Extracting the central charge from correlation functions and normalization of operators

Question

I have an elementary question about computing the central charge in a conformal field theory in dimension greater than 2. In principle, this is determined by the correlation functions involving the stress-tensor $T$; however, the precise formula would depend on the normalization of the stress tensor, and I am confused about how the normalization is defined.

Suppose I know a three-point correlation functions such as $\langle T T T \rangle$ or $\langle\phi \phi T \rangle$; the operators are not normalized conventionally but I also know $\langle \phi \phi \rangle$ and $\langle T T \rangle$. Where can I find a formula for the central charge in terms of a ratio of these correlation functions? Also how are these operators conventionally normalized in conformal field theory?

Answer 1

The most common convention is to have unit normalization for all operators that are not currents. For scalars, this looks like \begin{equation} \left< \phi(x_1) \phi(x_2) \right > = \frac{1}{|x_{12}|^{2\Delta}}. \quad (1) \end{equation} For currents, it is more natural to normalize them so that Ward identities are unit normalized, not the two-point functions themselves. In the case of the stress tensor, you want the operator equation \begin{align} \partial^1_\mu \left < T^{\mu\nu}(x_1) \phi(x_2) \phi(x_3) \right > = - \left [ \delta(x_1 - x_2) \partial_2^\nu + \delta(x_1 - x_3) \partial_3^\nu \right ] \left < \phi(x_2) \phi(x_3) \right > \quad (2) \end{align} to hold. So the steps are conceptually simple.

1. Rescale $\phi$ so that (1) holds.
2. You have $\left < \phi\phi \right >$ and $\left < T\phi\phi \right >$ so you can rescale $T$ so that (2) holds.
3. Compute $\left < TT \right >$ with this rescaled version to read off the central charge.

For the last step, the convention I just looked up in https://arxiv.org/abs/1203.6064 is \begin{align} \left < T_{\mu\nu}(x) T_{\rho\sigma}(0) \right > &= \frac{C_T / S_d^2}{|x|^{2d}} \left [ \frac{1}{2}(I_{\mu\rho} I_{\nu\sigma} + I_{\mu\sigma} I_{\nu\rho}) - \frac{1}{d} \delta_{\mu\nu} \delta_{\rho\sigma} \right ] \\ I_{\mu\nu} &= \delta_{\mu\nu} - 2x_\mu x_\nu / x^2 \\S_d &= s\pi^{d/2} / \Gamma(d / 2). \end{align}