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?)

  • 3
    $\begingroup$ (It's due to Poincare and Hopf and called their index theorem.) $\endgroup$ May 5, 2013 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. $\endgroup$ Jan 14, 2015 at 6:08
  • $\begingroup$ Related: math.stackexchange.com/questions/47370/… $\endgroup$
    – Seirios
    Jan 14, 2015 at 8:33

3 Answers 3


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? $\endgroup$ May 5, 2013 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.) $\endgroup$ May 5, 2013 at 19:01
  • 17
    $\begingroup$ 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 $D^n$. $\endgroup$ May 5, 2013 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.

  • 2
    $\begingroup$ 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, 2013 at 18:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.