Page 71 - Curvature and Homology

EXERCISES 53
By the usual 'polarization trick':
(2g(X,Y) = g(X + Y, X + Y) - g(X,X) - g(Y,Y)), a symmetric linear
transformation is determined by the quadratic form corresponding to it. Hence
L is determined by
&E), n

where the bivector f runs through Aa(T). It is sufficient to consider only
decomposable t. Consequently, L is determined by the sectional curvatures

of the planes spanned by X and Y for all X,Y E T.
2. Put
R(X, Y)Z = L(X A Y,Z)
and show that R(X,Y) is a tensor of type (1,l). The sectional curvature deter-
mined by the vectors X and Y may then be written as

For any set {xi, Xj, X,, X,} of orthonormal vectors, show that

3. Show that the curve C in the orthogonal group of Tp given by the matrix
(C(t)i) defining the parallel translation of Tp around the coordinate square
with corners (a) ui = udP), uj = uXP), (b) ui = uAP) + d< uj = u,(P),
(c) ui = uAP) + .\/i;, uj = uXP) + .\/i; (d) ui = ui(P), u, = uXP) + 46 all
other u's constant has derivative

J. Principal fibre bundles
1. Given a differentiable manifold M and Lie group G we define a new
differentiable manifold B = B(M,G) called a principal fibre bundle with base
space M and strwtwal group G as follows:
(i) The group G acts differentiably on B without fixed points, that is the
map (x,g) -+ xg, x E B, g E G from B x G + B is differentiable;
(ii) The manifold M is the quotient space of B by the equivalence relation
defined by G;
(iii) The canonical projection w : B -+ M is differentiable;