Page 64 - Curvature and Homology

46 I. RIEMANNIAN MANIFOLDS

Since abt" xy, these curves span a surface

If we think of a, b, c as variables and make the transformation x = a, y = b,
x = ce-*, it is seen that the integral surfaces are

Apply the above procedure to the form

and show that on the planes x = at, y = bt the surfaces 2 = f xy +y + c
are obtained whereas on the parabolic cylinders x = at, y = bP, the surfaces
obtained are 2 = xy + y + c. (This is not the case in the first example.)
Show that the reason integral surfaces are not obtained is given by 8 A do # 0.

1. Let P be a point of the n-dimensional differentiable manifold M of class
k and V, an r-dimensional subspace of the tangent space Tp at P. Put q - -n-r.
Let x(r, P) be a frame at P whose last r vectors eA(A, R, ... = q + 1, ..., n)
are in V,. Then, V, may be defined in terms of the vector3 81, ..., 8C of the dual
space T:, that is by the system of equations

The vectors of any other frame $(r,,P) satisfying these conditions may be
expressed in terms of the vectors of x(r, P) as follows:

It follows that af4 = 0 for i = 1, ..., q and A = q + 1, ..., n. Hence, the cor-
responding coframes (cf. D. 2) are given by

where the matrix (a:) c GL(q, R).
2. Conversely, let 81, ..., 89 be q linearly independent (over R) pfffin forms at P.
Let (P'), A = q + 1, ..., n be r pfaffian forms given in such a way that the
(P), u = 1, ..., n define a coframe (that is, the dual vectors form a frame).
The system of equations 8' = 0, ..., 69 = 0 then determines uniquely an