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By definition of a fibre bundle $F\hookrightarrow E\xrightarrow \pi B$ every point $p\in B$ has a neighbourhood $U$ such that there exists a diffeomorphism (local trivialization) $\phi: \pi^{-1} (U)\rightarrow U\times F$. Looking at local trivializations over overlapping open sets $U_i$, $U_j$ one can consider the map $\phi_j\circ\phi_i^{-1}:U_i\cap U_j\times F\rightarrow U_i\cap U_j\times F$, $\phi_j\circ\phi_i^{-1}(x,u)=(x,t_{ji}(x,u))$, with $t_{ji}(x,\cdot):F\rightarrow F$ a diffeomorphism. It follows from the definitions that $t_{ij}(x,t_{ji}(x,u))=u$. Similarly by considering triple intersections one has $t_{ij} (x,t_{jk}(x,u))=t_{ik}(x,u)$.

Usually in the literature a fiber bundle is called a $G$-bundle, or said to have structure group $G$, if it exists a group $G$ such that $t_{ij}(x,u)=\tau_{ij}(x)\cdot u$ with $\tau_{ij}(x)\in G$ and $\cdot$ denoting a left action of $G$ on $F$.

Take $G=\mathrm{Diff}(F)$, the group of diffeomorphisms of $F$, acting on $u\in F$ as $(\phi,u)\in \mathrm{Diff}(F)\times F\mapsto \phi(u)$. If $t_{ij}(x,u)=u^\prime$, define $\tau_{ij}(x)=\phi$, where $\phi$ is any element of $\mathrm{Diff}(F)$ such that $\phi(u)=u^\prime$. Doesn't this give $F\hookrightarrow E\xrightarrow \pi B$ the structure of a $G$-bundle? In other words, isn't every fiber bundle a $G$-bundle if I take $G$ to be the group of diffeomorphisms of the fibre?

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  • $\begingroup$ Apparently yes, since the action preserves the fibers. $\endgroup$
    – user40276
    Jan 23, 2014 at 22:00

1 Answer 1

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  1. I am not sure that your use to the terminology "$G$-bundle" is standard: In the literature I know one either says "principal $G$-bundle" or "bundle with the structure group $G$", the latter is what you mean by a $G$-bundle. With this caveat, yes, in the smooth category, every bundle with fiber $F$ is a $Diff(F)$-bundle. In the topological category, it is a $Homeo(F)$-bundle. For example, when one says "a circle bundle" this typically means "a bundle with circle fiber" rather than "a bundle with the structure group $S^1$".

  2. Addendum: There is yet another notion of $G$-bundle (different from yours but closely related in the principal bundle case): A bundle $\pi: E\to B$ is called a "$G$-bundle" if there exists an action of $G$ on $E$ by automorphisms: It commutes with $\pi$ in the sense that $\pi(g\cdot x)=\pi(x)$ for every $x\in E, g\in G$. A more complicated version of this definition requires $G$ to act on both $E$ and $B$ and $\pi$ to be $G$-equivariant. It seems clear that a general smooth bundle with typical $F$ does not admit a smooth action of $Diff(F)$ commuting with $\pi$, for instance, since this would require two commuting actions of $Diff(F)$ on $F$. One can see this most clearly in the case of zero-dimensional fibers (i.e., when $\pi$ is a covering map). Then there are examples when $F$ has more than 2 elements and $\pi$ has no nontrivial automorphisms at all.

The exceptional case happens when $\pi$ is a principal $G$-bundle, then you have two commuting actions of $G$ on itself, by left and right multiplication (one action, is a left action, the other is a right action). Then the bundle is a $G$-bundle in the sense of both definitions discussed in 1 and in 2.

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    $\begingroup$ G-bundle is a standard notation, the unique difference from principal G-bundle is that the action need not to be freely transitive, it just preserves fibers. $\endgroup$
    – user40276
    Jan 23, 2014 at 22:04
  • $\begingroup$ @user40276: Maybe, can you give some canonical examples? Mine is Kobayashi-Nomizu; Milnor-Stasheff. My guess, however, is that the name depends on the subfield in math/physics. $\endgroup$ Jan 23, 2014 at 22:31
  • $\begingroup$ I've just seen this term a lot in seminars and papers. $\endgroup$
    – user40276
    Jan 23, 2014 at 22:44
  • $\begingroup$ @user40276: I think, I understood what you meant by a G-bundle (it is different from the one which OP is asking about), see the edit. $\endgroup$ Jan 24, 2014 at 13:49

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