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

(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.

(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.

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Piotr Pstrągowski
Piotr Pstrągowski
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Action of U(1) on a sphere bundle, non-vanishing vector fields on odd-dimensional manifolds

(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.