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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)2012-03-08
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    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).2012-03-08
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    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.2012-03-08
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    A vectorbundle is noncompact (if the fibers have dimension $\geq 1$). The hopf fibration is a compact space. So it cannot be a vector bundle2012-03-08
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    Compact bundles can still have associated vector bundles (for example, if you have an $n$-sphere bundle over a compact space it will be compact, but if the structure group is in $O_n$ then it is clearly associated to a vector bundle).2012-03-08
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    @you: I am sorry, I completely misunderstood your question2012-03-09
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    In that case, do you know anything about the problem, properly interpreted?2012-03-10
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    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, 1970.2012-03-11
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    Every bundle has an associated structure group, pretty much always. If there is a finite dimensional representation of this bundle then you can construct an associated finite dimensional vector bundle.2012-03-12
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    @Max I looked 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.