# Showing that the Weyl tensor is invariant under conformal symmetries

• JD_PM
In summary, the Weyl tensor is a tensor obtained from the Riemann tensor by removing all traces and is given by a complicated expression involving the metric tensor and its derivatives. By considering infinitesimal rescaling of the metric, it is shown that the Weyl tensor is invariant under arbitrary rescaling. This is proven by using Riemann normal coordinates and showing that the terms involving the function f(x) cancel out, resulting in the Weyl tensor being a tensor of the same form as the original one.
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.

Last edited:
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 $$\delta C^{\mu}{}_{\nu\alpha\beta} = 0 .$$

JD_PM

## 1. How do conformal symmetries affect the Weyl tensor?

The Weyl tensor is a mathematical object that describes the curvature of space-time in general relativity. Conformal symmetries are transformations that preserve the angles between curves and therefore do not change the geometry of space-time. As a result, the Weyl tensor is invariant under conformal symmetries, meaning that it remains unchanged.

## 2. What is the significance of showing that the Weyl tensor is invariant under conformal symmetries?

Showing that the Weyl tensor is invariant under conformal symmetries is significant because it highlights the deep connection between conformal symmetries and the geometry of space-time. It also allows us to simplify calculations and better understand the behavior of the Weyl tensor in different situations.

## 3. How is the invariance of the Weyl tensor under conformal symmetries proven?

The invariance of the Weyl tensor under conformal symmetries can be proven using mathematical techniques such as tensor calculus and differential geometry. By applying conformal transformations to the Weyl tensor and using its properties, it can be shown that the resulting tensor is equivalent to the original Weyl tensor.

## 4. Can the invariance of the Weyl tensor under conformal symmetries be generalized to other tensors?

Yes, the invariance of the Weyl tensor under conformal symmetries can be generalized to other tensors. In general, any tensor that is invariant under conformal symmetries is said to have conformal weight zero. This includes tensors such as the metric tensor and the Riemann tensor.

## 5. How does the invariance of the Weyl tensor under conformal symmetries relate to the principle of general covariance?

The principle of general covariance states that the laws of physics should be independent of the choice of coordinates used to describe them. The invariance of the Weyl tensor under conformal symmetries is a manifestation of this principle, as conformal transformations do not change the physical laws described by the Weyl tensor. This highlights the important role of conformal symmetries in the framework of general relativity.

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