Page 118 - Curvature and Homology

100 111. RIEMANNIAN MANIFOLDS: CURVATURE, HOMOLOGY

Lemma 3.4.1. Let $ be a dzflerentiable map sending M into M' and X
a vector field on M. Then, the vector field $,(X) on M' generates the
I-parameter group $ y, $-l where yt is the I-parameter group generated
by x.
The proof is entirely straightforward.
A vector field X on M is said to be invariant by + : M + M if
$,(X) = X. Therefore, by the lemma, X is invariant by $, if and only
if $ commutes with cp, for every t.

Lemma 3.4.2. Let f be a dzrerentiable function (of class 2) defined in a
neighborhood of 0 E R. Assume f(0) = 0. Then, there is a dzrerentiable
function g defined in the same neQhborhood such that f(t) = tg(t) and
g(0) = f'(0) where f' = df/dt.
We remark that the lemma is trivial iff is analytic. The proof is
given by setting

g(t) = '(ts)dr.
0
The function g is of class one less than that off in general. It is
important that f be of class 2 at least. For, otherwise g may not be
differentiable. To see this, let

f(') = 1 t2, t 2 0,
Then, g(t) = I t 1. - t2, t 5 0.

Corollary. Let f be a dzrerentiable function on U x M where U is
a neighborhood of 0 E R and M is a d~flerentiable manifold. If f(0, P)
= 0 for every P E M, then there is a dz;f)cerentiable function g on
U x M with the property that f(t, P) = tg(t, P) and (af /at),,,p, = g(0, P)
for every P E M.
This is an immediate consequence of lemma 3.4.2.
For any two infinitesimal transformations X and Y of M, YX is
not in general an infinitesimal transformation. In fact, if M = En and
Xf = af/aul, Yf = af/au2, we have YXf == a2f/au2au1. Clearly, the map
f -+ (a2f/au2&& (P E En) is not a tangent vector on En. However,
one may easily check that the map XY - YX is a vector field on M.
We shall denote this vector field by [X, Y]. The bracket [X, Y] evidently
satisfies the Jacobi identity

and so the (diflerentiable) vector Jields on M form a Lie algebra over R.