I am currently reading some lecture notes on Riemannian geometry and it is stated that the fact that
$d|Rm|^2=2\langle Rm,\nabla Rm\rangle$
implies that
$\Delta|Rm|^2=2\langle Rm,\Delta Rm\rangle + 2|\nabla Rm|^2$.
Would someone mind clarifying why this is? To define the notation, $\Delta$ is the Laplace-Beltrami operator and $Rm$ is the Riemann curvature tensor. $\nabla$ is the covariant derivative.
Best Answer
$$\begin{align}\Delta|Rm|^2&=\nabla^\mu\nabla_\mu|Rm|^2\\ &=\nabla^\mu2\langle Rm,\nabla_\mu Rm\rangle\\ &=2g^{\mu\nu}\nabla_\nu\langle Rm,\nabla_\mu Rm\rangle\\ &=2g^{\mu\nu}\langle\nabla_\nu Rm,\nabla_\mu Rm\rangle+2\langle Rm,g^{\nu\mu}\nabla_\nu\nabla_\mu Rm\rangle\\ &=2|\nabla Rm|^2+2\langle Rm,\Delta Rm\rangle.\end{align}$$