Smoothness of transition maps of fiber bundle

A smooth fiber bundle with fiber a smooth manifold without boundary $F$ and structure group $G$ a Lie group of diffeomorphisms of $F$ is a smooth surjective map $p:E\to M$ between manifolds, with a family $\{(U_\alpha , u_\alpha)\}$ of diffeomorphisms $u_\alpha : p^{-1}(U_\alpha) \to U_\alpha \times F$ such that the first component is $p$ and for every $x\in U_\alpha \cap U_\beta$ the diffeomorphism $u_{\alpha \beta} (x):F\to F$ such that $u_\alpha \circ u_\beta ^{-1} (x,f) = (x,u_{\alpha \beta} (x)f)$ is an element of $G$.

I am trying without success to prove that $u_{\alpha \beta} : U_\alpha \cap U_\beta \to G$ is smooth. I am able to prove this only in the "vector bundle" case. Can you help me?

Thank you

• What is your definition of "a Lie group of diffeomorphisms of $F$"? – Eric Wofsey Dec 12 '15 at 9:37
• @EricWofsey I mean a subgroup $G$ of $Diff(F)$ with a finite-dimensional, second-countable and Hausdorff Lie group structure on it such that the natural left action $G \times F \to F$ is smooth. I think that there is at most one way to put such a Lie group structure on G, so the definition is well-posed. – Ervin Dec 12 '15 at 9:42
• Proving that there is at most one such Lie group structure on $G$ seems about as hard as answering this question, actually... – Eric Wofsey Dec 12 '15 at 9:58
• @EricWofsey Just a comment about your comment. I proved that "uniqueness" in this way: let $a:G×M→M$ be the natural left group action of $G$ on $M$, and let $G_1,G_2$ be two differentiable structures over $G$ such that $a:G_i×M→M$ is smooth for $i=1,2$. Then $a$ is a submersion in both cases, so if $U×V$ is open in $G_2×M$ then it is open also in $G_1×M$. Therefore, $id_G:G1→G2$ is a continuous bijective hom of Lie groups and so it is a Lie group Isom, so $G_1=G_2$. Did you have that in mind? – Ervin Dec 12 '15 at 12:09
• The argument I had in mind was just to derive the uniqueness from my answer to the question, which shows that if you have any smooth map $U\times F\to U\times F$ of the form $(x,f)\mapsto(x,\varphi(g)f)$, the induced map $U\to G$ is smooth. Taking $U=G_1$ and the obvious map given by the action of $G_1$ on $F$, you get that the identity $G_1\to G_2$ is smooth. – Eric Wofsey Dec 12 '15 at 19:35

For each $m\in\mathbb{N}$, let $G$ act on $F^m$ separately on each coordinate. For each $x\in F^m$, let $G_x\subseteq G$ denote the stabilizer, and let $V_m=\{x\in F^m:\dim G_x=0\}$. Note that $V_m$ is an open subset of $F^m$, and that the action of $G$ defines a foliation of $V_m$ whose leaves are orbits of $G$. In particular, for any $x\in V_m$, second-countability of $G$ implies there are only countably many leaves of this foliation in any neighborhood of $x$ that are in the orbit of $x$ under $G$. It follows that given any smooth map $\varphi:U\to V_m$ and a point $y\in U$ such that $\varphi(U)$ is contained in the orbit $G\varphi(y)$, we can lift $\varphi$ to a smooth map $W\to G$ in some neighborhood $W$ of $y$.
Now for each $f\in F$ write $\mathfrak{g}_f$ for the Lie algebra of the stabilizer $G_f$, and note that since $G$ acts faithfully on $F$, we must have $\bigcap_{f\in F}\mathfrak{g}_f=0$. It follows that there are finitely many points $f_1,\dots, f_m\in F$ such that $\bigcap_{i=1}^m \mathfrak{g}_{f_i}=0$. That is, $V_m$ is nonempty for some $m$. Let $x\in V_m$ and $d:G\to F^m$ be given by $g\mapsto gx$.
Now returning to your situation, smoothness of $u_\alpha\circ u_\beta^{-1}$ implies that the composition $d\circ u_{\alpha\beta}:U_\alpha\cap U_\beta\to F^m$ is smooth. Since $x\in V_m$, the first paragraph says that $d\circ u_{\alpha\beta}$ locally lifts to a smooth map to $G$. But this lift is none other than $u_{\alpha\beta}$, so this means $u_{\alpha\beta}$ is smooth.
It seems to me the following can help. First show that there are a finite number of points $f_1, f_2, \cdots , f_k$ in the fiber such that the map $d: G \to F \times F \times \cdots \times F$ given by $$g \to (g.f_1 , g.f_2, \cdots, g.f_k)$$ is an immersion. Now notice that the composition $d \circ u_{\alpha \beta}$ is smooth due to your assumption. Since $d$ is an immersion it seems to me that you can conclude that $u_{\alpha \beta}$ is smooth, isn'it?
• You can conclude that $u_{\alpha\beta}$ is smooth if you know that it is continuous, but it takes a bit more work to show that it is continuous. – Eric Wofsey Dec 12 '15 at 10:59
• You are right, I forgot the continuity. Could you provide the details showing that $u_{\alpha \beta}$ is indeed continuous?. – Holonomia Dec 12 '15 at 11:12