df )1
for all a ∈ A.
We are now in a position to prove (i). By Sard’s Theorem, the set of regular values of f
is dense in B. By assumption, B = f (A) and so there is b ∈ f (A) which is a regular value.
Hence, there is a0 ∈ A so that (df )a0 : Ta A → Tb B is onto. Thus, (df )a is onto for all a ∈ A.
To prove (ii), first suppose that (df )1 (X) = 0 for some X ∈ T1 A. Then, f (exp tX) =
exp(t(df )1 (X)) = 1 for all t ∈ R. Thus, {exp tX} ⊂ ker f = {1}. So, X = 0 and ker(df )a = 0
for all a ∈ A. ¤

We now prove Theorem 11.1:

 NOTES ON LIE GROUPS 29

Proof. Since f is continuous, its graph

Γf = {(a, f (a)) ∈ H × G | a ∈ H}
is a closed subgroup of H × G. By the Closed Subgroup Theorem, Γf is a Lie subgroup.
Consider the projections

Γf
@@
p1 }}} @@p2
}} @@
~}} @Ã
H G
Now, p1 is a group isomorphism and we can thus write f = p2 ◦ p−1 1 So, it’s enough to prove
p−1
1 is smooth. But, by Propostition 11.5, dp1 is everywhere onto and injective. By the Inverse
−1
Function Theorem, p1 is smooth. ¤
For another application of the Closed Subgroup Theorem, we look towards group actions.
Definition 11.1.6. Suppose a Lie group G acts on a manifold M . The stabilizer (or isotropy)
group of x ∈ M is

Gx = {a ∈ G | a · x = x}
The orbit of x ∈ M is

G · x = {a · x | a ∈ G} .
It is left as an exercise to prove that Gx is a subgroup of G for each x ∈ M .
Proposition 11.1.7. Suppose a Lie group G acts on a manifold M . For each x ∈ M , the
stabilizer group Gx is a Lie subgroup of G.
Proof. Choose x ∈ M . We will show that Gx is closed in G. Let A : G × M → M denote the
action of G on M . Define ιx : G → G × M by ιx (a) = (a, x). Then,

Gx = {a ∈ G | A(ιx (a)) = x}
That is, Gx = ι−1 −1
x (A (x)). Noting that both A and ιx are continuous, we’re done. ¤
We will denote the Lie algebra of Gx by gx . Note that

gx = {X ∈ g | (exp tX) · x = x, ∀t ∈ R}
Example 11.1.8. G = GL(n, R) acts on symmetric bilinear forms Sym2 ((Rn )∗ ) by (A ·
b)(v, w) = b(A−1 v, A−1 w). Let b0 = h, i be the standard inner product on Rn . Then,

Gb0 = {A ∈ GL(n, R) | hAv, Awi = hv, w, i} = O(n)

P
Example 11.1.9. G = GL(n, C) acts on H = hermitian forms on Cn . Let h0 (z, w) = z j wj
be the standard Hermitian form. By a computation similar to the previous example, Gh0 =
U (n).
30 NOTES ON LIE GROUPS

Example 11.1.10. Let B = all bilinear forms on Rn . Suppose n = 2k and examine

¿ · ¸ À Xk 2k
X
0 I
ω(v, w) = v, x = vi wi+k − vi wi−k
−I 0
i=1 i=k+1

The stabilizer group of ω is denoted Sp(R2k ) (or Sp(k, R)) and is called the symplectic group.
 NOTES ON LIE GROUPS 31

12. Lecture 12
12.1. Group Actions and Induced Vector Fields. In this lecture we will see how group
actions are used to find the Lie algebra of Lie groups given as the stabilizer of some group
action.
Suppose a Lie group G acts on a manifold M . Fix X ∈ g. Then, Φt (x) = (exp tX) · x is a
1-parameter group of diffeomorphisms of M :

Φt+s (x) = (exp (t + s)X) · x

= ((exp tX)(exp sX)) · x
= (exp tX) · (exp sX) · x
= Φt (Φs (x))
Given this 1-parameter group, the corresponding vector field (the induced vector field) is
given by
¯
d ¯¯
Xm (x) = ¯ (exp tX) · x
dt t=0
Remark 12.1.1. This gives us a map g → Γ(T M ), X 7→ XM . Is this a map of Lie algebras?
Lemma 12.1.2. Suppose a Lie group G acts on a manifold M and x ∈ M . Then,
½ ¯ ¾
d ¯¯
gx = {X ∈ g | XM (x) = 0} = X ∈ g | (exp tX) · x = 0
dt ¯t=0
Proof. If X ∈
¯ g, then exp tX ∈ Gx for all t and hence x = (exp tX)·x for all t. So, differentiating
d¯
gives 0 = dt t=0 (exp tX) · x = XM (x).
Conversely, suppose Y ∈ g and YM (x) = 0. Then, the curve γ(t) = x is an integral curve of
YM through x. By definition, the map t 7→ (exp tY · x is also an integral curve of YM through
x. But then, for all t ∈ R, (exp tY ) · x = x implies exp tY ∈ Gx and so Y ∈ gx . ¤
Example 12.1.3. Recall that GL(n, R) acts on the vector space B of all bilinear forms on Rn
by
(A · b)(v, w) = b(A−1 v, A−1 w)
Now, if b : Rn × Rn → R is a bilinear form, then
¯
d ¯¯
(db)(x,y) (v, w) = b(x + tv, y + tw)
dt ¯t=0
¯
d ¯¯
= (b(x, y) + tb(v, y) + tb(x, w) + t2 b(v, w))
dt ¯t=0
= b(x, w) + b(v, y).
What happens when we take the bilinear form to be b0 = h, i; the standard inner product on
Rn ? Recall that GL(n, R)b0 = O(n) and let’s compute the Lie Algebra o(n) of O(n). We have
32 NOTES ON LIE GROUPS

µ ¯ ¶
d ¯¯
0 = (exp tX) · b (v, w)
dt ¯t=0
¯
d ¯¯
= b(exp(−tX)v, exp(−tX)w)
dt ¯t=0
= (db)v,w (−Xv, −Xw)
= b(v, −Xw) + b(−Xv, w)
= −b(v, Xw) − b(v, X T w)
= −b(v, (X + X T )w).
© ª
Since v, w were arbitrary, we have o(n) = X ∈ gl(n, R) | X + X T = 0 .
Now we turn to examine the orbits of a Lie group action on a manifold. Recall that the
orbit of x ∈ M is G · x = {a · x | a ∈ G}.
Example 12.1.4. G = R+ acts on R by left multiplication. The orbit of any real number is
either (−∞, 0), {0} or (0, ∞).
Example 12.1.5. SO(n) act on Rn in the following fashion: if v ∈ Rn and A ∈ SO(n), then
A · v = Av (matrix-vector multiplication). The orbit of v under this action is SO(n) · v =
{w ∈ Rn | kwk = kvk}, i.e. the sphere of radius kvk.
An action of G on M defines an equivalence relation whose equivalence classes are orbits.
The relation is m ∼ m0 if and only if there exists g ∈ G so that m0 = g · m. Since this is an
equivalence relation (prove this!), if x, x0 ∈ M we have G · x ∩ G · x0 6= ∅ =⇒ G · x = G · x0 . In
particular, we may form the quotient space M/G := M/ ∼ from this relation. The elements of
this topological space are the orbits, but what is the topology?
Let π : M → M/G be defined by π(x) = G · x. This map is, of course, onto. Call a
set U ⊂ M/G open if and only if π −1 (U ) ⊂ M is open. Check that this does indeed define a
topology on M/G. We may now ask basic questions about the general topology of this quotient
space. For example, when is M/G Hausdorff?
Proposition 12.1.6. Let a Lie group G act on a space M . If the set
© ª
R = (m, m0 ) ∈ M × M | m0 = g · m for some g ∈ G
is closed in M × M , then M/G is Hausdorff.
Proof. Choose points x̄ 6= ȳ ∈ M/G. Since the canonical projection is onto, there exists
x, y ∈ M so that π(x) = x̄ and π(y) = ȳ. Since x̄ 6= ȳ, (x, y) 6∈ R. Since R is closed, there
exists (open) neighborhoods U of x ∈ M and V of y ∈ M so that (U × V ) ∩ R = ∅. Now, if
π(z) ∈ π(U ) ∩ π(V ), then there exists g1 , g2 ∈ G so that g1 z ∈ U and g2 z ∈ V . But then, we
must have (g1 z, g2 z) = (g1 z, (g2 g2−1 )g2 z) ∈ R. This is a contradiction since then

∅ = π −1 (π(U ) ∩ π(V )) = π −1 (π(U )) ∩ π −1 (π(V )).

¤
Corollary 12.1.7. Let G be a Lie group, H ⊆ G a closed subgroup. Then G/H is Hausdorff.
 NOTES ON LIE GROUPS 33

Proof. In this case, R = {(g1 , g2 ) | g2 = g1 h, h ∈ H}. The map ψ : G × G → G × G given by

ψ(a, b) = (a, ab) is a diffeomorphism and R = ψ(G×H). Since G×H is closed, R is as well. ¤
34 NOTES ON LIE GROUPS

13. Lecture 13
13.1. More on Group Actions and Principal Bundles I.
Definition 13.1.1. Let G be a Lie group and X a set. A right action of G on X is a map
X × G → X satisfying
(1) x · 1 = x,
(2) (x · g) · g 0 = x · gg 0 .
Example 13.1.2. Let X = Hom(V, W ). Then, GL(V ) acts on the left and GL(W ) acts on
the right.
Definition 13.1.3. Let H be a Lie group. A principal H-bundle is a manifold P equipped
with a right action of H such that
(1) B = P/H is a smooth manifold and the orbit map π : P → B is a submersion,
(2) For any b ∈ B, there is an open set U ⊂ B with b ∈ U and a smooth map
ψ : π −1 (U ) → U × H
so that
(i) the following diagram commutes:
ψ
π −1 (U ) / U ×H
FF xx
FF xx
F x
π FFF xx p1
F# x| x
U
(ii) ψ(p·a) = ψ(p)·a for all p ∈ P , a ∈ H where H acts on U ×H by (u, h)·a = (u, ha).
Example 13.1.4. 0) Let P = B × H, π = projection on the first factor and (b, h) · a =
(b, ha) for all (h, a) ∈ P , a ∈ H. This is called the trivial principal bundle.
1) S 2n−1 → CP n−1 is a principal S 1 bundle. S 1 acts on S 2n−1 ⊂ Cn by (z1 , . . . , zn ) · λ =
(λz1 , . . . , λzn ). The bundle projection π : S 2n−1 → CP n−1 maps (z1 , . . . , zn ) to the
equivalence class [z1 , . . . , zn ]. `
2) Let M be an n-dimensional manifold. The frame bundle of M is F r(M ) = x∈M Iso(Rn , Tx M ) ⊂
Hom(M × Rn , T M ). This is a principal GL(n, R) bundle: f · A = f ◦ A.
3) Let G be a Lie group, H ⊆ G a closed subgroup. Then, π : G → G/H is a principal
H-bundle.
Remark 13.1.5. If P is a principal H-bundle, P is called the total space of the bundle, B = P/H
is the base space, H is the structure group; for b ∈ B, π −1 (b) = Pb is the fiber above b. The
π
bundle is often written H → P → B.
Remark 13.1.6. A right action of a Lie group G on a manifold M is free if for all x ∈ M ,
x·g = x =⇒ g = 1. A right action is proper if the map f : M ×G → M ×M , (m, g) 7→ (m, m·g)
is proper; i.e. preimages of compact sets are compact. We have the following result:
Theorem 13.1.7. If a Lie group G acts freely and properly on a manifold P , then P/G is a
manifold and G → P → P/G is a principal G-bundle.
 NOTES ON LIE GROUPS 35

The last example above of a principal bundle was slipped in without proof. Let’s take care
of that now.
Theorem 13.1.8. let G be a Lie group, H ⊆ G a closed subgroup. Then, G/H is a smooth
π
manifold and H → G → G/H is a principal H-bundle. Moreover, T1H (G/H) ∼
= g/h.
Proof. We will show only that G/H is a smooth manifold and that T1H (G/H) ∼ = g/h. The
proof that G → G/H is a principal H−bundle is left to the reader. Recall from the proof of
the Closed Subgroup Theorem that if we choose a linear subspace m ⊂ g so that g = h × m
then the map φ : m × h → G, φ(Y, X) = exp Y exp X has the properties that
(1) φ is a diffeomorphism on a neighborhood V × U of (0, 0) in m × h,
(2) If U and V are sufficiently small, then
φ(V × U ) ∩ H = φ({0} × U ) = exp(U )
Choose V ⊂ V open so that Y1 , Y2 ∈ V 0 implies Y1 − Y2 ∈ V . Let Σ = exp(V 0 ). Then,
0

s1 , s2 ∈ Σ =⇒ s−12 s1 ∈ exp(V ).
Claim 1: The map ψ : Σ × H → G, ψ(s, h) = sh is injective.
To prove this claim, first suppose that s1 , s2 ∈ Σ, h1 , h2 ∈ H and s1 h1 = s2 h2 . Then
φ(V × U ) ⊃ φ(V × {0}) = exp(V ) 3 s−1 −1
2 s1 = h2 h1 ∈ H
But H ∩ φ(V × U = φ({0} × U ) and φ({0} × U ) ∩ φ(V × {0}) = φ(0, 0) = 1. Hence, s1 = s2
and h1 = h2 .
Claim 2: If V 0 is sufficiently small, then
ψ : exp(V 0 ) × H = Σ × H → G
is an open embedding.
Let’s compute (dψ)(1,1) : T1 Σ × T1 H → T1 G = g. Write T1 Σ = m and T1 H = h. For all
Y ∈ m and X ∈ h we have
¯
d ¯¯
(dψ)(1,1) (Y, 0) = ψ(exp Y · 1) = Y
dt ¯t=0
¯
d ¯¯
(dψ)(1,1) (0, X) = ψ(1 · exp X) = X
dt ¯
t=0
and hence (dψ)(1,1) (Y, X) = Y + X. So, if V 0 is sufficiently small, (dψ)(s,1) : Ts Σ × T1 H → Ts G
is an isomorhism for all s ∈ Σ.
Now, the diagram

ψ
Σ ×O H /G
O
id ×Rh Rn
ψ
Σ×H /G
commutes: i.e. ψ(s, ah) = sah = Rh sa = (Rh ◦ψ)(s, a) and it follows that, for all (s, h) ∈ Σ×H,
that

(dψ)(s,h) : Ts Σ × Th H → Tsh G
36 NOTES ON LIE GROUPS

is an isomorphism. Hence, ψ is an open embedding. ¤

 NOTES ON LIE GROUPS 37

14. Lecture 14
14.1. Principal Bundles II. In this lecture we continue the proof of Theorem 13.8. Recall
we have the embedded submanifold Σ = exp(V 0 ) of G and an open embedding ψ : Σ × H → G,
ψ(s, h) = sh. Note that ψ(s, ha) = sha = ψ(s, h)a. That is, ψ is equivariant.
Since H is a closed subgroup of G, we know that G/H is a Hausdorff topological space. What
remains is to produce charts on G/H and prove that it is a principal H-bundle (exercise).
Proof. Since ψ is an open embedding, for all a ∈ G ψa = La ◦ ψ : Σ × H → G, ψa (s, h) = ash
is also an open embedding which is H-equivariant. So, the following hold:
(1) For any orbit gH of H, gH ∩ aΣ (a ∈ G) is a single point.
(2) For any a ∈ G, π ◦ La : Σ → G/H is 1-1. Note that

ψa
Σ×H /G

p1 π
² π◦La
²
Σ / G/H
commutes.
(3) The map π ◦ La is open:
Suppose U ⊆ Σ is open. Then U × H ⊆ Σ × H is open. So, ψa (U × H) = aU H ⊆ G
is open and we can conclude that π(aU H) ⊂ G/H is open as well. But, π(aU H) =
π(aU ) = (π ◦ La )(U ).
We conclude from this that {π(aΣ) | a ∈ G} is an open cover of G/H and π ◦ La :
Σ → π(aΣ) ⊂ G/H are homeomorphisms.
(4) We define an atlas on G/H to be

