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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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You want the fibers to be linear spaces? I always think of fiberbundles as bundles with non-linear fibers (e.g. circles, and the Hopf fibration) – Thomas Rot Mar 8 '12 at 15:45
No, that could not be father from what I want. I want a bundle so wild that you CAN'T replace the fiber with a linear space (to form an associated bundle). – you Mar 8 '12 at 15:49
If you look at the beginning of section 11 in Bott and Tu, they make a passing mention (without citation, unfortunately) of the existence of sphere bundles whose structure group cannot be reduced to the orthogonal group. – NKS Mar 8 '12 at 16:08
A vectorbundle is noncompact (if the fibers have dimension $\geq 1$). The hopf fibration is a compact space. So it cannot be a vector bundle – Thomas Rot Mar 8 '12 at 16:14
Re: Bott and Tu - a little googling turns up - "Differentiable sphere bundles", SP Novikov - Amer. Math. Soc. Transl, 1967; and a paper "Some examples of differentiable sphere bundles" H Taniguchi - Journal of the Faculty of Science, Imperial University, 1970. – Max Mar 11 '12 at 6:04
up vote 3 down vote accepted

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

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.

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