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

$$\begin{align*} \frac 12 \Delta |\nabla u|^2 &= \frac 12 g^{kl} (g^{ij} u_i u_j)_{kl} \\ &= g^{kl} g^{ij} u_{i;k} u_{j;l} + g^{kl}g^{ij} u_{i;kl} u_j \\ &= |\nabla^2 u|^2 + g^{kl} g^{ij} u_{i;kl} u_j \\ &=|\nabla^2 u|^2 + g^{kl} g^{ij} u_{k;il} u_j \end{align*}$$

Then we use your second point:

$$\begin{align*} g^{kl} g^{ij} u_{k;il} u_j &=g^{kl} g^{ij}( u_{k;li} - {R_{lik}}^m u_m ) u_j \\ &= g^{ij} (g^{kl} u_{k;l})_i u_j + g^{kl} g^{ij}{R_{ilk}}^m u_mu_j \\ &= \langle \nabla \Delta u, \nabla u \rangle + g^{ij} {R_i}^m u_mu_j \\ &= \langle \nabla \Delta u, \nabla u \rangle + \operatorname{Rc} (\nabla u, \nabla u). \end{align*}$$

(We used $R_{ij} = g^{kl} R_{iklj}$)

  • $\begingroup$ In the second equality of the second equation, why do we have formula $$g^{kl}u_{k,li}=(g^{kl}u_{k,l})_{i}$$? $\endgroup$
    – Inuyasha
    Dec 8 '20 at 15:02
  • $\begingroup$ @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 ${g^{kl}}_{;i} = 0$. $\endgroup$ Dec 8 '20 at 16:03

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