{π(aΣ), τa = (π ◦ La )−1 : π(aΣ) → Σ}

We need to check that for any a, b ∈ G, if W := π(aΣ) ∩ π(bΣ) 6= ∅, then τb ◦ τa−1 :
τa (W ) → Σ is C ∞ . Consider the following commutative diagram:

ψb
/
Σ×H o bΣH ⊆ G
gOOO
ψb−1 OOO
OO
La OOO
p1 π τa (W ) ⊆ Σ
π◦Laoooo
o
o
π◦Lb ooooo
² / ² ow
Σo πbΣ
τb

Now, by definition, τa−1 = π ◦ La . Since π ◦ ψb = p1 ◦ (π ◦ Lb ), we must have τb ◦ π =

(π ◦ Lb )−1 ◦ π = p1 ◦ ψb−1 . Hence,

τb ◦ τa−1 = τb ◦ (π ◦ La ) = p1 ◦ ψb−1 ◦ La
which is a smooth map. Thus, G/H is a smooth manifold and π : G → G/H is a C ∞
submersion.
38 NOTES ON LIE GROUPS

Note finally that

ker{(dπ)1 : T1 G → Tπ(1) G/H} = T1 H

and so Tπ(1) G/H ∼ T
= 1 G/T1 H = g/h.
¤
Remark 14.1.1. If f : G/H → M is a C ∞ map of manifolds, then π ◦ f : G → M is also C ∞ .
The converse is true as well: if π ◦ f : G → M is C ∞ , then so is f : G/H → M . Why? Since f
is C ∞ , for all a ∈ G, f ◦ τa−1 : Σ → M is C ∞ . But, f ◦ τa−1 = (f ◦ π) ◦ La .
Similarly, f : N × G/H → M is smooth if and only if f ◦ (id ×π) : N × G → M is.
 NOTES ON LIE GROUPS 39

15. Lecture 15
15.1. Transitive Actions. In this section, we will use the notion of a transitive group action
to determine the uniqueness of the smooth structure on G/H.
Definition 15.1.1. An action of a Lie group G on a manifold M is transitive if for every
m, m0 ∈ M , there exists g ∈ G so that g · x = x0 . If the action of G on M is transitive, we say
M is a homogeneous G-space.
Example 15.1.2. The standard action of SO(n) on S n−1 ⊂ Rn is transitive.
Example 15.1.3. For a closed subgroup H of a Lie group G, the action of G on G/H given
by g · aH = gaH is transitive.
Example 15.1.4. The standard action of SU (n) on S 2n−1 ⊂ Cn is transitive:
given v ∈ S 2n−1 , extend to an orthonormal basis (v = v1 , v2 , ldots, vn ) of Cn . Let λ =
det[v1 | . . . |vn ]. Define A = [v1 | . . . |λ−1 vn ] ∈ SU (n) and note that Ae1 = v where e1 =
£ ¤T
1 0 ... 0 .
Theorem 15.1.5. Suppose a Lie group G acts transitively on a manifold M . Fix x ∈ M .
Then, the evaluation map
evx : G → M, evx (g) = g · x
induces a diffeomorphism β : G/H → M , β(aH) = a · x, where H = Gx (the stabilizer of x).
Proof. We first check that β is well-defined: suppose aH = a0 H for some a, a0 ∈ G. But, this
is if and only if a = a0 h for some h ∈ H. Now, β(aH) = a · x = (a0 h) · x = a0 · (hx) = β(a0 H).
Since the action of G on M is transitive, β is onto. Also, β is 1-1:

β(aH) = β(a0 H) ⇐⇒ a · a = a0 · x
⇐⇒ (a−1 a0 ) · x = x
⇐⇒ a−1 a0 ∈ H
⇐⇒ aH = a0 H
Next, observe that

G EE
EE evx
π EE
EE
² E"
β
G/H /M

commutes. Hence, for any open set U ⊂ M , π −1 (β −1 (U )) = ev−1 x (U ) is open. Since evx is
continuous, so it β. Finally, since evx = β ◦ π is C ∞ , β is a smooth map.
Since β is onto, Sard’s Theorem implies that

(dβ)aH : TaH (G/H) → Ta·x M

is onto for some aH. Now,
40 NOTES ON LIE GROUPS

T1 G K
KKK(d ev )
(dπ)1
KKK x 1
KKK
² %
T1H G/H / Tx M
(dβ)1H

commutes and (dπ)1 is onto. Thus,

ker (dβ)1H ∼
= ker d(β ◦ π)1 / ker (dπ)1
= ker (d evx )1 /h
Now,
½ ¯ ¾
d ¯¯
ker (d evx )1 = X ∈ g | 0 = ¯ evx (exp tX)
dt t=0
½ ¯ ¾
d ¯¯
= X ∈ g | 0 = ¯ (exp tX) · x
dt t=0
= h
Thus, (dβ)1 is 1-1. We conclude that β is a diffeomorphism. ¤
Remark 15.1.6. Whenever a Lie group acts on a manifold M , we get G/Gx → G · x ⊆ M for
any x ∈ M , 1-1 immersions. In general, these maps are not embeddings, but if G is compact
they are.
We now return to some examples.
Example 15.1.7. SO(n) acts transitively on S n−1 and by the theorem, S n−1 ∼
= SO(n)/H.
But what is H?

H = {A ∈ SO(n) | Ae1 = e1 }
   

 1 0 ... 0   


 0 ∗ ∗ ∗ ∗ ∗ 
 ∗
 

= A ∈ SO(n) | A =  ..  , ∗ ∗ ∗ ∈ SO(n − 1)

 . ∗ ∗ ∗ ∗ ∗ ∗ 


 

0 ∗ ∗ ∗
∼
= SO(n − 1)
Thus, S n−1 ∼
= SO(n)/SO(n − 1). In particular, S 1 ∼
= SO(2)/SO(1) = SO(2).
Example 15.1.8. In a similar fashion, we see that from the transitive action of SU (n) on
S 2n−1 , S 2n−1 ∼
= SU (n)/SU (n − 1). In particular, S 3 ∼
= SU (2).
What do these examples tell us about the topological properties of SO(n) and SU (n)? It
would seem that we can at least assert each is connected. First, a result from general topology.
Lemma 15.1.9. Suppose that f : X → Y is a continuous map of topological spaces which is
surjective and open. If the fiber f −1 (y) is connected for any y ∈ Y , and Y itself is connected,
then X is connected.
 NOTES ON LIE GROUPS 41

Proof. Suppose U, V are two open sets in X such that U ∪ V = X. We wish to show that
U ∩ V 6= ∅. Since U ∪ V = X and f is surjective, f (U ) ∪ f (V ) = Y . Also, both f (U ) and f (V )
are open, since f is an open mapping. Since Y is connected, f (U ) ∩ f (V ) 6= ∅. Let y be a point
in this intersection. Then, f −1 (y) ∩ U 6= ∅ and f −1 (y) ∩ V 6= ∅. Since f −1 (y) is connected,
(f −1 (y) ∩ U ) ∩ (f −1 (y) ∩ V ) ⊆ U ∩ V 6= ∅.
¤
π
Note now that is F → P → B is a principal bundle, π is a surjective, open map.
Proposition 15.1.10. SO(n) and SU (n) are connected.
Proof. We proceed by induction on n. The base cases are handles in the above examples, as
both S 1 and S 3 are connected. Now, suppose n − 1 ≥ 2 and consider the principal SO(n − 1)
π
bundle SO(n − 1) → SO(n) → SO(n)/SO(n − 1) ∼ = S n−1 . Assuming SO(n − 1) is connected,
1
using the lemma and the fact that S is connected, SO(n) is connected as well.
A similar argument works for SU (n). ¤
42 NOTES ON LIE GROUPS

16. Lecture 16
16.1. Fiber Bundles.
Definition 16.1.1. A (C ∞ ) fiber bundle π : E → B with total space E, base space B and
typical fiber F is a submersion such that for any b ∈ B, there exists an open neighborhood U
of b and a diffeomorphism φ : π −1 U → U × F so that
φ
π −1 (U ) / U ×F
FF y
FF yy
F
π FFF yyyp
y 1
F# y| y
U
commutes.
We already know some examples of fiber bundles:
(1) vector bundles are fiber bundles having finite dimensional vector spaces as typical fibers,
(2) principal bundles are fiber bundles having Lie groups as typical fibers,
(3) covering spaces of manifolds are fiber bundles having discrete sets of points as typical
fibers.
Our motivation for introducing the concept of a fiber bundle is to compute some topological
invariants of Lie groups. In fact, we have already done so when it was proved that both
SO(n) and SU (n) are connected via the fiber bundles SO(n − 1) → SO(n) → S n−1 and
SU (n − 1) → SU (n) → S 2n−1 .
Given a topological space X, we may (try) to compute it’s fundamental group π1 X. We will
show:
Proposition 16.1.2. π1 SU (n) is trivial for all n.
Proposition 16.1.3. π1 SO(n) ∼
= π1 SO(3) for all n ≥ 3.
Remark 16.1.4. Later, we will use the Adjoint representation to show that π1 SO(3) ∼
= Z/2Z.
16.2. Prerequisites from Homotopy Theory.
Definition 16.2.1. Let f0 , f1 : X → Y be continuous maps of topological spaces. We say f0
and f1 are homotopic if there exists a continuous map F : X × [0, 1] → Y satisfying

F (x, 0) = f0 (x)
F (x, 1) = f1 (x)
We may also fix a subset A ⊂ X and require F (x, t) = y0 for some y0 ∈ Y and all t ∈ [0, 1]
and all x ∈ A IN this case we say f0 and f1 are homotopic relative to A.
One can prove that continuous maps being homotopic is an equivalence relation. The equiv-
alence classes are written as [(X, A), (Y, y0 )].
Example 16.2.2. Let X = S 1 and A = {x0 } ∈ S 1 . Then, the equivalence classes are
[(S 1 , x0 ), (Y, y0 )] = π1 Y , the fundamental group of Y . The group operation is “concatenation”
of loops.
 NOTES ON LIE GROUPS 43

Letting I denote the unit interval and ∂I its boundary, we can also think of the fundamental
group as [(I, ∂I), (Y, y0 )]. At first, this does not seem too advantageous, but it allows us a way
to think of the higher homotopy groups: πk Y := [(I k , ∂I k ), (Y, y0 )].
Fact: πk S n is trivial for k < n.
The homotopy groups are functors on the category of pointed spaces and continuous maps;
in particular if f : (X, x0 ) → (Y, y0 ) is a continuous map, it induces a group homomorphism
f∗ : πk (X, x0 ) → πk (Y, y0 ). This comes into play in the following theorem.
π
Theorem 16.2.3. Let F ,→ E → B be a fiber bundle. Then, there is an exact sequence

∂2
... / π2 (B, b0 ) / π1 (F, f0 ) i∗
/ π1 (E, e0 ) π∗
/ π1 (B, b0 ) ∂1 / π0 (F )

where i∗ , π∗ , ∂2 . . . are group homomorphisms and ∂1 is a map of sets. Exactness at π1 (B, b0 )

means the fibers of ∂1 are cosets of π∗ (π1 (E, e0 )) in π1 (B, b0 ).
Idea of where ∂2 comes from in the exact sequence:
Suppose σ : (I 2 , ∂I 2 ) → (B, b0 ) represents a class in π2 (B, b0 ) Lift σ to a map σ̃ : I 2 → E
(we are using the fact that E → B is a fiber bundle here). Then, σ̃(∂I 2 ) ⊂ π −1 (b0 ) ⊂ F . This
gives a map α : ∂I 2 = S 1 → F , and hence an element of π1 F . Define ∂2 [σ] = [α].
Let’s apply this theorem and prove Propositions 16.2 and 16.3. Consider first the fiber
bundle SU (n − 1) → SU (n) → S 2n−1 . We proceed by induction on n. For n ≥ 3, the portion
of the homotopy long exact sequence we are interested in looks like

π1 SU (n − 1) → π1 SU (n) → π1 S 2n−1
By induction, and the fact listed above, we may rewrite this sequence as

{1} → π1 SU (n) → {1}

and hence π1 SU (n) ∼
= {1}
In a similar manner, considering the fiber bundle SO(n − 1) → SO(n) → S n−1 we look at

π2 S n−1 → π1 SO(n − 1) → π1 SO(n) → π1 S n−1

Since π1 S n−1 = {1} for n > 2 and π2 S n−1 = {1} for n > 3, we see that

π1 SO(n) ∼
= π1 SO(n − 1), n − 1 ≥ 3
44 NOTES ON LIE GROUPS

17. Lecture 17
17.1. Deformation Retracts of Classical Lie Groups.
Definition 17.1.1. A space deformation retracts onto a subspace A ,→ X if the identity map
on X is homotopic to a map r : X → X such that r(X) ⊆ A and r|A = idA .
Theorem 17.1.2. Each of the following Lie groups admits a deformation retraction onto the
indicated subspace:

GL(n, R) → O(n)
SL(n, R) → SO(n)
GL(n, C) → U (n)
SL(n, C) → SU (n)
Proof. (1) The main idea behind each of these proofs is the Gram-Schmidt method from
linear algebra. Let

T + (n, R) = {(aij ) ∈ GL(nR) | aii > 0, aij = 0 for i < j}

2
Then T + (n, R) is a Lie group and is diffeomorphic to (R>0 )n × R(n −n)/2 and is hence
contractible.
Claim 1: Any B ∈ GL(n, R) can be written uniquely as B = AT where A ∈ O(n)
and T ∈ T + (n, R).
To prove this, suppose that B = [v1 | . . . |vn ] and apply Gram-Schmidt to {v1 , . . . , vn }
to obtain an orthonormal set of vectors {w1 , . . . , wn }. Let W = [w1 | . . . |wn ] ∈ O(n).
Note that, by construction, we have

v1 = kv1 kw1
v2 = kv2 − hv2 , w1 ikw1 kw2 + hv2 , v1 iw1
..
.
That is,
v1 = a11 w1
v2 = a12 w1 + a22 w2
..
.
where aij > 0. But this says exactly that we have obtained the factorization B = W A
where  
a11 a12 . . . a1n
 0 a22 . . . a2n 
 
A= 0 0 . . . a3n 
 
.. .. .. ..
. . . .
is in T + (n, R).
 NOTES ON LIE GROUPS 45

What about uniqueness? Suppose W1 A1 = W2 A2 where Wi ∈ O(n) and Ai ∈

T + (n, R). Then, X := W2−1 W1 = A2 A−1 +
1 ∈ O(n) ∩ T (n, R). Since X is orthogonal,
X −1 = X T . Since X ∈ T + (n, R), so is X −1 and hence X T . But then X is a diagonal
matrix with positive entries so that X T = X −1 = X. Hence, X is the identity.
Now, existence and uniqueness of the factorization B = W A is equivalent to the map

φ : O(n) × T + (n, R) → GL(n, R)

(W, A) 7→ W A
Now, φ is a smooth map since it’s a polynomial. Is it a diffeomorphism? Note that

© ª
o(n) = X ∈ gl(n, R) | X + X T = 0
t+ = {Y ∈ gl(n, R) | Y is upper triangular}
Let Eij be the matrix with a 1 in the ijth entry and zeros elsewhere. Then,

Eji = (Eji − Eij ) + Eij =⇒ gl(n, R) = o(n) ⊕ t+

and the map δφ : TI O(n) × TI T + (n, R) → T1 GL(n, R) is an isomorphism. But φ is also
equivariant: for all W1 , W2 ∈ O(n) and A1 , A2 ∈ T + (n, R), we have

