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

Let $n$ be a positive integer.
Let $\langle M,\text{charts} \rangle$ be an $n$-dimensional $C^1$ manifold such that the
induced topological space is homeomorphic to the $n$-dimensional sphere.

Does it follow that ((all continuous vector fields on $\langle M,\text{charts} \rangle$ have a zero) if and only if ($n$ is even))?

share|cite|improve this question
Yes, this is a consequence of the Poincare-Hopf Index Theorem. – Ryan Budney Sep 26 '11 at 6:05
In particular, on an $n$-sphere (regardless of which differentiable structure it has) there is a vector field with precisely $|\chi(S^n)|$ zeros, each zero having index $1$, where $\chi(S^n)$ is the Euler characteristic of $S^n$, i.e. $0$ if $n$ odd and $2$ if $n$ even. – Ryan Budney Sep 26 '11 at 6:07
Is there any citation for that theorem only using $C^1$-ness? Everything I've found for it either does not say what amount of differentiability is being assumed or does not even state that it must be a differentiable manifold (which is necessary for the problem to even make sense). – Ricky Demer Sep 26 '11 at 6:14
If the manifold is $C^1$ it admits a compatible $C^\infty$ atlas. Similarly, if you have a continuous vector field without zeros, you can smooth it to a $C^\infty$ vector field without zeros. So I think the $C^0$ vector field case reduces directly to the $C^\infty$ case, provided the manifold admits any kind of smooth structure, like say a $C^1$ structure. – Ryan Budney Sep 26 '11 at 6:20
Is that basically a folk theorem? Do you know any reference for it? – Ricky Demer Sep 26 '11 at 6:27

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.