Generic uniqueness of optimal transportation networks

We prove that for the generic boundary, in the sense of Baire categories, there exists a unique minimizer of the associated optimal branched transportation problem.


Introduction
Let K ⊂ R d be a convex compact set.Given two Radon measures µ − (source) and µ + (target) on K with the same mass, i.e.M(µ − ) = M(µ + ), a transport path transporting µ − onto µ + is a rectifiable 1-current T on K whose boundary is the 0-current ∂T = µ + − µ − .For the precise definition of these objects we refer to Section 2. The current T can be identified with a vector valued measure on K which we denoted by T as well.It can be written as T = T θH 1 E, where E is a 1-rectifiable set, θ ∈ L 1 (H 1 E) and T is a unit vector field spanning the tangent Tan(E, x) at H 1 -a.e.x ∈ E. The constraint ∂T = µ + − µ − is equivalent to the condition that the vector valued measure T has distributional divergence which is a signed Radon measure and satisfies div(T ) = µ − − µ + .Given α ∈ (0, 1), the α-mass of a transport path T as above is defined as We denote by D k (K) the set of k-dimensional currents with support in K, and letting b := µ + −µ − ∈ D 0 (K), we denote by OTP(b) the set of minimizers of the optimal branched transportation problem with boundary b, namely the minimizers of the α-mass among rectifiable 1-currents T with boundary ∂T = b.
For α ≤ 1 − 1/d, there are boundaries b such that OTP(b) degenerates to the set of all currents T with boundary ∂T = b, since there is no 1-current T with ∂T = b and M α (T ) < ∞, see [25].In turn, it is well known that there are boundaries b such that OTP(b) contains more than one element of finite α-mass; for instance one can exhibit a non-symmetric minimizer T for which ∂T is symmetric, so that the network T ′ symmetric to T is a different minimizer (see Figure 1).
The boundary ∂T is symmetric (with respect to the horizontal axis) and T ∈ OTP(∂T ) is not symmetric, hence the symmetric copy T ′ is a different minimizer.
The aim of this paper is to prove that for the generic boundary, in the sense of Baire categories, the associated optimal branched transportation problem has a unique minimizer.To this purpose, we denote the set of boundaries by (1.1) We metrize A C with the flat norm F K , see (2.2), and we observe that the set A C endowed with the induced distance is a non-trivial complete metric space, see Lemma 2.1.Our main result is the following Theorem 1.1.The set of boundaries b ∈ A C , for which OTP(b) is a singleton, is residual.
1.1.Previous results on the well-posedness of the problem.The variational formulations of the optimal branched transportation problem were inspired by the discrete model introduced by Gilbert in [20] and are used to model supply and demand transportation systems which naturally show ramifications as a result of a transportation cost which favors large flows and penalizes diffusion.
In our paper, we adopt the Eulerian formulation proposed by Xia in [32].Due to a celebrated result by Smirnov, see [31], this is equivalent to the Lagrangian formulation, introduced by Maddalena, Morel and Solimini in [21], see [2,29].Existence results and some regularity properties of minimizers have been established for instance in [1,5,25,33,34].Recently, another helpful wellposedness property of the problem was established in [14]: the stability of minimizers with respect to variations of the boundary, see [15] for the Lagrangian counterpart.Slightly improving upon the main result of [14], see Theorem A.1, we advance on the study of the well-posedness properties of the branched transportation problem, as we establish the first result on the generic uniqueness of minimizers, in full generality, namely in every dimension d and for every exponent α ∈ (0, 1).
Prior to our work, we are aware of only one elementary result on the uniqueness of minimizing networks.It appeared in the original paper by Gilbert [20], and says that there exists at most one discrete minimum cost communication network with a given Steiner topology.
Several variants and generalizations of the branched transportation problem were proposed and studied by many authors in recent years, see for instance [3,4,6,7,8,9,10,11,13,22,23,28,35].For the sake of simplicity, we prove the generic uniqueness of minimizers only for the Eulerian formulation introduced in [32].
1.2.Strategy of the proof.Using a small modification of the stability property proved in [14], see Theorem A.1, we show that in order to prove Theorem 1.1, it suffices to prove the density of the set of boundaries b ∈ A C for which OTP(b) is a singleton, see Lemma 2.2.A similar reduction principle is used in [26,27] to prove that the generic (higher dimensional) boundary spans a unique minimal hypersurface.
