Weyl Tensor invariant under conformal transformations

[Al X]

Homework Statement

As the title says, I need to show this. A conformal transformation is made by changing the metric:
$g_{\mu\nu}\mapsto\omega(x)^{2}g_{\mu\nu}=\tilde{g}_{\mu\nu}$

Homework Equations

The Weyl tensor is given in four dimensions as:
$C_{\rho\sigma\mu\nu}=R_{\rho\sigma\mu\nu}-\left(g_{\rho\left[\mu\right.}R_{\left.v\right]\sigma}-g_{\sigma\left[\mu\right.}R_{\left.v\right]\rho}\right)+\frac{1}{3}g_{\rho\left[\mu\right.}g_{\left.\nu\right]\sigma}R$
where $R_{\mu\nu}$ is the Ricci tensor, $R$ is the Ricci scalar, and $R_{\rho\sigma\mu\nu}$ is the Riemann tensor

The Attempt at a Solution

$\begin{eqnarray*} \tilde{g}_{\rho\left[\mu\right.}R_{\left.v\right]\sigma}&=&\frac{1}{2}\left(\tilde{g}_{\rho\mu}R_{\nu\sigma}-\tilde{g}_{\rho\nu}R_{\mu\sigma}\right)=\frac{1}{2}\omega(x)^{2}\left(g_{\rho\mu}R_{\nu\sigma}-g_{\rho\nu}R_{\mu\sigma}\right)\\\tilde{g}_{\sigma\left[\mu\right.}R_{\left.v\right]\rho}&=&\frac{1}{2}\left(\tilde{g}_{\sigma\mu}R_{\nu\rho}-\tilde{g}_{\sigma\nu}R_{\mu\rho}\right)=\frac{1}{2}\omega(x)^{2}\left(g_{\sigma\mu}R_{\nu\rho}-g_{\sigma\nu}R_{\mu\rho}\right)\\\tilde{g}_{\rho\left[\mu\right.}\tilde{g}_{\left.\nu\right]\sigma}&=&\frac{1}{2}\left(\tilde{g}_{\rho\mu}\tilde{g}_{\nu\sigma}-\tilde{g}_{\rho\nu}\tilde{g}_{\mu\sigma}\right)=\frac{1}{2}\omega(x)^{4}\left(g_{\rho\mu}g_{\nu\sigma}-g_{\rho\nu}g_{\mu\sigma}\right) \end{eqnarray*}$

From here, I am lost. How do I make the $\omega(x)$ vanish?

 

[Twigg]

If $g_{\mu\nu} \to \omega (x)^{2}g_{\mu\nu}$, what about $\Gamma_{\mu\nu}^{\sigma}$? Is $\tilde{R}_{\rho\sigma\mu\nu}$ the same as $R_{\rho\sigma\mu\nu}$?

 

[Al X]

Thanks! That gave me the push in the right direction! Managed to solve it now.

 

[AmaroJuuulianoni]

Sorry to dig an old post, but I’m currently struggling with Weyl tensor conformal invariance as well.

I started with the following assumptions:

- the invariant tensor is not $C_{abcd}$ but $C^a\,_{bcd}$
- Connection was not metrically compatible
- the transformation I considered was slightly different but basically equivalent: $g_{\mu\nu}=e^{-2\omega}g_{\mu\nu}$

In this case, invariance was immediate as neither $\Gamma^\mu_{\nu\rho}$, nor the Riemann or the Ricci change under conformal rescaling, but only the scalar curvature and the Ricci with one index up ($R^\mu\,_\nu\equiv g^{\mu\lambda}R_{\lambda\nu}$).

Now, I was trying to prove conformal invariance with metrical connection, so with nontrivial modifications of connection coefficients, Riemann, Ricci tensor and scalar: as I’m stuck with huge formulas, could anyone confirm if it’s normal or if I’m missing some simplifying argument?
I’m not asking for detailed calculations but feel free to post them if you want.

Thanks!