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

Every symplectic form on a manifold $M^n$ determines a De Rham cohomology class in $H^2(M)$ (often a nontrivial class), and this in turn determines a class in $H_{n-2}(M)$. What in general can be said about this class? For example, over the rationals this class is represented by a submanifold of $M$; is it possible to explicitly describe such a submanifold in terms of the symplectic structure?

If there is a nice answer to this question, does it also shed light on the Poincare duals of $\omega^2$, $\omega^3$, etc?

share|improve this question
1  
I would wager a lot of money that the Poincare dual of the standard symplectic form on $\mathbb{C}P^n$ is $\mathbb{C}P^{n−1}$, but haven't yet done the calculation. Even if I can do it, I would still like some nice symplectic machinery that does the calculation for me. –  Paul Siegel Oct 15 '10 at 21:41
    
Some simple general nonsense tells you that $H^2(M) \equiv [M,\mathbb CP^\infty]$. $\mathbb CP^\infty$ contains a co-dimension $2$ sub-$\mathbb CP^\infty$ and the co-dimension $2$ submanifold of $M$ corresponding to a map $M \to \mathbb CP^\infty$ is the pre-image of this co-dimension $2$ sub-$\mathbb CP^\infty$, once the map is made to intersect it transversely. So provided you have this formulation of your symplectic structure you can "readily" get at this manifold. –  Ryan Budney Oct 15 '10 at 21:42
    
I don't think this really sheds much light on symplectic forms. Perhaps the only thing this is saying is that you can interpret a symplectic structure on a manifold $M$ to be Poincare dual precisely to an oriented co-dimension $2$ submanifold $N$ of $M$ such that you can take $n/2$ copies of $N$ in $M$, perturb them so that their total intersection is transverse, and the algebraic intersection number of the total intersection is a non-zero integer. –  Ryan Budney Oct 15 '10 at 22:01
    
it seems that we're assuming that the real cohomology class of the symplectic form comes from an integral class, in other words that the periods are integers. –  Tom Goodwillie Oct 15 '10 at 22:07
    
@Ryan: Yes, I must have slipped up on the TeX. It got so weird that I had a little trouble finding the X to click to delete the comment. –  Tom Goodwillie Oct 15 '10 at 22:13
add comment

2 Answers

up vote 23 down vote accepted

One of the big advances in symplectic topology in the 90s was Donaldson's theorem that when the symplectic class is integral, high multiples of its dual are represented by symplectic submanifolds.

These submanifolds behave like hyperplane sections in algebraic geometry; for instance, they satisfy the Lefschetz hyperplane theorem. They form the fibres of "symplectic Lefschetz pencils". Their intersections can be made to give symplectic submanifolds dual to multiples of wedge-powers of $\omega$.

Imagine first that $M$ is a compact complex manifold, $L\to M$ a hermitian, holomorphic line bundle, whose Chern connection has curvature $-2\pi i\omega$, a closed $(1,1)$-form. Then the zero-set of a $C^\infty$ section $s$ of $L^{\times k}$, if cut out transversely, is dual to $k[\omega]$. If $\omega$ is positive, the Kodaira embedding theorem then tells us that $L$ is ample: its high powers have enough holomorphic sections to embed $M$ into projective space. If $M$ is merely symplectic, with $-2\pi i\omega$ the curvature of some unitary connection in a hermitian line bundle, we can choose an almost complex structure $J$ on $M$ and consider transverse sections $s_k$ of $L^{\otimes k}$ for which, asymptotically, the $(0,1)$-part of $\nabla s_k$ along $s_k^{-1}(0)$ is much smaller than the $(1,0)$-part. Then $s_k^{-1}(0)$ will not quite be a $J$-holomorphic submanifold, but for $k \gg 0$ its tangent spaces will be close enough to being $J$-linear that it will still be a symplectic submanifold.

References: S. K. Donaldson, "Symplectic submanifolds and almost-complex geometry", J. Differential Geom. Volume 44, Number 4 (1996), 666-705; "Lefschetz pencils on symplectic manifolds", J. Differential Geom. Volume 53, Number 2 (1999), 205-236.

These papers are brilliant both geometrically and analytically: the analysis is mostly low-tech but extremely subtle.

share|improve this answer
1  
This sounds totally fascinating - thanks for the explanation and the references! –  Paul Siegel Oct 21 '10 at 12:56
    
on the 8th line, $L^{\times k}$ should be $L^{\otimes k}$, right? –  Yuhao Huang Oct 22 '10 at 6:20
    
Yuhao Huang: you are right. –  Tim Perutz Oct 22 '10 at 13:08
add comment

Paul Biran has also studied this situation where $(M, \omega, J)$ is Kahler, $[\omega] \in H^2(M, \mathbb{Z})$ and $\Sigma$ is a complex hypersurface Poincare dual to $k[\omega]$. He has proved that in this setting, $M$ can be decomposed symplectically as the unit normal disk bundle of $\Sigma$ and an isotropic CW-complex.

This decomposition result is then used to investigate various embedding questions into $M$.

Biran: "Lagrangian barriers and symplectic embeddings. Geom. Funct. Anal. 11 (2001), no. 3, 407–464"

Biran and Cieliebak: "Symplectic topology on subcritical manifolds. Comment. Math. Helv. 76 (2001), no. 4, 712–753"

Biran: "Lagrangian non-intersections. Geom. Funct. Anal. 16 (2006), no. 2, 279–326"

share|improve this answer
add comment

Your Answer

 
discard

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.