# Equation of a conformal Killing vector

Question 10(a) on pages 469-470 in the book "Spacetime and Geometry" by Sean Carroll asks:

Suppose that two metrics are related by an overall conformal transformation of the form:

$$\tilde{g}_{\mu \nu} = \exp(\alpha(x))g_{\mu \nu}$$

Show that if $$\xi^{\mu}$$ is a Killing vector for the metric $${g}_{\mu \nu}$$, then it is a conformal Killing vector for the metric $$\tilde{g}_{\mu \nu}$$· A conformal Killing vector obeys the equation

$$\nabla_{\mu} \xi_{\nu} + \nabla_{\nu} \xi_{\mu} = g_{\mu \nu} \xi^{\lambda} \nabla_{\lambda} \alpha$$

My solution starts with the statement

$$\mathcal{L}_{\xi} g_{\mu \nu} = 0 \implies g_{\mu \lambda}\nabla_{\nu} \xi^{\lambda} + g_{\nu \lambda}\nabla_{\mu} \xi^{\lambda} + \xi^{\lambda}\nabla_{\lambda}g_{\mu \nu} = 0$$

We can subsitiute $$g_{\mu \nu} = \exp(-\alpha(x))\tilde g_{\mu \nu}$$ on the third term and lower the indices of $$\xi$$ on the first and second terms to get the desired result. But isn't the third term $$0$$ because of the Levi Civita connection? And I've googled and found the equation of the conformal Killing vector to have a factor of $$\frac{2}{n}$$ on the right-hand side, where $$n$$ is the number of dimensions on the manifold. Where am I going wrong?

EDIT:

I started with the transformation of $$\tilde g'_{\mu \nu}(x)= \tilde g_{\mu \nu}(x) - \epsilon(\tilde g_{\mu \sigma}(x) \partial_{\nu} \xi^{\sigma} + \tilde g_{\rho \nu}(x) \partial_{\mu} \xi^{\rho} + \xi^{\lambda} \partial_{\lambda} \tilde g_{\mu \nu}(x)) + \mathcal{O}(\epsilon^{2})$$

Now we set $$\tilde g_{\mu \nu}(x) = \exp(\alpha(x))g_{\mu \nu}(x)$$ to get:

$$g'_{\mu \nu}(x)= g_{\mu \nu}(x) - \epsilon( g_{\mu \sigma}(x) \partial_{\nu} \xi^{\sigma} + g_{\rho \nu}(x) \partial_{\mu} \xi^{\rho} + \xi^{\lambda} \partial_{\lambda} g_{\mu \nu}(x) - \xi^{\lambda} (\partial_{\lambda} \alpha)g_{\mu \nu}(x)) + \mathcal{O}(\epsilon^{2})$$

Since $$\xi^{\mu}$$ is a Killing vector for $$g_{\mu \nu}$$, we must have

$$g_{\mu \sigma}(x) \partial_{\nu} \xi^{\sigma} + g_{\rho \nu}(x) \partial_{\mu} \xi^{\rho} + \xi^{\lambda} \partial_{\lambda} g_{\mu \nu}(x) - \xi^{\lambda} (\partial_{\lambda} \alpha)g_{\mu \nu}(x) = 0$$

This leads us to the desired equation.

• The starting point for your solution is incorrect. Apr 4 at 10:44
• @PraharMitra Is the part after the edit correct?
• no no the whole term is incorrect. Where did you manage to get a $\xi$ dependence there? Apr 5 at 6:07