# Extracting the central charge from correlation functions and normalization of operators

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?

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