There is no stationary cyclically monotone Poisson matching in 2d

We show that there is no cyclically monotone stationary matching of two independent Poisson processes in dimension d=2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$d=2$$\end{document}. The proof combines the harmonic approximation result from Goldman et al. (Commun. Pure Appl. Math. 74:2483–2560, 2021) with local asymptotics for the two-dimensional matching problem for which we give a new self-contained proof using martingale arguments.


Introduction
Consider two locally finite i point sets {X}, {Y } ⊂ R d in d-dimensional space.We are interested in their matching, which we think of as a bijection T from {X} onto {Y }.More specifically, we are interested in their matching by cyclically monotone maps T , which means that for any finite subset ii {X n } N n=1 we have It is elementary to see that (1.1) is equivalent to local optimality, meaning for any other bijection T that differs from T only on a finite number of points iii .This makes a connection to the optimal transportation between the measures related via T #µ = ν, which we shall explore in this paper.Note however that because of the (typically) infinite number of points, we cannot view T as a minimizer.
Before proceeding to the random setting, we make two simple observations that show that the set of ({X}, {Y }, T ) is rich: For d = 1, (1.1) is easily seen to be equivalent to plain  and thus countable ii which we momentarily enumerate iii so that the sum is effectively finite 1 monotonicity.Every single matching T (X 0 ) = Y 0 can obviously be extended in a unique way to a monotone bijection T of {X} and {Y }, so that for d = 1, the set of monotone bijections T has the same magnitude as {X} itself.Returning to general d, we note that for any monotone bijection T of {X} and {Y }, and for any two shift vectors x, ȳ ∈ R d , the map x → T (x − x) + ȳ is a monotone bijection of the shifted point sets {x + X} and {ȳ + Y }.
We are interested in the situation when the sets {X}, {Y } and their cyclically monotone bijection T are random.More precisely, we consider the case when {X} and {Y } are independent Poisson point processes of unit intensity.We assume that the σ-algebra for ({X}, {Y }, T ) is rich enough so that the following elementary observables are measurable, namely the number N U,V of matched pairs (X, Y ) ∈ U ×V for any two Lebesgue-measurable sets U, V ⊂ R d (with U or V having finite Lebesgue measure iv ): We now come to the crucial assumption on the ensemble: In view of the above remark, the additive group Z d ∋ x acts on ({X}, {Y }, T ) via ({X}, {Y }, T ) → ({x + X}, {x + Y }, T (• − x) + x).
We assume that this action is stationary and ergodic v .On the one hand, stationarity is a structural assumption; we shall only use it in following form: For any shift vector x, the random natural numbers N x+U,x+V and N U,V have the same distribution.On the other hand, ergodicity is a qualitative assumption vi ; we will only use it in form of the following application of Birkhoff's ergodic theorem: Theorem 1.1.For d ≤ 2, there exists no stationary and ergodic ensemble of ({X}, {Y }, T ), where {X}, {Y } are independent Poisson point processes and T is a cyclically monotone bijection of {X} and {Y }.
Our interest in this problem is motivated on the one hand by work on geometric properties of matchings by Holroyd [13] and Holroyd et al. [15,14], and on the other hand by work on optimally coupling random measures by the first author and Sturm [17] and the first author [16].In [14], Holroyd, Janson, and Wästlund analyze (stationary) matchings satisfying the local optimality condition (1.2) with the exponent 2 replaced by γ ∈ [−∞, ∞].They call matchings satisfying this condition γ-minimal and derive a precise description of the geometry of these matchings in dimension d = 1.In dimension d > 1 much less is known.In particular, in the critical dimension d = 2 they could only show existence of iv so that the following number is finite v Recall that this is automatic for the two first components ({X}, {Y }) by the properties of the Poisson point process so that this amounts to a property of T vi paraphrased by saying that spatial averages are ensemble averages stationary γ-minimal matchings for γ < 1.The cases γ ≥ 1 were left open, but see [15] and [13] for several open questions for d = 2 and in particular γ = 1.On the other hand, the first author and Sturm [17,16] develop an optimal transport approach to this (and related) problems.They identify the point sets {X}, {Y } with the counting measures (1.3) and seek a stationary coupling Q between µ and ν minimizing the cost (Note that any (stationary) bijection T : {X} → {Y } induces a (stationary) coupling between µ and ν by setting Q = (id, T ) # µ.)If this cost functional is finite there exists a stationary coupling which is necessarily locally optimal in the sense of (1.2) with the exponent 2 replaced by γ.In dimension 2 for µ and ν being two independent Poisson processes, this functional is finite if and only if γ < 1, which is in line with the results of [14].
