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

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 $\mathrm{Im}\Omega|_L=0$.

share|improve this question
add comment

2 Answers

up vote 4 down vote accepted

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


$L$ 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.

share|improve this answer
add comment

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.

share|improve this answer
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
add comment

Your Answer


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.