Show that $\mathcal L_X g_{ij} = \nabla_i X_j + \nabla_j X_i$

Question

I’m trying to understand equation (1.8) on p. 4 of Chow et al.’s “The Ricci flow: techniques and applications”. There, the authors say that, using indices, the equation

$$ -2\operatorname{Rc}(g) = \epsilon g + \mathcal L_X g$$

becomes

$$ -2R_{ij} = \nabla_i X_j + \nabla_j X_i + \epsilon g_{ij}. \tag{1.8}$$

But how come is $\mathcal L_X g_{ij} = \nabla_i X_j + \nabla_j X_i$?

Does this answer your question? How to see the Lie derivative of the tensor metric $g$ in terms of the Levi-Civita connection — Travis Willse, Mar 19 '24 at 19:04

score 1 · Accepted Answer · answered Mar 19 '24 at 19:02

This question has already been asked several times (in different phrasings), e.g. here and here.

The best answer in my opinion is this one: https://math.stackexchange.com/a/3189815/1060681

Its from page 14 of https://web.math.princeton.edu/~nsher/ricciflow.pdf.

Show that $\mathcal L_X g_{ij} = \nabla_i X_j + \nabla_j X_i$

1 Answers1