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This question was originaly posted in the stackexchange https://math.stackexchange.com/questions/1226701/a-measure-on-the-space-of-probability-measures but since it only got a comment I decided to post it here. I don't know if this is allow, please let me know if it's not. anyway...

I've been reading about optimal transport and it's connections to geometry. At some point one has to study a bit of the structure of the space of probability measures, $\mathcal{P}(X)$, (over a metric space X) given the 2-Wasserstein metric $W_2$. I was wondering if there is a 'canonical' way of endowing ($\mathcal{P}(X),W_2)$ with a 'nice' probability measure.

More specificaly, let $(X,d)$ be a metric space, and $(\mathcal{P}(X),W_2)$ it's space of probability measures with the 2-Wasserstein metric. Then,

  1. Can we set a 'nice' measure, $\mu$, on $(\mathcal{P}(X),W_2)$? (here by nice I guess I mean a non trivial measure that will maybe let us study $(\mathcal{P}(X),W_2,\mu)$ as a metric mesure space. I realize this is vague, a little guidance here would be appreciated)

  2. If we can, what conditions on $X$ are required?

  3. Is there any other measure that is usually given to $(\mathcal{P}(X),d)$? Where $d$ can be another metric distinct from $W_2$.

Thanks, for the time. Any comments and references are highly appreciated!

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    $\begingroup$ Any such measure would induce a probability measure on $X$, so there cannot be any 'canonical' choice whatever properties you impose on $X$. $\endgroup$
    – Simon Henry
    Commented Apr 21, 2015 at 15:55
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    $\begingroup$ @SimonHenry: in many cases, there are measures on $X$ that one can take as reference (e.g. Riemannian volume on a compact manifold, Lebesgue measure on $[0,1]$, etc.); also it makes more sense to study $\mathcal{P}(X)$ as a metric measure space as is asked here when $X$ itself is a metric measure space, in which case one does have by definition a reference proba on $X$. $\endgroup$
    – Benoît Kloeckner
    Commented Apr 22, 2015 at 8:34
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    $\begingroup$ For me "canonical" = "functorial with respect to isomorphisms". So this was a fact : there is no canonical choice of measure on $P(X)$. My point was just that one at least need to chose a measure on $X$ and that adding topological or metric property on $X$ is probably not enough (although the example of compact riemanian manifold can be considered as "topological properties"). $\endgroup$
    – Simon Henry
    Commented Apr 22, 2015 at 9:57

2 Answers 2

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von Renesse and Sturm have constructed a family "natural" measures $\mu$ making $(\mathcal{P}([0,1]),d_{W_2},\mu_\beta)$ into a metric measure space: http://www.ams.org/mathscinet-getitem?mr=2537551. They argue that their measure can be formally thought of as $$ \tag{*} \mu_\beta = C_\beta^{-1} e^{-\beta Ent(\cdot | \text{Leb})} \mu_{unif} $$ where $C_\beta$ is a normalization constant, $Ent(\rho \text{Leb} |\text{Leb}) = \int_{[0,1]}\rho \log \rho$ (and is infinite for non-absolutely continuous measures), and $\mu_{unif}$ is a sort of "uniform" measure.

Sturm later extended this to allow the underling space to be any $n$-dimensional Riemannian manifold: http://www.ams.org/mathscinet-getitem?mr=2857031.

However, (*) is not completely rigorous, and the measures do not behave quite as one would expect. There is a formal infinite dimensional Riemannian structure on the regular part of $(\mathcal{P}(M),d_{W_2})$ (i.e. on the subset of measures of the form $\rho dV_g$ where $\rho \in C^\infty(M)$) that goes back to Otto (http://www.ams.org/mathscinet-getitem?mr=1842429). One might hope that this Riemannian structure (formally) induced some sort of measure, and that one could use this formal argument to find an actual measure on $\mathcal{P}(M)$, which behaves nicely. However, it seems like while a good notion of "gradient" exists in this formal setting, a Laplacian does not. This suggests that there is no natural measure because otherwise the Laplacian could be defined by integration by parts. See Gigli's monograph http://www.ams.org/mathscinet-getitem?mr=2920736 (remark 5.6).

An obvious question given a measure on $(\mathcal{P}(M),d_{W_2})$ is whether or not it has lower Ricci curvature bounds in the sense of Lott--Villani--Sturm. When $M$ is the interval, this is not true: http://www.ams.org/mathscinet-getitem?mr=2900478. See also Section 8.3 in http://arxiv.org/pdf/1105.2883.pdf

In a slightly different direction. Erbar and Huesmann have recently studied http://link.springer.com/article/10.1007/s00526-014-0790-1 the curvature of the "configuration space" of a manifold and shown that there, there is a natural choice of measure which inherits lower Ricci curvature bounds from the underlying manifold. The configuration space is quite similar to the space of all probability measures (for example, the set of finite sums of dirac measures is dense in the space of probability measures with respect to weak convergence).

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  • $\begingroup$ a really complete overview. thanks a lot for the answer! $\endgroup$
    – Bruce Wayne
    Commented Apr 22, 2015 at 18:18
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What is a "natural measure" on a metric space? Nothing other than the Hausdorff measure (if it exists) comes to mind. However, in spite of a lot of literature on the geometry of the 2-Wasserstein metric, I have never seen anything about its Hausdorff measure (or even dimension).

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    $\begingroup$ Whenever $X$ is infinite, the Hausdorff dimension of its Wasserstein space is infinite. One can still define related invariants and compute them: I did precisely that in perso-math.univ-mlv.fr/users/kloeckner.benoit/papiers/… then improved in perso-math.univ-mlv.fr/users/kloeckner.benoit/papiers/… $\endgroup$
    – Benoît Kloeckner
    Commented Apr 22, 2015 at 8:40
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    $\begingroup$ However there is little hope to derive an equivalent to Hausdorff measure: for similar spaces, namely the set of closed subsets of the interval, it has been shown that any analogue of a Hausdorff measure is either zero or not $\sigma$-finite (see references in the above-mentioned papers). $\endgroup$
    – Benoît Kloeckner
    Commented Apr 22, 2015 at 8:42
  • $\begingroup$ I see - thank you - was actually waiting for a comment from you :) $\endgroup$
    – R W
    Commented Apr 22, 2015 at 9:20

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