Weitzenböck identity

Remarks

The idea of the proof: express $d$ and $\delta$ through $\nabla$, namely $d\omega = \sum_i e^i \wedge \nabla_{e_i} \omega$ and $\delta\omega = -\sum_i \iota_{e_i} \nabla_{e_i} \omega$. Expanding $d\delta + \delta d$ gives the second-order part $\nabla^* \nabla$ plus commutators $[\nabla_{e_i}, \nabla_{e_j}]$, which are curvature terms.

Example: for $\omega = du$ the identity, paired with $du$, gives Bochner's formula for the function $u$.

Bochner's vanishing theorem follows: let $M$ be compact. If $\omega$ is a harmonic 1-form, integrating $\ip{\Delta_H \omega}{\omega} = 0$ gives $\int_M \abs{\nabla \omega}^2 + \operatorname{Ric}(\omega^\sharp, \omega^\sharp) = 0$. So if $\operatorname{Ric} \ge 0$, every harmonic 1-form is parallel and the first Betti number is at most $n$; if $\operatorname{Ric} > 0$, the first Betti number is $0$.

Going deeper

For a student

Weitzenböck identities are the engine of the Bochner technique. Combined with the Hodge theorem, which identifies cohomology with harmonic forms, a sign of the curvature term forces harmonic forms to vanish or to be parallel. The same pattern holds for other Dirac-type operators; for the spin Dirac operator the curvature term is a quarter of the scalar curvature (the Lichnerowicz formula).