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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?

D Ford
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1 Answers1

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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 $R_{ij} = g^{kl} R_{iklj}$)

Arctic Char
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  • In the second equality of the second equation, why do we have formula $$g^{kl}u_{k,li}=(g^{kl}u_{k,l})_{i}$$? – Inuyasha Dec 08 '20 at 15:02
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    @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 ${g^{kl}}_{;i} = 0$. – Arctic Char Dec 08 '20 at 16:03
  • I’m wondering why the second derivative of the metric vanished in the second equality of the first equation. I.e. why does the term $$ g^{kl}g^{ij}_{;;;kl}u_iu_j$$ vanish? – Föölücks Apr 25 '25 at 15:12
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    Hi @Föölücks, note that the $g^{kl}_{;j}$ stands for the covariant differentiation, and this term is zero since $\nabla g = 0$. – Arctic Char Apr 25 '25 at 15:48
  • @Arctic Char First of all thanks for the quick response :) It took some time getting used to the indices, but this simply means everything here is to be interpreted using the product rule for covariant derivatives. So e.g. $$(g^{ij}u_iu_j){k}=(g^{ij}u_iu_j){;k}=g^{ij}{;;;k}u_iu_j+g^{ij}u{i;k}u_j+g^{ij}u_iu_{j;k}=2g^{ij}u_{i;k}u_j$$ and $g^{ij}_{;;;k}=0$ for the aforementioned reason, right? – Föölücks Apr 28 '25 at 21:52
  • Yes, exactly @Föölücks – Arctic Char Apr 29 '25 at 08:11
  • I'm wondering why $g^{kl} g^{ij} u_{i;k} u_{j;l} = |\nabla^2 u|$; after all, the coefficients of $\nabla^2 u$ are not $u_{i;j}$ but $u_{;ij}$, I feel like this is missing some Christoffel symbol terms to truly be the inner product. – Lukrau Apr 30 '25 at 02:09
  • In particular, we have the relationship $u_{i;j} = u_{;ij} + \Gamma_{ji}^k u_{;k}$, so that's what I mean when I say I feel the term $|\nabla^2 u|$ should have some christoffel symbol terms – Lukrau Apr 30 '25 at 02:30
  • Also, on the last line in the first calculation, why are we allowed to swap terms $g^{kl}g^{ij}u_{i;kl}u_j = g^{kl}g^{ij}u_{k;il}u_j$? – Lukrau Apr 30 '25 at 03:08
  • @Lukrau Can you tell me what is $u_{;ij}$? – Arctic Char May 02 '25 at 06:07
  • u_{;ij} is the i, jth term of the covariant Hessian of u (sorry, I am used to just putting the semicolon even for the first times taking the covariant derivatives of a smooth function, because that's what Lee does). – Lukrau May 02 '25 at 06:44
  • If $u_{;ij{$ denotes the Hessian, then we don't have $u_{i;j} = u_{;ij} + \Gamma_{ji} ^k u_{;k}$. Indeed $u_i = u_{;i}$ since $u$ is a function. @Lukrau – Arctic Char May 03 '25 at 06:05
  • Wait, but $\nabla^2(u) \neq \nabla(u_i) dx^i$ (where $\nabla u = u_i dx^i$), because $\nabla^2(u)(X, Y) \neq \nabla_Y \nabla_X(u)$. So when we do the Hessian, the coefficients aren't taking the coefficients of the covariant derivative of $u_i$, viewed as a function I don't think? You should have error terms coming from the involvement of $\nabla dx^i$ in the hessian formula, unless I'm misunderstanding something. – Lukrau May 03 '25 at 06:09
  • @Lukrau Yes $\nabla^2 u \neq (\nabla u_i) dx^i$, and the error terms is exactly the $\Gamma$ that you have in your expression. May be the following is the confusion: $u_{i;j}$ is not $\nabla j(u_i)$ (indeed this term depends on the coordinates and is not a tensor. Instead, $u{i;j}$ is the $(0,2)$ tensor $\nabla du$, which is the hessian. – Arctic Char May 03 '25 at 06:18
  • That makes sense notationally, but then my question becomes: we get these terms from taking covariant derivatives of $g^{ij}u_{i}u_{j}$, where in that expression everything is honest to god functions, no covectors around, we are just taking the hessian of functions. So why do the mixed terms from the Hessian product rule contain $\nabla (u_i dx^i)$ terms, since it seems to me that the mixed terms in the product rule correspond to just taking first derivatives of functions, not covectors? Thanks for helping clear up this confusion, the notation confuses me a lot. – Lukrau May 04 '25 at 00:41