I'm currently studying Carroll's GR book Spacetime & Geometry, and ran into some trouble understanding the text. When discussing Killing vectors, Carroll mentions that one can derive $$K^{\lambda}\nabla_{\lambda}R=0$$ That is, the directional derivative of the Ricci scalar along a Killing vector field vanishes (here, $K^{\lambda}$ is a Killing vector).

He remarks that the only necessary ingredients to derive this equation are: $$\cdot\;\;\text{The Killing equation:}\ \nabla_{(\mu}K_{\nu )}=0$$

$$\cdot\;\;\text{The Bianchi identity:}\ \nabla_{[\mu}R_{\nu\rho ]\sigma\lambda}=0$$

$$\cdot\;\;\ \nabla_{\mu}\nabla_{\nu}K^{\rho}=R^{\rho}_{\lambda\mu\nu}K^{\lambda}\longrightarrow \nabla_{\mu}\nabla_{\nu}K^{\mu}=R^{\mu}_{\lambda\mu\nu}K^{\lambda}=R_{\lambda\nu}K^{\lambda} $$

With $R_{\mu\nu\rho\sigma}$ the Riemann curvature tensor and $R_{\mu\nu}$ the Ricci tensor.

Although I was able to derive all of the equations involved, I don't see how to put them together to get the sought-after result.

The only way I saw to get started is the following: $$K^{\lambda}\nabla_{\lambda}R=\nabla_{\lambda}(K^{\lambda}R)-R\nabla_{\lambda}K^{\lambda}=g^{\mu\sigma}(\nabla_{\lambda}(R_{\sigma\mu}K^{\lambda})-R_{\sigma\mu}\nabla_{\lambda}K^{\lambda}) $$ which immediately runs me into trouble because I don't see how to simplify/manipulate either of the terms to even make use of any of the three 'ingredients'


2 Answers 2


Start with the following form of the Bianchi Identities $$ \nabla^\mu R_{\mu\nu} = \frac{1}{2} \nabla_\nu R $$ Contract both sides with $K^\nu$. We find $$ \frac{1}{2} K^\nu \nabla_\nu R = K^\nu \nabla^\mu R_{\mu\nu} = \nabla^\mu \left( K^\nu R_{\mu\nu} \right) - R_{\mu\nu} \nabla^\mu K^\nu $$ The second term vanishes due to symmetry of $R_{\mu\nu}$. Now, recall that $R_{\mu\nu}K^\nu = \nabla_\nu \nabla_\mu K^\nu$. We now use the following fact $$ \left[ \nabla_\rho, \nabla_\sigma \right] \tau^{\mu\nu} = R^\mu{}_{\lambda\rho\sigma} \tau^{\lambda\nu} + R^\nu{}_{\lambda\rho\sigma}\tau^{\mu\lambda} $$ This implies \begin{equation} \begin{split} \nabla^\mu \left( K^\nu R_{\mu\nu} \right) &= \nabla_\mu\nabla_\nu \nabla^\mu K^\nu = \nabla_{[\mu}\nabla_{\nu ]} \nabla^{[\mu} K^{\nu]} = \frac{1}{2}[\nabla_{\mu}, \nabla_{\nu }] \nabla^{[\mu} K^{\nu]} \\ &=\frac{1}{2}\left(R^\mu_{\;\;\lambda\mu\nu} \nabla^{[\lambda} K^{\nu]}-R^\nu_{\;\;\lambda\nu\mu} \nabla^{[\mu} K^{\lambda]}\right)\\ &=\frac{1}{2}\left(R_{[\lambda\nu ]}\nabla^{[\lambda} K^{\nu]}-R_{[\lambda\mu ]}\nabla^{[\mu} K^{\lambda]}\right) = 0 \end{split} \end{equation} which then implies $$ K^\nu \nabla_\nu R = 0 $$


This starts off pretty much the same as Prahar's answer. Use the Bianchi identities $ \nabla^\mu R_{\mu\nu} = \frac{1}{2} \nabla_\nu R$ and then contract with $K^\nu$ to obtain $\frac{1}{2} K^\nu \nabla_\nu R = K^\nu \nabla^\mu R_{\mu\nu} = \nabla^\mu \left( K^\nu R_{\mu\nu} \right) - R_{\mu\nu} \nabla^\mu K^\nu$. So far, completely identical.

But now, using $R_{\mu\nu} = R_{\nu\mu}$, we get

$$ R_{\mu\nu} \nabla^\mu K^\nu = R_{\nu\mu} \nabla^\mu K^\nu $$

Renaming the indices gives us

$$ R_{\mu\nu} \nabla^\mu K^\nu = R_{\mu\nu} \nabla^\nu K^\mu $$

Either $R^{\mu\nu}=0$, which implies $R=0$, and makes it obvious that $K^\lambda\nabla_\lambda R=0$, or $R\neq 0$, in which case

$$ \nabla_\mu K_\nu-\nabla_\nu K_\mu=0 $$

Adding that to the Killing equation $\nabla_\mu K_\nu+\nabla_\nu K_\mu=0$ implies

$$ \nabla_\mu K_\nu=0 $$

So we get

\begin{equation} \begin{split} \frac{1}{2} K^\nu \nabla_\nu R &= \nabla^\mu \left( K^\nu R_{\mu\nu} \right) - R_{\mu\nu} \nabla^\mu K^\nu \\ &=\nabla^\mu \left( K^\nu R_{\mu\nu} \right) \end{split} \end{equation}

Using the Ricci tensor equation, this equals

$$ \nabla^\mu\nabla_\nu\nabla_\mu K^\nu=\nabla_\mu\nabla_\nu\nabla^\mu K^\nu $$

But we already know that $\nabla^\mu K^\nu=0$, therefore, we get

$$ \frac{1}{2} K^\nu \nabla_\nu R=0 $$

which means

$$ K^\lambda \nabla_\lambda R=0 $$

  • $\begingroup$ This is incorrect. $\endgroup$
    – Prahar
    Commented May 12, 2020 at 21:13
  • 1
    $\begingroup$ $\nabla^\mu K^\nu$ is most definitely not equal $0$. $\endgroup$
    – Y2H
    Commented Jun 6, 2020 at 21:45

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