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.


These facts are a standard part of the lore of Lagrangian submanifolds of Kähler manifolds, but they are often not that explicitly explained in research papers. Probably your best source will be Calibrated Geometries, by Harvey and Lawson (Acta Mathematica 148 (1982), 47–157), which is where the original argument was given.

Here is the basic line of argument though:

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 $$ \lambda_P = \mathrm{e}^{i\theta}. $$


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