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Let $G$ be a compact Lie group which act on a manifold $M$. We fix this action throughout our question.

Assume that $E\to M$ is a vector bundle which has the potential of admitting a structure of a $G$-vector bundle.

Does the total space $E$ admit an action by $G$ such that the new equivariant bundle $p:E\to M$ satisfies $G_v=G_{p(v)}$, that is the projection map $p$ preserves the stabilizer.

This question is inspired by the following

Is this a submanifold?

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    $\begingroup$ Your condition is equivalent to asking that the action of $G_x$ on $E_x$ be trivial for every $x\in M$. This implies a necessary condition, that the restriction of $E$ to each orbit $Gx$ is a trivial bundle. But I doubt that's sufficient. $\endgroup$
    – Steve Costenoble
    Commented Mar 10, 2019 at 23:41
  • $\begingroup$ @SteveCostenoble very interesting point! What is an example of an equivariant vector bundle which fails the property "triviality on each orbit"? $\endgroup$
    – Ali Taghavi
    Commented Mar 12, 2019 at 6:13
  • $\begingroup$ One of the simplest would be $S^1\times_{{\mathbb Z}/2} {\mathbb R}$, with ${\mathbb Z}/2$ having the nontrivial action on $\mathbb R$. Thinking of this as a bundle over (the $S^1$-space) $S^1/({\mathbb Z}/2)$, it's the Moebius bundle. $\endgroup$
    – Steve Costenoble
    Commented Mar 12, 2019 at 16:38
  • $\begingroup$ @SteveCostenoble I think some thing is missing because it is well known that the mobius bundle over circle does not admit a nontrivial equivariant stricture.am i missing some thing? $\endgroup$
    – Ali Taghavi
    Commented Mar 12, 2019 at 20:36
  • $\begingroup$ $S^1$ here isn't acting on the circle in the usual way: $e^{i\theta}$ rotates by $2\theta$. $\endgroup$
    – Steve Costenoble
    Commented Mar 12, 2019 at 21:37

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