Research and Lecture notes

HW2 MA5311: Due February 3

Posted on January 28, 2017 by Fabrice Baudoin

Exercise 1. Let $g : \mathbf{R}^2 \to \mathbf{R}^4$ , $(u,v) \to (\cos u , \sin u, \cos v, \sin v)$ . Show that $g( \mathbf{R}^2)$ is a 2-dimensional smooth manifold homeomorphic to the torus $\mathbf{S}^1 \times \mathbf{S}^1$ .

Exercise 2. Let $h_+: \mathbf{S}^2-N \to \mathbf{C}$ be the stereographic projection from the north pole $N$ , and $h_-$ be the stereographic projection from the south pole $S$ .

Show that for $z \neq 0$ , $h_+ h_-^{-1} =\frac{1}{\bar z}$ .
Show that if $P$ is a non constant polynomial, the map $f=h_+^{-1} P h_+$ , $f(N)=N$ is smooth.
More generally, if $Q: \mathbf{C} \to \mathbf{C}$ is smooth, find a condition on $Q$ so that $f=h_+^{-1} Q h_+$ , $f(N)=N$ is smooth.
By following Milnor’s argument in the proof of the fundamental theorem of algebra, find sufficient conditions so that a smooth map $Q: \mathbf{C} \to \mathbf{C}$ is onto.

Exercise 3. By using Sard’s theorem, prove that the set of regular values of a smooth map $f : M \to N$ is dense in $N$ .

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MA5161. HW 1 due Wednesday 1/25

Posted on January 20, 2017 by Fabrice Baudoin

Exercise 1. Show that the $\sigma$ -algebra $\mathcal{T}(\mathbb{R}_{\ge 0},\mathbb{R}^d)$ is also generated by the following families:

$\{ f \in \mathcal{A}(\mathbb{R}_{\ge 0}, \mathbb{R}^d ), f(t_1) \in B_1,...,f(t_n) \in B_n \}$ where $t_1,...,t_n \in \mathbb{R}_{\ge 0}$ and where $B_1,...,B_n$ are Borel sets in $\mathbb{R}^d$ .
$\{ f \in \mathcal{A}(\mathbb{R}_{\ge 0}, \mathbb{R}^d), (f(t_1),...,f(t_n)) \in B \}$ where $t_1,...,t_n \in \mathbb{R}_{\ge 0}$ and where $B$ is a Borel set in $(\mathbb{R}^{d})^{\otimes n}$ .

Exercise 2. Show that the following sets are in $\mathcal{B} ([0,1],\mathbb{R})$ :

$\{ f \in \mathcal{C}([0,1], \mathbb{R}), \sup_{t\in [0,1]} f(t) <1 \}$
$\{ f \in \mathcal{C}([0,1], \mathbb{R}), \exists t\in [0,1] f(t) =0 \}$

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MA5311. About John Milnor

Posted on January 18, 2017 by Fabrice Baudoin

John Milnor is a renowned mathematician who made fundamental contributions to differential topology and was awarded the Fields medal in 1962. One of his most striking result is the existence of several distinct differentiable structures on the 7 dimensional sphere (!).

The following video is an introductory lecture to differential topology given in 1965. The first 26 minutes will give you a good overview of what differential topology is about.

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MA5311: Homework 1, due Wednesday 1/25

Posted on January 18, 2017 by Fabrice Baudoin

Exercise 1. Show that the sphere

$\mathbf{S}^n=\left\{ (x_1,\cdots,x_{n+1}) \in \mathbf{R}^{n+1}, x_1^2+\cdots+x_{n+1}^2=1 \right\}$

is a $n$ -dimensional smooth manifold.

Exercise 2. We consider the following two subsets of the plane

$A= \{ ( x , \sin (1/x)), x \neq 0 \}$

and

$B= \{ ( x , | x| ), x \in \mathbf{R} \}$ .

Are $A$ and $B$ smooth manifolds ? Of course, justify your answer with a proof.

Exercise 3. Let $M \subset\mathbf{R}^{k}$ be a n-dimensional smooth manifold. Let $x \in M$ . A smooth curve on $M$ is a smooth map $\gamma: \mathbf{R} \to M$ . We denote by $\Gamma_0$ the set of smooth curves on $M$ such that $\gamma (0)=x$ . Show that the set

$\left\{ \gamma'(0), \gamma \in \Gamma_0 \right\}$

is a linear space isomorphic to $TM_x$ .

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MATH 5311: Differential Topology

Posted on January 16, 2017 by Fabrice Baudoin

During the Spring, I will be teaching a class on differential topology. Lecture Notes will not be posted on this blog since I will be explicitly using several books. The course will mainly be organized in two parts.

Part 1. Introduction to differential topology

In this part, to simplify the presentation, all manifolds are taken to be infinitely differentiable and to be explicitly embedded in euclidean space. A small amount of point-set topology and of real variable theory is taken for granted. We shall follow the classical reference by John Milnor: Topology from the differentiable viewpoint.

