Derivation of Holomorphic Ward Identities in Franceso's CFT

Question

In equation 5.37 of francesco's CFT he writes the Ward Identities for traslation symmetry in the language of holomorphic functions. He goes from

\begin{equation} \frac{\partial}{\partial x^\mu} \langle T^\mu_{\ \ \ \nu}(x) X\rangle =-\sum_i^N\delta(x-x_i)\frac{\partial}{\partial x_i^\nu}\langle X\rangle \tag{5.32a}\end{equation}

to

\begin{align} 2\pi\partial_\bar{z}\langle T_{zz}X\rangle+2\pi\partial_z\langle T_{\bar{z}z}X\rangle=-\sum_i^N\partial_{\bar{z}}\frac{1}{z-\omega_i}\partial_{\omega_i}\langle X\rangle \tag{5.37a} \\ 2\pi\partial_\bar{z}\langle T_{z\bar{z}}X\rangle+2\pi\partial_z\langle T_{\bar{z}\bar{z}}X\rangle=-\sum_i^N\partial_{z}\frac{1}{\bar{z}-\bar{\omega}_i}\partial_{\bar{\omega}_i}\langle X\rangle. \tag{5.37b} \end{align}

I understand how to write the delta function as

\begin{equation} \delta(x-x_i)=\frac{1}{\pi}\partial_z\frac{1}{\bar{z}-\bar{\omega}_i}=\frac{1}{\pi}\partial_\bar{z}\frac{1}{z-\omega_i}\tag{5.33} \end{equation}

but the index manipulation on the left hand side is very obscure to me. When I try to perform the change of variables explicitly I get

\begin{align} \frac{\partial}{\partial x^\mu}T^\mu_{\ \ \ \nu}&=\frac{\partial}{\partial x^1} T^1_{\ \ \ \nu}+\frac{\partial}{\partial x^2} T^2_{\ \ \ \nu} \\ &= \Big\{\frac{\partial z}{\partial x^1}\partial_z +\frac{\partial \bar{z}}{\partial x^1}\partial_\bar{z} \Big\}T^1_{\ \ \ \nu}+\Big\{\frac{\partial z}{\partial x^2}\partial_z +\frac{\partial \bar{z}}{\partial x^2}\partial_\bar{z} \Big\} T^2_{\ \ \ \nu} \\ &=\partial_z\Big\{T^1_{\ \ \ \nu}+iT^2_{\ \ \ \nu}\Big\}+\partial_\bar{z}\Big\{T^1_{\ \ \ \nu}-iT^2_{\ \ \ \nu}\Big\} \end{align}

Since we are working in Euclidean metric I'm assuming indeces $\mu=1,2$ can be changed from upper to lower indistinctively (we can't do that if the indeces represent a $z$ or $\bar{z}$ since the metric is different in those coordinates). Using the energy-momentum tensor in complex coordinates

\begin{align} T_{zz}=\frac{1}{2}\big(T_{11}-iT_{12}\big) \\ T_{\bar{z}\bar{z}}=\frac{1}{2}\big(T_{11}+iT_{12}\big) \end{align}

Then I get $\partial_zT_{\bar{z}\bar{z}}+\partial_\bar{z}T_{zz}$ for $\nu=1$ and $-i\partial_zT_{\bar{z}\bar{z}}-i\partial_\bar{z}T_{zz}$ for $\nu=2$ (I'm using that the energy-momentum tensor is traceless symmetric to get these results). This two equations clearly don't resemble the two equations that Francesco gets, specially since he gets crossed terms in each equation. It seems like he is simply replacing the old indeces with complex indeces but I'd like to see how the equation is actually derived.

If anyone can point where I'm messing up the calculation it would be really helpful.

Answer 1

You have assumed that $T_{z\bar{z}}=0$ (i.e. $T_{11}+T_{22}=0$) in writing down the last pair of displayed equations. More generally, \begin{equation} \begin{aligned} T_{zz}&=\frac{1}{4}\big(T_{11}-T_{22}-2iT_{12}\big)\\ T_{\bar{z}\bar{z}}&=\frac{1}{4}\big(T_{11}-T_{22}+2iT_{12}\big)\\ T_{z\bar{z}}&=\frac{1}{4}\big(T_{11}+T_{22}\big) \end{aligned} \end{equation} Also, if you use $ds^2=dzd\bar{z}$ (rather than a general conformal metric $ds^2=2g_{z\bar{z}}dzd\bar{z}$) then you get factors of 2 when raising indices, e.g., $T^z_{\phantom{a}z}=g^{z\bar{z}}T_{\bar{z}z}=2T_{\bar{z}z}$. Finally, when you go through it in detail you'll see it is effectively correct to let $\mu$ range over $z,\bar{z}$ directly, so that provides a fast route to the Di Francesco result.

Answer 2

It's easiest to just use the $z,\bar{z}$ basis directly. The $\mu,\nu$ indices don't need to start out in the $1,2$ basis.

$$\frac{\partial}{\partial x^\mu} T^\mu_{\ \ \ \nu}=g^{\mu\lambda}\frac{\partial}{\partial x^\mu} T_{\lambda\nu}=2\frac{\partial}{\partial z} T_{\bar{z}\nu}+2\frac{\partial}{\partial \bar{z}} T_{z\nu},$$ since the inverse metric only has the $z\bar{z}$ components. Then you can choose $\nu$ as either $z$ (53.7a) or $\bar{z}$ (53.7b). Note on the RHS the $\nu$ becomes a $\omega_i,\bar{\omega}_i$, whereas the $\partial_z,\partial_\bar{z}$ are coming from the delta function.

So that was quite easy I think you'll admit.

---

Now I suspect the problem in comparing the way you did it to the RHS of (53.7a) and (53.7b) was that you substituted in $\nu=1,2$ which mixes together the two equations. I'll show that the last equation you wrote (trivially lowering indices as you mention) agrees with mine. \begin{align} \frac{\partial}{\partial x^\mu}T^\mu_{\ \ \ \nu}&=\partial_z\Big\{T_{1 \nu}+iT_{2 \nu}\Big\}+\partial_\bar{z}\Big\{T_{1 \nu}-iT_{2 \nu}\Big\} \end{align} Now rather than substituting in more $1,2$ for $\nu$, I'll transform the first index to the $z,\bar{z}$ basis. I'll just show the first term. $$\partial_z\Big\{T_{1 \nu}+iT_{2 \nu}\Big\}=\partial_z\Big\{\frac{\partial z}{\partial x^1}T_{z \nu}+\frac{\partial \bar{z}}{\partial x^1}T_{\bar{z} \nu}+i\left(\frac{\partial z}{\partial x^2}T_{z \nu}+\frac{\partial \bar{z}}{\partial x^2}T_{\bar{z} \nu}\right)\Big\}=2\partial_z T_{\bar{z} \nu}$$ This is exactly what I got for the first term