MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Milnor's book, Topology from the Differentiable Viewpoint, the definition of the index of a smooth vector field with isolated zeros on a manifold does not seem to require that the manifold is orientable. The definition is local i.e. defines the index around each zero in a small open ball.

So it seems then that the Poincare-Hopf theorem can be proved for an unorientable manifold even though the references I've found through Google state that the manifold is orientable.

The proof extends the vector field to a tube around the manifold so that the extended vector field points outward along the boundary of the tube and is homotopic to the Gauss mapping of the boundary into the unit sphere. I think that the sum of the indices of the vector field on the manifold ,even if it is not orientable ,therefore equals the degree of the Gauss mapping because the tube is orientable.

Is there a mistake in this reasoning?

share|cite|improve this question
Right. The statement makes sense without orientability, and the proof works, too, although there are surely many proofs that are easier to explain under the assumption of orientability. Besides, the general case follows from the orientable case by considering a two-sheeted covering space. – Tom Goodwillie Nov 19 '11 at 15:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.