Download Lie Groups and Their Lie Algebras One

Lie Groups and Their Lie Algebras
One-Parameter Subgroups
Lemma 1. Let X be a smooth vector field on M , F : M → N a diffeomorphism.
Let the the local 1-parameter group generated by X is given by θt , and the local
group generated by Y is ηt . Then Y = F∗ X iff F ◦ θt ◦ F −1 on the domain of θt :
F
M −−−−→


θt y
N

 ηt
y
M −−−−→ N
F
Proof 1. The commutativity of the diagram means that the following holds for all
(t, p) in the domain of θ:
ηt ◦ F (p) = F ◦ θt (p);
that is,
(1)
η F (p) (t) = F ◦ θ(p) (t),
for all t in the domain of definition of θ(p) .
(⇒) Suppose that Y = F∗ X. If we define γ by γ = F ◦ θp , then
γ 0 (t) = (F ◦ θ(p) )0 (t) = F∗ (θ(p)0 (t)) = F∗ Xθ(p) (t) = YF ◦θ(p) (t) = Yγ(t) ,
and hence γ is an integral curve of Y strating at F ◦ θ(p) (0) = F (p).
By uniqueness of integral curves, γ(p) = η F (p) (t) on the interval where θ(p) is
defined. This proves (3).
(⇐) If (3) holds, then ∀t ∈ M , we have
F∗ Xp = F∗ (θ(p)0 (0)) = (F θ(p) )0 (0) = η F (p)0 (0) = YF (p) .
Proof 2. F ◦ θt ◦ F −1 is a local 1-parameter group, and therefore, by uniqueness of
solutions of ODE, it suffices to show the claim near t = 0. Now
d
d
−1 −1 (F ◦ θt ◦ F )
(θt ◦ F ) t=0 ,
=dF
dt
dt
t=0
where dF is evaluated at θ0 ◦ F −1 (p) = F −1 (p)
=dFF −1 (p) X(F −1 (p)) = F∗ X(p).
Typeset by AMS-TEX
1
2
• Let G be a Lie group and θ be the folw of a left-invariant vector field X ∈ Lie(G).
Left-invariance of X means that X is Lg -related to itself for every g ∈ G, so
Lemma 0 implies that
(2)
Lg ◦ θt = θt ◦ Lg , (i.e. g · θt = θt (g))
on the domain of θt . In particular, for the integral curve θ(e) srarting at the
identity, this implies
θ(e) (s)θ(e) (t) =Lθ(e) (s) θt (e) = θt (Lθ(e) (s) (e)) = θt (θ(e) (s))
=θt (θs (e)) = θt+s (e) = θ(e) (t + s),
at least where both sides are defined.
• If θ(e) is defined for all t ∈ R, what we have shown is precisely that θ(e) : R → G
is a Lie group homomorphism.
• In fact, as the next lemma shows, this is always the case, because every integral
curve of X is defined for all time.
Lemma 2. Every left-invariant vector field on a Lie group is complete.
Proof. Let G be a Lie group, let X ∈ Lie(G), and let θ denote the flow of X.
— Suppose some maximum integral curve θ(g) is defined on an interval (a, b) ⊂ R,
and assume that b < ∞. (The case a > −∞ is handled similarly.)
— We use left-invariance to define an integral curve on a slightly larger interval,
contradicting the assumption that (a, b) was the maximal domain of θ(g) .
Observe that the integral curve θ(e) starting at the identity is defined at least on
some interval (−ε, ε) for ε > 0.
— Choose some c ∈ (b − ε, b), and define a new curve γ : (a, c + ε) → G by
(g)
θ (t),
t ∈ (a, b)
γ(t) =
(e)
Lθ(g) (c) (θ (t − c)),
t ∈ (c − ε, c + ε).
— By (2), when t ∈ (a, b) ∩ (c − ε, c + ε), we have
Lθ(g) (c) (θ(e) (t − c)) =Lθ(g) (c) (θt−c (e) = θt−c (Lθ(g) (c) (e))
=θt−c (θc (g)) = θt (g) = θ(g) (t),
so the two definitions of γ agree where they overlap.
Now, γ is clearly an integral curve of X on (a, b), and for t0 ∈ (c − ε, c + ε) we
use left-invariance of X to compute
d
γ 0 (t0 ) = Lθ(g) (c) θ(e) (t − c)
dt t=t0
d
= Lθ(g) (c) ∗ θ(e) (t − c)
dt
t=t0
=(Lθ(g) (c) )∗ Xθ(e) (t0 −c) = Xγ(t0 ) .
Thus γ is an integral curve of X defined for t ∈ (a, c + ε).
— Since c + ε > b, this contradicts the assumption that (a, b) was the maximal
domain of θ(g) . Definition. Given X ∈Lie(G), we will call the one-parameter subgroup determined in this way the one-parameter subgroup generated by X.
