Is there a fibre bundle with fibre homeomorphic to $\mathbb R^k$ which cannot be given the structure of a vector bundle?

Question

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.

Answer 1

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.