I start my question with a definition and some motivation. 


Let $M$ be a symplectic manifold. A subbundle $P\subset TM^{\mathbf{C}}$ of the complexified tangent bundle is called a complex polarization if 


1. $P$ is Lagrangian, i.e. Maximal isotroic, dim$P\_m=n$, $\forall m\in M$, and $\omega(P,P)=0$
2. P involutive, i.e. if $X,Y\in P$ then $[X,Y]\in P$
3. dim$P\cap\bar P \cap TM$ is constant


Now, Introduce an hermitian form on $P$ defined by 


$$b(X,Y)=i\omega(X,\bar Y).$$


Note that if $P$ is real, then $b$ is vanishes identically on $P$. Consequently, $b$ projects onto a non-degenerate form on the quotient $P/{(P\cap \bar P)}$ and we denote it by $\bar b$. $P$ is said to be of type $(r,s)$ if and only if $\bar b$ has signature $(r,s)$ i.e. its matrix is $$diag(\underbrace{1,1...,1}\_{r},\underbrace{-1,-1...,-1}\_{s} )$$ for $0\le r+s=n-dim\_{\mathbf C }P\cap \bar P$
. Then, $P$ is said to be positive if $s=0$. In the case if $r=s=0$ then $P$ is real



> 
> Let $(M,\omega, J)$ be a compact Kahler manifold with positive-definite
>  polarization $P$, and $(L,\nabla)$ be prequantum data. Let
>  $$M\_{quantum} = \left\{s\in \Gamma(L) \vert \nabla\_Xs=0 , \forall 
>  X\in \bar P \right\}$$ Then I can not see why $M\_{quantum}$ is
>  fnite-dimensional?. I am looking for a referrence for finding a proof for this assertion "the space of square integrable holomorphic sections is closed."
> 
> 
> 



The space of holomorphic sections are finite dimensional?
$\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\bR}{\mathbb{R}}$ $\DeclareMathOperator{\Hom}{Hom}$ Suppose that $V$ is a finite dimensional real space equipped with an almost complex structure $J$. Let $P\subset V^{\bC}=V\otimes\_{\bR}\bC$ be a **complex** vector subpace.


Then $P^\*=\Hom\_{\bC}(P,\bC)$ is a quotient of $V\_{\bC}^\*=\Hom\_{\bC}(V^{\bC},\bC)=\Hom\_{\bR}(V,\bC)$ . We denote by $\Pi\_P$ the projection $V\_{\bC}^\*\to P^\*$. We define $\Pi\_{\bar{P}}$ similarly. Now observe that $V^\*\_{\bR}=\Hom(V,\bR)$ is a **real** subspace of $V\_{\bC}^\*$. For $\xi\in V^\*\_{\bR}$ we set $\xi\_{\bar{P}}:=\Pi\_{\bar{P}}\xi.$


If the induced real linear map $\xi\mapsto \xi\_{\bar{P}}$ is injective, then the answer to your question is positive.


To see this consider the differential operator $\newcommand{\pap}{\partial\_{\bar{P}}}$ $\pap^\nabla$ defined as the composition


$$ C^\infty(L)\stackrel{\nabla}{\to} C^\infty(T^\*M\otimes\_{\bR} L)\stackrel{\Pi\_{\bar{P}}}{\to} C^\infty( \bar{P}^\*\otimes L). $$


Then 


$$M\_{quantum}=\ker \pap^\nabla=\ker (\pap^\nabla)^\*\pap^\nabla. $$


If the map $\xi\mapsto\xi\_{\bar{P}}$ is injective, then the operator $(\pap^\nabla)^\*\pap^\nabla$ is elliptic and it has finite dimensional kernel on the compact manifold $M$.


