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.

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    $\begingroup$ You might want to look up the Novikov conjecture. $\endgroup$ Apr 18, 2016 at 15:25
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    $\begingroup$ 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. $\endgroup$ Apr 18, 2016 at 16:21
  • $\begingroup$ Thanks Mckay for editing them to make them more aware and Belegradek for giving an enlightening answer. $\endgroup$ Apr 23, 2016 at 14:29

1 Answer 1


This is answered in [Crowley, Diarmuid J.; Zvengrowski, Peter D, On the non-invariance of span and immersion co-dimension 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 non-isomorphic. 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. 575-579].

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

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