CHAPTER Ⅱ
A. On the Geometry of Lie Groups
A.1.
note
(i) follows from Ad ( x)tX  x exp tXx 1  L( x) R( x 1 ) exp tX .for X  g , t  R
 ( x exp tX )  exp(tX ) x
1
For (ii) we
, so d  x (dL( x)e X )  dR( X 1 ) e X . For (iii) we observe for
X 0 , Y0  g
  g exp tX 0 , h exp sY0   g exp tX 0h exp sY0
 gh exp tAd (h 1 ) X 0 exp sY0 ,
(Continued on next page)
Exercises, ChapterⅡ
Solutions
so
d  (dL( f ) X 0 dL(h)Y0 )  dL( gh)( Ad (h 1 ) X 0  Y0 ) .
Putting X  dL( g ) X 0 , Y  dL(h)Y0 , the result follows from (i).
A.2. Suppose
 (t1 )   (t2 ) so  (t2  t1 )  e . Let L＞0 be the smallest number such that  ( L)  e .
Then  (t  L)   (t ) ( L)   (t ). If
T
L denotes the translation t→t+L, we have
A.3. The curve σ satisfies  (t  L)   (t ), , so as in A.2,

T
L   ,so
.
A.4 Let (pn) be a Cauchy sequence in G/H. Then if d denotes the distance, d ( pn , pm )  0 , if
m, n   . Let B (o) be a relatively compact ball of radius ∈ ＞0 around the origin o = {H} in G/H.
1
Select N such that d(pN, pm) ＜ ∈ for m ≥ N and select g ∈ G such that g • pN = o. Then ( g • pm) is a
2

Cauchy sequence inside the compact ball B (o) , hence it, together with the original sequence, is
convergent.
A.5. For X∈g, let
denote the corresponding left invariant vector field on G. From Prop. 1.4 we
for all Z ∈g. But by (2), §9 in Chapter Ⅰthis condition reduces
know that (i) is equivalent to
to
which is clearly equivalent to (ii). Next (iii) follows from (ii) by replacing X by X+Z. But (iii) is equivalent to
Ad(G)-invariance
  R( g
1
)

of
B
so
Q
is
right
invariant.
Finally,
the
map
: x  x
1
satisfies
L( g 1 ) , so d  g  dR( g 1 )e d  e dL( g 1 ) . Since d  e is automatically an isometry,
g
(v) follows.
1
2
A.6. Assuming first the existence of ▽, consider the affine transformation  : g  exp Yg
G which fixes the point exp
1
exp Y
2
1
exp Y of
2
1
Y and maps γ1, the first half of γ, onto the second half, γ2. Since
2
1
1
L( exp  Y ),

2
2
  I . Let X  (t )  Gexp tY (0  t  1) be the family of vectors parallel with respect to
  L(exp Y )
We have d
1
Exercises, ChapterⅡ
Solutions
 such that X  (0)  X . Then σ maps X＊(s) along γ1 into a parallel field along γ2 which must be the
1
2
1
2
field  X  (t ) because d ( X  ( ))   X  ( ). . Thus the map σ ○ ∫ = L(exp
1
1
Y) R(exp Y)
2
2
sends X into X＊(1), as stated in part (i). Part (ii) now follows from Theorem 7.1, Chapter Ⅰ, and part (iii)
from Prop. 1.4. Now (iv) follows from (ii) and the definition of T and R.
Finally,
we
prove
the
existence
of
▽ .
As
remarked
before
Prop.
1.4,
the
defined uniquely a left invariant affine connection
Since
equation
▽ on G.
, we get
～
this we generalize to any vector fields Z, Z’ by writing them in terms of X i (l≤i≤n). Next
(1)
.
Since both sides are right invariant vector fields, it suffices to verify the equation at e. Now
 X  X
where
X is right invariant, so the problem is to prove
For a basis X1, ．．． , Xn of g we write Ad(g
, it follows that
Chapter Ⅰ,§4,
Since ( Xfi )(e) = {(d/dt) f1 (exp tX)}t=0 and since
—
1
) Y, it follows that
Since
, so using ▽2 and Lemma 4.2. from
Exercises, ChapterⅡ
Solutions
d
{ Ad(exp(—tX))(Y)}t=0= —[X, Y].
dt
1
the expression on the right reduces to —[X, Y] + [X,Y], so (1) follows.
2
As before, (1) generalizes to any vector fields Z, Z’.
The connection ▽ is the 0- connection of Cartan- Schouten [1].
B. The Exponential Mapping
B.1. At the end of §1 it was shown that GL(2, R) has Lie algebra gl (2, R), the Lie algebra of all 2×2 real
matrices. Since det ( etx ) = et Tr ( X ), Prop. 2.7 shows that sI(2, R) consists of all 2×2 real matrices of trace 0.
Writing
1 0   0 1  0 0  a b 
X  a
  b
  c


