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We define a (rank $k$) vector bundle to be a fibre bundle with fibre $\mathbb R^k$ such that the local trivializations are fibre-by-fibre vector space isomorphisms. I'm curious whether this linearity condition is enough to stop certain fibre bundles from being turned into vector bundles.

Question: Given a fibre bundle $E\xrightarrow{p}B$ with fibre homeomorphic to $\mathbb R^k$, can we find

  • an open cover $(U_\alpha)$ of $B$ and
  • local trivializations $\phi_\alpha\colon p^{-1}(U_\alpha)\to U_\alpha\times \mathbb R^k$

such that the local trivializations are fibre-by-fibre vector space isomorphisms?

I expect that the answer is 'No', but I'd be very interested to see a counterexample.

A counterexample might take the form of a fibre bundle which demonstrably has no global section, or it might be a fibre bundle which cannot admit $GL_k(\mathbb R)$ as a structure group.

I'd be particularly interested if the example had the affine group $Aff(\mathbb R^k)$ as a structure group.

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    $\begingroup$ Any fiber bundle with contractible fiber has a section, so a counterexample of the first form does not exist. $\endgroup$ Commented Apr 1, 2015 at 21:15
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    $\begingroup$ Couldn't you get this with an unsmoothable topological manifold? You get an $\mathbb{R}^n$ bundle with structure group the homeomorphisms of $\mathbb{R}^n$ that fix the origin, that cannot be reduced to the structure group of diffeomorphisms fixing the origin, and so it certainly cannot be reduced to the general linear group. Of course, this doesn't work if you assume your base manifold is smooth. $\endgroup$
    – Carl
    Commented Apr 1, 2015 at 21:18
  • $\begingroup$ To put a fine point on it: so every such bundle with structure group $Aff(\mathbb{R}^k)$ can be turned into a vector bundle? $\endgroup$
    – Billy O.
    Commented Aug 19, 2018 at 22:43

1 Answer 1

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This question is answered here :https://mathoverflow.net/questions/104869/examples-for-open-disc-bundle-which-is-not-vector-bundle

Quoting the mathoverflow question:

William Browder showed in "Open and closed disc bundles", Ann. of Math. (2) 83 (1966), 218-230 that there are open disc bundles over some complex which cannot be isomorphic to a vector bundle.

Theorem 1 of the above paper gives the desired counterexample.

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