S YMMETRIC S PACES
P ETER H OLMELIN
Master’s thesis
2005:E3
CENTRUM SCIENTIARUM MATHEMATICARUM
Centre for Mathematical Sciences
Mathematics
Abstract
In this text we study the differential geometry of symmetric spaces. We
describe how a symmetric space (M , g) can be seen as a homogeneous space
G/K , the quotient of its isometry group G and a isotropy group K at a point.
We study the one-to-one correspondence between symmetric spaces and symmetric pairs. Furthermore we investigate the expressions for curvature on a
symmetric space. Finally we describe the notion of dual symmetric spaces.
To illustrate how well symmetric spaces lend themselves to explicit calculations we calculate the curvature of the real Grassmann manifold and find their
dual space.
Keywords: homogeneous spaces, symmetric spaces, symmetric pairs, the
Killing form, curvature of symmetric spaces, dual symmetric spaces.
Throughout this text it has been my intention to give reference to all the sources
that have been used.
Acknowledgements
I am very grateful to Sigmundur Gudmundsson for his support and substantial
amount of comments. I also wish to thank Martin Svensson his deep knowledge and judicious comments about the material.
Peter Holmelin
Contents
Overview
1
Chapter 1. Homogeneous Spaces
1. Group Actions
2. Homogeneous Spaces
3
3
3
Chapter 2. Symmetric Spaces
1. Definitions
2. The Isometry Group
3. G/K is Diffeomorphic to M
11
11
14
17
Chapter 3. Symmetric Pairs
1. From a Symmetric Space to a Symmetric Pair
2. The Tangent Space of G/K
3. From a Lie Group to a Symmetric Pair
4. From a Symmetric Pair to a Symmetric Space
19
19
20
23
25
Chapter 4. Curvature of a Symmetric Space
1. The Killing Form
2. The Curvature Formula
3. The Dual Space
29
29
36
40
Appendix A. The Hopf-Rinow Theorem
45
Appendix B. The Adjoint Representation
47
Appendix C. Inner Products From the Haar Measure
51
Appendix D. Lie Derivatives and Killing Fields
53
Bibliography
59
i
Overview
History[4]
At the end of the nineteenth century, after studying spaces of constant curvature, mathematicians wanted to classify all locally symmetric Riemannian
manifolds, i.e. Riemannian manifolds whose curvature tensor is parallel i.e.
satisfies ∇ R = 0. The whole issue was settled by Elie Cartan in 1932. Previously in his thesis he had classified all simple complex Lie algebras and he also
classified the simple real Lie algebras. Using this he gave a complete classification of all symmetric spaces which by the way were introduced by himself in
1926.
Symmetric Spaces
A symmetric space is a Riemannian manifold (M , g) such that for every
point p ∈ M there exist an isometry p of (M , g) called an involution such
that
(1) p (p) = p and
(2) d p = −idTp M .
By composing involutions one gets translations along geodesics, which can
be used to extend geodesics to the whole of R i.e. M is geodesically complete.
By the Hopf-Rinow theorem any two points in a geodesically complete Riemanninan manifold can be connected by a geodesic. Therefore the translations
along the geodesics makes the isometry group G acting on M transitive. Using
the theory of homogeneous spaces one can identify M with G/K where K is
the isotropy group at a point p ∈ M , i.e K = {k ∈ G : k(p) = p}.
Symmetric Pairs
The description of a symmetric space, in terms of a Lie group G, a closed
subgroup K and an involution , leads to the concept of a symmetric pair
which is defined as a Lie group G with a closed subgroup K and an involutive
automorphism s on G satisfying
(1) (Gs )0 ⊆ K ⊆ Gs ,
1
(2) Ad K is compact
where Gs is the set of elements left invariant by s. We show that symmetric
pairs lead to symmetric spaces.
Curvature on a Symmetric Space
We show that left invariant vector fields on the isometry group G are
mapped to Killing fields in the symmetric space (M , g), which generate Jacobi
fields and therefore provide the connection to the curvature tensor. This gives
a very simple formula for the curvature tensor on a symmetric space G/K ,
namely
R(X , Y )Z = −[[X , Y ], Z ] for X , Y , Z ∈ p
where p is the linear complement of k which is the Lie algebra of K .
Each symmetric space M has a dual space M † which is defined by the
existence of maps
(1) A Lie algebra isomorphism : k → k† such that
g † ( (X ), (Y )) = −g(X , Y )
†
(2) A linear isometry
that
ˆ : p → p such
†
(X ), ˆ (Y )]
[X , Y ] = −[ˆ
where k is the Lie algebra of K and p is a linear complement of k in the Lie
algebra g of G. Dual spaces have opposite curvature and if one is compact its
dual is noncompact and vice versa.
Although the path chosen in this paper is more along minimal differential
geometry, the standard description is more in the language of Lie groups and
Lie algebras.
For a reader who wants a more thorough description we refer to [5] Helgason, which is the standard reference on the theory of symmetric spaces.
2
CHAPTER 1
Homogeneous Spaces
In this chapter we create a manifold structure for the quotient of a Lie
group and a closed subgroup.
1. Group Actions
Definition 1.1. Let M be a smooth manifold and G be a Lie group with
identity element e ∈ G. A smooth map : G × M → M is called a group
action on M if
(g1 , (g2 , p)) = (g1 g2 , p)
(e, p) = p
for all g1 , g2 ∈ G and p ∈ M . The action is said to be effective if
(g, p) = p for all p ∈ M implies that g = e.
The action is said to be transitive if for all p, q ∈ M there exists a g ∈ G such
that (g, p) = q. The isotropy group Kp0 of at p0 ∈ M is given by
Kp0 = {g ∈ G : (g, p0 ) = p0 }.
If : G × M → M is a group action on M , then the group G is said to
act on M and it is customary to write gp for (g, p).
Example 1.2. Let M = S n be the unit sphere in Rn+1 and G = SO(n + 1)
be the special orthogonal group. Then we have the action
: SO(n + 1) × S n → S n , (A, p) 7→ A · p
For p0 = (1, 0, . . . , 0) we get Kp0 = SO(1) × SO(n).
2. Homogeneous Spaces
We need the following lemma to prove that G/K can get a manifold structure.
Lemma 1.3. [10] Let G be a Lie group and let K be a closed subgroup of G.
Denote the Lie algebra of G by g and the Lie algebra of K by k. If m is a linear
complement to k, i.e. g = k ⊕ m and we give G/K the quotient topology, let
: G → G/K be the canonical projection onto the quotient, then the map
◦ exp : m → G/K
3
is a local homeomorphism at 0.
P ROOF. [10] Let be the map
: g = k ⊕ m → G,
(X , Y ) = expX expY
then d (0,0) (X , Y ) = X + Y , so by the inverse function theorem is a diffeomorphism from an open neighborhood U0 × V0 of (0, 0).
We define some sets that will help to show that ◦ is one-to-one. The
set exp(V0 ) is an open neighborhood of e in K with the subspace topology so
exp(V0 ) = H ∩ K for some open set H in G. So there exists an open set
U1 × V 1 ⊆ U 0 × V 0
such that
(U1 × V1 ) ∩ K ⊆ H ∩ K = expV0
If X ∈ U1 , Y ∈ V1 and (X , Y ) ∈ K then
expX expY = expY 0 for some Y 0 ∈ V0
But is a diffeomorphism on U1 × V1 so X = 0 and Y = Y 0 , thus
(U1 × V1 ) ∩ K = exp(V1 ).
Now we will show the injectivity. Let U2 ⊆ U1 be a neighborhood of 0 in
m such that
exp(−U2 )exp(U2 ) ⊆ (U1 × V1 )
Then ◦ exp|U2 is injective, since if X 0 , X 00 ∈ U2 and
(exp(X 0 )) = (exp(X 00 ))
then
exp(−X 0 )exp(X 00 ) ∈ (U1 × V1 ) ∩ K so X 0 = X 00 .
The surjectivity on a neighborhood follows since is surjective from U 2 × {0}
onto (U2 , 0).
By definition ◦ exp|m is continuous.
If N is an open subset of U2 then (exp(N )) = ◦ (N , V1 ) which is open
since is a diffeomorphism here and we have the quotient topology. So the
inverse is continuous. Thus ◦ exp : m → G/K is a local homeomorphism at
0 in the quotient topology of G/K .
Theorem 1.4. [10] Let G be a Lie group and K a closed subgroup of G, then
the quotient space G/K has a unique manifold structure such that
(1) the projection : G → G/K , g 7→ gK is smooth and
(2) has smooth local lifts such that for every gK ∈ G/K there is a neighborhood U and a map l : U → G such that ◦ l = id .
4
With this manifold structure a map
is smooth if and only if
f : G/K → N
f ◦ :G →N
is smooth. Furthermore the action
is smooth.
(g, g 0 K ) 7→ gg 0 K
G × G/K → G/K ,
P ROOF. [10] If one chooses the quotient topology on G/K then we get a
Hausdorff space by the following. Consider the map
: G × G → G,
(g1 , g2 ) = g1−1 g2
which is continuous, and since K is closed the inverse image −1 (K ) is closed.
Now if g1 K 6= g2 K then g1−1 g2 6∈ K so there are open sets W1 , W2 such that
(g1 , g2 ) ∈ W1 × W2 and W1 × W2 ∩ −1 (K ) = ∅. Now if gK ∈ Wi K then
there is a ki ∈ K such that gki ∈ Wi . So if gK ∈ W1 K ∩ W2 K then gk1 ∈ W1
and gk2 ∈ W2 . Therefore (gk1 , gk2 ) ∈ W1 × W2 is mapped to k1−1 k2 ∈ K
which contradicts that W1 × W2 ∩ −1 (K ) = ∅. So G/K is Hausdorff.
To get coordinates we need a linear complement m to k. If we use the same
sets as in Lemma 1.3, the maps U2 → G/K , X 7→ gexp(X )K , for g ∈ G are
homeomorphisms. Then we can choose local coordinates on G/K at gK as the
inverses of these maps. We identify a chart with g. Suppose two charts g, g 0
intersect then gexp(X0 )K = g 0 exp(X00 )K for X0 , X00 ∈ U2 . Then exp(X0 ) =
g −1 g 0 exp(X00 )k0 and exp(X0 ) ⊆ (U2 × {0}) ⊆ (U2 × V1 ). So if X 0 is close
to X00 then there must be an X ∈ U2 and Y ∈ V1 such that g −1 g 0 exp(X 0 )k0 =
0
(X , Y ) and X depends smoothly on X since is a diffeomorphism on U2 ×
V1 . Which finally gives g 0 exp(X 0 )K = gexp(X )K , which shows that points in
chart g are mapped smoothly to points in chart g 0 and vice versa. So G/K has
a manifold structure.
(1) The projection is smooth since
(gexp(X )exp(Y )) = gexp(X )K .
(2) The lifts l are given by l(gexp(X )K ) = gexp(X ), which are smooth.
Uniqueness: If (G/K )0 and (G/K )00 satisfy the above then ◦ l 0 : U 0 →
(G/K )00 is the identity, also ◦ l 00 : U 00 → (G/K )0 is the identity. Therefore
one can make the identity mapping (G/K )0 → (G/K )00 a diffeomorphism, so
they are identical as manifolds.
If f ◦ is smooth then f is locally given by f ◦ ◦ l and is smooth. The
reverse is trivial.
The map (g, g 0 K ) 7→ gg 0 K is smooth since it is ◦ (group op. on G) ◦ l
which is a composition of smooth maps.
5
The following result tell us important things about the kernel of d .
Corollary 1.5. [10] Let G be a Lie group, K be a closed subgroup in G, g be
the Lie algebra of G and k be the Lie algebra of K . If m is a linear complement to
k in g then
d e : g → TeK G/K
is a surjective vector space homomorphism with ker d
e
= k.
P ROOF. By the proof of Theorem 1.4 the map ◦exp |m : m → G/K is local diffeomorphism, so the differential d( ◦ exp|m ) = d |m is an isomorphism.
Therefore d e is surjective.
Since g = k ⊕ m and d |k = 0 we get that ker d = k.
We now give some examples that illustrates the use of the previous theory.
Example 1.6. Following Example 1.2 let G = SO(n+1), K = SO(n) then
the quotient SO(n + 1)/SO(n) has manifold structure such that the action