3
• The one-parameter subgroups of the general liner group GL(n, R) are not hard
to compute explicitly.
Proposition 3. For any A ∈ gl(n, R), let
eA =
(3)
∞
X
1 k
1
A = In + A + A2 + · · · .
k!
2
k=0
The series converges to an invertible matrix eA ∈ GL(n, R), and the one-parameter
subgroup of GL(n, R) generated by A ∈ gl(n, R) is generated by A ∈ gl(n, R) is
F (t) = etA .
Proof. (I) First we verify convergence.
— Since matrix multiplication satisfies |AB| ≤ |A||B|, where the norm is the Eu2
clidean norm on gl(n, R), under its obvious identification with Rn .
k
k
— It follows by induction that |A | ≤ |A| .
— The Weierstrass M -tset shows that (3) converges
P uniformly on any bounded
subset of gl(n, R) (by comparison with the series k (1/k!)ek = ec ).
(II) Fix A ∈ gl(n, R). The one-parameter subgroup generated by A is an integral
e on GL(n, R), and therefore the ODE
curve of the left-invariant vector field A
initial value problem
eF (t) , F (0) = In .
F 0 (t) = A
Since
j ∂ i ∂ i
e
= Fj (t)Ak
.
AF (t) = (LF (t) )∗ A = (LF (t) )∗ Aj
i
i
∂Xj In
∂Xk F (t)
the condition for F to be an integral curve can be written as
(Fki )0 (t) = Fji (t)Ajk ,
or in matrix notation
F 0 (t) = F (t)A.
Claim: F (t) = etA satisfies this equation.
e starting
Since F (0) = In , this implies that F is the unique integral curve of A
at the identity and is therefore the desired one-parameter subgroup.
— To see that F is differentiable, we note that differentiating the series (3) formally
term by term yields the result
F 0 (t) =
∞
∞
X
X
k k−1 k
1
t
tk−1 Ak−1 A = F (t)A.
A =
k!
(k − 1)!
k=1
k=1
— Since the differentiated series converges uniformly on bounded sets , the termby-term differentiation is justified.
By smoothness of solutions to ODEs, F is a smooth curve.
4
It remains only to show that F (t) is invertible for all t, so that F actually
takes its values in GL(n, R).
— If we let σ(t) = F (t)F (−t) = etA e−tA , then σ is a smooth curve in gl(n, R), and
by the previous computation and the product rule it satisfies
σ 0 (t) = (F 0 (t))F (−t) − F (t)(F 0 (−t)).
— A similar computation as above shows that F 0 (t) = AF (t), and hence
σ 0 (t) = (F (t)A)F (−t) − F (t)(AF (−t)) = 0.
— Therefore, σ is the constant curve σ(t) ≡ σ(0) = In , that is F (t)F (−t) = In .
— Substituing −t for t, we obtain F (−t)F (t) = In , which shows that F (t) is invertible and (F (t))−1 = F (−t). 5
The Exponential Map
• We saw in the preceeding section that the matrix exponential maps gl(n, R)
to GL(n, R) and takes each line through the origin to a one-parameter
subgroup.
— This has a powerful generalization to arbitrary Lie groups.
Definition. Given a Lie group G with Lie algebra g, define a map exp : g → G,
called the exponential map of G, by letting
exp X = F (1),
where F is the one-parameter subgroup generated by X, or equivalently the integral
curve of X starting at the identity.
Example. The results of the preceeding section shows that the exponential map
of GL(n, R) (or any Lie subgroup of it) is given by exp A = eA .
— This, obviously, is the reason for the term exponential map.
Example. If V is a finite-dimensional real vector space, a choice of basis for V
yields isomorphisms GL(V ) ∼
= GL(n, R) and gl(V ) ∼
= gl(n, R).
The analysis of the GL(n, R) can then shows that the exponential map of GL(V )
can be written in the form
∞
X
1 k
exp A =
A ,
k!
k=1