φ(W1 W2 , A1 A−1 −1 −1
2 ) = W1 W2 A1 A2 = W1 φ(W2 , A1 )A2
Hence, φ is a diffeomorphism and GL(n, R) deformation retracts onto O(n). A similar
proof works to show that GL(n, C) retracts onto U (n).
(2) Now, suppose that B ∈ SL(n, R). Then, as above there is a unique factorization
B = W A. Note that 1 = det B = det W det A = ±1 det A. and det A is positive.
Hence, A ∈ SL(n, R) ∩ T + (n, R). Now,
n Y o
T + (n, R) ∩ SL(n, R) = (aij ) | aii > 0, aii = 1
n X o 2
≈ (ex1 , . . . , exn ) | xi ∈ R, xi = 0 × R(n −n)/2
≈ Rn+(n−2)/2
Therefore, SL(n, R) ≈ SO(n) × Rn+(n−2)/2 and from this we see SO(n) is a deforma-
tion retract of SL(n, R). A similar proof works in showing that SL(n, C) deformation
retracts onto SU (n).
¤
Remark 17.1.3. From the above computations we can say a few more things about the classical
Lie groups:
 
−1 0 . . . 0
 ..  `
(1) O(n) =  0 1 . 0 SO(n) SO(n) and hence O(n) has exactly two connected
0 ... ... 1
components.
(2) SL(n, R) is connected and π1 SL(n, R) = π1 SO(n).
46 NOTES ON LIE GROUPS

(3) The map

φ : U (1) × SU (n) → U (n)
 
λ
 .. 
(λ, A) 7→  . A
λ
is onto and ker φ ∼
= {λ | λn = 1}. Hence, U (n) is connected and π1 U (n) = π1 U (1) ∼
= Z.
Furthermore, GL(n, C) is connected and π1 GL(n, C) ∼ = Z as well.
(4) SL(n, C) is connected and π1 SL(n, C) = π1 SU (n) = {1}.
f π
Theorem 17.1.4. Let SL(2, R) → SL(2, R) denote the double cover of SL(2, R). Then, any
f
Lie group homomorphism ρ : SL(2, R) → GL(n, R) has a nontrivial kernel.
Proof. Consider the following commutative diagram:

(δρ)C
sl(2, C) / gl(n, C)
O O

Â? δρ Â?
sl(2, R) / gl(n, R)

exp exp
² ²
ρ
exp f
SL(2, R) / GL(n, R) exp
;; Ä_
;;
π ;;
² ;;
; τ ◦π
SL(2,Ä R) ;;;
_ ;;
;;
;;
 ² À ² Ä
τ /
SL(2, C) GL(n, C)
where
(i) sl(2, C) = sl(2, R) ⊕ isl(2, R) (similarly for gl(2, C)),
(ii) (δρ)C is the extension of δρ with respect to the decomposition in (i), and
(iii) since π1 sl(2, C) = {1}, there is a unique τ : SL(2, C) → GL(n, C) such that (dτ )I = δρ.
f
Now, since SL(2, R) is connected, τ ◦ π = ρ. But π is a covering map with two sheets and
∼
ker ρ ⊇ ker π = Z/2Z. ¤
 NOTES ON LIE GROUPS 47

18. Lecture 18
18.1. Compact Connected Abelian Groups. In this lecture, we prove that all compact,
connected, abelian Lie groups are isomorphic to tori. This will aid us greatly when we examine
representations of Lie groups.
Theorem 18.1.1. Any compact, connected, abelian Lie group is isomorphic to Tn := (S 1 )n ∼ =
Rn /Zn for some n.
Before we prove this theorem, we’ll need a few results about discrete subgroups of finite
dimensional real vector spaces.
Lemma 18.1.2. Let V be a finite dimensional real vector space and Λ ⊂ V a discrete subgroup.
Then, there is a set {v1 , . . . , vn } P
which is linearly independent such that for all λ ∈ Λ, there
exists n1 , . . . , nk ∈ Z so that λ = ni vi . In other words, Λ = span Z {v1 , . . . , vn }.
Proof. We proceed by induction on dim V . Suppose that dim V = 1. We may assume V = R.
Let v1 = inf {λ ∈ Λ | λ > 0}. Since Λ is discrete, v1 6= 0 and Λ = Zv1 .
Now, fix an inner product on V . Since Λ is discrete, Λ has a shortest nonzero vector, call it
v1 . Then,

0 < kv1 k ≤ kλk for all λ ∈ Λ, λ 6= 0.

Let W = (Rv1 )⊥ , the orthogonal complement so that V = Rv1 ⊕ W . Let p1 : V → Rv1 and
p2 : V → W be the corresponding orthogonal projections. If we can show that p2 (Λ) is discrete,
we are done by induction.
Since p2 (Λ) is a subgroup of W , it is enough to show: for any λ ∈ Λ with p2 (λ) 6= 0,
kp2 (λ)k ≥ 1/2kv1 k. Suppose not, that is there exists λ ∈ Lambda so that 0 < kp2 (λ)k <
1/2kv1 k. Now, for some N ∈ Z,

N v1 ≤ p1 (λ) ≤ (N + 1)v1
and thus there exists an integer m such that

kp1 (λ − mv1 )k = kp1 (λ) − mv1 k ≤ 1/2kv1 k

This implies

kλ − mv1 k = (kp1 (λ − mv1 k)2 + kp2 (λ − mv1 )k2 )1/2

≤ (kp2 (λ)k2 + 1/4kv1 k2 )1/2
< ((1/2kv1 k)2 + (1/2kv1 k)2 )1/2
√
= 1/ 2kv1 k < kv1 k
But this contradicts the choice of v1 . ¤
Corollary 18.1.3. Let V be an n-dimensional real vector space and λ ⊂ V a discrete subgroup.
Then, V /Λ ∼
= (Rk /Zk ) × Rn−k .
Proof. Choose {v1 , . . . , vk } ⊂ V such that span Z {v1 , . . . , vk } = Λ. Complete
P this to a basis
k
{v1 , . . . , vn } of V . The isomorpism R × Rn−k → V , (a1 , . . . , an ) 7→ ai vi identifies Zn with
∼
Λ. Hence, V /Λ = R /Z × R k k n−k . ¤
48 NOTES ON LIE GROUPS

We are now in a position to prove Theorem 18.1:

of Theorem 18.1. We first argue that exp : g → G is a surjective Lie group homomorphism.
Since G is abelian, the multiplication map m : G × G → G is a homomorphism. Since exp is
natural, the diagram

δm /g
g×g
exp × exp exp
² m ²
G×G /G
commutes. Since δm(X, Y ) = (dm)(1,1) (X, Y ) = X + Y we get
exp(X + Y ) = exp X exp Y
and so exp : g → G is a homomorphism. Now, exp(g) is a connected subgroup containing an
open neighborhood of 1 ∈ G. Since G is connected, it follows that exp is onto.
Since (d exp)0 = id, we see from the above that exp is a covering map. Hence, Λ =
ker{exp} ⊂ g is a discrete subgroup.
Since g/Λ = G is compact, there is a basis {v1 , . . . , vn } of g so that Λ = span Z {v1 , . . . , vn }.
Hence, G ∼
= g/Λ ∼ = Rn /Zn = Tn . ¤
 NOTES ON LIE GROUPS 49

19. Lecture 19
19.1. Representations of Lie Groups. Recall that a representation of a Lie group G is a
homomorphism from G to the general linear group of some finite dimensional vector space. Let
ρ : G → GL(V ) be such a representation.
Definition 19.1.1. A subrepresentation is a subspace W ⊂ V such that for any g ∈ G, w, ∈ W
we have ρ(g)w ∈ W . In other words, W is a G-invariant subspace.
if ρ1 : G → GL(V1 ) and ρ2 : G → GL(V2 ) are two representations of G, then we have

ρ1 ⊕ ρ2 : G → GL(V1 ⊕ V2 )
(ρ1 ⊕ ρ2 )(g(v1 + v2 ) = ρ1 (g)v1 + ρ2 (g)v2
A representation (real or complex) ρ : G → GL(V ) is irreducible if it has no nontrivial
invariant subspaces: if W ⊂ V is such that ρ(g)W ⊂ W for all g ∈ G, then either W = {0} or
W =V.
Example 19.1.2. ρ : S 1 → GL(1, C) given by ρ(λ) = lambda is irreducible since dim C = 1.
Example 19.1.3. ρ : SU (2) → GL(2, C) is irredducible: SU (2) acts transitively on S 3 and
span C (S 3 ) = C2 .
· ¸
1 t
Example 19.1.4. ρ : R → G : (2, R) given by ρ(t) = is not irreducible: R × {0} is
0 1
invariant.
Definition 19.1.5. A complex representation ρ : G → GL(V ) is unitary if there is a Hermitian
inner product h, i on V such that

hρ(g)v, ρ(g)wi = hviw, ∀g ∈ G, ∀v, w, ∈ V

That is, the representation of G on V preserves h, i.
Example 19.1.6. ρ : S 1 → GL(2, C) given by ρ(eiθ )z = eiθ z is unitary but not irreducible
(Cz is invariant for any z).
Lemma 19.1.7. Let ρ : G → GL(V ) be a unitary represenatation. Then, ρ is a direct sum of
irreducible representations.
Proof. We proceed by induction on dimC V . If dimC V = 1, then ρ is irreducible. Suppose
dimC V > 1 and V is not irreducible. Then, there exists an G-invariant subspace W , dimC W 6=
dimC V . Let W ⊥ denote the orthogonal complement of W with respect to the Hermitian inner
product on V .
claim: W ⊥ is an invariant subspace.
To prove this, take v ∈ W ⊥ . Then, for any w ∈ W and any g ∈ G

hρ−1 (G)w, vi = hρ(g)ρ(g −1 )w, ρ(g)vi

= hw, ρ(g)vi
50 NOTES ON LIE GROUPS

Since W is invariant, ρ(g −1 )w ∈ W and so 0 = hρ(g −1 )w, vi = hw, ρ(g)vi. Hence, ρ(g)v ∈ W ⊥
for all g ∈ G. Thus, V = W ⊕ W ⊥ where (by induction and assumption) both W and W ⊥ are
invariant. ¤
Definition 19.1.8. A representation ρ : G → GL(V ) is completely reducible if it is a direct
sum of irreducible representations.
Remark 19.1.9. We have thus proved that any unitary representation is completely reducible.
Example 19.1.10. The representation rho : R → GL(2, C) from Example 19.4 is neither
irreducible nor completely reducible. Note that C × {0} is invariant. If w 6∈ C × {0} say
w = (w1 , w2 ) then w2 6= 0. So
· ¸· ¸ · ¸
1 t w1 w1 + tw2
=
0 1 w2 w2
· ¸
t
Set t = −w1 /w2 so that ρ(t)2 w = w2 . Hence
1
½ · ¸ ¾
w
span C ρ(t) 1 | t ∈ C = C2
w2
The following proposition is a glance at results to come:
Proposition 19.1.11. Any complex representation of a finite group G is unitary.
Proof. let ρ : G → GL(V ) be a representation. Pick a hermitian inner product h, i on V . It
need not be invariant. Now, define

1 X
¿ v, w À= hρ(g)v, ρ(g)wi
|G|
g∈G

where |G| is the number of elements in G.

Then, for any a ∈ G, v, w ∈ V

1 X
¿ ρ(a)v, ρ(a)w À = hρ(g)ρ(a)v, ρ(g)ρ(a)wi
|G|
g∈G
1 X
= hρ(ga)v, ρ(ga)wi
|G|
g∈G

But, Ra : G → G is a bijection. So, let g 0 = ga. Then, the last equality becomes

1 X
hρ(g 0 )v, ρ(g 0 )wi =¿ v, w À
|G| 0
g ∈G

It follows that ¿, À is invariant. It is clear that ¿, À is sesquilinear and moreover, for v 6= 0,

¿ v, v À> 0. Hence, ¿, À is an invariant Hermitian inner product.
¤
 NOTES ON LIE GROUPS 51

We’d like to do the same thing for Lie groups: if h, i is a Hermitian inner product on a
representation ρ : G → GL(V ) of G, then for fixed v, w ∈ V g 7→ hρ(g)v, ρ(g)wi is a function
on G.
Also, for v 6= 0, fv (g) = hρ(g)v, ρ(g)vi > 0 for all g. Thus, for an appropriate measure dµg ,
we have
Z
fv (g)dµg > 0
Z G

If |G| := µg < ∞, then

G
Z
1
¿ v, w À= hρ(g)v, ρ(g)widµg
|G| G
makes sense and is a Hermitian inner product. Hence,
Theorem 19.1.12. Any representation of a compact Lie group is completely reducible.
52 NOTES ON LIE GROUPS

20. Lecture 20
20.1. Schur’s Lemma.
Definition 20.1.1. let V1 , V2 be two representations of Lie group G. A linear map

T : V1 → V2
intertwines the two representations if T (g · v) = g · T (v) for all g ∈ G, v ∈ V1 . We write

HomG (V1 , V2 ) = {T : V1 → V2 | T intertwines V1 , V2 }

Proposition 20.1.2 (Schur’s Lemma, Version 1). Suppose T1 , T2 are two irreducible represen-
tations and T ∈ HomG (V1 , V2 ). Then, either T = 0 or T is an isomorphism.
Proof. Note that since T is intertwining, ker T and im T are invariant subspaces of V1 and V2
respectively.
(i) T v = 0 =⇒ T (g · 0) = g · (T V ) = 0
(ii) g · (T v) = T (g · v) ∈ im T .
Suppose T 6= 0. Them im T 6= 0 and hence im T = V2 . Since im T 6= 0, ker T 6= V1 and T is an
isomorphism. ¤
Proposition 20.1.3 (Schur’s Lemma, Version 2). Suppose V1 and V2 are two irreducible rep-
resentations. Suppose further V is complex. Then,
HomG (V, V ) ∼
=C
That is to say, if T : V → V intertwines then there is λ ∈ C, depending on T , such that
T v = λv for all v ∈ V .
Proof. Let λ be an eigenvalue of T . Then, T − λ id : V → V is an intertwining map and
ker(T − λ id) 6= 0. Thus, T − λ id = 0. ¤
Corollary 20.1.4. Any complex irreducible representation of a compact abelian group is 1-
dimensional.
Proof. Let ρ : G → GL(V ) be a complex irreducible representation. For any a ∈ G, ρ(a) : V →
V is an intertwining map. Thus, ρ(a) = λa for some λa ∈ C. Since a is arbitrary, ρ(a) = λa
for all a ∈ G. Hence, any 1-dimensional subspace is ireeducible and dim V = 1. ¤
20.2. Irreducible Representations of Tn .
Lemma 20.2.1. Any complex irreducible representation ρ : Tn → GL(V ) is of the form

ρ(exp v) = ρ(v mod Zn ) = e2πiδρ(v)

where δρ : Rn → Rn satisfies δρ(Zn ) ⊂ Zn (i.e. δρ ∈ HomZ (Zn , Z)).
Proof. If µ ∈ HomZ (Zn , Z) ⊂ HomR (Rn , R), then

ξµ : Rn /Zn → R/Z
ξµ (v mod Zn ) = µ(v) mod Z
 NOTES ON LIE GROUPS 53

is well-defined. We identify R/Z with S 1 : a mod Z 7→ e2πia . So, ξµ is a representation

ξµ : Tn → Gl(C)
mod Zn ) = e2πiµ(v)
ξµ (v
Conversely, suppose that ρ : Tn → GL(C) ∼
= C \ {0} is a representation. Since ρ(Tm ) ⊆ C×
n 1
is compact, ρ(T ) ⊆ S . We have a commuting diagram

δρ
Rn /R

exp exp
² ρ ²
Tn / S1
so δρ(ker exp) ⊆ ker exp = Z. Therefore, δρ ∈ HomZ (Zn , Z) ¤
Remark 20.2.2. If G is any compact, connected abelian Lie group (i.e. a torus),

ZG := ker{exp : g → G}
is called the integral lattice. The set

Z∗G := HomZ (ZG , Z)

is called the weight lattice
The above lemma proves: irreducible, unitary represenations of G are in one-to-one corre-
spondence with elements of the weight lattice.
20.3. Representations of SU (2). We start by constructing complex irreducible representa-
tions of SU (n). Let Vn be the set of all complex homogeneous polynomials of degree n in two
variables. That is

vn = span C {z1n , z1n−1 z2 , . . . , z2n }

Note that V0 = C and V1 ∼
= C2 .
We have an action of GL(2, C) on Vn .

