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?


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.

  • $\begingroup$ The starting point for your solution is incorrect. $\endgroup$
    – Prahar
    Commented Apr 4, 2021 at 10:44
  • $\begingroup$ @PraharMitra Is the part after the edit correct? $\endgroup$
    – saad
    Commented Apr 5, 2021 at 6:00
  • $\begingroup$ the last term is incorrect. $\endgroup$
    – Prahar
    Commented Apr 5, 2021 at 6:01
  • $\begingroup$ @PraharMitra There should be a plus sign? $\endgroup$
    – saad
    Commented Apr 5, 2021 at 6:06
  • $\begingroup$ no no the whole term is incorrect. Where did you manage to get a $\xi$ dependence there? $\endgroup$
    – Prahar
    Commented Apr 5, 2021 at 6:07

2 Answers 2


Instead of using Lie derivatives, I will stick to the basics and try to confine the proof to materials presented in the chapter itself.

First of all, there is some ambiguity in the phrasing of the the exercise itself. The author means that with the new $\tilde{g}_{\mu\nu} = \omega^2 g_{\mu\nu}$, $\xi^\mu$, as individual components, have the same numerical values as before when we had $g_{\mu\nu}$. This also means that now $\xi_\mu$, as individual components of the corresponding one-form, are numerically scaled by $\omega^2$ compared to the original $\xi_\mu$. (So the length of $\xi$ is scaled by $\omega$.)

For clarity, I will add a tilde to everything pertaining to the new metric $\tilde{g}_{\mu\nu}$. Also, I will write $\partial$ instead of $\nabla$ whenever I want to stress that it is about numerical values, rather than some tensors. Our goal is to prove the following.

$$\tilde{\nabla}_\mu \tilde{\xi}_\nu + \tilde{\nabla}_\nu \tilde{\xi}_\mu = (\tilde{\nabla}_\lambda \alpha) \tilde{\xi}^\lambda \tilde{g}_{\mu\nu}.$$

The conversions between the two versions, one with tilde and one without, are as follows.

  • $\tilde{g}_{\mu\nu} = \omega^2 g_{\mu\nu}$
  • $\tilde{g}^{\mu\nu} = \omega^{-2} g^{\mu\nu}$
  • $\tilde{k}^\mu = k^\mu$
  • $\tilde{k}_\mu = \omega^2 k_\mu$
  • $\tilde{\Gamma}^\rho_{\mu\nu} = \Gamma^\rho_{\mu\nu} + \omega^{-1}(\delta^\rho_\mu \partial_\nu \omega + \delta^\rho_\nu \partial_\mu \omega - g_{\mu\nu}g^{\rho\lambda}\partial_\lambda \omega)$

Now we have $$ \begin{equation} \begin{split} && \tilde{\nabla}_\mu \tilde{\xi}_\nu + (\mu \leftrightarrow \nu) \\ &=& \partial_\mu (\tilde{\xi}_\nu) - \tilde{\Gamma}^\rho_{\mu\nu} \tilde{\xi_\rho} + (\mu \leftrightarrow \nu) \\ &=& \partial_\mu (\xi_\nu \omega^2) - \left(\Gamma^\rho_{\mu\nu} + \omega^{-1}(\delta^\rho_\mu \partial_\nu \omega + \delta^\rho_\nu \partial_\mu \omega - g_{\mu\nu}g^{\rho\lambda}\partial_\lambda \omega)\right)(\xi_\rho \omega^2) + (\mu \leftrightarrow \nu) \\ &=& \omega^2 \left(\partial_\mu \xi_\nu - \Gamma^\rho_{\mu\nu} \xi_\rho \right) + 2 \xi_\nu \omega \partial_\mu \omega - \omega(\xi_\mu \partial_\nu \omega + \xi_\nu \partial_\mu \omega - g_{\mu\nu}\xi^\lambda\partial_\lambda \omega) + (\mu \leftrightarrow \nu) \\ &=& \omega^2 \nabla_\mu \xi_\nu + \omega \left( \xi_\nu \partial_\mu \omega - \xi_\mu \partial_\nu \omega + g_{\mu\nu}\xi^\lambda\partial_\lambda \omega \right) + (\mu \leftrightarrow \nu). \end{split} \end{equation} $$

Notice that the first term, after being symmetrized with $\mu \leftrightarrow \nu$, vanishes because $\xi^\mu$ is a Killing vector with the original $g_{\mu\nu}$. Also $\xi_\nu \partial_\mu \omega - \xi_\mu \partial_\nu \omega$ is antisymmetric in $\mu$ and $\nu$ and thus vanishes as well. So we have, with $\omega^2 = e^\alpha$,

$$ \begin{equation} \tilde{\nabla}_\mu \tilde{\xi}_\nu + (\mu \leftrightarrow \nu) = 2 \omega g_{\mu\nu}\xi^\lambda\partial_\lambda \omega = \frac{2}{2} \omega^2 g_{\mu\nu}\xi^\lambda\partial_\lambda \alpha = \tilde{g}_{\mu\nu} \tilde{\xi}^\lambda \tilde{\nabla}_\lambda \alpha, \end{equation} $$

our desired result.


Using $L_\xi g_{\mu\nu}=\nabla_\mu \xi_\nu+\nabla_\nu \xi_\mu$ is not the easiest way to proceed. Indeed, it is possible to give a two-line proof of the statement without ever writing the Lie derivative of the metric explicitly. Let $g$ be a metric and $\widetilde{g} = e^{\alpha}g$ a Weyl rescaling. Let $\xi$ be Killing for $g$ so that $L_\xi g=0$. We evaluate $L_\xi \widetilde{g}$. To do so use the Liebnitz rule obeyed by the Lie derivative

$$L_\xi \widetilde{g}= (L_\xi e^\alpha)g+e^\alpha L_\xi g.$$

Now the last term vanishes because $\xi$ is Killing for $g$. Moreover $L_\xi e^\alpha = \xi^\mu \partial_\mu e^\alpha$ since $e^\alpha$ is a $C^\infty(M)$ function. This is $$L_\xi e^\alpha = (\xi^\mu \partial_\mu \alpha)e^\alpha$$

Combining everything we have $$L_\xi \widetilde{g} = (\xi^\mu \partial_\mu \alpha)e^\alpha g = (\xi^\mu \partial_\mu \alpha)\widetilde{g},$$

so that $\xi$ is conformal Killing for $\widetilde{g}$.


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