Lie groups. Here are a few simple examples:

1. Let An be the set of diagonal n-by-n matrices with positive entries on the diagonal.

2. Let Nn be the set of upper triangular n-by-n matrices with all diagonal entries all

equal to 1.

3. (n = 2 only) Let C• = a ’b

| a2 + b2 > 0 . Then C• is a matrix Lie group di¬eo-

ba

morphic to S 1 — R. (You should check that this is actually a subgroup of GL(2, R)!)

4. Let GL+ (n, R) = {a ∈ GL(n, R) | det(a) > 0}

There are more interesting examples, of course. A few of these are

SL(n, R) = {a ∈ GL(n, R) | det(a) = 1}

O(n) = {a ∈ GL(n, R) | ta a = In }

SO(n, R) = {a ∈ O(n) | det(a) = 1}

which are known respectively as the special linear group , the orthogonal group , and the

special orthogonal group in dimension n. In each case, one must check that the given

subset is actually a subgroup and submanifold of GL(n, R). These are exercises for the

reader. (See the problems at the end of this lecture for hints.)

A Lie group can have “wild” subgroups which cannot be given the structure of a Lie

group. For example, (R, +) is a Lie group which contains totally disconnected, uncountable

subgroups. Since all of our manifolds are second countable, such subgroups (by de¬nition)

cannot be given the structure of a (0-dimensional) Lie group.

However, it can be shown [Wa, pg. 110] that any closed subgroup of a Lie group G is

an embedded submanifold of G and hence is a Lie subgroup. However, for reasons which

will soon become apparent, it is disadvantageous to consider only closed subgroups.

L.2.2 13

Example: A non-closed subgroup. For example, even GL(n, R) can have Lie

subgroups which are not closed. Here is a simple example: Let » be any irrational real

number and de¬ne a homomorphism φ»: R ’ GL(4, R) by the formula

«

’ sin t

cos t 0 0

¬ sin t ·

cos t 0 0

φ» (t) =

’ sin »t

0 0 cos »t

0 0 sin »t cos »t

Then φ» is easily seen to be a one-to-one immersion so its image is a submanifold G» ‚

GL(4, R) which is therefore a Lie subgroup. It is not hard to see that

±«

’ sin t

cos t 0 0

¬ sin t 0·

cos t 0

s, t ∈ R .

G» =

’ sin s

0 0 cos s

0 0 sin s cos s

Note that G» is di¬eomorphic to R while its closure in GL(4, R) is di¬eomorphic to S 1 —S 1 !

It is also useful to consider matrix Lie groups with complex coe¬cients. However,

complex matrix Lie groups are really no more general than real matrix Lie groups (though

they may be more convenient to work with). To see why, note that we can write a complex

n-by-n matrix A + Bi (where A and B are real n-by-n matrices) as the 2n-by-2n matrix

A ’B

B A . In this way, we can embed GL(n, C), the space of n-by-n invertible complex

matrices, as a closed submanifold of GL(2n, R). The reader should check that this mapping

is actually a group homomorphism.

Among the more commonly encountered complex matrix Lie groups are the complex

special linear group, denoted by SL(n, C), and the unitary and special unitary groups,

denoted, respectively, as

U(n) = {a ∈ GL(n, C) | —a a = In }

SU(n) = {a ∈ U(n) | detC (a) = 1 }

where —a = ta is the Hermitian adjoint of a. These groups will play an important role in

¯

what follows. The reader may want to familiarize himself with these groups by doing some

of the exercises for this section.

Basic General Properties. If G is a Lie group with a ∈ G, we let La , Ra : G ’ G

denote the smooth mappings de¬ned by

La (b) = ab and Ra (b) = ba.

Proposition 1: For any Lie group G, the maps La and Ra are di¬eomorphisms, the map

µ: G — G ’ G is a submersion, and the inverse mapping ι: G ’ G de¬ned by ι(a) = a’1

is smooth.

L.2.3 14

Proof: By the axioms of group multiplication, La’1 is both a left and right inverse to

La . Since (La )’1 exists and is smooth, La is a di¬eomorphism. The argument for Ra is

similar.