Smooth manifolds and smooth maps
Sard-Brown theorem
Smooth homotopy and smooth isotopy
Brouwer degree
Vector fields and Euler number

Part 2. Riemannian geometry and differential topology

In this part we plan to show how Riemannian geometry can be used to study topological properties of a manifold. Depending on the pace of the class, all of the following topics may not be covered.

Abstract manifolds
Whitney embedding theorem
Differential forms on manifolds
De Rham cohomology
Riemannian metrics
The Laplacian on forms and Hodge theory
Connections and curvature
Weitzenbock formula
The Chern-Gauss-Bonnet formula
Introduction to index theory and characteristic classes

General references for this part of the course will be

Riemannian Geometry, by Gallot-Hulin-Lafontaine
The Laplacian on a Riemannian manifold, by Rosenberg

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Take home questions

Posted on December 13, 2016 by Fabrice Baudoin

Let
$L=\Delta +\langle \nabla U, \nabla \cdot\rangle,$
where $U$ is a smooth function on $\mathbb{R}^n$ and $\Delta$ the usual Laplace operator on $\mathbb{R}^n$ . Show that with respect to the measure $\mu(dx)=e^{U(x)} dx$ , the operator $L$ is essentially self-adjoint on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ .
Compute $\Gamma_2(f)=\frac{1}{2}( L\Gamma(f,f)-2\Gamma(f,Lf))$ for the previous operator.
Show that if, as a bilinear form, $\mathbf{Hess} U \ge \rho$ for some $\rho \in \mathbb{R}$ , then the semigroup $P_t=e^{tL}$ is stochastically complete.
Let now $V_1,\cdots,V_n$ be smooth vector fields on $\mathbb{R}^n$ such that $V_1(x),\cdots, V_n(x)$ is a basis of $\mathbb{R}^n$ for every $x$ . We denote by $\Delta^g$ the Laplace-Beltrami operator associated with the corresponding Riemannian metric and the diffusion operator $L^g f =\Delta^g f +\sum_{i=1}^n V_i U V_i f,$ where $U$ is a smooth function. Show that $L^g$ is symmetric with respect to a measure that shall be computed.
Show that if the vector fields $V_i$ ‘s are globally Lipschitz, then $L^g$ is essentially self-adjoint.
Compute $\Gamma^g_2(f)=\frac{1}{2}( L^g\Gamma^g(f,f)-2\Gamma^g(f,L^gf))$ where $\Gamma^g$ is the carre du champ of $L^g$ .
Deduce a criterion for the stochastic completeness of the semigroup $P^g_t=e^{tL^g}$ .
Let $K$ be an elliptic diffusion operator with smooth coefficients on $\mathbb{R}^n$ . Show that $K$ can be written as $K=L^g+Z$ , where $Z$ is a smooth vector field and $\Delta^g$ is the Laplace Beltrami operator of some Riemannian metric.
Show that if $K$ is symmetric with respect to a measure equivalent to the Lebesgue measure, then it can be written as $L^g$ (see question 4) for some $g$ and some $U$ .

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Lecture 13. The Bochner’s formula

Posted on December 13, 2016 by Fabrice Baudoin

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 $\mathbb{R}^n$ .

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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Lecture 12. The distance associated to subelliptic diffusion operators

Posted on October 29, 2016 by Fabrice Baudoin

In this lecture we prove that most of the results that were proven for Laplace-Beltrami operators may actually be generalized to any locally subelliptic operator.
Let $L$ be a locally subelliptic diffusion operator defined on $\mathbb{R}^n$ . For every smooth functions $f,g: \mathbb{R} \rightarrow \mathbb{R}$ , we recall that the carre du champ operator is the symmetric first-order differential form defined by:

$\Gamma (f,g) =\frac{1}{2} \left( L(fg)-fLg-gLf \right).$
A straightforward computation shows that if

$L=\sum_{i,j=1}^n \sigma_{ij} (x) \frac{\partial^2}{ \partial x_i \partial x_j} +\sum_{i=1}^n b_i (x)\frac{\partial}{\partial x_i},$
then,

$\Gamma (f,g)=\sum_{i,j=1}^n \sigma_{ij} (x) \frac{\partial f}{\partial x_i} \frac{\partial g}{\partial x_j}.$
As a consequence, for every smooth function $f$ , $\Gamma(f) \ge 0.$

Definition: An absolutely continuous curve $\gamma: [0,T] \rightarrow \mathbb{R}^n$ is said to be subunit for the operator $L$ if for every smooth function $f : \mathbb{R}^n \to \mathbb{R}$ we have $\left| \frac{d}{dt} f ( \gamma(t) ) \right| \le \sqrt{ (\Gamma f) (\gamma(t)) }$ . We then define the subunit length of $\gamma$ as $\ell_s(\gamma) = T$ .