The proof of the density result is based on the following perturbation argument.Firstly, we prove that we can reduce to a finite atomic boundary b with integer multiplicities, exploiting the fact that multiples of such boundaries are dense in A C , see Lemma 3.1.For these boundaries, we prove that the solutions to the optimal branched transportation problem are multiples of polyhedral integral currents, see Lemma 3.2.Then we improve the uniqueness result of [20] to suit the discrete branched transportation problem, obtaining as a byproduct that for every finite atomic boundary b as above the set OTP(b) is finite, see Lemma 3.7.We deduce the existence of a set of points {p 1 , . . ., p h } in the regular part of the support of a fixed transport path T ∈ OTP(b) with the property that T is the only element in OTP(b) whose support contains {p 1 , . . ., p h }, see Lemma 3.8.
Next, we aim to "perturb" the boundary b close to the points p 1 , . . ., p h in order to obtain boundaries with unique minimizers, keeping in mind the fact that the perturbed boundaries should not escape from the set A C .More in detail, we define a sequence (b n ) n≥1 ⊂ A C of boundaries for the optimal branched transportation problem with the property that F K (b n − b) → 0 as n → ∞.Moreover, each b n has points of its support (with small multiplicity) in proximity of p 1 , . . ., p h , so that every minimizing transport path S n with boundary ∂S n = b n is forced to have such close-by points in its support.Exploiting again the stability property of Theorem A.1, we deduce that for every choice of S n ∈ OTP(b n ) there exists S ∈ OTP(b) such that, up to subsequences, it holds F K (S n − S) → 0 and we can infer the Hausdorff convergence of the supports of the S n 's to the union of the support of S and the points p 1 , . . ., p h , see Lemma 4.2.Notice that at this stage we cannot deduce from Lemma 3.8 that S = T , since the portion of S n which is in proximity of some of the p i 's might vanish in the limit.In order to exclude this possibility, we perform a fine analysis of the structure of the network S n around the points p 1 , . . ., p h , see §4.3: this allows us to exclude all possible local topologies except for two, see (4.18), proving that p 1 , . . ., p h are contained in the support of S (so that in particular S = T by Lemma 3.8) and that OTP(b n ) = {S n }, for n sufficiently large, see Lemma 4.3, which concludes the proof of Theorem 1.1.

Remark.
It is much easier to prove just density in C>0 A C of the boundaries b for which OTP(b) is a singleton.Indeed, it is significantly simpler to perform the strategy outlined above if one is allowed to choose b n simply satisfying F K (b n − b) → 0 and M α (S n ) ≤ C, but possibly with M(b n ) > C: for instance it suffices to choose the perturbation b n as in (4.1) with k = 1, in which case it is easy to prove that OTP(b n ) is a singleton.Obviously such type of perturbation is not admissible in order to prove the residuality result of Theorem 1.1, since such boundaries b n do not belong to A C .One of the challenges in our proof is therefore to find suitable perturbations b n of b which are internal to the set A C and such that for the boundary b n there exists a unique minimizer of the optimal branched transportation problem, for n sufficiently large.
1.4.Remark.Following [26,27], it would be tempting to adopt a seemingly simpler strategy to prove Theorem 1.1.Indeed the density result would be an easy consequence of the following unique continuation principle: if b is a finite atomic boundary with integer multiplicities, then any two elements of OTP(b) which coincide on a neighbourhood of the support of b necessarily coincide globally.One reason to believe that such a statement could be true is the fact that, knowing the directions and the multiplicities of all the edges colliding at a branch point except for one, it is possible to deduce the information on the last edge, by exploiting a balancing condition which is due to the stationarity of the network for the α-mass.The main obstruction to prove the statement is the following.If for a minimizer T in OTP(b) two or more edges emanating from the boundary collide at some branch point, it is not obvious that for another minimizer T ′ which coincides with T on a neighbourhood of the support of b the same edges still collide: it might happen that T ′ has some branch point in the interior of one of these edges.We do not exclude that the statement could be true, but we believe that this cannot be proved only by local properties, which would make a potential proof quite involved.This is the reason why we opted for a completely different strategy, which is based ultimately on local arguments only.The presence of singularities is not an issue in the framework of minimal surfaces, because the singular set is too small to disconnect the regular part of the surface.We believe that the strategy which we devised is of general interest and can be adapted to prove generic uniqueness of solutions to other variational problems with large singular sets, see [17,18].

