A version of the "hairy ball" theorem, due probably to Chern, says that the Euler-characteristic of a closed (i.e. compact without boundary) manifold $M$ can be computed as follows. Choose any vector field $\vec v \in \Gamma(\mathrm T M)$. If $p$ is a zero of $\vec v$, then the matrix of first derivatives at $p$ makes sense as a linear map $\partial\vec v|\_p : \mathrm T\_p M \to \mathrm T\_p M$. By perturbing $\vec v$ slightly, assume that at every zero, $\partial\vec v|\_p$ is invertible. Then $\chi(M) = \sum\_{\vec v(p)=0} \operatorname{sign}\bigl( \det \bigl(\partial\vec v|\_p\bigr)\bigr)$.


I am curious about the following potential converse: "If $M$ is closed and connected and $\chi(M) = 0$, then $M$ admits a nowhere-vanishing vector field."


Surely the above claim is false, or else I would have learned it by now, but I am not sufficiently creative to find a counterexample. Moreover, I can easily see an outline of a proof in the affirmative, which I will post as an "answer" below, in the hopes that an error can be pointed out. Thus my question:



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> What is an example of a compact, connected, boundary-free manifold with vanishing Euler characterstic that does not admit a nowhere-vanishing vector field? (Or does no such example exist?)
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Does a connected manifold with vanishing Euler characteristic admit a nowhere-vanishing vector field?
Yes, if M is closed and connected and χ(M)=0, then M admits a nowhere-vanishing vector field.


Start with generic vector field, it has zero of index $\pm 1$.
Two zeros of opposite sign can kill each other (maybe it is called Whitney trick?).


So you get a field with zeros of the same sign.
The result follows since the sum of the indexes is the Euler characteristic.


