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.
Best Answer
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.