# Constraint on conformal transformation (Ketov)

1. Sep 14, 2014

### maverick280857

Hi,

First of all, I'm not sure if this thread belongs to the BSM forum because the question I'm posing here is a simple CFT question which could well be posed in the forum on GR or Particle Physics/QFT. I will defer to the judgment of the moderator to put this in the right place if it already isn't.

So, while working through the book on Conformal Field Theory by Ketov, I came across a differential equation that must be satisfied by the conformal parameter $\xi$ defined through

$$x^\mu \rightarrow {\tilde{x}}^\mu = x^\mu + \xi^\mu$$

Now,

$$\tilde{g}_{\mu\nu}(\tilde{x}) = \frac{\partial x^\lambda}{\partial \tilde{x}^\mu}\frac{\partial x^\rho}{\partial \tilde{x}^\nu}g_{\lambda\rho}(x)$$

For the infinitesimal transformation in question,

$$\frac{\partial x^\mu}{\partial \tilde{x}^\rho} = \delta^{\mu}_{\rho} - \frac{\partial x^\lambda}{\partial \tilde{x}^\rho}\frac{\partial \xi^{\mu}}{\partial x^\lambda}$$

using the chain rule in the second term. Now, to lowest order in $\xi$, this becomes

$$\frac{\partial x^\mu}{\partial \tilde{x}^\rho} = \delta^{\mu}_{\rho} - \partial_{\rho}\xi^\mu$$

But this means,

$$\tilde{g}_{\mu\nu}(\tilde{x}) = g_{\mu\nu}(x) - g_{\mu\rho}(x)\partial_\nu\xi^\rho - g_{\nu\lambda}\partial_\mu \xi^\lambda$$

which, with the identification $\Omega(x) = e^{\omega(x)}$ and $\tilde{g}(\tilde{x}) = \Omega(x)g_{\mu\nu}(x)$, yields

$$(1 + \omega)g_{\mu\nu} = g_{\mu\nu} - \partial_\nu \xi_\mu - \partial_\mu \xi_\nu$$

The right hand side has the wrong signs for the second and third terms, when compared to the result given in Ketov's book. Now this looks like a fairly straightforward application of the chain rule and Taylor expansion, to me. Is there something I'm missing?

Solved. There is a sign error but this can be fixed by replacing $\omega$ with $-\omega$ and taking the trace of both sides.