Bargmann–Fock Sheaves on Kähler Manifolds

Abstract

Fedosov used flat sections of the Weyl bundle on a symplectic manifold to construct a star product $\star$ which gives rise to a deformation quantization. By extending Fedosov’s method, we give an explicit, analytic construction of a sheaf of Bargmann–Fock modules over the Weyl bundle of a Kähler manifold X equipped with a compatible Fedosov abelian connection, and show that the sheaf of flat sections forms a module sheaf over the sheaf of deformation quantization algebras defined $(C^\infty _X[[\hbar ]], \star )$. This sheaf can be viewed as the $\hbar$-expansion of $L^{\otimes k}$ as $k \rightarrow \infty$, where L is a prequantum line bundle on X and $\hbar = 1/k$.

Appendices

Appendix A. Proof of Theorem 2.15

Let $\Phi := 2\sqrt{-1}\left( -\omega _{i\bar{j}}y^i\bar{y}^j+\Phi _\omega \right) -\Phi _\alpha$. It is clear that

$$\begin{aligned}e^{\Phi /\hbar } := 1+\Phi /\hbar +\frac{1}{2!}(\Phi /\hbar )^2+\cdots \end{aligned}$$

is an invertible section in $\mathcal {W}_{X,\mathbb {C}}^+$ under the Wick product, and we denote by $\left( e^{\Phi /\hbar }\right) ^{-1}$ its inverse.

Lemma A.1

Let O be any section of the Weyl bundle. Then we have

$$\begin{aligned} (\nabla ^{0,1}-\delta ^{0,1})\left( e^{\Phi /\hbar }\star O\star (e^{\Phi /\hbar })^{-1}\right) =e^{\Phi /\hbar }\star D_{F,\alpha }^{0,1}(O)\star (e^{\Phi /\hbar })^{-1}. \end{aligned}$$

(A.1)

In other words, the operators $D_{F,\alpha }^{0,1}$ and $\nabla ^{0,1}-\delta ^{0,1}$ differ by the gauge action by $e^{\Phi /\hbar }$.

Proof

We will restrict our attention to the case where $\alpha =0$; the general case is similar. The operator $\left( \nabla ^{0,1}-\delta ^{0,1}\right)$ is a derivation with respect to both the classical and quantum product on $\mathcal {W}_{X,\mathbb {C}}$, there is

$$\begin{aligned}&\left( \nabla ^{0,1}-\delta ^{0,1}\right) e^{\Phi /\hbar }\\&\quad =\frac{1}{\hbar }\left( \nabla ^{0,1}\Phi -\delta ^{0,1}\Phi \right) \cdot e^{\Phi /\hbar }\\&\quad =\frac{1}{\hbar }\sum _{k\ge 2}\left( \nabla ^{0,1}\Phi _{*,k}-\delta ^{0,1}\Phi _{*,k}\right) \cdot e^{\Phi /\hbar }\\&\quad =-\frac{1}{\hbar }(\delta ^{0,1}\Phi _{*,2})\cdot e^{\Phi /\hbar }+\frac{1}{\hbar }\sum _{k\ge 2}\left( \nabla ^{0,1}\Phi _{*,k}-\delta ^{0,1}\Phi _{*,k+1}\right) \cdot e^{\Phi /\hbar }\\&\quad =-\frac{1}{\hbar }e^{\Phi /\hbar }\star \left( \delta ^{0,1}(\Phi _{*,2})\right) +\frac{\sqrt{-1}}{2 \hbar }\omega ^{i\bar{j}}\frac{\partial \Phi }{\partial y^i}\cdot \frac{\partial }{\partial \bar{y}^j}\left( \delta ^{0,1}(\Phi _{*,2})\right) \cdot e^{\Phi /\hbar }\\&\qquad +\frac{1}{\hbar }\sum _{k\ge 2}\left( \nabla ^{0,1}\Phi _{*,k}-\delta ^{0,1}\Phi _{*,k+1}\right) \cdot e^{\Phi /\hbar }\\&\quad =-\frac{1}{\hbar }e^{\Phi /\hbar }\star \left( \delta ^{0,1}(\Phi _{*,2})\right) \\&\qquad +\frac{1}{\hbar }\sum _{k\ge 2}\left( \nabla ^{0,1}\Phi _{*,k}-\delta ^{0,1}\Phi _{*,k+1}+\frac{\sqrt{-1}}{2}\omega ^{i\bar{j}}\frac{\partial \Phi _{*,k}}{\partial y^i}\cdot \frac{\partial }{\partial \bar{y}^j}\left( \delta ^{0,1}(\Phi _{*,2})\right) \cdot e^{\Phi /\hbar }\right) \cdot e^{\Phi /\hbar }\\&\quad =-\frac{1}{\hbar }e^{\Phi /\hbar }\star \left( \delta ^{0,1}(\Phi _{*,2})\right) . \end{aligned}$$

