Optimal Transport and Ricci curvature

Introduction and Motivation

This article provides a brief introduction into a connection of optimal transport and the curvature tensor of a Riemannian manifold. In fact, we are going to study the transport map $T_{t}(x):={\text{exp}}_{x}(t\xi (x)),$ where $\xi$ denotes a $C^{1}$ vector field on the manifold $(M,g).$

These kind of maps appear very naturally in the context of optimal transport. Recall that in optimal transport one is particularly interested in the Monge Problem, being the following optimization problem: Let $(M,g)$ be a compact and connected Riemannian manifold. Let furthermore, $\mu =fdV,\nu =gdV$ denote two probability measures on $M$ which are absolutely continuous with respect to the measure on the manifold, induced by the metric. the Monge Problem is then given by

               $\inf _{T\#\mu =\nu }\int d(x,T(x))^{2}\,dV,$

where the infimum is taken among all measurable maps $T:M\rightarrow M$ and $d$ denotes the Metric on $(M,g)$ induced by $g.$ Then the Monge Problem admits a unique solution $T.$ Moreover, in that case

         $T(x)=\exp _{x}(\nabla \psi (x))$

for some $\psi$ (see for more details of this).

To conclude the introductory part of this article, let us also mention that these kind of transport maps, turned out to be useful in the area of geometric analysis. In fact, Simon Brendle could prove a Sobolev inequality on non compact Riemannian manifolds with nonnegative Ricci curvature, the proof of which makes use of defining a map which is of the type $T(x)=\exp _{x}(\nabla u(x))$ ( see proof of Theorem 1.1 in for more details).

Curvature and Optimal Transport

Let $(M,g)$ be a Riemannian manifold. In this article we assume basic knowledge about the notions of curvature and geodesics on a manifold. For some background information on these topics, we refer the reader to Chapter three to five in .

The Goal of this article is to show the follwing

Proposition

Let $T_{t}(x)=\exp _{x}(t\nabla \psi (x)),$ where $\psi$ denotes a $C^{2}$ function on $M$ and let ${\mathcal {J}}(t):={\text{log}}({\text{det}}[d_{x}T_{t}]).$ Then the following inequality holds true:

  ${\mathcal {J}}''+{\frac {1}{n}}{\mathcal {J}}'+{\text{Ric}}_{\gamma (t)}(\gamma '(t),\gamma '(t))\leq 0,\quad {\text{ for all }}t{\text{ such that }}{\mathcal {J}}(t)>0,$

where $\gamma$ is defined to be the mapping $t\mapsto T_{t}(x),$ which is a geodesic.

Notice that this inequality involves the transport map $T$ and the Ricci curvature tensor and therefore constitutes a connection of the curvature and the optimal transport problem.

Remarks

Before we prove the Proposition, let us do some remarks: let $x\in M$ be given and let $e_{i}$ for $i=1,\dots ,n$ be an orthonormal basis. After doing parallel transport along $\gamma$, we have an orthonormal basis also in $T_{T_{t}(x)}M$. Let ${\textbf {J}}$ denote the matrix representation of $d_{x}T_{t}:T_{x}M\rightarrow T_{T_{t}(x)}M.$ Then we have that ${\textbf {J}}''+R{\textbf {J}}=0$ where $R$ denotes the matrix $(R(\gamma ',e_{i},\gamma ',e_{j}))_{i,j}.$ Indeed, this follows right away from the fact that $J_{i}(t):=d_{x}T_{t}(e_{i})$ is a Jacobi field along $\gamma$ for each $i=1,\dots ,n.$

Proof of the Proposition

With the notation of the proposition stated above, we compute the derivative of ${\mathcal {J}}$ as follows: ${\frac {d}{dt}}{\mathcal {J}}(t)={\text{trace}}({\textbf {J}}'{\textbf {J}}^{-1})$ which follows right away from Jacobi's formula. Let now $U(t):={\textbf {J}}'{\textbf {J}}^{-1}$. Since ${\textbf {J}}$ satisfies the matrix Jacobi equation as noticed in the preceding remark, we may infer a Ricatti equation for $U$. Indeed, taking derivative of $U(t)$ we obtain the following equation

$U'={\textbf {J}}''{\textbf {J}}^{-1}+{\textbf {J}}'{\frac {d}{dt}}{\textbf {J}}^{-1}.$

Now observe that

${\frac {d}{dt}}{\textbf {J}}^{-1}=-{\textbf {J}}^{-1}{\textbf {J}}'{\textbf {J}}^{-1}.$

In fact, notice that we only consider $t$ such that $\det({\textbf {J}})>0$. Thus the inverse matrix exists for all $t$ in a small neighborhood of that $t$. Then we have that

$0={\frac {d}{dt}}1_{n}={\frac {d}{dt}}{\textbf {J}}{\textbf {J}}^{-1}={\textbf {J}}'{\textbf {J}}^{-1}+{\textbf {J}}{\frac {d}{dt}}{\textbf {J}}^{-1}$

from which, after rearranging and multiplying with the inverse from the left, one gets the desired equality. Plugging this into the equation concerning the derivative of $U$, we get that

$U'={\textbf {J}}''{\textbf {J}}^{-1}-{\textbf {J}}'{\textbf {J}}^{-1}{\textbf {J}}'{\textbf {J}}^{-1}={\textbf {J}}''{\textbf {J}}^{-1}-U^{2}.$

Then, using the matrix Jacobi equation, we get that

$U'=-R{\textbf {J}}{\textbf {J}}^{-1}-U^{2}=-R-U^{2},$

so that

$U'+R+U^{2}=0.$

Notice that this is a first order ODE. Since $U(0)={\text{Hess}}\psi ,$ we get that

This being an equation of matrices, we can now take the trace in that equation. We therefore obtain that

${\text{tr}}(U')+{\text{Ric}}_{\gamma }(\gamma ',\gamma ')+{\text{tr}}(U^{2})=0.$