## Lecture 13. The Bochner’s formula

The goal of this lecture is to prove the Bochner formula: A fundamental formula that relates the so-called Ricci curvature of the underlying Riemannian structure to the analysis of the LaplaceBeltrami operator. The Bochner’s formula is a local formula, we therefore only need to prove it on $\mathbb{R}^n$.

Let $(V_1,\cdots,V_n)$ be an elliptic system of smooth vector fields on $\mathbb{R}^n$. As usual, we introduce the structure constants of the underlying Riemannian metric:
$[V_i,V_j]=\sum_{k=1}^n \omega_{ij}^k V_k.$

We know that the Laplace-Beltrami operator is given by $L=\sum_{i=1}^n V_i^2 +V_0,$ where
$V_0= -\sum_{i,k}^n \omega_{ik}^k V_i.$
We first introduce the Ricci curvature, which is seen in this lecture as a first order differential bilinear form.

If $f$ is a smooth and compactly supported function on $\mathbb{R}^n$, we define
$\mathcal R(f,f) = \sum_{k,l=1}^n \mathcal{R}_{k,l} V_k f V_l f$
where
$\mathcal{R}_{k,l} = \sum_{j=1}^n (V_l\omega^j_{kj} - V_j\omega^k_{l j}) + \sum_{i,j=1}^n \omega_{ji}^i \omega^l_{k j} - \sum_{i=1}^n\omega_{k i}^i \omega_{l i}^i$
$+ \frac{1}{2} \sum_{1\le i

Though, it is not apparent, it is actually an intrinsic Riemannian invariant. That is, $\mathcal{R}$ only depends on the Riemannian metric $g$ induced by the vector fields.

In the sequel, we will use the following differential bilinear form that already has been widely used throughout these lectures:
$\Gamma(f,g) =\frac{1}{2}(L(fg)-fLg-gLf)=\sum_{i=1}^n V_i f V_i g,$
and we now introduce its iteration
$\Gamma_{2}(f,g) = \frac{1}{2}\left(L\Gamma(f,g) - \Gamma(f, Lg)-\Gamma (g,Lf)\right).$

$f_{,ij} = \frac{V_i V_j f + V_j V_i f}{2}$
for the entries of the symmetrized Hessian of $f$ with respect to the vector fields $V_1,...,V_d$. Noting that $V_iV_j f\ =\ f_{,ij}\ +\ \frac{1}{2}\ [V_i,V_j] f$ and using the structure constants we obtain the useful formula
$V_iV_jf = f_{,ij} + \frac{1}{2} \sum_{l=1}^n \omega^l_{ij} V_l f$

Our principal result of this lecture is the following:

Theorem: (Bochner’s identity) For every smooth function $f :\mathbb{R}^n \rightarrow \mathbb{R}$,
$\Gamma_{2}(f,f)$
$= \sum_{l=1}^n \left(f_{,ll} -\sum_{i=1}^n \omega_{il}^l V_i f \right)^2 +2 \sum_{1 \le l
where $\mathcal R(f,f)$ is the quadratic form defined above.

Proof: We begin by observing that for any smooth function $F$ on $\mathbb{R}^n$ $L(F^2) = 2 F LF + 2 \Gamma(F,F)$.
This and the definition of $\Gamma$ gives
$L \Gamma(f,f) = \sum_{i=1}^n L((V_i f)^2) = 2 \sum_{i=1}^n V_i f L(V_i f) + 2 \sum_{i=1}^n \Gamma(V_i f,V_if) .$
We now have
$L(V_i f) = V_0 V_i f + \sum_{j=1}^n V_j^2 V_i f = V_i V_0 f + [V_0,V_i]f + \sum_{j=1}^n V_j (V_iV_j f) + V_j[V_j,V_i]f$
$= V_i V_0 f + [V_0,V_i]f + \sum_{j=1}^n \big\{V_i (V_jV_j f) + [V_j,V_i]V_j f + V_j[V_j,V_i]f\big\}$
$= V_i(Lf) + [V_0,V_i]f + \sum_{j=1}^n \big\{[V_j,V_i]V_j f + V_j[V_j,V_i]f\big\}$
$= V_i(Lf) + [V_0,V_i]f + 2 \sum_{j=1}^n [V_j,V_i]V_j f +\sum_{j=1}^n [V_j,[V_j,V_i]]f.$
Using this identity we find
$L \Gamma(f,f) = 2 \sum_{i=1}^n V_i f \left\{V_i(Lf) + [V_0,V_i]f + 2 \sum_{j=1}^n [V_j,V_i]X_j f + \sum_{j=1}^n [V_j,[V_j,V_i]]f\right\} + 2 \sum_{i,j=1}^n (V_jV_i f)^2$
$= 2 \Gamma(f,Lf) + 2 \sum_{i=1}^n V_i f [V_0,V_i]f + 4 \sum_{i,j=1}^n V_if [V_j,V_i]V_j f + 2 \sum_{i,j=1}^n V_i f [V_j,[V_j,V_i]]f + 2 \sum_{i,j=1}^n (V_jV_i f)^2.$
Since, thanks to the skew-symmetry of the matrix $\{[V_i,V_j]f\}_{i,j=1,...,n}$, we have $\sum_{i,j=1}^n f_{,ij} [V_i,V_j] f = 0$, we find
$\sum_{i,j=1}^n (V_jV_i f)^2 = \sum_{i,j=1}^n f_{,ij}^2 + \frac{1}{4} \sum_{i,j=1}^n ([V_i,V_j] f)^2 +\sum_{i,j=1}^n f_{,ij} [V_i,V_j] f$
$= \sum_{i,j=1}^n f_{,ij}^2 + \frac{1}{4} \sum_{i,j=1}^n ([V_i,V_j] f)^2.$
We thus obtain

