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Let $F\rightarrow E\stackrel{\pi}{\rightarrow} B$ be a fiber bundle with structure group $G$. We know that if we can reduce the structure group to a subgroup of $GL_n$ for some $n$ (Diffeo$(\mathbb{R}^n)$ suffices since it retracts to $GL_n$) then we can replace $F$ with $\mathbb{R}^n$ to form an associated vector bundle.

Question: For which bundles can we NOT reduce the structure group to a group acting linearly on $\mathbb{R}^n$ (for some $n$)?

I have a vague idea which involves choosing a sufficiently complicated manifold $Y$ and trying to construct a bundle with fiber $Y$ which in some sense "uses the whole structure group" Diffeo$(Y)$. If this "idea" is successful I imagine the resulting bundle will be very large. Is there a simpler example that I am missing?

I haven't had luck yet, but to be honest I haven't worked that hard on it. I was hoping someone here had already been exposed to this problem. (If you want my motivation, I'm wondering if the theory of fiber bundles has any characteristic classes that don't come from vector bundles)

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    @Max I loo$k$ed at the Novikov reference. Heavy stuff, but looks like it does provide an example of the phenomenon I'm looking for. I'll take a closer look.2012-03-12

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Consider the spaces $Homeo(\mathbb{R}^n)$ and $Diff(\mathbb{R}^n)$. The latter is contained in the former, and it is known that the inclusion is not a weak homotopy equivalence (and not even onto on homotopy groups). Unfortunately, the only decent source for this I found is https://mathoverflow.net/questions/96670/classification-of-surfaces-and-the-top-diff-and-pl-categories-for-manifolds

Anyway, start with a map $S^k \rightarrow Homeo(\mathbb{R}^n)$ which we cannot homotope into $Diff(\mathbb{R}^n)$. Then the usual gluing construction gives a bundle with fiber $\mathbb{R}^n$ over $S^{k+1}$ which does not allow a $Diff$-structure and hence no vector bundle structure.