# What is geometric intuition of special Lagrangian manifolds?

Let $M$ be (for example) a Calabi-Yau threefold with Kaehler form $\omega$ and holomorphic 3-form $\Omega$. We say that a submanifold $L$ of $M$ is a special Lagrangian submanifold if $L$ is Lagrangian with respect to symplectic form $\omega$ and also $\mathrm{Im}\Omega|_L=0$. I would like to know geometric intuition of the latter condition. I am aware that Lagrangian condition roughly corresponds to $L$ having only position coordinate, momentum coordinate, or certain mixture of them (Lagrangian formulation of classical mechanics). I wonder if there is a good explanation about the extra condition $\mathrm{Im}\Omega|_L=0$.

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Let $(M,g,J,\Omega)$ be a Calabi-Yau $n$-fold. Then $\textrm{Re }\Omega$ is a calibration on $(M,g)$. Let $L\subset M$ be a real submanifold with $\dim_{\mathbb{R}}L=n$. You have the following

Proposition

$L$ is a special Lagrangian if and only if it admits an orientation making it into a calibrated (for $\textrm{Re }\Omega$) submanifold of $(M,g)$. In that case, it is volume-minimising in its homology class.

See Propositions 10.1 and 7.1 in the "Calabi-Yau manifolds..." book by Gross, Huybrechts and Joyce.

Here $\textrm{Re }\Omega$ being a calibration means that at each point $p\in M$, and for every oriented tangent $n$-plane $V\subset T_{M,p}$, one has $\left.\textrm{Re }\Omega\right|_V\leq vol_V$, where $vol$ is the volume form of $g$. Then $L$ being a calibrated submanifold for $\textrm{Re }\Omega$ means that on the tangent spaces $T_{L,p}$ of $L$ the above inequality becomes equality.

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Since nobody else answered this question, I'll give it a try from the point of view of mirror symmetry. This is probably a very backwards way of telling the story and there are maybe better answers from, say, the point of view of calibrated geometry. A Bridgeland stability condition on a triangulated category $D$ consists of a group homomorphism:

$$Z : K(D) → \mathbb{C}$$

and a "slicing" $P$ of $D$ such that if $E \neq 0$ in $P(\phi)$ then Z(E) = $m(E)e^{iπφ}$ for some $m(E) ∈ R>0$. See Definition 5.1. of the paper which began this subject: http://annals.math.princeton.edu/wp-content/uploads/annals-v166-n2-p01.pdf

Just as for ordinary stability conditions in geometric invariant theory, there are notions of stability and semi-stability which give us a notion of moduli spaces of objects. There is a conjecture that the derived Fukaya category of the symplectic manifold X carries a Bridgeland stability condition such that the semi-stable objects are Floer theoretically unobstructed special Lagrangians.

Edit: Another important theorem that I totally forgot to mention is that which is due to Thomas and Yau http://arxiv.org/pdf/math/0104197.pdf and which says that there can be at most one special Lagrangian in a Hamiltonian deformation class. We can therefore say that, roughly speaking, the special Lagrangian condition picks out a nice geometric representative in a semi-stable isomorphism class in the Fukaya category. Unfortunately, this is probably only a partial truth since there will probably be semi-stable objects which do not correspond to smooth Lagrangians, but singular ones.

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So is it the analogue of a harmonic representative, in the Hodge decomposition? –  Allen Knutson Nov 9 '12 at 15:33
It is a nice analogy, but I didn't make it clear enough that you get a representative of the semi-stable isomorphism classes, not necessarily all of the isomorphism classes. –  Daniel Pomerleano Nov 9 '12 at 16:57
So is it the analogue of the condition "be in the zero level set of the moment map for the compact group"? –  Allen Knutson Nov 12 '12 at 1:54
Actually recently Jake Solomon defined certain functional (of Calabi type) on a Hamiltonian deformation class of Lagrangians. The minima of this functional should be special Lagrangians and they should be the mirror of the Hermitian-Einstein metrics on a stable vector bundle. –  Guangbo Xu Dec 10 '12 at 20:27