# How many Lagrangian submanifolds?

An $n$-dimensional submanifold $L$ of a symplectic manifold $(M^{2n}, \omega)$ is called Lagrangian if $\omega|_L = 0$. I want to get some feeling about how many Lagrangian submanifolds are.

For each $\alpha \in H_n(M)$, is there a Lagrangian submanifold representing $\alpha$? Maybe it's not a good idea to distinguish Lagrangians by its homology classes. Floer homology is the same if we move Lagrangians by Hamiltonian isotopy. Is there a notion of space of Lagrangian submanifolds modulo Hamiltonian isotopy?

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## 2 Answers

The answer to the first question is 'no'. For example, if $M = S^2\times S^2$ is given the product symplectic structure, then there is no Lagrangian submanifold in the homology class of $S^2\times\{x\}$.

The answer to your second question is 'yes, the notion exists', but the question is whether this set can be endowed with any good properties that makes it into a 'useful space'. For example, just look at the closed curves on a surface $S$ endowed with a nonvanishing $2$-form $\omega$. This is the space of (closed) Lagrangian submanifolds in this case. What can you say about the set that you get by regarding two that differ by Hamiltonian isotopy as equivalent?

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Thank you. I should've thought about it carefully before I post the question. For the second question, my naive hope was to find suitable classes to do intersection theory. It seems to me that people think about Lagrangian intersections, but intersection of two Lagrangians is not Lagrangian. So I was curious if we can consider larger classes modulo Hamiltonian isotopy containing Lagrangians. I have no idea whether this makes sense at all due to my ignorance. –  Hwang Sep 4 '13 at 12:27
@Hwang Are you aware of displaceability issues of Lagrangians? Take for example the 2-sphere and consider two closed curves, the equator and some other closed curve $C$ which lies fully in the northern hemisphere. Under any Hamiltonian isotopy of the sphere, the image of the equator will intersect itself, but for the curve $C$ you can find an isotopy such that the image of $C$ is disjoint from $C$. Notice that both the equator and $C$ represent the same homology class - I'm trying to point out that intersections of Lagrangians are not a purely topological issue, there is some geometry hidden. –  Oldřich Spáčil Sep 4 '13 at 15:51
@Oldřich Spáčil. Thank you for the comment. I know Lagrangian intersection is not a topological issue. What I meant by classes was modulo Hamiltonian isotopy, not modulo homologous manifolds, so that self-intersection of nondisplaceable Lagrangian is not zero. –  Hwang Sep 5 '13 at 0:32

There's an idea of Hitchin's that (all Lagrangian submanifolds)/(Hamiltonian isotopy) should be, at least approximately, the same as (all "special" Lagrangian submanifolds). "Special" is defined with respect to some background structure on the symplectic manifold, like a metric. Special Lagrangians make a finite-dimensional space at least locally.

A good analogy is the Hodge theorem (but Hitchin's idea doesn't work as well): the space (closed differential forms)/(exact differential forms) is the same as (harmonic differential forms), again the meaning of "harmonic" depends on some background structure, and also harmonic forms make a finite-dimensional space.

The Hitchin paper is here: http://arxiv.org/abs/dg-ga/9711002

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