A Geometric Study of Wasserstein Spaces: An Addendum on the Boundary

We extend the geometric study of the Wasserstein space W2(X) of a simply connected, negatively curved metric space X by investigating which pairs of boundary points can be linked by a geodesic, when X is a tree. Let X be a Hadamard space, by which we mean that X is a complete globally CAT(0), locally compact metric space. Mainly, X is a space where triangles are “thin”: points on the opposite side to a vertex are closer to the vertex than they would be in the Euclidean plane. This assumption can also be interpreted as X having non-positive curvature, in a setting more general than manifolds; it has a lot of consequences (the distance is convex, X is contractible, it admits a natural boundary and an associated compactification, . . . ) An important example of Hadamard space, on which we shall focus in this paper, is simply an infinite tree. The set of Borel probability measures of X having finite second moment can be endowed with a natural distance defined using optimal transportation, giving birth to the Wasserstein space W2(X). It is well-known that W2(X) does not have non-positive curvature even when X is a tree. This note is an addendum to [BK12], where we defined and studied the boundary of W2(X). We refer to that article and references therein for the background both on Hadamard space and optimal transportation, as well as for notations. Note that a previous (long) version of [BK12] contained the present content, but has been split after remarks of a referee. Let us quickly sum up the content of [BK12]. The boundary of X can be defined by looking at geodesic rays, and identifying rays that stay at bounded distance one to another (“asymptote” relation). We showed that there is a natural boundary ∂ W2(X) of the Wasserstein space that is both close to the traditional boundary of Hadamard spaces (a boundary point can be defined as an asymptote class of rays) and relevant to optimal transportation (a boundary point can be seen as a measure on the cone over ∂X , encoding the asymptotic direction and speed distribution of the mass along a ray). This boundary can ⋆ Both authors where partially supported by ANR grant ANR-11-JS01-0011. ha l-0 07 85 68 5, v er si on 2 13 M ay 2 01 3 be given a topology consistent with both points of view, and an angular metric; unsurprisingly, it carries geometric information about W2(X). Here we adress the visibility, or lack thereof, of W2(X). A Hadamard space satisfies the visibility condition if any pair of boundary points can be linked by a geodesic (e.g. all trees have the visibility property), and the same definition makes sense for its Wasserstein space. It is easily seen that even when X has the visibility condition, W2(X) does not; our result is a complete characterization of pairs of asymptotic measures that are the ends of a complete geodesic when X is a tree (Theorem 1 in Section 4). Our motivation is twofold: first this result shows how much more constrained complete geodesics of W2(X) are compared to complete rays; second the method of proof involves cyclical monotonicity in an interesting way, because we have to deal with an optimal transport problem that needs not have a finite infimum. 1 A first necessary condition: antipodality A complete geodesic (μt) in W2(X) defines two rays and one therefore gets two asymptotic measures, denoted by μ−∞ and μ+∞, also called the ends of the geodesic. We recall that these measures are probability measures on the cone c∂X over the geodesic boundary of X . But by Proposition 5.2 of [BK12], these measures are in fact concentrated on ∂X , viewed as a subset of c∂X . In particular, W2(X) is already far from satisfying the visibility condition. Note that we shall need to consider measures μ on the set of unit complete geodesics G R 1 (X) that satisfy the cyclical monotonicity, but such that et#μ need not have finite second moment. We still call such maps dynamical transport plan and we say that e±∞#μ are its ends. Such a measure μ defines a complete unit geodesic in W2(X) if and only if et#μ ∈ W2(X) for some, hence all t ∈ R. In this section, we only consider unit geodesics even if it is not stated explicitly. The asymptotic formula (Theorem 4.2 of [BK12]) gives us a first necessary condition valid for any Hadamard space. Let us say that two points ζ, ξ ∈ ∂X are antipodal if they