SO(n + 1) × SO(n + 1)/SO(n) → SO(n + 1)/SO(n)
(A1 , A2 K ) 7→ A1 A2 K
is smooth.
The Lie algebra g = so(n+1) of SO(n+1) consists of the skew symmetric
matrices. The Lie algebra of k can be identified with the matrices
0 0T
0 A
where A ∈ so(n) and 0 is the n × 1 matrix of zeros. The linear complement m
corresponds to the matrices
0 vT
−v 0
where v is a n × 1 matrix. By Lemma 1.5 the trivial map m → TeK G is an
isomorphism.
Example 1.7. For x, y ∈ Rn+1 define
f (x, y) = x1 y1 −
using matrix multiplication we write this as
T
f (x, y) = x Qy
n+1
X
x i yi
i=2
where Q =
1 0T
0 −I
where 0 is the n × 1 matrix of zeros and I is the n × n identity matrix.
6
Now define
O(1, n) = {A ∈ GLn+1 (R) : AT QA = Q}
We find the Lie algebra g = o(1, n) of O(1, n) by taking the derivative of a
curve at e, in the defining expression for O(1, n). So if X =Ȧ(0) where A(t) ∈
O(1, n) then
(1.1)
0=
d
(A(t)T QA(t)|t=0 = X T Q + QX
dt
if we write X as
X =
we get for Equation (1.1)
So
o(1, n) =
E F
G H
E T + E −G T + F
F T − G −H t − H
0 FT
F H
=0
: F ∈ M(n,1) , H ∈ o(n)
If A ∈ O(1, n) then det(AT QA) = det(Q) which implies det(A)2 = 1 so
det(A) = ±1. We turn our attention to the subgroup SO(1, n) with determinant 1. The hyperboloid
H1,n = {x ∈ Rn+1 : f (x, x) = 1}
+
−
has two connected components H1,n
with x1 > 0 and H1,n
with x1 < 0.
There are A ∈ SO(1, n) which interchange the two components of H1,n , so we
restrict our attention to the subgroup
+
+
Lor(1, n) = {A ∈ SO(1, n) : AH1,n
= H1,n
}
If we consider the vector e1 = (1, 0, . . . , 0) which is left unchanged by
matrices of the form
1 0T
where B ∈ SO(n)
0 B
and therefore correspond to the isotropy group of e1 . Therefore
Lor(1, n)/SO(n) has a manifold structure.
+
+
We will now show that Lor(1, n) acts transitively on H1,n
. Let u ∈ H1,n
and consider the vectors v ∈ Rn+1 such that
T
0 = u Qv = u1 v1 −
7
n+1
X
i=2
u i vi
This is an n-dimensional subspace V , since by choosing v i , i = 2 . . . n + 1
arbitrarily we get v1
u·v
+
, this is well defined since u ∈ H1,n
have u1 > 0
v1 =
u1
It also follows that the vector v satisfies f (v, v) < 0 since
|u · v|2
− |v|2
u21
|u · v|2
− |v|2
=
2
1 + |u|
|u · v|2
<
− |v|2 by Cauchy-Schwartz
|u|2
≤ |v|2 − |v|2 = 0
f (v, v) =
So −fV is a positive definite inner product on V and by the Gram-Schmidt
process we can construct a basis {w i }n+1
i=2 such that
f (wi , wj ) =
Then the matrix
ij
, f (u, wi ) = 0
u1 w11 . . . w1n
A=
u w1 . . . wn
is in Lor(1, n) and A(e1 ) = u. We decompose the Lie algebra of O(1, n) as
0 0
k=
: A ∈ o(n)
0 A
and
m=
0 AT
A 0
: A ∈ M(n,1)
Example 1.8. Let Gk (Rn ) denote the set of all k-dimensional subspaces in
Rn . This space is called the real Grassman manifold. The orthogonal group
O(n) acts transitively on Gk (Rn ), since if V is spanned by the k first elements
of the canonical basis e1 , . . . , en of Rn let e10 , . . . , en0 be another basis of Rn . The
matrix A ∈ O(n) with e10 , . . . ek0 as the k first columns has the effect
AV = W
where W is the space spanned by e10 , . . . ek0 . The isotropy group of V consists
of matrices of the form
B 0
where B ∈ O(k), C ∈ O(n − k)
0 C
so we have that
O(n)/(O(k) × O(n − k))
8
is a manifold. Since the Lie algebra of O(n) consists of the skew-symmetric
matrices we get
A 0
k=
: A ∈ o(k), D ∈ o(n − k)
0 D
and
m=
0
B
T
−B 0
: B ∈ M(k,n−k)
Example 1.9. Let K = R, C or H, then if one considers Kn+1 as a left
K-vector space with the action of
K∗ = {a ∈ K : a 6= 0}
on Kn+1
= Kn+1 − 0 given by
0
−1
: K∗ × Kn+1
→ Kn+1
0
0 , z (x) = xz
Then set of orbits is denoted by KPn and is called the n-dimensional Kprojective space. The orbit is denoted as
[x] = {xz −1 : z ∈ K∗ }
Let us now restrict ourselves to CPn . Let A ∈ U(n + 1) then
A [x]
= [Ax]
is well defined since A(xz −1 ) = (Ax)z −1 . Now [e1 ] is stabilized by
i 0
e
: ∈ R and B ∈ U(n)
0 B
So U(n + 1)/(U(1) × U(n)) has a manifold structure. Similarly to the case of
the sphere the Lie algebra decomposes as
ir 0
k=
: r ∈ R, B ∈ u(n)
0 B
and
m=
0 A
−A∗ 0
9
:
A ∈ M(1,n) .
CHAPTER 2
Symmetric Spaces
In this section we define the notion of a symmetric space and study some
of their properties.
1. Definitions
Here it is assumed that the reader is familiar with the concept of a geodesically complete Riemannian manifold and the Hopf-Rinow theorem. If not,
the reader is referred to Appendix A.
Definition 2.1. A Riemannian manifold (M , g) is said to be a symmetric
space if for every point p ∈ M there exists an isometry p of (M , g) such that
(1) p (p) = p, and
(2) d p = −idTp M .
Such an isometry is called an involution at p ∈ M .
Lemma 2.2. [6] Let (M , g) be a symmetric space and let p : (M , g) →
(M , g) be an involution at p ∈ M . Then p reverses the geodesics through p, i.e.
p ( (t)) = (−t) for all geodesics ∈ M such that (0) = p.
P ROOF. [6] A geodesic : I → M is uniquely determined by the initial
data (0) and ˙ (0). Both the geodesics t 7→ p ( (t)) and t 7→ (−t) take the
value (0) and have the tangent − ˙ (0) for t = 0.
The following lemma entails the core features of a symmetric space.
Lemma 2.3. [6] Let (M , g) be a symmetric space. If : I → M is a geodesic
with (0) = p and ( ) = q then q ◦ p ( (t)) = (t + 2 ). For v ∈ T (t) M ,
d q (d p (v)) ∈ T (t+2
) M is the vector at (t + 2 ) obtained by parallel transport
of v along .
P ROOF. [6] Let ˜(t) = (t + ) then ˜ is a geodesic with ˜ (0) = q. So by
Lemma 2.2 it follows that
q ( p (
(t))) = q ( (−t))
= q (˜ (−t − ))
= ˜(t + )
= (t + 2 ).
11
If v ∈ Tp M and V is a parallel vector field along with V (p) = v, then d p (V )
is parallel, since p is an isometry. Also
d q ◦ d p (V ( (t))) = V ( (t + 2 ))
by the above and since d applied twice cancels direction reversals.
As a display of the power of Lemma 2.3 we have the following
Corollary 2.4. [6] Every symmetric space (M , g) is geodesically complete and
thus any two points p, q ∈ M in the same path component of M can be connected
by a geodesic.
P ROOF. [6] By repeatedly composing as in Lemma 2.3 we can get to
l
l
(2
) until 2 greater than any real number. This shows that (M , g) is
geodesically complete, so by the Hopf-Rinow theorem any two points can be
connected by a geodesic.
Definition 2.5. Let (M , g) be a symmetric space and let : R → M be a
geodesic with p = (0) and v = ˙ (0). Then for t ∈ R the isometries
tv
given by
: (M , g) → (M , g)
tv
=
(t/2)
are called transvections.
◦
(0)
It is easily seen, by using Lemma 2.3, that for the particular geodesic
R → M in Definition 2.5 the transvection tv satisfies
:
tv ( (s)) = (s + t)
Proving things about isometries are facilitated by
Lemma 2.6. [10] Let (M , g) be a connected Riemannian manifold and p ∈
M . If , 0 : (M , g) → (M , g) are isometries such that (p) = 0 (p) and
d p = d 0p then = 0 .
P ROOF. [10] Let B (p) be a normal ball around p ∈ M . Then if v (t) =
Expp (tv) is a geodesic then ( v (t)) = 0 ( v (t)) are the same geodesics since
geodesics are given by initial data. So = 0 on an open set but, this set is
also closed, hence it is M .
Here we have some nice properties of transvections
Proposition 2.7. [10] Let (M , g) be a symmetric space, p ∈ M , v ∈ Tp M
and define : R → M by (t) = Expp (tv) as the unique geodesic with (0) =
p, ˙ (0) = v. Let bv be the transvections corresponding to the involution p , then
(1) av = (t+ 2a ) ◦ (t) for all t ∈ R
12
(2)
(3)
(4)
(5)
depends only on av
av ( (t)) = (t + a)
av ◦ bv = (a+b)v
p ◦ av ◦ p = −av
av
P ROOF. [10] (1) Note that is the composition of two isometries so it is
actually an isometry.
For (s) ∈ M , if ˜ (r) = (r + t) then (t) = ˜ (0) and (s) = ˜(s − t) so
(t+ a )
2
◦
(t) ( (s)) =
=
◦ ˜(0) (˜ (s − t)) = ˜(s − t + a)
(s + a)
˜( a )
2
Now we also have ( 2a ) ◦ (0) ( (s)) = (s + a). So the isometries agree at (s).
If w ∈ T (s) M and V is the parallel vector field along such that V ( (s)) = w.
Then d( (t+ 2a ) ◦ (t) )w is the parallel transport of w along from (s) to
(s + a). The same is true for d av w = d( ( a ) ◦ (0) )w.
2
So since their initial data agree we have av = (t+ 2a ) ◦ (t) by Lemma 2.6.
(2) av = ( 2a ) ◦ (0) = Expp ( av2 ) ◦ p0 , so it depends only on the value of av
0
(3) This follows directly from the proof of 1
(4) By (1) av ◦ bv = ( a + b ) ◦ ( b ) ◦ (0) = ( a + b ) ◦ (0) = (a+b)v
2 2
2
2
2
(5) (0) ◦ ( 2a ) ◦ (0) ◦ (0) = (0) ◦ ( 2a ) = −av
Here are some concrete examples of involutions and transvections
Example 2.8. At p0 ∈ Rn define the involution
p0
: x 7→ −(x − p0 ) + p0
In Rn geodesics are straight lines, so consider the geodesic
(t) : R → Rn , t 7→ p0 + tv
where v ∈ Tp0 Rn = Rn
Then we get the transvection
tv (x)
= (t/2) ◦ (0) (x)
= (t/2) (−(x − (0)) + (0))
= −(−(x − (0)) + (0) − (t/2)) + (t/2)
= −(−(x − p0 ) + p0 − (p0 + tv)) + (p0 + tv)
= x + tv.
Now we can verify the translational properties of the transvection
tv (s) = (p0 + sv) + tv = p0 + (s + t)v = (s + t)
13
Example 2.9. If we have a submanifold M in (Rm , g) for some m ∈ Z+
+
and M = {x ∈ Rm : g(x, x) = r} for example S n or H1,n
, then we have the
involution at p0
:
M →M
g(p0 , x)
p0 − x
x →
7
2
r
then
g(p0 , p0 )
p0 − p 0 = p 0
r
since Tp0 M ⊥p0 we get for v ∈ Tp0 M
p0 (p0 )
=2
g(p0 , v)
p0 − v = −v
r
so we have an involution on the submanifold M .
d
p0 (v)
=2
2. The Isometry Group
Definition 2.10. Let I (M ) be the isometry group of the symmetric space
(M , g), and let G be the connected component of I (M ) containing the neutral
element e ∈ G, i.e. G is the identity component of I (M ). Furthermore let K p0
be the isotropy group of G at p0 ∈ M .
It should be noted that by continuity, all the transvections belong to the
group G.
Now we can prove the a very important thing concerning the isometry
group.
Theorem 2.11. [6] Let (M , g) be a symmetric space and G the identity component of I (M ), then G acts transitively on M
P ROOF. [6] By Corollary 2.4 any p, q ∈ M can be connected by a geodesic
. If (0) = p, ˙ (0) = v and (s) = q and tv is the family of translations along
then q = sv (p). So the action is transitive.
Lemma 2.12. [10] Let N be a smooth manifold and let f : N → I (M ) be
a map such that f (n)(p) depends smoothly on (n, p) for n ∈ N and p ∈ M is in a
neighborhood of p0 ∈ M . Then the dependence is smooth for all p ∈ M .
P ROOF. [10] If we prove the claim for p ∈ M in a normal ball around
q ∈ M , then we can cover M with intersecting balls from p0 to an arbitrary
point in M . So if p ∈ B (q) then there exists a v ∈ Tq M such that p = Expq v
and v depends smoothly on p. Then
f (n)(p) = f (n)(Expq v) = Expf (n)(q) (df (n)(q))v
where the right side depends smoothly on (n, p).
14
Now we have a very important lemma for showing that charts and actions
on our manifold are smooth.
Lemma 2.13. [10] Let (M , g) be a symmetric space with involutions p ,
transvections v and let Kp0 be the isotropy group of the isometries I (M ) acting at
a point p0 ∈ M , then
