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I have already seen a similar question but I have not sure to have understand completely so I hope you can help me.

If I write the killing equation $\cal{L}_X g=0$ as $X_{\alpha;\beta}+X_{\beta;\alpha}=0$, the only way to obtain this last expression is to consider the normal coordinates in order to have that the partial derivatives are nothing but covariant derivatives? Or this holds in whatever local coordinates I can choose?

If not how it is possibile to write the following? $$\color{red}{Xg_{\sigma\beta}}-g([X,\partial_\sigma],\partial_\beta)-g(\partial_\sigma], [X,\partial_\beta])=\color{red}{X^{\alpha}\nabla_{\partial_\alpha} g_{\sigma\beta}}+...$$

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    $\begingroup$ You don't need to use normal coordinates. The other questions on here explicitly show how to go from the usual Lie derivative expression (with partial derivatives) to the covariant derivatives in the question. $\endgroup$ – Eletie Apr 7 at 16:27
  • $\begingroup$ Ok thanks! so this physics.stackexchange.com/q/625571 is for me misleading! Can you indicate me the answers where the passages are shown please? $\endgroup$ – willie Apr 7 at 16:31
  • $\begingroup$ Using normal coordinates is completely valid. I've added an answer using a different method though. $\endgroup$ – Eletie Apr 7 at 16:50
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As stated in my comment, you can do this without making reference to normal coordinates. E.g. see Confusion about Lie derivative on metric. Here's a quick proof in local coordinates which I haven't seen answered here though.

Take the standard form of Killing's equation, $$ \begin{align} \tag{1} \mathcal{L}_{\xi} g_{\mu \nu} &= 0 \\ &= \xi^{c} \partial_{c} g_{ab} + g_{c b} \partial_{a} \xi^{c} + g_{c a} \partial_{b} \xi^{c} \ . \end{align} $$ Rewrite the first term using the metricity condition (with a Levi-Civita connection) $$ \tag{2} \xi^{c} \nabla_{c} g_{ab} = 0 = \xi^{c} \partial_{c} g_{ab} - \xi^{c} \Gamma_{c a}^{d} g_{db} - \xi^{c} \Gamma_{d a}^{d} g_{cb} \\ \implies \xi^{c} \partial_{c} g_{ab} = \xi^{c} \Gamma_{c a}^{d} g_{db} +\xi^{c} \Gamma_{cb}^{d} g_{ad} $$ which you can plug into (1) to obtain $$ \tag{3} \xi^{c} \Gamma_{c a}^{d} g_{db} +\xi^{c} \Gamma_{cb}^{d} g_{ad} + g_{c b} \partial_{a} \xi^{c} + g_{c a} \partial_{b} \xi^{c} = 0 \ . $$ You can then easily verify that this is equivalent to $$ \begin{align} \tag{4} \nabla_{a} \xi_b + \nabla_{b} \xi_{a} &= g_{cb} \nabla_{a} \xi^{c} + g_{ca} \nabla_{b} \xi^{c} \\ &= g_{cb} \partial_{a} \xi^{c} + g_{cb} \Gamma^{c}_{a d} \xi^{d} + g_{ca} \partial_{b} \xi^{c} + g_{ca} \Gamma_{bd}^{c} \xi^{d} \ . \end{align} $$ Hence $ \mathcal{L}_{\xi}g_{\mu \nu} = 2 \nabla_{(a} \xi_{b)} $ holds. Note that you can do a similar thing for a general tensor, except in (2) you'll have an extra term (the covariant derivative of the tensor which won't vanish in general).

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  • $\begingroup$ Thanks a lot! Now I will try to understand all passages! One thing: in (3) I can rename c with d in the last two terms in such a way that I can rewrite the entire expression as $\nabla_a\xi^b+\nabla_b\xi^a$ as suggested by you in the follwing passage? $\endgroup$ – willie Apr 7 at 17:06
  • $\begingroup$ Yes, any summed indices are merely placeholds and you can relabel them. The expression you wrote above is incorrect though (the indices are inconsistent), I think you meant to have $\xi$ with downstairs indices. $\endgroup$ – Eletie Apr 7 at 17:27
  • $\begingroup$ Oh yes It was a typo! Bu if I rename c with d I have $\xi^{c} \Gamma_{c a}^{d} g_{db} +\xi^{c} \Gamma_{cb}^{d} g_{ad} + g_{d b} \partial_{a} \xi^{d} + g_{d a} \partial_{b} \xi^{d}=\xi^{c} \Gamma_{c a}^{d} g_{db} +\xi^{c} \Gamma_{cb}^{d} g_{ad} +\xi_{a,b}+\xi_{b,a}$ and in first two terms I can't lower the index $c$ in order to write so the covariant derivatives of $\xi_a$ and $\xi_b$...how can I do? $\endgroup$ – willie Apr 7 at 17:37
  • $\begingroup$ You can't lower the indices of $\xi^{a}_{,d}$ because it's the partial derivative (you can't simply move the metric through it). Look at my Eq (4), this is exactly what you need to do. Perhaps it's more clear if you re-write (3) as $ g_{cb} (\partial_{a} \xi^{c} + \Gamma_{a d}^{c} \xi^{d}) + g_{ca} (\partial_{b} \xi^{c} + \Gamma^{c}_{bd} \xi^{d} )$. (I've grouped terms 1 & 3 and 2 & 4, and relabelled some indices, which you should verify). You should now be able to see the terms in the brackets are clearly just the covariant derivatives of $\xi$. $\endgroup$ – Eletie Apr 7 at 17:49
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    $\begingroup$ @willie correct, because $\nabla g=0$ so we can move $g$ through $\nabla$. No worries! $\endgroup$ – Eletie Apr 7 at 18:41

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