I saw this come up in a lecture on conformal field theory, and was a bit skeptical of the claim. So, given some conformal Killing field,

$$ \mathcal{L}_{\xi}~g = \lambda g $$ it was said the following is true $$ \mathcal{L}_{\nabla^2\xi}g=g\nabla^2\lambda $$ Where $\nabla^2=\nabla^a\nabla_a$ (a contraction of covariant derivatives). I have tried writing things out but can't get anywhere. I end up with a bunch of Riemann tensor terms that don't cancel. So, first off is this even true, and if so how can I show it?

  • $\begingroup$ Have you tried using the metric property of the Levi-Civita connection $\nabla g=0$ to add the missing contributions you need to form the "total covariant derivatives"? I mean something like $(\nabla_a\xi^b)g_{dc}\rightarrow \nabla_a(\xi^bg_{dc})$. $\endgroup$ – green.onion Sep 14 '18 at 21:02
  • $\begingroup$ yea of course... $\endgroup$ – ClassicStyle Sep 14 '18 at 22:17

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