(1) p (q) depends smoothly on (p, q) ∈ M × M .
(2) v (q) depends smoothly on (v, q) ∈ Tp M × M .
(3) p (q), depends smoothly on (p, q) ∈ M × M , where p is v such that
v (p0 ) = p.
(4) k(q) depends smoothly on (k, q) ∈ Kp0 × M .
P ROOF. [10] (1) By Lemma 2.12 we only need to show the claim for
a point q in a neighborhood of p. Then q = Exp p (v) where v depends
smoothly on (p, q). But Expp (tv) goes through p so p (q) = p (Expp (tv))|t=1 =
Expp (−v) which again depends smoothly on (p, q).
(2) The claim follows from the definition of v and 1)
(3) Since v in the geodesic Expp0 (tv) connecting p0 to p depends smoothly
on p close to p0 , then v depends smoothly on p close to p0 . Then by Lemma
2.12 this is true for all p ∈ M .
(4) By Lemma 2.12 we show it for q in a neighborhood of p0 . Then
q = Expp0 (v) so k(q) = k(Expp0 (v)) = Expk(p0 ) (dk)v = Expp0 (dk)v which
depends smoothly on (k, q).
Note that Kp0 immediately can be defined as a Lie group using Lemma
2.6 and the fact that for all f ∈ Kp0 , f (p0 ) = p0 , i.e. an isometry on a
connected manifold is identified by its value at one point and its differential at
the same point. So since df (p0 ) is an element of the Lie group of isometries
Tp0 M → Tp0 M which is O(dim M ). Since K0 is closed we can identify K0
with a Lie-subgroup of O(dim M ).
Theorem 2.14. [10] Let (M , g) be a symmetric space. Then the isometry
group I (M ) has the structure of a Lie group such that
(1) The map I (M ) × M → M , with (g, p) 7→ g(p) is smooth.
(2) If N is a manifold then f : N → I (M ) is smooth if and only if
is smooth.
N × M → M,
(n, q) 7→ f (n)(q)
P ROOF. [10] We will define charts in a neighborhood around g 0 ∈ I (M ).
Let U be a neighborhood of p0 ∈ M . We’ll define a map that will help us with
the charts. Let U × Kp0 → I (M ), (p, k) 7→ g0 p k, where p is defined as in
Lemma 2.13. This map is injective since we can retrieve both p, v by
g0−1 g0 p kp0 =
15
p p0
=p
−1 −1 p g0 g0 p k
=k
So we can talk about the inverse image of this map around g0 .
If we choose local coordinates for M by p 7→ (x1 , . . . xm ) at p0 and local
coordinates k 7→ (y1 , . . . yl ) at k0 in Kp0 , then we get local coordinates around
g0 in I (M )
g0 p k 7→ (x1 (p), . . . , xm (p), y1 (k), . . . , yl (k)).
These coordinates are smooth, since if the image of two such charts intersect
we get g1 p1 k1 = g2 p2 k2 so
p1 = g2−1 g2 dk1 (p0 ) =
p2 k2 (p0 )
−1
(d −1
p1 (p1 ))d(g1 g2 )(p2 )(d p2 (p0 ))(dk2 (p0 ))
these depend smoothly on each other by Lemma 2.13.
The topology is Hausdorff since if , 0 are two different isometries then
we either have (p0 ) 6= 0 (p0 ) or (d (p0 ) 6= (d (p0 )). If the first one holds we
can separate the isometries since M is Hausdorff. If the second hold we must
have that k1 6= k2 so we can separate in O(dim M ).
Now the group operation is smooth since if f , g ∈ I (M ) then f −1 g =
(g1 p1 k1 )−1 (g2 p2 k2 ) = (g3 p3 k3 ) in local coordinates. So
p3 = p3 k3 (p0 )
= g3−1 k1−1 −1 −1 p1 g1 g2 p2 k2 (p0 )
and
dk3 = d
−1
−1
−1 −1
−1
p3 dg3 dk1 d p1 dg1 dg2 d p2 dk2
which is smooth by Lemma 2.13. Next
(1) Since g is an isometry then (g, p) 7→ g(p) is smooth
(2) Suppose f (n) is smooth. Let n0 ∈ N then f (n) = g0 p k where (p, k)
depends smoothly on n. So f (n)(q) = g0 p k(q) depends smoothly on (n, q).
Conversely if the assignment (n, q) 7→ f (n)(q) is smooth then
f (n) = g0 p k
where p = g0−1 f (n)(p0 ) and dk = d
on n.
−1 −1
p dg0 df
(n). So (p, k) depends smoothly
Example 2.15. If we consider the isometry groups in Examples 1.6-1.9 of
Section 2 then we see that
SO(n + 1) acts smoothly on S n ,
+
Lor(1, n) acts smoothly on H1,n
,
n
O(n) acts smoothly on Gk (R ) and
U(n + 1) acts smoothly on CPn
16
3. G/K is Diffeomorphic to M
Here we show that the quotient of the isometry group and the isotropy
group can be identified with the symmetric space.
Theorem 2.16. [10] Let (M , g) be a symmetric space and let G = I (M ) act
transitively on M . Let K be the isotropy group of G at p0 ∈ M . Then the map
: G/K → M with gK 7→ g(p0 )
is a diffeomorphism such that
◦ (g) = g ◦ where (g 0 )gK = g 0 (gK ) = g 0 gK as in Theorem 1.4.
If the action is not effective one can consider the groups G/N and K /N
instead, where N is the kernel of the action. So if M is a symmetric space then
I0 (M )/K0 is diffeomorphic to M .
P ROOF. [10] Clearly
k ∈ K . Also
is well defined and bijective since
◦ (g 0 )gK = (g 0 gK ) = g 0 g (p0 )
=
g 0 g (p0 )
=
g 0 (
k (p0 )
= p0 for
(gK ))
is smooth since ◦ (g) = g and
are smooth.
We will show that −1 is smooth. This is done by showing that dim (G/K ) ≥
dim M . Then we show d is injective, so d is an isomorphism and by the
inverse function theorem locally −1 is smooth. But −1 exists globally so by
some form of gluing lemma −1 is smooth.
Now M = ∪∞
j=1 Uj , where Uj are the images of the charts on G/K . Since
M is locally compact, at least one Ūj contains an open subset of M by the
Baire category theorem. This open subset on M is in bijective correspondence
with an open set V in G/K since G/K acts smoothly. So using the charts we
would get an onto map Rn → V → Uj → Rm , so we cannot have n < m i.e.
dim (G/K ) ≥ dim M .
Thus we only have to show the injectivity which will imply that
dim (G/K ) ≤ dim M . Thus dim (G/K ) = dim M . But ◦ (g) = g ◦ so
−1
). If we differentiate at g 0 we get
= g ◦ ◦ (g
(d )g 0 K = (d
g
◦ d )g −1 g 0 K ◦ (d (g −1 ))g 0 K
So if d is injective at aK then it is injective at g 0 K by choosing g such that
g −1 g 0 K = aK in = g ◦ ◦ (g −1 . Therefore it is enough to show the
injectivity at eK . Let v ∈ TeK (G/K ) with d (v) = 0. By Corollary 1.5 d
is surjective and has kernel k. So there is a X ∈ g such that d (X ) = v. The
curve (t) = exp(tX ) (p0 ) satisfies
d (t)
d
=
◦ (exp(tX ))
dt
dt
17
=
=
=
=
=
=
d ◦ d exp(tX ) (X )
d ◦ d ◦ dLexp(tX ) (X (e))
d ◦ d (exp(tX )) ◦ d (X (e))
d exp(tX ) ◦ d ◦ d (X (e))
d exp(tX ) ◦ d (v)
0
so ∈ K and X ∈ k, i.e. v = 0.
Example 2.17. We have seen in Examples 1.6-1.9 of section 2 groups G
that act transitively on manifolds and their isotropy groups K at certain points.
Therefore we have the following diffeomorphisms
SO(n + 1)/SO(n) ∼
= Sn
+
Lor(1, n)/SO(n) ∼
= H1,n
O(n)/O(k) × O(n − k) ∼
= Gk (Rn )
U(n + 1)/U(1) × U(n) ∼
= CPn
18
CHAPTER 3
Symmetric Pairs
In this chapter we introduce the concept of a symmetric pair and show
how they lead to symmetric spaces.
Definition 3.1. A pair (G, K ) is said to be a Riemannian symmetric pair
if G is a Lie group, K a closed subgroup of G, and sp0 an involutive automorphism on G such that
(1) (Gsp0 )0 ⊆ K ⊆ Gsp0
(2) Ad (K ) is a compact subset of GL(g).
Here Gsp0 are the elements of G that are left invariant by sp0 , i.e
Gsp0 = {g ∈ G : sp0 g = g}.
The involutive automorphism sp0 is often uniquely defined, so it is common to only write (G, K ) for a symmetric pair.
1. From a Symmetric Space to a Symmetric Pair
As before let G, Kp0 denote the identity component of the isometry and
isotropy groups on M . Then by Theorem 2.16 there is a bijective correspondence
G/K ↔ M , gK 7→ g(p0 )
We define an involution corresponding to
s p0 : G → G
sp0 (g) =
p0
p0
◦g ◦
on G by
p0
=
p0
◦g ◦
−1
p0
Then sp20 = id and sp0 (g)p = p0 ◦ g ◦ −1
p0 (p), so by Lemma 2.13 it depends
smoothly on (g, p). Thus sp0 is an involutive Lie group automorphism of G.
Note also that
(dsp0 )e : g → g
is a involutive Lie algebra automorphism.
Let Gsp0 = {g ∈ G : sp0 g = g}, and let gsp0 be its Lie algebra. Then Gsp0 is
a Lie group since it is a closed subgroup.
We have the following theorem
19
Theorem 3.2. [10] Let (M , g) be a symmetric space with a fixed point p0 , G
be the identity component of the isometry group and let K be the isotropy group of
G at p0 . Then the map G/K → M with K 7→ g(p0 ) is a bijection. The group G
has an involutive automorphism sp0 given by
sp0 (g) =
p0
with stabilizer Gsp0 such that
◦g ◦
p0
(Gsp0 )0 ⊆ K ⊆ Gsp0
This says that (G, K ) is a symmetric pair because Ad (K ) is compact, since
K is closed and bounded and Ad is a homeomorphism.
P ROOF. [10] It only remains to prove the last statement.
If X ∈ gp0 then exp(tX ) ∈ Gp0 so sp0 (exp(tx)) = exp(tX ) therefore
dsp0 = X . Also if dsp0 X = X then exp(tX ) = exp(tdsp0 X ) = sp0 exp(tX ),
the implications work in the opposite direction as well, so
Let k ∈ K then
gp0 = {X ∈ g : dsp0 X = X }.
sp0 (k)p0 = p0 ◦ k ◦ p0 (p0 ) = p0 = k(p0 ) and
dsp0 (k) = d p0 ◦ dkp0 ◦ d p0
= (−idTp0 M ) ◦ dkp0 ◦ (−idTp0 M )
= dkp0 .
This means that Lemma 2.6 implies that sp0 (k) = k so K ⊆ Gp0 . Further
if X ∈ gp0 then p0 (exp(tX ))p0 = p0 (exp(tX )) p0 (p0 ) = sp0 (exp(tX ))p0 =
exp(tX )p0 , so exp(tX )p0 is a fixed point of p0 . But if t is small then exp(tX )p0 is
close to p0 and this is the only fixed point so exp(tX )p0 = p0 . Thus exp(tX ) ∈
K0 = K and X ∈ k. So gp0 ⊆ k ⇒ (Gp0 )0 ⊆ K .
2. The Tangent Space of G/K
Here we will split up the Lie algebra g so that the complement of k can be
identified with the tangent space of M . We know that
and define
k = {X ∈ g : dsp0 (X ) = X }
p = {X ∈ g : dsp0 (X ) = −X }
Then since dsp0 is an automorphism k ∩ p = {0}. We also have that g = k + p
for all X ∈ g since
1
1
X = (X + dsp0 (X )) + (X − dsp0 (X ))
2
2
20
where the first term is in k and the second is in p. Therefore
g=k⊕p
Since dsp0 is a Lie algebra automorphism, i.e.
dsp0 [X , Y ] = [dsp0 X , dsp0 Y ]
we have
[k, k] ⊆ k
[p, p] ⊆ k
[k, p] ⊆ p
We now reveal the connection between the isometry group and the symmetric space.
Theorem 3.3. [10] Let (M , g) be a symmetric space, G be the identity component of I (M ), K the isotropy group of G at p0 ∈ M , g the Lie algebra of
G, k the Lie algebra of K and p a linear complement of k in g. As usual let
: G → M , g 7→ g (p0 ) then
If X ∈ p then exp(X ) =
d |k (p0 ) = 0
d |p (p0 ) ∼
= Tp0 M
d
X (p0 )
and exp(X ) (p0 )
= Expp0 (d X (p0 )).
P ROOF. [10] d |k (p0 ) = 0 since (p0 ) is constant on K . Now let v ∈
Tp0 M then the map t 7→ tv is a smooth group homomorphism R → G.
There is a unique X ∈ g such that tv = exp(tX ). For this X we have
exp(tdsp0 X ) = sp0 (exp(tX )) =
p0
◦
This means that dsp0 (x)) = −X so X ∈ p. Also
tv
◦
p0
=
−tv .
d
exp(tX )|t=0 (p0 )
dt
d
=
exp(tX )(p0 )|t=0
dt
d
=
tv (p0 )|t=0
dt
d
=
Expp0 (tv)|t=0 = v
dt
This shows that d (p0 ) : p → Tp0 M is a surjective map.
Since the coordinates of G were the coordinates for K and Tp0 ,
d X (p0 ) =
dim G = dim K + dim M
21
Also since g = k + p, we get dim M =dim p. Therefore d (p0 ) : p → Tp0 M
must be injective as well.
Now the last formulas of the claim follow immediately
expX =
v
=
d
X (p0 )
and
expX (p0 )
= exp(X )p0 = d
X (p0 )
p0 = Expp0 (d X (p0 ))
So the map : G/K → M in Theorem 2.16 defined by gK 7→ g(p) is a
submersion. We apply this theorem in some of the symmetric spaces we know.
Example 3.4. By Theorem 3.3 for G = SO(3), K = SO(2) acting on S 2
at