In the last line, we have used the following identity:

$$\begin{aligned} \delta ^{0,1}(\Phi _{m,n+1})=\nabla ^{0,1}(\Phi _{m,n})+\frac{\sqrt{-1}}{2}\omega ^{i\bar{j}}\frac{\partial \Phi }{\partial y^i}\cdot \frac{\partial }{\partial \bar{y}^j}\left( \delta ^{0,1}(\Phi _{*,2})\right) . \end{aligned}$$

Since $e^{\Phi /\hbar }\star (e^{\Phi /\hbar })^{-1}=1$, it is easy to show that:

$$\begin{aligned} (\nabla ^{0,1}-\delta ^{0,1})(e^{\Phi /\hbar })^{-1}=\frac{1}{\hbar }\delta ^{0,1}(\Phi _{*,2})\star (e^{\Phi /\hbar })^{-1}. \end{aligned}$$

Using the fact that $\nabla ^{0,1}-\delta ^{0,1}$ is a derivation with respect to $\star$, we have

$$\begin{aligned}&(\nabla ^{0,1}-\delta ^{0,1})\left( e^{\Phi /\hbar }\star O\star (e^{\Phi /\hbar })^{-1}\right) \\&\quad =e^{\Phi /\hbar }\star \left( -\frac{1}{\hbar }\delta ^{0,1}(\Phi _{*,2})\right) \star O\star (e^{\Phi /\hbar })^{-1}+e^{\Phi /\hbar }\star (\nabla ^{0,1}O-\delta ^{0,1}O)\star (e^{\Phi /\hbar })^{-1}\\&\qquad + e^{\Phi /\hbar }\star O\star \frac{1}{\hbar }(\delta ^{0,1}(\Phi _{*,2})) \star (e^{\Phi /\hbar })^{-1}\\&\quad =e^{\Phi /\hbar }\star \left( \nabla ^{0,1}O-\delta ^{0,1}O+\frac{1}{\hbar }[-\delta ^{0,1}(\Phi _{*,2}), O]_{\star }\right) \star (e^{\Phi /\hbar })^{-1}\\&\quad =e^{\Phi /\hbar }\star D_{F,\alpha }^{0,1}(O)\star (e^{\Phi /\hbar })^{-1}. \end{aligned}$$

In the last line, we have used Eq. (2.6) to obtain:

$$\begin{aligned} -\delta ^{0,1}(\Phi _{n,2})=-2\sqrt{-1}\cdot \delta ^{0,1}(\Phi _\omega )_{n,2}=-2\sqrt{-1}\frac{\sqrt{-1}}{2}I_n=I_n. \end{aligned}$$

$\square$

Proposition A.2

Let O be any section of the Weyl bundle $\mathcal {W}_{X,\mathbb {C}}$, and let $O_q$ (q for quantization) be the unique solution of the following equation:

$$\begin{aligned} O\cdot e^{\Phi /\hbar }=e^{\Phi /\hbar }\star O_q. \end{aligned}$$

Here $\Phi$ is the same as the previous part. Then there is the following identity describing an explicit relation between the classical and quantum (Fedosov) connections:

$$\begin{aligned} D_C(A)\cdot e^{\Phi /\hbar }=e^{\Phi /\hbar }\star D_{F,\alpha }(A_q). \end{aligned}$$

(A.2)