(A · f )(z1 , z2 ) = f ((z1 , z2 )A)

where (z1 , z2 )A is regarded as matrix multiplication.
It is left as an exercise to the reader to prove this indeed defines an action. This also gives
us a representation:

A · (λf + µg) = λ(A · f ) + µ(A · g)

for all λ, µ ∈ C, f, g ∈ Vn and A ∈ GL(2, C).
Theorem 20.3.1. Let Vn be as above. Then,
(i) Vn is an irreducible representation of SU (n) for all n ≥ 0.
(ii) If V is an irreducible representation of SU (n) of dimension n + 1 then V ∼
= Vn as
representations.
54 NOTES ON LIE GROUPS

We defer the proof of this theorem for now.

 NOTES ON LIE GROUPS 55

21. Lecture 21
21.1. Complexification.
Definition 21.1.1. let V be a finite dimensional vector space over R. The complexification
VC of V is V ⊗ C.
Note that VC is a complex vector space: for any a, b ∈ C, v ∈ V we have a(v ⊗ b) = v ⊗ ab.
Also, V is embedded in VC as a real subspace

V ,→ VC , v 7→ v ⊗ 1
We now identify V with V ⊗ 1 ⊂ VC and we write av for v ⊗ a, v ∈ V , a ∈ C.
As a real vector space, VC = V ⊕ iV , where iV = {v ⊗ i | v ∈ V }.
Remark 21.1.2. If {v1 , . . . , vn } is a basis of V , then it is also a complex basis of VC . Considered
as a real vector space, VC has as a basis the set {v1 , . . . , vn , iv1 , . . . , ivn }.
Lemma 21.1.3. let V be a real vector space, W a complex vector space and T : V → W an
R-linear map. Then, there exists a unique C-linear map TC : VC → W extending T .
Proof. (Uniqueness) For any v, w ∈ V , we have TC (v + iw) = TC (v) + iTC (w) = T (v) + iT (w).
(Existence) Let {v1 , . . . , vn } be a basis for V . Define
³X ´ X
TC a i vi = ai T (vi )
for ai ∈ C. Then, TC is complex linear and extends T . By uniqueness, TC doesn’t depend on
the choice of basis. ¤
Lemma 21.1.4. Let V be a real vector space, W a complex vector space and b : V × V → W
and R-bilinear map. Then, there exists a unique C-bilinear map bC : VC × VC → W extending
b.
The proof of this lemma is left as an exercise. As a corollary, we get
Corollary 21.1.5. If g is a real Lie algebra, then the Lie bracket on g extends to a unique
C-bilinear map

[, ]C : gC × gC → gC
such that (gC , [, ]C ) is a Lie algebra.
Example 21.1.6. su(2)C ∼ = sl(2, C). To see this, first let T : su(2) → sl(2, C) denote the
inclusion. Then, there exists a unique TC : su(2)C → sl(2, C) extending T Now
½· ¸ · ¸ · ¸¾
i 0 0 1 0 i
su(2) = span R , ,
0 −i −1 0 i 0
so that su(2)·C is the
¸ complex
· span
¸ of the same
· matrices.
¸
1 0 0 1 0 0
Let H = ,E= and F = . Then, sl(2, C) = span C {H, E, F }. But,
0 −1 0 0 1 0
56 NOTES ON LIE GROUPS

· ¸
i 0
H = (−i) ∈ TC (su(2)C )
0 −i
µ· ¸ · ¸¶
1 0 1 0 i
E = −i ∈ TC (su(2)C )
2 −1 0 i 0
µ · ¸ · ¸¶
1 0 i 0 1
F = (−i) − ∈ TC (su(2)C )
2 i 0 −1 0
Thus, TC is onto and hence an isomorphism (by dimension count).
21.2. Representations of sl(2, C) I.
Lemma 21.2.1. There is a bijection between complex representations of SU (2) and of sl(2, C).
Proof. Since π1 SU (2) is trivial, there is a bijection between representations of SU (2) and
su(2). Any complex representation of su(2) extends to a unique complex representation of
sl(2)C = sl(s, C).
Conversely, a representation of sl(2, C) restricts to a representation of su(2) ⊂ sl(2, C). ¤
Corollary 21.2.2. Any finite dimensional complex representation of sl(2, C) is completely
reducible.
Proof. representations of SU (2) are completely reducible. ¤
We now start proving:
Theorem 21.2.3. Irreducible representations of sl(2, C) are classified by non-negative integers:
for any n = 0, 1, 2, . . . there exists a unique representation of sl(2, C) of dimension n + 1
We first observe

[H, E] = 2E
[H, F ] = −2F
[E, F ] = H
Lemma 21.2.4. Let τ : sl(2, C) → gl(V ) be a representation. Suppose τ (H)v = cv for some
c ∈ C. Then,

τ (H)(τ (E)v) = (c + 2)(τ (E)v)

τ (H)(τ (F )v) = (c − 2)(τ (F )v)
Proof.
2τ (E)v = τ ([H, E])v
= τ (H)τ (H)τ (E)v − τ (E)τ (H)v
= τ (H)(τ (E)v) − cτ (E)v
and so, τ (H)(τ (E)v) = (c + 2)τ (E)v
Similarly,
 NOTES ON LIE GROUPS 57

−2τ (F )v = τ ([H, F ]v)

= τ (H)(τ (F )v) − cτ (F )v
and thus τ (H)(τ (F )v) − (c − 2)τ (F )v ¤
Remark 21.2.5. By induction, for k = 1, 2, . . . we see that

τ (H)(τ (E)k v) = (c + 2k)(τ (E)k v).

But, τ (H) has only finitely many eigenvalues and so there exists k ≥ 1 such that τ (E)k v = 0
, τ (E)k−1 v 6= 0. We conclude that there exists a v0 ∈ V such that v0 6= 0, τ (E)v0 = 0 and
τ (H)v0 = λv0 for some λ ∈ C.
58 NOTES ON LIE GROUPS

22. Lecture 22
22.1. Representations of sl(2, C) II.
Lemma 22.1.1. Let τ : sl(2, C) → gl(V ) be a representation, v0 ∈ V be as above. Let

1
vk = τ (F )k v0
k!
v−1 ≡ 0
Then,
(i) τ (H)vk = (λ − 2k)vk ,
(ii) τ (F )vk = (k + 1)vk+1 ,
(iii) τ (E)vk = (λ − k + 1)vk−1 .
Proof. (i) This is a restatement of Lemma 21.10.
(ii) τ (F )vk = τ (F )(1/k!τ (F )k v0 ) = (k + 1)(1/(k + 1)!τ (F )k+1 v0 ).
(iii) Proceed by induction on k: τ (E)v0 = (λ + 1)v−1 = 0. Now,

kτ (E)vk = τ (E)(τ (F )vk−1 )

= ((τ (E)τ (F ) − τ (F )τ (E)) + τ (F )τ (E))vk−1
= τ (H)vk−1 + τ (F )(τ (E)vk−1 )
= (λ − 2(k − 1))vk−1 + τ (F )((λ − (k − 1) + 1)vk−2 )
= ((λ − 2k + 2) + (λ − k + 2)(k − 1))vk−1
= k(λ − k + 1)vk−1
So, τ (E)vk = (λ − k + 1)vk−1 .
¤
· ¸
1 0
Lemma 22.1.2. Let τ : al(2, C) → gl(V ) be a representation and H = . Then, τ (H)
0 −1
is diagonalizable, i.e. V has a basis of eigenvectors.
Proof. The vector v is an eigenvector of τ (H) if and only if v is an eigenvector of τ (iH) which
tiHt
happens if and only if etτ (tiH))=ρ(e where ρ : SL(2, C) → GL(V ) is a representation with
(dρ)1 = τ .
But,
½· it ¸¾
iHt} e 0 ∼
{e = = S1
0 e−it
and any representation of S 1 is completely reducible. Moreover, all irreducible representations
of S 1 are 1-dimensional. Hence, ρ(eitH ) has a basis of eigenvalues. ¤
Since the list of eigenvalues of τ (H) is finite, there exists an n ≥ 0 such that vn+1 = 0 and
vv =
6 0. Then, 0 = τ (E)vn+1 = (λ − n)vn . and so λ = n. Also, if we let

W = span C {vi | 1 ≤ i ≤ n}
 NOTES ON LIE GROUPS 59

we know that W is a representation of sl(2, C).

Comclusion: Suppose τ : sl(2, C) → gl(V ) is a complex irreducible representation, n =
dimC V − 1. Then, there is a basis {v0 , . . . , vn } such that the action of τ (H), τ (E) and τ (F ) is
given by Lemma 22.1 (i)-(iii) with λ = n. Hence, if V1 , V2 are two representations of sl(2, C),
and dim V1 = dim V2 , then V1 ∼= V2 as represenations. That is, there exists an isomorphism
T : v1 → V2 such that for any X ∈ sl(2, C) we have

T (X · V ) = X · T (v)
To complete the proof that irreducible representations of sl(2, C) are classified by non-
negative integers, we check that
n o
Vn = span C z1n−k z2k | 0 ≤ k ≤ n
is an irreducible representation of sl(2, C).
We note that

· ¸
tH et 0
e =
0 e−1
· ¸
1 t
etE =
0 1
· ¸
1 0
etF =
t 1

So, (z1 z2 )etH = (et z1 , e−t z2 ). Thus, we get

etH · (z1n−k z2n−k = (et )n−k (e−t )k z1n−k z2k

= e(n−2k)t z − 1n−k z2k
which implies
¯
d ¯¯
H· (z1n−k z2k ) = etz · (z1n−k z2k )
dt ¯t=0
= (n − 2k)z1n−k z2k .
Also, the fact that (z1 , z2 ) · etE = (z1 , tz1 + z2 ) gives
¯
d ¯¯
E· (z1n−k z2k ) = z n−k (tz1 + z2 )k
dt ¯t=0 1
= z1n−k k(tz1 + z2 )k−1 z1
= kz1n−k+1 z2k−1 .
Finally, since (z1 , z2 )(eT f ) = (z1 + tz2 , z2 ), we see that
60 NOTES ON LIE GROUPS

¯
d ¯¯
F· (z1n−k z2k ) = (z1 + tz2 )n−k z2k
dt ¯t=0
n−(k+1) k+1
= (n − k)z1 z2
Letting

v0 = z1n
v1 = F · z1n = nz1n−1 z2
µ ¶
1 n−2 2 n n−2 2
v2 = n(n − 1)z1 z2 = z z2
2! 2 1
.. .. ..
. . .µ ¶
n n−k k
vk = z z2
k 1
.. .. ..
. . .
thus completes the classification.
 NOTES ON LIE GROUPS 61

23. Lecture 23
23.1. Representation Theory of Compact Lie Groups. Let’s list some generalities and
give a quick review:
(1) If V1 , V2 are two complex representations of a Lie group G, then so are
(i) V1 ⊕ V2 ; g · (v1 ⊕ v2 ) = (g · v1 ) ⊕ (g · v2 ),
(ii) V1 ⊗C V2 ; g · (v1 ⊗ v2 ) = (g · v1 ) ⊗ (g · v2 ),
(iii) HomC (V1 , V2 ); (g · T )v = g · (T (g −1 · v)).
In particular,
(iv) V ∗ = Hom(V, C) is the dual representation (g · l)v = l(g −1 · v).
(2) If V is a representation of G, we define

V G = {v ∈ V | g · v = v, ∀g ∈ G}
the subspace of G-fixed vectors. It’s a subrepresentation.
In particular, if V1 , V2 are two representations of G, then

© ª
Hom(V1 , V2 )G = T : V1 → V2 | g · T (g −1 · v) = T (v)
= intertwining maps
= HomG (V1 , V2 )
= morphisms of representations from V1 to V2
(3) Schur’s Lemma: Let V, W be two complex irreducible representations of G. Then,
(i) A morphism T : V → W is either indentically the zero map or an isomorphism,
(ii) every morphism T : V → V has the form

T (v) = λT v
for some λT ∈ C.
(iii) we have
½
C V ∼
=W
HomG (V, W ) ∼
=
0 V ∼
6 W
=
(4) Let U, V be two representations ofG. If U is isomorphic to a G-invariant subspace of
W , we say that U is contained in W .
Example 23.1.1. Let G =· S 1 . Then ¸ ρ1 : S 1 → GL(1, C), ρ1 (λ) = λ2 is contained in
λ2 0
ρ2 : S 1 → GL(2, C), ρ2 (λ) = .
0 λ−1
Definition 23.1.2. let W be a representation of G, U an irreducible representation of G. We
call

dimC HomG (U, W )

the multiplicity of U in W .
62 NOTES ON LIE GROUPS

The reason for this definition: write W = ⊕dj=1 Wj where Wj is irreducible. Then,
M
HomG (U, W ) = HomG (U ⊕ Wj ) = HomG (U, Wj )
By Schur’s Lemma,
½
1 U∼
= Wj
dimC HomG (U, Wj ) = .
0 U∼
6= Wj
So,

dimC HomG (U, W ) = |{Wj | Wj ∼ = U }|

We will write IrrC (G) for the irreducible representations of G.
Example 23.1.3. Irr(Tn ) ∼
= Zn .
Suppose that W is a complex representation of a compact
P Lie group G. Then, since W
is completely reducible, we have a decomposition W = j Wj where the summands are irre-
ducible representations.
We can rewrite this as
 
M M M
(1) W =  Wj  = WU
U ∈Irr(G) Wj ∼
=U U ∈Irr(G)

where we call WU a U -isotypical summand of W .

Example 23.1.4. Let W = C3 , G = S 1 and let G act on W by λ·(z1 , z2 , z3 ) = (λz1 , λz2 , λ−1 z3 ).
Then,

C3 ∼
= {λ · z = λz} ⊕ {λ · z = λ−1 z} = C2 ⊕ C
Lemma 23.1.5. The outer direct sum in (1) is a canonical decomposition.
Proof. Given U ∈ Irr(G), consider

dU : (HomG (U, W )) ⊗C U → W
T ⊗ u 7→ T (u).
Then, dU is an intertwining map:

dU (g · (T ⊗ u)) = dU ((g · T g −1 )(·) ⊗ g · u)

= g · T (g −1 · g · u)
= g · T (u)
if V ⊆ W is an irreducible representation, then HomG (U, V ) ,→ HomG (U, W ). By Schur’s
Lemma, HomG (U, V ) ∼= C if and only if U ∼ = V . So, V = dU (HomG (U, V ) ⊗ U ) ⊆ im dU . In
other words, im dU contains all irreducible represenataions isomorphis to U .
Since WU = ⊕Wj ∼ =U , im dU ⊇ WU .
 NOTES ON LIE GROUPS 63

Coinversely, if 0 6= v ∈ im dU , then v = T (u) for some u ∈ U and T ∈ HomG (U, W ). Since u

is nonzero, T 6= 0 and so T : U → W is injective and T (U ) is a subrepresentation isomorphic
to U . Thus, im dU ⊆ WU .
Since we have shown im dU = WU , WU is canonically defined. ¤
Remark 23.1.6. The map
M
d := ⊕dU : HomG (U, W ) ⊗ U → W
U ∈Irr(G)
is an isomorphism.
Proof. By the previous lemma, d is onto. Also, if W = W1 ⊕ W2 , then HomG (U, W1 ⊕ W2 ) =
HomG (U, W1 ) ⊕ HomG (U, W2 ). So, d = d(1) + d(2) . By induction, both of these maps are
isomorphisms, and it follows that d is as well. ¤
64 NOTES ON LIE GROUPS

24. Lecture 24
24.1.
Vn ∗Invariant
V Integration. Recall that if G is a Lie group of dimension n, 0 6= µ1 ∈
(T1 G) = n (g∗ ), then we may define a left invariant volume form µ ∈ Ωn (G) by

µg = (Lg−1 )∗ (µ1 ).
That is,

µg (v1 , . . . , vn ) = µ1 (dLg−1 v1 , . . . , dLg−1 vn )

for all v1 , . . . , vn ∈ Tg G. Then, it’s clear that µ is left G-invariant: for any h ∈ G (L∗h µ)g = µg
for all g ∈ G.
Question: Are there any bi-invariant (i.e., left and right invariant) volume forms?
We want: L∗h µ = µ and Rh∗ µ = µ for all h ∈ G. This would imply that

(Lh ◦ Rh−1 )∗ µ = µ
and hence for all v1 , . . . , vn ∈ T1 G we would have

µ1 (v1 , . . . , vn ) = µ1 (d(Lh ◦ Rh−1 )v1 , . . . , d(Lh ◦ Rh−1 )vn )

= µ1 (Ad(h)v1 , . . . , Ad(h)vn )
= det(Ad(h))µ1 (v1 , . . . , vn )

We conclude that a necessary condition for bi-invariant forms to exist is if det(Ad(h)) = 1

for all h ∈ G. It can be shown that this is also a sufficient condition; the proof is left to the
reader.