In view of these results, it is natural to conjecture that Theorem 1.1 also holds for γ-minimal matchings with γ ≥ 1.However, our proof crucially relies on the harmonic approximation result of [10] which so far is only available for γ = 2.
Before we explain the main steps of the proof of Theorem 1.1 in Section 1.1 we would like to give a few remarks on extensions and variants of Theorem 1.1.
Remark 1.2.Theorem 1.1 remains true if we replace the bijection T by the a priori more general object of a stationary coupling Q.This can be seen by either using that matchings are extremal elements in the set of all couplings of point sets or by directly writing the proof in terms of couplings which essentially only requires notational changes.
Remark 1.3.Very well studied siblings of stationary matchings are stationary allocations of a point process {X}, i.e. a stationary map T : R d → {X} such that Leb(T −1 (X)) equals E[#{X ∈ (0, 1) d }] −1 .There are several constructions of such an allocation, for instance by using the stable marriage algorithm in [12], the flow lines of the gravitational force field exerted by {X} in [8], an adaptation of the AKT scheme in [21], or by optimal transport methods in [17].
By essentially the same proof as for Theorem 1.1, one can show that in d = 2 there is no cyclical montone stationary allocation to a Poisson process.The only place where we need to change something in the proof is the L ∞ estimate Lemma 2.2.
Remark 1.4.We do not use many particular features of the Poisson measures µ and ν in the proof of Theorem 1.1 since ergodicity and stationarity allow us to argue on a pathwise level via the harmonic approximation result (cf.Section 1.1).
We use two properties of the Poisson measure.The first property is concentration around the mean.The second property is more involved.Denote by W p the L p Wasserstein distance.We use that 1 for some ε > 0 (here we use ε = 1).
As the last remark indicates stationary matchings are closely related to the bipartite matching problem, which is the natural variant of the problem studied in this paper with only a finite number of points {X 1 , . . ., X n }, {Y 1 , . . ., Y n }.Note that then the local optimality condition (1.2) turns into a global optimality condition vii .This (finite) bipartite matching problem has been the subject of intense research in the last 30 years, see e.g.[1] for the first proof of the rate of convergence in the case of iid points in dimension d = 2, [23] for sharp integrability properties for matchings in d ≥ 3, [7] for a new appraoch based on the linearization of the Monge-Ampère equation, and [5] for a thorough analysis of the bipartite matching problem in d = 1.
We rely on this connection in two ways.On the one hand we use the asymptotics of the cost of the bipartite matching which are known since [1].However, we need a local version for which we give a self-contained proof using martingale techniques (see Section 2.5) which is new and interesting on its own.On the other hand we exploit a large scale regularity result for optimal couplings, the harmonic approximation result, developed in [10].This regularity result was inspired by the PDE approach proposed by [7], see also [3,2,20,11] for remarkable results for the bipartite matching problem using this approach.
1.1.Main steps in the proof of Theorem 1.1.In the following we will describe the main steps in the proof of Theorem 1.1.For the detailed proofs we refer to Section 2.
We argue by contradiction.We assume that there is a locally optimal stationary matching T between {X} and {Y }.On the one hand, we will show that On the other hand, it is known (we will prove the local version needed for our purpose) that any bipartite matching satisfies leading to the desired contradiction.