Preliminaries
Through the paper K ⊂ R d denotes a convex compact set.We denote by M (K) the space of signed Radon measures on K and by M + (K) the subspace of positive measures.The total variation measure associated to a measure µ ∈ M (K) is denoted by µ and µ + := 1/2( µ + µ) and µ − := 1/2( µ − µ) denote respectively the positive and the negative part of µ.The mass of µ is the quantity M(µ) := µ (K).We say that a measure is finite atomic if its support is a finite set.
We adopt Xia's Eulerian formulation [32] of the optimal branched transportation problem.This employs the theory of currents, for which we refer the reader to [19].We recall that a k-dimensional current on R d is a continuous linear functional on the space D k (R d ) of smooth and compactly supported differential k-forms and we denote by D k (K) the space of k-dimensional currents with support in K.The space D k (K) is endowed with a norm which is called mass and denoted by M. By the Riesz representation theorem, a current T with M(T ) < ∞ can be identified with vectorvalued Radon measures T µ T where T is a unit k-vector field and µ T a positive Radon measure.The mass of the current T coincides with the mass of the measure µ T .We denote by supp(T ) the support of a current T , which coincides with the support of the measure µ T , if T has finite mass.The boundary of a current A current T such that M(T ) + M(∂T ) < ∞ is called a normal current.The space of k-dimensional normal currents with support in K is denoted by N k (K).We say that a current T ∈ D k (K) is rectifiable and we write T ∈ R k (K) if we can identify T with a triple (E, τ, θ), where E ⊂ K is a k-rectifiable set, τ (x) is a unit k-vector spanning the tangent space Tan(E, x) at H k -a.e.x and θ ∈ L 1 (H k E), where the identification means that Those currents T = (E, τ, θ) which are normal and rectifiable with integer multiplicity θ are called integral currents.The subgroup of integral currents with support in K is denoted by I k (K).
A k-dimensional polyhedral current is a current P of the form where θ i ∈ R \ {0}, σ i are nontrivial k-dimensional simplexes in R d , with disjoint relative interiors, oriented by k-vectors τ i and σ i = (σ i , τ i , 1) is the multiplicity-one rectifiable current naturally associated to σ i .The subgroup of polyhedral currents with support in K is denoted P k (K).A polyhedral current with integer coefficients θ i is called integer polyhedral.
Given α ∈ (0, 1) and a 1-current T ∈ N 1 (K) ∪ R 1 (K) we define the α-mass We define the set of optimal transport paths with boundary b by Let A C be the set of boundaries defined in (1.1).Due to the Baire category theorem, the next lemma ensures that a residual subset of A C (namely a set which contains a countable intersection of open dense subsets) is dense.We recall that the flat norm F K (T ) of a current T ∈ D k (K) is the following quantity, see [19, §4.1.12], Proof.The second part of the statement follows from the first part and from the F K -compactness of 0-currents with support in K and mass bounded by C, see [19, §4.2.17].
In order to prove that A C is F K closed, let (b j ) j∈N be a sequence of elements of A C and let b be such that F K (b j − b) → 0 as j → ∞.We want to prove that b ∈ A C .By the lower semicontinuity of the mass (with respect to the flat convergence), we have M(b) ≤ C. For any j ∈ N, let T j ∈ OTP(b j ).By [12,Proposition 3.6], we have M(T j ) ≤ C 1−α M α (T j ) ≤ C 2−α .By the compactness theorem for normal currents, there exists T ∈ N 1 (K) such that, up to (non relabeled) subsequences F K (T j − T ) → 0. By the continuity of the boundary operator we have ∂T = b and by the lower semicontinuity of the α-mass, see [16], we have M α (T ) ≤ C and hence b ∈ A C .
Consider the following subset of A C , which represents the set of boundaries admitting non-unique minimizers: We have the following: As in the proof of Lemma 2.1, we deduce that there exist 3. Density of boundaries with unique minimizers: preliminary reductions Since we also have Now apply [24,Theorem 5] to obtain, possibly after rescaling, a current and in particular as in (2.1).Up to changing the orientation of σ i , we may assume θ ′ i > 0 for every i.Fix η := ε/(16N ) and denote (where ⌊x⌋ is the largest integer smaller than or equal to x) for every i ∈ {1, . . ., N }.