we have for

1
p0 =  0  ∈ S 2
0


0 1 0
X =  −1 0 0  ∈ p
0 0 0

  

cos t
cos t sin t 0
1
 − sin t cos t 0   0  =  − sin t 
exptX p0 =
0
0
1
0
0
is the geodesic in S 2 starting at p0 in direction

  

0 1 0
1
0
d X p0 =  −1 0 0   0  =  −1 
0 0 0
0
0
To get a geodesic going in an arbitrary direction we apply any element of K to
the above geodesic which gives



1
0
0
cos t
 0 cos sin   − sin t 
0 − sin cos 0
Example 3.5. By Theorem 3.3 for G = Lor(1, 2), K = SO(2) acting on
+
H1,2
at
 
1
+
p0 =  0  ∈ H1,2
0
22
we have for


0 1 0
X = 1 0 0 ∈p
0 0 0

  

cosh t sinh t 0
1
cosh t
 sinh t cosh t 0   0  =  sinh t 
exptX p0 =
0
0
1
0
0
+
is the geodesic in H1,2
starting at p0 in direction

   
0
0 1 0
1
d X p0 =  1 0 0   0  =  1 
0
0 0 0
0
To get a geodesic going in an arbitrary direction we apply any element of K to
the above geodesic which gives



1
0
0
cosh t
 0 cos sin   sinh t 
0 − sin cos 0
3. From a Lie Group to a Symmetric Pair
Now we are ready for the theorem that tell us if a Lie group and a closed
subgroup can be made into a symmetric pair.
Lemma 3.6. Let G be a Lie group and K a closed subgroup of G. Let k ∈ K .
If we denote by In g(g 0 ) = gg 0 g −1 . Then
and by differentiating at e
(k) ◦ =
d (k)eK ◦ d
e
◦ In k
= d e ◦ Ad k.
P ROOF. [10] Let g ∈ G and k ∈ K then (k)(gK ) = kgK = kgk −1 K =
In (k)gK = (In (k)g), i.e. (k) ◦ = ◦ In k.
Theorem 3.7. [10] Let G be a Lie group and let K be a closed subgroup, with
identity components G0 , K0 . Form the quotient manifold G/K . Let denote the
action (g 0 )gK = g 0 gK . Let J ∈ Z (K ) ⊆ K be such that (J )eK = eK and
suppose
(1) d (J )eK = −idG/K
Then for every compact subset K 0 in K where
K0 ⊆ K 0 ⊆ K ∩ G 0
(G0 , K 0 ) is a Riemannian symmetric pair with
seK (g) = JgJ −1 = In(J )g
23
The assumption (1) is equivalent to
(2) X + Ad J (X ) ∈ k
for all X ∈ g.
condition (2) holds if k has a linear complement m in g, i.e. g = k ⊕ m, on which
Ad J = −id
P ROOF. [10] By Lemma 3.6 (k) ◦ =
d (k)eK ◦ d
e
◦ In k and
= d e ◦ Ad k
If we put k = J we get −d = d ◦ Ad J . So d (X + Ad (J )X ) = 0 and thus
X + Ad (J )X ∈ k for all X ∈ g. Thus (1) implies (2). By working backwards
we get −d e = d ◦ Ad J = d (J )eK ◦ d e so d (J )eK = −idG/K so also (2)
implies (1).
Now we prove that we have a symmetric pair. Since J ∈ Z (K ),
In J|K = idK and Ad J|k = idk .
If X ∈ m where m is the linear complement of k in g. Then
X + Ad (X ) = Y
where Y ∈ k but Y depends linearly on X so we write Y = T (X ) where
T : m → k is a linear map. Then
(Ad J )2 (X ) = Ad J (−X + T (X )) = −(−X + T (X )) + T (X ) = X .
Now since (Ad J|k )2 = idk we have Ad (J )2 = idg . So if X ∈ g then
In (J )2 (exp(x)) = exp(dIn (J )2 X ) = exp(Ad (J )2 X ) = exp(X ).
So In is an involutive automorphism on G0 .
Now we’ll show ((G0 )In J )0 ⊆ K 0 ⊆ (G0 )In J . First ((G0 )In J )0 = (GIn J )0
since (GIn J )0 = (G0 )In J . Also (G0 )In J = G0 ∩ GIn J . So
K 0 ⊆ (G0 )In J
since K 0 ⊆ G0 , K 0 ⊆ GIn J because J ∈ Z (K ) and K 0 ⊆ K . Secondly if
gIn J ⊆ k0 = k then (GIn J )0 ⊆ K 0 . So let X ∈ gIn J then
X = dIn (J )X = Ad J (X )
and
d (X ) = d (Ad (J )X ) = d (J )d (X ) = −d (X ).
Thus d(X ) = 0 and so X ∈ k.
Finally Ad K 0 is compact since Ad is a homeomorphism.
24
Example 3.8. Let G = O(n + 1) and K = O(n) in Theorem 3.7 with


1


−1

J =
.


