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Are Chern classes topological invariants? To be more precise: Given two complex manifolds $M$ and $N$.

Does a homeomophism $f:M\to N$ map Chern classes to Chern classes?

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I don't think so. The Chern classes of a vector bundle are certainly a topological invariant of the vector bundle. The issue here is that we're taking the Chern class of the (complexified) tangent bundle, and a homeomorphism isn't guaranteed to have the property that the pullback of the complexified tangent bundle is the complexified tangent bundle. Indeed up to homeomorphism one cannot even recover the smooth structure in general. –  Qiaochu Yuan Apr 17 at 5:10
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Of course, however Pontriyagin classes (complexify the tangent bundle and take only the even chern classes multiplied by $\pm 1$, but you probably know what they are) are topological invariants. So I also think the statement above is not true but I am not able to produce a counterexample. Best –  O. Straser Apr 17 at 7:44
    
It's been a while, but I think I remember the answer is yes and that it has to do with microbundles. –  Najib Idrissi Apr 17 at 19:24

1 Answer 1

up vote 4 down vote accepted

No. Even Chern numbers are not topologically invariant — an example was given by Borel and Hirzebruch in 1959 (Characteristic classes and homogeneous spaces, §24.11).

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Perfect. Thanks. –  O. Straser Apr 18 at 15:39

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