Remark 24.1.1. Note that det ◦ Ad : G → GL(g) → R× . If G is compact and connected, then
det(Ad(G)) ⊂ R× is a compact, connected subgroup. Hence, det ◦ Ad ≡ 1.
What if G is not connected?

Example 24.1.2. Let

½· ¸ ¾ ½· ¸ ¾
a b 2 2 a −b 2 2
G = O(2) = |a +b = 1 ∪ |a +b = 1
b −a b a
(note that the second set in the above union is SO(2)). We have
½· ¸ ¾ · ¸
0 x 0 1
g = o(2) = |x ∈ R = R
−x 0 −1 0
Since SO(2) is abelian, Ad(g) = id for all g ∈ SO(2). Thus, SO(2) ⊆ ker{Ad : O(2) →
GL(g) = GL(1, R)}. passing to the quotient we get a map Ad : O(s)/SO(2) → GL(1, R).
What is this map? Well,
 NOTES ON LIE GROUPS 65

· ¸· ¸ · ¸· ¸· ¸
0 1 0 1 0 1 0 1 0 1
Ad =
1 0 −1 0 1 0 −1 0 1 0
· ¸· ¸
−1 0 0 1
=
0 1 1 0
· ¸
0 −1
=
1 0
· ¸
0 1
= −
−1 0
· ¸
0 1
from which we conclude that Ad = −1. Thus, O(2) has no bi-invariant 1-forms. It does,
1 0
however have a bi-invariant measure: “dθ00 .
24.2. Densities - a Crash Course. Recall that V volume forms on a manifold M of dimension
d are nowhere zero sections of the line bundle d (T ∗ M ). If {(Uα , φα )} are coordinate charts
V
on M , then the transition maps for d (T ∗ M ) are det(d(φβ ◦ φ−1
α )).
We want objects that transform by

(2) | det(d(φβ ◦ φ−1

α ))|.

These objects are called “densities”. they are sections of a line bundle over M with transi-
V
tion maps given by (2). More concretely, The fibers of d (T ∗ M ) are linear functionals µx :
Vd V
(Tx M ) → R. The fibers of the density bundle D(M ) → M are functions θx : d (Tx M ) → R
satisfying

θx (cv1 ∧ . . . ∧ vn ) = |c|θx (v1 ∧ . . . ∧ vn ).

Densities can be integrated in the same way as top dimensional forms. We do not need,
however, to worry about orientations. Moreover, if dg is a bi-invariant density on a Lie group
G, then | det Ad(g)| = 1 for all g ∈ G. This is also sufficient.
Bi-invariant densities are unique up to scaling. We normalize a density dg by requiring
Z
1 dg = 1
G
The upshot of all this is: given a compact Lie group G, we have a linear map
Z
· dg : C 0 (G, V ) → R
G
Z
f 7→ f (g) dg
G
satisfying
Z Z
(i) f (ag) dg = f (g) dg and
G G
66 NOTES ON LIE GROUPS
Z Z
(ii) f (ga) dg = f (g) dg.
G G
for all a ∈ G. The measure dg is also known as the Haar measure.
Remark 24.2.1. If V is a finite dimensional vector space and f : G → V is continuous, we can
define
Z
f (g) dg ∈ V
G
So, if T : V1 → V2 is a linear map, then
µZ ¶ Z
T f (g) dg = T (f (g)) dg.
G G
 NOTES ON LIE GROUPS 67

25. Lecture 25
25.1. Group Characters.
Definition 25.1.1. let G be a Lie group and ρ : G → GL(V ) a complex representation. The
character of the representation is the function

χρ = χV : G → C
χV (g) = tr(ρ(g))
Remark 25.1.2. If A, B are complex matrices such that tr(AB) = tr(BA), then tr(ABA−1 ) =
tr(A). So tr is independent of the chosen basis.
Also, if T : V → V is linear,{v ∗ ∗
∗
P ∗ 1 , . . . , vn } is a basis of V , v1 , . . . , vn the corresponding dual
basis of V , then tr(T ) = i vi (T (vi )).
If V is a representation of G, then V ∗ = Hom(V, C) is the dual representation of G. If
G is compact, we may choose a G-invariant Hermitian inner product h, i on V . This gives a
G-equivariant complex antilinear map

V → V∗
v 7→ h, v, ·i
This gives an isomorphism V ∗ = ∼ V where V is the complex vector space with the same
addition as V and scalar multiplication is given by λ · V = λv for λ ∈ C, v ∈ V .
Proposition 25.1.3. Let G be a Lie group. Then,
a) a character of a representation of G is a C ∞ function on G,
b) if V and W are isomorphic representations of G, then χV = χW ,
c) χV (ghg −1 ) = χV (h), for all g, h ∈ G,
d) χV ⊕W = χV + χW ,
e) χV ⊗W = χV χW ,
f) χV ∗ (g) = χV (g) = χV (g) = χV (g −1 ),
g) χV (1) = dimC V .
Proof. a) This is left as an exercise for the reader.
b) If ρ1 , ρ2 : G → GL(n, C) are two representations and

T / Cn
Cn
ρ1 (g) ρ2 (g)
² T ²
Cn / Cn
commutes, then tr(ρ2 (g)) = tr(T ρ2 (g)T −1 ) = tr(ρ1 (g)).
c) tr(ρ(ghg −1 )) = tr(ρ(g)ρ(h)ρ(g −1 )) = tr(ρ(h)).
d),e) recall from linear algebra that if A : V → W and B : V → V are linear, then

tr(A ⊕ B) = tr(A) + tr(B)

tr(A ⊗ B) = tr(A) tr(B).
68 NOTES ON LIE GROUPS

f) If ρ : G → GL(V ) is a representation, {v1 , . . . , vn } is a basis for V and v1∗ , . . . , vn∗ is the

associated dual basis, then

χρ∗ (g) = tr(ρ∗ (g))

X
= vi (ρ∗ (g)vi∗ )
i
X
= vi∗ (ρ(g −1 )vi )
i
= χρ(g−1 )

If h, i is an invariant Hermitian inner product, and {vi } is an orthonormal basis, then

X
tr ρ∗ (g) = hvi , ·i ◦ ρ(g −1 )(vi )
i
X
= hvi , ρ(g)−1 vi i
i
X
= hρ(g)vi , vi i
i
X
= hρ(g)ji vj , vi i
X
= ρ(g)ji hvj , vi i
X
= ρ(g)ii
= χρ (g)

g) χV (1) = tr(id) = dimC V . ¤

Proposition 25.1.4. Let ρ : G → GL(V ) be a representation of G and

V G = {g ∈ V | g · v = v} .

Then,

Z
χV (g) dg = dimC V
G
Z
Proof. Consider P : V → V given by P (v) = ρ(g)v dg. We claim that P is a linear G-
G
equivariant map such that P (V ) ⊆ V G and P |V G = idV G .
It’s clear that P is linear. Now,
 NOTES ON LIE GROUPS 69

Z
P (ρ(a)v) = ρ(g)ρ(a)v dg
ZG
= ρ(ag)v dg
G
Z
= ρ(g)v dg
G
= P (v)
Z
= ρ(ag)v dg
G
Z
= ρ(a) ρ(g) dg
G
= ρ(a)P (v)
and so P (V ) ⊆ V G and P (ρ(a) · v) = P (v) for all g ∈ G and v ∈ V . Also, if v ∈ V G , we have
Z Z
P (v) = ρ(g)c dg = v dg = v
G G
Z
(since 1 dg = 1).
G
This claim implies that tr(P ) = dim V G . On the other hand,

X
tr(P ) = vi∗ (P (vi ))
X µZ ¶
∗
= vi ρ(g)vi dg
G
Z ³X ´
= vi∗ (ρ(g)vi ) dg
ZG
= χV (g) dg
G

This completes the proof. ¤

25.2. Orthogonality of Characters.

Theorem 25.2.1. Let V, W be two compact representations of a compact Lie group G. Then,
Z
hχV , χW i := χV (g)χW (g) dg = dimC HomG (V, W )
G
In particular, if V, W are irreducible, then
½
1 V ∼
=W
hχV , χW i =
0 V ∼
6= W
70 NOTES ON LIE GROUPS

Proof. χV χW = χV ∗ χW = χW ⊗V ∗ = χHom(V,W ) . Since W ⊗ V ∗ ∼

= Hom(V, W ), Hom(V, W )G =
HomG (V, W ). By Proposition 25.4,
Z Z
(χV χW )(g) dg = χHom(V,W ) (g) dg
G G
= Hom(V, W )G
= HomG (V, W ).
The result now follows from Shur’s Lemma. ¤
Example 25.2.2. Let G = S 1 , ρn : S 1 → GL(1, C) be given by ρn (eiθ ) = einθ . Then,
dg = 1/(2π)dθ and
Z 2π ½
1 −imθ inθ 1 m=n
e e dθ =
2π 0 0 m 6= n
 NOTES ON LIE GROUPS 71

26. Lecture 26
26.1. Maximal Tori I. Recall: We have proved that the characters of complex irreducible
representations of compact Lie groups are lineraly independent as elements of C ∞ (G).
We also have the notion of the multiplicity mU,W of an irreducible representation U in a
representation W :
M
W = mU,W U
U ∈Irr(G)
where nU is the direct sum of n copies of U .
Corollary 26.1.1. Let G be a compact Lie group. Then, the map from representations of G
to characters of G, W 7→ χW .
Proof. Suppose W, W 0 are two representations of G such that χW = χW 0 . Write
M M
W = mU,W U, W0 = mU,W 0 U
U ∈Irr(G) U ∈Irr(G)
Then,
X X
χW = mU,W χU = mU,W 0 χU
U ∈Irr(G) U ∈Irr(G)

Since {χU | U ∈ Irr(G)} is linearly dependent, mU,W = mU,W 0 and hence W = W 0 . ¤

Example 26.1.2. Take G = SU (2). Let
½· ¸ ¾
λ 0
T= | |λ| = 1 ⊂ SU (2)
0 λ−1
Recall that
( n )
X
Vn = ak z1n−k z2k | an ∈ C
k=0
is an irreducible representation of SU (2) and A ∈ SU (2) acts on f (z1 , z2 ) ∈ Vn by (A ·
f )(z1 , z2 ) = f ((z1 , z2 )A). So
· ¸
λ 0
· z1n−k z2k = (λz1 )n−k (λ−1 z2 )k
0 λ−1
= λn−2k z1n−k z2k
µ· ¸¶ Xn
λ 0
Hence, χn = λn−2k .
0 λ−1
k=0

What about χn (A) where A is an arbitrary element of SU (2)? Well, any unitary matrix is
diagonalizable, χ(BAB −1 ) = χ(A) and the eigenvalues of any unitary matrix lie on S 1 . Hence,
χn is completely determined by its values on T.
72 NOTES ON LIE GROUPS

Since for any character χ : G → C of a Lie group G, χ(aga−1 ) = χ(g) for all a, g ∈ G, in
order to understand characters of G it’s important to understand the congugacy classes of G,
i.e. the orbit space G/G where G acts on itself by congugation. Let’s consider an example to
see what we should expect.
Example 26.1.3. Consider the Lie group U (n). As mentioned above, every element of U (n)
is congugate to a diagonal matrix whose nonzero entries have norm 1. Note, however, that the
set of eigenvalues for a particular element of U (n) is an unordered set! So

U (n)/U (n) ∼
= {D ∈ U (n) | D is diagonal} /Σn = Tn /Σn
where Σn is the group of permutations of n letters.
Definition 26.1.4. A torus is a compact, connected abelian Lie group. A maximal torus T
of a Lie group G is a torus subgroup of G such that if T 0 is any other torus subgroup of G and
T ⊆ T 0 then T = T 0 .
Example 26.1.5. Consider again G = U (n). The torus
  
 λ1
 

 . 
T =  ..  | |λj | = 1 ∼= Tn

 

λn
is a maximal torus.
Reason: Suppose A is a matrix in U (n) that commutes with all elements of T . But every
element of T is diagonal, and matrix theory tells us that a matrix that commutes with an
arbitrary diagonal matrix is itself diagonal. Hence, A ∈ T .
We will prove over the next two lectures:
(1) maximal tori exist,
(2) if G is compact and T1 , T2 are two maximal tori, then T1 and T2 are conjugate, i.e.
there exists a ∈ G so that aT1 a−1 = T2 ,
(3) if G is compact and connected, then for any g ∈ G there is a maximal torus T such
that g ∈ T ,
(4) if T ≤ G is a maximal torus, G compact and connected, then
© ª
N (T ) = g ∈ G | gT g −1 ⊆ T ,
the normalizer of T in G satisfies:
a) W := N (T )/T is finite (N (T )/T is called the Weyl group of T ),
b) T /W ∼= G/G.
Proposition 26.1.6. Let G be a Lie group, K ⊆ G a connected abelian Lie subgroup. Then,
the closure K of K in G is also a connected abelian subgroup.
Proof. Since the closure of a connected set is connected, K is connected. We have only to argue
that K is an abelian subgroup. Since f : G × G → G, f (a, b) = ab−1 is continuous and since
f (K × K) ⊂ K, f (K × K) = f (K × K) ⊂ f (K) ⊂ K, we see that K is a closed subgroup of
G.
 NOTES ON LIE GROUPS 73

It remains to show that K is abelian. For any x, y ∈ K, xyx−1 = y which also certainly
holds if y ∈ K. But then, yxy −1 holds as well. By the same argument, this relation holds for
all x, y ∈ K. ¤
Corollary 26.1.7. Any compact Lie group G with dim G > 0 has a torus T with dim T > 0.
Proof. Pick any X ∈ g, then {exp tX | t ∈ R} ⊆ G is a connected abelian group. Hence, its
closure is a closed connected abelian subgroup of G. Since G is compact, {exp tX | t ∈ R} is a
torus. ¤
So, at least tori exist as subgroups of compact Lie groups.
74 NOTES ON LIE GROUPS

27. Lecture 27
27.1. Maximal Tori II. In this lecture we begin by proving that maximal tori exist in a
compact, connected Lie group. We then examine the geometry of the adjoint representation
to begin our proof that maximal tori are conjugate.
Lemma 27.1.1. let G be a compact Lie group and g its Lie algebra. The maximal tori of G
are in one to one correspondence with maximal abelian Lie subalgebras of g (such subalgebras
are called Cartan subalgebras).
Proof. Suppose a ⊂ g is a maximal abelian subalgebra. Then, A := exp a ⊆ G is a connected,
abelian Lie subgroup. Hence, the closure A of A is a torus and a ⊃ a. Since a is maximal,
a = a. Since A is a torus, A = exp a = exp a = A. This proves that if a is a maximal abelian
subalgebra, then A is a torus. We now argue that A is maximal. Let T ⊃ A be another torus.
Then, a ⊆ Lie(T ) and since a is maximal, a = Lie(T ). Hence, A = T .
Conversely, suppose T ⊆ G is a maximal torus. We argue that the Lie algebra t is a
maximal abelian subalgebra of g. If a ⊃ t is an abelian subalgebra of g, then by the above
exp a ⊃ exp a ⊃ T . Since exp a is a torus, and T is maximal, T = exp a and hence t = a. ¤
As an exercise, prove that Cartan subalgebras, and hence maximal tori, exist.
Lemma 27.1.2. Let G be a torus, G = g/ZG , where ZG is the integral lattice. Let Z∗G denote
the weight lattice. For X ∈ g, {exp tX | t ∈ R} = G if and only if η(X) 6= 0 for any 0 6= η ∈ ZG .
Proof. Suppose first that η(X) = 0 for some 0 6= η ∈ ZG . Then ξη : G → S 1 given by
ξη (exp Y ) = e2πiη(Y ) is a well-defined Lie group map. Since η is nonzero, ξη 6= 1, i.e. ker ξη 6= G.
If X = ker η, then ξη (exp tX) = 1 for all t. But then, {exp tX | t ∈ R} ⊆ ker ξη implies that
{exp tX} ≤ ker ξη 6= G.
Conversely, suppose H := {exp tX | t ∈ R} = 6 G. Note that H, being a compact, connected
abelian subgroup of G is a torus. Since H 6= G, G/H is a nontrivial compact connected abelian
Lie group, i.e. another torus. Since G/H 6= {1}, there exists ρ : G/H → S 1 , ρ 6≡ 1 (why?).
Therefore, if π : G → G/H denotes the quotient map, ρ ◦ π : G → S 1 is a nontrivial Lie group
map whose kernel contains H.
Set η = d(ρ ◦ π)1 : g → R. Then η ∈ Z∗G and is nonzero. Since H ∈ ker(ρ ◦ π), X ∈ ker η. ¤
Corollary 27.1.3. Let G be a torus. Then,
 
[
X ∈g\ ker η  ⇐⇒ {exp tX | t ∈ R} = G
06=η∈Z∗G

Hence, “for almost all” X ∈ g, {exp tX | t ∈ R} = G.

