## What goes wrong in the above reasoning

This is a good question, the answer (as far as I can see) is a little subtle.
The OP reasons that because $\delta z(z)$ is holomorphic in $z$, and also,$^{\dagger}$
$$
\delta \phi(z,\bar{z}) = \partial_z \delta z(z)+(\partial_z \delta z(z))^*,
\tag{1}\label{1}
$$
that therefore holomorphic derivatives of $\delta\phi$ must also be holomorphic in the coordinate value, say $z=z_1$, at which the energy-momentum variation is inserted. This is generically incorrect (i.e. $\partial_z^n\delta\phi|_{z=z_1}$ is not holomorphic for $n\geq1$) because in order for this holomorphic change of coordinates,
$$
w(z) = z + \delta z(z),
$$
to specifically generate a *Weyl transformation* of a local operator inserted at $z=z_1$, it is necessary that there is implicit $z_1,\bar{z}_1$ dependence in $\delta z(z)$ in addition to the holomorphic $z$ dependence; more fully we could therefore write $\delta z(z) = \delta z(z;z_1,\bar{z}_1)$ if the local operator whose Weyl variation we wish to compute is inserted at $z=z_1$. Let us derive this conclusion (after making a couple of comments).

**Comment 1:** In terms of an auxiliary real coordinate, $\sigma^a$, with $a=1,2$, another way of saying the same thing is that a holomorphic frame, $z$, depends implicitly on the point, say $\sigma_1^a$, at which the frame might be centred, as well as the point, $\sigma^a$, at which the coordinate frame $z$ is evaluated; that is, we could more fully write $z=z_{\sigma_1}(\sigma)$ and $z_1=z_{\sigma_1}(\sigma_1)$ (or $z_1=0$ if the frame is centred at $\sigma_1$).

**Comment 2:** Unless $\delta z(z_1;z_1,\bar{z}_1) =0 $ there will be an additional obstruction which forbids $\delta z(z;z_1,\bar{z}_1)$ and its holomorphic derivatives from being holomorphic in $z_1$, the obstruction being local Ricci curvature (if present). This obstruction is irrelevant here, because for a Weyl transformation we are to solve (\ref{1}) subject to $\delta z(z_1;z_1,\bar{z}_1)=0$, which in turn ensures that the Weyl variation does not shift the location of the inserted operator (namely the energy momentum tensor, its variation, or any other local operator whose Weyl variation we are computing).

$^\dagger$ I hope you will forgive me for writing the equations in terms of $\delta\phi\equiv 2\delta \omega$ and $\delta z(z) \equiv \epsilon v^z(z)$ rather than $\delta\omega$ and $v^z(z)$.

## Derivation

(I am preparing a detailed derivation that I will post here asap.)

## Best Answer

When both "space" ($x$) and "time" ($y$) directions are periodic, the Laplacian on torus with coodinate $z=x+iy$ has a normalized zero mode $$ \varphi_0(z) = \frac 1 {\sqrt{{\rm Im}(\tau)}} $$ (Here $\tau$ is the modular parameter defining the torus.)

As $$ -\nabla^2 \varphi_0=0. $$ the zero mode means that the Laplace operator is not 1-1 and so prevents the Laplacian with periodic boundary conditions from having an inverse. There is therefore no actual Green function. Instead we must therefore resort to a

modified Green function. We can make use of a theta function with characteristics defined by $$ \theta\left[ \matrix{a\cr b}\right] (z|\tau)= \sum_{m=-\infty}^\infty \exp\{i\pi \tau(m+a)^2 +2\pi i (m+a)(z+b)\}, \quad {\rm Im}(\tau)>0\quad a,b \in {\mathbb R}. $$Observe that $$ F(x,y) \equiv e^{-\pi y^2/ {\rm Im}(\tau)} \theta\left[ \matrix{\textstyle{\frac 12}\cr \textstyle{\frac 12}}\right] (z|\tau) $$ obeys $$ F(x+1,y) =-F(x,y), \quad F(x+ {\rm Re}(\tau), y+{\rm Im}(\tau)) =({\rm phase}) F(x,y) $$ so $$ G_0(x,y)) = -\frac{1}{2\pi} \ln |F| = -\frac{1}{2\pi} \ln\left| \theta\left[ \matrix{\textstyle{\frac 12}\cr \textstyle{\frac 12}}\right] (z|\tau)\right| + \frac 12 y^2 / {\rm Im}(\tau)\\ = -\frac 1{2\pi} \ln |E(z)| + \frac 12 y^2 / {\rm Im}(\tau)+const. $$ is both periodic on the torus and obeys $$ -\nabla^2 G_0(x,y) = \delta^2(x,y) - 1/{\rm Im}(\tau)\\ = \sum_n \varphi_n(z) \varphi^*_n(0)- \varphi_0(z)\varphi_0(0) $$ Here the sum is over all $n$ including $n=0$. The $\varphi_n(z)$, $n>0$ are the eigenfuctions of the Laplacian with non-zero eigenvalues.

The modified Green function can be used to solve $$ -\nabla^2 \phi= f(x,y) $$ as $$ \phi(x,y) = \int_{\rm torus} G_0(x-x', y-y'))f(x',y') dx' dy'=0. $$ provided that $f$ is perpendicular to the the zero mode, i.e. $$ \int_{\rm torus} f(x,y) dx dy=0. $$

This is consistent with the Fredholm alternative for linear operators.