Functional derivative for boson CFT on torus Let us consider a bosonic CFT on torus:
\begin{eqnarray}
S=\int dzd\bar{z}\frac{1}{2}\partial X\bar{\partial} X.\tag{2.1.10}
\end{eqnarray}
From Page 35 of Polchinski Vol. 1, I do the same function derivative trick:
\begin{eqnarray}
0&=&\int[dX]\frac{\delta}{\delta X(z,\bar{z})}\left[X(w,\bar{w})\exp(-S[X])\right]\\
&=&\int[dX][\delta^2(z-w,\bar{z}-\bar{w})+X(w,\bar{w})\partial\bar{\partial}X(z,\bar{z}),\tag{2.1.18}
\end{eqnarray}
to obtain
\begin{eqnarray}
\partial_z\bar{\partial}_{\bar{z}}\langle X(z,\bar{z})X(w,\bar{w})\rangle=-\delta^2(z-\bar{z},w-\bar{w}). 
\end{eqnarray}
However,
we know that the correct answer should be
\begin{eqnarray}
\partial_z\bar{\partial}_{\bar{z}}\langle X(z,\bar{z})X(w,\bar{w})\rangle=-\left[\delta^2(z-\bar{z},w-\bar{w})-\frac{1}{A}\right],
\end{eqnarray}
where $A$ is the area of the torus due to the elimination of the zero mode of $\partial\bar{\partial}$.
Another understanding is that on a compact manifold, the Poisson function must have zero net charge; the integration of right-hand side above must be zero.
My question is where I had made mistake in the above derivative trick?
 A: I have found the problem and the formal functional derivative should be treated carefully.
First,
we define a complete mode expansion by
\begin{eqnarray}
\partial\bar{\partial}\phi_I(z)=-\omega_I^2\phi_I(z);\\
\sum_I\phi_I(z)\phi_I(w)=\delta^2(z-w);\\
\int dzd\bar{z}\phi_I(z)\phi_J(z)=\delta_{I,J}.
\end{eqnarray}
We can also define the zero mode to be $\phi_0$.
Furthermore,
\begin{eqnarray}
\int[dX]=\prod_I\int da_I;
\end{eqnarray}
where
\begin{eqnarray}
X(z)=\sum_Ia_I\phi_I(z);\\
a_I[X]\equiv\int dzd\bar{z}X(z)\phi_I(z),
\end{eqnarray}
so
\begin{eqnarray}
\frac{\delta}{\delta X(z)}=\frac{\delta a_I[X]}{\delta X(z)}\frac{\partial}{\partial a_I}.
\end{eqnarray}
Put all of these into the functional derivative before,
we would find
\begin{eqnarray}
&&\int[dX]\frac{\delta}{\delta X(z,\bar{z})}\left[X(w,\bar{w})\exp(-S[X])\right]\\
&=&\prod_{I\neq0}\int da_I\int da_0\phi_0(z)\partial_{a_0}\left[a_0\phi_0(w)\exp[-\sum_{M\neq0}\omega^2a_Ma_M]\right]\\
&=&\phi_0(z)\phi_0(w)\langle 1\rangle\\
&=&\langle 1/A\rangle.
\end{eqnarray}
Here the main reason is that the zero mode does not participate in the decay from $\exp[-S]$.