Proof.
S The set Z∗G is countable and for 0 6= η ∈ Z∗G , ker η ⊆ g is nowhere dense. Hence,
06=η∈Z∗ ker η is nowhere dense.
G
¤

27.2. Geometry of the Adjoint Representation.

Example 27.2.1. Consider G = SO(3). We have seen that so(3) is the set of 3 × 3 skew-
symmetric real matrices. For w = (w1 , w2 , w3 ) ∈ R3 and x ∈ R3 , define
 NOTES ON LIE GROUPS 75

 
0 −w3 w2
w × x =  w3 0 −w1  .
−w2 w1 0
For A ∈ SO(3), we have A(w × x) = Aw × Ax. So,
 
0 −w3 w2
(Aw) × x = A  w3 0 −w1  A−1 x.
−w2 w1 0
 
0 −w3 w2
Thus, the map φ : R3 → so(3), φ(w) =  w3 0 −w1  satisfies: φ(AW ) = Aφ(w)A−1 =
−w2 w1 0
Ad(A)φ(w). Therefore the orbirts of the action of SO(3) are 2-spheres (and 0).
Now, consider the Adjoint action of G on g given by g · X = Ad(g)X. If xi ∈ g, we have
Ad(exp tξ)X = et ad(ξ) X and so, denoting the induced vector field of this action by ξg , we have
¯
d ¯¯
ξg (X) = ¯ et ad(ξ) X = ad(ξ)X = [ξ, X]
dt t=0
We then calculate the following:

GX = {g ∈ G | Ad(g)X = X}
gX = {ξ ∈ g | [ξ, X] = 0}
= ker{ad(X) : g → g}
Tx (G · X) = {ξM (X) | ξ ∈ g}
= {[ξ, X] | ξ ∈ g}
= im{ad(X) : g → g}
Lemma 27.2.2. Let g be the Lie algebra of a Lie group G. Suppose there is an Ad(G)-invariant
inner product (·, cdot) on g. Then

(ker ad(X))⊥ = im ad(X)

for any X ∈ g.
Proof. Since the inner product is Ad(G)-invariant, for any X, Y, Z ∈ g we have

(Ad(exp tX)Y, Ad(exp tX)Z) = (Y, Z)

and so
¯
d ¯¯
0 = ¯ (Ad(exp tX)Y, Ad(exp tX)Z) = ([X, Y ], Z) + (Y, [X, Z])
dt t=0
(the converse is also true if G is connected, i.e. ([X, Y ], Z) + (Y, [X, Z]) = 0 implies (, ) is
Ad(G)-invariant).
Now, Y ∈ ker ad(X) if and only if for all Z ∈ g, 0 = (ad(X)Y, Z) = −(Y, ad(X)Z). But,
this happens if and only if Y ∈ (im ad(X))⊥ . ¤
76 NOTES ON LIE GROUPS

28. Lecture 28
28.1. Maximal Tori III. Our current goal is to prove the following:
Theorem 28.1.1. Let G be a compact Lie group, T, T 0 two maximal tori in G. Then, there
exists g ∈ G such that gT g −1 = T .
Recall from the last lecture we proved that if we have an Ad(G)-invariant inner product on
the Lie algebra g of a Lie group G (such an inner product always exists if G is compact), then
we get a splitting

g = ker ad X ⊕ im ad X.
Notation: For X ∈ g, we write gX for the Lie group of GX = {g ∈ G | Ad(g)X = X}. We’ve
seen that gX = ker ad X.

Lemma 28.1.2. Suppose G is a compact Lie group, X ∈ g is such that T := {exp tX} is a
maximal torus. Denote by t the Lie algebra of T . Then, gX = t. Hence, if h ⊆ g is a maximal
abelian subalgebra, then for a “generic” X ∈ h, ker ad X = h.
Proof. Since T is a torus, t is abelian. So, X ∈ t implies that [Y, X] = 0 for all Y ∈ t. Thus,
t ⊆ gX .
Conversely, suppose that Y ∈ gX . We want to show that Y ∈ t. Since Y ∈ gX , [Y, X] = 0
and the flows φX Y
t and φt of the corresponding left invariant vector fields commute. But,
Y Y
φ1 (a) = a exp tY , φ1 (b) = b exp tY and so

φYt (φX X Y
s (1)) = φs (φt (1))

and hence exp sX exp tY = exp tY exp sX.

But, by definition of T , for all a ∈ T ,

a exp tY = (exp tY )a =⇒ a(exp tY )a−1 = exp tY

=⇒ Ad(a)Y = Y
=⇒ [Z, Y ] = 0 ∀Z ∈ t
which implies t + RY is an abelian subalgebra of g. But T is a maximal torus, and t is a
maximal abelian subalgebra of g. Therefore, t + RY = t, i.e. Y ∈ t. So we have proved that
gX ⊆ t. ¤

We are now in a position to prove Theorem 28.1:

Proof. Pick X, X 0 ∈ g so that {exp tX} = T and {exp tX 0 } = T 0 Then, ker ad X = t

and ker ad X 0 = t0 . Suppose we can find g ∈ G such that Ad(g)X ∈ ker ad X 0 . Then,
exp(t Ad(g)X) = g(exp tX)g −1 ∈ T 0 for all t.
So, gT g −1 = g{exp tX}g −1 ⊆ T 0 = T 0 . Since both T and T 0 are maximal, we actually have
equality: gT g −1 = T 0 .
Now we must find such a g ∈ G. Fix an Ad(G)-invariant inner product (, ) on g. Consider
 NOTES ON LIE GROUPS 77

f : Ad(G)X = G · X → R
f (Y ) = kY − X 0 k2 = kY k2 − 2(Y, X 0 ) + kX 0 k2
We claim that Y ∈ Ad(G)X is a critical point of f if and only if Y ∈ Ad(G)X ∩ ker ad X 0 .
To show this, note that Y is critical if and only if for all Z ∈ g
¯
d ¯¯
0 = f (Ad(exp Z)Y )
dt ¯t=0
¯
d ¯¯
= (k Ad(exp tZ)Y k2 − 2(Ad(exp tZ)Y, X 0 ) + kXk2 )
dt ¯t=0
¯
d ¯¯
= −2 ¯ (Ad(exp tZT, X 0 )
dt t=0
= (−2)([Z, Y ], X 0 )
= 2(Y, [Z, X 0 ]).
We conclude that Y is critical for f if and only if Y ∈ (im ad X 0 )⊥ = ker ad X 0 . Since G is
compact, Ad(G)X is compact and so f does indeed have critical points. This establishes the
theorem. ¤
78 NOTES ON LIE GROUPS

29. Lecture 29
29.1. Maximal Tori IV. Our next goal is to prove that every element of a compact, connected
Lie group lies in some maximal torus. Suppose we know that exp : g → G is onto. Then, if
g ∈ G, we see that g = exp X for some X ∈ g. Now, RX is an abelian subalgebra of g
and therefore lies in a maximal abelian subalgebra h. Then, exp h is a maximal torus in G
containing g. To prove that exp is onto, we will appeal to familiar tools from Riemannian
geometry.
Lemma 29.1.1. let G be a compact Lie group. Then G has a bi-invariant Riemannian metric.
Proof. On a Lie group G, bi-invariant metrics correspond to Ad-invariant inner products on g:
if g is a bi-invariant metric, g1 on T1 G is Ad-invariant. If g1 is an Ad-invariant inner product
on T1 G, then its left translation is a bi-invariant metric. If G is compact, then T1 G has an
Ad-invariant inner product: take an arbitrary positive definite inner product and average it
over G. ¤
Before we proceed, we shall review some facts about Riemmanian manifolds. In particular,
we need to use the notions of connections and geodesics on a manifold.
Definition 29.1.2. A connection ∇ on a manifold M is an R-bilinear map

∇ : Γ(T M ) × Γ(T M ) → Γ(T M )

(X, Y ) 7→ ∇X Y
such that
(1) ∇f X Y = f ∇X Y and
(2) ∇X (f Y ) = (Xf )Y + f ∇X Y for any f ∈ C ∞ (M ) and X, Y ∈ Γ(T M ).
Theorem 29.1.3 (Levi-Civita). Let (M, g) be a Riemannian manifold. Then, there is a unique
connection ∇ = ∇g on M such that
(1) X(g(Y, Z)) = g(∇X Y, Z) + g(Y, ∇X Z) and
(2) ∇X Y − ∇Y X = [X, Y ].
Moreover,
2g(X, ∇Z Y ) = Z(g(X, Y )) + Y (g(X, Z)) − X(g(Y, Z))
+g(Z, [X, Y ]) + g(Y, [X, Z]) − g(X, [Y, Z])
Theorem 29.1.4. Let M be a manifold with a connection ∇ and γ : (a, b) → M a curve.
Then there exists a unique R-linear map
∇
: Γ(γ ∗ T M ) → Γ(γ ∗ T M )
dt
such that
∇
(1) dt (f V ) = df ∇ ∞
dt V + f dt V for all f ∈ C (a, b) and V ∈ Γ(T M ).
(2) if X ∈ Γ(T M ), then
∇
(X ◦ Y ) = ∇Ẏ X
dt
 NOTES ON LIE GROUPS 79

Definition 29.1.5. A curve γ : (a, b) → M is a geodesic for a connection ∇ if

∇
γ̇ = 0
dt
Recall that if x ∈ M , v ∈ Tx M , then there is a unique geodesic γ such that γ(0) = x and
γ̇(0) = v.
Theorem 29.1.6. Let G be a Lie group, g a bi-invariant metric on G and ∇ the corresponding
Levi-Civita connection. Then, for any left invariant vector fields Z and Y
1
∇Z Y = [Y, Y ]
2
Proof. Let X, Y, Z be left invariant vector fields. Then, (g(X, Y ))(a) = (g(X, Y ))(1) for any
a ∈ G. Consequently, the map a 7→ (g(X, Y ))(a) is a constant function. Also, since g is
bi-invariant, we see that g([X, Y ], Z) + g(X, [Y, Z]) = 0. These two facts together, along with
the formula for the Levi-Civita connection in the above theorem show that 2g(X, ∇Z Y ) =
g(X, [Z, Y ]). Since X is arbitrary and the metric is nondegenerate, 2∇Z Y = [Z, Y ] ¤
Corollary 29.1.7. For any X ∈ g, a ∈ G, γ(t) = a exp tX is a geodesic. Moreover, all the
geodesics are of this form.
Proof. If γ(t) = a exp tX, then

γ̇(t) = (dLa exp tX )X(1)

= X(γ(t))
and so

∇
γ̇ = ∇X X
dt
1
= [X, X] = 0.
2
Thus, γ(t) is a geodesic. Moreover, for all a ∈ G and for all v ∈ Ta G, there is X ∈ g such that
X(a) = v. Therefore, γ(t) = a exp tX is a geodesic with γ(0) = a, γ̇(0) = X(a) = v. ¤
The following theorem from Riemannian geometry is integral to our goal:
Theorem 29.1.8 (Hopf-Rinow). If (M, g) is a complete, connected Riemannian manifold,
then any two points can be joined by a geodesic.
As a consequence, we recover
Theorem 29.1.9. Let G be a compact, connected Lie group. Then, exp : g → G is onto.
Proof. Any point g ∈ G can be connected to 1 ∈ G by a geodesic which is of the form t 7→ exp tX
for some X ∈ g. ¤
By the remark at the beginning of this lecture, we see that we have proved any element of
a compact, connected Lie group lies in a maximal torus.
80 NOTES ON LIE GROUPS

30. Lecture 30
30.1. The Weyl Group.
Definition 30.1.1. Let H ≤ G be a subgroup of a group G. The normalizer NG (H) = N (H)
of H in G is
© ª
N (H) = g ∈ G | gHg −1 = H
Note that H ⊆ N (H) and N (H) is a subgroup of G.
Proposition 30.1.2. Let G be a Lie group and H ≤ G a closed subgroup. Then, N (H) is
closed in G and is hence a Lie subgroup of G.
Proof. For b ∈ H consider the map

ψb : G → G
ψb (g) = gbg −1
Since H is closed and ψb is smooth, ψb−1 (H) is closed for all b ∈ H. Now,
\© ª
N (H) = g ∈ G | gbg −1 ∈ H
b∈H
\
= ψb−1 (H)
b∈H
and so N (H) is closed. ¤
Definition 30.1.3. Let G be a compact Lie group, T ⊆ G a maximal torus. The Weyl group
W = W (T, G) is W = N (T )/T .
Note that W acts on T : (nT ) · a = nan−1 for all a ∈ T , nT ∈ W . We will see that

T /W = G/ ∼
where G/tilde is the quotient of G by the conjugation action.
Theorem 30.1.4. let G be a compact Lie group, T ⊆ G a maximal torus. Then, the Weyl
group W is finite.
Proof. We will argue that the connected component N (T )0 of 1 in N (T ) is T . This will be
enough since |N (T )/N (T )0 | is the number of connected components of N (T ).
Now, N (T )0 acts on T by conjugation” for all g ∈ N (T )0 and a ∈ T , we have gag −1 =
cg (a) ∈ T . Hence,

dgg1 (t) ⊆ t.
In other words, Ad(g)(t) ⊆ t. Thus, we get a Lie group map

Ad(·)|t : N (T0 ) → GL(t)

g 7→ Ad(g)|t
 NOTES ON LIE GROUPS 81

Also, for any g ∈ N (T )0 ,

Ad(g)
t /t
exp exp
² cg ²
T /T
commutes since cg is a Lie group map. Therefore, Ad(g)(ker exp) ⊂ ker exp for all g ∈ N (T )0
Recall that

