Isometric rigidity of Wasserstein spaces over Euclidean spheres

We study the structure of isometries of the quadratic Wasserstein space $\mathcal{W}_2\left(\mathbb{S}^n,\|\cdot\|\right)$ over the sphere endowed with the distance inherited from the norm of $\mathbb{R}^{n+1}$. We prove that $\mathcal{W}_2\left(\mathbb{S}^n,\|\cdot\|\right)$ is isometrically rigid, meaning that its isometry group is isomorphic to that of $\left(\mathbb{S}^n,\|\cdot\|\right)$. This is in striking contrast to the non-rigidity of its ambient space $\mathcal{W}_2\left(\mathbb{R}^n,\|\cdot\|\right)$ but in line with the rigidity of the geodesic space $\mathcal{W}_2\left(\mathbb{S}^n,\sphericalangle\right)$. One of the key steps of the proof is the use of mean squared error functions to mimic displacement interpolation in $\mathcal{W}_2\left(\mathbb{S}^n,\|\cdot\|\right)$. A major difficulty in proving rigidity for quadratic Wasserstein spaces is that one cannot use the Wasserstein potential technique. To illustrate its general power, we use it to prove isometric rigidity of $\mathcal{W}_p\left(\mathbb{S}^1, \|\cdot\|\right)$ for $1 \leq p<2.$

In recent years, there has been considerable activity in characterising isometries of various metric spaces of probability measures.See e.g.[2-4, 6, 7, 9-21, 24, 27] for results about the total variation, Lévy, Kuiper, Lévy-Prokhorov, Kolmogorov-Smirnov, and Wasserstein metrics.Among these, an interesting result is due to Kloeckner.In [18, Theorem 1.1 and Theorem 1.2], he shows that the quadratic Wasserstein space W 2 R n+1 , ∥ • ∥ , where ∥ • ∥ stands for the metric induced by the norm, exhibits the rare phenomenon of not being isometrically rigid, meaning that not all isometries of W 2 R n+1 , ∥ • ∥ are induced by an isometry of R n+1 , ∥ • ∥ .In this paper, we consider the metric subspace (S n , ∥ • ∥) of the base space R n+1 , ∥ • ∥ and prove that the non-rigidity does not carry over: the exotic isometries of W 2 R n+1 , ∥ • ∥ send measures supported on S n to measures supported also outside of S n , while we gain no new exotic isometries by restricting to this smaller metric space.In general, when H is an arbitrary Borel subset of R n+1 , then W p (H, ∥ • ∥) embeds isometrically into W p R n+1 , ∥ • ∥ , but this does not necessarily imply that there exists such a natural embedding for their isometry groups.To see an example, we mention the case of the real line (R, | • |) with the subset H = [0, 1] (see [12,Theorem 2.5 and Theorem 3.7] for details): the isometry group of W 1 ([0, 1], | • |) is the Klein group, which cannot be embedded by a group homomorphism into the isometry group of W 1 (R, | • |), which is isomorphic to the isometry group of the real line.
Finally, we draw attention to Santos-Rodríguez's paper [24] and our recent work [15].In [24], the author considers (among others) Wasserstein spaces with p > 1 whose underlying metric space is a rank-one symmetric space, which class contains the sphere S n with the spherical distance ∢, while in [15], we considered finite-dimensional tori and spheres with their geodesic distances for all parameters p ≥ 1. Together, these two papers show that W p (S n , ∢) is isometrically rigid for all p ≥ 1.As explained above, in this paper, we replace the angular distance ∢ with another natural metric: the distance inherited from the norm of R n+1 .We focus on the case of p = 2 because this is the only parameter value for which the ambient space W p R n+1 , ∥ • ∥ is not rigid.We expect that for p ̸ = 2, techniques similar to the ones used in [14] would lead to a proof of isometric rigidity.The situation is analogous to the case of the real line and the unit interval: the quadratic Wasserstein space is not rigid over R but it is rigid over the compact subset [0, 1], see [18,Theorem 1.1] and [12,Theorem 2.6].Our main result reads as follows.
In our recent works [14,15], recovering measures from their Wasserstein potentialssee (2.3) for precise definition -turned out to be a powerful method to prove isometric rigidity.However, this method cannot be used in the case of W 2 (S n , ∥ • ∥), as shown by the following simple example.Let δ x denote the Dirac measure concentrated at x ∈ S n , let µ z := 1 2 (δ z + δ −z ) for z ∈ S n , and note that for any x ∈ S n we have of probability measures endowed with the p-Wasserstein metric where the infimum is taken over the set Π(µ, ν) of all couplings of µ and ν.
