Showing that the Weyl tensor is invariant under conformal symmetries

[JD_PM]

Homework Statement

Show that the Weyl tensor $C^{\mu}_{ \ \nu \sigma \rho}$ is left invariant under a conformal transformation of the metric i.e.

\begin{equation*}
g_{\mu \nu}(x) \to \tilde g_{\mu \nu}(x) = \Omega^2(x) g_{\mu \nu}(x)
\end{equation*}

Relevant Equations

N/A

The Weyl tensor is given by (Carroll's EQ 3.147)

\begin{align*}
C_{\rho \sigma \mu \nu} &= R_{\rho \sigma \mu \nu} - \frac{2}{n-2}\left(g_{\rho [\mu}R_{\nu]\sigma} - g_{\sigma [\mu}R_{\nu]\rho}\right) \\
&+ \frac{2}{(n-1)(n-2)}g_{\rho [\mu}g_{\nu]\sigma}R
\end{align*}

Where $n$ are the number of dimensions

Rising the first index (i.e. $C^{\mu}_{ \ \nu \rho \sigma}=g^{\mu \lambda}C_{\lambda \nu \sigma \rho}$) yields

\begin{align*}
C^{\rho}_{ \ \sigma \mu \nu} &= R^{\rho}_{ \ \sigma \mu \nu} - \frac{2}{n-2}\left(\delta^{\rho}_{ [\mu}R_{\nu]\sigma} - g_{\sigma [\mu}R_{\nu]}^{\rho}\right) \\
&+ \frac{2}{(n-1)(n-2)}\delta^{\rho}_{ [\mu}g_{\nu]\sigma}R
\end{align*}

Let us work out the individual terms that explicitly have the $g_{\mu \nu}$ term

\begin{equation*}
\delta^{\rho}_{[\mu}g_{\nu] \sigma} = \frac{1}{2} \left(\delta^{\rho}_{\mu} \tilde g_{\nu \sigma} - \delta^{\rho}_{\nu} \tilde g_{\mu \sigma} \right) = \frac{1}{2} \Omega^2(x) \left(\delta^{\rho}_{\mu} g_{\nu \sigma} - \delta^{\rho}_{\nu} g_{\mu \sigma} \right)
\end{equation*}

\begin{equation*}
\tilde g_{\sigma[\mu} R_{\nu]}^{ \ \ \rho} = \frac{1}{2}\left( \tilde g_{\sigma \mu} R_{\nu}^{ \ \ \rho} - \tilde g_{\sigma \nu} R_{\mu}^{ \ \ \rho} \right) = \frac{1}{2} \Omega^2 (x) \left( g_{\sigma \mu}R_{\nu}^{ \ \ \rho} -g_{\sigma \nu}R_{\mu}^{ \ \ \rho} \right)
\end{equation*}

Now the question are

1) How to work out the Riemann tensor $R^{\rho}_{ \ \sigma \mu \nu}$? I am quite sure that we should not go for the brute force method i.e. apply the definition of the Riemann tensor straightaway. I have been thinking that using Riemann normal coordinates could be a good idea, as the $\Gamma \Gamma$ terms would vanish. Is this the right approach?

2) What to do with $\delta^{\rho}_{ [\mu}R_{\nu]\sigma}$ term?

I have been trying to perform the transformation but I do not get rid of the terms with $\Omega(x)$

Thank you!

PS: The exact same question was asked here but I happen to have the same main doubt: how do the $\Omega(x)$ terms cancel each other out.

PS2: I have been thinking a lot about this one. Any little help will be much appreciated.