..
−1
with J ∈ Z (K ). If we form G/K = O(n+1)/O(n) and since for X ∈ o(n+1)
A
B
A
B
A −B
X =
, Ad J
=
−B T D
−B T D
BT D
we have
A 0
∈k
X + Ad J (X ) = 2
0T D
so J satisfies 2) in Theorem 3.7 so we have the symmetric pair
(SO(n + 1), SO(n)) with s = In J
Example 3.9. Let G = O(n) and K = O(k) × O(n − k) in Theorem 3.7
with
Ik
0
J=
0T −In−k
with J ∈ Z (K ). If we form G/K = O(n)/O(k) × O(n − k) and since for
X ∈ o(n + 1)
A
B
A
B
A −B
X =
, Ad J
=
−B T D
−B T D
BT D
we have
A 0
X + Ad J (X ) = 2
∈k
0T D
so J satisfies 2) in Theorem 3.7 so we have the symmetric pair
SO(n), S(O(k) × O(n − k)) with s = In J
4. From a Symmetric Pair to a Symmetric Space
We will soon prove a theorem which shows that symmetric pairs induce
symmetric spaces. With this it is possible to show that several familiar group
quotients produce symmetric spaces.
First we need to produce some nice metrics for the intended symmetric
space.
Lemma 3.10. [10] Let G be a Lie group and K be a closed subgroup of G. If
Ad K is compact then there exists a G-invariant inner product on p ∼
= TeK G/K ,
such that the action of G on G/K is an isometry. Here p is the linear complement
of k in g, i.e. g = k ⊕ p.
25
P ROOF. [10] If we use Lemma C.2 with H = Ad K , V = p and =
id , we get an Ad K -invariant inner product {·, ·} on p. We transfer this to
TeK G/K by d (eK ) by Corollary 1.5 and get an Ad K -invariant inner product
h·, ·ieK on TeK G/K . Denote the action of G on G/K by
(g 0 )gK = g 0 (gK ) = g 0 gK .
Define
(·, ·)gK = hd (g −1 ) ·, d (g −1 ) ·ieK
This is well defined since if aK = bK then a = bk so we get
(·, ·)aK =
=
=
=
=
=
hd (a−1 ) ·, d (a−1 ) ·ieK
hd (k−1 b−1 ) ·, d (k−1 b−1 ) ·ieK
hd (k−1 )d (b−1 ) ·, d (k−1 )d (b−1 ) ·ieK
hAd (k−1 )d (b−1 ) ·, Ad (k−1 )d (b−1 ) ·ieK by Lemma 3.6
hd (b−1 ) ·, d (b−1 ) ·ieK since h·, ·i is Ad (K )-invariant
(·, ·)bK
Note that the submersion of Theorem 3.3 becomes a isometric submersion.
Finally we have the long awaited theorem.
Theorem 3.11. [10] Let (G, K ) be a symmetric pair with involutive automorphism s. Denote the action of G on G/K by (g 0 )gK = g 0 gK . Then there is a
G-invariant metric g on M = G/K which makes (M , g) a symmetric space with
involution eK such that
eK
◦ = ◦s
eK ( i.e.
(g)eK ) = s(g)eK
For X ∈ p ≡ {x ∈ g : ds(X ) = −X } we have
(exp(X )) = d (X )eK
and (exp(X ))eK = ExpeK (d (X ))
P ROOF. For a symmetric pair Ad K is compact so by Lemma 3.10 we can
pick a G-invariant inner product on the tangent space of G/K , as metric for
G/K . Define the involution by
eK ( (g)eK ) = (s(g))eK
this is well defined since
eK ( (gk)eK ) = (s(g)) (s(k))eK = (s(g))eK =
It is smooth since
Further
◦ = ◦ s which is smooth.
2
eK ( (g)eK ) =
26
eK ( (s(g))eK )
eK ( (g)eK ).
=
=
=
(s(s(g)))eK
(s2 (g))eK
(e)eK = eK
so 2eK = e so it is involutive. Therefore −1
eK is also smooth, so eK is a
diffeomorphism. To show that it is an isometry we note that by differentiating
(d
so on p we have d
eK
|p
◦ (d ) = (d )e ◦ (ds)e
◦ d = −d . But at p0 = eK
so since d
eK )eK
: g → geK
is an isomorphism so d (d
(g)eK = geK
|p
is also that. Therefore
eK )eK
= −id .
So since dpi|p is an isomorphism by Corollary 1.5 d (s(g))
eK ( (g 0 )eK ) =
=
=
=
|p
is also that. Moreover
(s(g)) (s(g 0 ))eK
(s(gg 0 ))eK
0
ek ( (gg )eK )
0
eK ( (g) (g )eK )
So
(3.1)
(s(g)) ◦
Let u, v ∈ TgK M = T
hd
= hd
(g)eK M
eK
=
eK (g)
then
eK (u), d eK (v)i eK ( (g))eK
eK (u), d eK (v)i (s(g))eK
hd (s(g)−1 )d eK (u), d (s(g)−1 )d eK (v)ieK by G-invariance
hd eK ◦ d (g −1 )u, d eK ◦ d (g −1 )vieK by eq (3.1)
h−d (g −1 )u, −d (g −1 )vieK since d ek = −id
So
=
=
=
= h−u, −vigK
= hu, vigK
ek
by G-invariance
is an isometry and
(d
eK )eK
= id
eK (eK ) = seK ( (e)eK )
= (s(e))eK
= eK .
So we have shown that (M , g, eK ) is a symmetric space.
The last claims follow from Theorem 3.3.
27
Example 3.12. By applying Theorem 3.11 to Examples 3.8-3.9 we get the
following symmetric spaces from the corresponding symmetric pairs
SO(n + 1), SO(n) → SO(n + 1)/SO(n) = S n
SO(n), S(O(k) × O(n − k)) → SO(n)/S(O(k) × O(n − k))
= Gk (Rn ).
In a similar fashion one can get the symmetric spaces
+
Lor(1, n)/SO(n) = H1,n
SU(n + 1)/SU(1) × SU(n) = CPn
28
CHAPTER 4
Curvature of a Symmetric Space
The aim of this chapter is to show how to transfer the curvature calculations from the symmetric space to the Lie group which is “behind” the symmetric space. So we need something like an inner product on the Lie algebra.
1. The Killing Form
The Killing form will provide us with a metric on the Lie algebra g of the
isometry group G on the symmetric space.
Definition 4.1. Let g be a Lie algebra and let be an involutive automorphism on g with fixed point set k, which is a Lie subalgebra or g. If Int k(see
Appendix B) with Lie algebra ad (k) is compact, then (g, ) is called an orthogonal symmetric algebra. The orthogonal symmetric algebra is called effective if
z(g) ∩ k = {0}.
Example 4.2. Let
g = R, = −idR , k = {0}
then we have an orthogonal symmetric algebra which corresponds to
and also
G = R, K = {0}, s = −idR
G = T = R/Z, K = {0, 1/2}, s = −idT
Definition 4.3. Let g be a Lie algebra, then the Killing form B of g over a
field F is the bilinear form
B : g × g → F, (X , Y ) 7→ tr(ad X ◦ ad Y ).
The Lie group G and its Lie algebra g are called semisimple if B is nondegenerate, i.e. if B(X , Y ) = 0 for all Y ∈ g then X = 0.
Example 4.4. To calculate1 the trace of linear maps from a vector space
to itself we calculate the trace of the corresponding matrix representations.
Then we will have an expression that can be used for an arbitrary vector. It
1To save space we use the convention to sum over repeated indices without explicit summation expressions, i.e.
Pn
i=1 xi yi .
x i yi =
29
becomes especially easy if we have an inner product h·, ·i, so that we have an
orthonormal basis {ei }i∈I where I is some finite set. Then the trace is given by
trace = hei , (ei )i
since has the matrix representation ij = hei , (ej )i. We do an explicit
calculation for so(n) and display some others afterwards.
For so(n) we chose the basis to be √12 (e ij − e ji ) for 1 ≤ i < j ≤ n, where
e ij is the matrix with a 1 at position (i, j) and zeros elsewhere.
trace (ad X ◦ ad Y ) = hfk , [X , [Y , fk ]]i by Lemma B.4
where fk are the basis elements. This gives
1 ij
he − e ji , [X , [Y , e ij − e ji ]]i
2
1 ij
1
=
(4.1)
he , [X , [Y , e ij ]]i − he ij , [X , [Y , e ji ]]i
2
2
1 ji
1
− he , [X , [Y , e ij ]]i + he ji , [X , [Y , e ji ]]i
2
2
we only need to calculate the first two on the right hand side of Equation (4.1)
since the last two are the first with ij switched. The inner product we use is
trace (ad X ◦ ad Y ) =
hA, Bi = trace (AT B)
and the basis elements are orthogonal for this choice. The first expression in
Equation (4.1) is
1 n ij T ij
ij
ij
(e ) X Y e −(e ij )T X e Y − (e ij )T Y e X 2
o
ij
+(e ij )T e Y X ji
Now (e ij )T = e so we get
1 n ji ij
ji ij
ji ij
=
e X Y e −e X e Y − e Y e X
2
o
ji ij
+e e Y X
ij
ij
j
ij
= i Xj and X e = X i , e ekl = i jk l so we get
o
1 n j
j
j
i
j
i
j ii j
=
− Xi Yj − Yi Xj + Y X
Xi Y i 2
n
o
1 jj
ii
=
Xi Y i − Xii Yjj − Yii Xjj + Yj X j
2
Also e X
30
The fourth term of Equation (4.1) gives
o
1 n ii
jj
Xj Y j − Xjj Yii − Yjj Xii + Yi X i
=
2
adding these gives
1 jj
( Xi Y i + ii Xj Y j ) − Xjj Yii −
2
1
Yjj Xii + ( ii Yj X j + jj Yi X i )
2
Since so(n) is skew symmetric the two middle terms in Equation (4.2) vanish
and summing the other term over 1 ≤ i < j ≤ n gives
1
(ntrace (XY ) + ntrace (YX )) = ntrace (XY )
2
The second term in Equation (4.1) becomes
1 n ij T ji
ji
ji
− (e ) X Y e − (e ij )T X e Y − (e ij )T Y e X
2
o
(4.2)
ji
+(e ij )T e Y X
1 n ji ji
ji ji
ji ji
= − e X Y e − e X e Y − e Y e X
2
o
ji
ji
+e e Y X
o
1n
= − j Xi Y j i − j Xi j Yi − j Yi j Xi + j ij i Y X
2
o
1n
= − 0 − Xij Yij − Yij Xij + 0
since i 6= j we get the zeros
2
= −trace (XY ) after summing over i, j
the last equality follows since Xij = −Xji for X ∈ so(n). The third term of
equation (4.1) similarly gives
−trace (XY )
So adding it all up gives the expression for the Killing form on so(n) as
B(X , Y ) = (n − 2)trace (XY )
Some other explicit expressions of Killing forms are[1]
B(X , Y ) = (n − 2)trace (XY )
B(X , Y ) = 2ntrace (XY )
B(X , Y ) = 2ntrace (XY ) − 2trace X trace Y
The Killing form has several nice symmetry properties.
31
on o(n)
on su(n)
on u(u)
Lemma 4.5. [6] The Killing form B of g is symmetric. Also B is invariant
under automorphisms of g. In particular
also
B((Ad g)X , (Ad g)Y ) = B(X , Y ) for all X , Y ∈ g and g ∈ G
B((ad X )Y , Z ) + B(Y , (ad X )Z ) = 0 for all X , Y , Z ∈ g
P ROOF. [6] The symmetry follows from trAB = trBA.
If is an automorphism of g then
(ad X )Y
So
= [ X , Y ] by Lemma B.4
= [ X , −1 Y ]
= [X , −1 Y ]
◦ ad X ◦ −1 Y
=
tr(ad X ◦ ad Y ) = tr( ad X
−1
by the cyclic property of the trace.
Now if = Ad exp(tX ) then
◦ ad Y
−1
) = tr(ad X ◦ ad Y )
d
B(Ad (exp(tX ))Y , Z )|t=0
dt
d
B(Y , Ad (exp(−tX ))Z )|t=0
=
dt
= B(Y , (ad X )Z )
B((ad X )Y , Z ) =
Theorem 4.6. [10] Let (g, ) be an effective orthogonal symmetric algebra,
then the Killing form B is negative definite on k.
P ROOF. [10] Since (g, ) is an orthogonal symmetric algebra Int g is compact. By the proof of Lemma 3.10 there is an Ad G0 -invariant inner product
h·, ·i on g so
hAd (exp(tX ))Y , Z i = hY , Ad (exp(−tX ))Z i
therefore by differentiating at t = 0 we get
h(ad X )Y , Z i = −hY , (ad X )Z i.
If we choose an orthonormal basis with respect to h·, ·i for g, then the matrix
representation of ad X is skew symmetric i.e. (ad X )ij = aij = −aji . Thus
X
B(X , X ) = tr(ad X ◦ ad X ) = aij aji =
−aij2 ≤ 0
i,j
32
P
Now aij = hei , (ad X )ej i so if i,j −aij2 = 0 then ad X = 0 so X ∈ z(g) but
k ∩ z(g) = {0} so X = 0. Thus B|k is negative definite.
The sign of the Killing form on p leads to the following classifications,
Definition 4.7. An orthogonal symmetric algebra (g, ) is said to be of
(1) compact type if B is negative definite on p,
(2) noncompact type if B is positive definite on p,
(3) Euclidean type if B is identically zero on p.
Example 4.8. We make show that so(n) is a compact Lie algebra. Let
X ij = √12 (e ij − e ji ) be an arbitrary basis element in p of so(n) then
(n − 2)
trace (e ij − e ji )(e ij − e ji )
2
(n − 2)
= −
trace 2e ij e ji
2
= −(n − 2).
B(X ij , X ij ) =
Hence it is negative definite.
It is also true that su(n), u(n) and o(n) are compact, while o(p, q) 2 and
u(p, q) are noncompact Lie algebras.
We now want to prove that the Lie algebra of a compact Lie group is
compact i.e. agrees with the above classifications. To do so we need a lemma.
Lemma 4.9. [10] Let g be a Lie algebra, Aut g be the Lie algebra automorphisms of g, ∂g be the Lie algebra of Aut g and let Int g be the identity component
of Aut g. If g is semisimple then
ad g = ∂g
and therefore
Int g = (Aut g)0
For the reader unfamiliar with this terminology we refer to Appendix B.
i.e
P ROOF. [10] Let D ∈ ∂g then (D ◦ ad X )Y = ad DX (Y ) + ad X ◦ DY
[D, ad X ] = ad DX
so [∂g, ad g] ⊆ ad g.
The bi-linear form C (F1 , F2 ) = tr(F1 F2 ) is nondegenerate on ad g by the
assumptions of the lemma. Let a be the orthogonal complement of ad g in
∂g with respect to C . We will show that a = 0, then the claims follows. By
construction ∂g ⊆ ad g + a and (ad g) ∩ a = {0}. Now
tr([F1 , F2 ]F3 ) = tr(F1 F2 F3 − F2 F1 F3 )
= tr(F1 F2 F3 − F1 F3 F2 )
2see Example 4.21 for the definition
33
= tr(F1 [F2 , F3 ])
so for A ∈ a, D ∈ ∂g and X ∈ g
tr([A, D]ad X ) = tr(A[D, ad X ]) = tr(Aad DX ) = 0
so [A, D] ∈ a, thus a is a Lie ideal.
Let A ∈ a, X ∈ g then ad AX = [A, ad X ] ⊆ a ∩ ad g so ad AX = 0
therefore AX = 0 since C is semisimple on ad g, so A = 0 and thus a =
{0}.
Theorem 4.10. [10] Let B be the Killing form on the Lie algebra g then then
following are equivalent
(1) g is the Lie algebra of a compact Lie group,
(2) g = z(g) ⊕ g0 where B is negative definite on the ideal g0 .
P ROOF. [10] Let G be a compact Lie group with Lie algebra g. We can
assume that G is connected otherwise we consider G0 . Then Int g = Ad G
since G is connected and so Int g is compact.
By Lemma C.2 we can pick an Int g-invariant inner product h·, ·i on g.
Let g0 be the orthogonal complement of z(g) with respect to h·, ·i. Then for
X ∈ g, Y ∈ g0 , Z ∈ z(g) we have
h[X , Y ], Z i = hY , [Z , X ]i = hY , 0i = 0
so [g, g0 ] ⊆ g0 , and g0 is an ideal.
With an orthonormal basis of g w.r.t. h·, ·i, the elements of Int g are orthogonal matrices i.e exp(t ad X ) for X ∈ g is an orthogonal matrix so ad X is
a skew symmetric i.e. aij = −aji so for X ∈ g0
X
B(X , X ) = tr(ad X )2 = aij aji =
−aij2 ≤ 0
i,j
which is zero if and only if ad X = 0 is X ∈ z(g) i.e. X = 0. Thus g =
z(g) ⊕ g0 with B|g0 ×g0 negative definite.
Conversely if g = z(g) ⊕ g0 with B negative definite on g0 . Note that g0 is
semisimple, since if B(X , Y ) = 0 for all Y ∈ g0 then B(X , X ) = 0 so X = 0.
We will show that g0 is the Lie algebra of a compact Lie group H, then
g0 ⊕ z(g) correspond to H × T dim z(g) , where T n is the n dimensional torus,
which is compact.
Since B is Ad G invariant, B is invariant under Int g0 , so Int g0 is represented by orthogonal matrices, i.e. Int g0 ⊆ O(dim Int g0 ). Also by Lemma
4.9 Int g0 = (Aut g)0 so Int g0 is closed and thus compact. So g0 is the Lie
algebra of a compact Lie group.
It turns out that the Killing form decomposes the orhogonal symmetric
algebra of a symmetric pair into a very convenient form.
34
Theorem 4.11. [6] Let (g, ) be an effective orthogonal symmetric algebra of
a symmetric pair (G, K ). Then we can decompose g as
g = k ⊕ p 1 ⊕ . . . ⊕ pm
where pi ⊥pj , i 6= j with respect to B. We can also find an Ad K -invariant inner
product on g given by
g(·, ·) = −B|k +
1
1
B|p1 + . . . +
1
m
B|pm
We also have [pi , pj ] = 0 for i 6= j.
P ROOF. [6] By the proof of Lemma 3.10 we get a Ad K invariant inner
product (·, ·) on p. Define