For every i ∈ {1, . . ., N }, we denote by x i and y i respectively the first and second endpoint of the oriented segment σ i , so that we can write which we can rewrite as where all points z j are distinct and Analogously, we define Thus we obtain and by (3.1) and (3.3) we deduce Proof.Combining the good decomposition properties of optimal transport paths [12, Proposition 3.6] and their single path property [2, Proposition 7.4] with the assumption ∂T ∈ P 0 (K), we deduce that there are finitely many Lipschitz simple paths γ 1 , . . ., γ N of finite length such that T can be written as a T = N i=1 a i γ i , where a i > 0 for every i and . Moreover, again by [2, Proposition 7.4], one can assume that Im(γ i ) ∩ Im(γ j ) is connected for every i, j, which in turn implies that T ∈ P 1 (K).Hence we can write where σ ℓ are non-overlapping oriented segments and θ ℓ ∈ R. We want to prove that θ ℓ ∈ Z, ∀ℓ.We say that T and T ′ have the same topology if there exist two ordered sets, each made of distinct points, {x 1 , . . ., x M } and {x ′ 1 , . . ., x ′ M } with the following properties: (i) for every p ∈ supp(b) there exists i such that x i = p = x ′ i ; (ii) denoting σ ij the segment with first endpoint x i and second endpoint x j and σ ′ ij the segment with first endpoint x ′ i and second endpoint x ′ j , T and T ′ can be written respectively as (iii) the representations in (3.8), restricted to the nonzero addenda, is of the same type as (2.1).
In particular, if a ij and a kl (resp.a ′ ij and a ′ kl ) are nonzero, then σ ij and σ kl (resp.σ ′ ij and σ ′ kl ) have disjoint interiors.Moreover, the number of nonzero addenda in the representation of T (resp.T ′ ) given in (3.8) coincides with the smallest number N for which T (resp.T ′ ) can be written as in (2.1).(iv) a ij = 0 if and only if a ′ ij = 0.In particular, the number N of the previous point is the same for T and T ′ .One can check that the above conditions define an equivalence relation on the set of polyhedral currents.We call the topology of a polyhedral current T the corresponding equivalence class.Notice that the number M depends only on the equivalence class and for every T the (unordered) set {x 1 , . . ., x M } is uniquely determined, by property (iii).The set {x 1 , . . ., x M } \ supp(b) is called the set of branch points of T and denoted by BR(T ).By Lemma 3.2, for every T ∈ OTP(b) the topology of T and the set BR(T ) are well defined.Proof.Suppose without loss of generality that H 0 (BR(T )) > 0. Assume by contradiction that the lemma is false and let n be the minimal number such that there exist b ∈ I 0 (K) and T ∈ OTP(b) such that Notice that n > 2. Fix p ∈ BR(T ) and let ε > 0 be such that Denote by T 1 , . . ., T m the restriction of T to the connected components of supp(T ) \ B ε (p).We notice that m ≥ 3. Indeed, if m = 1 we would have the contradiction p ∈ supp(b) and if m = 2, writing T as in (3.8), the only two segments with nonzero coefficient having p as an endpoint cannot be collinear by property (iii): this contradicts the the fact that T ∈ OTP(b).Observe that for every i we have that supp(∂T i ) \ supp(b) consists of exactly one point p i , so that By minimality of n and the fact that m ≥ 3, we have with properties (i)-(iv) and with the same orientation on each segment.Then a ij = a ′ ij for every i, j.
Proof.By contradiction, let T, T ′ be nonzero currents with the same topology, ∂T = b = ∂T ′ , and minimizing the quantity M in Definition 3.3 among all pairs for which the lemma is false.We claim that there exists a point p ∈ supp(b) and (up to reordering) indexes i, j ∈ {1, . . ., M } such that (a) a lj = 0 = a ′ lj for every l = i; By [2, Proposition 7.8] the support of T and T ′ does not contain loops, hence we can apply Lemma 3.5 and we can assume that T and T ′ can be written as in (3.8) with a ij = a ′ ij for every i, j = 1, . . ., M .This means that the set of competitors for the branched transportation problem with boundary b and a given topology can be reduced to a family of polyhedral currents T ∈ P 1 (K) whose only unknown is the position of the points {x 1 , . . ., x M } \ supp(b).Accordingly, we denote n := H 0 (supp(b)) and we order the points {x 1 , . . ., x M } in such a way that BR(T ) = {x 1 , . . ., x M −n }.The α-mass of such T is computed as and by the previous discussion, since the vector (x M −n+1 , . . ., x M ) is fixed, this is a functional of the vector (x 1 , . . ., x M −n ) only, which can be written as where (3.11), as a function of the variable x j , is strictly convex on a segment [s, t] whenever x i , s and t are not collinear.
