lie derivative clarification

Question

Suppose you have a (1,3) tensor $R^{\mu}_{\alpha\beta\gamma}$ where $R^{\mu}_{\alpha\beta\gamma}$ is the Riemann curvature tensor. I want to take the lie derivative $L_CR^{\mu}_{\alpha\beta\gamma}$ of this tensor, where C is a vector field. Computing this lie derivative we get $$L_CR^{\mu}_{\alpha\beta\gamma}=C^{\sigma}\nabla_{\sigma}R^{\mu}_{\alpha\beta\gamma}-\nabla_{\sigma}C^{\mu}R^{\sigma}_{\alpha\beta\gamma}+\nabla_{\alpha}C^{\sigma}R^{\mu}_{\sigma\beta\gamma}+\nabla_{\beta}C^{\sigma}R^{\mu}_{\alpha\sigma\gamma}+\nabla_{\gamma}C^{\sigma}R^{\mu}_{\alpha\beta\sigma}$$. This can be simplified by computing all the covariant derivatives. My question is, did I get all the index placements right for this lie derivative expression?

Answer 1

For any torsion-free connection $\nabla$, the Lie derivative of a tensor field $T^{a_1 \dots a_k}{}_{b_1 \dots b_l}$ along a vector field $V^c$ can be computed using the following formula: $$ L_V T^{a_1 \dots a_k}{}_{b_1 \dots b_l} = V^c \nabla_c T^{a_1 \dots a_k}{}_{b_1 \dots b_l} - \sum^{k}_{i = 1} T^{a_1 \dots a_{i-1} c a_{i+1} \dots a_k}{}_{b_1 \dots b_l} \nabla_c V^{a_i} \\ + \sum^{l}_{j = 1} T^{a_1 \dots a_k}{}_{b_1 \dots b_{j-1} c b_{j+1} \dots b_l} \nabla_{b_j} V^{c} $$

See R.M.Wald, General Relativity (1984), p.441.