Given $x, y\in \mathbb{R}^n$ , we indicate with

$S(x,y) =\{\gamma:[0,T]\to \mathbb{R}^n \mid \gamma\ \text{is subunit for}\ \Gamma, \gamma(0) = x,\ \gamma(T) = y\}.$
In these lectures we always assume that

$S(x,y) \not= \emptyset,\ \ \ \ \text{for every}\ x, y\in \mathbb{R}^n.$
If $L$ is an elliptic operator or if $L$ is a sum of squares operator that satisfies Hormander’s condition, then this assumption is satisfied.

Under such assumption it is easy to verify that

$d(x,y) = \inf\{l_s(\gamma)\mid \gamma\in S(x,y)\},$
defines a true distance on $\mathbb{R}^n$ . This is the intrinsic distance associated to the subelliptic operator $L$ . A beautiful result by Fefferman and Phong relates the subellipticity of $L$ to the size of the balls for this metric:

Theorem: Let $x \in \mathbb{R}^n$ . There exist constants $\varepsilon >0$ , $r_0 >0$ and $C_1,C_2 >0$ such that for $0 \le r < r_0$ ,

$B(x,r) \subset B_d (x,C_1 r^\varepsilon) \subset B (x,C_2 r^\varepsilon) ,$
where $B_d$ denotes here the ball for the metric $d$ and $B$ the ball for the Euclidean metric on $\mathbb{R}^n$ .

A corollary of this result is that the topology induced by $d$ coincides with the Euclidean topology of $\mathbb{R}^n$ . The distance $d$ can also be computed using the following definition:

Proposition: For every $x,y \in \mathbb{R}^n$ ,
$d(x,y)=\sup \left\{ |f(x) -f(y) | , f \in C^\infty(\mathbb{R}^n) , \| \Gamma(f) \|_\infty \le 1 \right\},\ \ \ \ x,y \in \mathbb{R}^n.$

Proof: Let $x,y \in \mathbb{R}^n$ . We denote
$\delta (x,y)=\sup \{ | f(x)-f(y) |, f \in C_0^\infty(\mathbb{R}^n), \| \Gamma(f) \|_\infty \le 1 \}.$
Let $\gamma: [0,T] \to \mathbb{R}^n$ be a sub-unit curve such that
$\gamma(0)=x, \gamma(T)=x.$
We have $\left| \frac{d}{dt} f ( \gamma(t) ) \right| \le \sqrt{ (\Gamma f) (\gamma(t)) }$ , therefore, if $\Gamma(f) \le 1$ ,

$\left| f(y)-f(x) \right| \le T.$
As a consequence
$\delta (x,y) \le d(x,y).$

We now prove the converse inequality which is trickier. We already know that if $L$ is elliptic then $d(x,y)=\delta(x,y)$ . If $L$ is only subelliptic, we consider the sequence of operators $L_k=L+\frac{1}{k} \Delta$ where $\Delta$ is the usual Laplacian. We denote by $d_k$ the distance associated to $L_k$ . It is easy to see that $d_k$ increases with $k$ and that $d_k(x,y) \le d(x,y)$ . We can find a curve $\gamma_k:[0,1] \to \\mathbb{R}^n$ , such that $\gamma(0)=x,\gamma(1)=y$ and for every $f \in C^\infty(\mathbb{R}^n)$ ,

$\left| \frac{d}{dt} f(\gamma_k(t)) \right|^2 \le \left(d^2_k(x,y)+\frac{1}{k} \right)\left( \Gamma(f) (\gamma_k(t)) + \frac{1}{k} \Gamma_\Delta(f)(\gamma_k(t)) \right),$
where $\Gamma_\Delta$ is the carre du champ operator of $\Delta$ . Since $d_k \le d$ , we see that the sequence $\gamma_k$ is uniformly equicontinuous. As a consequence of the Arzela-Ascoli theorem, we deduce that there exists a subsequence which we continue to denote $\gamma_k$ that converges uniformly to a curve $\gamma:[0,1] \to \mathbb{R}^n$ , such that $\gamma(0)=x,\gamma(1)=y$ and for every $f \in C^\infty(\mathbb{R}^n)$ ,

$\left| \frac{d}{dt} f(\gamma (t)) \right|^2 \le \sup_k d^2_k(x,y) \Gamma(f) (\gamma_k(t)).$
By definition of $d$ , we deduce $d(x,y) \le \sup_k d_k(x,y)$ . As a consequence, we proved that $d(x,y)=\lim_{k \to \infty} d_k(x,y)$ . Since it is clear that

$d_k(x,y)= \sup \left\{ |f(x) -f(y) | , f \in C^\infty(\mathbb{R}^n) , \left\| \Gamma(f) + \frac{1}{k} \Gamma_\Delta(f) \right\|_\infty \le 1 \right\} \le \delta(x,y),$
we finally conclude that $d(x,y) \le \delta(x,y)$ , hence $d(x,y)=\delta(x,y)$ .
$\square$

A straightforward corollary of the previous proposition is the following useful result:

Corollary: If $f \in C^\infty(\mathbb{R}^n)$ satisfies $\Gamma(f)=0$ , then $f$ is constant.
The Hopf-Rinow theorem is still true with an identical proof in the case of subelliptic operators.

