By Molk J. (ed.)

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Proof. In particular, if M does not admit a continuous k-plane field, then M does not possess a k-dimensional foliation. For example, the sphere S 5 does not 36 Geometric Theory of Foliations have a continuous 2-plane field (see 1531 page 142); so, there do not exist foliations of dimension 2 on S s . A natural question then is the following. Given a k-plane field P on M, under what conditions does there exist a k-dimensional foliation a such that, for each q E M, Tq g = P (q)? This question is answered by the theorem of Frobenius whose statement follows.

A fibered space (E, B such consists of differentiable manifolds E, B, F and a submersion r : E that for every b E B there is an open neighborhood Ub of b and a diffeomorphism : 7r- ( ) 11h x F which makes the following diagram commute: ( Ub ) Sob Ub X F / PI Ur, In the above diagram P 1 is the projection onto the first factor. The fibers ( b ), b E B. of the fibration are the submanifolds An important example of this situation is given by the following theorem, whose proof the reader can find in [27].

In particular, M is orientable. §7. Orientable and transversely orientable foliations In the theory of foliations, more useful than the notion of orientability is that of transverse orientability. Foliations 39 Let P be a k-plane field on M. We say that I is a field complementary to P or transverse to P if for every x E M we have P(x) + I3 (x) = Titi and P(x) fl 15 (x) = 101. It is a plane field of codimension k. If P is CT it is possible to define a complementary CT field as follows. Fix on M a Riemannian metric < , ) .