ZT := ker{exp : t → T } ∼
= Zn
where n = dim T . So, the image of Ad |t in GL(ZT ) ∼ = GL(n, Z) is discrete. But, N (T )0 is
connected and so for all g ∈ N (T )0 , Ad(g)|t = id. Thus, for all X ∈ Lie(N (T )0 ) and Y ∈ t
we have Ad(exp X)Y = Y and so [X, Y ] = 0. Since t is maximal abelian, we must have
Lie(N (T )0 ) ⊆ t.
On the other hand, T ⊆ N (T )0 and so Lie(N (T )0 ) = t. Since both N (T )0 and T are
connected, they must therefore be equal. ¤
Remark 30.1.5. In fact, Aut(T ) = {φ : T → T | φ is a Lie group map} = GL(ZT ).
Definition 30.1.6. let G be a Lie group. A function f ∈ C ∞ (G) is a class function if

f (x) = f (gxg −1 )
for all x, g ∈ G. We denote the space of all class functions by C 0 (G)G .
Lemma 30.1.7. Let G be a compact Lie group. Two elements x1 , x2 of a maximal torus T
are conjugate in G if and only if there is w ∈ W = N (T )/T so that w · x1 = x2 .
Before we prove this lemma, let’s make a couple observations.
Remark 30.1.8. Recall that W acts on T “by conjugation”: for all gT ∈ W and a ∈ T we have
(gT ) · a = gag −1 .
Definition 30.1.9. let G be a group and H ⊆ G a subgroup. The centralizer Z(G) of H in
G is
© ª
Z(H) = g ∈ G | ghg −1 = H, ∀h ∈ H
Remark 30.1.10. If G is a Lie group and H ⊆ G is closed, then
\
Z(H) = ψh−1 (h)
h∈H
where ψh : G → G is given by ψh (g) = ghg −1 .
proof of Lemma 30.7. Suppose x, y ∈ T and y = gxg −1 for some g ∈ G. Then, Z(y) =
gZ(y)g −1 = cg (Z(x)). Since x ∈ T and T ⊆ Z(x) we have cg (T ) ⊆ Z(y).
Now, Z(y)0 is compact and connected and T, cg (T ) ⊆ Z(y)0 are tori. Since both tori are
maximal in G, they are maximal in Z(y)0 . Thus, there exists h ∈ Z(y)0 such that ch (gg (T )) = T
and so hg ∈ N (T ).
82 NOTES ON LIE GROUPS

Also, chg (x) = hgxg −1 h−1 = hyh−1 = y since h ∈ Z(y). We conclude that hgT ∈ N (T )/T
and (hgT ) · x = y. If x1 , x − 2 ∈ T and w = gT ∈ W with w · x1 = x2 , then gx1 g −1 = x2 . So
x1 and x2 are conjugate in G. ¤
The previous lemma implies that the induced map T /W → G/ ∼ is a continuous bijection.
Since T /W is compact and G/ ∼ is Hausdorff, this map is actually a homeomorphism. Hence,
Lemma 30.1.11. Let G be a compact Lie group, T ⊆ G a maximal torus and W the Weyl
group. Then,
(i) G/ ∼≈ T /W and
(ii) the map C 0 (G)G → C 0 (T )W given by f 7→ f |T is an isomorphism.
Example 30.1.12. The Weyl group of U (n) is Σn , the permutation group on n letters. To
see this, we first remark that
  

 λ1 

 . 
T =  ..  | |λj | = 1 ∼= Tn

 

λn
is a maximal torus. Now, two matrices are conjugate if and only if they have the same set of
eigenvalues and so the Weylgroup W is a subset of Σn . Moreover, if σij is the permutation
−1
matrix that swaps the ith and jth columns of the identity matrix, and A ∈ T , then σij Aσij
transposes the eigenvalues λi and λj of A. Hence, all two-cycles are in W . But Σn is generated
by two-cycles and we conclude that W = Σn .
 NOTES ON LIE GROUPS 83

31. Lecture 31
31.1. The Peter-Weyl Theorems I. Our final goal will be to prove the Peter-Weyl theo-
rem(s):
Theorem 31.1.1 (Peter-Weyl I). Any compact Lie group G is isomorphic to a closed subgroup
of U (n) for some choice of n.
Another way to state this is that a compact Lie group G can be realized as a subset of
Mn (C), i.e. is a matrix group. Yet another way to look at this is there is a faithful (kernel
= {1}) representation ρ : G → Gl(n, C).
We will need a few preliminary definitions before we can state version II of this theorem.
Definition 31.1.2. Suppose ρ : G → GL(n, C) is a representation of a Lie group G. Let
rij : Mn (C) → C denote the standard coordinate functions. The functions

rij ◦ ρ : G → C
are called the matrix coefficients of the representation ρ.
More abstractly, the representation coefficients may be realized as

(rij ◦ ρ)(g) = he∗i , ρ(g)ej i

where {e1 , . . . , en } is a basis of Cn and {e∗1 , . . . , e∗n } is the associated dual basis.
Definition 31.1.3. Let G be a Lie group. A function f : G → C is a representative function
(an abstract matrix coefficient) if there is a representation ρ : G → GL(V ) such that

f (g) = hl, ρ(g)ξi

∗
for some ξ ∈ V , l ∈ V and for all g ∈ G. We will usually denote such a function by fV,l,ξ .
Lemma 31.1.4. Representative functions of a Lie group G form a subalgebra of C 0 (G).
Proof. The following equalities are left as an exercise:

fV1 ,l1 ,ξ1 + fV2 ,l2 ,ξ2 = fV1 ⊕V2 ,l1 ⊕l2 ,ξ1 ⊕ξ2
fV1 ,l1 ,ξ1 fV2 ,l2 ,ξ2 = fV1 ⊗V2 ,l1 ⊗l2 ,ξ1 ⊗ξ2
λfV,l,ξ = fV,λl,ξ ∀λ ∈ C
¤
0 (G) the algebra of representative functions of G.
Notation We will denote by Calg

Theorem 31.1.5 (Peter-Weyl II). Let G be a compact Lie group. Then, Calg 0 (G) is dense in

C 0 (G) with respect to the supremum norm. That is, for all f ∈ C 0 (G) and for all ² > 0, there
0 (G) such that
exists h ∈ Calg

sup |f (g) − h(g)| < ²

g∈G
84 NOTES ON LIE GROUPS

We claim that the two versions of the Peter-Weyl Theorem are equivalent. Version I implies
Version II by the Stone-Weierstrass Theorem:
Theorem 31.1.6 (Stone-Weierstrass). Let X be a compact topological space. Suppose A ⊆
C 0 (X) is a subalgebra such that
(1) A separates points, i.e. for all x1 , x2 ∈ X with x1 6= x2 there is f ∈ A so that
f (x1 ) 6= f (x2 ),
(2) 1 ∈ A (here, 1 denotes the constant function 1(x) = 1 for all x ∈ X) and
(3) if f ∈ A, then the complex conjugate f ∈ A.
Then, A is dense in C 0 (X).
Assuming the Stone-Weierstrass Theorem, let’s see why version I of the Peter-Weyl Theorem
implies the second version:
Proof. Suppose G ∈ GL(n, C) is a compact Lie subgroup. Consider the subalgebra A of C 0 (G)
0 (G) so
generated by rij |G , rij |G and 1. The functions rij separate points. Also, rij , rij , 1 ∈ Calg
0 (G). By Stone-Weierstrass, A is dense in C 0 (G) and so C 0 (G) is dense in C 0 (G). ¤
A ⊆ Calg alg

We now endeavor to prove that the second version of the Peter-Weyl Theorem implies the
first. We will need a considerable amount of background material to complete the proof.
Proposition 31.1.7. Any descending chain

G1 + G2 ) . . . ) Gk ) . . .
of compact Lie groups is finite.
Proof. Suppose A and B are two compact Lie groups with A ( B. We claim that either dim A <
dim B or the connected components A0 and B0 of the identities are equal and A/A0 ( B/B0 .
To prove this claim, first suppose that dim A = dim B. Then, since dim Lie(A) = dim Lie(B)
and Lie(A) ⊆ Lie(B) they must be equal. Hence, A0 = B0 and A/A0 ⊆ B/B0 . Since A 6= B
A/A0 6= B/B0 .
Now that the claim has been established, note that at eash step of the chain, say Gi ) Gi+1 ,
either dim Gi+1 dim Gi or |(Gi /Gi+1 )| < |Gi /(Gi )0 |. So, eventually dim Gi = 0 and then
|Gi+k /(Gi+k )0 | = 1. ¤
 NOTES ON LIE GROUPS 85

32. Lecture 32
32.1. A Bit of Analysis. We will now review some properties of normed vector spaces that
we will need to prove the Peter-Weyl Theorems via a little bit of functional analysis.
Definition 32.1.1. A norm k · k on a vector space V is a map

k · k : V → [0, ∞)
such that
(1) kf k = 0 if and only if f = 0 for all f ∈ V ,
(2) kαf k = |α|kf k for all α ∈ C, f ∈ V ,
(3) kf + gk ≤ kf k + kgk for all f, g ∈ V .
A vector space V with a norm is called a normed vector space.
Recall that a normed vector space V is a Banach space if it is complete, i.e. every Cauchy
sequence is a convergent sequence.
Fact: (C 0 (G), k · kC 0 ) is a Banach space, where G is a compact Lie group and

kf kC 0 = sup |f (g)|
g∈G
A Banach space is a Hilbert space if there is a Hermitian inner product (·, ·) : V × V → C
such that

kf k = (f, f )1/2
for all f ∈ V .
Example 32.1.2. Let G be a compact Lie group. Define
½ Z ¾
2 2
L (G) = f : G → C | |f (x)| dx < ∞
G
Then, L2 (G) is a Hilbert space with
µZ ¶1
2
2
kf kL2 = |f (x)| dx
G
and Z
(f, g) = f (x)g(x) dx.
G

Suppose now that (V1 , k · k1 ) and V2 , k · k2 ) are Banach spaces. A linear map T : V1 → V2 is
continuous if there exists A > 0 so that kT (v)k2 ≤ Akvk1 .
Example 32.1.3. If f ∈ C 0 (G), then
µ ¶2
2
sup |f (x)| = sup |f (x)| = kf k2C 0
x∈G x∈G
and so
86 NOTES ON LIE GROUPS

Z Z
kf k2L2 = |f (x)|2 dx
ZG
≤ sup |f (x)|2 dx
G x∈G
Z
≤ kf k2C 0 dx
G
Z
= kf k2C 0 1 dx
G
= kf k2C 0 .

That is, kf kL2 ≤ kf kC 0 . Hence, the identity map id : C 0 (G) → L2 (G) (with the above
indicated norms) is continuous.

Theorem 32.1.4 (Cauchy Schwarz Inequality). Suppose (X, dx) is a measure space. Then,
for all f, g ∈ L2 (X) we have

|(f, g)| ≤ kf k2 kgk2

In particular, we see that if X = G is a compact Lie group, then

¯Z ¯ µZ ¶ 1 µZ ¶1
¯ ¯ 2 2
¯ f (x)g(x) dx¯ ≤ |f (x)|2
dx |g(x)|2
dx
¯ ¯
X X X

and

¯Z ¯ ¯Z ¯
¯ ¯ ¯ ¯
¯ |f (x)| dx¯ = ¯ |f (x)| · 1 dx¯
¯ ¯ ¯ ¯
G G
µZ ¶ 1 µZ ¶1
2 2
2 2
≤ |f (x)| dx |1 dx
X X
≤ kf kL2 .

From this we conclude that if f ∈ L2 (G), then f ∈ L1 (G) and the inclusion map L2 (G) ,→
L1 (G) is continuous.

Example 32.1.5. A function k ∈ C 0 (G × G) defines a linear map

K : C 0 (G) → functions on G
Z
(Kf )(x) := k(x, y)f (y) dy.
G

It is left as an exercise to the reader to prove that Kf ∈ C 0 (G). Moreover,

 NOTES ON LIE GROUPS 87

Z
|(Kf )(x)| ≤ |k(x, y)||f (y)| dy
G
Z
≤ (sup |k|)|f (y)| dy
G
Z
= kkkC 0 |f (y)| dy
G
≤ kkkC 0 kf kL2 .

So, K : (C 0 (G), k · kL2 ) → (C 0 (G), k · kC 0 ) is continuous.

Definition 32.1.6. A linear map K : V1 → V2 of normed vector spaces is compact if it maps

bounded sets B ⊆ V1 to precompact sets K(B) ⊆ V2 .

Definition 32.1.7. Let X be a topological space. A subset L ⊂ C 0 (X) is equicontinuous at

x0 ∈ X if for all ² > 0 there is a neighborhood U of x0 such that

|f (x) − f (x0 )| < ² ∀x ∈ X, ∀f ∈ L.

L is equicontinuous if it is equicontinuous at every point of X.

Theorem 32.1.8 (Ascoli). A subset L ⊆ (C 0 (G), k · kC 0 ) is precompact if and only if it is

bounded and equicontinuous.

We now present an application of the Ascoli Theorem:

Example 32.1.9. Let k ∈ C 0 (G × G) and K : (C 0 (G), k · kL2 ) → (C 0 (G), k · kC 0 ) be as above.

We claim that K, and hence K : (C 0 (G), k · kL2 ) → (C 0 (G), k · kL2 ), is compact.

Proof. Suppose B ⊆ C 0 (G) is L2 -bounded by C > 0, i.e. for all f ∈ B, kf kL2 < C. We want
to show that K(B) is precompact in the C 0 -norm. By Ascoli, it is enough to show that K(B)
is C 0 -bounded and equicontinuous. Since |(Kf )(x)| ≤ kkkC 0 kf kL2 ,

kKf kC 0 = sup |(Kf )(x)| ≤ kkkC 0 · C

x∈G

To continue, we’ll need the following lemma:

Lemma 32.1.10. If h ∈ C 0 (G), then for all ² > 0 there is a neighborhood U of 1 in G so that

xy −1 ∈ U =⇒ |h(x) − h(y)| < ²

In light of this lemma, given ² > 0 we can find a neighborhood V of 1 in G such that

|k(x, y) − k(x0 , y)| < ²C −1

as long as x0 x−1 ∈ V . Then,
88 NOTES ON LIE GROUPS

¯Z ¯
¯ ¯
|(Kf )(x) − (Kf )(x )| = ¯¯ (k(x, y) − k(x y))f (y) dy ¯¯
0 0
G
Z
−1
≤ ²C |f (w)| dy
G
−1
≤ ²C kf kL2
= ²
¤
 NOTES ON LIE GROUPS 89

33. Lecture 33
33.1. The Peter-Weyl Theorems II. We continue to apply analysis to our study of compact
Lie groups. In the last lecture, we used a lemma which we will now prove.
Lemma 33.1.1. Let H be a compact Lie group and f ∈ C 0 (H). Then, for any ² > 0 there is
a neighborhood V of 1 in H so that

yx−1 ∈ V =⇒ |f (x) − f (y)| < ²

Proof. Since f is continuous, for any x ∈ H there is a neighborhood Ux of 1 such that for any
y ∈ Ux x
²
|f (y) − f (x)| <
2
Choose Vx ⊆ H so that 1 ∈ Vx and Vx2 ⊆ Ux . Then, {Vx x} is an open cover of H. Since H
is compact, there is a finite subcover: there exist x1 , . . . , xn so that
n
[
Vxi xi
i=1
Set V = ∩i Vxi and suppose yx−1 ∈ V . Then, x ∈ Vxi xi for some i. So, y = (yx−1 )x ∈
V Vxi xi ⊆ Uxi xi . We then have
² ²
|f (y) − f (xi )| < , |f (x) − f (xi )| < .
2 2
Therefore,

|f (y) − f (x)| ≤ |(f (y) − f (xi )) + (f (xi ) − f (x))|

² ²
< + = ².
2 2
¤
Remark 33.1.2. We may further assume that V = V −1 by replacing V with V ∩ V −1 .