Theorem: The metric space $(\mathbb{R}^n,d)$ is complete (i.e. Cauchy sequences are convergent) if and only the compact sets are the closed and bounded sets.
Similarly, we also have the following key result:
Proposition: There exists an increasing sequence $h_n\in C^\infty_0(\mathbb{R}^n)$ , $0 \le h_n \le 1$ , such that $h_n\nearrow 1$ on $\mathbb{R}^n$ , and $\| \Gamma(h_n) \|_{\infty} \to 0$ , as $n\to \infty$ if and only if the metric space $(\mathbb{R}^n,d)$ is complete.

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Lecture 11. Laplace-Beltrami operators on Rn

Posted on October 29, 2016 by Fabrice Baudoin

In this lecture we define Riemannian structures and corresponding Laplace–Beltrami operators. We first study Riemannian structures on Rn to avoid technicalities in the presentation of the main ideas and then, in a later lecture, will extend our results to the manifold case.

We start with the following basic definition:

Definition: A Riemannian structure on $\mathbb{R}^n$ is a smooth map $g$ from $\mathbb{R}^n$ to the set of symmetric positive matrices.

In other words, a Riemannian structure induces at each point $x \in \mathbb{R}^n$ an inner product $g_x$ , and the dependence $x \rightarrow g_x$ is asked to be smooth.

A natural way to define Riemannian structures, is to start from a family of smooth vector fields $V_1,\cdots, V_n$ such that for every $x \in \mathbb{R}^n$ , $(V_1(x),\cdots, V_n(x))$ is a basis of $\mathbb{R}^n$ . It is then easily seen that there is a unique Riemannian structure on $\mathbb{R}^n$ that makes for $x \in \mathbb{R}^n$ , $(V_1(x),\cdots, V_n(x))$ an orthonormal basis.

Conversely, given a Riemannian structure on $\mathbb{R}^n$ , it is possible to find smooth vector fields $V_1,\cdots, V_n$ on $\mathbb{R}^n$ such that for $x \in \mathbb{R}^n$ , $(V_1(x),\cdots, V_n(x))$ an orthonormal basis for this Riemannian structure.

In this course, we shall mainly deal with such a point of view on Riemannian structures and use as much as possible the language of vector fields. This point of view is not restrictive and will allow more easy extensions to the sub-Riemannian case in a later part of the course.

Let us consider a family of smooth vector fields $V_1,\cdots, V_n$ such that for every $x \in \mathbb{R}^n$ , $(V_1(x),\cdots, V_n(x))$ is a basis of $\mathbb{R}^n$ . Without loss of generality we may assume that $\mathbf{det} ( V_1(x),\cdots, V_n(x)) > 0.$ Our goal is to associate to this Riemannian structure a canonical diffusion operator.

As a first step, we associate with the vector fields $V_1,\cdots, V_n$ a natural Borel measure $\mu$ which is the measure with density $d\mu =\frac{1}{ \mathbf{det} ( V_1(x),\cdots, V_n(x))} dx$ with respect to the Lebesgue measure. This is the so-called Riemannian measure. The diffusion operator we want to consider shall be symmetric with respect to this measure.

Remark: Let $(U_1,\cdots,U_n)$ be another system of smooth vector fields on $\mathbb{R}^n$ such that for every $x \in \mathbb{R}^n$ , $(U_1(x),\cdots,U_n(x))$ is an orthonormal basis with respect to the inner product $g_x$ . The systems of vector fields $(U_1,\cdots,U_n)$ and $(V_1,\cdots,V_n)$ are related one to each other through an orthogonal mapping. This implies that $|\mathbf{det} ( V_1(x),\cdots, V_n(x))|=| \mathbf{det} ( U_1(x),\cdots, U_n(x))|.$ In other words, the Riemannian measure $\mu$ only depends on the Riemannian structure $g$ .

Due to the fact that for every $x \in \mathbb{R}^n$ , $(V_1(x),\cdots, V_n(x))$ is a basis of $\mathbb{R}^n$ , we may find smooth functions $\omega^{k}_{ij}$ ‘s on $\mathbb{R}^n$ such that $[V_i,V_j]=\sum_{k=1}^n \omega_{ij}^k V_k.$ Those functions are called the structure constants of the Riemannian structure. Every relevant geometric quantities may be expressed in terms of these functions. Of course, these functions depend on the choice of the vector fields $V_i$’s and thus are not Riemannian invariants, but several combinations of them, like curvature quantities, are Riemannian invariants.

The following proposition expresses the formal adjoint with respect to the Lebesgue measure of the vector field $V_i$ which is seen as an operator acting on the spave of smooth and compactly supported functions.

