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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.

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What do you mean the standard action of $U(1)$ on $S^{n}$? Rotation is only defined to some axis in dimension higher than 2. – Kerry Dec 9 '11 at 16:15
I'm guessing he means multiplication by an element in $U(1)$ (viewing $S^{2k-1}$ as a subspace of $\mathbb C^{k}$). – Eric O. Korman Dec 9 '11 at 16:28
I think yes. Thank you. – Kerry Dec 9 '11 at 16:44
This is exactly what I mean. – Piotr PstrÄ…gowski Dec 9 '11 at 16:45
For me, this seems to be the principle fiber bundle under the action of U(1), right? – bonnnnn2010 Dec 9 '11 at 17:24

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