For more details about Wasserstein spaces, we refer the reader to the comprehensive textbooks [1,8,23,26].Now we only mention that optimal couplings always exist, and the infimum in (2.1) becomes minimum [1, Theorem 1.5].Furthermore, finitely supported measures are dense in Wasserstein spaces, see, e.g., [26,Theorem 6.18].An isometric embedding between metric spaces (X, d) and (Y, ρ) is a map ϕ : (X, d) → (Y, ρ) which preserves distances, i.e., a map such that d(x, x ′ ) = ρ (ϕ(x), ϕ(x ′ )) for all x, x ′ ∈ X.We shall use the term isometry for a surjective isometric embedding from a metric space onto itself.It is important to note that if (X, d) is a compact metric space, then every isometric embedding ϕ : (X, d) → (X, d) is surjective and hence an isometry [5,Theorem 1.6.14].
For a Borel-measurable map ψ : Y → Y , its push-forward In particular, when ψ : Y → Y is an isometry, then so is ψ # by the very definition of the Wasserstein distance, giving rise to a canonical embedding of the isometries of (Y, ρ) to the isometries of W p (Y, ρ).
In this paper, we consider the compact metric space (S n , ∥ • ∥), where is the unit sphere of R n+1 equipped with the distance inherited from the Euclidean norm of R n+1 , that is, dist(x, y) = ∥x − y∥.The point −x is called the antipodal of x.Since (S n , ∥ • ∥) is bounded, the Wasserstein space W p (S n , ∥ • ∥) is the entire set P (S n ) endowed with the distance We write W p (S n , ∥ • ∥) instead of the usual W p (S n ) notation to avoid any confusion with the results in [15,24].As the Wasserstein distance metrizes the weak convergence of probability measures over bounded metric spaces (see, e.g., [25,Theorem 7.12]), by Prokhorov's theorem, (S n , ∥ • ∥) being compact tells us that W p (S n , ∥ • ∥) is compact too -see also Remark 6.19 in [26].This implies that every isometric embedding of W p (S n , ∥ • ∥) into itself is an isometry.
For a measure µ ∈ P (S n ), its support supp(µ) is the set of all points x ∈ S n for which every open neighbourhood of x has positive measure.As usual, δ x denotes the Dirac measure supported on the single point x ∈ S n .
The question arises whether it is possible to identify a measure if we know its distance from all Dirac measures.(Recall that d Wp (δ x , δ y ) = ∥x − y∥ for all x, y ∈ S n and thus the set of all Dirac measures is an isometric copy of the underlying metric space.)To answer this question, we first introduce the notion of Wasserstein potential T (p) µ .For a given µ ∈ W p (S n , ∥ • ∥), the Wasserstein potential is the function Now, the question above can be rephrased as follows: does the Wasserstein potential determine the measure uniquely?
3. Does the Wasserstein potential determine the measure uniquely?
The answer to this question is no, in general.A prominent example is W 2 (S n , ∥ • ∥) where measures supported on antipodal points with both weights equal to 1  2 have the same (constant) potential function -see the example in Section 1, after Theorem 1.1.Beyond this, exotic isometries of W 2 (R, | • |) (see [18, Section 5.1 and Section 5.2]) are also counterexamples.
However, we will now prove that it does in the case of S 1 ≃ T = {z ∈ C : |z| = 1} equipped with the distance function r(z, ω) = 1 2 (z − ω) for 1 ≤ p < 2. This normalization of the distance is consistent with the one used in [27].In this section, we assume that 1 ≤ p < 2, and we recall that the p-Wasserstein distance of µ, ν ∈ W p (T) in this case is , and therefore for any z ∈ T, the Wasserstein potential is of the form We showed in [15] that Fourier analytic methods can sometimes solve the problem of rigidity in a very elegant way.For example, we showed that isometric rigidity of W 2 (T, ∢) can be proved by using the Fourier transform of the Wasserstein potential, however, the same method fails in the case W 1 (T, ∢).As we will see, if we endow T with the distance r(z, ω) = 1 2 (z − ω) , then the situation changes: the same method works to prove isometric rigidity of W 1 (T, r), but fails in the case W 2 (T, r).Now we recall the very basics of Fourier analysis on the abelian group T. The main reason for doing so is to fix the notation.The continuous characters of T are exactly the power functions with an integer exponent.That is, if φ k (z) = z k for all k ∈ Z and Γ is the dual group (i.e., the group of all continuous characters), then Γ = {φ k : T → C | k ∈ Z}, and Γ ∼ = Z.The group T is compact, hence it admits a unique Haar probability measure λ, which can be expressed explicitly as dλ (z) = dz 2πiz .
