Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is a bit of a stab in the dark but I was wondering if anyone has defined cohomology classes on a hyperkähler manifold which pull back to the Stiefel-Whitney classes on any submanifold which is Lagrangian in two of the symplectic structures, which implies that it is symplectic in the third (and thus holomorphic in the compatible complex structure).

As evidence that this is not completely insane, let me note that

if two submanifolds Lagrangian in a symplectic structure $A$ and $B$ intersect cleanly, then the pull-backs of $w_i(A)$ and $w_i(B)$ to $A\cap B$ are equal.

Proof. Consider the symplectic form on the restrictions of the tangent bundles of $A$ and $B$ to the intersection. Its kernel is the tangent bundle of the intersection, and it pairs the normal bundles of the intersection in A and B non-degenerately. Since dual vector bundles have the same Stiefel-Whitney classes, the Whitney product formula shows that the Stiefel-Whitney classes of the pull-backs (and thus the pull-backs of the Stiefel-Whitney classes) agree.

I don't think that this is enough to show that a class like I have asked for exists, but it does show that one method of trying to prove that one doesn't will always fail.

EDIT: As Eric Zaslow points out, this can't work with just an arbitrary symplectic manifold, but at least the example he provides can't be done with holomorphic submanifolds.

share|improve this question
1  
What am I missing ? T^*R^N = R^{2N} has no cohomology classes in positive degrees, but the conormal bundle of a nonorientable submanifold of R^N has nonvanishing first Stiefel-Whitney class. –  Eric Zaslow Jan 7 '11 at 3:46
    
A related thing that you may already know, Ben: the Chern class $c_{2k}(TM)$ of a symplectic manifold $M$ restricts to $(-1)^k p_k(TL)$ for each Lagrangian $L\subset M$. This is because $TM|_L\cong TL\otimes \mathbb{C}$. –  Tim Perutz Jan 7 '11 at 3:55
    
@Eric- That's a good point. As you might guess I'm interested in applications that have a lot more structure than a general symplectic manifold. Since no one had answered yet, I shifted the goalposts and added the hyperkähler hypothesis, which avoids your counterexample. –  Ben Webster Jan 7 '11 at 4:32
    
@Tim- Sadly, I'm most interested in $w_2$, which is, of course, trivial on anything hyperkähler. –  Ben Webster Jan 7 '11 at 4:37
    
@Eric: The tangent bundle of the ENTIRE conormal bundle (which is the actual Lagrangian in question) is trivial. So this is not a counter example to the original question. –  Thomas Kragh Feb 4 '11 at 20:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.