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I need to prove the following identity for scalar field ($\phi:M\rightarrow R$) in curved spacetime without torsion called $M$

$\nabla_{\mu}[\Box \phi \nabla^{\mu}\phi-\frac{1}{2}\nabla^{\mu}(\nabla \phi)^{2}]=(\Box \phi)^{2}- R_{\mu\nu}\nabla^{\mu}\phi\nabla^{\nu}\phi- \nabla^{\mu}\nabla^{\nu}\phi \nabla_{\mu}\nabla_{\nu}\phi$.

I opened up the total derivative and the term $(\Box \phi)^{2}$ appears but the others I cannot get the right combination. I tried to use the following identity to make appears the terms $R_{\mu\nu}\nabla^{\mu}\phi\nabla^{\nu}\phi$, namely, $[\nabla_\mu, \nabla_\nu]X_{\alpha}=-R^{\kappa}_{\alpha\mu\nu}X_{\kappa}$, but I got stuck at some point, and the third term does not appear at all.

Any help would be much appreciated.

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  • $\begingroup$ Is your left-most term missing a $\phi $? $\endgroup$
    – Michael Engelhardt
    Commented Sep 7, 2020 at 22:14
  • $\begingroup$ Would the Bianchi identity be of some help? $\endgroup$
    – Sylvain JULIEN
    Commented Sep 7, 2020 at 22:19
  • $\begingroup$ Dear Michael, yes there was a typo. $\endgroup$
    – gustavo
    Commented Sep 7, 2020 at 22:22
  • $\begingroup$ Dear Sylvain, I tried to use the Bianchi identity but I could not get the right answer. $\endgroup$
    – gustavo
    Commented Sep 7, 2020 at 22:24
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    $\begingroup$ “Used in general relativity” suggests that you’ve seen someone else use this before, in which case it’d be helpful to cite that source, and they’ve probably provided a name or a citation for the identity. $\endgroup$
    – user44143
    Commented Sep 8, 2020 at 0:23

1 Answer 1

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This is the differential form of the Reilly formula. It holds for a function on any pseudo-Riemannian manifold. (Robert C. Reilly. Applications of the Hessian operator in a Riemannian manifold, Indiana Univ. Math. J. 26 (1977), no. 3, 459–472, doi:10.1512/iumj.1977.26.26036)

Use the product rule to say $$(\Delta f)^2=\operatorname{div}(\Delta f\cdot\nabla f)-\langle\nabla f,\nabla\Delta f\rangle.$$ Use the commutation formula for covariant derivatives to replace the last term by $$\langle\nabla f,\nabla\Delta f\rangle=\langle\nabla f,\Delta\nabla f\rangle-\operatorname{Ric}(\nabla f,\nabla f).$$ Use the product rule to replace the second to last term by $$\langle\nabla f,\Delta\nabla f\rangle=\operatorname{div}\big(\nabla^2f(\nabla f,\cdot)\big)-|\nabla\nabla f|^2.$$ Finally $\nabla^2f(\nabla f,\cdot)=\frac{1}{2}\nabla|\nabla f|^2$. This gives your formula.

Edit. As pointed out by Jeffrey Case below, this also follows from the Bochner formula $$\frac{1}{2}\Delta|\nabla f|^2=|\nabla\nabla f|^2+\langle\nabla\Delta f,\nabla f\rangle+\operatorname{Ric}(\nabla f,\nabla f),$$ where you just need to use the very first line above to replace the middle term on the RHS. The proof of the Bochner formula is by the other lines above

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    $\begingroup$ I think it is better to call this (a variant of) the Bochner formula. The formula in the original post will (essentially) appear in any textbook under this name. "Reilly formula" usually refers to the result after integrating on a manifold with boundary and performing further algebraic manipulations on the boundary. $\endgroup$
    – Jeffrey Case
    Commented Sep 8, 2020 at 12:49
  • $\begingroup$ That's fair. When I called it the Reilly formula I was thinking of the case of a closed manifold $\endgroup$
    – Quarto Bendir
    Commented Sep 8, 2020 at 16:50
  • $\begingroup$ Dear Quarto Bendir, thank you so much for your help and specially for proving the mathematical reference. $\endgroup$
    – gustavo
    Commented Sep 8, 2020 at 17:38

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