Let us now describe the different steps leading to (1.4) and (1.5) in more detail.Our starting point is the observation that by stationarity and ergodicity the following L 0estimate on the displacement T (X) − X holds see Lemma 2.1 for a precise statement.Since T is locally optimal, its support is in particular monotone, which means that for any X, X ′ ∈ {X} we have vii which in measure theoretic formulation used in optimal transport is nothing but the cyclical mononotonicity of the support of the corresponding coupling.
By (1.6), we know that most of the points in (−R, R) d are not transported by a large distance.Combining this with monotonicity allows us to also shield the remaining points from being transported by a distance of order R, so that see Lemma 2.2 for a precise statement.By concentration properties of the Poisson process we may assume that Now comes the key step of the proof.We want to exploit regularity of T to upgrade (1.8) to an O(ln R) bound.The tool that allows us to do this is the harmonic approximation result [10,Theorem 1.4] which quantifies the closeness of the displacement T (X) − X to a harmonic gradient field ∇ϕ, taking as input only the local energy and the distance of where , and . By (1.8) (together with its counter part arising from exchanging the roles of {X} and {Y }) we have E(R) ≤ o(R 2 ).By the well-known bound for the matching problem in d = 2 and by concentration properties of the Poisson process we have D(R) ≤ O(ln R), see Lemma 2.6.Iteratively exploiting the harmonic approximation result on an increasing sequence of scales we obtain that the local energy E inherits the asymptotic of D: It remains to establish the lower bound (1.5), which is essentially known.However, for our purpose we need the local version (1.5).Our proof is very similar to the proof of the lower bound in the seminal paper [1].Both approaches construct candidates for the dual problem based on dyadic partitions of the cube.However, instead of using a quantitative embedding result into a Gaussian process as in [1] we use a natural martingale structure together with a concentration argument.More precisely, we show that there exists ζ with suppζ ∈ (0, R) see Lemma 2.5.Note that this is sufficient for the contradiction stated at the beginning of the subsection, indeed O(ln
Lemma 2.1.For any ε > 0 there exist a deterministic L and a random radius and consider the number of points in Q R which are transported by a distance greater than L, namely We show that stationarity together with the ergodic theorem implies as R → ∞ that Note that (2.2) together with (2.3) imply the existence of a random radius r * and a deterministic L as in the Lemma, indeed for any fixed ε > 0 we can choose L large enough so that and then choose r * large enough so that for We now turn to (2.2).For i ∈ Z d note that Since µ is stationary and ergodic, by the Birkhoff-von Neumann ergodic theorem [18,Theorem 9.6] for any divergent sequence of integer radii Note that for integer R the term on the left hand side amounts to 1 . Since for every real R there exists an integer R such that R ≤ R ≤ R + 1 and the ratio between R and R + 1 goes to 1 as R → ∞, (2.4) holds also for any divergent sequences of real radii.
Finally we turn to (2.3).By dominated convergence it suffices to show that L is finite for every realization of {X}, then there exists L large enough such that c is finite by the properties of the Poisson process.
2.2.The L ∞ -estimate.The proof is very similar to the L ∞ estimate in [10, Lemma 2.9].However, note that in [10, Theorem 1.4] a local L 2 estimate was turned into a L ∞ estimate whereas in the current setup we want to turn a L 0 estimate into a L ∞ estimate.The key property that allows us to do this is the monotonicity of the support of T .This translates the partial control given by (2.1) into the claimed L ∞ estimate.Lemma 2.2.For every ε > 0 there exists a random radius r * < ∞ a. s. such that for every Proof.
Step 1. Definition of r * = r * (ε) given 0 < ε ≪ 1 as the maximum of three r * 's.First, by Lemma 2.1, there exists a (deterministic) length L < ∞ and the (random) length r * < ∞ such that for 4R ≥ r * , the number density of the Poisson points in (−2R, 2R) d transported further than the "moderate distance" L is small in the sense of Second, by Lemma 2.6 we may also assume that r * is so large that for R ≥ r * , the nondimensionalized transportation distance of µ to its number density n is small, and that n ≈ 1, in the sense of Third, w. l. o. g. we may assume that r * is so large that L ≤ εr * . (2.9) We now fix a realization and R ≥ r * .
