[ad(x), ad(y)], where ad: g ’ gl(g) is de¬ned by ad(x)(y) = [x, y] A Lie subalgebra of g is

a linear subspace h ‚ g which is closed under bracket. A homomorphism of Lie algebras

is a linear mapping of vector spaces •: h ’ g which satis¬es

• [x, y] = •(x), •(y) .

At the moment, our only examples of Lie algebras are the ones provided by Proposition

6, namely, the Lie algebras of Lie groups. This is not accidental, for, as we shall see, every

¬nite dimensional Lie algebra is the Lie algebra of some Lie group.

L.2.12 23

Lie Brackets of Vector Fields. There is another notion of Lie bracket, namely

the Lie bracket of smooth vector ¬elds on a smooth manifold. This bracket is also skew-

symmetric and satis¬es the Jacobi identity, so it is reasonable to ask how it might be

related to the notion of Lie bracket that we have de¬ned. Since Lie bracket of vector ¬elds

commutes with di¬eomorphisms, it easily follows that the Lie bracket of two left-invariant

vector ¬elds on a Lie group G is also a left-invariant vector ¬eld on G. The following result

is, perhaps then, to be expected.

Proposition 8: For any x, y ∈ g, we have [Xx , Xy ] = X[x,y] .

Proof: This is a direct calculation. For simplicity, we will use the following character-

ization of the Lie bracket for vector ¬elds: If ¦x and ¦y are the ¬‚ows associated to the

vector ¬elds Xx and Xy , then for any function f on G we have the formula:

√ √ √ √

f(¦y (’ t, ¦x (’ t, ¦y ( t, ¦x ( t, a))))) ’ f(a)

([Xx , Xy ]f)(a) = lim .

t

t’0+

Now, as we have seen, the formulas for the ¬‚ows of Xx and Xy are given by ¦x (t, a) =

a exp(tx) and ¦y (t, a) = a exp(ty). This implies that the general formula above simpli¬es

to

√ √ √ √

f a exp( tx) exp( ty) exp(’ tx) exp(’ ty) ’ f(a)

([Xx , Xy ]f)(a) = lim .

t

t’0+

Now √ √ √

exp(± tx) exp(± ty) = exp(± t(x + y) + 2 [x, y] + · · ·)

t

√ √ √ √

so exp( tx) exp( ty) exp(’ tx) exp(’ ty) simpli¬es to exp(t[x, y] + · · ·) where the omit-

ted terms vanish to higher t-order than t itself. Thus, we have

f a exp(t[x, y] + · · ·) ’ f(a)

([Xx , Xy ]f)(a) = lim .

t

t’0+

Since [Xx , Xy ] must be a left-invariant vector ¬eld and since

f a exp(t[x, y]) ’ f(a)

(X[x,y] f)(a) = lim ,

t

t’0+

the desired result follows.

We can now prove the following fundamental result.

Theorem 2: For each Lie subgroup H of a Lie group G, the subspace h = Te H is a Lie

subalgebra of g. Moreover, every Lie subalgebra h ‚ g is Te H for a unique connected Lie

subgroup H of G.

L.2.13 24

Proof: Suppose that H ‚ G is a Lie subgroup. Then the inclusion map is a Lie group

homomorphism and Theorem 1 thus implies that the inclusion map h ’ g is a Lie algebra

homomorphism. In particular, h, when considered as a subspace of g, is closed under the

Lie bracket in G and hence is a subalgebra.

Suppose now that h ‚ g is a subalgebra.

First, let us show that there is at most one connected Lie subgroup of G with Lie

algebra h. Suppose that there were two, say H1 and H2 . Then by Theorem 1, expG (h) is

a subset of both H1 and H2 and contains an open neighborhood of the identity element

in each of them. However, since, by Proposition 2, each of H1 and H2 are generated by

¬nite products of the elements in any open neighborhood of the identity, it follows that

H1 ‚ H2 and H2 ‚ H1 , so H1 = H2 , as desired.

Second, to prove the existence of a subgroup H with Te H = h, we call on the Global

Frobenius Theorem. Let r = dim(h) and let E ‚ T G be the rank r sub-bundle spanned

by the vector ¬elds Xx where x ∈ h. Note that Ea = La (Ee ) = La (h) for all a ∈ G, so E

is left-invariant. Since h is a subalgebra of g, Proposition 8 implies that E is an integrable

distribution on G. By the Global Frobenius Theorem, there is an r-dimensional leaf of E

through e. Call this submanifold H.

