Is there a fiber bundle $(E,B, \mathbb{R}^{n})$, with typical fiber $\mathbb{R}^{n}$, such that there is no any $n$-dimensional vector bundle structure on the pair $(E,B)$? That is there is no a continuous map $p:E \to B$ such that the triple $(E,B,P)$ would be a $n$ dimensional vector bundle.
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asked Oct 26, 2016 at 12:03
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$\begingroup$ You might want to look for the universal case $BTop(n)$ in the literature. $\endgroup$– user95545Commented Oct 26, 2016 at 14:05
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$\begingroup$ see "Open and Closed Disk Bundles" by William Browder, jstor.org/stable/1970428. $\endgroup$– Igor BelegradekCommented Nov 23, 2016 at 4:02
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2 Answers
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Such a fibre bundle does not exists if you suppose that it is endowed with a differentiable structure. Stewart has shown that the group of diffeomorphisms of $R^n$ retract to $O(n)$. So every $Diff(R^n)$-bundle has an $O(n)$-reduction.
Stewart, T. E. (1960). On groups of diffeomorphisms. Proceedings of the American Mathematical Society, 11(4), 559-563.
answered Oct 26, 2016 at 21:07
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$\begingroup$ Thank you very much for your interesting answer and helpful reference. $\endgroup$ Commented Oct 29, 2016 at 7:11
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Yes, there a lots of them.
The topological tangent bundle $\tau$ of any topological non-smoothable manifold $M$ gives you an example since a vector bundle reduction of $\tau$ is by smoothing theory equivalent to a smoothing of $M$.
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$\begingroup$ Thank you very much for your interesting answer. Unfortunately I can not accept two answers simultaneously. $\endgroup$ Commented Oct 29, 2016 at 7:13