PROOF. This follows from the Frobenius Theorem in differential geometry. For a basis Xi of a Lie subalgebra h of g, we defined a subspace distributionHgwhich is spanned by Xi gfor all g ∈ G. The distribution H = Sg∈GHg is integrable. Indeed, for any two C∞ vector fields V =∑ fiXi and W=∑ giXi, we compute
[V, W] =
∑
figj[Xi, Xj] +∑
fi(Xigj)Xj−∑
gj(Xjfi)Xi ∈ H.We then take H to be the maximal integral submanifold passing through e∈ G.
To check that H is a group, let g ∈ H. The map Lg maps the manifold H to gH. The left invariance says that dLgHh = Hgh, hence gH is also an integral submanifold. Now H and gH both contain the element g, hence the maximality (uniqueness) implies that H = gH.
This implies that H is closed under multiplication and also g−1 ∈ H (since e ∈ H). So H is a subgroup of G.
Finally, H is a Lie groups simply because the map H×H → H sending(g, h)to gh−1is the restriction of the given C∞map G×G→
G.
Remark 5.5. For any Lie group G, the tangent bundle TG is a trivial vector bundle with global frame given by any basis of g.
More generally, a Lie group homomorphism ρ : G→ H is a C∞map which is also a group homomorphism. The tangent map dρ : TG → TH is compatible with l.i.v.f.’s. To see this, ρ(gg0) = ρ(g)ρ(g0)means ρ◦Lg= Lρ(g) ◦ρ, so
dρ◦dLg =dLρ(g)◦dρ.
Thus dρ : g → h. dρ is indeed a Lie algebra homomorphism in the sense that
dρ[X, Y] = [dρ(X), dρ(Y)], which is easily verified from the definitions.
2. EXPONENTIAL MAP 141
2. Exponential map
We call a nontrivial Lie group homomorphismR → G a one pa-rameter subgroup, even though it may not be injective. The exponential map links Lie algebras with Lie groups through the consideration of all one parameter subgroups. Before treating the abstract setting, we look at the case for matrix groups.
Example 5.6. For A ∈ Mn×n(C), t ∈ C, we define the absolutely convergent series
etA =1+tA+t
2
2!A2+ · · · + t
k
k!Ak+ · · · .
It is easily checked that if AB =BA then eAeB =eA+B. Hence eAhas inverse e−A and so eA ∈ GL(n,C). Moreover γ(t) = etA is the one parameter subgroup with
γ0(t) =etAA=dLγ(t)A = Aγ(t).
That is, etA is the integral curve of the l.i.v.f. determined by A ∈ gl(n,C).
The discussion works forC being replaces by R. Also if we take A be in a Lie subalgebra, the eA lies in the corresponding Lie sub-group. This follows from the previous theorem. But we can also see how it works explicitly: For example,
tr A=0 =⇒ det eA =etr A =1.
Also
A∗ = −A =⇒ (eA)∗eA=eA∗eA =e−AeA =In.
Now we turn to a general Lie group G. Let X ∈ g. SinceRX <g is a one dimensional Lie subalgebra, by the previous theorem its inte-gral curve is a one dimensional subgroup H. By taking the universal coverR→ H if necessary, we get a one parameter subgroup which we denote by t 7→ exp tX. We shall give a direct proof of this with stronger conclusions.
Let φt be the flow generated by X. That is, φt(g) is the curve with φ0(g) = g and
d
dtφt(g) = Xφt(g).
Theorem 5.7. The range of t isR for all g∈ G. Moreover, φt : G→ G is a one-parameter group of diffeomorphisms as right translations φt =Rφt(e).
PROOF. Consider the curve gφt(e). Since gφ0(e) = g and d
dt gφt(e)=dLg dLφt(e)Xe
=dLgφt(e)Xe
= Xgφt(e), we conclude that φt(g) = gφt(e) = Rφt(e)g.
By substituting g = φs(e) we find φs(e)φt(e) = φt(φs(e)) = φt+s(e). This shows that for g = e, the range of t can be extended to allR and φt(e)is a one parameter subgroup. The theorem is proved by using the relation φt(g) = gφt(e)again.
Now we define the exponential map exp : g→G
by exp tX =φt(e)where φt is the flow generated by X. Since (d exp)0(X) = d
dt t=0
exp tX =X, we get(d exp)0 =Idgand exp is invertible near 0∈ g.
Corollary 5.8. If H <G is a Lie subgroup, then H is generated by exp h.
However, exp is not necessarily surjective, hence exp g is not nec-essarily a group.
Exercise 5.1. Let X ∈sl(2,R)and d=p|det X|. Then (i) eX = (cosh d)I2+1d(sinh d)X if det X <0.
(ii) eX = (cos d)I2+1d(sin d)X if det X>0.
(iii) eX = I2+X if det X=0.
3. ADJOINT REPRESENTATION 143
Let ga = a 0 0 a−1
!
∈ SL(2,R). Then ga lies in a unique one pa-rameter subgroup if a >0. ga lies in infinitely many one parameter subgroup if a= −1. If a 6= −1 and a<0, then ga 6∈ exp sl(2,R).
