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through, the reader can simply imagine that all of the Lie groups mentioned are matrix
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

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