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 Laplace–Beltrami operator. The Bochner’s formula is a local formula, we therefore only need to prove it on .

Let be an elliptic system of smooth vector fields on . As usual, we introduce the structure constants of the underlying Riemannian metric:

We know that the Laplace-Beltrami operator is given by where

We first introduce the Ricci curvature, which is seen in this lecture as a first order differential bilinear form.

If is a smooth and compactly supported function on , we define

where

Though, it is not apparent, it is actually an intrinsic Riemannian invariant. That is, only depends on the Riemannian metric 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:

and we now introduce its iteration

Henceforth, we adopt the notation

for the entries of the symmetrized Hessian of with respect to the vector fields . Noting that and using the structure constants we obtain the useful formula

Our principal result of this lecture is the following:

**Theorem: (Bochner’s identity)*** For every smooth function ,
where is the quadratic form defined above.*

**Proof:** We begin by observing that for any smooth function on .

This and the definition of gives

We now have

Using this identity we find

Since, thanks to the skew-symmetry of the matrix , we have , we find

We thus obtain

Since we have

, we conclude

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

.

If we now complete the squares we obtain

Next, we have

and also

.

Using the structure constants we find

Next we have

.

We obtain therefore

where we have let

.

Simplifying the expression we obtain

.

To complete the proof we need to recognize that the monster coincides with . This simple computation is let to the reader

Let be a smooth function. The matrix with coefficient given by is a Riemannian invariant. This matrix is called the Riemannian Hessian of and denoted by or . As a consequence of this and of the Bochner's identity, is seen to be a Riemannian invariant.

**Definition:*** Let be a Riemannian manifold. The bilinear form ( a (0,2) tensor) locally defined by is called the Ricci curvature of . *

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

As a consequence, it should come as no surprise that a lower bound on translates into a lower bound on .

**Theorem:*** Let be a Riemannian manifold. We have, in the sense of bilinear forms, if and only if for every ,
*

**Proof:** Let us assume that . In that case, from Bochner’s formula we deduce that From Cauchy-Schwartz inequality, we have the bound

Since , we conclude that

Conversely, let us now assume that for every ,

Let and . It is possible to find a function such that, at , and . We have then, by using Bochner’s identity at ,

The inequality is called the curvature-dimension inequality. It is an intrinsic property of the operator .

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

**Lemma:** * Let be a Riemannian manifold such that . For every ,
*

**Proof:** It follows from the fact that and Cauchy-Schwartz inequality implies that . Details are let to the reader