Step 2. There are enough Poisson points on mesoscopic scales.We claim that for any cube Q ⊂ (−2R, 2R) d of "mesoscopic" side length Indeed, it follows from the definition of W (−2R,2R) (µ, n) that for any Lipschitz function η with support in Q we have We now specify to an η ≤ 1 supported in Q, to the effect of ´ηdµ ≤ ´dµ = #{ X ∈ Q }, so that by Young's inequality At the same time, we may ensure ´(−2R,2R) d η r d and Lipη r −1 , so that by (2.8), which in particular ensures n ≈ 1, (2.12) turns into Thanks to assumption (2.10) we obtain (2.11).
Step 3. Iteration.There are enough Poisson points of moderate transport distance on mesoscopic scales.We claim that for any cube Q ⊂ (−2R, 2R) d of side-length satisfying (2.10) we have there exists We suppose that (2.13) were violated for some cube Q.By (2.11), there are r d of such points.By assumption (2.10), there are thus ≫ (εR) d Poisson points in (−2R, 2R) d that get transported by a distance > L, which contradicts (2.7).
Step 4. At mesoscopic distance around a given point X ∈ (−R, R) d , there are sufficiently many Poisson points that are transported only over a moderate distance.More viii The notation A B means that there exists a constant C > 0, which may only depend on d, such that A ≥ CB.Similarly A B means that there exists a constant C > 0, which may only depend on d, such that A ≤ CB.
precisely, we claim that provided (2.10) holds, there exist d + 1 Poisson points {X n } d+1 n=1 that are transported over a moderate distance, i. e.

|T (X
but on the other hand lie in "sufficiently general" directions around X, meaning that the convex hull of { Xn−X |Xn−X| } d+1 n=1 contains B ρ (2.15) for ρ ≪ 1, while the distances to X are of order r Indeed, this can be seen as follows: Consider the symmetric tetrahedron ix with barycenter at X.For each of its d + 1 vertices, consider a rotationally symmetric cone with apex at X and with axis passing through the vertex.Provided the opening angles are ≪ 1, by continuity, any selection e n of unit vectors in this cones still has the property that their convex hull contains B ρ for ρ ≪ 1.Consider the intersection of these cones with the (dyadic) annulus centered at X of radii r and 2r.These d + 1 intersections are contained in (−2R, 2R) d , and each contains a cube of side-length ∼ r.Hence (2.13) applies and we may pick a Poisson point X n with (2.14) in each of these intersections, see Figure 1.Condition (2.16) is satisfied because of the annulus, and condition (2.14) is satisfied because of the cones.
Step 5.All Poisson points are transported over distances ≪ R. We claim that for all Poisson points X n=1 be as in Step 4. By cyclical monotonicity (1.1), which implies monotonicity of the map T , we have (T (2.18) ix using the d = 3-language, for d = 2 it is the equilateral triangle We now appeal to (2.14), which by (2.9) and and (2.10) implies Inserting this and (2.16) into (2.18),we obtain Since by (2.15), any unit vector e can be written as a linear combination of { Xn−X |Xn−X| } d+1 n=1 with non-negative weights ≤ 1 ρ ∼ 1, this implies |T (X) − X| r.Since (2.10) was the only constraint on r, we obtain (2.17).
2.3.Key step: Harmonic approximation.Lemma 2.3.There exist a constant C x and a random radius r * < ∞ a. s. such that for every R ≥ r * we have Proof of Lemma 2.3.The proof relies on the harmonic approximation result from [10, Theorem 1.4].This result establishes that for any 0 < τ ≪ 1, there exists an ε > 0 and a C τ < ∞ such that provided for some R ) and (1.10) for the definition) there exists a harmonic gradient field Φ such that 1 (2.21) The fraction τ will be chosen at the end of the proof.Note that in (1.10) D (R) is defined on boxes (−R, R) d while [10, Theorem 1.4] (2.20) requires balls.Since B 6R ⊆ (−6R, 6R) d we may assume that (2.20) holds also for B 6R ′ for R ′ close to R, see [10,Lemma 2.10] for this type of restriction property.
