# If a manifold M has zero Euler characteristic, there is a non-vanishing vector field on it

There is hint: if M has isolated singular points, find a diffeomorphism to make these singular points in a any neighborhood which you want. How can we do next?

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@Javier That explains why zero Euler char is necessary condition for existence of a non-vanishing vector field. But OP asks why it is sufficient — and AFAICS the linked post doesn't help much. – Grigory M Jun 25 '11 at 18:45
@Grigory M: Indeed, you are right!! Sorry, I read the problem backwards as I had just posted about index theorems giving as an amusing consequence the hairy ball theorem and thought it may be of conceptual help. Your obstruction solution is of much more help anyway. – Javier Álvarez Jun 25 '11 at 19:01
@Javier Actually... you were right, it's just degree theory, essentially. – Grigory M Jun 26 '11 at 12:00

// Let me begin with an obstruction-theoretic solution — in hope that someone might find it of interest.

(For simplicity, let $\pi_1(M)=0$.) Non-vanishing vector field is a section of the spherization of $TM$, a bundle with fiber $S^{n-1}$. Obstructions to finding a section of a bundle with fiber $S^{n-1}$ lie in groups $H^k(M;\pi_{k-1}S^{n-1})$. These groups are trivial for $k<n$ (since coefficients are trivial) and for $k>n$ (since $M$ is $n$-dimensional). So the only non-trivial obstruction is the principal obstruction $\chi\in H^n(M;\pi_{n-1}(S^{n-1}))=\mathbb Z$.

And it's not hard to show, that it coincides with Euler char (indeed, the value of the obstruction on an $n$-cell is the degree of vector field on the bounding sphere — which coincides with the sum of indices of singular points of an extension of the field inside the cell).

// Reference (obstruction-theoretic approach to char. classes): Milnor-Stasheff, section 12.

This also explains, how to solve the problem directly (actually, it's the same solution in slightly different language). Take any vector field $v$ on $M$, and choose some sphere, containing all singular points. Degree (aka index) of the field on the sphere is exactly $\chi(M)=0$ — which means exactly that there is a non-vanishing extension of the field from the sphere to the ball (coinciding with $v$ on the boundary, but not inside the ball).

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I don't see how having $\chi(M)=0$ means the extension is non-vanishing. – Joe Johnson 126 Jul 27 '12 at 2:43