# Proving Bochner's formula with coordinates

I'm working on Problem 7-7 in Lee's "Introduction to Riemannian Manifolds", which asks us to prove Bochner's formula: for a Riemannian manifold $$(M,g)$$ and $$u \in C^\infty(M)$$, $$\Delta \left(\frac 1 2 |\mathrm{grad}\: u|^2\right) = \left|\nabla^2 u\right|^2 + \left\langle \mathrm{grad}\:(\Delta u), \mathrm{grad}\:u\right\rangle + Rc(\mathrm{grad}\:u, \mathrm{grad}\:u)$$ where $$\Delta u = \mathrm{div}\:\mathrm{grad}\:u$$ is the Laplacian of $$u \in C^\infty(M)$$, $$\nabla^2 u = u_{;ij} dx^i \otimes dx^j$$ is the covariant Hessian (where $$u_{;ij} = \partial_j\partial_i u - \Gamma_{ji}^k \partial_k u$$), and $$Rc = R_{ij} dx^i \otimes dx^j$$ is the Ricci curvature, where $$R_{ij} = R_{kij}^{\:\:\:\:k}$$ and $$R_{ijk}^{\:\:\:\:l}$$ are the coefficients of the curvature endomorphism $$R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]}Z.$$ Lee suggests using the following two facts:

1. $$\Delta u = g^{ij} u_{;ij} = u_{;i}^{\:\,i}$$
2. If $$\beta$$ is a smooth 1-form on $$M$$, then $$\nabla^2_{X,Y}\beta - \nabla^2_{Y,X} \beta = -R(X,Y)^*\beta,$$ or in coordinates, $$\beta_{j;pq} - \beta_{j;qp} = R_{pqj}^{\:\:\:\,m}\beta_m$$ where $$\beta_{j;pq}$$ are the coefficients of $$\nabla^2\beta$$.

I've tried deriving Bochner's formula from a variety of calculations, mostly involving Riemannian normal coordinates $$(x^i)$$ at a point $$x \in M$$. I've used the first fact to expand both sides but the right side especially gets pretty hairy even with normal coordinates. I am really not sure where the second fact comes into play. Any suggestions?

Brutal force: Note that $$g_{ij;k} = 0$$, we have
\begin{align*} \frac 12 \Delta |\nabla u|^2 &= \frac 12 g^{kl} (g^{ij} u_i u_j)_{kl} \\ &= g^{kl} g^{ij} u_{i;k} u_{j;l} + g^{kl}g^{ij} u_{i;kl} u_j \\ &= |\nabla^2 u|^2 + g^{kl} g^{ij} u_{i;kl} u_j \\ &=|\nabla^2 u|^2 + g^{kl} g^{ij} u_{k;il} u_j \end{align*}
\begin{align*} g^{kl} g^{ij} u_{k;il} u_j &=g^{kl} g^{ij}( u_{k;li} - {R_{lik}}^m u_m ) u_j \\ &= g^{ij} (g^{kl} u_{k;l})_i u_j + g^{kl} g^{ij}{R_{ilk}}^m u_mu_j \\ &= \langle \nabla \Delta u, \nabla u \rangle + g^{ij} {R_i}^m u_mu_j \\ &= \langle \nabla \Delta u, \nabla u \rangle + \operatorname{Rc} (\nabla u, \nabla u). \end{align*}
(We used $$R_{ij} = g^{kl} R_{iklj}$$)
• In the second equality of the second equation, why do we have formula $$g^{kl}u_{k,li}=(g^{kl}u_{k,l})_{i}$$? Dec 8 '20 at 15:02
• @Inuyasha That's the Leibniz rule for the tensor $(g^{kl} u_{k;l})_i = {g^{kl}}_{;i} u_{k;l} + g^{kl} u_{k;li}$ together with ${g^{kl}}_{;i} = 0$. Dec 8 '20 at 16:03