>&=&\sum_{m}c_{m}|i+4n>  \nonumber \\
&=&\frac{1}{\pi }\sum_{m}\int_{-\infty }^{\infty
}dxdyc_{m}|z><z|i+4m>, \label{eq:44}\end{aligned}$$$$\begin{aligned}
<\phi _{2}|&=&\sum_{m=0}^{\infty }d_{m}<j+4m|  \nonumber \\
&=&\frac{1}{\pi }\sum_{n=0}^{\infty }\int_{-\infty }^{\infty
}dxdyd_{n}<j+4n|z><z|.  \label{eq:45}\end{aligned}$$We let $R_{N}$ act on $|\phi _{1}>$ and $<\phi _{2}|$ and get $$R_{N}|\phi _{1}>=\omega ^{i}|\phi _{1}>,$$

$$<\phi _{2}|R_{N}^{-1}=<\phi _{2}|\omega ^{-j}.$$

Subsequently, we substitute (\[eq:43\])(\[eq:44\])(\[eq:45\]) into (\[eq:42\]) and make use of the formula

$$<k,q|n>=\frac{1}{\sqrt{n!}}\frac{d^{n}}{dz^{n}}\left( e^{\frac{1}{2}z\bar{z}%
}<k,q|z>\right) \Vert _{z=0}$$

to obtain$$\begin{aligned}
F_{st}(k,q_{0}) &=&\sum_{m,n}\sum_{h=0}^{A-1}c_{m}d_{n}<k,q_{0}+\frac{l(h+s)%
}{A}|i+4m>\times <j+4n|k,q_{0}+\frac{lh}{A}>\times e^{2\pi
i(q_{0}/l+h/A)t}
\nonumber\\
&=&\sum_{m,n}\sum_{h=0}^{A-1}c_{m}d_{n}\frac{1}{\sqrt{(i+4m)!(j+4n)!}}\frac{%
d^{n+m}}{dz_{1}^{m}\bar{z_{2}}^{n}}(e^{\frac{1}{2}(z_{1}\bar{z}_{1}+z_{2}\bar{z}%
_{2})}\sum_{h=0}^{A-1}<k,q_{0}+\frac{l(h+s)}{A}|z_{1}> \nonumber\\
  &&\times<z_{2}|k,q_{0}+\frac{lh}{A}>\times e^{2\pi
i(q_{0}/l+h/A)t})\Vert
_{z_{1}=\bar{z}_{2}=0} \nonumber\\
&=&\sum_{m,n}\sum_{h=0}^{A-1}c_{m}d_{n}\frac{1}{\sqrt{(i+4m)!(j+4n)!}}\frac{%
d^{n+m}}{dz_{1}^{m}\bar{z_{2}}^{n}}\left( e^{\frac{1}{2}(z_{1}\bar{z}_{1}+z_{2}%
\bar{z}_{2})}g_{st}(k,q_{0},z_{1},z_{2})\right) \Vert _{z_{1}=\bar{z}%
_{2}=0}.\nonumber\\\end{aligned}$$So, we get the projector in the case of $z_{4}$ $$\begin{aligned}
P &=&\sum_{m,n}c_{m}d_{n}\frac{1}{\sqrt{(i+4m)!(j+4n)!}}\frac{d^{n+m}}{dz_{1}^{m}%
\bar{z}_{2}^{n}}(e^{z_{1}\bar{z}_{1}+z_{2}\bar{z}_{2}}\times e^{2\pi i(-%
\frac{3st}{2A})}\times e^{4\pi ^{2}iz_{1}\bar{z}_{2}}\times e^{-2\pi i(%
\frac{sd}{2}+\frac{ta}{2})} \nonumber\\
&&\times \sum_{a,d=0}^{1}(-1)^{ad}\theta \left[
\begin{array}{l}
\frac{s}{A} \nonumber\\
\frac{t}{2}%
\end{array}%
\right] (-\frac{l\hat{y}_{2}}{2\pi }+\frac{1}{2}Ad+\frac{\sqrt{2}A}{2l}%
(z_{1}+\bar{z}_{2}),\frac{Ai}{2}) \nonumber\\
&&\times \theta \left[
\begin{array}{l}
\frac{t}{A} \nonumber\\
\frac{s}{2}%
\end{array}%
\right] (-\frac{l\hat{y}_{1}}{2\pi }+\frac{1}{2}Aa+\frac{i\sqrt{2}A}{2l}%
(z_{1}-\bar{z}_{2}),\frac{Ai}{2}))\Vert _{z_{1}=\bar{z}_{2}=0} \nonumber\\
&&\times \{A\sum_{m,n}\sum_{a,d=0}^{1}(-1)^{ad}c_{m}d_{n}\frac{d^{n+m}}{%
dz_{1}^{m}\bar{z}_{2}^{n}}(e^{z_{1}\bar{z}_{1}+z_{2}\bar{z}_{2}}\times
e^{4\pi
^{2}iz_{1}\bar{z}_{2}}\times \theta (-\frac{l\hat{y}_{2}}{2\pi }+\frac{1%
}{2}Ad+\frac{\sqrt{2}A}{2l}(z_{1}+\bar{z}_{2}),\frac{Ai}{2}) \nonumber\\
&&\times \theta (-\frac{l\hat{y}_{1}}{2\pi }+\frac{1}{2}Aa+\frac{i\sqrt{2}A}{%
2l}(z_{1}-\bar{z}_{2}),\frac{Ai}{2}))\}^{-1}\Vert _{z_{1}=\bar{z}%
_{2}=0}.\end{aligned}$$

Thus, We derive two forms of explicit expressions of the projector $P$ in terms of the integration and derivative of the classical theta functions.

Discussion
==========

In this paper, $P$ is represented by a form of fraction which make sense only when the denominator has inverse. The formula demands: $$D=A\int d^{2}\mu d^{2}\nu f_{1}(\mu )f_{2}(\nu 