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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!

George
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2 Answers2

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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. $\square$

George
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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.

Yuri Vyatkin
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  • Sorry for taking too long to answer, but I tried understand the topic linked and the Lie Derivatives's section in Wald's book. I think Wald's it's a little bit more clear about the relation between Lie derivative and Levi-Civita connection, but I don't be sure if I understood Wald since the notation of the book is confuse for me, for example, on page $441$ he assumes that $v^a$ denote a vector field, but in the previous page he states that $v^a = \left( \frac{\partial}{\partial x_1} \right)^a$, but I don't know what this means. Furthermore, – George Apr 09 '19 at 12:48
  • he quotes in the beginning of page $441$ the equation $3.1.2$ (it's the final of the page $31$), which is the local expression for $[v,w]$, and there he states that $v^a$ denotes a vector field, but it seems that the $v^a$ it's the $a^{th}$-component of the vector field, so I really don't understand what is $v^a$ exactly. Other strange notation for me it's $\mu_a w^a$, is it means that $\mu_a (w^a)$? Because $w^a$ and $\mu_a$ it's the vector and dual vector, respectively, according Wald, but there is an index $a$ above and other below, then, probably, we need assume an contraction here, but – George Apr 09 '19 at 12:56
  • I really don't understand why we have a contraction here (if it is a contraction in fact). By the same reason that I explained before for $v^a$, I understood that $w^a$ is the $a^{th}$-component of the vector $w$ and $\mu_a$ is the $a^{th}$-component of the dual vector $\mu$. Am I right? If not, what exactly the notations $w^a$ and $\mu_a$ and the sum $\mu_aw^a$ means? – George Apr 09 '19 at 13:00
  • @George Wald uses the abstract index notation. I wrote a brief outline of this notation in my answer here. So, $w^a$ means that $w$ is a vector, and $\mu_a$ means that $\mu$ is a 1-form. No more than that. – Yuri Vyatkin Apr 09 '19 at 19:49
  • Abstract indices can be dropped. By convention, the abstract indices are placed where the coordinate indices would be placed for that objects. With this in mind, $\mu_aw^a$ means just $\mu(w)$, the natural action of 1-forms on vectors, which in the coordinate indices would be the sum of products $\mu_aw^a$ (the Einstein summation). This is somewhat tricky, I know :) – Yuri Vyatkin Apr 09 '19 at 19:49
  • I understood, thanks! Just more one question: do you know what this notation for a conformal vector field that is used in the second paper that I linked? Because what the Wald did don't seem the same that the author of the paper did if I consider in the paper the abstract index notation, because if I denote by $\omega$ the $1$-form associated to the vector field $X$ by the metric $g$, then $X_i = \omega = X_j$ by the abstract index notation (if I understood correctly), then $\nabla_j \omega + \nabla_i \omega = \nabla_j X_i + \nabla_i X_i = 2 \lambda g_{ij}$, but – George Apr 11 '19 at 17:44
  • it seems that he assumes a symmetry of $ \nabla_j X_i$ in the following and I don't know what this symmetry would mean if we have $\nabla_j \omega + \nabla_i \omega = \nabla_j X_i + \nabla_i X_i$, which led me to think that the notation of the author is not the same of the Wald's notation. – George Apr 11 '19 at 17:44
  • @George 1) both papers you've cited don't use abstract indices: their indices denote components; 2) formulas in components look identical to abstract indices, when they make sense; in particular, the indices must be balanced: $X_i = \omega = X_j$ does not make sense neither in abstract index notation, not in the classical component notation ($X_i$ is a function); 3) Vector field $X$ in a choice of coordinates has components $X^i$, but using the metric we can convert them to $X_i = X^k g_{k i}$ and call the latter components of $X$ too. You need to be careful with the notation :) – Yuri Vyatkin Apr 12 '19 at 08:50