It remains is to show that H is closed under multiplication and inverse. Inverse is

easy: Let a ∈ H be ¬xed. Then, since H is path-connected, there exists a smooth curve

±: [0, 1] ’ H so that ±(0) = e and ±(1) = a. Now consider the curve ± de¬ned on [0, 1]

¯

by ±(t) = a’1 ±(1 ’ t). Because E is left-invariant, ± is an integral curve of E and it joins

¯ ¯

e to a’1 . Thus a’1 must also lie in H. Multiplication is only slightly more di¬cult: Now

suppose in addition that b ∈ H and let β: [0, 1] ’ H be a smooth curve so that β(0) = e

and β(1) = b. Then the piecewise smooth curve γ: [0, 2] ’ G given by

if 0 ¤ t ¤ 1;

±(t)

γ(t) =

aβ(t ’ 1) if 1 ¤ t ¤ 2,

is an integral curve of E joining e to ab. Hence ab belongs to H, as we wished to show.

Theorem 3: If H is a connected and simply connected Lie group, then, for any Lie group

G, each Lie algebra homorphism •: h ’ g is of the form • = φ (e) for some unique Lie

group homorphism φ: H ’ G.

Proof: In light of Theorem 1 and Proposition 6, all that remains to be proved is that for

each Lie algebra homorphism •: h ’ g there exists a Lie group homomorphism φ satisfying

φ (e) = •.

We do this as follows: Suppose that •: h ’ g is a Lie algebra homomorphism. Con-

sider the product Lie group H — G. Its Lie algebra is h • g with Lie bracket given by

[(h1 , g1 ), (h2 , g2 )] = ([h1 , h2 ], [g1, g2 ]), as is easily veri¬ed. Now consider the subspace

h ‚ h • g spanned by elements of the form (x, •(x)) where x ∈ h. Since • is a Lie algebra

homomorphism, h is a Lie subalgebra of h • g (and happens to be isomorphic to h). In

L.2.14 25

particular, by Theorem 2, it follows that there is a connected Lie subgroup H ‚ H — G,

whose Lie algebra is h. We are now going to show that H is the graph of the desired Lie

group homomorphism φ: H ’ G.

Note that since H is a Lie subgroup of H — G, the projections π1 : H ’ H and

π2 : H ’ G are Lie group homomorphisms. The associated Lie algebra homomorphisms

1 : h ’ h and 2 : h ’ g are clearly given by 1 (x, •(x)) = x and 2 (x, •(x)) = •(x).

Now, I claim that π1 is actually a surjective covering map: It is surjective since

1 : h ’ h is an isomorphism so π1 (H) contains a neighborhood of the identity in H and

hence, by Proposition 2 and the connectedness of H, must contain all of H. It remains to

show that, under π1 , points of H have evenly covered neighborhoods.

Let Z = ker(π1 ). Then Z is a closed discrete subgroup of H. Let U ‚ H be a

neighborhood of the identity to which π1 restricts to be a smooth di¬eomorphism onto

a neighborhood U of e in H. Then the reader can easily verify that for each a ∈ H the

map σa : Z — U ’ H given by σa (z, u) = azu is a di¬eomorphism onto (π1 )’1 (Lπ1 (a)(U))

which commutes with the appropriate projections and hence establishes the even covering

property.

Finally, since H is connected and, by hypothesis, H is simply connected, it follows

’1

that π1 must actually be a one-to-one and onto di¬eomorphism. The map φ = π2 —¦ π1 is

then the desired homomorphism.

As our last general Theorem, we state, without proof, the following existence result.

Theorem 4: For each ¬nite dimensional Lie algebra g, there exists a Lie group G whose

Lie algebra is isomorphic to g.

Unfortunately, this theorem is surprisingly di¬cult to prove. It would su¬ce, by

Theorem 2, to show that every Lie algebra g is isomorphic to a subalgebra of the Lie

algebra of a Lie group. In fact, an even stronger statement is true. A theorem of Ado

asserts that every ¬nite dimensional Lie algebra is isomorphic to a subalgebra of gl(n, R)

for some n. Thus, to prove Theorem 4, it would be enough to prove Ado™s theorem.

Unfortunately, this theorem also turns out to be rather delicate (see [Po] for a proof).

However, there are many interesting examples of g for which a proof can be given by

elementary means (see the Exercises).

On the other hand, this abstract existence theorem is not used very often anyway.

It is rare that a (¬nite dimensional) Lie algebra arises in practice which is not readily

representable as the Lie algebra of some Lie group.

The reader may be wondering about uniqueness: How many Lie groups are there

whose Lie algebras are isomorphic to a given g? Since the Lie algebra of a Lie group

G only depends on the identity component G, it is reasonable to restrict to the case of

connected Lie groups. Now, as you are asked to show in the Exercises, the universal cover

˜

G of a connected Lie group G can be given a unique Lie group structure for which the

covering map G ’ G is a homomorphism. Thus, there always exists a connected and

˜

simply connected Lie group, say G(g), whose Lie algebra is isomorphic to g. A simple

L.2.15 26

application of Theorem 3 shows that if G is any other Lie group with Lie algebra g, then

there is a homomorphism φ: G(g) ’ G which induces an isomorphism on the Lie algebras.

