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?

See Torsten's comment above for an answer to the question you asked... And here's an answer to a question you didn't ask:

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.

This takes care of the stably parallelizable case... I don't know anything about the general case.

possiblydifferent and it sounds like you want to know if there's ever a case where they'reactuallydifferent. As Johannes showed in the question linked above, for the case of spheres the answer is 'no.' $\endgroup$ – Dylan Wilson Apr 26 '11 at 13:59