Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider a compact simply-connected smooth $4$-manifold $M$ and its exotic copy $M'$. Identify the underlying topological manifolds. Dold-Whitney theorem guaranties ($\mathbb R$-linear) isomorphism of their tangent bundles. Can we say that these bundles are isomorphic over $\mathbb C$ in cases where the manifolds has almost complex structures?

More precisely, let be $b_2^+$ be odd. Then both manifolds admit almost complex structures. Assume that $M$ (and $M'$) is not spin, in this case there is at most one almost complex structure for a given first Chern class (see the answer). So the space of almost complex structures for both manifolds is parametrized exactly by characteristic elements $\underline{w}_2$ satisfying $\underline{w}_2^2=3sign(M)+2\chi(M).$

Is it true that for any choice of almost complex structures on $M$ and $M'$ with the same Chern class the tangent bundles are actually $\mathbb C$-isomorphic?

I'm not sure but it seems that in fact the argument of Peter Teichner in the mentioned answer still works here perfectly well, since the obstruction to the uniqueness is purely topological. If so, then I apologize for wasting your time.

share|cite|improve this question

1 Answer 1

Dold-Whitney is already a complete answer to every question of this form: once you know that the tangent bundles are isomorphic, any question you can ask about one tangent bundle has the same answer for the other one.

share|cite|improve this answer

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.