Expository Suggestions
While the exposition of Lie group theory in this book is connected to some
parts of Chapter I (including affine connections) we shall now indicate how the
book can be used more directly as a text for a one-term course on Lie groups.
The idea is to introduce the exponential mapping for a Lie group at the
beginning and use it to develop the correspondence between Lie subalgebras
and Lie subgroups.
The prerequisites are linear algebra, point set topology and some familiartity
with topological groups.
Introduction to Lie Groups.
(I) Examples of Lie Groups.
Discussion of Example 1, p.98 and GL(n, R).
(II) Introduction of basic notions of differential geometry. Manifolds, C∞
functions, tangent spaces, vector fields, C∞ mappings. Pages 1–7, 9–10,
22-24 from the book.
(III) Definition of a Lie group and its Lie algebra. Abstract Lie algebra and
its universal enveloping algebra. Pages 97–101 (including Corollary 1.3).
(IV) Integral curves of a vector field. Let Y be a vector field on a manifold X.
In a local chart
(U, ϕ) ,
ϕ(q) = (x1 (q), . . . , xm (q)) .
Y can be written (cf. p. 10)
Y =
m
X
i=1
Yi
∂
.
∂xi
Given a point p ∈ U , an integral curve to Y through p is a curve t → γ(t),
(t ∈ I) satisfying
(A)
γ(0) = p, Yγ(t) = γ(t) (tangent vector to γ at γ(t) ).
In the local chart above let
ai = xi (p) ,
xi (t) = xi (γ(t)) , (1 ≤ i ≤ m) .
Then (A) takes the form
(B)
dxi
dt
= (Y i ◦ ϕ−1 )(x1 (t), . . . , xm (t)), xi (0) = ai (1 ≤ i ≤ m).
By standard theory (B) has a locally unique solution, smooth (analytic)
in t and (a1 , . . . , am ).
1
(V) The exponential mapping (replacing pages 102–104).
Take X in (IV) to be a Lie group G (as defined on page 98). Let Lg
and Rg denote the translations h → gh and h → hg, respectively, and
(dLg )h , (dRg )h their differentials at h. For each Y ∈ Ge consider the
vector field Ye on G given by Yeg = (dLg )e (Y ). Using the result of (IV) let
Γ(t) (0 ≤ t ≤ ǫ) be the local integral curve of Ye through e with Γ(0) = e.
Thus,
Γ(t) = YeΓ(t) for 0 ≤ t ≤ ǫ .
Defining successively for n ∈ Z
Γ(t) = Γ(nǫ)Γ(−tǫ) ,
nǫ ≤ t ≤ (n + 1)ǫ ,
Γ extends to R. On the interval nǫ ≤ t ≤ (n + 1)ǫ we have (L referring
to translations both on G and on R)
Γ
= LΓ(nǫ) ◦ Γ ◦ L−nǫ
so
Γ(t)
d
= dLΓ(nǫ) ◦ dΓ ◦ dL−nǫ
dt
t
t
d
= dLΓ(nǫ) ◦ dΓ
= dLΓ(nǫ) YeΓ(t−nǫ) = YeΓ(t) .
dt t−nǫ
= dΓ
d
dt
Thus Γ(t) = YeΓ(t) for all t ∈ R. Fix s ∈ R and consider the curve
Γs : t → Γ(s)Γ(t − s) = LΓ(s) ◦ Γ ◦ L−s (t) .
Again we have
d
dΓs
dt t
=
d dt
t−s
=
YeΓ(s)Γ(t−s) = YeΓs (t)
=
YeΓs (t) .
so
Γs (t)
dLΓ(s) ◦ dΓ
= dLΓ(s) YeΓ(t−s)
Thus Γs is another integral curve of Ye . Since it agrees with Γ at s = t
we have Γs ≡ Γ so Γ is a one-parameter subgroup. We put
exp Y = Γ(1) .
For each s ∈ R, t → Γ(st) is a one-parameter group with tangent vector
sY at t = 0. Thus
exp sY = Γ(s) .
2
The map exp : g → G therefore satisfies
exp(s + t)Y = exp sY exp tY .
The mapping exp : g → G is analytic: For Z ∈ g let gZ denote the tangent
space to g at Z. Consider the product group G × g and the vector field
Y on G × g given by
Y (σ, Z) = (Zeσ , 0) ∈ Gσ ⊕ gZ .
Because of the formula p. 99, Y is analytic. By (IV) the integral curves
of Y are analytic. The one through (σ, Z) is given by
η(t) = (σ exp tZ, Z) ,
which we write as Yt (σ, Z). Writing τ for the projection of G × g on the
first factor we have exp Z = τ ◦ Y1 (e, Z) so exp is analytic. Its differential
at 0 is I. This then proves Proposition 1.6, p. 104 in the text.
(VI) Finally we give an independent proof of Theorem 1.7 in the text (p. 105).
P∞ (−A)n
−A
In the statement, 1−eA stands for 0 (n+1)!