 0 1  0 0   1 0   c a 
a direct computation gives for the Killing form
B( X , X )  8(a 2  bc)  4Tr ( XX ) ,
Where B( X , Y )  4Tr ( XY ) , and semisimplicity follows quickly.
Part (i) is obtained by direct computation. For (ii) we consider the equation
 0 
eX  
1 
0  
(λ∈R, λ≠1).
Case 1: λ＞0. Then det X ＜0. In fact det X＝0 implies
 0 
,
I  X 
1 
0  
So b=c=0, so a=0, contradicting λ≠1. If det X＞0, we deduce quickly from (i) that b＝c＝0, so det X＝－
a2, which is a contradiction. Thus det X＜0 and using (i) again we find the only solution
0 
 log 
X 
.
 log  
 0
1
Case 2: λ＝－1. For det X ＞0 put   (det X ) 2 . Then using (i) the equation amounts to
cos   ( 1 sin  )a  1,
(  1 s i n b)
0,
cos   ( 1 sin  )a  1,
(  1 s i n c )
0.
These equations are satisfied for
  (2n  1)
(n  Z ) ,
d e tX  a2  b c  ( 2n  21) 2
.
This gives infinitely many choices for X as claimed.
Case 3: 
0,   1. If det X =0, then (i) shows b=c=0, so a=0; impossible. If det X＞0 and we put
Exercises, ChapterⅡ
Solutions
1
  (det X ) 2 , (i) implies
cos   (  1 sin  )a   ,
(  1 s i n b)
cos   ( 1 sin  )a   1 ,
0,
(  1 s i n c )
0.
Since    , we have sin   0 . Thus b=c=0, so det X  a , which is impossible. If det X＜0, and
1
2
we put   ( det X )
1
2
, we get from (i) the equations above with sin and cos replaced by sinh and cosh.
Again b=c=0, so det X  a 2    2 ; thus a    , so
cosh   sinh    ,
c o s h
s i nh 1
,
Contradicting λ＜0. Thus there is no solution in this caase, as stated.
B.2. The Killing form on sl(2,R) provides a bi-invariant pseudo-Riemannian structure with the properties of
Exercise A.5. Thus (i) follows from Exercise B.1. Each gwhere g  SL(2, R) can be written g=kp where
k  SO(2) and p is positive definite. Clearly K= exp T where T  s l (2, R) ; and using diagonalization, p=
exp X where X  s l (2, R) . The formula g  exp T exp X proves (ii).
B.3．Follow the hint.
B.4. Considering one-parameter subgroups it is clear that g consists of the matrices
0

c
X (a, b, c)  
0

0
c 0 a

0 0 b
0 0 c

0 0 0
(a , b , c R.)
Then [ X (a, b, c), X (a1 , b1 , c1 )]  X (cb1  c1b, c1a  ca1 ,0), so g is readily seen to be solvable. A direct
computation gives
 cos c sin c

 sin c cos c
exp X (a, b, c)  
 0
0

0
 0
0 c 1 ( a sin c  b cos c  b) 

0 c 1 (b sin c  a cos c  a ) 
.