In particular, La : T G ’ T G induces an isomorphism of tangent spaces Tb G ’ Tab G

˜

for all b ∈ G and Ra : T G ’ T G induces an isomorphism of tangent spaces Tb G ’ Tba G

˜

for all b ∈ G. Using the natural identi¬cation T(a,b)G — G Ta G • Tb G, the formula for

µ (a, b): T(a,b) G — G ’ Tab G is readily seen to be

µ (a, b)(v, w) = La (w) + Rb (v)

for all v ∈ Ta G and w ∈ Tb G. In particular µ (a, b) is surjective for all (a, b) ∈ G — G, so

µ: G — G ’ G is a submersion.

Then, by the Implicit Function Theorem, µ’1 (e) is a closed, embedded submanifold

of G — G whose tangent space at (a, b), by the above formula is

T(a,b)µ’1 (e) = {(v, w) ∈ Ta G — Tb G La (w) + Rb (v) = 0}.

Meanwhile, the group axioms imply that

µ’1 (e) = (a, a’1 ) | a ∈ G ,

which is precisely the graph of ι: G ’ G. Since La and Ra are isomorphisms at every

point, it easily follows that the projection on the ¬rst factor π1 : G — G ’ G restricts to

µ’1 (e) to be a di¬eomorphism of µ’1 (e) with G. Its inverse is therefore also smooth and

is simply the graph of ι. It follows that ι is smooth, as desired.

For any Lie group G, we let G—¦ ‚ G denote the connected component of G which

contains e. This is usually called the identity component of G.

Proposition 2: For any Lie group G, the set G—¦ is an open, normal subgroup of G.

Moreover, if U is any open neighborhood of e in G—¦ , then G—¦ is the union of the “powers”

U n de¬ned inductively by U 1 = U and U k+1 = µ(U k , U) for k > 0.

Proof: Since G is a manifold, its connected components are open and path-connected,

so G—¦ is open and path-connected. If ±, β: [0, 1] ’ G are two continuous maps with

±(0) = β(0) = e, then γ: [0, 1] ’ G de¬ned by γ(t) = ±(t)β(t)’1 is a continuous path from

e to ±(1)β(1)’1 , so G—¦ is closed under multiplication and inverse, and hence is a subgroup.

It is a normal subgroup since, for any a ∈ G, the map

’1

Ca = La —¦ (Ra ) :G ’ G

(conjugation by a) is a di¬eomorphism which clearly ¬xes e and hence ¬xes its connected

component G—¦ also.

Finally, let U ‚ G—¦ be any open neighborhood of e. For any a ∈ G—¦ , let γ: [0, 1] ’ G

be a path with γ(0) = e and γ(1) = a. The open sets {Lγ(t)(U) | t ∈ [0, 1]} cover γ [0, 1] ,

L.2.4 15

so the compactness of [0, 1] implies (via the Lebesgue Covering Lemma) that there is a ¬nite

subdivision 0 = t0 < t1 · · · < tn = 1 so that γ [tk , tk+1 ] ‚ Lγ(tk ) (U) for all 0 ¤ k < n.

But then each of the elements

ak = γ(tk )’1 γ(tk+1 )

lies in U and a = γ(1) = a0 a1 · · · an’1 ∈ U n .

An immediate consequence of Proposition 2 is that, for a connected Lie group H, any

Lie group homomorphism φ: H ’ G is determined by its behavior on any open neighbor-

hood of e ∈ H. We are soon going to show an even more striking fact, namely that, for

connected H, any homomorphism φ: H ’ G is determined by φ (e): Te H ’ Te G.

The Adjoint Representation. It is conventional to denote the tangent space at

the identity of a Lie group by an appropriate lower case gothic letter. Thus, the vector

space Te G is denoted g, the vector space Te GL(n, R) is denoted gl(n, R), etc.

For example, one can easily compute the tangent spaces at e of the Lie groups de¬ned

so far. Here is a sample:

sl(n, R) = {a ∈ gl(n, R) | tr(a) = 0}

so(n, R) = {a ∈ gl(n, R) | a + ta = 0}

u(n, R) = {a ∈ gl(n, C) | a + ta = 0}

¯

De¬nition 4: For any Lie group G, the adjoint mapping is the mapping Ad: G ’ End(g)

de¬ned by

’1

Ad(a) = La —¦ (Ra) (e): Te G ’ Te G.

As an example, for G = GL(n, R) it is easy to see that

Ad(a)(x) = axa’1

for all a ∈ GL(n, R) and x ∈ gl(n, R). Of course, this formula is valid for any matrix Lie

group.

The following proposition explains why the adjoint mapping is also called the adjoint

representation.

Proposition 3: The adjoint mapping is a linear representation Ad: G ’ GL(g).