. For the proof we use the
Taylor formula (6) on p. 105. We have for f analytic near exp X,
=
(d expX (Y )f )(exp X) = Y (f ◦ exp)(X)
)
(
∞
d X 1 e
d
n
e
f (exp(X + tY ))
(X + tY ) f (e)
=
dt
dt 0 n!
t=0
t=0
=
∞
X
0
1
en + X
e Ye X
e n−1 + · · · + X
e n Ye )f (e) .
(Ye X
(n + 1)!
e Ye − Ye X
e so we define
Now (ad X(Y ))∼ = X
Thus
e Ye − Y X
e = L e Ye − R e Ye .
ad X(Ye ) = X
X
X
RXe = LXe − ad X
and they commute. Thus
e m = (R e )m (Ye ) = (L e − ad X)m (Ye )
Ye X
X
X
m
X
m
e m−p (ad X)p (Ye )
X
=
(−1)p
p
p=0
3
so
=
n
X
p=0
ep
X
en + · · · + X
e n Ye =
Ye X
n−p
X
k=0
n
X
p=0
e p Ye X
e n−p
X
k n − p e n−p−k
X
(ad X)k (Ye )
(−1)
k
n n−p
X
X
k n − p e n−k
(−1)
X
(ad X k )Ye )
=
k
p=0
k=0
=
k=o
Now what is
to tk .
n − p e n−k
X
(ad X)k (Ye ) .
(−1)
k
p=0
n n−k
X
X
Pn−k
p=0
k
Pn−k
? Look at p=0
(1+t)n−p and collect coefficient
n−p
k
(1 + t)n + · · · + (1 + t)k = (1 + t)k (1 + · · · + (1 + t)n−k )
(1 + t)n+1 − (1 + t)k
(1 + t)n−k+1 − 1
=
= (1 + t)k
(1 + t) − 1
t
and the coefficient to tk equals n+1
k+1 . Thus
n−k
X
n−p
k
p=0
so our sum is
n
X
(−1)k
k=0
Thus
(d expX (Y )f )(exp X)=
=
n+1
=
,
k+1
n + 1 e n−k
X
(ad X)k (Ye ) .
k+1
n
X
1
n + 1 e n−k
(X
(ad X)k (Ye )f )(e)
(−1)k
(n
+
1)!
k
+
1
n=0
∞
X
k=0
∞ X
n
X
1
(−1)k
e n−k (ad X)k (Ye )f )(e)
(X
(k
+
1)!
(n
−
k)!
n=0
k=0
=
∞
∞ X
X
k=0 n=k
=
1
(−1)k
e n−k (ad X)k (Ye )f (e)
X
(k + 1)! (n − k)!
∞ X
∞
X
(−1)k 1 e r
X (ad X)k (Ye )f (e)
(k
+
1)!
r!
r=0
k=0
=
∞ e X
∞
X
Xr
(−1)k
((ad X)k (Ye )f )(e)
r!
(k
+
1)!
r=0
0
4
=
=
=
1 − e− ad X e
(Y )f (exp X)
ad X
∼ 1 − e− ad X
(Y )
f (exp X)
ad X
dL(exp X) ◦
1 − e− ad X
(Y )(f )(exp X) .
ad X
Thus
dLexp(−tX) ◦ d exptX (Y ) =
1 − e− ad (tX)
(Y )
ad (tX)
for all t in some interval |t| < δ. Both sides of this equation are analytic
in t with values in g. Since they agree for |t| < δ they must agree for all
t ∈ R.
(VII) After Theorem 1.7, p. 105, continue with Chapter II. The end of §6 (Theorem 6.9 and Proposition 6.10) uses Riemannian geometry results proved
in Ch. I.
Alternative proof of Proposition 6.10 comes from Theorem 2.5 in Ch. VII
(with T reduced to the identity element). Similarly, Theorem 6.9 (Ch. II)
is contained in Theorem 6.1 in Ch. VII.
(VIII) Section 7 on Invariant Differential Forms is rather optional for an introductory course. If included it would require the material on differential
forms in Ch. I, that is pp. 20–25.
(IX) Here §8 is of historical nature. Lie’s theorem (pp. 144-145) was one of
his original results. Since this theorem seems to have disappeared from
recent books on ordinary differential equations we give here a sketch of a
proof.
Suppose ϕt is a 1-parameter group leaving the equation
Y (x, y)
dy
=
dx
X(x, y)
stable. If U (x, y) = c is a solution we have with ϕt (x, y) = (xt , yt ),
U (xt , yt )
= c(t)
(all t)
so
∂U dyt
∂U dxt
+
= c′ (t)
∂x dt
∂y dt
and by definition of ξ and η
(1)
5
∂U
∂U
ξ+
η = c′ (0) .
∂x
∂y
Secondly
∂U
∂U dy
+
=0
∂x
∂y dx
so
(2)
∂U
∂U
X +
Y = 0.
∂x
∂y
If c′ (0) 6= 0 we can normalize U such that c′ (0) = 1. Then (1) and (2)
imply
∂U
−Y
∂U
X
=
,
=
∂x
Xη − Y ξ ∂y
Xη − Y ξ
so (Xη − Y ξ)−1 is an integrating factor for Xdy − Y dz = 0 as claimed.
On the other hand, if c′ (0) = 0, (1)–(2) imply dy/dx = η/ξ so the integral
curves are just the orbits of ϕt .
6