Proof

Let A and B be sections of $\mathcal {W}_X$ and $\overline{\mathcal {W}}_X$ respectively, then so are the $D_C(A)$ and $D_C(B)$ as the classical connection $D_C$ does not change the type in $\mathcal {W}_{X,\mathbb {C}}$. For A, there is $A_q=A$ by type reason, and there is

$$\begin{aligned} D_C(A)\cdot e^{\Phi /\hbar }=e^{\Phi /\hbar }\star D_C(A_q)=e^{\Phi /\hbar }\star D_{F,\alpha }(A_q). \end{aligned}$$

The last equality follows from the fact that the Fedosov connection $D_{F,\alpha }$ equals $D_C$ when restricted to $\mathcal {W}_X$. For B, there is

$$\begin{aligned} B\star e^{\Phi /\hbar }=B\cdot e^{\Phi /\hbar }=e^{\Phi /\hbar }\star B_q. \end{aligned}$$

By Lemma A.1, there is

$$\begin{aligned} D_C^{0,1}(B)&=(\nabla ^{0,1}-\delta ^{0,1})(B)=(\nabla ^{0,1}-\delta ^{0,1})\left( e^{\Phi /\hbar }\star B_q\star (e^{\Phi /\hbar })^{-1}\right) \\&=e^{\Phi /\hbar }\star D_{F,\alpha }^{0,1}(B_q)\star (e^{\Phi /\hbar })^{-1}. \end{aligned}$$

In a similar fashion, we can show that $D_C^{1,0}(B)=e^{\Phi /\hbar }\star D_{F,\alpha }^{1,0}(B_q)\star (e^{\Phi /\hbar })^{-1}$. A general monomial of $\mathcal {W}_{X,\mathbb {C}}$ must be a sum of the forms $A\cdot B$. We first have the following:

$$\begin{aligned} (A\cdot B)\cdot e^{\Phi /\hbar }&= A\cdot \left( B\cdot e^{\Phi /\hbar }\right) =A\cdot \left( e^{\Phi /\hbar }\star B_q\right) = e^{\Phi /\hbar }\star \left( B_q\star A\right) , \end{aligned}$$

which implies that $(A\cdot B)_q=B_q\star A$. And there is

$$\begin{aligned} D_C(A\cdot B)\cdot e^{\Phi /\hbar }&= (D_C(A)\cdot B+A\cdot D_C(B))\cdot e^{\Phi /\hbar }\\&= D_C(A)\cdot B\cdot e^{\Phi /\hbar }+A\cdot D_C(B)\cdot e^{\Phi /\hbar }\\&= D_C(A)\cdot (e^{\Phi /\hbar }\star B_q)+A\cdot (e^{\Phi /\hbar }\star D_{F,\alpha }(B_q))\\&= (e^{\Phi /\hbar }\star B_q)\star D_C(A)+A\cdot (e^{\Phi /\hbar }\star D_{F,\alpha }(B_q))\\&= (e^{\Phi /\hbar }\star B_q)\star D_{F,\alpha }(A)+(e^{\Phi /\hbar }\star D_{F,\alpha }(B_q))\star A\\&= e^{\Phi /\hbar }\star (B_q\star D_{F,\alpha }(A)+D_{F,\alpha }(B_q)\star A)\\&= e^{\Phi /\hbar }\star D_{F,\alpha }(B_q\star A)\\&= e^{\Phi /\hbar }\star D_{F,\alpha }(A\cdot B)_q. \end{aligned}$$

$\square$

This proposition reduces the proof of Theorem 2.15 to showing that $\sigma (O_f)=f$. This follows the definition of $O_f$ and the fact that the section of $\Phi$ does not contain any non-trivial purely holomorphic or anti-holomorphic components. Thus we complete the proof of Theorem 2.15.

Appendix B. Proof of Theorem 4.4

Using Lemma 3.11 and the fact that $D_{F,\alpha }|_{\mathcal {W}_X}=D_K$, we have

$$\begin{aligned} D_{B,\alpha }(A\cdot e^{\beta /\hbar }\otimes e_{x_0})&= D_{B,\alpha }(A\circledast e^{\beta /\hbar }\otimes e_{x_0})\\&= D_{F,\alpha }(A)\circledast (e^{\beta /\hbar }\otimes e_{x_0})+A\circledast D_{B,\alpha }(e^{\beta /\hbar }\otimes e_{x_0})\\&= D_K(A)\cdot (e^{\beta /\hbar }\otimes e_{x_0})+A\circledast D_{B,\alpha }(e^{\beta /\hbar }\otimes e_{x_0}). \end{aligned}$$

Hence, to prove the theorem, we only need to show that $D_{B,\alpha }(e^{\beta /\hbar }\otimes e_{x_0})=0$. We first recall that $\alpha =-\hbar \cdot R_{i\bar{j}k}^kdz^i\wedge d\bar{z}^j$.

