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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.

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  • $\begingroup$ The starting point for your solution is incorrect. $\endgroup$
    – Prahar
    Apr 4 at 10:44
  • $\begingroup$ @PraharMitra Is the part after the edit correct? $\endgroup$
    – saad
    Apr 5 at 6:00
  • $\begingroup$ the last term is incorrect. $\endgroup$
    – Prahar
    Apr 5 at 6:01
  • $\begingroup$ @PraharMitra There should be a plus sign? $\endgroup$
    – saad
    Apr 5 at 6:06
  • $\begingroup$ no no the whole term is incorrect. Where did you manage to get a $\xi$ dependence there? $\endgroup$
    – Prahar
    Apr 5 at 6:07

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