Bochner's formula

If ${\displaystyle u\colon M\rightarrow \mathbb {R} }$ is a smooth function, then

${\displaystyle \Delta {\bigg (}{\frac {|\nabla u|^{2}}{2}}{\bigg )}=\langle \nabla \Delta u,\nabla u\rangle +|\nabla ^{2}u|^{2}+{\mbox{Ric}}(\nabla u,\nabla u)}$,

where ${\displaystyle \nabla u}$ is the gradient of ${\displaystyle u}$ with respect to ${\displaystyle g}$ and ${\displaystyle {\mbox{Ric}}}$ is the Ricci curvature tensor.[1] If ${\displaystyle u}$ is harmonic (i.e., ${\displaystyle \Delta u=0}$, where ${\displaystyle \Delta =\Delta _{g}}$ is the Laplacian with respect to the metric ${\displaystyle g}$), Bochner's formula becomes

${\displaystyle \Delta {\frac {1}{2}}|\nabla u|^{2}=|\nabla ^{2}u|^{2}+{\mbox{Ric}}(\nabla u,\nabla u)}$.

Bochner used this formula to prove the Bochner vanishing theorem.

As a corollary, if ${\displaystyle (M,g)}$ is a Riemannian manifold without boundary and ${\displaystyle u\colon M\rightarrow \mathbb {R} }$ is a smooth, compactly supported function, then

${\displaystyle \int _{M}(\Delta u)^{2}\,d{\mbox{vol}}=\int _{M}{\Big (}|\nabla ^{2}u|^{2}+{\mbox{Ric}}(\nabla u,\nabla u){\Big )}\,d{\mbox{vol}}}$.

This immediately follows from the first identity, observing that the integral of the left-hand side vanishes (by the divergence theorem) and integrating by parts the first term on the right-hand side.

• Bochner identity
• Weitzenböck identity