How to see the Lie derivative of the tensor metric $g$ in terms of the Levi-Civita connection

Question

According the first line on page $2$ of this paper,

A smooth vector field $\xi$ on a Riemannian manifold $(M, g)$ is said to be a conformal vector field if its flow consists of conformal transformations or, equivalently, if there exists a smooth function $f$ on $M$ (called the potential function of the conformal vector field $\xi$) that satisfies $\mathscr{L}_{\xi} g = 2fg$, where $\mathscr{L}_{\xi} g$ is the Lie derivative of $g$ with respect $\xi$.

By the other hand, this paper that I'm reading define in a different way:

A vector field $X$ is conformal if $\nabla_j X_i + \nabla_i X_j = 2 \lambda g_{ij}$ for a function $\lambda$.

I would like to know how can I see the Lie derivative of the tensor metric $g$ in terms of Levi-Civitta connection.

I'm not familiar with the Lie's derivative, then I saw in Lee's Introduction to Smooth Manifold the following corollary:

$\textbf{Corollary 12.33.}$ If $V$ is a smooth vector field and $A$ is a smooth covariant $k$-tensor field, then for any smooth vector fields $X_1, \cdots, X_k$ ,

$$\mathscr{L}_V A = V(A(X_1, \cdots, X_k)) - A([V,X_1], X_2, \cdots, X_k) - \cdots - A(X_1, \cdots, X_{k-1}, [V, X_k]).$$

Denoting by $\partial_i := \frac{\partial}{\partial x_i}$, defining $X = X^k \partial_k$ and applying this corollary to the tensor metric, I obtained

\begin{align*} \mathscr{L}_X g &= X(g(\partial_i,\partial_j)) - g([X,\partial_i], \partial_j) - g(\partial_i, [X,\partial_j])\\ &= X^k \frac{\partial g_{ij}}{\partial x_k} + g \left( \frac{\partial X^k}{\partial x_i} \partial_i, \partial_j \right) + g \left( \partial_i, \frac{\partial X^k}{\partial x_j} \partial_j \right)\\ &= X^k \frac{\partial g_{ij}}{\partial x_k} + \frac{\partial X^k}{\partial x_i} g_{ij} + \frac{\partial X^k}{\partial x_j} g_{ij}. \end{align*}

I'm stuck here.

I also read on this Wikipedia's article that

$\mathscr{L}_X g = (X^c g_{ab \ ; \ c} + g_{cb} X_{; \ a}^c + g_{ac} X_{; \ b}^c ) dx^a \otimes dx^b = (X_{b \ ; a} + X_{a \ ; b})dx^a \otimes dx^b$. (This is the last example of the section of Coordinate expressions and was explained in the beginning of this section the notation "$;$")

I didn't understand how this computation was done, but it seems that the notation "$;$" is the same of "$\nabla$" given in the second paper linked, which lead me to think that $\nabla_i X^j$ it's just a notation for the covariant derivative of a coordinate $X^j$ of the vector field $X^k \partial x_k$ in the direction $\partial x_i$, if I'm right, then the work it's just understand why $\mathscr{L}_X g = (X_{b \ ; a} + X_{a \ ; b})dx^a \otimes dx^b$. Am I right? If I'm right, then how can I deduce the expression above?

Thanks in advance!

Answer 1

Look at this question, and answers therein, where it is explained why one can replace partial derivatives with covariant ones in this case (provided that $\nabla$ is torsion-free). Thus, the term with the derivative of the metric drops off, and you come to your second definition of conformal Killing fields.

For a general reference, see e.g. R.Wald, General Relativity, p.441.

Answer 2

I found on the final of the page $14$ of this thesis how to prove that $\nabla_i X_j + \nabla_j X_i = (\mathscr{L}_X g)_{ij}$, which is more clear than in Wald's book. I will put here the development:

$\textit{Proof.}$ Let $\omega$ be the $1$-form dual to the vector field $X$, $\omega(Y) = \langle X, Y \rangle$. Using the product rule (from Lemma 1.6) and the metric compatibility and torsion-free conditions on the Levi-Civita connection we have

\begin{align*} \mathscr{L}_X g(Y,Z) &= X(g(Y,Z)) - g(\mathscr{L}_X Y, Z) - g(Y, \mathscr{L}_X Z)\\ &= \langle \nabla_X Y, Z \rangle + \langle Y, \nabla_X Z \rangle - \langle [X, Y], Z \rangle - \langle Y, [X, Z] \rangle\\ &= \langle \nabla_X Y - [X,Y], Z \rangle + \langle Y, \nabla_X Z - [X,Z] \rangle\\ &= \langle \nabla_Y X, Z \rangle + \langle Y, \nabla_Z X \rangle\\ &= Y \langle X, Z \rangle - \langle X, \nabla_Y Z \rangle + Z \langle Y, X \rangle - \langle \nabla_Z Y, X \rangle\\ &= Y(\omega(Z)) - \omega(\nabla_Y Z) + Z(\omega(Y)) - \omega(\nabla_Z Y)\\ &= (\nabla_Y \omega) (Z) + (\nabla_Z \omega) (Y), \end{align*}

which is the coordinate-free way of expressing the identity we wanted. Note that we use the product rule again to get the last line. $\square$