The Fourier transform of a (complex-valued) function f ∈ L 1 (T, λ) is defined by Let us denote the set of all (complex-valued) measures of finite total variation by M (T).
The Fourier transform of µ ∈ M (T) is defined by We note that L 1 functions can be naturally identified with absolutely continuous measures (with respect to the Haar measure), see [22,Subsection 1.3.4.].The convolution of L 1 functions f and g is defined by and the convolution of f ∈ L 1 (T, λ) and µ ∈ M(T) is defined by It is a key identity that the Fourier transform factorizes the convolution, that is, Now we are ready to state and prove the main result of this section.It says that if 1 ≤ p < 2, then the Wasserstein space W p (T, r) is isometrically rigid.Theorem 3.1.Let p ∈ [1, 2) be a real number and let Ψ : W p (T, r) → W p (T, r) be an isometry.Then there exists an isometry ψ : (T, r) → (T, r) such that Ψ = ψ # .
Proof.First observe that the diameter of W p (T, r) is 1, and d Wp (µ, ν) = 1 if and only if µ = δ x and ν = δ −x for some x ∈ T. Since Ψ is an isometry, we have for all x ∈ T, which implies that Ψ (δ x ) is a Dirac measure as well.
Let us define the map ψ : T → T via the identity Ψ(δ x ) = δ ψ(x) -this means that Ψ coincides with ψ # on the set of Dirac measures.The map ψ : (T, r) → (T, r) is in fact an isometry: for all x, y ∈ T, and (T, r) is compact.These together combine into that ψ −1 # • Ψ is an isometry which fixes all Dirac measures.If we now prove that any isometry of W p (T, r) which fixes all Dirac measures must be the identity, we are done: in that case, ψ −1 # •Ψ = id Wp(T,r) , i.e., Ψ = ψ # as claimed.
From now on, let us assume that Φ : W p (T, r) → W p (T, r) is an isometry such that Φ(δ z ) = δ z for all z ∈ T. Then we have ) for all z ∈ T and µ ∈ W p (T, r).The question is whether this implies µ = Φ(µ).The proof will be done once we prove that a measure µ ∈ W p (T, r) is uniquely determined by its Wasserstein potential.To this end, assume that µ and ν are two measures such that We need to show that (3.6) implies µ = ν.Let us introduce the map Then by (3.1) and (3.3) one can observe that T (p) holds for all z ∈ T and µ ∈ W p (T).Indeed, we have The key observation is that the Fourier transform of f p does not vanish anywhere, that is, fp (n) ̸ = 0 for all n ∈ Z.For n = 0, we have while for n ̸ = 0, we use that and by the binomial series expansion we get that where . Using that the sign of p 2 k (−1) k is negative for all k ≥ 1, equality (3.9) can be written as It is a useful feature of the group T that the Fourier series of a function coincides with its power series.Therefore, the above binomial expansion gives us useful information about fp , namely, fp (k) coincides with the coefficient of z k in the expansion (3.9).Let us note that for n ̸ = 0, the coefficient of z n must be strictly negative because the expressions p 2 , 1 − p 2 , 2 − p 2 , . . .are all positive -here we use the assumption that p < 2. So we obtained that fp (0) > 0 and fp (n) < 0 for n ̸ = 0 which means that fp (n) ̸ = 0 for all n ∈ Z.By (3.8), the assumption that T The assumption p < 2 was crucial in the previous section, and therefore the quadratic case cannot be handled with the same Fourier-analytic technique.In this section, we use a method that allows us to prove isometric rigidity in the quadratic case not only over the circle but over higher-dimensional spheres too.We start this section with three propositions which will be utilized later in the proof of Theorem 1.1.
The first proposition, which can be found also in the Appendix of [14] (see the proof of Lemma 3.13 there), helps us understand how a translation affects the Wasserstein distance.For µ ∈ P R n+1 and v ∈ R n+1 , the translation of µ by v is the measure Note that the barycenter m(µ) defined above is also the unique point in R n+1 satisfying ⟨m(µ), z⟩ = R n+1 ⟨x, z⟩ dµ(x) for all z ∈ R n+1 , which equation is used to define the barycenter in an infinite-dimensional setting.However, in finite dimensions, the above direct definition, which does not refer to the dual space, is available.