Step 1. Definition of r * depending on τ .For 0 < τ ≪ 1 let ε = ε (τ ) be as above.By Lemma 2.6 we may assume that r * is large enough so for any dyadic x here and throughout the paper C denotes a positive constant, which may only depend on the dimension, the value of which may change from line to line.
Note that only the bound (2.22) is specific to d = 2. From now on, we restrict ourselves to dyadic R, which we may do w.l. o. g. for (2.19).Note that by the bound on D (6R) in (2.23) and the second and fourth term in the definition of D (R) in (1.10) Moreover, we may assume that r * is large enough so that (2.6) holds.Since B 6R ⊂ (−6R, 6R) d we may sum (2.6) over B R to obtain for By symmetry, potentially enlarging r * , we may also assume that (2.6) holds with X replaced by T (X) so that both and in particular (2.20) holds.Finally by Lemma 2.1 we may assume, possibly enlarging r * , that there exists a deterministic constant L τ and for R ≥ r * we both have

and
(2.27) Step 2. Application of harmonic approximation.For all R ≥ r * (2.28) We split the sum according to whether the transportation distance is moderate or large.
On the latter we use the harmonic approximation: The last estimate combines to 1 Relabeling τ and C τ , this implies (2.28).
Step 3. Iteration.Iterating (2.28), we obtain for any k ≥ 1 We now fix τ such that 36τ < 1 to the effect of
Lemma 2.4.For every ε > 0 there exists a random radius r * < ∞ a. s. such that Proof.By Lemma 2.3, we know that there exists a random radius r * such that for R ≥ r * we have (2.29) Let 0 < ε ≪ 1. Possibly enlarging r * , we may also assume by Lemma 2.1 that there exists a deterministic constant L such that for R ≥ r * Furthermore, note that by Lemma 2.6 and the second and fourth term in the definition of D (R) in (1.10) we may also assume possibly enlarging r * again that for R ≥ r * (2.24) holds.Finally, we may also assume possibly enlarging r * that for R ≥ r * (2.31) R. We split again the sum into moderate and large transportation distance and apply Cauchy-Schwarz: 1 (2.31) ≤ Cε this suggests a transportation cost per particle that is logarithmic in the ratio between the macroscopic scale R and the microscopic scale 1.However, by the independence properties of the Poisson point process, the displacements on every dyadic scale are essentially independent, so that there are the usual cancellations when adding up the dyadic scales.Hence the displacement behaves like the square root of the logarithm.
The proof of the lower bound is similar to the original proof in [1].However, in the final step we use martingale arguments instead of Gaussian embeddings.The proof of the upper bound has some similarities to the proof in [3].Again we use martingale arguments and crucially the Burkholder-Davis-Gundy inequality.It would be interesting to see whether our technique for the upper bound allows to get the precise constant as in [3].
Lemma 2.5.Let µ, ν denote two independent Poisson point processes in R 2 of unit intensity.There exists a constant C, and a random radius r * < ∞ a. s. such that for any Lemma 2.6.Let µ, ν be as in Lemma 2.5.Then there exists a constant C, and there exists a random radius r * < ∞ a. s. such that for any dyadic radii For the lower as well as the upper bound, our proof proceeds in two stages.First we show the desired estimate in expectation.Then we use concentration arguments to lift it to a pathwise estimate.
