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:

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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    $\begingroup$ (It's due to Poincare and Hopf and called their index theorem.) $\endgroup$ – Chris Gerig May 5 '13 at 19:36
  • $\begingroup$ For the nuclear option, Thurston proved that every manifold of Euler characteristic zero admits a co-dimension one foliation. Choosing a Riemannian metric, a unit normal vector field to this foliation defines a nowhere vanishing vector field. Though, I'll admit that I'm not sure that this argument isn't circular. $\endgroup$ – Andy Sanders Jan 14 '15 at 6:08
  • $\begingroup$ Related: math.stackexchange.com/questions/47370/… $\endgroup$ – Seirios Jan 14 '15 at 8:33

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.

  • $\begingroup$ Don't we need high dimensions to do a Whitney trick? $\endgroup$ – Dylan Wilson May 5 '13 at 18:55
  • $\begingroup$ @Dylan, formally you can get into trouble if $\dim M=2$, so $\dim TM=4$, but it is easy to construct embedded disc in our case. (Take a curve connecting the points and take the ruled disc over this curve.) $\endgroup$ – Anton Petrunin May 5 '13 at 19:01
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    $\begingroup$ This is more elementary than the Whitney trick. Two zeroes of opposite index can be taken to lie in a coordinate chart (because the manifold is connected) and then they can be eliminated using that fact that a map $S^{n-1}\to S^{n-1}$ of degree zero extends to $D^n$. $\endgroup$ – Tom Goodwillie May 5 '13 at 21:56

That a compact manifold M with vanishing Euler characteristic has a nonvanishing vector field was proved by Heinz Hopf, Vektorfelder in Mannifaltigkeiten, Math. Annalen 95 (1925), 340-367. A pretty convincing "intuitive proof" was outlined by Norman Steenrod in his book Fibre Bundles (Theorem 39.7), using a smooth triangulation and obstruction theory. For a complete proof he refers to page 549 of the 1935 book Topologie by P. Alexandroff and H. Hopf


Here is an outline of a proof that a compact manifold $M$ with vanishing Euler characteristic has a nonvanishing vector field. I'll post it as an "answer" to provide a convenient place for comments where errors might be pointed out (or, you know, a reference given that "that's the proof that ABC used in their paper LMNOP").

Call a vector field $\vec v$ with only regular zeros (i.e. at every $p$ with $\vec v|_p = 0$, the linear map $\partial \vec v|_p : \mathrm T_p M \to \mathrm T_p M$ is invertible) divergence-free if for every zero $p$, $\partial \vec v|_p$ has only real eigenvalues. A zero $p$ of a divergence-free vector field $\vec v$ has a Morse index $\mu(p) = \#\lbrace$negative eigenvalues of $\partial\vec v|_p\rbrace$. Of course, $\operatorname{sign}(\det(\partial \vec v|_p)) = (-1)^{\mu(p)}$. By choosing a Morse function and a metric, our manifold $M$ certainly has a divergence-free vector field.

Pick some vector field $\vec v$ on $M$, and some little neighborhood without any zeros. I claim I can modify $\vec v$ by some vector field with compact support in that neighborhood to introduce two new zeros, with Morse indexes $\mu$ and $\mu+1$, for any $\mu = 0,\dots,\dim M - 1$. In one dimension, which is trivial: in a neighborhood, $\vec v = \nabla(x^3+x)$ for some coordinate $x$, and I can perturb this to $x^3 - x$. In higher dimensions, it is not much harder, and I could probably write out formulas is necessary.

Conversely, and here's the crux of the argument, where I'm not sure it's correct: Suppose I have a divergence-free vector field $\vec v$, with nearby zeros at consecutive Morse index. Then I think I can cancel them, by undoing the insertion step in the previous paragraph.

If so, then choose any divergence-free vector field $\vec v$ on $M$. Since $M$ by assumption has vanishing Euler characteristic, $\vec v$ has the same number of zeros with odd Morse index as with even Morse index. If $\vec v$ has no zeros, we're done; otherwise, choose two with Euler characteristics $\mu$ and $\mu+k$, for $k\geq 1$ odd. By assumption, $M$ is connected; choose a simply path between the two zeros, and a small neighborhood thereof. Now, along this path, insert in pairs zeros with Morse index $\mu +1$, $\mu + 2$, ... $\mu + (k-1)$. Now cancel the zeros in pairs but this time cancel the zero with Morse index $\mu$ with the new one with index $\mu + 1$, and so on. After all this, you end up with a vector field with two fewer zeros than you started with.

Then the result follows by induction, provided the crux in the 4th paragraph is correct.

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    $\begingroup$ While this argument is, in the main, correct, you can see from Anton's proof that it really boils down to something quite a bit simpler. You might find a complete argument along these lines written out with care in differential topology textbooks such as Guilleman and Pollack. $\endgroup$ – Lee Mosher May 5 '13 at 18:58

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