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

(This question has been asked on Stack Exchange:

Suppose $N$ is a closed, $n$-dimensional, orientable smooth manifold. Moreover, $n$ is odd.

Consider the tangent bundle $TN$ of $N$. By adding a point at infinity to each tangent space I define a sphere bundle $E$ over $N$.

Notice, that every fiber is homeomorphic to $S^{n}$ and that odd-dimensional spheres admit a free action of $U(1)$ (coming from standard embedding into $\mathbb{C}^k$).

Is it possible to make $U(1)$ act on $E$, such that the action is free on every fiber? Can you define the action in such that locally over $N$ there exist trivializing charts for $E$ where the action is isomorphic to the standard action of $U(1)$ on $S^n$? (Isomorphism in this case probably should mean equivariant homeomorphism.)

I do not care if the action is smooth. Since this is MO I will explain the motivation for asking this question. I am trying to construct a non-vanishing vector field on a manifold $N$ as above (with possibly some more assumptions). This would imply that it has Euler number $0$. Is there any "geometric" way of doing this? By "geometric" I mean a way that makes it clear that odd-dimensionality is crucial.

share|cite|improve this question
How about starting with a generic vector field with finitely many zeroes? For example, use a Morse function on $N$ and then there would be a source and a sink and some other zeroes in between. The importance of odd lets you conclude that a source and sink contribute $+1$ and $-1$ respectively when you calculate the index. Moreover, since you can use $-f$ for $f$, this means that the sum of the indices cancel out. Now, one may try to gather the zeroes together; the final result would be a vector field with exactly one zero with zero index which perhaps may be modified locally to have no zeroes. – Somnath Basu Dec 11 '11 at 6:27
It looks (to me) that what you're asking is stronger than what you need, and that there should be obstructions to a positive answer to the first question. You're done if you can find an almost-complex structure on a codimension-1 sub-bundle of the tangent bundle to $N$ (you have this for all 3-manifolds, and in general for all manifolds with a contact structure). But there could (should?) be an obstruction to having this acs on codimension-1 sub-bundles. In any case, this wouldn't be an explicit construction. – Marco Golla Dec 11 '11 at 12:12
On the other hand, if you only care about constructing non-vanishing vector fields, you can use open books+induction on the dimension, just mimicking the Thurston-Winkelnkemper construction. – Marco Golla Dec 11 '11 at 12:19

Here is a proof that a compact oriented manifold $M$ of dimension $n$ with zero Euler characteristic has a nowhere vanishing vector field. There a vector field $X$ on $M$ with all zeroes non-degenerate. Since $\chi(M)=0$, we can split the zeroes into pairs such that in each pair there are points of indices $1$ and $-1$.

Take a ball $B$ containing one such pair in its interior. Identify $B$ with the unit ball in $\mathbb{R}^n$ and $TM|B$ with $TB$. We can write the field restricted to $S=\partial B$ as $x\mapsto f(x)$ where $x\in S$ and $f(x)\in\mathbb{R}^n$; since $X$ no zeroes on $S$, $f$ gives after normalizing a map $g:S\to S$, which has degree 0, since the vector field has two zeroes of the opposite signs inside $B$. By Hopf theorem (= maps between spheres of the same dimension are classified by their degrees up to homotopy; this can be proven with or without obstruction theory), $g$ is homotopic to a constant map $S\to S$. We can use the homotopy to extend $X|S$ to a nowhere vanishing field in $B$.

After doing this for each pair of zeroes of $X$ we are done.

Regarding the original question: introducing a $U(1)$-action seems (to me) to be a very expensive way of getting a non-zero vector field! I'm not sure this is always possible, but can't think of any examples either.

share|cite|improve this answer

A very closely related question is discussed here:

nowhere vanishing vector field on a manifold

share|cite|improve this answer
This is of course very nice but it only states that there exists such a vector field by obstruction theory if the Euler characterstic is 0, which it is in this case. I am interested in a direct construction of such a vector field. – Piotr Pstrągowski Dec 10 '11 at 17:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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