Proof.W. l.o. g. we may assume that R ∈ 2 N , and will consider the (finite) family of all dyadic cubes Q ⊂ (0, R) 2 of side-length ≥ 1.For such Q, we call Q r and Q l the right and left half of Q and consider the integer-valued random variable For later purpose we note that by standard properties of the Poisson process (see [19, equation (4.23)]) we have and note that for a fixed point x ∈ (0, R) d , we have observing that the sum over Q restricts to the one cube of dyadic level/side-length ρ that contains x.We now argue that (2.38) is a martingale, where r = R, R 2 , R 4 , • • • , plays the role of a discrete time.More precisely, it is a martingale with respect to the filtration generated by {{N Q } Q of level r } r .It suffices to show that even when conditioned on {N Q} Q of level ρ for ρ ≥ 2r, the expectation of N Q for every Q of level r vanishes.This can be seen by appealing to a Lebesgue-measure preserving transformation of R 2 that swaps the left and right half of Q; when applied to point configurations, it preserves the law of the Poisson point process, it leaves N Q for Q of level ρ ≥ 2r invariant, and it converts As the sum over Q in (2.38) reduces to a single summand, this martingale has (total) quadratic variation xi and we claim that it satisfies Indeed, by (2.36) we obtain that the l.h.s. of (2.39) is given by

40)
xi here we consider the total quadratic variation and not the more involved quadratic variation since it is sufficient for our purposes.
Indeed, the l.h.s. of (2.40) is again given by Q |Q|.We now appeal to the (quadratic) Burkholder inequality [22,Theorem 6.3.6](exchanging E and ´dx) to obtain from (2.39) We now are in the position to define ζ via a stopping "time".Given an M < ∞, which we think of as being large and that will be chosen later, we keep subdividing the dyadic cubes (recall that we restrict to those of side length ≥ 1) as long as This defines a nested sub-family of dyadic cubes, and thereby a random (spatially piecewise constant) stopping scale r * = r * (x) (note that r * 2 is a stopping time but we will not need that in this proof).We then set We first argue that we have the Lipschitz condition M ln R.This is a geometric series that is estimated by r −1 * (x) √ M ln R, as desired.
We now argue that the "exceptional" set where the dyadic decomposition stops before reaching the minimal scale r = 1, has small volume fraction in expectation: Indeed, by definitions (2.37) and (2.42), E is the disjoint union of dyadic cubes Q that have at least one (out of four) children which combines to Summing over all Q covering E, this yields Taking the expectation we obtain from (2.36) and (2.41) which yields (2.46).
We finally argue that Together with (2.40) this implies E ´ζdµ R 2 ln R for M ≫ 1, which we fix now.In combination with (2.44) this implies the claim of the lemma.The dyadic decomposition defines a family of exceptional cubes Q (a cube Q is exceptional iff for any x ∈ Q we have that its level r satisfies r < r * (x)).In view of (2.37) and (2.43), we thus have to estimate Q I(Q exceptional)N Q ´ζQ dµ.Applying E| • | and using Hölder's inequality in probability we obtain As for (2.36), we have for the last factor E( ´ζQ dµ) For the middle factor, we recall that the two numbers in (2.33) are Poisson distributed with mean 1  2 |Q|, so that EN 4 Q |Q| 2 by elementary properties of the Poisson distribution.Hence we gather As we noted before, we obtain for the second factor Q |Q| R 2 ln R. For the first factor, we note which proves (2.49).
Step 2. Concentration around expectation.We claim that Indeed, by the triangle inequality, we split the statement into two: By a Borel-Cantelli argument it suffices to show that for any fixed ε > 0 and R ≫ 1 the following statements hold (2.52) We first turn to (2.52).For z ∈ R 2 we consider the difference operator Since the ζ's in the definition (2.50) satisfy sup |ζ| ≤ R, we have Applying [25, Proposition 3.1] we obtain for R ≫ 1 The argument for (2.53) is almost identical: For arbitrary w ∈ R 2 we need to consider Then applying once more [25, Proposition 3.1] with β = R and α = R 2 implies (2.53).
is the (random) number density.