 −B(X , Y ) X , Y ∈ k
(X , Y )
X,Y ∈ p
g(X , Y ) =

0
X ∈ k, Y ∈ p or vice versa.
Then g(·, ·) is positive definite and Ad K invariant.
Now for a fixed X ∈ p consider the functional f (Y ) = B(X , Y ) for all
Y ∈ p. Then the Riesz-theorem implies
(4.3)
B(X , Y ) = g(Y , T (X ))
Since B is symmetric T is self adjoint with respect to g(·, ·), so we can find an
orthonormal
basis of eigenvectors {Xj } of T such that T (Xj ) = j Xj . Note that
all j 6= 0 since B is nondegenerate. Also since B is symmetric, the eigenspaces
corresponding to different eigenvalues are orthogonal.
In fact since {X k } are
orthonormal with respect to g(·, ·) we have that k = B(Xk , Xk ). So
p = p1 ⊕ . . . ⊕ pm
The expression for g(·, ·) follows by Equation (4.3).
For the last claim let Yi ∈ pi , Yj ∈ pj then
B([Yi , Yj ], [Yi , Yj ]) = B(Yi , [Yj , [Yi , Yj ]]) =
i g(Yi , [Yj , [Yi , Yj ]])
but also
B([Yi , Yj ], [Yi , Yj ]) = −B([Yj , Yi ], [Yi , Yj ])
= −B(Yj , [Yi , [Yi , Yj ]])
= − j g(Yj , [Yi , [Yi , Yj ]])
= j g(Yi , [Yj , [Yi , Yj ]])
where the last equality follows from Theorem
4.13 and the symmetries of the
i
curvature tensor Rjkl
. So if i 6= j then i 6= j so B([Yi , Yj ], [Yi , Yj ]) = 0 but g
is semisimple hence [Yi , Yj ] = 0. Thus [pi , pj ] = 0 for i 6= j.
35
With the metric g in Theorem 4.11 the symmetric pair (G, K ) makes the
quotient G/K into a nice Riemannian manifold (G/K , g).
2. The Curvature Formula
To do calculations on the symmetric space we turn our attention from
G to the symmetric space M = G/K . By Theorem 3.3 we get the vector
fields on M by parallel translation and the map d p p0 → Tp M . The images
X ∗ (p0 ) = d X (p0 ) are Killing fields (see Appendix D) i.e. their local flows are
isometries.
We start with some properties of these Killing fields.
Lemma 4.12. [6] Let G be the isometry group on the symmetric space (M , g),
K ⊆ G be the isotropy group at p0 ∈ M . If g is the Lie algebra of G with the
standard decomposition g = k ⊕ p, then
X ∗ (p0 ) = 0
∇vX ∗ (p0 ) = 0
for all X ∈ k
for all X ∈ p, v ∈ Tp M
P ROOF. [6] Let X ∈ k then
d
d
(exp(tX )p0 )|t=0 = (p0 )|t=0 = 0.
dt
dt
: t → M be a curve such that ˙ (0) = v ∈ Tp M then
X ∗ (p0 ) = d X (p0 ) =
Next let X ∗ (p0 ) =
d
dt
tX ∗ (p0 ) (p0 )|t=0
by Theorem3.3
so
∇vX ∗ (p0 ) = ∇∂ ∂t∂ ∂s
=
=
tX ∗ (p0 ) ( ∇∂ ∂s∂ tX ∗ (p0 ) (
∂t
∇∂ d v|s=t=0
∂t
(s))
|s=t=0
h∂
∂i
(s))
since
=0
,
∂s ∂t
|s=t=0
but by Lemma 2.3 v is a parallel
transport of v along (t), so
= 0
Since we can parallel translate all vector fields on M to the origin it is
enough to be able to do the curvature calculations for vector fields at the origin.
36
Theorem 4.13. [6] Let (M , g) be a symmetric space. With the identification
p∼
= Tp M X 7→ d X (p)
in Theorem 3.3, the curvature tensor of M satisfies
R(X ∗ , Y ∗ )Z ∗ = −[[X ∗ , Y ∗ ]Z ∗ ](p)
for all X ∗ , Y ∗ , Z ∗ ∈ d p (p) = Tp M
P ROOF. [6] We denote the image d p (p) by p∗ . Let X ∗ ∈ p∗ , Y ∗ ∈ p∗ .
Then X ∗ (p) = d X (p). The geodesic c(t) = Expp tY ∗ (p) satisfies Y ∗ (c(t)) =
ċ(t), since by theorem 3.3
d
exp(sY (c(t)))c(t)|s=0
ds
d
=
sY (c(t))∗ (c(t))|s=0
ds
d
=
c(t + s)|s=0 = ċ(t)
ds
By Lemma D.10 the Killing field X ∗ is a Jacobi field along c so we have
the Jacobi equation
Y ∗ (c(t)) =
∇Y ∗∇Y ∗X ∗ + R(X ∗ , Y ∗ )Y ∗ = 0.
But p∗ is a subspace so for Y ∗ , Z ∗ ∈ p∗ we have Y ∗ + Z ∗ ∈ p∗ . Inserting this
into the above equation we obtain
0 = ∇(Y ∗ + Z ∗ )∇(Y ∗ + Z ∗ )X ∗ + R(X ∗ , Y ∗ + Z ∗ )(Y ∗ + Z ∗ )
= ∇Y ∗∇Y ∗X ∗ + ∇Y ∗∇Z ∗X ∗ + ∇Z ∗∇Y ∗X ∗ + ∇Z ∗∇Z ∗X ∗ +
R(X ∗ , Y ∗ )Y ∗ + R(X ∗ , Z ∗ )Y ∗ + R(X ∗ , Y ∗ )Z ∗ + R(X ∗ , Z ∗ )Z ∗
= ∇Y ∗∇Z ∗X ∗ + ∇Z ∗∇Y ∗X ∗ + R(X ∗ , Z ∗ )Y ∗ + R(X ∗ , Y ∗ )Z ∗
(But R(X ∗ , Z ∗ )Y ∗ = −R(Z ∗ , X ∗ )Y ∗ )
= ∇Y ∗∇Z ∗X ∗ + ∇Z ∗∇Y ∗X ∗ − R(Z ∗ , X ∗ )Y ∗ + R(X ∗ , Y ∗ )Z ∗
(By the Bianchy identity
−R(Z ∗ , X ∗ )Y ∗ = R(X ∗ , Y ∗ )Z ∗ + R(Y ∗ , Z ∗ )X ∗ )
= ∇Y ∗∇Z ∗X ∗ + ∇Z ∗∇Y ∗X ∗ + 2R(X ∗ , Y ∗ )Z ∗ + R(Y ∗ , Z ∗ )X ∗
(Now R(Y ∗ , Z ∗ )X ∗ = ∇Y ∗∇Z ∗X ∗ − ∇Z ∗∇Y ∗X ∗ − ∇[Y ∗ , Z ∗ ]X ∗
But for Y ∗ , Z ∗ ∈ p∗ [Y ∗ , Z ∗ ] ∈ k∗ since
[Y ∗ , Z ∗ ] = −[Y , Z ]∗ by Appendix D so [Y ∗ , Z ∗ ](p) = 0
Thus R(Y ∗ , Z ∗ )X ∗ = ∇Y ∗∇Z ∗X ∗ − ∇Z ∗∇Y ∗X ∗ )
= 2∇Y ∗∇Z ∗X ∗ + 2R(X ∗ , Y ∗ )Z ∗ .
37
Hence
∇Y ∗∇Z ∗X ∗ + R(X ∗ , Y ∗ )Z ∗ = 0.
Therefore starting with the Bianchy identity
(4.4)
R(X ∗ , Y ∗ )Z ∗ (p) = −R(Y ∗ , Z ∗ )X ∗ (p) + R(X ∗ , Z ∗ )Y ∗ (p)
= ∇Z ∗∇X ∗Y ∗ (p) − ∇Z ∗∇Y ∗X ∗ (p)
by eq. (4.4)
= ∇Z ∗[X ∗ , Y ∗ ](p) since ∇ is torison free
= ∇[X ∗ , Y ∗ ]Z ∗ (p) − [[X ∗ , Y ∗ ], Z ∗ ](p)
again since ∇ is torision free
= −[[X ∗ , Y ∗ ], Z ∗ ](p) since [X ∗ , Y ∗ ](p) = 0
For M = G/K we give M the Riemann structure induced by Theorem
3.11. There are a few details about Lie algebra operations in p ∗ versus p, for
our purposes it is the fact that [X ∗ , Y ∗ ] = −[X , Y ]∗ , see Appendix D.
Corollary 4.14. [6] Let (G/K , g) be a symmetric space. Then the sectional
curvature of X ∗ , Y ∗ ∈ Tp G/K is
K (X ∗ , Y ∗ )(p) =
−g([[X , Y ], Y ]∗ , X ∗ )
g(X ∗ , X ∗ )g(Y ∗ , Y ∗ ) − g(X ∗ , Y ∗ )2
P ROOF. This follows from the definition of the sectional curvature and
that [X ∗ , Y ∗ ] = −[X , Y ]∗ .
Now consider the curvature tensor R(X ∗ , Y ∗ )Z ∗ = −[[X ∗ , Y ∗ ], Z ∗ ] which
can be rewritten as R(X ∗ , Y ∗ )Z ∗ = −[[X , Y ], Z ]∗ . Since we know that
R(X ∗ , Y ∗ )Z ∗ ∈ Tp M we formally denote its corresponding vector in p by
R(X , Y )Z therefore we formally have
R(X , Y )Z = −[[X , Y ], Z ] for X , Y , Z ∈ p
Since g(·, ·) on TM derives from h·, ·i on p we can always write
−g([[X , Y ], Y ]∗ , X ∗ )
−h[[X , Y ], Y ], X )
=
g(X ∗ , X ∗ )g(Y ∗ , Y ∗ ) − g(X ∗ , Y ∗ )2
hX , X ihY , Y i − hX , Y i2
in p. So in terms of this Corollary 4.14 becomes
K (X , Y ) =
−h[[X , Y ], Y ], X )
hX , X ihY , Y i − hX , Y i2
38
for X , Y ∈ p
Example 4.15. By Corollary 4.14 and an Ad g invariant metric g the formula for the sectional curvature can be expressed as
g([X , Y ]∗ , [X , Y ]∗ )
g(X ∗ , X ∗ )g(Y ∗ , Y ∗ ) − g(X ∗ , Y ∗ )2
K (X ∗ , Y ∗ ) =
If we define a metric in the same way as in Lemma 3.10 we can write the
formula in p as
h[X , Y ], [X , Y ]i
hX , X ihY , Y i − hX , Y i2
K (X , Y ) =
where we use the same notation for the metrics. If we work with the real Grassmann manifolds we can choose a basis e ij − e ji . Before we start the calculations
we produce some intermediate results.
[e ij , e kl ]
=
i
jk l
this gives
−
k li j
=
jk il
e − li e kj
[e ij − e ji , e kj − e lk ] = [e ij , e kl ] − [e ij , e lk ] − [e ji , e kl ] + [e ji , e lk ]
= jk e il − li e kj − jl e ik + ki e lj − ik e jl + lj e ki
+ il e jk − jk e il − kj e li
= jk (e il − e li ) + il (e jk − e kj ) + ik (e lj − e jl )
+ jl (e ki − e ik )
Now if we choose the metric
1
hX , Y i = − trace (XY )
2
which is proportional to the Killing form on so(n). Therefore
1
he ij − e ji , e kl − e lk i = − trace ( jk e il − jl e ik − ik e jl + il e jk )
2
1
= − ( jk il − jl ik − ik jl + il jk )
2
= − jk il + ik jl but i < j, k < l so
=
ik jl
So it is indeed an orthogonal basis. Therefore
g [e ij − e jk , e kl − e lk ], [e ij − e jk , e kl − e lk ] =
jk
+
il
+
ik
+
Note that only none or one of these can be satisfied, so

 1 if (i = k, j 6= l), (i = l, j 6= k),
, (i 6= k, j = l) or (i 6= l, j = k)
K (e ij − e ji , e kl − e lk ) =

0 else
39
jl
So the sphere S n has constant sectional curvature 1, while the real Grassmannian manifolds have sectional curvature 0 or 1.
3. The Dual Space
Definition 4.16. Let (G/K , g, ) be a symmetric space such that the inner
product g on g is both Ad G and d invariant. Such that we can decompose
the Lie algebra as
g=k⊕p
The symmetric space is then called a normal symmetric space.
Note that when g is semisimple the Killing form B makes
(G/K , B, )
a normal symmetric space.
The concept of dual spaces makes it possible to pair up symmetric spaces
to that if one knows properties of one in the pair, one also knows the same
properties of the dual and vice versa.
Definition 4.17. Two normal symmetric spaces M = G/K and M † =
G † /K † are said to be dual provided
(1) There is a Lie algebra isomorphism : k → k† such that
g † ( (X ), (Y )) = −g(X , Y )
(2) A linear isometry ˆ : p → p† such that
[ˆ (X ), ˆ (Y )]† = − [X , Y ],
where all the corresponding entities in M † have been denoted with an †.
Remark 4.18. Because
is a Lie algebra isomorphism we have
[ (X ), (Y )]† = [X , Y ]
and since ˆ is an isometry
g † (ˆ (X ), ˆ (Y )) = g(X , Y ).
The above isometry ˆ induce an isometry
TeK M → TeK † M †
via the following commutative diagram
p