$\frac{1}{2} \big[L \Gamma(f,f) - 2\Gamma(f,Lf)\big] = \sum_{i=1}^n V_i f [V_0,V_i]f + 2 \sum_{i,j=1}^n V_i f [V_j,V_i]V_j f + \sum_{i,j=1}^n V_i f [V_j,[V_j,V_i]]f + \sum_{i,j=1}^n f_{,ij}^2 +\frac{1}{4} \sum_{i,j=1}^n ([V_i,V_j] f)^2.$

Since we have

$\Gamma_{2}(f,f) = \frac{1}{2}\left(L\Gamma(f,f) - 2 \Gamma(f, Lf)\right)$, we conclude $\Gamma_{2}(f,f) = \sum_{i,j=1}^n f_{,ij}^2 - 2 \sum_{i,j=1}^n V_i f [V_i,V_j]V_j f + \frac{1}{4} \sum_{i,j=1}^n ([V_i,V_j] f)^2 + \sum_{i=1}^n V_i f [V_0,V_i]f + \sum_{i,j=1}^n V_i f [[V_i,V_j],V_j]f$

To complete the proof we need to recognize that the right-hand side coincides with that in the statement of our result.

With this objective in mind, using the structure constants we obtain
$\sum_{i,j=1}^n f_{,ij}^2 - 2 \sum_{i,j=1}^n V_i f [V_i,V_j]V_j f$
$= \sum_{l=1}^n f_{,ll}^2 + 2 \sum_{1 \le l
$= \sum_{l=1}^n f_{,ll}^2 + 2 \sum_{1 \le l
$= \sum_{l=1}^n f_{,ll}^n + 2 \sum_{1 \le l
$= \sum_{l=1}^n f_{,ll}^2 + 2 \sum_{1 \le l
$= \sum_{l=1}^n \left(f_{,ll}^2 - 2 \left(\sum_{i=1}^n \omega^l_{il} V_i f\right) f_{,l l}\right) + 2 \sum_{1 \le l.
If we now complete the squares we obtain

$\sum_{i,j=1}^n f_{,ij}^2 - 2 \sum_{i,j=1}^n V_i f [V_i,V_j]V_j f$
$= \sum_{l=1}^n \left( f_{,ll} -\sum_{i=1}^n \omega_{il}^l V_i f \right)^2 + 2 \sum_{1 \le l
$- \sum_{l=1}^n \left(\sum_{i=1}^n \omega_{il}^l V_i f \right)^2 - 2 \sum_{1 \le l

Next, we have
$\sum_{i=1}^n V_i f [V_0,V_i]f = \sum_{i,j,k,l=1}^n \omega^k_{jk} \omega^l_{ij} V_l f V_i f + \sum_{i=1}^n \sum_{j,k=1}^n (V_i \omega^k_{jk}) V_if V_j f,$
and also
$\sum_{i,j=1}^n V_i f [[V_i,V_j],V_j]f = \sum_{i,j=1}^n \sum_{l= 1}^n [\omega^l_{ij} V_l,V_j]f V_if$
$= \sum_{i,j=1}^n \sum_{l=1}^n \omega^l_{ij} V_i f [V_l,V_j] f - \sum_{i,j=1}^n \sum_{l=1}^n (V_j\omega^l_{ij}) V_if V_l f$.
Using the structure constants we find
$\sum_{i,j=1}^n V_i f [[V_i,V_j],V_j]f = \sum_{i,j=1}^n \sum_{l,k=1}^n \omega^l_{ij} \omega^k_{l j} V_i f V_kf - \sum_{i,j=1}^d \sum_{l=1}^d (X_j\omega^l_{ij}) X_if X_l f.$
Next we have
$\frac{1}{4} \sum_{i,j=1}^n ([V_i,V_j]f)^2 = \frac{1}{2} \sum_{1\le i.
We obtain therefore
$\Gamma_2(f,f) = \sum_{l=1}^n \left( f_{,ll} -\sum_{i=1}^n \omega_{il}^l V_i f \right)^2 + 2 \sum_{1 \le l
where we have let

$\mathcal{M}onster = - \sum_{l=1}^n \left(\sum_{i=1}^d \omega_{il}^l V_i f \right)^2 - 2 \sum_{1 \le l
$+ \sum_{i,j,k,l=1}^n \omega^k_{jk} \omega^l_{ij} V_l f V_i f - \sum_{i,j,k,l=1}^n \omega_{ij}^k \omega^l_{k j} V_l f V_i f + \sum_{i=1}^n \sum_{j,k=1}^n (V_i \omega^k_{jk}) V_if V_j f + \sum_{i,j=1}^n \sum_{l,k=1}^n \omega^l_{ij} \omega^k_{l j} V_i f V_kf$
$- \sum_{i,j=1}^n \sum_{l=1}^n (V_j\omega^l_{ij}) V_if V_l f+ \frac{1}{2} \sum_{1\le i.

