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Is there any difference between the equivalent classes of $\mathbb R^n$ vector bundles and $\mathbb R^n$-fiber bundles? The first one is related to $K$ group. What is the second one? I am thinking they are the same for $\mathbb R^n$.

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I assume that by $\mathbb{R}^n$-fiber bundles you mean differentiable bundles with fiber diffeomorphic to $\mathbb{R}^n$. So you are forgetting the linear structure of $\mathbb{R}^n$, keeping only the smooth structure. It is a fact that $\textit{Diff}(\mathbb{R}^n)$ deformation retracts to $\textit{GL}(n,\mathbb{R})$, hence yes, the classification of rank $n$ real vector bundles is essentially the same as that of $\mathbb{R}^n$-fiber bundles. I don't know whether this holds at the topological level already (that is, whether $\textit{Homeo}(\mathbb{R}^n)$ deformation retracts to $\textit{GL}(n,\mathbb{R})$).

EDIT: it seems that at the topological level the two beasts are different, at least for some $n$. The result is contained in a paper by William Browder entitled "Open and closed disc bundles", Ann. of Math. (2) 83 (1966), 218-230. It is freely available on JSTOR. As a consequence of the results in that paper, there exists some $n$ for which $\textit{Homeo}(\mathbb{R}^n)$ does not deformation retract onto $\textit{GL}(n,\mathbb{R})$. As far as I know, $n>2$.

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Thanks John, I am considering topological $\mathbb R^n$ bundle so $Homeo$ is the one I was asking. –  J. GE Aug 7 '12 at 14:52
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As far as I understand, the difference between "Homeo(R^n)" and O(n) is studied quite a bit. See the introduction to the following paper of A. Ranicki & M. Weiss: arxiv.org/pdf/0901.0819v2.pdf. In particular, they claim that it is known that rational cohomology of BHomeo and BO are the same, but I've heard Weiss talk about how this is a difficult problem integrally (if I remember right). –  Ilya Grigoriev Aug 8 '12 at 6:06

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