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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."

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Can you define precisely what you mean by Lagrangian and involutive? –  Liviu Nicolaescu Dec 31 '13 at 14:31
    
Involutive, means, if $X,Y\in P$ then $[X,Y]\in P$. and Lagrangian means maximal isotropic –  Hassan Jolany Dec 31 '13 at 14:37
    
Maximal isotropic with respect to what bilinear form? –  Liviu Nicolaescu Dec 31 '13 at 14:56
    
I am asking all these question because there is another notion of polarization used in a similar context. In that case $P\cap \bar{P}=0$, by definition. –  Liviu Nicolaescu Dec 31 '13 at 15:15
    
in the case $P\cap \bar{P}=0$ we say kahler polarization. and is special case of polarization –  Hassan Jolany Dec 31 '13 at 15:53

1 Answer 1

up vote 1 down vote accepted

$\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\bR}{\mathbb{R}}$ $\DeclareMathOperator{\Hom}{Hom}$ 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 $M$.

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In your solution you have assumed $P\subset V^{\bC}=V\otimes_{\bR}\bC$ while we tae $P$ as subbundle of $TM\otimes \bC$ –  Hassan Jolany Dec 31 '13 at 16:09
    
At every point $x$ in $M$ the fiber $P_x$ is a subspace of $V^{\mathbb{C}}$ where $V=T_xM$. –  Liviu Nicolaescu Dec 31 '13 at 16:39
    
Thanks a lot, for your nice answer –  Hassan Jolany Dec 31 '13 at 16:42
    
I am still waiting for second part of answer. –  Hassan Jolany Dec 31 '13 at 17:18
1  
Once you have the corresponding ellipticity, the statement about the closedness is immediate: all holomorphic sections are smooth in particular square integrable. They form a finite dimensional subspace of a Hilbert space, hence it is a closed subspace. I reiterate my comment that your question needs to be more carefully formulated because the meaning of isotropic is trickier than you think. –  Liviu Nicolaescu Dec 31 '13 at 17:40

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