Proposition: If $f,g\in \mathcal{C}_c(\mathbb{R}^n, \mathbb{R})$ are smooth and compactly supported functions, then we have $\int_{\mathbb{R}^n} (V_i f ) g d\mu =\int_{\mathbb{R}^n} f (V_i^* g) d\mu,$ where $V_i^*=-V_i+ \sum_{k=1}^n \omega_{ik}^k.$

Proof: Let us denote $V_i=\sum_{j=1}^n v_i^j \frac{\partial}{\partial x_j},$ and $m(x)= \mathbf{det} ( V_1(x),\cdots, V_n(x)).$ We have $\int_{\mathbb{R}^n} (V_i f ) g d\mu =\int_{\mathbb{R}^n}\sum_{j=1}^n v_i^j \frac{\partial f}{\partial x_j} g \frac{dx}{m} =-\sum_{j=1}^n\int_{\mathbb{R}^n} f \left( m \frac{\partial}{\partial x_j} \frac{1}{m} g v_i^j \right) d\mu.$ We now compute $\sum_{j=1}^n m \frac{\partial}{\partial x_j} \frac{1}{m} v_i^j=-\sum_{j=1}^n v_i^j \frac{1}{m} \frac{\partial m}{\partial x_j} +\sum_{j=1}^n\frac{\partial v_i^j}{\partial x_j}$ . We then observe that $\frac{\partial m}{\partial x_j} = \frac{\partial}{\partial x_j} \mathbf{det} ( V_1(x),\cdots, V_n(x)) =\sum_{k=1}^n \mathbf{det} \left( V_1(x),\cdots,\frac{\partial V_k}{\partial x_j}(x),\cdots, V_n(x)\right).$
Thus, we obtain
$-\sum_{j=1}^n v_i^j \frac{\partial m}{\partial x_j} +\sum_{j=1}^n m \frac{\partial v_i^j}{\partial x_j} = \sum_{k=1}^n \mathbf{det} \left( V_1(x),\cdots,-\sum_{j=1}^n v_i^j \frac{\partial V_k}{\partial x_j}(x),\cdots, V_n(x)\right)$
$+\sum_{k=1}^n\mathbf{det} \left( V_1(x),\cdots,\frac{\partial v_i^j}{\partial x_j} V_k(x) ,\cdots, V_n(x)\right)$
$=-\sum_{k=1}^n\mathbf{det} \left( V_1(x),\cdots,[V_i,V_k](x) ,\cdots, V_n(x)\right)$
$=-\sum_{k=1}^n\mathbf{det} \left( V_1(x),\cdots,\sum_{j=1}^n \omega_{ik}^j(x) V_j(x) ,\cdots, V_n(x)\right)$
$=-\sum_{k=1}^n \omega_{ik}^k (x)\mathbf{det} \left( V_1(x),\cdots,V_k(x) ,\cdots, V_n(x)\right)=-m\sum_{k=1}^n \omega_{ik}^k$ . $\square$

With this integration by parts formula in hands we are led to the following natural definition
Definition: The diffusion operator $L =-\sum_{i=1}^n V_i^* V_i =\sum_{i=1}^n V_i^2 -\sum_{i,k}^n \omega_{ik}^k V_i$ is called the Laplace-Beltrami operator associated with the Riemannian structure $g$ .

The following straightforward properties of $L$ are let as an exercise to the reader:

$L$ is an elliptic operator;
The Riemannian measure $\mu$ is invariant for $L$ ;
The operator $L$ is symmetric with respect to $\mu$ .

Exercise: Let $(U_1,\cdots,U_n)$ be another system of smooth vector fields on $\mathbb{R}^n$ such that for every $x \in \mathbb{R}^n$ , $(U_1(x),\cdots,U_n(x))$ is an orthonormal basis with respect to the inner product $g_x$ . Show that $\sum_{i=1}^n V_i^* V_i=\sum_{i=1}^n U_i^* U_i.$ In other words, the Laplace-Beltrami operator is a Riemannian invariant: It only depends on the Riemannian structure $g$ .

In order to apply the diffusion semigroup theory developed in the first lectures and construct without ambiguity the semigroup associated to the Laplace-Beltrami operator L, we need to know if L is essentially self-adjoint. Interestingly, this property of essential self-adjointness is clodely related to a metric property of the underlying Riemannian structure: The completeness of the associated distance.

