# Parallelizability of exotic structure

I'd like to discuss a little bit about the problem I asked Diarmuid Crowley that whether all smooth structures on $S^7$ are parallelizable. He first came to using semi-characteristic but then came up with trivial answer which was : Because by chance we know $\pi_7(BO(7))=0$ so all smooth structures are parallelizable. But still I am curious to knpw why smooth structure does not play role here. Can we have trivial bundle in category of topological bundles but not in smooth category?

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–  Daniel Litt Apr 26 '11 at 3:54
On smooth manifolds there is no difference between smooth and topological vector bundles (up to isomorphism). This is seen by smoothing, the approximation of continuous sections by smooth ones. –  Torsten Ekedahl Apr 26 '11 at 3:55
The question you actually asked is not, I think, the question you really want to be answered. The tangent bundles on two homeomorphic spaces with different smooth structures are possibly different and it sounds like you want to know if there's ever a case where they're actually different. As Johannes showed in the question linked above, for the case of spheres the answer is 'no.' –  Dylan Wilson Apr 26 '11 at 13:59

Suppose you have homeomorphic manifolds $M^n$ and $\widetilde{M}^n$ that have distinct differentiably structures but are both stably parallelizable (i.e. they embed into a high-dimensional Euclidean space with a trivial normal bundle.) From the same paper that Hatcher cited in his answer to this question, namelt "Vector fields on $\pi$-manifolds" by Bredon and Kosinski, we have the following result:
Define a number $\chi^*(M^n)$ to be $\frac{1}{2}\chi(M^n)$ if $n$ is even and $\sum_{i=0}^rH_{i}(M, Z_2)$ (mod 2) for $n = 2r+1$. Then $M^n$ is parallelizable if $n=1,3,7$ or if $1-\chi^*(M) = 1$. (Is this the 'semi-characteristic' you mentioned in your answer?)
In particular, since $\chi^*$ is clearly a topological invariant, $M^n$ is parallelizable if and only if $\widetilde{M}^n$ is.