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This should surely be well-known by I have not been able to find a good reference to the following question: Given a smooth fiber bundle $\pi\colon P \longrightarrow M$ over a smooth manifold $M$ with typical fiber $F$, one has the group of fiber-preserving automorphisms of $P$: a diffeomorphism $\Phi\colon P \longrightarrow P$ is called fiber-preserving if $\pi \circ \Phi = \phi \circ \pi$ for some smooth map $\phi\colon M \longrightarrow M$, which then turns out to be a diffeomorphism of $M$. If $\phi = \mathrm{id}_M$ then one calls $\Phi$ a gauge transformation. Clearly they form a normal subgroup $\mathrm{Gau}(P) \subseteq \mathrm{Aut}(P)$, being the kernel of the group morphism $\Phi \mapsto \phi$. Hence we get a subgroup of the diffeomorphism group as the image of this quotient $\mathrm{Aut}(P) / \mathrm{Gau}(P) \subseteq \mathrm{Diffeo}(M)$. Of course, the case of principal fiber bundles is of particular interest here.

It is now well-known and not too hard to show that all the small diffeomorphisms of $M$ are contained in this image: this can be done by using a (complete) connection and it's parallel transport.

My question is about the large diffeomorphisms: are they also in the image, i.e. is the whole diffeomorphism group isomorphic to this quotient $\mathrm{Aut}(P) / \mathrm{Gau}(P)$? What conditions of $P$ would guarantee this (beside being trivial...)?

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    $\begingroup$ If $P$ is a natural fibre bundle, then - by definition - any diffeomorphism is covered by a bundle automorphism. $\endgroup$
    – Johannes Ebert
    Commented Jun 28, 2015 at 14:10
  • $\begingroup$ Is there a specific $P$ you care about? I suspect that in most cases your group morphism $\mathrm{Aut}(P)\to\mathrm{Diff}(M)$ isn't onto, but things very much depend on $P$ and $M$. The case $P\to M$ is a covering map seems the easiest to analyse, what is the answer there? $\endgroup$
    – Igor Belegradek
    Commented Jun 28, 2015 at 14:15
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    $\begingroup$ @StefanWaldman: if you just want a counterexample, consider the selfcover of $T^2$ given by $(z,w)\to (z^p, w^q)$ for coprime $p,q>1$. The flip diffeomorphism $(z,w)\to (w,z)$ of $T^2$ doesn't lift to the cover becaus on the $\pi_1$ level the diagonal element $(a,a)$, where $a$ is a generator pf $\pi_1(S^1)$, maps by the cover to $(pa,qa)$, and then by the flip to $(qa, pa)$, so it is not in the image of the cover. $\endgroup$
    – Igor Belegradek
    Commented Jun 28, 2015 at 15:13
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    $\begingroup$ Here's an even simpler example: Let $M$ be two (disjoint) copies of the circle $S^1$ and let $P\to M$ be the $\mathbb{Z}/2$-covering that is trivial over one copy of the circle and nontrivial over the other copy. Thus, $P$ has three connected components. Then any fiber-preserving diffeomorphism $\Phi:P\to P$ must preserve the connected component that double covers its base. Hence $\phi$ cannot exchange the two circles in $M$. $\endgroup$
    – Robert Bryant
    Commented Jun 29, 2015 at 4:00
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    $\begingroup$ @Bence Racskó: Small diffeos are those which are generated by flows of vector fields. If one topologizes the diffeo group appropriate, they are the connected component of the identity. $\endgroup$
    – Stefan Waldmann
    Commented Sep 10, 2019 at 6:34

1 Answer 1

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Any fiber bundle $\pi:P \longrightarrow M$ with fiber $F$ is classified by a map $f_{\pi}: M \longrightarrow B Diff(F)$ where $B Diff(F)$ is the classifying space of the diffeomorphism group of $F$. If an automorphism $\phi: M \longrightarrow M$ lifts to $P$ then the bundle $\phi^*P$ is isomorphic to $P$ and hence the classifying map $f_{\pi}$ is homotopic to the classifying map $f_{\pi} \circ \phi$ of the bundle $\phi^*P$. This observation can be used to produce easy examples of diffeomorphisms $\phi: M \longrightarrow M$ which do not lift to diffeomorphisms of $P$. For example, take $F$ be the discrete manifold consisting of two points. Then $Diff(F) = \mathbb{Z}/2$ and fiber bundles with fiber $F$ are classified by maps $M \longrightarrow B\mathbb{Z}/2$, or equivalently, by elements in $H^1(M,\mathbb{Z}/2)$. Now take $M = \mathbb{T}^2 = \mathbb{R}^2/\mathbb{Z}^2$ to be the $2$-dimensional torus and let us identify $H^1(\mathbb{T}^2,\mathbb{Z}/2)$ with $\mathbb{Z}/2 \oplus \mathbb{Z}/2$. Let $P \longrightarrow \mathbb{T}^2$ be the $F$-bundle corresponding to the class $\alpha = (1,0) \in H^1(\mathbb{T}^2,\mathbb{Z}/2)$. Let $\phi: \mathbb{T}^2 \longrightarrow \mathbb{T}^2$ be the automorphism of $\mathbb{T}^2$ induced by the matrix $\left(\begin{matrix} 0 & 1 \\ 1 & 0 \\ \end{matrix}\right)$. Then $\phi^*(\alpha) \neq \alpha$ and hence the classifying map of $\phi^*P$ is not homotopic to the classifying map $P$. As a result the diffeomorphism $\phi$ does not lift to a diffeomoprhism of $P$.

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