Proposition 4.1.Let µ, ν ∈ W 2 R n+1 and v ∈ R n+1 .Then we have In particular, substituting v = m(ν) − m(µ) gives Proof.For any π ∈ Π(µ, ν) and v ∈ R n+1 , we have (t (v,0) ) # π ∈ Π ((t v ) # µ, ν), and vice versa.(Here, 0 stands for 0 ∈ R n+1 .)Hence ⟨y, v⟩ dν(y), which gives (4.1).The identity (4.2) follows if we translate both arguments in the left-hand side by the vector m(ν).□ In quadratic Wasserstein spaces over uniquely geodesic spaces, the α-weighted mean squared-error function ρ → (1 − α)d 2 W 2 (µ, ρ) + αd 2 W 2 (ν, ρ) defined by µ and ν has a unique minimizer -provided that the optimal coupling of µ and ν is unique -which is the displacement convex combination or displacement interpolation of µ and ν with weights (1−α) and α [25,26].Intuitively, this is the measure that we obtain if we start moving µ to ν according to the optimal transport plan, but stop at proportion α of the journey.A great challenge concerning (S n , ∥ • ∥) is that it has no geodesics at all, and hence the quadratic Wasserstein space W 2 (S n , ∥ • ∥) has no geodesics either.Still, mean squared-error functions make perfect sense on W 2 (S n , ∥ • ∥), they are invariant under isometries in an appropriate sense, and hence if the measures µ and ν defining them are fixed by an isometry Φ, then so are the unique minimizers -if they exist.We will prove in Proposition 4.2 that on (S n , ∥ • ∥), the minimizer of the α-weighted squared-error function is the projection of the displacement interpolation onto the sphere.This is similar to how for a measure µ ∈ P (S n ), its closest Dirac measure supported on a point in R n+1 is δ m(µ) , while among those supported on S n , it is the projection of δ m(µ) .We are going to exploit this characterisation in Step 6 of the proof of Theorem 1.1.
We will use the following projection of the α-weighted mean of two points x, y onto S n frequently: Note that p α (x, y) is not defined when α = π ∈ W 2 (S n , ∥ • ∥) be defined to be the displacement interpolation in R n+1 at time α between µ and ν according to the plan π, projected to S n .Formally, Now let µ, ν ∈ W 2 (S n , ∥ • ∥) and consider the α-weighted mean squared error be such that there is a unique optimal transport plan π * for them with respect to the cost c α defined in (4.3).Suppose that α ̸ = 1 2 or α = 1 has infinitely many minimizers.
Proof.We proceed by establishing a lower bound for (4.5) and taking care of the case of equality.Let ρ ∈ W 2 (S n , ∥ • ∥) be arbitrary, and let π µ,ρ and π ρ,ν be optimal transport plans (w.r.t. the quadratic distance) between µ and ρ, and ρ and ν, respectively.Let π µ,ρ,ν ∈ P (S n × S n × S n ) be the gluing of π µ,ρ and π ρ,ν -see [25,Lemma 7.6] for the precise definition.Then π µ,ν := (π µ,ρ,ν ) 1,3 ∈ P (S n × S n ) is a coupling of µ and ν.Now The inequality (4.6) is saturated if and only if z = p α (x, y) for π µ,ρ,ν -a.e.(x, z, y) πµ,ν .Moreover, the right-hand side of (4.6) is minimal if and only if π µ,ν = π * .Consequently, and the only ρ realizing this minimum is ρ π * .On the other hand, if π * puts on antipodal points, that is, π * ({(z, −z) : z ∈ S n }) > 0, and α = 1 2 , then we have an infinite collection of minimizing measures by the theorems of Thales and Pythagoras -or by a simple direct computation.□ In the next proposition, we consider the case when the first argument of p α is fixed and clarify injectivity and surjectivity properties of p α as α varies from 0 to 1. Proposition 4.3.Let α ∈ 0, 1  2 ∪ 1 2 , 1 and let N ∈ S n be arbitrary but fixed -it may be considered as the "north pole".Let p α (N, •) : S n → S n be the map sending u to is injective, and its range is the open "upper" hemisphere {z ∈ S n | ⟨z, N ⟩ > 0}.