1
c

0
1

Thus exp X (a, b, 2 ) is the same point in G for all a,b∈R, so exp is not injective. Similarly, the points in G
with   n 2 ( n  Z )
2  2
0 are not in the range of exp. This example ocurs in Auslander and
MacKenzie [1]; the exponential mapping for a solvable group is systematically investigated in Dixmier [2].
B.5. Let N0 be a bounded star-shaped open neighborhood of 0∈g which exp maps diffeomorphically onto an
Exercises, ChapterⅡ
Solutions
1
2
open neighborhood Ne of e in G. Let N  exp( N 0 ) . Suppose S is a subgroup of G contained in N*, and let
*
1
1
s  e in S. Then s  exp X ( X  N 0 ) . Let k  Z  be such that X , 2 X ,..., kX  N 0 but
2
2
1
( k  1) X  N 0 . Since N0 is star-shaped, (k  1) X  N 0 ; but since s k 1  N * , we have
2
1
1
S k 1  exp Y Y  N 0 . Since exp is one-to-one on N0, (k  1) X  Y  N 0 which is a contradiction.
2
2
C. Subgroups and Transformation Groups
C.1. The proofs given in Chapter X for SU*(2n) and Sp(n,C) generalize easily to the other subgroups.
C.2. Let G be a commutative connected Lie group,
during the proof of Theorem 1.11,
its universal covering group. By facts stated
p
is topologically isomorphic to a Euclidean group R . Thus G is
p
topologically isomorphic to a factor group of R and by a well-known theorem on topologically isomorphic to
R n  T m . Thus by Theorem 2.6, G is analytically isomorphic to R n  T m .
For the last statement let  be the closure of  in H. By the first statement and Theorem 2.3,
  R n  T m for some n, m  Z  . But γ is dense in  , so either n=1 and m=0 (γ closed) or n=0 (  compact).
C.3. By Theorem 2.6, is analytic and by Lemma 1.12, dI is injective. Q.E.D.
C.4. The mapping  g turns g N0 into a manifold which we denote by ( g N 0 ) x . Similarly,
turns
 g
g  N0 into a manifold ( g  N 0 ) y . Thus we have two manifolds ( g N0  g  N0 ) x and
( g N 0  g  N 0 ) y and must show that the identity map from one to the other is analytic. Consider the analytic
section maps
 g : ( g N 0 ) x  G,
 g  : ( g  N0 ) y  G
defined by
 g ( g exp( x1 X 1  ...  x X  ) p0 )  g exp( x1 X 1  ...  x X  ),
 g ( g  exp( y1 X 1  ...  y X  ) p0 )  g  exp( y1 X 1  ...  y X  ),
and the analytic map

g
:  1 (g N 0 )  ( g N 0 ) x  H
given by

g
( z )  ( ( Z ),[ g ( ( Z ))]1 z ).
Furthermore, let P : ( g N0 ) x  H  ( g N0 ) x  H denote the projection on the first component. Then the
identity mapping
Exercises, ChapterⅡ
Solutions
I : ( g N0  g  N0 ) y  ( g N0  g  N0 ) x
can be factored:
P
g
 ( g N )  H 
( g N0  g  N0 ) y 
 1 ( g N0 ) 
( g N0 ) x
0 x

g
↑ See Some details. P. 586.
In fact if p  g N0  g  N0 , we have
p  g exp( x1 X 1  ...  x X  ) p0  g  exp( y1 X 1  ...  y X  ) p0 ,
So for some h  H ,
P(  g ( g  ( p)))  P(  g ( g  exp( y1 X 1  ...  y X  )))
 P( ( g  exp( y1 X 1  ...  y X  )), h)
 P( ( g exp( x1 X 1  ...  x X  )), h)
 P exp( x1 X 1  ...  x X  )) p0 .
Thus I is composed of analytic maos so is analytic, as desired.
C.5. The subgroup H  G p of G leaving p fixed is closed, so G / H is a manifold. The map
I : G / H  M given by I ( gH )  g p gives a bijection of G / H onto the orbit G p .Carring the
differentiable structure over on G p by means of I , it remains to prove that I : G / H  M is everywhere
regular. Consider the maps on the diagram.
G
β

