Action of U(1) on a sphere bundle, non-vanishing vector fields on odd-dimensional manifolds (This question has been asked on Stack Exchange: https://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. 
 A: 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.
A: A very closely related question is discussed here:
nowhere vanishing vector field on a manifold