Lemma B.1

We have $(J_\alpha )_n=-(n+1)\hbar \cdot R_{ii_1\ldots i_{n},\bar{l}}^id\bar{z}^l\otimes y^{i_1}\ldots y^{i_n}$

Proof

The proof is by induction on n. For $n=1$, we have

$$\begin{aligned} (J_\alpha )_1=(\delta ^{1,0})^{-1}\left( -\hbar \cdot R_{i\bar{j}k}^kdz^i\wedge d\bar{z}^j\right) =-2\hbar \cdot \left( \frac{1}{2} R_{i\bar{j}k}^kd\bar{z}^j\otimes y^i\right) . \end{aligned}$$

Then by the induction hypothesis for $n-1$, we have

$$\begin{aligned} \nabla ^{1,0}(J_\alpha )_{n-1}&=\nabla ^{1,0}\left( -n\hbar \cdot R_{ii_1\ldots i_{n-1},\bar{l}}^id\bar{z}^l\otimes y^{i_1}\ldots y^{i_{n-1}}\right) . \end{aligned}$$

On the other hand,

$$\begin{aligned}&\nabla ^{1,0}\left( n\hbar \cdot R_{i_1\ldots i_{n},\bar{l}}^jd\bar{z}^l\otimes y^{i_1}\ldots y^{i_{n}}\otimes \partial _{y^j}\right) \\&\quad =(n+1)\cdot n\hbar \cdot R_{i_1\ldots i_{n+1},\bar{l}}^jdz^{i_{n+1}}\wedge d\bar{z}^l\otimes y^{i_1}\ldots y^{i_n}\otimes \partial _{y^j}. \end{aligned}$$

Since $\nabla ^{1,0}$ is compatible between the contraction between TX and $T^*X$, the above computation shows that

$$\begin{aligned} (J_\alpha )_n=(\delta ^{1,0})^{-1}(\nabla ^{1,0}(J_\alpha )_{n-1})=-(n+1)\hbar \cdot R_{i_1\ldots i_{n+1},\bar{l}}^{i_1} d\bar{z}^l\otimes y^{i_1}\ldots y^{i_n}y^{i_{n+1}}. \end{aligned}$$

$\square$

Lemma B.2

The section $\beta$ satisfies $D_K(\beta )=2\sqrt{-1}\omega _{i\bar{j}}d\bar{z}^j\otimes y^i-\partial \rho .$

Proof

The function $\rho$ satisfies the condition that $\partial \bar{\partial }(\rho )=-2\sqrt{-1}\omega$. Recall that $\beta =\sum _{k\ge 1}(\tilde{\nabla }^{1,0})^k(\rho )$. A straightforward computation shows that

$$\begin{aligned} D_K(\beta )&=(-\delta ^{1,0}+\bar{\partial })(\tilde{\nabla }^{1,0}\rho )=-\partial \rho +\bar{\partial }\circ (\delta ^{1,0})^{-1}(\nabla ^{1,0}\rho )\\&=-\partial \rho +(\delta ^{1,0})^{-1}(\bar{\partial }\partial \rho )=2\sqrt{-1}\omega _{i\bar{j}}d\bar{z}^j\otimes y^i-\partial \rho . \end{aligned}$$

