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 $\mathrm{Im}\Omega|_L=0$.
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
|
5
3
|
|
|
|
|
4
|
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. |
|||
|
|
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
|
7
|
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: $$ Z : K(D) ā \mathbb{C} $$ and a "slicing" $P$ of $D$ such that if $E \neq 0$ in $P(\phi)$ then Z(E) = $m(E)e^{iĻĻ}$ for some $ m(E) ā 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. |
||||||||||||||||||
|

