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<j\le n} \bigg(\omega^l_{ij} \omega^k_{ij} - (\omega_{l j}^i +\omega_{li}^j)(\omega^i_{kj} + \omega^j_{ki})\bigg)

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).

Henceforth, we adopt the notation
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<j \le n} \left( f_{,l j}-\sum_{i=1}^n \frac{\omega_{il}^j +\omega_{ij}^l}{2} V_i f \right)^2 +\mathcal{R}(f,f).
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<j \le n} f_{,jl}^2 - 2 \sum_{i,j=1}^n V_i f \left(\sum_{l=1}^d \omega_{ij}^l V_l \right)V_j f
= \sum_{l=1}^n f_{,ll}^2 + 2 \sum_{1 \le l<j \le n} f_{,jl}^2 - 2 \sum_{i,j=1}^n \sum_{l=1}^n \omega_{ij}^l V_l V_j f\ V_i f
= \sum_{l=1}^n f_{,ll}^n + 2 \sum_{1 \le l<j \le n} f_{,jl}^2 - 2 \sum_{i,j=1}^n \sum_{l=1}^n \omega_{ij}^l f_{,l j}V_i f - \sum_{i,j=1}^n \sum_{l, k=1}^n \omega_{ij}^l \omega^k_{l j} V_k f V_i f
= \sum_{l=1}^n f_{,ll}^2 + 2 \sum_{1 \le l<j \le n} f_{,jl}^2 - 2 \sum_{l,j=1}^n \left(\sum_{i=1}^n \omega_{ij}^l V_i f\right) f_{,l j} - \sum_{i,j=1}^n \sum_{l, k=1}^n \omega_{ij}^l \omega^k_{l j} V_k f V_i f
= \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<j \le n}\left( f_{,jl}^2 - 2 \sum_{1\le l<j\le n} \left(\sum_{i=1}^n \frac{\omega_{ij}^l + \omega_{il}^j}{2} V_i f\right) f_{,l j} \right)-\sum_{i,j=1}^n \sum_{l, k=1}^n \omega_{ij}^l \omega^k_{l j} V_k f V_i f.
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<j \le n} \left( f_{,jl} -\sum_{i=1}^n \frac{\omega_{ij}^l +\omega_{il}^j}{2} V_i f \right)^2
- \sum_{l=1}^n \left(\sum_{i=1}^n \omega_{il}^l V_i f \right)^2 - 2 \sum_{1 \le l<j \le n} \left(\sum_{i=1}^n \frac{\omega_{ij}^l +\omega_{il}^j}{2} V_i f \right)^2-\sum_{i,j,k,l=1}^n \omega_{ij}^l \omega^k_{l j} V_k f V_i f.

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<j\le n}\left(\sum_{l=1}^n \omega^l_{ij} V_l f\right)^2 .
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<j \le n} \left( f_{,jl} -\sum_{i=1}^n \frac{\omega_{ij}^l +\omega_{il}^j}{2} V_i f \right)^2 + \mathcal{M}onster
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<j \le n} \left(\sum_{i=1}^n \frac{\omega_{ij}^l +\omega_{il}^j}{2} V_i f \right)^2
+ \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<j\le n}\left(\sum_{l=1}^n \omega^l_{ij} V_l f\right)^2.

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<j \le n} (\omega_{l j}^i +\omega_{l i}^j)(\omega^i_{kj} + \omega^j_{ki}) V_k f V_l f + \sum_{k,l=1}^n \sum_{j=1}^n (V_l\omega^j_{kj} - V_j\omega^k_{l j}) V_kf V_l f
+ \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<j\le n} \omega^l_{ij} \omega^k_{ij} V_k f V_l f.

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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