The Skew-Torsion Holonomy Theorem 255
i ≥ 1. Moreover, such groups satisfy the assumptions of Theorem 9.2.3. Then, if M
is irreducible, we either have
I
o
(M)
p
= H
0
and T
p
M = V
0
or
I
o
(M)
p
= H
1
and T
p
M = V
1
.
If I
o
(M)
p
= H
0
,thenallg(Θ)=0 and in particular Θ = 0andso∇
c
=
¯
∇
c
.
Let us analyze the remaining case I
o
(M)
p
= H
1
. From the Skew-Torsion Holonomy
Theorem, there are only two cases:
(a) H
1
= SO(T
p
M). In this case M has constant curvature. Then M = S
n
or its
symmetric dual M = H
n
. The latter case will be excluded in Proposition 9.6.2, except
for n = 3, in which case the hyperbolic space H
3
is the dual space of the Lie group
S
3
= Spin
3
.
(b) H
1
acts on T
p
M as the adjoint representation of a compact simple Lie group.
Then M is isometric to a comp act simple gr oup with a bi-invariant Riemannian metric
(a classification free and geometric proof of this fact is given in Proposition 9.6.9)
We have proved the following result (see [261, 262]).
Theorem 9.6.1 Let M be a simply connected, irreducible, naturally reductive Rie-
mannian homogeneous space. Assume that M is neither isometric to a sphere nor to
a compact simple Lie group with a bi-invariant Riemannian metric or its symmetric
dual. Then the canonical connection is unique. (In particular, any isometry of M is
affine with respect to the canonical connection).
Proposition 9.6.2 The real hyperbolic space H
n
,n= 3 , admits a unique naturally
reductive decompositon: the Cartan decomposition of H
n
= SO
o
n,1
/SO
n
.
Proof Let G be a connected Lie subgroup of I
o
(H
n
)=SO
o
n,1
that acts transitively
on H
n
andsuchthatH
n
= G/H is a naturally reductive space. If G is semisimple, it
is standard to show that G = SO
o
n,1
. In fact, let K be a maximal compact subgroup
of G.ThenK has a fixed point, say p . We may assume that H is the isotropy group
at p and so H = K since K is maximal. Hence (G, H) is a presentation of H
n
as an
effective Riemannian symmetric pair, and therefore G = SO
o
n,1
(otherwise, H
n
would
have two different presentations as an effective Riemannian symmetric pair). We
will prove for n = 3 that there is only one reductive decompo sition (not necessarily,
a priori, naturally reductive) of the pair (SO
o
n,1
,SO
n
). Using Remark 9.6.3 we can
transfer this question to the sphere and we are done by Remark 9.6.4.
If G is non-semisimple, then G contains a nontrivial normal abelian Lie subgroup
A. It is a well known fact that either A fixes a unique point at infinity or A translates
a unique geodesic. If A translates a unique geodesic, then G leaves this geodesic
invariant since A is a normal subgroup of G and so G cannot be transitive, which is
a contradiction. So, let q
∞
be the unique point at infinity that is fixed by A.Observe
that q
∞
must be fixed by G. In fact, since A is a normal subgroup of G, any element
of G leaves invariant the fixed set {q
∞
} of A at infinity. Let F be the foliation of
H
n
by parallel horospheres centered at q
∞
.ThenG leaves invariant F .Letp ∈ H
n