# Lagrangian submanifold of a Calabi-Yau manifold

In the paper 'Special Lagrangians, stable bundles and mean curvature flow' by R. P. Thomas and S.-T. Yau, page 2. They said

A Lagrangian submanifold $L$ of the Calabi-Yau manifold $(X,\Omega)$, we get an induced volume form $vol$ on $L$, and by a short calculation $\Omega_L=e^{i\theta}dvol_L$. And By Lagrangian we will always mean graded Lagrangian (thus the Maslov class of the Lagrangian, which is the class of $d\theta$ in $H^1(L; 2\pi Z)$, is assumed to vanish, and we have chosen a lift of $\theta$).

Can anyone tell we how a short calculation to getting $\Omega_L=e^{i\theta}dvol_L$. How to lift and getting the zero Maslov class.

The first part is really a linear algebra fact: Let $V = \mathbb{C}^n$, endowed with the standard symplectic form $$\omega = \tfrac{\sqrt{-1}}{2}\left(\mathrm{d}z_1\wedge\mathrm{d}\overline{z_1} + \cdots + \mathrm{d}z_n\wedge\mathrm{d}\overline{z_n}\right),$$ which is also the Kähler form for the Kähler metric $$g = \mathrm{d}z_1\circ\mathrm{d}\overline{z_1} + \cdots + \mathrm{d}z_n\circ\mathrm{d}\overline{z_n}\,.$$ Let $\mathrm{Sp}(\omega) \simeq \mathrm{Sp}(n,\mathbb{R})$ be the group of linear transformation of $V$ that preserve $\omega$. Then, by the usual symplectic linear algebra, $\mathrm{Sp}(\omega)$ acts transitively on the manifold $\mathrm{Lag}^+(\omega)\subset \mathrm{Gr}^+_n(V)$ of oriented Lagrangian $n$-planes in $V\simeq\mathbb{R}^{2n}$ and hence its maximal compact subgroup does as well.
Now, the subgroup $\mathrm{U}(\omega,g)\subset \mathrm{Sp}(\omega)$ consisting of the linear transformations that preserve both $\omega$ and $g$ is a maximal compact subgroup of $\mathrm{Sp}(\omega)$ and $\mathrm{U}(\omega,g)$ is isomorphic to $\mathrm{U}(n)\subset \mathrm{O}(2n)$. Thus, $\mathrm{U}(n)$ acts transitively on $\mathrm{Lag}^+(\omega)$.
Now, while $\mathrm{U}(n)$ preserves the volume form on an oriented $n$-planes (since it preserves the metric), it does not preserve the complex valued $n$-form $$\Omega = \mathrm{d}z_1\wedge\cdots\wedge\mathrm{d}z_n$$ on the nose, but only up to a 'phase factor', i.e., $$g^*\Omega = \det(g)\,\Omega$$ for $g\in\mathrm{U}(n)$, and, for such $g$, we have $\bigl|\det(g)\bigr| = 1$, i.e., $\det(g)\in S^1$. Since $L = \mathbb{R}^n\subset\mathbb{C}^n$ is a Lagrangian $n$-plane and since, on it, we have that $$\Omega_{\mathbb{R}^n} = \mathrm{d}x_1\wedge\cdots\wedge\mathrm{d}x_n = dvol_{\mathbb{R}^n}$$ when $\mathbb{R}^n$ is given its standard orientation, it follows that when an oriented Lagrangian plane $L$ is written in the form $L = g(\mathbb{R}^n)$ for some $g\in \mathrm{U}(n)$, we must have $$\Omega_{L} = \det(g)\,dvol_{L}\,.$$ In particular, there is a (smooth) map $\lambda:\mathrm{Lag}^+(\omega)\to S^1$ for which $$\Omega_{L} = \lambda(L)\,dvol_{L}\,.$$
For the second fact, if you now assume that your manifold $X$ has an $\mathrm{SU}(n)$ structure defined by a nondegenerate $2$-form $\omega$ and a compatible complex-valued $n$-form $\Omega$, then, on the bundle $\mathrm{Lag}^+(X,\omega)$ of oriented Lagrangian $n$-planes, there is a well-defined function $\lambda:\mathrm{Lag}^+(X,\omega)\to S^1$ such that $$\Omega_{L} = \lambda(L)\,dvol_{L}\,$$ for all $L\in \mathrm{Lag}^+(X,\omega)$. In particular, for any oriented Lagrangian submanifold $P\subset X$, we have a well-defined map $\lambda_P:P\to S^1$ that satisfies $\lambda_P(p) = \lambda(T_pP)$ for all $p\in P$. If $\mu\in H^1(S^1,\mathbb{R})$ is the fundamental generator (i.e., the dual to the fundamental class), then the Maslov class of $P$ is the element $$\mu_P = \lambda_P^*\bigl(\mu\bigr)\in H^1(P,\mathbb{R}).$$ It vanishes if and only if there is a smooth function $\theta:P\to \mathbb{R}$ such that $$\lambda_P = \mathrm{e}^{i\theta}.$$