Given an absolutely continuous curve $\gamma: [0,T] \rightarrow \mathbb{R}^n$ , we define its Riemannian length by $L_g (\gamma)=\int_0^T \sqrt{g_{\gamma(s)} (\gamma'(s),\gamma'(s) )} ds.$ If $x,y \in \mathbb{R}^n$ , let us denote by $\mathcal{C}(x,y)$ the set of absolutely continuous curves $\gamma: [0,1] \rightarrow \mathbb{R}^n$ such that $\gamma(0)=x, \gamma(1)=y.$ The Riemannian distance between $x$ and $y$ is defined by $d(x,y)=\inf_{\gamma \in \mathcal{C}(x,y)} L_g(\gamma).$ By using reparametrization, we may define the Riemannian distance in a equivalent way by using the notion of sub-unit curve. Let $\gamma: [0,T] \to \mathbb{R}^n$ be an absolutely continuous curve. Since the vector fields $V_1,\cdots,V_n$ ‘s form a basis of $\mathbb{R}^n$ at each point, we may find continuous functions $\alpha_i:[0,T] \to \mathbb{R}^n$ such that $\gamma'(t)=\sum_{i=1}^n \alpha_i(t) V_i (\gamma(t)).$ The curve $\gamma$ is then said to be sub-unit if for almost every $t \in [0,T]$ , $\sum_{i=1}^n \alpha_i(t)^2 \le 1$ . By using reparametrization, it is easily seen that for $x,y \in \mathbb{R}^n$ , $d(x,y)=\inf \left\{ T \text{ such that there exists a sub-unit curve } \gamma:[0,T] \to \mathbb{R}^n, \gamma(0)=x, \gamma(1)=y \right\}$ .

Exercise: Let $\theta_1, \cdots , \theta_n \in \mathbb{R}$ . Show that for every $x \in \mathbb{R}^n$ , $d \left(x, e^{\sum_{i=1}^n \theta_i V_i } x \right) \le \sum_{i=1}^n \theta_i^2.$

An important fact is that $d$ hence defined is indeed a distance that induces the usual topology of $\mathbb{R}^n$ .

Definition: The function $d$ defined above is a distance that induces the usual topology of $\mathbb{R}^n$ .

Proof: Since any curve can be parametrized backwards and forwards, we have $d(x,y)=d(y,x)$ . The triangle inequality is easily proved by using juxtaposition of curves. Plainly $d(x,x)=0$ , so it remains to prove that if $x \neq y$ then $d(x,y) > 0$ . Let $x,y \in \mathbb{R}^n$ such that $x \neq y$ . Let us denote $R=\| x-y \|$ . The closed Euclidean ball $\bar{B}_e (x,R)$ is compact, therefore there exist two constants $\alpha, \beta > 0$ such that for every $z \in \bar{B}_e (x,R)$ and $u \in \mathbb{R}^n$ , $\alpha^2 \| u \|^2 \le g_z(u,u) \le \beta^2 \| u \|^2.$ Let now $\gamma:[0,1]\to \mathbb{R}^n$ be an absolutely continuous curve such that $\gamma(0)=x, \gamma(1)=y$ . Let $\tau =\inf \{ t, \| \gamma(t) \|=R \}.$
We have
$L_g(\gamma)$
$=\int_0^1 \sqrt{g_{\gamma(s)} (\gamma'(s),\gamma'(s) )} ds$
$\ge \int_0^\tau \sqrt{g_{\gamma(s)} (\gamma'(s),\gamma'(s) )} ds$
$\ge \alpha \int_0^\tau \| \gamma'(s) \| ds$
$\ge \alpha \int_0^\tau \| \gamma'(s) \| ds$
$\ge \alpha \| \gamma(\tau) - \gamma (0) \|$
$\ge \alpha \| x -y \|.$
As a consequence, we deduce that $d(x,y) \ge \alpha^2 \| x -y \| > 0.$ Therefore $d$ is indeed a distance. Moreover, it is shown as above that for every $z \in \mathbb{R}^n, R> 0$ , there are constants $C_1,C_2> 0$ such that for every $x,y \in \bar{B}_e (z,R)$ , $C_1 \| x-y \| \le d(x,y) \le C_2 \| x-y \|.$ This implies that $d$ induces the usual topology of $\mathbb{R}^n$ $\square$

As shown in the following proposition, the distance $d$ is intrinsically associated to the Laplace-Beltrami operator.

Proposition: For $x,y \in \mathbb{R}^n$ , we have $d(x,y) =\sup \{ | f(x)-f(y) |, f \in \mathcal{C}_c^\infty(\mathbb{R}^n,\mathbb{R}), \| \Gamma(f,f) \|_\infty \le 1 \}.$

Proof: Let $x,y \in \mathbb{R}^n$ . We denote $\delta (x,y)=\sup \{ | f(x)-f(y) |, f \in \mathcal{C}_c^\infty(\mathbb{R}^n,\mathbb{R}), \| \Gamma(f,f) \|_\infty \le 1 \}.$ Let $\gamma: [0,T] \to \mathbb{R}$ be a sub-unit curve such that $\gamma(0)=x, \gamma(T)=x.$ We can find $\alpha_1,\cdots,\alpha_n:[0,T] \to \mathbb{R}^n$ such that $\gamma'(t)=\sum_{i=1}^n V_i (\gamma(t)) \alpha_i (t),$ and $\sum_{i=1}^n \alpha_i^2 \le 1$ . Let now $f\in \mathcal{C}^\infty(\mathbb{R}^n,\mathbb{R}), \| \Gamma(f,f) \|_\infty \le 1$ . From the change of variable formula we have, $f(\gamma(T))= f(\gamma(0)) + \sum_{i=1}^n \int_0^T V_i f(\gamma(s)) \alpha_i(s) ds.$ Therefore, from Cauchy-Schwarz inequality, $\left| f(y)-f(x) \right| \le T.$ As a consequence $\delta (x,y) \le d(x,y).$

