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.mathwonk said: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.
There's one obvious case without isolated zeros :)lavinia said: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.
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).mathwonk said:nice examples! can you give us two integrable almost complex structures on the same smooth manifold with different chern classes?
zhentil said: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.
In some dimensions, say 2^k, where k is odd. But if k is even, the Euler class is preserved.lavinia said:But isn't the conjugate bundle just changing orientation?
zhentil said:In some dimensions, say 2^k, where k is odd. But if k is even, the Euler class is preserved.