3. Adjoint representation
3.0.1. Three adjoints Ig, Adgand adX. For g∈ G, let Ig : G →G be the inner automorphism Ig(h) = LgRg−1(h) = Rg−1Lg(h) = ghg−1. Since Ig(e) = e, we get its differential
Adg :=dIg : g→g
as a Lie algebra automorphism. From dIgg0 =d(Ig◦Ig0) = dIg◦dIg0, we get the adjoint representations
Ad : G →Aut g and
ad :=d(Ad) : g→End g.
For G a matrix group, g is a matrix Lie algebra and it is clear that Adg(Y) = gYg−1. For g(t)a curve with g(0) = e and g0(0) = X we then compute
adX(Y) = (g(t)Yg(t)−1)0(0) = XY−YX = [X, Y]. This property holds true in general:
Theorem 5.9. For X, Y ∈ g,
adXY = [X, Y].
PROOF. Let f ∈ C∞(G) and φ, ψ be the flows generated by X, Y.
Then
(adXY)f = d dt
t=0
(Adexp tXY)f
= d dt
t=0
d ds
s=0
f(Iexp tX(exp sY))
= d ds
s=0
d dt
t=0
f(exp tX·exp sY·exp(−tX))
= d ds
d
dt f ◦φ−t◦ψs◦φt(e)(0, 0)
= d
dsd f(−Xψs(e)) +d(f ◦ψs)Xe
s=0
= − d ds
s=0
Xψs(e)f +Xe d ds
s=0
f ◦ψs
= − d ds
s=0
(X f) ◦ψs(e) +XeY f
= −YeX f +XeY f = [X, Y]ef .
Remark 5.10. Readers with experience in differential geometry may observe that the proof is identical with the one for Lie derivative LXY= [X, Y]. Indeed,
adXY= d dt
t=0
(Adexp tXY)
= d dt
t=0
dRexp(−tX)dLexp tXY
= d dt
t=0
dφ−tY =LXY
by the left invariance of Y and the definition of LXY.
It is harder to get explicit formula for Adgin the abstract setting.
We have such a formula in two special cases, both are based on the
3. ADJOINT REPRESENTATION 145
commutative diagram
G ρ
//OO
exp
HOO
exp
g dρ // h
To see this, simply notice that ρ exp tX and exp dρ(tX) are both one parameter subgroups in H with the same tangent vector dρ(X) at t =0.
By applying the diagram to ρ = Ig, we get:
exp(AdgX) = g(exp X)g−1. (For matrix groups this is obvious).
By applying the diagram to H = Aut g, ρ = Ad and g = exp X, we get
Adexp XY=eadXY.
With these preparation, we give some applications of the adjoint representation:
3.0.2. Center of a Lie group. A Lie algebra is called abelian if[X, Y] = 0 for all X, Y.
Proposition 5.11. Let G be a connected Lie group, then Center(G) = Ker Ad. In particular, G is abelian if and only if g is abelian.
PROOF. If g is in the center, then for all t ∈ R and X ∈g, exp tX =g(exp tX)g−1=exp AdgtX=exp tAdgX.
Hence X =AdgX for all X. That is, Adg =idg.
Conversely, g∈ Ker Ad implies that exp X =g(exp X)g−1. Hence g commutes with all elements in a neighborhood of e in G. By the connectedness of G we conclude that g commutes with every
ele-ments in G.
Corollary 5.12. [X, Y] = 0 implies that exp X·exp Y =exp(X+Y).
PROOF. Let h be the two dimensional abelian Lie subalgebra of g spanned by X and Y. Consider the Lie group H generated by exp h.
The proposition show that H is abelian and so the curve γ(t) = exp tX·exp tY is an one parameter subgroup. Since γ0(0) = X+Y, we conclude that exp tX·exp tY=exp t(X+Y). Corollary 5.13. If G is a connected Lie groups with trivial center, then
Ad : G ,→Aut g=GL(g)
is a faithful representation. In particular, G is a matrix subgroup.
3.0.3. Normal Lie subgroups. A subspace h of g is a Lie ideal if [h, g] ⊂h. In this case we denote by hCg. It is clear that h is at least a subalgebra.
Proposition 5.14. Let H <G be a connected Lie subgroup of a connected Lie group. Then
HCG⇐⇒ h :=Lie H C g. PROOF. Let g =exp X with X ∈gand Y∈ h, If h is a Lie ideal of g, then
g(exp Y)g−1 =exp AdgY
=exp(eadXY)
=exp
I+adX+ 1
2!ad2X+ · · ·
Y
∈ exp h⊂H.
Since H is generated by h, this proves that H is normal.
Conversely, if H is normal, then the above computation shows that
γ(t):=exp(eadtXY) ∈ H.
Hence h 3γ0(0) = adXY = [X, Y]and h is a Lie ideal. 3.1. Fundamental correspondences.
4. DIFFERENTIAL GEOMETRY ON LIE GROUPS 147
3.1.1. Equivalence of categories.
Theorem 5.15. Let G and H be connected Lie groups with Lie algebras g