Corollary 33.1.3. If G is a compact Lie group and k ∈ C 0 (G × G), then for all ² > 0 there
exists a neighborhood U of 1 in G such that if xy −1 ∈ U ,

|k(x, h) − k(y, h)| < ²

for all h ∈ G.
Proof. Let H = G × G. We may assume that U ⊆ G × G is of the form U × U . Then, xy −1 ∈ U
implies that (x, h)(y, h)−1 ∈ U × U for any h ∈ G. ¤
Lemma 33.1.4. Let G be a compact Lie group, k ∈ C 0 (G × G). Then, for any f ∈ L2 (G),
Z
(Kf )(x) := k(x, y)f (y) dy
G
is a continuous function of x. In fact, for all ² > 0 there exists a neighborhood U of 1 in G so
that
90 NOTES ON LIE GROUPS

xy −1 ∈ U =⇒ |(Kf )(x) − (Kf )(y)| < ²

Proof. By Corollary 33.3, there exists a neightborhood U of 1 such that if xy −1 ∈ U , then
²
|k(x, h) − k(y, h)| <
C
where C will be chosen later. Then,
¯Z ¯
¯ ¯
|(Kf )(x) − (Kf )(y)| ≤ ¯ (k(x, h) − k(y, h))f (h) dh¯
¯ ¯
ZG
≤ |k(x, h) − k(y, h)||f (h)| dh
G
Z
²
≤ |f (h)| dh
G C
Z
²
= |f (h)| dh
C G
²
≤ kf kL2 .
C
Now, simply take C = kf kL2 . ¤
Lemma 33.1.5. let G, k and K be as above. Then, for all f ∈ L2 (G), kKf kC 0 ≤ kkkC 0 kf kL2 .
Hence, K : (L2 (G), k · kL1 ) → (C 0 (G), k · kC 0 ) is bounded.
Proof. For any x ∈ G,
¯Z ¯
¯ ¯
|(Kf )(x)| = ¯ k(x, y)f (y) dy ¯
¯ ¯
ZG
≤ |k(x, y||f (y)| dy
ZG
≤ sup |k(x, y)||f (y)| dy
G x,y∈G
Z
= kkkC 0 |f (y)| dy
G
≤ kkkC 0 kf kL2 .
Therefore, |Kf |C 0 = supx∈G |(Kf )(x)| ≤ kkkC 0 kf kL2 ¤
It is a well-known fact, and left as an exercise for the reader, that bounded linear maps
between normed vector spaces are continuous, and we have seen that K : L1 (G) → L2 (G) is
compact. The question is: how does this help us with the Peter-Weyl Theorem: the density of
0 (G) in C 0 (G)?
Calg
Two Ideas:
(1) G acts on functions on G,

(a · f )(x) = f (a−1 x)
 NOTES ON LIE GROUPS 91

and we will prove that

Lemma 33.1.6. f ∈ C 0 (G) is a representative function if and only if there is a finitw
dimensional subspace W ⊆ C 0 (G) with f ∈ W .
(2) Let f ∈ C 0 (G) be a function. Fix ² > 0. We want to find h ∈ Calg
0 (G) such that

kf − hkC 0 < ²
Since f ∈ C 0 (G) there exists a neighborhood U of 1 in G such that
²
xy −1 ∈ U =⇒ |f (x) − f (y) < .
2
Pick ρ ∈ C 0 (G) with ρ ≥ 0 such that supp ρ ⊆ U ,
(i) ρ(x)
Z = ρ(x−1 ) and
(ii) ρ(x) dx = 1.
G
To get (ii), take any ρ̃ with supp ρ̃ ⊆ U and let
1
ρ(x) = R ρ̃(x)
G ρ̃(x) dx
To get (ii), take any ρ̃ and let
1
ρ(x) = (ρ̃(x) + ρ̃(x−1 ))
2
Now, let k(x, y) = ρ(x−1 y) = ρ((x−1 y)−1 ) = ρ(y −1 x) and consider

K : L2 (G) → C 0 (G) ,→ L2 (G)

Z
(Kf )(x) = k(x, y)f (y) dy
G
Then, we have
epsilon
kKf − f kC 0 <
2
On the other hand, since we know K is compact, and we also see that g · (Kf ) =
K(g · f ) (that is, K is G-equivariant). Hence, the eigenspaces of K are G-invariant and
finite dimensional. Furthermore, the span of eigenfunctions is dense in (L2 (G), k · kL2 ).
That is, there exist v1 , . . . , vk ∈ L2 so that KVi = λi vi and
X ²
k vi − f kL2 < kkkC 0 .
2
Hence,
X X
k λi vi − Kf kC 0 = kK( vi − f )kC 0
X
≤ kkkC 0 k vi − f kL2
92 NOTES ON LIE GROUPS

34. Lecture 34
34.1. The Peter-Weyl Theorems III.
Proposition 34.1.1. Let H be a Hilbert space and K : H → H be a symmetric, compact
operator. Then,

kKk = sup |(Kf, f )|

kf k=1

Proof. For f ∈ H with kf k = 1, we have by Cauchy-Schwarz

|(Kf, f )| ≤ kKf kkf k ≤ kKk12

and so

M := sup |(Kf, f )| ≤ kKk

kf k=1
We now argue that kKk ≤ M . For this, it’s enough to show that kKf k ≤ M for all f ∈ H
and all kf k = 1. Note that if Kf = 0, we’re done. So, assume x ∈ H, kxk = 1 and Kx 6= 0.
Let y = Kx/kKxk. We then see that kyk = 1 as well. Also,

(x, Ky) = (Kx, y)

= (Kx, Kx/kKxk)
1
= kKxk2
kKxk
= kKxk
It is also easy to verify that

(K(x + y), x + y) − (K(x − y), x − y) = 2(Kx, y) + 2(Ky, x) = 4kKxk

On the other hand, for any z ∈ H, z 6= 0 we have

|(Kz, z)| = (|(K(z/kzk), z/kzk|)kzk2 ≤ M kzk2

and so

(K(x + y), x + y) ≤ M kx + yk2

−(K(x − y), x − y) ≤ M kx − yk2 .
We therefore conclude that

4kKxk = (K(x + y), x + y) − (K(x − y), x − y)

≤ M (kx + yk2 + kx − yk2 )
= M (2kxk2 + 2kyk2
= 4M
and so kKxk ≤ M . ¤
 NOTES ON LIE GROUPS 93

Proposition 34.1.2. Let K : H → H be a compact symmetric operator. Then, either kKk or

−kKk is an eigenvalue of K.
Proof. By the last proposition, there is a sequence {xn } ⊆ H with kxn k = 1 such that

lim |(Kxn , xn )| = kKk

n→∞
By passing to a subsequence. if necessary, we may assume

lim (Kxn , xn ) = α
n→∞
where α is either kKk or −kKk. Note, if α = 0 there is nothing to prove. So, assume α 6= 0.
Then,

0 ≤ kKxn − αxn k2 = (Kxn − αxn , Kxn − αxn )

= kKxn k2 − (αxn , Kxn ) − (Kxn , αxn ) + |α|2 kxn k2
≤ kKk2 kxn k2 − 2α(xn , Kxn ) + α2 kxn k2
= 2α(α − (xn , Kxn ))
Since (Kxn , xn ) → α as n → ∞. (α − (xn , Kxn )) → 0. Thus,

kKxn − αxn k2 → 0
Since K is compact and xn is bounded, by passing to a subsequence we may assume that
Kxn → y. But, kKxn − αxn k2 → 0 implies that αxn → y as n → ∞. Now,

Ky = K( lim αxn )
n→∞
= lim K(αxn )
n→∞
= α lim (Kxn )
n→∞
= αy.
So, y is an eigenvector of K with eigenvalue α, provided y 6= 0. But y cannot be zero since

kyk = k lim αxn k = lim kαxn k = |α|

n→∞ n→∞
¤
Suppose that H is a Hilbert space and K : H → H is compact and symmetric. If v is an
eigenvector with eigenvalue α, then
(1) (Kv, v) = (λv, v) = λ(v, v) = λkvk2 and
(2) (v, Kv) = (v, λv) = λ(v, v) = λkvk2 .
We conclude that all eigenvalues of K are real.
Also, if Kv = λv, Kw − µw and µ 6= λ, then

λ(v, w) = (Kv, w) = (v, Kw) = µ(v, w).

So, (v, w) = 0 and eigenspaces associated to distinct eigenvalues are mutually orthogonal.
94 NOTES ON LIE GROUPS

Theorem 34.1.3. Let H be a Hilbert space, K : H → H a compact symmetric operator. Then,

L
(1) For each ² > 0, the direct sum of eigenspaces |λ|>² Hλ is finite dimensional and
L
(2) λ Hλ is dense in H.
L
Proof. (1) Suppose that |λ|>² Hλ is infinite dimensional. Then, there is a sequence {xn } of
orthornormal eigenvectors with eigenvalues λn such that |λn | > ² for each n. Since {xn } is
bounded and K is compact, {Kxn } contains a convergent subsequence.
On the other hand, for all positive integers m, n we have

kKxn − Kxm k2 = kλn xn − λm xm k2

= |λn |2 kxn k2 + |λm |2 kxm k2
= λ2n + λ2n > 2².
L L L L
To prove (2), let E = λ Hλ , the closure of λ Hλ in H Since K ( λ Hλ ) ⊆ Hλ , K ( λ Hλ ) ⊆
L ⊥
λ Hλ . Since K is continuous, K(E) ⊆ E. Let F = E . Since K(E) ⊆ E, K(F ) ⊆ F . So,
for any e ∈ E, f ∈ F ,

(K(f ), e) = (f, K(e)) = 0

K(e) ∈ E. Moreover, F is closed. So, F is a Hilbert space.
The inclusion map F ,→ H is continuous and so K|F : F → F is compact, and K|F is still
symmetric. By Proposition 34.2, K|F has an eigenvector v 6= 0. So, v ∈ Hλ for some λ. But
then, v ∈ E which is a contradiction. ¤
 NOTES ON LIE GROUPS 95

35. Lecture 35
35.1. The Peter-Weyl Theorems IV. Let’s first recall a few facts we have discussed earlier:
(1) A function f ∈ C 0 (G) is a representative function if and only if there exists a represen-
tation ρ : G → Gl(V ), v ∈ V and l ∈ V ∗ such that

f (g) = hl, ρ(g)ci.

(2) G acts on C 0 (G) by

(g · f )(x) = f (g −1 x).
We also owe the reader a proof of the following lemma:
Lemma 35.1.1. The function f ∈ C 0 (G) is a representative function of and only if there
exists a finite dimensional subspace W ⊆ C 0 (G) which is a subrepresentation and f ∈ W .
Proof. Suppose ρ : G → GL(V ) is a finite dimensional representation. Take a basis {v1 , . . . , vn }
of V and let {v1∗ , . . . , vn∗ } be the associated dual basis. Then,
X
aij (x) := vj∗ (ρ(x)vi )
j
are representative functions. P
Since ρ(gx) = ρ(g)ρ(x), (aij (gx)) = (aij (g))(akj (x)) and (g·aij )(x) = aij (g −1 x) = k aik (g −1 )akj (x)
That is, g · aij ∈ span C {akj for all g ∈ G and aij .
Also, for any l ∈ V ∗ ,v ∈ V

f (g) = l(ρ(g)v) ∈ span C {aij (g)}

Therefore, W = span C {aij }.
Conversely, suppose that W ⊆ C 0 (G) is finite dimensional and for any f ∈ W , g · f ∈ W .
Consider l : W → C given by l(f ) = f (1). Then, l ∈ W ∗ . Also,

l(g −1 · f ) = f (g(x))|x=1 = f (g).

Let V = W ∗ and ρ : G → GL(W ∗ ) be given by

ρ = (g · f )|W )∗
Then, f ∈ V = (W ∗ )∗ and f (g) = hf, ρ(g)li. ¤
We now finally begin the proof of
Theorem 35.1.2 (Peter-Weyl Version II). Let G be a compact Lie group. Then, representative
functions are dense in C 0 (G).
Proof. Fix f0 ∈ C 0 (G). For any ² > 0, we want to find a representative function h so that

kf0 − hkC 0 < ².

Since f0 ∈ C 0 (G), there exists a neighborhood U of 1 in G such that
 96 NOTES ON LIE GROUPS

²
xy −1 ∈ U =⇒ |f0 (x) − f0 (y)| <
2
Pick τ1 ∈ C 0 (G) such that τ1 ≥ 0 and supp τ1 ⊆ U and
Z
τ1 (x) dx = 1
G Z
Let τ (x) = (1/2)(τ1 (x) + τ1 (x−1 )). Then, τ (x) dx = 1, supp τ ⊆ U and τ (x) = τ (x−1 ).
Now, let

k(x, y) = τ (x−1 y),

K : L2 (G) → C 0 (G) ,→ L2 (G)
Z
(Kf )(x) = τ (x−1 y)f (y) dy
G
Claim 1: K is a compact, symmetric operator.
Claim 2: kKf0 − f0 kC 0 < ²/2.
Claim 3: g · (Kf ) = K(g · f ) for all f ∈ L2 (G).
Let’s see, assuming these claims, how to finish the proof. Since k is compact, the nonzero
eigenspaces of K are finite dimensional. So, if Kf = λf for some nonzero λ, then f is a
representative function.
On the other hand, eigenfunctions of K are dense in L2 (G). Thus, there exist h − 1, . . . , hn ∈
2
L (G) such that Khi = λi hi and
X ²
k hi − f0 kL2 ≤ (kkkC 0 )−1 ( )
2
i
So, we get
X X
k λi hi − Kf0 kC 0 = kK( hi − f0 kC 0
i i
X
≤ kkkC 0 k hi − f0 kL2
i
² ²
= kkkC 0 (kkkC 0 )−1 ( ) =
2 2
P
Now, let h = i hi . Then,

kf0 − hkC 0
≤ kf0 − Kf0 kC 0 + kKf0 − hkC 0
² ²
= + = ².
2 2
Modulo the proof of the above three claims, this completes the proof of the theorem.
¤
 NOTES ON LIE GROUPS 97

36. Lecture 36
36.1. Finale. We still need to prove the claims from the last lecture.

Proof of Claim 1: Since τ (z) = τ (z −1 ), τ (x−1 y) = τ (y −1 x). So, k(x, y) = τ (x−1 y) = k(y, x). Therefore,

Z
(Kf1 , f2 ) = Kf1 (x)f2 (x) dx
G
Z Z
= k(x, y)f1 (y) dyf2 (x) dx
G G
Z Z
= f1 (y)k(x, y)f2 (x) dy dx
ZG G µZ ¶
= f1 (y) k(y, x)f2 (x) dx dy
G G
Z
= f1 (y)Kf2 (y) dy
G
= (f1 , Kf2 )

and it follows that K is symmetric.

Proof of Claim 2: Note first that if we write y 0 = x−1 y, we have

Z Z
Kf0 (x) = τ (y 0 )f0 (y) dy = τ (y 0 )f0 (xy 0 ) dy.
G G

This implies

¯Z Z ¯
¯ ¯
¯ 0
|Kf0 (x) − f0 (x)| = ¯ τ (y)f0 (xy ) dy − f0 (x) τ (y) dy ¯¯
ZG G

≤ |τ (y)||f0 (xy) − f0 (x)| dy.

G

If τ (y) 6= 0, then y ∈ U is so that y = x−1 xy and hence |f0 (xy) − f0 (x)|, ²/2. Then,

kKf0 − f0 k = sup |Kf0 (x) − f0 (x)|

Zx
² ²
≤ τ (y)( ) dy =
G 2 2

Proof of Claim 3: Claim 3 follows from a quick calculation:

98 NOTES ON LIE GROUPS

(g · Kf )(x) = Kf (g −1 x)
Z
= τ ((g −1 x)−1 y)f (y) dy
G
Z
= τ (x−1 y 0 )f (g −1 y 0 ) dy 0
ZG
= τ (x−1 y)(g · f )(y) dy
G
= (K(g · f ))(x)
Finally, we present an application of this theorem:
Example 36.1.1. Let G = Tn = Rn /Zn . Any representation of Tn is a direct sum of irre-
ducibles. A matrix coefficient of an irreducible representation of Tn is of the form
Pn
am (x mod Zn ) = e2πi( j=1 mj xj )

where n = (m1 , . . . , mn ) ∈ Zn .
We can think of f ∈ C 0 (Tn ) as f ∈ C 0 (Rn ) with f (x1 , . . . , xm ) = f (x1 + k1 , . . . , xn + kn ) for
ki ∈ Z. Then,
( )
X P
0 n 2πi( n m x )
Calg (T ) = cm e j=1 j j
| cm ∈ C ,
m∈Zn
the space of trigonometric polynomials. In this context, the Peter-Weyl Theorem can be stated
as follows:
Theorem 36.1.2. Any continuous Zn -periodic function on Rn can be approximated by a
trigonometric polynomial.