23 August 2022
Notes on the Bochner formula for functions. Section 3 of (Li 2012) provides a more general account of the Bochner formula for differential forms, while a few exercises and facts found in (Chow, Lu, and Ni 2006) provide a shorter account for the Bochner formula for functions. For now, we only give the account of the Bochner formula for functions and may add in the account of the Bochner formula for differential forms at a later date. The more general Bochner technique is important because it relates the Laplacian on a manifold to its curvature and can be used to show a number of basic results in Riemannian geometry.
Let \((M, g)\) be a Riemannian manifold. First recall that for a smooth function \(f \in C^{\infty}(M)\), smooth vector fields \(X, Y\) on \(M\), the covariant Hessian of \(f\), is given by (See Exercise 4.5 of (Lee 1997))
\[ \nabla^2f(X,Y) = Y(Xf) - (\nabla_YX)f. \]
Then taking coordinate vector fields \(\{\partial_i\}\), we have that in local coordinates, the covariant Hessian of \(f\) is given by
\[\begin{aligned} \nabla^2f(\partial_i, \partial_j) &= \partial_j\partial_i f - \nabla_{\partial_j}\partial_i f \\ &= \frac{\partial^2 f}{\partial_j \partial_i} - \Gamma^k_{ji}\frac{\partial f}{\partial_k} \\ &= \frac{\partial^2 f}{\partial_i\partial_j} - \Gamma^k_{ij}\frac{\partial f}{\partial_k}, \end{aligned}\]
Where we used the symmetry of mixed partial derivatives and the Christoffel symbols in the last equality above. This is also Exercise 1.14 of (Chow, Lu, and Ni 2006).
Then we define the also called as the trace of the covariant Hessian (equivalently as the divergence of the covariant derivative). In other words, we have
\[ \Delta := \text{trace}_g\nabla^2 = g^{ij}\nabla_i\nabla_j. \]
Note that this definition of the Laplacian is not just restricted to functions, but can also be extended to more general tensors.
Next we want to establish the Ricci identity for a 1-form ((1.30) of (Chow, Lu, and Ni 2006)), which shows how commuting covariant derivatives can be expressed in terms of the curvature. Recall that the Riemann curvature \((3,1)\)-tensor is given by
\[ R(X,Y)Z = \nabla_X\nabla_Y Z - \nabla_Y\nabla_X Z - \nabla_{[X,Y]} Z \]
and the components of the Riemann curvature tensor are given by
\[ R(\partial_i,\partial_j)\partial_k =: {R_{ijk}}^{\ell}\partial_\ell,\qquad R_{ijk\ell} = g_{\ell m}{R_{ijk}}^m. \]
Then we wish to show that for a 1-form \(\alpha: TM\to\mathbb{R}\)$, we have
\[ (\nabla_i\nabla_j - \nabla_j\nabla_i)\alpha_k = -{R_{ijk}}^\ell\alpha_\ell. \]
Note we have
\[\begin{aligned} \nabla^2\alpha(Z,X,Y) &= \nabla_Y(\nabla \alpha)(Z,X) \\ &= \nabla_Y(\nabla \alpha(Z,X)) - \nabla\alpha(\nabla_YZ,X) - \nabla\alpha(Z,\nabla_YX) \\ &= \nabla_Y\left(\nabla_X\alpha\right)(Z) - (\nabla_X\alpha)(\nabla_YZ) - \nabla_{\nabla_YX}\alpha(Z) \\ &= \nabla_Y\nabla_X\alpha(Z) + (\nabla_X\alpha)(\nabla_YZ) - (\nabla_X\alpha)(\nabla_YZ) - \nabla_{\nabla_YX}\alpha(Z) \\ &= \nabla_Y\nabla_X \alpha(Z) - \nabla_{\nabla_YX}\alpha(Z). \end{aligned}\]
So we see that
\[\begin{aligned} \nabla^2\alpha(Z,X,Y) - \nabla^2\alpha(Z,Y,X) &= \nabla_Y\nabla_X\alpha(Z) - \nabla_{\nabla_YX}\alpha(Z) - \nabla_X\nabla_Y\alpha(Z) + \nabla_{\nabla_XY}\alpha(Z) \\ &= -\left(\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]}\right)\alpha(Z) \\ &= -R(X,Y)\alpha(Z) \end{aligned}\]
which in local coordinates becomes
\[ \nabla_i\nabla_j\alpha_k - \nabla_j\nabla_i\alpha_k = -{R_{ijk}}^\ell\alpha_\ell \]
as required. Here we used the fact that \(\nabla\) is torsion free in the first equality above to get that \(\nabla_{\nabla_XY} - \nabla_{\nabla_YX} = \nabla_{[X,Y]}\).
