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Is it possible to find a specific example of two fiber bundles with the same base, group, fiber and homeomorphic total spaces but these bundles are not equivalent/isomorphic, if so

should I find a bundle map F between two bundles, inducing identity on the common base but F does not preserve fibers? (I don't know what it means, got mixed up) or

should I define the action of group on fibers differently?

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  • $\begingroup$ for each (h,j) in Z+Z there corresponds an S^3 bundle over S^4 and when h+j=1 the total spaces are all homeomorphic to S^7 and the group is SO(4). I try to understand where the difference come from? $\endgroup$
    – wqr
    Commented Jun 29, 2012 at 23:33
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    $\begingroup$ Why do you ask wether such bundles exist if you already know examples of this behavior? $\endgroup$ Commented Jun 29, 2012 at 23:38

2 Answers 2

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EDIT: This answer is wrong: see Dan Ramras's comment.

The circle (thought of as the set of norm one elements in the complex plane) is a fiber bundle over itself in many ways, via the maps $z \mapsto z^k$ where $k \neq 0$ is an integer. These are all non-isomorphic as fiber bundles (which follows from the computation that the fundamental group of $S^1$ is $\mathbb{Z}$), but the fiber has $|k|$ points, so the bundles corresponding to $\pm k$ have the same fiber, total space, and base.

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  • $\begingroup$ (The group here is just $\mathbb{Z}/k$.) $\endgroup$
    – user29743
    Commented Jun 29, 2012 at 23:44
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    $\begingroup$ These bundles are of different group, though. $\endgroup$ Commented Jun 30, 2012 at 3:20
  • $\begingroup$ I don't understand - I am fixing $k$ and looking at the two bundles corresponding to $\pm k$. Surely both these bundles have structure group $\mathbb{Z}/k$, right? $\endgroup$
    – user29743
    Commented Jun 30, 2012 at 17:23
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    $\begingroup$ But the map $z\mapsto z^{-1}$ is a bundle isomorphism between the bundles for $k$ and $-k$. Covering spaces of $S^1$, as you say, are classified by subgroups of $\pi_1 (S^1) = \mathbb{Z}$, by sending a covering space to the image of its fundamental group under the projection map. Both of these covering spaces correspond to the subgroup $k\mathbb{Z} < \mathbb{Z}$. $\endgroup$
    – Dan Ramras
    Commented Jul 3, 2012 at 3:58
  • $\begingroup$ hard to argue with that... $\endgroup$
    – user29743
    Commented Jul 4, 2012 at 18:31
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In the paper "K-theory doesn't exist," (J. Pure Appl. Alg., 1978 vol 12) Akin explains that if $p: P\rightarrow B$ is a non-trivial principal $G$-bundle, the map $P\times G\rightarrow B$ (project to the first factor and then apply $p$) can be made into a principal $G\times G$-bundle in two different ways, by specifying two different actions of $G\times G$ on $P\times G$.

In general, these are not isomorphic as principal bundles (Akin shows that if they were always isomorphic, the $K$-theory of $B$ would be trivial, which is the origin of the paper's title).

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