Page 148 - Curvature and Homology

130 III. RIEMANNIAN MANIFOLDS: CURVATURE, HOMOLOGY
Then, the oi are 1-forms in R" and

The linear independence of the oi is shown by making use of the fact that
when a' = 0, i = 1, -, n,

E. The homogeneous space SU(3)/S0(3)
1. Show that a compact symmetric space admitting a vector field generating
globally a 1-parameter group of non-isometric conformal transformations is
isometric with a sphere.
Hint: Apply the following theorem: If a compact simply connected symmetric
space is a rational homology sphere, it is isometric with a sphere except for
SU(3)/S0(3) [82]. The exceptional case may be disposed of as follows: Let G
be a compact simple Lie group, o # identity an involutary automorphism of G
(cf. VI.E.1) and H the subgroup of G consisting of all elements fixed by a. Then,
there exists a unique (up to a constant factor) Riemannian metric on G/H
invariant under G. With respect to this metric, GIH is an irreducible symmetric
space (that is, the linear isotropy group is irreducible). Hence, G/H is an Einstein
space. But a compact Einstein space admitting a non-isometric conformal transforma-
tion is isometric with a sphere [77].
Let G be the Lie algebra of SU(3) consisting of all skew-hermitian matrices
of trace 0 and H the Lie algebra of SO(3) consisting of all real skew-hermitian
matrices of trace 0. Let o denote the map sending an element of SU(3) into
its complex conjugate. Since SU(3)/S0(3) is symmetric and simply connected,
its homogeneous holonomy group is identical with G/H. It follows that the
action of SO(3) on G/H is irreducible. Hence SU(3)/S0(3) is irreducible.
That SU(3)/S0(3) does not admit a non-isometric conformal transformation
is a consequence of the fact that it is not isometric with a sphere in the given
metric.

F. The conformal transforination group [79]
1. Show that a compact homogeneous Riemannian manifold M of dimension
n > 3 which admits a non-isometric conformal transformation, that is, for
which Co(M) # Io(M) (cf. 93.7) is isometric with a sphere.
To see this, let G = Io(M) and M = GIK. The subgroup K need not be
connected. Since G is compact, it can be shown that the fundamental group