 

[samalkhaiat]

I hate these type of problems which involve backward reasoning. The Weyl tensor does not drop down from the sky. It is obtained from R^{\mu}{}_{\nu\rho\sigma} by subtracting the piece that is not invariant under arbitrary rescaling of the metric, or, equivalently, by removing all the traces from the Riemann tensor. That is how Weyl obtained his tensor. Anyway, your problem involves lengthy but straightforward algebra. So, you should do it on a rainy day. I will only explain the important steps in the proof. Also, to make life easier, consider infinitesimal rescaling, i.e., set \Omega^{2}(x) = \left( 1 + \frac{\epsilon}{2} f(x) \right)^{2} \approx 1 + \epsilon \ f(x), with 0 < \epsilon \ll 1 being the infinitesimal parameter, and f(x) is an arbitrary real function. So, the infinitesimal variation of the metric tensor is \delta g_{\mu\nu}(x) = g^{\prime}_{\mu\nu}(x) - g_{\mu\nu}(x) = \epsilon \ f(x) \ g_{\mu\nu}(x) . From now on, we will treat the variation symbol \delta as an operator satisfying the following properties: \delta (A + B) = \delta A + \delta B,\delta (AB) = (\delta A)B + A (\delta B),\delta (\partial A) = \partial (\delta A) , \ \ \ \ \ \ \delta (\mbox{const.}) = 0 . From these rules, you find \delta g^{\rho \sigma} = - g^{\rho \mu}g^{\sigma \nu} \delta g_{\mu\nu} = - \epsilon \ f(x) \ g^{\rho\sigma}. Next, calculate the infinitesimal change (i.e., up to first order in \epsilon) in the connection (this is lengthy but easy calculation using the above rules): \delta \Gamma^{\sigma}_{\mu\nu} = \frac{\epsilon}{2}\left( \delta^{\sigma}_{\mu} \ \partial_{\nu}f + \delta^{\sigma}_{\nu}\ \partial_{\mu}f - g_{\mu\nu} \ \partial^{\sigma}f \right) . This tells you that \delta \Gamma is a tensor (why?). Now, we can calculate the infinitesimal change in the Riemann tensor: \delta R^{\tau}{}_{\nu\alpha\beta} = \delta \left( \partial_{\beta} \Gamma^{\tau}_{\nu \alpha} - \partial_{\alpha} \Gamma^{\tau}_{\nu \beta} + \Gamma^{\tau}_{\rho \beta} \Gamma^{\rho}_{\nu \alpha} - \Gamma^{\tau}_{\rho \alpha} \Gamma^{\rho}_{\nu \beta} \right). Using the above rules together with the fact that \delta \Gamma is a tensor, we find \delta R^{\tau}{}_{\nu\alpha\beta} = \nabla_{\beta}\left(\delta \Gamma^{\tau}_{\nu \alpha}\right) - \nabla_{\alpha} \left( \delta \Gamma^{\tau}_{\nu \beta} \right) . Substituting this in \delta R_{\mu\nu\alpha\beta} = \delta \left( g_{\mu \tau}R^{\tau}{}_{\nu \alpha \beta} \right) = g_{\mu \tau} \ \delta R^{\tau}{}_{\nu \alpha \beta} + \delta g_{\mu \tau} \ R^{\tau}{}_{\nu \alpha \beta} , we get \delta R_{\mu \nu \alpha \beta} = \frac{\epsilon}{2} \left( 2 f R_{\mu \nu \alpha \beta} + g_{\mu \alpha} \nabla_{\nu}(\partial_{\beta}f) - g_{\mu \beta} \nabla_{\nu} (\partial_{\alpha}f) + g_{\beta \nu} \nabla_{\alpha}(\partial_{\mu}f) - g_{\alpha \nu} \nabla_{\beta} (\partial_{\mu}f) \right). The same boring thing allows you to calculate \delta R_{\mu\nu} = \delta \left( g^{\rho \sigma} R_{\rho \mu \sigma \nu}\right) = \frac{\epsilon}{2} \left( g_{\mu\nu}g^{\rho\sigma}\nabla_{\sigma}(\partial_{\rho}f) + ( n - 2) \nabla_{\mu}(\partial_{\nu}f ) \right) ,\delta R = \delta (g^{\mu\nu}R_{\mu\nu}) = \epsilon \ (n - 1) \ g^{\mu\nu}\nabla_{\mu}(\partial_{\nu} f ) - \epsilon \ f \ R . Now, if you substitute everything in the (God-giving) Weyl tensor, you will get \delta C_{\mu\nu\alpha\beta} = \epsilon \ f \ C_{\mu\nu\alpha\beta}, which is equivalent to \delta C^{\mu}{}_{\nu\alpha\beta} = 0 .