4
$\begingroup$

In the article "Wasserstein Geometry Of Gaussian Measures" by Asuka Takatsu the author shows how the space of d-dimensional Gaussian probability measures with non-singular covariance matrices are a Riemannian submanifold of the $2$-Wasserstein space on $\mathbb{R}^d$; by showing that the Wasserstein distance thereon can be induced by a Riemannian metric.

Are there any other examples of parameteric families of $d$-dimensional probability measures for which this is true? I.e. for which their Wasserstein (metric) geometry coincides with a Riemmannian geometry?

$\endgroup$
2
  • 2
    $\begingroup$ There is the simple example of delta masses concentrated at one point. This seems trivial, but gives the insight that Wasserstein space has some negative curvature whenever the underlying space does. $\endgroup$
    – Gabe K
    Commented Jan 26, 2023 at 9:19
  • 1
    $\begingroup$ just slightly more advanced than @GabeK 's answer: fix a model distribution $\rho_0\in P_2(R^d)$. Then the family $\{\rho_\tau(\cdot)=\rho_0(\cdot-\tau),\quad\tau\in R^d\}$ is trivially a Riemannian manifold with Euclidean metrics, since $W_2^2(\rho_\tau,\rho_{\tau'})=|\tau-\tau'|^2$ $\endgroup$ Commented Apr 3 at 13:43

1 Answer 1

1
$\begingroup$

Other than point masses in a length space or Gaussian measures in $\mathbb{R}^n$, I don't know of examples of families which are totally geodesic in Wasserstein geometry.

However, it is possible to take a smooth family of probability measures and induce it with a Riemannian metric coming from the Wasserstein geometry. This idea is sometimes known as "Wasserstein Information Geometry" and was studied by Wuchen Li and Jiaxi Zhao [1]. The general framework is to consider a smooth parametrized family $\mathcal{F}$ of probability measures and assume the underlying sample space $S$ has some metric structure. In several important cases (e.g., when $S$ is a finite set or a Riemannian manifold), the 2-Wasserstein space $\mathbb{P}_2(S)$ admits a formal Riemannian structure known as the Otto metric. As such, $\mathcal{F}$ can be understood as a sub-manifold of a "Riemannian manifold." (I put "Riemannian manifold" in quotes here because $\mathbb{P}_2(S)$ is generally not a Banach manifold.)

The same way that a surface in $\mathbb{R}^3$ inherits a Riemannian metric from the ambient Euclidean space, we can induce $\mathcal{F}$ with a Riemannian metric from the ambient Otto metric, and in most cases this will be a genuine Riemannian metric (even though $\mathbb{P}_2(S)$ will often be singular). In general, $\mathcal{F}$ will not be a totally geodesic submanifold of $\mathbb{P}_2(S)$, so you cannot compute distance between points in $\mathcal{F}$ in terms of their distances in $\mathbb{P}_2(S)$ or vice-versa.

Li, Wuchen; Zhao, Jiaxi, Wasserstein information matrix, Inf. Geom. 6, No. 1, 203-255 (2023). ZBL1521.94014.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .