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



What is geometric intuition of special Lagrangian manifolds?
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.