(p)y
ˆ
−−−→
p†


(p† )y
TeK M −−−→ TeK † M †
For compact spaces the following choice of metric is common
40
Definition 4.19. Let G/K be a homogeneous space with G compact and
g semisimple then the standard homogeneous Riemannian metric on G/K is
chosen as the negative Killing form i.e. g = −B.
As promised duality provides important information about the dual spaces.
Theorem 4.20. [9] Let (M , g) and (M † , g † ) be dual spaces. Then (M , g and
(M , g † ) have opposite sectional curvature i.e.
†
KM† † (ˆ X , ˆ Y ) = −KM (X , Y ) for all X , Y ∈ p
P ROOF.
g † ([ˆ X , ˆ Y ], [ˆ X , ˆ Y ])
g(ˆ X , ˆ X )g(ˆ
Y,ˆ
Y ) − g(ˆ
X , ˆ Y )2
g † (− [X , Y ], − [X , Y ])
=
g(X , X )g(Y , Y ) − g(X , Y )2
−g([X , Y ], [X , Y ])
=
g(X , X )g(Y , Y ) − g(X , Y )2
= −KM (X , Y )
KM † (ˆ X , ˆ Y ) =
To illustrate duality we find the the dual space of the real Grassmannian
manifold.
Example 4.21. If we now generalize Example 1.7 by considering
Ip 0 T
(4.5)
Q=
0 −Iq
Then we define
O(p, q) = {A ∈ GLp+q (R) : AT QA = Q}
Then O(p, q) represents orthogonal matrices with respect to the inner product
defined by Equation (4.5). We find the Lie algebra g = o(p, q) of O(p, q)
by taking the derivative of a curve at the origin in the defining expression for
O(p, q). Then as in Example 1.7
A B
o(p, q) =
: A ∈ o(p), B ∈ M(p,q) , C ∈ o(q)
BT C
If we let O(p, q) act on the real Grassmann manifold Gp (Rp+q ) as in Example
1.8, then the isotropy group consists of matrices of the form
B 0
where B ∈ O(p), C ∈ O(q)
0 C
41
so we have that
O(p, q)/(O(p) × O(q))
is a manifold. Since the Lie algebra of O(k) consists of the skew symmetric
matrices we get
A 0
: A ∈ o(p), D ∈ o(q)
k=
0 D
and
m=
0 B
BT 0
: B ∈ M(p,q)
Example 4.22. Let G = O(p, q) and K = O(p) × O(q) be as in Theorem
3.7 with
Ip
0
J=
0T −Iq
with J ∈ Z (K ). If we form G/K = O(p, q)/O(p) × O(q) and observe that for
X ∈ o(p, q)
A B
A B
A
−B
X =
, Ad J
=
BT D
BT D
−B T D
we have
X + Ad J (X ) = 2
A 0
0T D
∈k
so J satisifies 2) in Theorem 3.7 so we have the symmetric pair
SO0 (p, q), S(O(p) × O(q)) with s = In J
Example 4.23. Let (G, K ) be the symmetric pair
(G, K ) = (SO(p + q), (SO(p) × SO(q))
with involutive automorphism s given by Example 3.9
Ip 0
−1
(A) = In S(A) = SAS where S =
0 −Ip
on the manifold G/K and the Ad G-invariant metric
1
B(X , Y ) = − trace (XY )
2
at the origin. Then by extending the inner product B to all of M as in Theorem
3.11 becomes an isometry, so
M = SO(p + q)/(SO(p) × SO(q))
becomes a normal symmetric space. Note that B is positive definite on p.
42
On the symmetric pair (G † , K † ) = (SO0 (p, q), SO(p) × SO(q)) we pick
the same involutive automorphism s as in Example 4.22. And on the manifold
G † /K † we take the metric
1
B † (X , Y ) = trace (XY )
2
at the origin, which again is Ad G † invariant and is an isometry after extending the metric. So we have the normal symmetric space
M † = SO0 (p, q)/(SO(p) × SO(q))
Again B † is positive definite on p† .
Let
: k → k† = idk
which certainly is a Lie algebra isomorphism such that
1
1
B(X , Y ) = − trace XY = −(− trace XY ) = −B † (X , Y )
2
2
and
ˆ
†
:p→p ,
0
x
−x T 0
7→
0 x
xT 0
which is an isometry such that
[ˆ (X ), ˆ (Y )]
†
=
=
=
=
=
0 y
0 y
0 x
−
yT 0
yT 0
xT 0
T
T
xy
0
yx
0
−
0 xT y
0 yT x
0 x
0
y
−
+
−x T 0
−yT 0
0 x
0 y
−x T 0
−yT 0
−[X , Y ]
− ([X , Y ])
0 x
xT 0
Therefore M , M † are dual.
43
Example 4.24. Further examples of dual pairs of symmetric spaces can be
found in the following table,
Noncompact
compact
Dimension
SL(n, R)/SO(n)
SU(p, q)/S(U(p) × U(q))
SO0 (p, q)/SO(p) × SO(q)
Sp(n, R)/U(n)
Sp(p, q)/Sp(p) × Sp(q)
SU(n)/SO(n)
SU(p + q)/S(U(p) × U(q))
SO(p + q)/SO(p) × SO(q)
Sp(n)/U(n)
Sp(p + q)/Sp(p) + Sp(q)
1
(n
2
see also [5].
44
− 1)(n + 2)
2pq
pq
n(n + 1)
4pq
APPENDIX A
The Hopf-Rinow Theorem
Definition A.1. A Riemannian manifold (M , g) is said to be geodesically
complete if for all p ∈ M the exponential map Expp is defined on all of Tp M ,
i.e. all geodesics ∈ M can be defined for all t ∈ R.
Definition A.2. A normal ball B (p) at p ∈ M on a Riemanninan manifold (M , g) is the image of an open ball B̃ (p) in the tangent space Tp M of the
exponential map Expp .
Definition A.3. Let p, q ∈ M then define the distance d (p, q) as the infimum of the lengths of all curves on M from p to q.
Remark A.4. It can be shown that the distance function is continuous and
that geodesics locally minimize the distance between two points, also that if a
curve is a minimum then it must be a geodesic.
Theorem A.5. [3] Hopf-Rinow
Let (M , g) be a Riamannian manifold if (M , g) is geodesically complete then any
two points p, q ∈ M can be connected by a geodesic.
P ROOF. [3] If M is geodesically complete then the exponential map Exp p
is defined on all of Tp M .
Let p, q ∈ M and r = d (p, q). Let B (p) be a normal ball in M and
let S (p) be the boundary of B (p). Let x0 be a point where the continuous
function d (q, x), x ∈ S (p) takes on its infimum(It does so since S (p) is
compact). Then x0 = Expp v where v ∈ Tp M and |v| = 1. Let be the
geodesic given by (s) = Expp sv. Now let
A = {s ∈ [0, r] : d ( (s), q) = r − s}.
We will show that r ∈ A and (r) = q.
Now A is not empty since 0 ∈ A and A is closed by continuity of the
distance. We are going to show that if s0 ∈ A and s0 < r then there exists a
> 0 such that s0 + ∈ A. This will imply that supA = r and r ∈ A since A
is closed. Thus d ( (r), q) = r − r = 0 so (r) = q.
Let B 0 ( (s0 )) be a normal ball at (s0 ) and S 0 be the boundary of B 0 ( (s0 )).
Let x00 be the point where d (x, q), x ∈ S 0 takes on its minimum. So
d ( (s0 ), q) =
0
+ min{d (x, q) : x ∈ S 0 } =
45
0
+ d (x00 , q)
But d ( (s0 ), q) = r − s0 so r − s0 = 0 + d (x00 , q). Now if x00 = (s0 + 0 ) we
get
r − s0 = 0 + d ( (s0 + 0 , q) so s0 + 0 ∈ A.
We only have to show that x00 = (s0 + 0 ). But by the triangle inequality
d (p, x00 ) ≥ d (p, q) − d (q, x00 ) = r − (r − s0 − 0 ) = s0 + 0
which is no surprise. But more importantly the broken curve joining p and x 00
by going along from p to (s0 ) and along the exponential ray from (s0 ) to
x00 in B 0 ( (s0 )), has length s0 + 0 so d (p, x00 ) = s0 + 0 . By minimality of the
distance that broken curve must be a geodesic. But at it is starting out as it
must remain , so x00 = (s0 + 0 ).
46
APPENDIX B
The Adjoint Representation
Definition B.1. Let G be a Lie group. For each h ∈ G we define the inner
automorphism of G by
In h : G → G
g 7→ hgh−1
Denote by GL(g) the group of vector space automorphisms of g.
Definition B.2. Let G be a Lie group, then the adjoint representation of G
is given by
Ad : G → GL(g) h 7→ de In(h)
The map In g is a Lie group automorphism so for fixed g ∈ G Ad g is a
Lie algebra automorphism, thus it satisfies
and
Ad g([X , Y ]) = [Ad g(X ), Ad g(Y )] for all X , Y ∈ g
exp(Ad g(X )) = In g(exp(X ))
But we can also view Ad as a map Ad : G → GL(g). Then
Ad g1 ◦ Ad g2 = Ad g1 g2
so it is a Lie group homomorphism. Therefore it induces a Lie algebra automorphism.
Definition B.3. Let g be a Lie algebra, then the adjoint representation of g
is defined as
ad : g → End(g), X 7→ (de Ad )(X )
where End(g) is the space of linear maps g → g
So we have
ad [X , Y ] = [ad X , ad Y ]
Lemma B.4. [10] Let g be the Lie algebra of the Lie group G then
(adX )Y = [X , Y ]
for X , Y ∈ g.
47
P ROOF. For g ∈ G, Y ∈ g we have
Ad gYe = d(In g)(Ye ) = d(Rg −1 Lg )(Ye ) = dRg −1 Yg
Let exp(tX ) be an integral curve of X then
1
ad X (Y ) = lim (Ad exp(tX )(Ye ) − Ad idG (Ye ))
t→0 t
1
= lim (dRexp(−tX ) YexptX − Ye )
t→0 t
1
= lim (dRexp(−tX ) YRexp(tX ) e − Ye )
t→0 t
= LX Y = [X , Y ]
since RexptX is the flow of X in G by Lemma D.15.
Proposition B.5. [6] Let g be a Lie algebra, then for all X ∈ g
exp(ad X ) = Ad exp(X )
P ROOF. [6]
d
exp(tad X )|t=0 = ad X
dt
= (dAd )X
d
= = Ad exp(tX )|t=0 .
dt
So the claim follows by the uniqueness of ordinary differential equations with
initial data.
Let Aut g be the set of Lie algebra automorphisms on g. Then Aut g is a
closed subgroup of the Lie group GL(g). Let Int g be the identity component
of Aut g, with Lie algebra ad g. By exponentiating ad g we get the entire Int g
which is contained in Ad g ⊆ Aut g.
Definition B.6. Let ∂g be the Lie algebra of Aut g. Then
Aut g ⊆ GL(g) and ∂g ⊆ End(g).
Definition B.7. And endomorphism D ∈ End(g) is called a derivation if
D[X , Y ] = [DX , Y ] + [X , DY ]
Theorem B.8. [10] Let g be a Lie algebra and ∂g be the Lie algebra of Aut g
then ∂g is the set of derivations on g.
P ROOF. [10] Let D ∈ ∂g then for t ∈ R, exp(tD) ∈ Aut g so
exp(tD)[X , Y ] = [exp(tD)X , exp(tD)Y ].
48
Therefore
d
exp(tD)[X , Y ]|t=0
dt
d
=
[exp(tD)X , exp(tD)Y ]|t=0
dt
= [DX , Y ] + [X , DY ],
D[X , Y ] =
so D is a derivation.
Conversely let D be a derivation. Then
D[X , Y ] = [DX , Y ] + [X , DY ]
so inductively
k X
k
k
D [X , Y ] =
i
i=0
So we get
exp(tD)[X , Y ] =
∞ l l
X
tD
l=0
=
l!
i=0
l!
[Di X , Dl−i Y ]
(l − i)!i!
l h i
∞ X
X
l=0 i=0
=
[X , Y ]
l
∞ l X
X
t
l=0
=
l!
[Di X , Dk−i Y ].
t Di X t l−i Dl−i Y i
,
i!
(l − i)!
if we switch summation order we get
t Di X t l−i Dl−i Y i
,
let k = l − i
i!
(l − i)!
∞ X
∞ h i
X
i=0 l=i
∞ X
∞ h i i
X
t D X t k Dk Y i
,
=
i!
k!
i=0
k=0
= [exp(tD)X , exp(tD)Y ].
The above manipulations of the summations are legitimate since exp(tD) converges absolutely so can replace upper limit ∞ with N and do manipulations.
Thus exp(tD) ∈ Aut g so D ∈ ∂g.
Theorem B.9. [10] Let G be a Lie group and let Z (G) be the center of G. If
G is connected then Z (G) = kerAd g.
P ROOF. [10] Obviously Z (G) ⊆ kerAd g. Now suppose Ad g = id then
exp(X ) = gexp(X )g −1
by exponentiating,
so g ∈ Z (G0 ) = Z (G) since G is connected.
49
We define z(g) = {X ∈ g : ad X = 0}
Theorem B.10. Let G be a Lie group with Lie algebra g. If G is connected
then z(g) is the Lie algebra of Z (G).
P ROOF. Let X ∈ g then if X is belongs to the Lie algebra of Z (G) then
exp(tX ) ∈ Z (G) so
Ad exp(tX ) = idGL(g) .
This is equivalent to that
exp(ad tX ) = idGL(g)