$\square$

We also have the following:

$$\begin{aligned}&\frac{1}{\hbar }I_n\circledast (e^{\beta /\hbar }\otimes e_{x_0})\\&\quad =-2\sqrt{-1}\cdot R_{i_1\ldots i_n,\bar{l}}^j\omega _{j\bar{k}}d\bar{z}^l\otimes (y^{i_1}\ldots y^{i_n}\bar{y}^k)\circledast (e^{\beta /\hbar }\otimes e_{x_0})\\&\quad =-2\sqrt{-1}\cdot R_{i_1\ldots i_n,\bar{l}}^j\omega _{j\bar{k}}d\bar{z}^l\otimes \left( \frac{\omega ^{i\bar{k}}}{2\sqrt{-1}}\frac{\partial }{\partial y^i}\right) (y^{i_1}\ldots y^{i_n}e^{\beta /\hbar }\otimes e_{x_0})\\&\quad =R_{i_1\ldots i_n,\bar{l}}^id\bar{z}^l\otimes y^1\ldots y^n\frac{\partial (\beta /\hbar ))}{\partial y^i}\cdot (e^{\beta /\hbar }\otimes e_{x_0})+n\cdot R_{ii_1\ldots i_{n-1},\bar{l}}^id\bar{z}^l\otimes y^{i_1}\ldots y^{i_{n-1}}\cdot (e^{\beta /\hbar }\otimes e_{x_0})\\&\quad =\tilde{R}_n^*(\beta /\hbar )+n\cdot R_{ii_1\ldots i_{n-1},\bar{l}}^id\bar{z}^l\otimes y^{i_1}\ldots y^{i_{n-1}}\cdot (e^{\beta /\hbar }\otimes e_{x_0}). \end{aligned}$$

Summarizing the above computations, we have

$$\begin{aligned}&D_{B,\alpha }(e^{\beta /\hbar }\otimes e_{x_0})\\&\quad =\left( \nabla +\frac{1}{\hbar }\gamma _\alpha \circledast \right) (e^{\beta /\hbar }\otimes e_{x_0})+e^{\beta /\hbar }\otimes \nabla _{L_{\omega /\hbar }}e_{x_0}\\&\quad =\left( \nabla (\beta /\hbar )+\frac{2\sqrt{-1}}{\hbar }\omega _{i\bar{j}}(dz^i\otimes \bar{y}^i-d\bar{z}^j\otimes y^i)\circledast +\frac{1}{\hbar }(I+J_\alpha )\circledast \right) (e^{\beta /\hbar }\otimes e_{x_0})\\&\qquad +e^{\beta /\hbar }\otimes \nabla _{L_{\omega /\hbar }}e_{x_0}\\&\quad =\left( \nabla (\beta /\hbar )-\frac{2\sqrt{-1}}{\hbar }\omega _{i\bar{j}}d\bar{z}^j\otimes y^i+\frac{1}{\hbar }\partial \rho \right) (e^{\beta /\hbar }\otimes e_{x_0})\\&\qquad +\frac{1}{\hbar }(2\sqrt{-1}\omega _{i\bar{j}}dz^i\otimes \bar{y}^j+I+J_\alpha )\circledast (e^{\beta /\hbar }\otimes e_{x_0})\\&\quad =\left( \nabla (\beta /\hbar )+\sum _{n\ge 2}\tilde{R}_n^*(\beta /\hbar )-\frac{2\sqrt{-1}}{\hbar }\omega _{i\bar{j}}d\bar{z}^j\otimes y^i+\frac{1}{\hbar }\partial \rho +2\sqrt{-1}\omega _{i\bar{j}}dz^i\frac{\omega ^{k\bar{j}}}{2\sqrt{-1}}\frac{\partial (\beta /\hbar )}{\partial y^k}\right) \\&\qquad (e^{\beta /\hbar }\otimes e_{x_0})\\&\quad =\left( \nabla (\beta /\hbar )+\sum _{n\ge 2}\tilde{R}_n^*(\beta /\hbar )-\frac{2\sqrt{-1}}{\hbar }\omega _{i\bar{j}}d\bar{z}^j\otimes y^i+\frac{1}{\hbar }\partial \rho -\delta ^{1,0}(\beta /\hbar )\right) (e^{\beta /\hbar }\otimes e_{x_0})\\&\quad =\frac{1}{\hbar }\left( D_K(\beta )-2\sqrt{-1}\omega _{i\bar{j}}d\bar{z}^j\otimes y^i+\partial \rho \right) (e^{\beta /\hbar }\otimes e_{x_0})\\&\quad =0. \end{aligned}$$

This completes the proof of Theorem 4.4.