Assume by contradiction that T = T ′ ∈ OTP(b) have the same topology and consider the corresponding sets . By Lemma 3.5 there exists j ∈ {1, . . ., M − n} such that x j = x ′ j .As in the proof of Lemma 3.4 we infer that x j is an endpoint of at least three segments in the support of T which are not collinear.We deduce by the discussion after (3.11) that the function F is strictly convex in the j-th variable.Since F (x 1 , . . ., x M −n ) = F (x ′ 1 , . . ., x ′ M −n ) we deduce that there exists a point (y 1 , . . ., y M −n ) with (3.12) Denote and let S be the current where σij is the segment with first endpoint z i and second endpoint z j .Notice that in principle it might happen that S does not have the same topology as T and T ′ , since (3.13) might fail to have property (iii) of Definition 3.3.However we have ∂S = b and by (3.12)Moreover, the p i 's can be chosen so that if p i ∈ supp(T ) ∩ supp(S) for some S ∈ OTP(b), then there exists ρ > 0 such that supp(T ) ∩ B ρ (p i ) = supp(S) ∩ B ρ (p i ).
Proof.By Lemma 3.7 we have that OTP(b) consists of finitely many polyhedral currents T 1 , . . ., T h and, by Lemma 3.6, the symmetric difference supp(T i )△ supp(T j ) is a relatively open set of positive length for every i = j.Up to reordering, we assume T 1 = T and for every i ∈ {2, . . ., h} we consider the set U i := supp(T ) \ (supp(T i ) ∪ BR(T )).We observe, recalling that BR(T ) is finite by Lemma 3.4, that each U i is relatively open with positive length.Define the subset BR(T j ) ∪ p ∈ U i : supp(T j ) intersects U i transversally at p and observe that V i is finite since every T j is polyhedral.Then choose p i ∈ U i \ V i .Clearly p i ∈ supp(T ) \ (BR(T ) ∪ supp(b)) and p i ∈ supp(T i ); moreover if p i ∈ supp(T j ) then locally supp(T j ) agrees with supp(T ).
Observe that by [12, Proposition 3.6], the multiplicity of T is bounded from above by 2 −1 M(b) and moreover, for n sufficiently large, the closed balls B n −1 (p i ) are disjoint and do not intersect supp(b) ∪ BR(T ), so that we have and For every n, we choose S n ∈ OTP(b n ) and we apply Lemma 3.2 to the boundaries kb n to deduce that kS n ∈ P 1 (K) ∩ I 1 (K).By (4.1) we have The aim of this section is to prove the following:  For every subsequence (S n j ) j∈N and current S such that F K (S n j − S) → 0 as j → ∞ we have S ∈ OTP(b) and moreover for every ρ > 0 we have supp(S n j ) ⊂ B ρ (supp(S) ∪ {p 1 , . . ., p h }), for j sufficiently large.
Proof.The first part of the proposition is a direct consequence of Theorem A.1.Towards a proof by contradiction of the second part, assume that there exists r > 0 and, for every j, a point q j ∈ supp(S n j ) \ B 2r (supp(S) ∪ {p 1 , . . ., p h }).On the other hand, by (4.1) and (4.5) the current S n j has no boundary in B r (q j ), for j sufficiently large.Moreover, by [2, Proposition 7.4] the restriction R j of S n j B r (q j ) to the connected component of its support containing q j has non-trivial boundary, and more precisely applying [30, Lemma 28.5] with f (x) = |x−q j | we deduce that ∅ = supp(∂R j ) ⊂ ∂B r (q j ).We conclude that supp(R j ) contains a path connecting q j to a point of ∂B r (q j ).By Lemma 3.2 such path has multiplicity bounded from below by k −1 .This allows to conclude that Combining (4.5),(4.6),and (4.7), we conclude which contradicts (4.4).
