Page 149 - Curvature and Homology
P. 149
EXERCISES 131
of M is finite. Indeed, the first betti number of M is zero by theorem 3.7.5.
Secondly, M is conformally flat provided n > 3. For, if X is an infinitesimal
conformal transformation
4
= - 86 (C, C)
n
where C is the conformal curvature tensor. This formula is an immediate
consequence of (3.7.4) and the fact that B(X) C = 0. The manifold M being
homogeneous, and the tensor C being invariant by Io(M), (C, C) is a constant.
Therefore, if X is not an infinitesimal isometry, 86 # 0, from which (C, C) = 0,
that is, C must vanish. Hence, if n > 3, M is conformally flat.
Let I@ be the universal covering space of M. Since, the fundamental group
of M is finite, i@ is compact. Since M is conformally flat, so is I@. Thus, I@
is isometric with a sphere. We have invoked the theorem that a compact, simply
connected, conformally fit Riemannian manfold is conformal with a sphere [83].
The manifold M is consequently an Einstein space. It is therefore isometric
with a sphere (cf. 111.32.1).