are linked by a geodesic, that two sets A−, A+ ⊂ ∂X are antipodal if all pairs (ζ, ξ) ∈ A−×A+ are antipodal, and that two measures ν−, ν+ on ∂X are antipodal when they are concentrated on antipodal sets. Morever, let us call uniformly antipodal a pair of measures whose supports are antipodal. Given a complete unit geodesic μ, the asymptotic formula readily implies that the ends of any complete unit geodesic of W2(X) must be antipodal. When X is a tree, every pair of boundary points is antipodal and this condition simply reads that the ends must be concentrated on disjoint sets. 2 Flows and antagonism From now on, X is assumed to be a tree, described as a graph by a couple (V,E) where: V is the set of vertices; E is the set of edges, each endowed with one or two endpoints in V and a positive length. Since X is assumed to be complete, ha l-0 07 85 68 5, v er si on 2 13 M ay 2 01 3 the edges with only one endpoint are exactly those that have infinite length. It is assumed that vertices are incident to 1 or at least 3 edges, so that the combinatorial description of X is uniquely determined by its metric structure. Since X is locally compact, as a graph it is then locally finite. We fix a base point x0 ∈ X and use d to denote the distance on X . We say that two geodesics are antagonist if there are two distinct points x, y such that one of the geodesics goes through x and y in this order, and the other goes through the same points in the other order. We add to each infinite end a formal endpoint at infinity to unify notations. Each edge e has two orientations (xy) and (yx) where x, y are its endpoints. The complement in X̄ of the interior of an edge e of endpoints x, y has two components Cx(xy) ∋ x and Cy(xy) ∋ y. An oriented edge (xy) has a future (xy)+ := Cy(xy) ∩ ∂X and a past (xy)− := Cx(xy) ∩ ∂X . Assume ν− and ν+ are antipodal measures on ∂X . Define a signed measure by ν = ν+−ν− and note that ν(∂X) = 0. The flow (defined by (ν−, ν+)) through an oriented edge (xy) is φ(xy) := ν((xy)+). The flow gives a natural orientation of edges: an oriented edge is positive if its flow is positive, neutral if its flow is zero, and negative otherwise. Given a vertex x, let y1, . . . , yk be the neighbors of x such that (xyi) is positive, and z1, . . . , zl be the neighbors of x such that (xzj) is negative. Then ∑ i φ(xyi) = ∑ j φ(zjx) is called the flow through x and is denoted by φ(x). If x 6= x0, then there is a unique edge starting at x along which the distance to x0 is decreasing. If this edge is a positive one, (xyi0) say, then define the specific flow through x as φ(x) = ∑ i6=i0 φ(xyi). If this edge is a negative one, (xzj0 ), then let φ(x) = ∑ j 6=j0 φ(zjx). If this edge is neutral or if x = x0, then let φ(x) = φ(x). Note that φ(xy) = −ν((xy)−) = −φ(yx). Given a dynamical transport plan μ, we denote by μ(xy) the μ-measure of the set of geodesics that go through an edge (xy) in this orientation, by μ(x) the μ-measure of the set of geodesics that pass at x, and by μ(x) the μ-measure of those that are moreover closest to x0 at this time. Lemma 1. If μ is any dynamical transport plan with ends ν±, then: 1. for all edge (xy) we have μ(xy) > max(φ(xy), 0), 2. for all vertex x we have μ(x) > φ(x). and each of these inequality is an equality for all (xy), respectively all x, if and only if μ contains no pair of antagonist geodesics in its support. In this case, we moreover have μ(x) = φ(x) for all x. Proof. We prove the first point, the other ones are similar. Denote by μ(Cy(xy)) the measure of the set of geodesic that lie entirely in Cy(xy). We have μ(xy) + μ(Cy(xy)) = ν+((xy)+) = φ(xy) + ν−((xy)+) and ν−((xy)+) = μ(Cy(xy)) + μ(yx). ha l-0 07 85 68 5, v er si on 2 13 M ay 2 01 3 It follows that φ(xy) = μ(xy)−μ(yx) so that μ(xy) > φ(xy). Moreover the case of equality μ(xy) = max(φ(xy), 0) implies that μ(yx) = 0 whenever μ(xy) > 0, and we get the conclusion. Lemma 2. A dynamical transport plan μ is d-cyclically monotone if and only if μ⊗ μ-almost no pairs of geodesics are antagonist. Proof. Assume that the support of μ contains two antagonist geodesics γ, β and let x, y be points such that γt = x, γu = y where u > t and βv = y, βw = x where w > v. Let r = min(t, v) and s = max(u,w). Then d(γr, βs) 2 + d(γs, βr) 2 < d(γr, γs) 2 + d(βr, βs) ) so that the transport plan (er, es)#μ between μr and μs would not be cyclically monotone (see Figure 1).