Simplifying the expression we obtain
$\mathcal{M}onster = - \sum_{k,l=1}^n \sum_{i=1}^n \omega_{k i}^i \omega_{l i}^i V_k f V_l f - \frac{1}{2} \sum_{k,l=1}^n \sum_{1 \le i
$+ \sum_{i,j,k,l=1}^n \omega_{ji}^i \omega^l_{k j} V_k f V_l f+ \frac{1}{2} \sum_{k,l=1}^n \sum_{1\le i.

To complete the proof we need to recognize that the monster coincides with $\mathcal R(f,f)$. This simple computation is let to the reader $\square$

Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be a smooth function. The matrix with coefficient $(l,j)$ given by $f_{,l j} -\sum_{i=1}^n \frac{\omega_{il}^j +\omega_{ij}^l}{2} V_i f$ is a Riemannian invariant. This matrix is called the Riemannian Hessian of $f$ and denoted by $\mathbf{Hess} f$ or $\nabla^2 f$. As a consequence of this and of the Bochner’s identity, $\mathcal{R}$ is seen to be a Riemannian invariant.

Definition: Let $\mathbb{M}$ be a Riemannian manifold. The bilinear form ( a (0,2) tensor) locally defined by $\mathbf{Ricc} (\nabla f, \nabla f)= \mathcal{R}(f,f)$ is called the Ricci curvature of $\mathbb{M}$

On a Riemannian manifold, the Bochner’s formula can therefore synthetically be written

$\Gamma_2 (f,f)=\| \mathbf{Hess} f \|^2_{HS}+ \mathbf{Ric} (\nabla f, \nabla f).$

As a consequence, it should come as no surprise that a lower bound on $\mathbf{Ric}$ translates into a lower bound on $\Gamma_2$.

Theorem: Let $\mathbb{M}$ be a Riemannian manifold. We have, in the sense of bilinear forms, $\mathbf{Ric} \ge \rho$ if and only if for every $f \in C^\infty(\mathbb{M})$,
$\Gamma_2(f,f) \ge \frac{1}{n} (Lf)^2 + \rho \Gamma(f,f).$

Proof: Let us assume that $\mathbf{Ric} \ge \rho$. In that case, from Bochner’s formula we deduce that $\Gamma_2 (f,f) \ge \| \mathbf{Hess} f \|^2_{HS}+ \rho \Gamma(f,f).$ From Cauchy-Schwartz inequality, we have the bound
$\| \mathbf{Hess} f \|^2_{HS} \ge \frac{1}{n} \mathbf{Tr} \left(\mathbf{Hess} f \right)^2.$
Since $\mathbf{Tr} \left(\mathbf{Hess} f \right)=Lf$, we conclude that
$\Gamma_2(f,f) \ge \frac{1}{n} (Lf)^2 + \rho \Gamma(f,f).$

Conversely, let us now assume that for every $f \in C^\infty(\mathbb{M})$,
$\Gamma_2(f,f) \ge \frac{1}{n} (Lf)^2 + \rho \Gamma(f,f).$

Let $x \in \mathbb{M}$ and $v \in \mathbf{T}_x \mathbb{M}$. It is possible to find a function $f \in C^\infty(\mathbb{M})$ such that, at $x$, $\mathbf{Hess} f =0$ and $\nabla f=v$. We have then, by using Bochner’s identity at $x$, $\mathbf{Ric} (v,v) \ge \rho \| v \|^2$ $\square$

The inequality $\Gamma_2(f,f) \ge \frac{1}{n} (Lf)^2 + \rho \Gamma(f,f)$ is called the curvature-dimension inequality. It is an intrinsic property of the operator $L$.

We finally mention another consequence of Bochner’s identity which shall be later used.

Lemma: Let $\mathbb{M}$ be a Riemannian manifold such that $\mathbf{Ric} \ge \rho$. For every $f \in C^\infty(\mathbb{M})$,
$\Gamma(\Gamma(f)) \le 4 \Gamma (f) \left( \Gamma_2(f)-\rho\Gamma(f)\right).$

Proof: It follows from the fact that $\Gamma_2 (f,f) \ge \| \mathbf{Hess} f \|^2_{HS}+ \rho \Gamma(f,f)$ and Cauchy-Schwartz inequality implies that $\Gamma(\Gamma(f)) \le 4 \| \mathbf{Hess} f \|^2_{HS} \Gamma(f)$. Details are let to the reader $\square$

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