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Let $p:E \longrightarrow M$ be a smooth fibre bundle, with standard fibre space $F$ and $G$ a Lie group acting effectively on $F$ as a structure group.

Then, are the transition functions always smooth?

I mean, if for every open $U\subseteq M$ and function $g:U \longrightarrow G$ such that the function: $$ H:U\times F \longrightarrow U \times F$$ $$ H(x,s)=(x,g(x)\cdot s)$$ It's a diffeomorphism. Then, does it follow that $g$ is smooth?

PS: In Michor's book, natural operators in differential geometry, I read a similar fact in the step 5 of 9.11, but I don't see why it's true. https://www.mat.univie.ac.at/~michor/kmsbookh.pdf

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  • $\begingroup$ Do you mean $f=g$? $\endgroup$
    – Ben McKay
    Commented Jun 9, 2020 at 11:24
  • $\begingroup$ Yes, my fault sorry. Thanks. I edited it $\endgroup$
    – alexpglez98
    Commented Jun 9, 2020 at 16:21

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It is true but tricky to prove; I can't think of a reference off hand. You want to know that that an effective Lie group action allows one to smoothly determine each element $g$ by knowing ``how it acts''. You can prove this by thinking about the action of $G$ on the bundle of jets of local coordinates. The effectiveness, and analyticity of a Lie group acting on a homogenous space, ensures that if you take high enough jets, the stabilizer is trivial. That ensures that you can solve for an element $g$ uniquely and smoothly if you are giving the information of what $g$ does to a high enough jet.

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  • $\begingroup$ Thanks. I'm not familiar with Jets enough to undesrtand it, but I think I catch the idea. Is it like calculate the "Taylor expansion" on $s$ of $g(x)\cdot s$ and recover $g(x)$? $\endgroup$
    – alexpglez98
    Commented Jun 9, 2020 at 16:35
  • $\begingroup$ That is the idea. But you only need to compute Taylor series up to some finite order to get this to work. The group action cannot fix a point and fix an analytic coordinate system at that point without fixing all nearby points. The Hilbert basis theorem can prove that the subgroup fixing that point can only fix a finite number of Taylor coefficients of a coordinate system without fixing all points in the connected component of that point. At least when there are finitely many components to the homogeneous space, this should do the trick. I think there is a way to deal with infinite components. $\endgroup$
    – Ben McKay
    Commented Jun 9, 2020 at 16:51
  • $\begingroup$ Mmm, I see that I have to study more to completely understand it. The analyticity property of the action means that the action can be recovered by the Taylor coefficients? Is it necessary the action to be transtitive? You've mentioned that $(G,F)$ is a homogeneus space. Do you have any recommendation to read these properties about Jets and group actions? $\endgroup$
    – alexpglez98
    Commented Jun 9, 2020 at 17:22
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About step 5 in 9.11:

The first line of the display in step 5 gives a formula for the transition functions involving mappings $x\mapsto c^x_{\alpha}$ which are smooth enough (see 4 lines above this), and parallel transport which is smooth even in the choice of curves (if they depend on finitely many parameters) by theorem 8.9, 5 pages before.

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