Now we are in a position to prove the Bochner formula for \(f\), which states that
\[ \Delta|\nabla f|^2 = 2|\nabla^2f|^2 + 2\langle\nabla\Delta f,\nabla f\rangle + 2\text{Rc}(\nabla f,\nabla f). \]
We compute directly
\[\begin{aligned} \Delta|\nabla f|^2 &= \Delta g^{k\ell}\nabla_k f \nabla_\ell f \\ &= g^{ij}\nabla_i\nabla_j\left(g^{k\ell}\nabla_k f \nabla_\ell f\right) \\ &= g^{ij}g^{k\ell}\nabla_i\left(\nabla_j\nabla_k f \nabla_\ell f + \nabla_k f \nabla_j\nabla_\ell f\right) \\ &= g^{ij}g^{k\ell}\nabla_i\nabla_j\nabla_k f \nabla_\ell f + g^{ij}g^{k\ell}\nabla_j\nabla_k f \nabla_i\nabla_\ell f \\ &\qquad + g^{ij}g^{k\ell}\nabla_i\nabla_k f \nabla_j\nabla_\ell f + g^{ij}g^{k\ell}\nabla_k f \nabla_i\nabla_j\nabla_\ell f \\ &= 2|\nabla^2 f|^2 + 2 g^{ij}g^{k\ell} \nabla_i\nabla_j\nabla_k f \nabla_\ell f, \end{aligned}\]
where the last equality is by symmetry of the metric inverse and remembering that \(|\nabla^2 f|^2 = g^{ik}g^{j\ell}\nabla_i\nabla_j f \nabla_k\nabla_\ell f\). Then by symmetry of the Hessian followed by using the Ricci identity established above, we have
\[\begin{aligned} \Delta|\nabla f|^2 &= 2|\nabla^2 f|^2 + 2 g^{ij}g^{k\ell} \nabla_i\nabla_j\nabla_k f \nabla_\ell f \\ &= 2|\nabla^2f|^2 + 2g^{ij}g^{k\ell}\nabla_i\nabla_k\nabla_j f \nabla_\ell f \\ &= 2|\nabla^2f|^2 + 2g^{ij}g^{k\ell}\nabla_k\nabla_i\nabla_j f \nabla_\ell f - 2g^{ij}g^{k\ell}{R_{ikj}}^m\nabla_m f\nabla_\ell f \\ &= 2|\nabla^2f|^2 + 2g^{k\ell}\nabla_kg^{ij}\nabla_i\nabla_j f \nabla_\ell f + 2g^{ij}g^{k\ell}{R_{kij}}^m\nabla_m f \nabla_\ell f \\ &= 2|\nabla^2f|^2 + 2\langle\nabla\Delta f,\nabla f\rangle + 2g^{k\ell}{R_k}^m\nabla_m f \nabla_\ell f \\ &= 2|\nabla^2f|^2 + 2\langle\nabla\Delta f,\nabla f\rangle + 2\text{Rc}(\nabla f,\nabla f), \end{aligned}\]
as required. In particular, note that if \(f\) is harmonic, then the Bochner formula becomes
\[ \Delta|\nabla f|^2 = 2|\nabla^2 f|^2 + 2\text{Rc}(\nabla f,\nabla f). \]