We now prove the converse inequality which is trickier. The idea is to use the function $f(y)=d(x,y)$ that satisfies $| f(x) -f(y)|=d(x,y)$ and $"\Gamma(f,f) =1"$ . However, giving a precise meaning to $\Gamma(f,f) =1$ is not so easy, because it turns out that $f$ is not differentiable at $x$ . It suggests to use an approximation of the identity to regularize $f$ and avoid the discussion of this differentiability issue. More precisely, fix $x_o ,y_o\in \mathbb{R}^n$ , and for $N \ge 1$ , consider the function $\Psi_N(y)= \eta \left( \int_{\mathbb{R}^n} \rho_N (t) d(x_o,y-t) dt \right),$ where $\rho \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ , $\rho \ge 0$ , $\int_{\mathbb{R}^n} \rho=1$ , $\rho_N(t)=N^n \rho(N t)$ and $\eta \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ , $\eta \ge 0$ , has the property that $\| \eta ' \|_\infty \le 1$ and $\eta (u)=u$ for $u \in [0,1+d(x_o,y_o)]$ . Since for any $\theta \in \mathbb{R}^{n}$ , $\| \theta \|=1$ , $t \ge 0$ $| d(x_o,y)-d\left(x_o, e^{t \sum_{i=1}^n \theta_i V_i} y\right)| \le d\left(y, e^{t \sum_{i=1}^n \theta_i V_i} y\right) \le t,$ it is easy to see that $\Gamma(\Psi_N,\Psi_N) \le 1+\frac{C}{N}$ , for some constant $> 0$ . Hence $\delta(x_o,y_o) \ge \lim \inf_{N \to +\infty} | \Psi_N(y_o) -\Psi_N(x_o) |=d(x_o,y_o)$ $\square$

The following theorem is known as the Hopf–Rinow theorem, it provides a necessary and sufficient condition for the completeness of the metric space $(\mathbb{R}^n,d)$ .

Proposition: The metric space $(\mathbb{R}^n,d)$ is complete (i.e. Cauchy sequences are convergent) if and only the compact sets are the closed and bounded sets.

Proof: It is clear that if closed and bounded sets are compact then the metric space $(\mathbb{R}^n,d)$ is complete; It comes from the fact that Cauchy sequence are convergent if and only if they have at least one cluster value. So, we need to prove that closed and bounded sets for the distance $d$ are compact provided that $(\mathbb{R}^n,d)$ is complete. To check this, it is enough to prove that closed balls are compact. Let $x \in \mathbb{R}^n$ . Observe that if the closed ball $\bar{B}(x, r)$ is compact for some $r> 0$ , then $\bar{B}(x, \rho)$ is closed for any $\rho < r$ . Define $R=\sup \{ r> 0, \bar{B}(x, r) \text{ is compact } \}.$ Since $d$ induces the usual topology of $\mathbb{R}^n$ , $R>0$ . Let us assume that $R < +\infty$ and let us show that it leads to a contradiction. We first show that $\bar{B}(x, R)$ is compact. Since $(\mathbb{R}^n,d)$ is assumed to be complete, it suffices to prove that $\bar{B}(x, R)$ is totally bounded: That is, for every $\varepsilon > 0$ there is a finite set $S_\varepsilon$ such that every point of $\bar{B}(x, R)$ belongs to a $\varepsilon$ -neighborhood of $S_\varepsilon$ .

So, let $\varepsilon > 0$ small enough. By definition of $R$ , the ball $\bar{B}(x, R-\varepsilon / 4)$ is compact; It is therefore totally bounded. We can find a finite set $S=\{ y_1,\cdots,y_N\}$ such that every point of $\bar{B}(x, R-\varepsilon / 4)$ lies in a $\varepsilon / 2$ -neighborhood of $S$ . Let now $y \in \bar{B}(x, R)$ . We claim that there exists $y' \in \bar{B}(x, R-\varepsilon / 4)$ such that $d(y,y') \le \varepsilon /2$ . If $y \in \bar{B}(x, R-\varepsilon / 4)$ , there is nothing to prove, we may therefore assume that $y \notin \bar{B}(x, R-\varepsilon / 4)$ . Consider then a sub-unit curve $\gamma: [0,R+\varepsilon / 4]$ such that $\gamma(0)=x$ , $\gamma(R+\varepsilon/4)=y$ . Let $\tau =\inf \{t, \gamma(t) \notin \bar{B}(x, R-\varepsilon / 4) \}.$ We have $\tau \ge R-\varepsilon / 4$ . On the other hand, $d(\gamma(\tau), \gamma(R+\varepsilon/4)) \le R+\varepsilon/4 -\tau.$ As a consequence, $d(\gamma(\tau),y) \le \varepsilon /2.$ In every case, there exists therefore $y' \in \bar{B}(x, R-\varepsilon / 4)$ such that $d(y,y') \le \varepsilon /2$ . We may then pick $y_k$ in $S$ such that $d(y_k,y') \le \varepsilon / 2$ . From the triangle inequality, we have $d(y,y_k) \le \varepsilon$ . So, at the end, it turns out that every point of $\bar{B}(x, R)$ lies in a $\varepsilon$ -neighborhood of $S$ . This shows that $\bar{B}(x, R)$ is totally bounded and therefore compact because $(\mathbb{R}^n,d)$ is assumed to be complete. Actually, the previous argument shows more, it shows that if every point of $\bar{B}(x, R)$ lies in a $\varepsilon /2$ -neighborhood of a finite $S$ , then every point of $\bar{B}(x, R+\varepsilon/4)$ will lie $\varepsilon$ -neighborhood of $S$ , so that the ball $\bar{B}(x, R+\varepsilon/4)$ is also compact. This contradicts the fact the definition of $R$ . Therefore every closed ball is a compact set, due to the arbitrariness of $x$ $\square$

