1

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?

Arctic Char
  • 16,972
aygx
  • 200
  • 1
    Do you have a formula? Looks like a direct checking to me. – Arctic Char Oct 23 '21 at 14:57
  • 1
    Yes I had formulas for 2 index tensors but none for 4 index tensors. I'm just making sure that I'm correct in extending the definition out to any number of indices. – aygx Oct 23 '21 at 15:09
  • Your formula is correct but the usual convention is that the covariant derivative acts on the rest of the term to the right, so you should either use parentheses, or, place the differentiated factor as it is done in my answer below. – Yuri Vyatkin Oct 24 '21 at 12:58

1 Answers1

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.

Yuri Vyatkin
  • 11,679