Proof.Clearly, the intuition mentioned at the beginning of Section 2.5 suggests to uncover a martingale structure.W.l.o.g.we may assume that R ∈ 2 N , and will consider the family of all dyadic squares Q.For such Q, we consider the number density Note that if, for a fixed point x, we consider the sequence of nested squares Q that contain x, the corresponding random sequence n Q is a martingale.We want to stop the dyadic subdivision just before the number density leaves the interval [ 1  2 , 2] of moderate values.This defines a scale: (2.57) Since on a square Q of side length r = Thus we may estimate We now distinguish the case of r * ≤ R on (0, R) 2 and its complement.In the last case, there exists a y ∈ (0, R) 2 such that r * (y) > R, which by (2.57) means that there exists a dyadic cube Q ∋ y of sidelength r Q ≥ R and n Q / ∈ 1 2 , 2 , which in turn implies that r * > R on the entire (0, R) 2 .Now fix a deterministic y ∈ (0, R) 2 .Since n (0,R) 2 R 2 is the number of particles we use the brutal estimate of the transportation distance Since by assumption, (0, R) 2 ⊂ (0, r * (y)) 2 , this yields W 2 (0,R) 2 (µ, n (0,R) 2 ≤ n (0,r * (y)) 2 r 2 * (y)2R 2 .Hence by definition of r * (y), see (2.57) we obtain W 2 (0,R) 2 µ, n (0,R) 2 ≤ 2r 2 * (y)2R 2 r 4 * (y).By (2.59), taking the expectation yields as desired EW 2 (0,R) 2 µ, n (0,R) 2 I (r * (y) > R) 1, where I (r * (y) > R) denotes the indicator function of the event {r * (y) > R}.
In the more interesting case of r * ≤ R, we define a partition of (0, R) 2 .A cube Q * is an element of the partition if and only if its sidelength r Q * satisfies (2.61) In words, this means that the number density of Q * is still within [ 1 2 , 2] but the number density of at least one of its four children leaves the interval [ 1  2 , 2].This implies the following reverse of (2.61) Equipped with this partition, we define the density λ by We consider the transportation distance between µ and the measure λdx.By definition of λ, there exists a coupling where the mass is only redistributed within the partition {Q * }.This implies the inequality Taking the expectation, by (2.59) and Jensen's inequality, Hence in view of (2.59) and the triangle inequality for the transportation distance it remains to show The purpose of the stopping (2.57) is that we have the lower bound λ, n (0,R) 2 ≥ 1 2 on the two densities, which implies the inequality (2.65) Inequality (2.64) can be easily derived from the Eulerian description of optimal transportation (see for instance [4, Proposition 1.1] or [24,Theorem 8.1]).Indeed, the couple (ρ, j), with ρ = tn (0,R) 2 + (1 − t) λ ≥ 1 2 and j satisfying (2.65) is an admissible candidate for the Eulerian formulation of W 2 (0,R) 2 (λ, n (0,R) 2 ).We now construct a j: For a dyadic square Q with its four children Q ′ , we (distributionally) solve the Poisson equation with piecewise constant r. h.s. and no-flux boundary data (2.66) Note that (2.66) admits a solution since the integral of the r.h.s. over Q vanishes by definition of {n Q }.Since the no-flux boundary condition allows for concatenation of −∇ϕ Q without creating singular contributions to the divergence, which means that the sum xii extends over all dyadic cubes Q ⊂ (0, R) 2 containing x and being strictly coarser than {Q * }, defines indeed a distributional solution of (2.65).
xii with the understanding that j(x) = 0 when the sum is empty.
The Poincaré inequality gives the universal bound Appealing to the trivial estimate Inserting definition (2.67) into the l.h.s., using triangle inequality, integrating over x ∈ (0, R) 2 , and using (2.58) on the r.h.s. we obtain Using (2.64) on the l.h.s. and (2.68) on the r.h.s. yields (2.63).
Remark 2.9.The same argument in d > 2 yields the bound EW (0,R) d (µ, n) 1.However, the interesting (well known) information from the proof is that in d > 2 the main contribution comes from the term W 2 (0,R) d (µ, λ) which collects the contributions on the small scales.
Proof of Lemma 2.6.W.l.o.g. it suffices to prove that for the Poisson point process µ there exists a random radius r * such that for any dyadic radii R ≥ r * The statement will follow by choosing the maximum of this random radius and the one pertaining to ν.