We dedicate the rest of this section to prove the following: Lemma 4.3.There exists k 0 = k 0 (α) with the following property.Let T and (T n ) n be as in (4.1) with k ≥ k 0 and let S and (S n j ) j be as in Lemma 4.2.Then S n j = T n j , for j sufficiently large and in particular S = T .
Proof.We divide the proof in three steps.In §4.2 we prove that locally in a box around each point p ∈ {p 1 , . . ., p h } ∩ supp(S) for j sufficiently large the restriction to the box of the current S n j is a minimizer of the α-mass for a certain boundary whose support is a set of four almost collinear points.In §4. 3 we analyze all the possible topologies for the minimizers with such boundary and we are able to exclude all of them except for two.In §4.4 we combine the local analysis with a global energy estimate to conclude.

4.2.
Local structure of S n j .Let ρ be sufficiently small, to be chosen later (see (2a), (3a) and (3b) in §4.3).For every i = 1, . . ., h and for p i ∈ supp(S), by Lemma 3.8 we can choose orthonormal coordinates (x, y) ∈ R × R d−1 such that, up to a dilation with homothety ratio c with The choice of Q is illustrated in Figure 2. By Lemma 4.2, for this ρ, we may choose j large enough such that supp(S n j ) ∩ Q ⊂ B ρ (σ).(4.8)For x ∈ R we denoting by S x n j the slice of S n j Q at the point x with respect to the projection π : R × R d−1 → R, see [30, §28] or [19, §4.3].We infer from the flat convergence of S n j to S that for H 1 -a.e.x ∈ [−8, 8] it holds and moreover by Lemma 3.2 the multiplicities of S x n j are integer multiples of k −1 .We aim to prove that for j sufficiently large there are points y To this aim we seek points for some points y i (j) ∈ B d−1 ρ (0).If so, by [30,Lemma 28.5], (4.11) and (4.8) imply, denoting In turn, by [2, Proposition 7.4] the latter implies (4.10).
In order to prove (4.11), we focus on the interval I := [1,2] as the argument for the remaining intervals is identical.Firstly, we observe that by (4.9) we have lim inf Next, denoting Ω := I × B d−1 ρ (0), we claim that for j sufficiently large and for every C > 0 it holds that H 1 ({x ∈ I : M α (S x n j ) ≤ θ α + C}) > 0, (4.13)where for a 0-current Z := ℓ∈N θ ℓ δ z ℓ we denoted M α (Z) := ℓ∈N |θ ℓ | α .Assume by contradiction that (4.13) is false for infinitely many indices j.By [14, equation (3.11)], for those indices we have The latter, combined with (4.4), implies that for the same indices we have which contradicts [16,Proposition 2.6].From (4.12) and (4.13) we deduce that for j sufficiently large there exists x 1 (j) ∈ I such that Lastly we prove that if C is sufficiently small, then (4.15) implies for some points y 1 (j) ∈ B d−1 ρ (0), thus completing the proof of (4.11).Towards a proof by contradiction of (4.16), observe that for every 0-current This contradicts (4.15), by the arbitrariness of C.

Analysis of the possible topologies of
In general, we will denote by σ P R the oriented segment from the point P to the point R. We aim to prove that for k ≥ k 0 (α) and for ρ ≤ ρ(k) sufficiently small it holds S n j Q ′ ∈ {W j , Z j }, for j large enough, where see Table 1.We will do this by excluding every other topology comparing angle conditions which are given by the multiplicities of the segments (which depend on k) and contradict the choice of ρ.Thus, when we say for ρ small enough, we mean implicitly to choose j large enough such that by Lemma 4.2, we have supp(S n j ) ⊂ B ρ (supp(S) ∪ {p 1 , . . ., p h }) for the desired ρ .Write S n j Q ′ = i<j a ij σ ij as in (3.8) and observe that by Lemma 3.4, as H 0 (∂(S n j Q ′ )) = 4, then H 0 (BR(S n j Q ′ )) ∈ {0, 1, 2}.We thus analyze the three cases separately and we recall that, by Lemma 3.6, in order to prove (4.18) it suffices to prove that Table 1.Representation of W j and Z j .From now on δ := θ/k and we remove the subscript j from the points.
