Why does Chern class belong to INTEGER cohomology class?

[lavinia]

so it seems (total) chern classes are not smooth invariants on the category of almost complex manifolds, but it is still not clear to me whether they are smooth invariants on the category of complex manifolds, or maybe of non singular complex projective varieties.

That's what the paper says about almost complex structures. For complex manifolds I would be surprised if all of the Chern classes were the same because the number of complex structures seems large for any given manifold. Even for non-simply connected Riemann surfaces there is a continuum of complex structures although for surfaces the only Chern class is the Euler class.

 

[mathwonk]

back to the original question. it seems from reading milnor and stasheff that the reason a chern class is integral is that the euler class, i.e. the top chern class, is integral. Then the other chern classes are defined inductively as euler classes of other induced bundles. since the euler class of every bundle is integral, every chern class is integral.

 

[mathwonk]

here is a related result. the canonical class of an algebraic surface is determined by the diffeomorphism type.

http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.bams/1183554176

 

[zhentil]

so your example either proves that chern classes are not smooth invariants, or that the two bundles are not smoothly equivalent.

so i guess you are saying it proves they are not smooth invariants? yes it seems so.

In general the bundles will be smoothly equivalent (perhaps not orientation-preserving), but may not be equivalent as U(n) bundles. It's quite easy to construct counterexamples: given an almost complex structure on the tangent bundle with non-trivial first Chern class, the conjugate complex structure will also yield an almost complex structure, but these can't be isomorphic as complex bundles, since they have different first Chern classes.

 

[zhentil]

here is a related result. the canonical class of an algebraic surface is determined by the diffeomorphism type.

http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.bams/1183554176

This is a very strong result, though. Deformation equivalence clearly implies diffeomorphism, but the other way is nowhere near as obvious.

 

[zhentil]

that interesting paper seems to be at least derived from brown's 2006 phd thesis with segert at mizzou, in columbia, mo.

are you a topologist? i always wanted to be a differential topologist but ultimately went in a different direction (algebraic geometry). It seems complex algebraic geometry yields a lot of interesting examples in topology though, and not just for 4 manifolds.

Haha, I'm looking at it from the other side. I'm a differential topologist (in training) who looks at algebraic surfaces because of all the interesting examples.

 

[zhentil]

Every oriented vector bundle has an Euler class. Every complex vector bundle is canonically orientable.
Every holomorphic section of a complex line bundle over a Riemann surface must have isolated zeros - I think.

I think this is the two ways that holomorphic comes in.

There's one obvious case without isolated zeros :)

A non-zero holomorphic section will have isolated zeros. Convergent power series, and all that.

 

[mathwonk]

nice examples! can you give us two integrable almost complex structures on the same smooth manifold with different chern classes?

 

[zhentil]

nice examples! can you give us two integrable almost complex structures on the same smooth manifold with different chern classes?

I'll have to think on this for a bit, assuming you mean that we're not allowed to change the orientation on the tangent bundle. Constructing such an example is basically a problem in cohomology (Noether's formula and forcing the Nijenhuis tensor to be zero).

 

[lavinia]

In general the bundles will be smoothly equivalent (perhaps not orientation-preserving), but may not be equivalent as U(n) bundles. It's quite easy to construct counterexamples: given an almost complex structure on the tangent bundle with non-trivial first Chern class, the conjugate complex structure will also yield an almost complex structure, but these can't be isomorphic as complex bundles, since they have different first Chern classes.

But isn't the conjugate bundle just changing orientation?

 

[zhentil]

But isn't the conjugate bundle just changing orientation?

In some dimensions, say 2^k, where k is odd. But if k is even, the Euler class is preserved.

 

[lavinia]

In some dimensions, say 2^k, where k is odd. But if k is even, the Euler class is preserved.

yes I know that. I am just saying that while the Chern classes change - they only change in sign so their topological content is unchanged.