# Lie derivative of Riemann tensor along killing vector ( = 0 )

I'm currently learning the mathematical framework for General Relativity, and I'm trying to prove that the Lie derivative of the Riemann curvature tensor is zero along a killing vector.

With the following notation for covariant differentiation, $A_{a||b}$ (instead of $\nabla_b A_a$ ), I have the following:

$\it\unicode{xA3}_\xi R_{amsq} = R_{amsq||x} \xi ^x + R_{xmsq} \xi^x{}_{||a} + R_{axsq} \xi^x{}_{||m} + R_{amxq} \xi^x{}_{||s} + R_{amsx} \xi^x{}_{||q}$.

I suspect that I need to invoke the second Bianchi identity. However, before I can do this, I need to somehow get this into a different form. There has to be some property of either killing vectors or maybe covariant derivatives that I'm forgetting/failed to learn. Any help would be appreciated.

• This was done as part of the following answer: (physics.stackexchange.com/q/88655) Commented Dec 3, 2013 at 6:46
• Yes it was. However, I had difficulty understanding that answer and would like to understand how to do it this way. That is to say, I'd really like to know what property or identity that I'm missing before I can use use the Bianchi identities to show that it is manifestly zero. Commented Dec 3, 2013 at 20:52
• (LfX )(p) = f(p)(LX )(p) + (df) ^ (iX )(p)
– user38412
Commented Jan 28, 2014 at 6:06

The other proof uses the first Bianchi identity. That's where the starting assumption $R^a{}_{bcd}\xi^d = \xi^a{}_{;bc}$ comes from. If you want to use the second Bianchi identity, it is $$(\nabla_\xi R)(X,Y) + (\nabla_X R)(Y,\xi) + (\nabla_Y R)(\xi,X) = 0\text{,}$$ and therefore applying it and the Leibniz rule produces: \begin{align} \underbrace{\nabla_\xi[R(X,Y)]+\nabla_X[R(Y,\xi)]+\nabla_Y[R(\xi,X)]}_\mathrm{foo} = \underbrace{R(\mathcal{L}_\xi X,Y) + R(\mathcal{L}_XY,\xi) + R(\mathcal{L}_Y\xi,X)}_\mathrm{bar}\text{,} \end{align} where it was assumed that the torsion vanishes, so that $\mathcal{L}_AB = \nabla_AB-\nabla_BA$. Additionally, $$\begin{eqnarray*} (\mathcal{L}_\xi R)(X,Y) &=& \mathcal{L}_\xi[R(X,Y)] - R(\mathcal{L}_\xi X,Y) - R(X,\mathcal{L}_\xi Y)\\ &=&\mathcal{L}_\xi[R(X,Y)] - R(\mathcal{L}_\xi X,Y) - R(\mathcal{L}_Y\xi,X)\\ &=&\underbrace{\mathcal{L}_\xi[R(X,Y)] - [\mathrm{foo}] + R(\mathcal{L}_XY,\xi)}_\mathrm{qux}\text{.} \end{eqnarray*}$$ So the objective is to show that the right-hand side, $\mathrm{qux}$, is identically zero whenever $\xi$ is a Killing vector field.
Let's write $S^a{}_b = [R(X,Y)]^a{}_b = R^a{}_{bcd}X^cY^d$, and just crank it out: $$\begin{eqnarray*} \mathcal{L}_\xi S^a{}_b &=& \nabla_\xi S^a{}_b - S^e{}_b\xi^a{}_{;e} + S^a{}_e\xi^e{}_{;b}\\ &=& \nabla_\xi S^a{}_b + X^cY^d(R^a{}_{ecd}\xi^e{}_{;b} - R^e{}_{bcd}\xi^a{}_{;e})\\ &=& \nabla_\xi S^a{}_b + X^cY^d(\nabla_c\nabla_d-\nabla_d\nabla_c)\xi^a{}_{;b}\text{,} \end{eqnarray*}$$ where the last step is actually valid for arbitrary $Z^a{}_b$, not just $\xi^a{}_{;b}$. The first term of this cancels with the first term of $\mathrm{foo}$. So far we have not used the fact that $\xi$ is a Killing vector field. Let's do so now by considering the other two terms of $\mathrm{foo}$: $$\nabla_X[R(Y,\xi)]^a{}_b - \nabla_Y[R(X,\xi)]^a{}_b = \nabla_X\nabla_Y\xi^a{}_{;b} - \nabla_Y\nabla_X\xi^a{}_{;b}\text{,}$$ where the starting identity $R^a{}_{bcd}\xi^d = \xi^a{}_{;bc}$ was used. The same identity also gives: $$R(\mathcal{L}_XY,\xi)^a{}_{b} = \nabla_{[X,Y]}\xi^a{}_{;b}\text{.}$$ Therefore, we have shown that for any vector fields $X,Y$, $$\begin{eqnarray*} X^cY^d(\mathcal{L}_\xi R^a{}_{bcd}) &=& \left[X^cY^d(\nabla_c\nabla_d-\nabla_d\nabla_c)-(\nabla_X\nabla_Y-\nabla_Y\nabla_X) + \nabla_{[X,Y]}\right]\xi^a{}_{;b}\\ &=& 0\text{.}\end{eqnarray*}$$ (If you have trouble with the last step, check Christoph's answer to the other question and modify appropriately.) Thus $\mathcal{L}_\xi R^a{}_{bcd} = 0$, QED.
• Ah, now I see! The first line you posted has exactly what I needed, that is, $R^a{}_{bcd} \xi^d = \xi^a{}_{;bc}$. I've got the rest under control. Thanks. Commented Dec 4, 2013 at 20:21