MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathbb R^n$.

share|cite|improve this question
up vote 11 down vote accepted

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, $n>2$.

share|cite|improve this answer
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
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: 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.