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In proving that it is possible to write the killing equation in coordinates as $$L_X g=0\iff X_{\alpha;\beta}+X_{\beta;\alpha}=0$$ I have read that the key observation, to write the equation in coordinates as above, is that when we consider the Levi-Civita connection we can replace the partial derivative with the covariant ones, i.e $Xg_{\sigma\beta}=\nabla_X g_{\sigma\beta}$.

This is my work, with $X=X^\alpha\partial_\alpha$: $$L_X g=Xg_{\sigma \beta}-g([X,\partial_\sigma]\partial_\beta)-g(\partial_\sigma, [X,\partial_\beta])$$ But I can't understand why...can you help me?

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  • $\begingroup$ Why don't you compute the Lie derivative and then plug in $\frac\partial{\partial^\alpha}$ and $\frac\partial{\partial^\beta}$ i.e $L_Xg(\frac\partial{\partial^\alpha}, \frac\partial{\partial^\beta})=0$. $\endgroup$
    – MBN
    Mar 29 at 11:17
  • $\begingroup$ @MBN This is exactly what I have done...see my edit $\endgroup$
    – pawel
    Mar 29 at 12:38
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The point is that, for the Levi-Civita connection, we can find Riemann normal coordinates around any point P. In those coordinates $\partial_\alpha g_{\mu\nu}=0$ and the ${\Gamma^\alpha}_{\mu\nu}=0$ at P. Now the formula for the Lie derivative $$ [{\mathcal L}_Xg]_{\mu\nu}= X^\alpha \partial_\alpha g_{\mu\nu} + g_{\mu\alpha}\partial_\nu X^\alpha + g_{\alpha\nu }\partial_\mu X^\alpha $$ and the covariant derivative expression $$ \nabla_{\mu}X_\nu + \nabla_\nu X_\mu $$ coincide if both $\partial_\alpha g_{\mu\nu}=0$ and ${\Gamma^\alpha}_{\mu\nu}=0$. But both of these quanties are doubly-covariant tensors and if two tensors of the same type coincide in one system of coordinates, they coincide in all coordinate systems.

We conclude that
$$ [{\mathcal L}_Xg]_{\mu\nu} =\nabla_{\mu}X_\nu + \nabla_\nu X_\mu $$ at the point P --- but P is an arbitrary point, so the two expressions coincide everywhere.

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  • $\begingroup$ Thanks! So the fact that I can consider to take the normal coordinates around an arbitrary point $p$, tells me that since the christoffel's symbols are zero, then the covariant derivative can be replaced by the partial one and viceversa? $\endgroup$
    – pawel
    Mar 29 at 12:56
  • $\begingroup$ Exactly! It is important to remember that being able to set the ${\Gamma^\alpha}_{\mu\nu}=0$ at an arbitrary point is a property only of the Levi-Cita connection. It does not hold for connections with torsion. This, and he fact that we have local intertial frames (on a scale such that we can ignore tidal forces) in which ${\Gamma^\alpha}_{\mu\nu}=0$ is why GR uses the torsion-free LC connection. $\endgroup$
    – mike stone
    Mar 29 at 13:01
  • $\begingroup$ Thanks a lot now it is all clear! $\endgroup$
    – pawel
    Mar 29 at 13:20

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