Step 1.We claim that there exists a constant C and a random radius r * < ∞ a. s. such that for any dyadic radii R ≥ r * By Lemma 2.8 we may assume that (2.55) holds with (0, R) 2 replaced by (0, 2R) 2 .Furthermore by stationarity of the Poisson point process we may assume that (2.55) holds in the form provided that R ≫ 1.By [9, Proposition 2.7] it follows that for any ε > 0, Then (2.70) follows by arbitrarity of ε together with a Borel-Cantelli argument.
Step 2. We show that there exists a constant C and random radius r * < ∞ a. s. such that for any dyadic radii R ≥ r * Finally, by a Borel-Cantelli argument (2.72) holds for any dyadic radii R ≥ r * .

Date:
September 29, 2021.All authors are supported by the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) through the SPP 2265 Random Geometric Systems.MH and FM have been funded by the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) under Germany's Excellence Strategy EXC 2044 -390685587, Mathematics Münster: Dynamics-Geometry-Structure . i

2 . 5 . 1 √ 1 √
Relabeling ε proves the claim.Asymptotics for the bipartite matching problem in dimension d = 2.In this section we give a self-contained proof of the upper and lower bounds on the asymptotics in the matching problem in the critical dimension d = 2.The intuition why d = 2 is critical for optimal transportation is clear: the fluctuations of the number density of the Poisson point process of unit intensity, i. e. its deviation from 1 on some mesoscopic scale r ≫ 1 are of size O( r d ).To compensate these fluctuations, one has to displace particles by a distance O(r × r d ), which is O(1) iff d = 2. Hence taking care of the fluctuations on every dyadic scale r requires a displacement of O(1).Naively,

2 E
, and to the standard estimate on the variance of the Poisson process, namelyE(n Q − 1) ˆQ |∇ϕ Q | 2 |Q|.(2.68)Now comes the crucial observation: We momentarily fix a point x and consider all squares Q ∋ x.We note that ∇ϕ Q (x) depends on the Poisson point process only through {n Q ′ } where Q ′ runs through the children of Q.Moreover, since the expectation of the r.h.s. of the Poisson equation (2.66) vanishes, also the expectation of ∇ϕ Q (x) vanishes.Hence the sumQ∋x,R≥r Q ≥r ∇ϕ Q (x) is a martingale in the scale parameter r = R, R 2 , R 4 , • • • w. r. t.to the filtration generated by the {n Q } Q .Since r * ≥ 1, we thus obtain by the Burkholder inequality[22, Theorem  6.3.6]
we have that the sum in r of the l.h.s. of (2.48) is E|E| ln R.Proof of Lemma 2.5.Step 1.From one to two measures.We claim that for R ≫ 1 the Indeed, write F = ζ : suppζ ⊂ (0, R) 2 , |∇ζ| ≤ 1, ´ζdx = 0 and S = sup ζ∈F ´ζdµ, where the supremum has to be understood as an essential supremum.By basic results on essential suprema, there exists a countable subset {ζ n , n ≥ 1} ⊂ F such that S = sup n≥1 ´ζn dµ.Setting, S n = max k≤n ´ζk dµ we have S n ր S and by monotone convergence ES n ր ES.Hence, there is n such that ES n R 2 ln 1 2 R by Lemma 2.7.By definition of S n , S n = ´ζdµ for some F -valued random variable ζ (that takes the finitely many values ζ 1 , . . ., ζ n ), which is admissible in (2.49).Clearly, ζ is measurable w. r. t. µ alone.However, ζ is measurably only dependent on µ and not on ν.Since ν has intensity measure dx we obtain by independence of µ and ν 1  2we have n Q ∈ 4N 0 we trivially obtain r * (x) ≥ 1.(2.58)As we shall argue now, it follows from the properties of the Poisson distribution that (the stationary) r * (x) is O(1) with overwhelming probability, in particular Indeed, by the concentration properties of the Poisson distribution we have for any ρ ∈ 2 N