where we consider A ∈ gl(V ) as a linear map from V to itself, and Ak = A ◦ · · · ◦ A
is the k-fold composition of A with itself.
Proposition 4.1 (Properties of the Exponential Map). Let G be a Lie group
and let g be its Lie algebra.
(a) The exponential map is a smooth map from g to G.
(b) ∀X ∈ g, F (t) = exp tX is the one-parameter subgroup of G generated by X.
(c) ∀X ∈ g, exp(s + t)X = exp sX exp tX.
(d) The pushforward
exp∗ : To g → Te G
is the identity map, under the canonical identification of both To g and Te G
with g itself.
(e)The exponential map restricts to a diffeomorphosm from some nbhd of 0 in g
to a nbhd of e in G.
Proof. In this proof, for any X ∈ g be arbitrary, and let θ(X) denote the flow of X.
(e)
(a) To prove (a), we need to show that the expression θ(X) (1) depends smoothly
on X, which amounts to showing that the flow varies smoothly as the vector
field varies.
— This is a situation not covered by the fundamental theorem on flows, but we can
reduce it to that theorem by the following simple trick.
6
— Define a smooth vector field Ξ on the product manifold G × g by
Ξ(g,X) = (Xg , 0) ∈ Tg G ⊕ TX g ∼
= T(g,X) (G × g).
It is easy to verify that the flow Θ of Ξ is given by
Θt (g, X) = (θ(X) (t, g), X).
By the fundamental theorem on flows, Θ is a smooth map.
— Since exp X = π1 (Θ1 (e, X)), where π1 : G × g → G is the projection, it follows
that exp is smooth.
(b) Since the one-parameter subgroup generated by X is equal to the integral curve
(e)
of X starting at e, to prove (b), it suffices to show that exp tX = θ(X) (t), or in
other words that
(e)
(4)
(e)
θ(tX) (1) = θ(X) (t).
In fact, we will prove that for all s, t ∈ R,
(e)
(5)
(e)
θ(tX) (s) = θ(X) (st),
which clearly implies (4).
— To prove (5), fix t ∈ R and define a smooth curve γ : R → G by
(e)
γ(s) = θ(X) (st).
By the chain rule,
(e)
γ 0 (s) = t(θ(X) )0 (st) = tXγ(s) ,
so γ is an integral curve of the vector field tX. Since γ(0) = e, by uniqueness of
(e)
integral curves we must have γ(s) = θtX (s), which is (5).
(c) (c) follows immediately from (b), since t 7→ exp tX is a group homomorphism.
(d) Let X ∈ g be arbitrary, and let σ : R → g be the curve σ(t) = tX. Then
σ 0 (0) = X, and (b) implies
exp∗ X = exp∗ σ 0 (0) = (exp ◦ σ)0 (0) =
d exp tX = X.
dt t=0
(e) (e) follows immediately from (d) and the inverse function theorem.
7
Proposition 4.2 (Properties of the Exponential Map). Let G be a Lie group
and let g be its Lie algebra.
The flow θ of a left-invariant vector field X is given by θt = Rexp tX (right multiplication by exp tX).
Proof. We use (b)in Proposition 4.1 and (2) to show that
Rexp tX(g) =g exp tX = Lg (exp tX) = Lg ((θ(X) )t (e))
=(θ(X) )t (Lg (e)) = (θ(X) )t (g).
Proposition 4.3 (Properties of the Exponential Map). Let G be a Lie group
and let g be its Lie algebra.
If H is another Lie group and h is its Lie algebra, for any Lie group homomorphism
F : G → H, the following diagram commutes:
F
g −−−∗−→


expy
h

exp
y
G −−−−→ H.
F
Proof. We need to show that exp(F∗ X) = F (exp X) for every X ∈ g.
In fact, we will show that for all t ∈ R,
exp(tF∗ X) = F (exp tX).
The left hand side is, by (b), the one-parameter subgroup generated by F∗ X.
Thus if we set σ(t) = F (exp tX), it suffices to show that σ : R → H is a group
homomorphism satisfying σ 0 (0) = F∗ X. Indeed, we compute
d d σ (0) = F (exp tX) = F∗ exp tX = F∗ X
dt t=0
dt t=0
0
and
σ(s + t) =F (exp(s + t)X)
=F (exp sX exp tX)
by (c)
=F (exp sX)F (exp tX) since F is a homomorphism
=σ(s)σ(t).