Proof: For any a ∈ G, let Ca = La —¦ Ra’1 . Then Ca: G ’ G is a di¬eomorphism which

satis¬es Ca (e) = e. In particular, Ad(a) = Ca(e): g ’ g is an isomorphism and hence

belongs to GL(g).

L.2.5 16

The associative property of group multiplication implies Ca —¦ Cb = Cab, so the Chain

Rule implies that Ca (e) —¦ Cb (e) = Cab(e). Hence, Ad(a)Ad(b) = Ad(ab), so Ad is a

homomorphism.

It remains to show that Ad is smooth. However, if C: G — G ’ G is de¬ned by

C(a, b) = aba’1 , then by Proposition 1, C is a composition of smooth maps and hence

is smooth. It follows easily that the map c: G — g ’ g given by c(a, v) = Ca (e)(v) =

Ad(a)(v) is a composition of smooth maps. The smoothness of the map c clearly implies

the smoothness of Ad: G ’ g — g— .

Left-invariant vector ¬elds. Because La induces an isomorphism from g to Ta G

for all a ∈ G, it is easy to show that the map Ψ: G — g ’ T G given by

Ψ(a, v) = La (v)

is actually an isomorphism of vector bundles which makes the following diagram commute.

Ψ

G—g ’’ TG

¦ ¦

¦ ¦π

π 1

id

’’

G G

Note that, in particular, G is a parallelizable manifold. This implies, for example, that the

only compact surface which can be given the structure of a Lie group is the torus S 1 — S 1.

For each v ∈ g, we may use Ψ to de¬ne a vector ¬eld Xv on G by the rule Xv (a) =

La (v). Note that, by the Chain Rule and the de¬nition of Xv , we have

La (Xv (b)) = La (Lb (v)) = Lab (v) = Xv (ab).

Thus, the vector ¬eld Xv is invariant under left translation by any element of G. Such

vector ¬elds turn out to be extremely useful in understanding the geometry of Lie groups,

and are accorded a special name:

De¬nition 5: If G is a Lie group, a left-invariant vector ¬eld on G is a vector ¬eld X on

G which satis¬es La (X(b)) = X(ab).

For example, consider GL(n, R) as an open subset of the vector space of n-by-n ma-

trices with real entries. Here, gl(n, R) is just the vector space of n-by-n matrices with real

entries itself and one easily sees that

Xv (a) = (a, av).

(Since GL(n, R) is an open subset of a vector space, namely, gl(n, R), we are using the

standard identi¬cation of the tangent bundle of GL(n, R) with GL(n, R) — gl(n, R).)

The following proposition determines all of the left-invariant vector ¬elds on a Lie

group.

Proposition 4: Every left-invariant vector ¬eld X on G is of the form X = Xv where

v = X(e) and hence is smooth. Moreover, such an X is complete, i.e., the ¬‚ow ¦ associated

to X has domain R — G.

L.2.6 17

Proof: That every left-invariant vector ¬eld on G has the stated form is an easy exercise

for the reader. It remains to show that the ¬‚ow of such an X is complete, i.e., that for each

a ∈ G, there exists a smooth curve γa : R ’ G so that γa (0) = a and γa (t) = X (γa (t)) for

all t ∈ R.

It su¬ces to show that such a curve exists for a = e, since we may then de¬ne

γa (t) = a γe (t)

and see that γa satis¬es the necessary conditions: γa (0) = a γe (0) = a and

γa (t) = La (γe (t)) = La (X (γe (t))) = X (aγe (t)) = X (γa (t)) .

Now, by the ode existence theorem, there is an µ > 0 so that such a γe can be de¬ned

on the interval (’µ, µ) ‚ R. If γe could not be extended to all of R, then there would be a

maximum such µ. I will now show that there is no such maximum µ.

For each s ∈ (’µ, µ), the curve ±s : (’µ + |s|, µ ’ |s|) ’ G de¬ned by

±s (t) = γe (s + t)

clearly satis¬es ±s (0) = γe (s) and

±s (t) = γe (s + t) = X (γe (s + t)) = X (±s (t)) ,

so, by the ode uniqueness theorem, ±s (t) = γe (s)γe (t). In particular, we have

γe (s + t) = γe (s)γe (t)

for all s and t satisfying |s| + |t| < µ.

Thus, I can extend the domain of γe to (’ 3 µ, 3 µ) by the rule