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I have a group action of a group $G$ on a fiber bundle that "plays well" with the fiber structure (i.e. the action of each element maps fibers to fibers diffeomorphically). Can I assume that the diffeomorphism on the fiber is given by a group action of $G$ on the fiber? To make things maybe more clear consider the trivial bundle $M=\mathbb{T}^2\times\mathbb{S}^1$ with group action

\begin{eqnarray} \mathbb{S}^1\times M \rightarrow M\\ a\cdot (\phi,\psi, \theta)\rightarrow (\phi+a,\psi, \theta+a) \end{eqnarray}

Then there is an induced action of $\mathbb{S}^1$ on the fiber. But perhaps if $M$ is not a trivial bundle there are some pathological examples where such a thing is not true? Is such an action of a group on a fiber bundle necessarily given by (an action of the group on the base) composed with (an action of the group on the fiber)?

Note that I am obviously not asking that this be a priciple bundle or anything like that, just a general (smooth) fiber bundle.

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  • $\begingroup$ You write of "the diffeomorphism on the fiber". What diffeomorphism would that be? $\endgroup$
    – Lee Mosher
    Mar 15, 2018 at 17:43

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Consider $p:\mathbb{R}\times\mathbb{R}\longrightarrow\mathbb{R}$ defined by $p(x,y)=x$ and define the action of $\mathbb{R}$ by $t.(x,y)=(x,y+tx)$, on the fiber over $0$ the action is the identity, and it is not the identity on the others fibers, so the action is not induced by an action of $\mathbb{R}$ on $\mathbb{R}$.

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