G/H
I
where  ( g )  gH ,  ( g )  g p so   I  . If we restrict
M
 to a local cross section, we can write
I    1 on a neighborhood of the origin in G / H . Thus I is C  near the origin, hence everywhere.
Moreover, the map d  e : g  M p has kernel
, the Lie algebra of H (cf. proof of Prop. 4.3). Since
Exercises, ChapterⅡ
Solutions
d e  dI H d e , wee that dI H is one-to-one. Finally, If T ( g ) denotes the diffeomorphism m  g m of
M , we have I  T ( g ) I  ( g 1 ) , whence
dI gH  dT ( g ) dI H d ( g 1 ) gH ,
so I is everywhere regular.
C.6. By local connectedness each component of G is open. It acquires an analytic structure from that of G 0
by left translation. In order to show the map
 : ( x, y)  xy 1 analytic at a point ( x0 , y0 )  G  G let G1
and G2 denote the components of G containing x0 and y0 , respectively. If
0   G0  G0 and
   G1  G2 then
  L( x0 y01 ) I ( y0 ) L( x01 , y01 ),
1
where I ( y0 )( x)  y0 xy0 ( x  G0 ) . Now I ( y0 ) is a continuous autoorphism of the Lie group G0 , hence
by Theorem 2.6, analytic; so the expression for  shows that it is analytic.
C.8. If N with the indicated properties exists we may, by translation, assume it passes through the origin
o  H  in M. Let L be the subgroup
g  G : g
N  N . If g  G maps o into N , then gN  N   ;
1
so by assumption, gN  N . Thus L   ( N ) where
 : G  G / H is the natural map. Using theorem 15.5,
Chapter Ⅰwe see that L can be given the structure of a submanifold of G with a countable basis and by the
transitivity of G on M, L o  N . By C.7., has the desired property. For the converse, define N  L o and use
Prop. 4.4 or Exercise C.5. Clearly, if gN  N   , then g  L , so gN  N .
For more information on the primitivity notion which goes back to Lie see e.g. Golubitsky [1].
D. Closed Subgroups
2
D.1. R T is a torus (Exercise C.2), so it suffices to take a line through 0 in R2 whose image in the torus is
dense.
D.2. g has an Int(g)-invariant positive definite quardratic form Q. The proof of Prop. 6.6 now shows
g    g '(  center of g, g’=[g,g] compact and semisimple). The groups Int(g) and Int(g’) are analytic
subgroups of GL(g) with the same Lie algebra so coincide.
D.3 We have
 1 (c1 , c2 , s)  (c1 , e 2 i 3c2 , s)
0,
3
(a1 , a2 , r )(c1 , c2 , s)(a1 , a2 , r ) 1
 (a1 ( 1 e2 i s ) c1 e2 i r , a2 ( 1 2e
ihs
) 2c 2e
ihr
,s )
Exercises, ChapterⅡ
So 
0,
1
3
is not an inner automorphism, and A
0,
sequence (nk )  Z such that hnk 
1
3
Solutions
 Int ( g ). Now let Sn  0 and let tn  hsn  hn . Select a
1
(mod 1) (Kronecker’s theorem), and let
3
T
k be the unique point in
T
[0,1) such that tn  k  Z . Putting sk  sn , tk tn , we have
k
k
k
s t   T   1
k k
sk k
0,
3
1 0
 , c  C ,   1.
c a
Note: G is a subgroup of H  H where H  
E. Invariant Differential Forms
is torsiion free; and by (5), §7, Chapter Ⅰ,
E.1. The affine connection on G given by
 is a left invariant 1-form,
if
for all left invariant forms  . Now use Exercise C.4. in Chapter Ⅰ.
so
t
t
t
E.2. The first relation is proved as (4), §7. For the other we have g g  I , so (dg ) g  g (dg )  0 .
Hence ( g
1dg )  t (dg )( t g )1  0 and .   t   0
For U(n) we find similarly for   g
1dg ,
d      0,
  t  0 .
For Sp(n)  U (2n) we recall that g  Sp ( n) if and only if
gt g  I
(cf. Chapter Ⅹ). Then the form   g
2n
,
g t g   .
n
n
1dg statisfies
d      0,
  t  0
  n   n t   0.
E.3. A direct computation gives
 0 dx dz  xdy 


g dg   0 0
dy 
0 0
0 

1
and the result follows.