How obtain the last expression of the Killing equation? In order to write down the Killing equation, if by definition a vector field $X$ is said to be Killing $\iff$ $L_X g=0$, then I can rewrite this condition as:
$$L_X g=X g(U, V)-g(L_XU, V)-g(U,L_XV)=g(\nabla_XU, V)+g(U, \nabla_XV)-g(L_XU,V)-g(U,L_XV)=g(\nabla_XU-L_XU, V)+g(U, \nabla_X V-L_XV)=g(\nabla_U X, V)+g(U, \nabla_V X)=0$$
Now my question is: I have read always the Killing equation as $\nabla_U X+\nabla_V X=0$, but what is the passage from "$g(\nabla_U X, V)+g(U, \nabla_V X)=0$" to "$\nabla_U X+\nabla_V X=0$"?
And how it can be translated in coordinates as $X_{\alpha;\beta}+X_{\beta;\alpha}=0$ (this is another expression I have found out)?
$\textbf{EDIT with my work:}$ I choose to write all in coordinates, so $X=X^{\alpha}\partial_\alpha$ and then
$$L_X g_{\sigma \beta}=X^{\alpha}\partial_\alpha g_{\sigma \beta}-g( \partial_\sigma, [X^{\alpha}\partial_\alpha, \partial_\beta])=X^{\alpha}\color{red}{\nabla_{\alpha} g_{\sigma \beta}}-g(X^\alpha[\partial_\alpha, \partial_\sigma]-(\partial_\sigma X^\alpha)\partial_\alpha,\partial_\beta)-g(\partial_\sigma, x^\alpha [\partial_\alpha, \partial_\beta]-(\partial_\beta X^\alpha)\partial_\alpha)=\color{lightgreen} 0+g_{\alpha \beta}\partial_\sigma X^\alpha+g_{\sigma\alpha}\partial_\beta X^\alpha=\color{pink}{\partial_\sigma g_{\alpha \beta}X^\alpha}+\color{violet}{\partial_\beta g_{\sigma\alpha} X^\alpha}=\nabla_\sigma X_{\beta}+\nabla_{\beta}X_{\sigma}=X_{\beta;\sigma}+X_{\sigma;\beta}$$
So my questions become:
$\textbf{1)}$ The passages are right?
$\textbf{2)}$What I have written in red is right? I have thought that $g_{\sigma\beta}$ can be seen as a scalar function for which so the covariant derivative coincides with the partial one.
$\textbf{3)}$In the pink and violet terms I have put into the partial derivatives the terms $g_{\alpha\beta}$ and $g_{\sigma\alpha}$, since the partial derivative is linear, right?
$\textbf{4)}$ The expression obtained is $X_{\beta;\sigma}+X_{\sigma;\beta}$ and not $X_{\beta;\alpha}+X_{\alpha;\beta}$ as in my book...where I am failing?
 A: You got the result but some steps are not correct. Here is how I would have done it:
The components of the Lie derivative of a 2-covariant tensor are
$$(L_X T)_{\mu\nu}=X^\lambda \partial_\lambda T_{\mu\nu}+T_{\lambda\nu}\partial_\mu X^\lambda+T_{\mu\lambda}\partial_\nu X^\lambda.\tag{1}$$
It can be proved that if there is no torsion, i.e. the connection is symmetric, all partial derivatives in $(1)$ can be replaced by covariant derivatives.
$$\Gamma^\mu_{\nu\lambda}=\Gamma^\mu_{\lambda\nu}\Leftrightarrow(L_X T)_{\mu\nu}=X^\lambda \nabla_\lambda T_{\mu\nu}+T_{\lambda\nu}\nabla_\mu X^\lambda+T_{\mu\lambda}\nabla_\nu X^\lambda \tag{2}$$
Besides, the covariant derivative of the metric tensor is $0$, so
$$(L_X g)_{\mu\nu}=X^\lambda \nabla_\lambda g_{\mu\nu}+g_{\lambda\nu}\nabla_\mu X^\lambda+g_{\mu\lambda}\nabla_\nu X^\lambda=\nabla_\mu(g_{\lambda\nu}X^\lambda)+\nabla_\nu(g_{\mu\lambda}X^\lambda)=\\=\nabla_\mu X_\nu+\nabla_\nu X_\mu,$$
then $$L_X g=0\Leftrightarrow\nabla_\mu X_\nu+\nabla_\nu X_\mu=0.$$
Regarding your last questions:


*$g_{\mu\nu}$ can't be seen as a scalar function.


*The partial derivative is linear, right.


*You started with $\sigma$ and $\beta$ as free indices, I chose $\mu$ and $\nu$, the book chose $\alpha$ and $\beta$...
