We know that the tangent bundles of the sphere arising from different smooth structures are equivalent as vector bundles. Is it right in general? I want to know the relationship between the set of smooth structures and these tangent bundles.

3$\begingroup$ You might want to look up the Novikov conjecture. $\endgroup$– Mikhail KatzApr 18, 2016 at 15:25

1$\begingroup$ It seems like a negative answer to the (open) Novikov conjecture would give an example here where the tangent bundles are different. That doesn't directly imply that the original question is open, however. $\endgroup$– Dylan ThurstonApr 18, 2016 at 16:21

$\begingroup$ Thanks Mckay for editing them to make them more aware and Belegradek for giving an enlightening answer. $\endgroup$– Jialong DengApr 23, 2016 at 14:29
1 Answer
This is answered in [Crowley, Diarmuid J.; Zvengrowski, Peter D, On the noninvariance of span and immersion codimension for manifolds, Arch. Math. (Brno) 44 (2008), no. 5, 353–365], see here.
Specifically, in each dimension $>8$ there is a closed PL manifold admitting two smooth structures whose tangent bundles are nonisomorphic. One tangent bundle is trivial and the other one has nonzero second Pontryagin class. See remark 1.3.
Such examples do not exist in dimensions $\le 8$ by Corollary 2.6.
In dimensions $\ge 18$ this was known since 1969 and due to Roitberg in [On the PL noninvariance of the span of a smooth manifold, Proceedings of the American Mathematical Society Vol. 20, No. 2 (Feb., 1969), pp. 575579].

1$\begingroup$ I think the first example was actually given by Milnor in his 1963 paper Microbundles, Part I, namely Theorem 9.2. $\endgroup$ Mar 2, 2019 at 4:37

1$\begingroup$ @MichaelAlbanese: True, even though Milnor's example is for tangent bundles of open manifolds. $\endgroup$ Mar 2, 2019 at 12:17