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(This question has been asked on Stack Exchange: http://math.stackexchange.com/questions/89920/action-of-u1-on-a-sphere-bundle)

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.

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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
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2 Answers 2

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.

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A very closely related question is discussed here:

nowhere vanishing vector field on a manifold

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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
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