Checking that the metric space $(\mathbb{R}^n,d)$ is complete is not always easy in concrete situations. From the Hopf-Rinow theorem, it suffices to prove that the closed balls are compact. The following proposition is therefore useful.

Proposition: Suppose that the vector fields $V_1,\cdots,V_n$ ‘s have globally Lipschitz coefficients . Then the closed ball $\bar B(x,R)$ is compact for every $x \in \mathbb{R}^n$ and $R > 0$ . As a consequence the metric space $(\mathbb{R}^n,d)$ is complete.
Proof: By the hypothesis on the $V_j$ ‘s there exists a constant $M > 0$ such
that $\|V(x)\|=\left(\sum_{j=1}^n \|V_j(x)\|^2\right)^\frac{1}{2}\leq M(1+\|x\|)$ for any $x\in\mathbb{R}^n$ . Fix $x_o,y\in\mathbb{R}^n$ and let $\gamma:[0,T]\to\mathbb{R}^n$ , be a sub-unit curve such that $\gamma(0)=x_o, \gamma(T)=y.$ Letting $y(t)=\|\gamma(t)\|^2$ we obtain $y'(t)=2\leq 2 \|\gamma(t)\|\|\gamma'(t)\| \leq 2\|\gamma(t)\|\|V(\gamma(t))\|.$ We infer that $y'(t)\leq C\left(\sqrt{y(t)}+y(t) \right)$ , for some $C>0$ depending only on $M$ . Integrating the latter inequality one has $\|\gamma(t)\| \leq (A+\| x_o \|) e^{\frac{C}{2}T},\quad t\in [0,T]$ . for some constant $A >0$ . The previous estimate shows in particular, that $B(x_o,R)\subset B_e(0, (A+\| x_o \|) e^{CR}).$ We conclude that $\bar B(x_o,R)$ is Euclidean compact. Since the metric and the Euclidean topology coincide, it is also $d$ -compact $\square$

Completeness of the metric space $(\mathbb{R}^n,d)$ is related to the essential self-adjointness of the Laplace-Beltrami operator.

Theorem: If the metric space $(\mathbb{R}^n,d)$ is complete, then the Laplace-Beltrami operator $L$ is essentially self-adjoint.

Proof: We know that if there exists an increasing sequence $h_n\in C_c(\mathbb{R}^n,\mathbb{R})$ such that $h_n\nearrow 1$ on $\mathbb{R}^n$ , and $||\Gamma(h_n,h_n)||_{\infty} \to 0$ , as $n\to \infty$ , then the operator $L$ is essentially self-adjoint. We are therefore reduced to prove the existence of such a sequence. Let us fix a base point $x_0\in\mathbb{R}^n$ . We can find an exhaustion function $\rho\in C^\infty(\mathbb{R}^n,\mathbb{R})$ such that $|\rho - d(x_0,\cdot)| \le L,\ \ \ \ \ \ |\Gamma(\rho,\rho)|\le L \ \ \text{on } \mathbb{R}^n.$ By the completeness of $(\mathbb{R}^n,d)$ and the Hopf-Rinow theorem, the level sets $\Omega_s = \{x\in \mathbb{M} \mid \rho(x) < s\}$ are relatively compact and, furthermore, $\Omega_s \nearrow \mathbb{R}^n$ as $s\to \infty$ . We now pick an increasing sequence of functions $\phi_n\in C^\infty([0,\infty))$ such that $\phi_n\equiv 1$ on $[0,n]$ , $\phi_n \equiv 0$ outside $[0,2n]$ , and $|\phi_n'|\le \frac{2}{n}$ . If we set $h_n(x) = \phi_n(\rho(x))$ , then we have $h_n\in C_c(\mathbb{R}^n,\mathbb{R})$ , $h_n\nearrow1$ on $\mathbb{R}^n$ as $n\to \infty$ , and $||\Gamma(h_n,h_n)||_{\infty} \le \frac{2L}{n}$ $\square$