which happens if and only if ad X = 0 so X ∈ z(g).
50
APPENDIX C
Inner Products From the Haar Measure
Fact C.1. Let G be a compact topological group and let C (G) be the set of
continuous real valued functions on G. Then there is a unique map
Z
I : C (G) → R, I (f ) 7→
f (g)dg
G
such that
(1) I (e) = 1,
(2) I (f ) ≥ 0 for f ≥ 0,
(3) I ( f + g) = I (f ) + I (g) for , ∈ R,
(4) I is G − invariant.
The function I is called the Haar-integral.
Lemma C.2. [1] Let H be a Lie group, V be a vector space and : H →
Aut(V ) be a representation of the group H on V . If H is compact there exists an
H invariant inner product on V .
P ROOF. Since H is compact, we can define the Haar-integral
Z
I (f ) =
f (h)dh
H
The Haar-integral is H -invariant i.e.
I (f ) = I (f ◦ Rg ) = I (Lg ◦ f ).
So if h·, ·i is an inner product on V then we define the H invariant inner
product on V
Z
(u, v) = h (h)u, (h)vidh.
H
51
APPENDIX D
Lie Derivatives and Killing Fields
We state the following result from the theory of ordinary differential equations
Lemma D.1. [6] Let X be a vector field on the manifold M . Then for every
point p ∈ M there exists an open neighborhood U of p and an open open interval
I of R such that 0 ∈ I , with the property that for all q ∈ U there is a curve
cq : I → M satisfying
ċq (t) = X (cq (t)),
cq (0) = q.
Then map (t, q) → cq (t) from I × U to M is smooth.
Definition D.2. Let X , c be as in Lemma D.1, then the map (t, q) 7→ cq (t)
is called the local flow of the vector field X. The curve cq is called the integral
curve of X through q.
For a fixed t and varying q we write t (q) = cq (t). Note that
(D.1)
since
t
◦ s (q) =
t+s (q)
if s, t, s + t ∈ Iq
d
d
s+t (q)|t=0 = X ( s (q)) =
t ( s (q))|t=0
dt
dt
and they have the same initial data for t = 0 at q.
It can be shown that if t is defined on an open subset U of M , it maps U
diffeomorphically onto its image.
Definition D.3. A family ( t )t∈I of diffeomorphisms from M to M satisfying equation (D.1) is called a local 1-parameter group of diffeomorphisms.
Definition D.4. Let : M → N be a diffeomorphism between differentiable manifolds and let X be a vector field on M. Then we define the following
vector field on N
( ∗ X )(p) = d (X ( −1
(p))) for p ∈ N
The following lemma follows from the definition of the differential map.
Lemma D.5. [6] Let N be a differentiable manifold. For any differentiable
function f : N → R we have
( ∗ X )(f )(p) = X (f ◦ )( 53
−1
(p)).
Now consider the special case when N = M so −t = −1
t .
Let f be a real valued function defined in the codomain of
pull-back of f via t is defined by
∗
tf
∂
ak ∂x
k
For a vector field X =
≡f ◦
t,
then the
t
on the codomain of t
∂ i−t ∂
evaluated at t (p)
∂x k ∂yi
where x,y are local coordinates in the codomain and domain and
∗
t X (p)
≡ (
−t )∗ X (p)
i
−t
= ak
= (y ◦ −t )
i
.
For a 1-form = i dyi in the codomain of t
we define the pull-back by
∂ it k
dx
∂x k
The action of ∗t on higher order tensors is obtained by its action on the atomic
tensors.
∗
t (
)(p) =
i ( t (p))
Definition D.6. Let X be a vector field on the differentiable manifold M
with local 1-parameter group t of local diffeomorphisms and S a tensor field
on M . Then the Lie-derivative of S in the direction of X is defined by
d
LX S = ( ∗t S)|t=0
dt
Definition D.7. Let X be a vector field and S be a tensor of type (0, r)
then the inner product X S is the contraction
( X S)(X1 , X2 , . . . , Xr−1 )(p) ≡ C1,1 (X ⊗ S)(X1 , X2 , . . . , Xr−1 )(p)
= S(X , X1 , . . . , Xr−1 )
where C1,1 is the contraction operator.
Theorem D.8. [6] Let M be a differenitiable manifold and X a vector field
on M .
(1) If f : M → R be a differentiable function, then
LX (f ) = df (X ) = X (f )
(2) If Y be a vector field on M , then
LX Y = [X , Y ]
(3) Let =
j dx
j
∂
be a 1-form on M . If X = X i ∂x
i , then
LX = d( X ) + x (d ).
For higher order tensors one simply uses the product rule for differentiations on the
atomic tensors.
54
P ROOF. [6] 1)
LX (f ) =
d
d ∗
f ◦
t f|t=0 =
dt
dt
t|t=0
=
∂f i
X = X (f )
∂x i
2) Let Y = Y i ∂x∂ i
LX Y
=
=
=
=
=
=
d ∗ i ∂ Y
dt t
∂x i |t=0
∂ d
( −t )∗ Y i i
dt
∂x |t=0
j
∂ −t ∂ d
i
Y ( t ) i
dt
∂x ∂x j |t=0
∂X j ∂
∂Y i k j ∂
i
X i j +Y − i
since
∂x k
∂x
∂x ∂x j
o
n ∂ j
∂(x j ◦ −t )
∂x j ( 0 )
j
−t
=
=
= i
∂x i |t=0
∂x i
∂x i
|t=0
o
nd
−t|t=0 = −X
dt
∂Y j
∂X j ∂
Xk k − Yk k
∂x
∂x ∂x j
[X , Y ]
3)
j
d ∗
d
∂ t k
( t )|t=0 =
dx
j ( t )
dt
dt
∂x k
|t=0
j
j
∂
∂X
j
=
X i k dx k + j k dx k as in the previous
i
∂x
∂x
∂
∂X i j
j i
X + i j dx .
=
∂x i
∂x
Now if we expand
LX =
∂ j i
dx ∧ dx j )
∂x i
∂ j j i ∂X i
∂ j
=
X dx + j i dx j + i X i dx j
i
∂x
∂x
∂x
∂ j j i
− i X dx
∂x
∂ j i j
∂X i
j
=
X dx
i dx +
j
∂x
∂x i
d( X ) + X (d ) = d( j X j ) + X (
55
and
So indeed
LX = d( X ) + X (d ).
Definition D.9. Let (M , g) be a Riemannian manifold with a metric g =
gij dx i ⊗dx j then a vector field X on M is called a Killing field or an infinitesimal
isometry if
LX (g) = 0.
Lemma D.10. [6] A vector field X on a Riemannian manifold (M , g) is a
Killing field if and only if the local 1-parameter group generated by X consists of
local isometries.
P ROOF. [6] By definition
d ∗
( g)|t=0 = 0
dt t
holds at every point of M . Hence ∗t g = g for all t ∈ I , so the diffeomorphisms t are isometries.
Conversely if t are isometries then
so
g(( t )∗ (X ), ( t )∗ (Y )) = g(X , Y )
d ∗
g = 0.
dt t
Lemma D.11. [8] Let M be a differentiable manifold. Two derivations
D1 , D2 agree if they agree on vector fields and functions on M .
P ROOF. [8] Since the product rule applies for derivations on tensors, we
only have to show the claim for one forms, since a general tensor can be written
as a tensor product of vector fields, functions and one forms.
Let = i dx i be an arbitrary one form on M and let V = V i ∂x∂ i be an
arbitrary vector field on M . (V ) is a function on M so if D is a derivation on
M then
D( (V )) = D(C11 ⊗ V ) = C11 (D ⊗ V + ⊗ DV ) = (D )(V ) + (DV )
where C11 is the contraction operating on and V . This defines the derivation
on one forms if one knows how the derivation operates on vector fields and
functions.
Corollary D.12. [8] Let M be a differentiable manifold. Then the Lie derivative satisfies
L[U ,V ] = [LU , LV ]
for arbitrary vector fields U , V on M .
56
P ROOF. [8] Since both [LU , LV ] and L[U ,V ] are derivations, Lemma D.11
implies that it is enough that they agree on functions and vector fields on M .
Let f be a function on M then
[LU , LV ]f
=
=
=
=
L U LV f − L V LU f
U ◦ V (f ) − V ◦ U (f )
[U , V ](f )
L[U ,V ] f .
Similarly if W is a vector field on M then
[LU , LV ]W
=
=
=
=
L U LV W − L V LU W
[U , [V , W ]] − [V , [U , W ]]
[[U , V ], W ]
L[U ,V ] W ,
where we have used the Jacobi identity and part 2 of Theorem D.8.
Lemma D.13. [6] The Killing fields of a Riemannian manifold (M , g) is a
Lie-algebra.
P ROOF. [8],[6] Since vector fields on a differentiable manifold constitute a
Lie-algebra, we only have to show that if X , Y are Killing fields then L [X ,Y ] g =
0. But by Corollary D.12 the equation L[X ,Y ] = [LX , LY ] and the fact that
X , Y are Killing fields LX g = 0, LY g = 0 so L[X ,Y ] g = 0.
Proposition D.14. [6] Every Killing field X on a Riemannian manifold
(M , g) is a Jacobi field along any geodesic ∈ M .
P ROOF. [6] By Lemma D.10 each Killing field X generates a local 1parameter group of isometries. Isometries map geodesics to geodesics, so X
generates a variation of geodesics c(t, s) = t ( (s)), where is the local 1parameter group of X . Since every variation of geodesics generates a Jacobi
field by X = ∂∂t c(t, s)|t=0 , the claim follows.
Lemma D.15. [9] Let G be a Lie group with Lie algebra g. Let
one-parameter subgroup of X ∈ g then the flow of X is R t
be the
P ROOF. [9] Here Lg is the left translation, i.e it has nothing to do with the
Lie derivative. If g ∈ G then Lg ◦ is the integral curve of dLg X = X starting
at g, so Lg t = g t = R t g. Thus R t is the flow of X.
Proposition D.16. [9] Let G be an isometry group that acts on a Riemannian manifold (M , g) by p (g) = gp. Denote the Lie algebra of G by g. Let
X , Y ∈ g and define the Killing field
d
X ∗ = d p (X ) = exp(tX )p
dt
57
and similarly for Y ∗ then
[X ∗ , Y ∗ ] = −[X , Y ]∗
P ROOF. [9] Again Lg is the left translation. Let Then for Y ∈ G we have
−t
◦ exp(Y ) ◦ t (p) = ( p R t L
−t
t
be the local flow of X ∗ .
)exp(Y )
so by the left-invariance of Y we have
d
∗
−t (Y t (p) )
= (d p dR t dL
−t
)Ye = d p dR t Y
−t
= d p dR t YR
Furthermore
−t
e.
[X ∗ , Y ∗ ] = LX ∗ Y ∗
1
= lim (dR −t Y ∗t − Yp∗ )
t→0 t
1
= lim (d p dR t YR −t e − d p Ye )
t→0 t
1
= d p lim (dR t YR −t e − Ye )
t→0 t
= d p L−X Y
= [−X , Y ]∗
The second last equality follows from the formula that if t is the flow of X
then
d
LX Y = −t Y t .
dt
By Lemma D.15 if t is the integral curve of X then R t is the flow of X in
G.
58
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Library vol. 22, AMS (2003).
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[3] M. P. DoCarmo, Riemannian Geometry, Birkhäuser (1992).
[4] S. Hassani, Mathematical Physics, A Modern Introduction to Its Foundations, Springer (2002)
[5] S. Helgason, Differential Geometry, Lie Groups and Symmetric Spaces, Academic Press (1978).
[6] J. Jost, Riemannian Geometry and Geometric Analysis, Universitext, Springer (2002).
[7] S. Kobayashi and K. Nomizu, Foundations of Differential Geometry vol. II, John Wiley & Sons (1963).
[8] M. Kriele, Spacetime, Foundations of General Relativity and Differential Geometry, Lecture Notes in Physics Monographs, Springer (1999).
[9] B. O’Neill, Semi Riemannian Geometry, Academic Press (1983).
[10] P. Sjögren, Riemannska Symmetriska Rum, Matematiska Institutionen, Chalmers Tekniska Högskola och
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59
Master’s Theses in Mathematical Sciences 2005:E3
ISSN 1404-6342
LUTFMA-3110-2005
Mathematics
Centre for Mathematical Sciences
Lund University
Box 118, SE-221 00 Lund, Sweden
http://www.maths.lth.se/