□
Now we turn to the proof of Theorem 1.1 which, for the sake clarity, we divide into six steps.
Step 1. Similarly as in the proof of Theorem 3.1, we first understand the action of Ψ on the set of Dirac measures.The maximal distance in W 2 (S n , ∥ • ∥) is 2, and is attained only on pairs of Dirac measures that are concentrated on antipodal points.Since we get that Ψ(δ x ) is a Dirac measure for all x ∈ S n .Since Ψ and Ψ −1 are both isometries, the map ψ : S n → S n defined by Ψ(δ x ) = δ ψ(x) is a bijection, and furthermore, since it is in fact an isometry.Just as before, we will be done once we prove that an isometry of W 2 (S n , ∥ • ∥) which fixes all Dirac measures is necessarily the identity, because then in particular, , and so Ψ = ψ # as claimed.From now on, we assume that Φ is an isometry of W 2 (S n , ∥ • ∥) such that Φ(δ x ) = δ x , and our aim is to show that Φ(µ) = µ for all µ ∈ W 2 (S n , ∥ • ∥).
Step 2. Next we claim that Φ preserves the barycenter of measures.
For any µ ∈ W 2 (S n , ∥ • ∥) and x ∈ S n , we have This implies that Since for a fixed µ, equation (4.8) holds for every x ∈ S n , we conclude that m (Φ (µ)) = m (µ).
Step 3. Now we prove that measures supported on two points are mapped to measures supported on two points.We first show that for all µ, ν ∈ W 2 (S n , ∥ • ∥), holds if and only if affspan (supp(µ)) ⊥ affspan (supp(ν)) .
Kloeckner proved in [18,Lemma 6.2] that orthogonality of supports can be characterized by the metric in the ambient space W 2 R n+1 , ∥ • ∥ .Namely, ) ν, δ m(ν) holds if and only if there exist two orthogonal affine subspaces L, M ⊂ R n+1 such that supp(µ) ⊆ L and supp(ν) ⊆ M .We proceed by showing that the isometries of W 2 (S n , ∥ • ∥) leave the W 2 (R n+1 )-distance of a measure from the Dirac mass concentrated on its barycenter invariant, that is, for any µ ∈ W 2 (S n , ∥ • ∥).Indeed, a direct computation very similar to (4.7) shows that , meaning that orthogonally supported measures must be mapped to orthogonally supported measures by Φ.
A maximal set of measures whose supports are one-dimensional and pairwise orthogonal must therefore be mapped to a set of measures whose supports are zero-or one-dimensional.But zero-dimensionally supported measures are exactly the Dirac masses, to which only Dirac masses can be mapped by Φ, and so one-dimensionally supported measures must be mapped to one-dimensionally supported measures.Continuing similarly, we would see more generally that the affine dimension of the support is preserved by Φ, but since on the sphere, one-dimensionally supported measures are exactly the two-point supported measures, the one-dimensional case is enough to prove our statement.
Step 4. We proceed with showing that measures supported on two points are fixed by Φ.Let us introduce the notation ∆ ′ 2 (S n ) for the set of all elements in W 2 (S n , ∥ • ∥) with a two-point support, set µ := t −m(µ) # µ for all µ ∈ ∆ ′ 2 (S n ), and ∆ ′ 2,0 (S n ) := ) is an isometric embedding.By Proposition 4.1 we know that for all µ, ν ∈ ∆ Consequently, µ = ν holds if and only if Φ(µ) = Φ(ν), in other words, Φ(ν) is a translate of Φ(µ) if and only if ν is a translate of µ.
The proof of this claim relies on preserving the mass of bisectors which are defined as follows: for u, v ∈ S n , the corresponding bisector is B(u, v) := {y ∈ S n : ∥u − y∥ = ∥v − y∥} ∼ = S n−1 .
For every x ∈ {x 1 , . . ., x m } = supp(µ), there exists a sequence (S j ) j∈N of n − 1dimensional subspheres of S n such that S j ∩ supp(µ) = { x} for every j, and the intersection of any n subspheres is trivial, that is, n k=1 S j k = {x, −x} for any choice of j 1 < j 2 < • • • < j n .Therefore, Step 6.A crucial consequence of the claim made in Step 5 is that the isometry Φ fixes all measures that are supported within an open hemisphere of S n .Indeed, we learned from Step 2 that Φ preserves the barycenter of measures, and from Step 5 that the only possible action Φ can do is to send some mass from a point to its antipodal point.But if a measure