It follows easily that, up to isomorphism, there is only one simply connected and connected

Lie group with Lie algebra g. Moreover, every other connected Lie group with Lie algebra

G is isomorphic to a quotient of G(g) by a discrete subgroup of G which lies in the center

of G(g) (see the Exercises).

The Structure Constants. Our work so far has shown that the problem of classify-

ing the connected Lie groups up to isomorphism is very nearly the same thing as classifying

the (¬nite dimensional) Lie algebras. (See the Exercises for a clari¬cation of this point.)

This is a remarkable state of a¬airs, since, a priori, Lie groups involve the topology of

smooth manifolds and it is rather surprising that their classi¬cation can be reduced to

what is essentially an algebra problem. It is worth taking a closer look at this algebra

problem itself.

Let g be a Lie algebra of dimension n, and let x1 , x2 , . . . , xn be a basis for g. Then

there exist constants ck so that (using the summation convention)

ij

[xi , xj ] = ck xk .

ij

(These quantities c are called the structure constants of g relative to the given basis.) The

skew-symmetry of the Lie bracket is is equivalent to the skew-symmetry of c in its lower

indices:

ck + ck = 0.

ij ji

The Jacobi identity is equivalent to the quadratic equations:

cij cm + cjk cm + cki cm = 0.

k i j

Conversely, any set of n3 constants satisfying these relations de¬nes an n-dimensional Lie

algebra by the above bracket formula.

Left-Invariant Forms and the Structure Equations. Dual to the left-invariant

vector ¬elds on a Lie group G, there are the left-invariant 1-forms, which are indispensable

as calculational tools.

De¬nition 9: For any Lie group G, the g-valued 1-form on G de¬ned by

for v ∈ Ta G

ωG (v) = La’1 (v)

is called the canonical left-invariant 1-form on G.

It is easy to see that ωG is smooth. Moreover, ωG is the unique left-invariant g-valued

1-form on G which satis¬es ωG (v) = v for all v ∈ g = Te G.

By a calculation which is left as an exercise for the reader,

φ— (ωG ) = •(ωH )

for any Lie group homomorphism φ: H ’ G with • = φ (e). In particular, when H is a

subgroup of G, the pull back of ωG to H via the inclusion mapping is just ωH . For this

reason, it is common to simply write ω for ωG when there is no danger of confusion.

L.2.16 27

Example: If G ‚ GL(n, R) is a matrix Lie group, then we may regard the inclusion

g: G ’ GL(n, R) as a matrix-valued function on G and compute that ω is given by the

simple formula

ω = g ’1 dg.

From this formula, the left-invariance of ω is obvious.

In the matrix Lie group case, it is also easy to compute the exterior derivative of ω:

Since g g ’1 = In , we get

dg g ’1 + g d g ’1 = 0,

so

d g ’1 = ’g ’1 dg g ’1 .

This implies the formula

dω = ’ω § ω.

(Warning: Matrix multiplication is implicit in this formula!)

For a general Lie group, the formula for dω is only slightly more complicated. To

state the result, let me ¬rst de¬ne some notation. I will use [ω, ω] to denote the g-valued

2-form on G whose value on a pair of vectors v, w ∈ Ta G is

[ω, ω](v, w) = [ω(v), ω(w)] ’ [ω(w), ω(v)] = 2[ω(v), ω(w)].

Proposition 9: For any Lie group G, dω = ’ 1 [ω, ω].

2

Proof: First, let Xv and Xw be the left-invariant vector ¬elds on G whose values at e

are v and w respectively. Then, by the usual formula for the exterior derivative

dω(Xv , Xw ) = Xv ω(Xw ) ’ Xw ω(Xv ) ’ ω [Xv , Xw ] .

However, the g-valued functions ω(Xv ) and ω(Xw ) are clearly left-invariant and hence are

constants and equal to v and w respectively. Moreover, by Proposition 8, [Xv , Xw ] =

X[v,w] , so the formula simpli¬es to

dω(Xv , Xw ) = ’ω X[v,w] .

The right hand side is, again, a left-invariant function, so it must equal its value at the

identity, which is clearly ’[v, w], which equals ’[ω(Xv ), ω(Xw )] Thus,

dω(Xv , Xw ) = ’ 1 [ω(Xv ), ω(Xw )]

2

for any pair of left-invariant vector ¬elds on G. Since any pair of vectors in Ta G can be

written as Xv (a) and Xw (a) for some v, w ∈ g, the result follows.

L.2.17 28