Recalling [2, Proposition 7.4], supp(S n j Q ′ ) must be one of the following sets, sorted alphabetically: independently of the position of the points, the support either contains a loop or does not contain one of the four points in the support of the boundary.The only exceptions to this behaviour are (ii) and (iii) only when the four points are collinear, which is not relevant, as we discuss in Sub-case 1-1 below.
Sub-case 1-1.Firstly we observe that when the points A j , B j , C j , D j are collinear the only admissible competitor is Z j .
Sub-case 1-2.Next, we analyze the case in which no triples among the points A j , B j , C j , D j are contained in a line.
We immediately exclude those cases for which the corresponding set is not the support of any current with boundary ∂(S n j Q ′ ).Hence we can exclude (1i) and (1n), because the endpoints of the two segments in the support have different multiplicities.Moreover we exclude (1d), (1j), (1q) and (1r) as well, because the segment σ A j ,D j should have multiplicity θ, being either for A j or D j the only segment in the support containing it.On the other hand, the remaining point (respectively D j or A j ) is an endpoint also for a different segment of the support, from which we deduce that the multiplicity of the latter segment should be 0 (see Table 2).
Lastly, we exclude case (1e) by direct comparison with the α-mass of Z j .For j sufficiently large and for k ≥ k 0 (α), the α-mass corresponding to (1e) is (see Table 5) Table 5. Representation of (1e).
Sub-case 1-3.The last situation which we need to take into account is when exactly three points are collinear.We will discuss the case in which the collinear points are A j , B j and C j or A j , C j and D j .The remaining cases in which the collinear points are B j , C j and D j or A j , B j and D j are symmetric and can be treated analogously, therefore we leave the analysis to the reader.
We exclude case (1a), since it coincides with case (1j), which we have already excluded and we exclude cases (1k) and (1o) because they contain a loop (see Table 8).
We do not need to exclude cases (1c) and (1m), since the current coincides with Z j (see Table 9).9. Representation of (1c), (1m) in the collinear case.
Lastly, cases (1e), (1h), (1s) can be excluded with the same argument used in Sub-case 1-2, since the segments in the corresponding support are in general position also when A j , B j , and C j are collinear (see Table 10).10.Representation of (1e), (1h), (1s) in the collinear case.
Sub-case 1-3-2: A j , C j and D j are collinear.
Recalling that E j is the endpoint of at least three segments in the support of S n j Q ′ , see the proof of Lemma 3.4, the only possibilities are that supp(S n j Q ′ ) is one of the following sets (see Table 16): We exclude case (2a), indeed by [2, Lemma 12.1 and Lemma 12.2], E j ∈ conv({A j , B j , C j }), hence we have for ρ small and j sufficiently large This contradicts [2, Lemma 12.2] for ρ ≤ ρ(k), since the modulus of the multiplicity of σ A j E j , σ E j B j and σ Cases (2b), (2c) and (2d) are excluded with a similar argument as in case (2a), where the angle ∠A j E j C j in (4.23) is replaced respectively by ∠A j E j D j , ∠A j E j D j and ∠B j E j D j .
We exclude case (2e) by direct comparison with the α-mass of Z j .The α-mass corresponding to (2e) is where the inequality is strict unless {E j } = σ A j D j ∩ σ B j C j , namely unless the current is Z j , which of course we do not need to exclude.
and the fact that both E j and F j are the endpoints of at least three segments in the support of S n j Q ′ , see the proof of Lemma 3.4, up to switching between E j and F j , the only possibilities are that supp(S n j Q ′ ) is one of the following sets (see Table 17 ) Denote by π 0 the affine 2-plane passing through A j , B j and F j (and therefore containing E j as well).By [2, Lemma 12.2] the line ℓ containing σ E j F j divides π 0 \ ℓ into two open half-planes π − 0 and π + 0 containing respectively A j and B j .Let C ′ j and D ′ j denote the orthogonal projections onto π 0 of C j and D j respectively and observe that C ′ j ∈ π + 0 .This follows from the fact that by Table 16.Representation of (2a), (2b), (2c), (2d), (2e).In (2a), (2b), (2c), (2d) we do not represent the segment σ.
[2, Lemma 12.2] there exists a positive constant κ (depending on k) such that ∠A j E j F j ≤ π − κ and assuming C ′ j / ∈ π + 0 would lead to ∠A j B j C ′ j ≤ ∠A j E j F j ≤ π − κ, which is a contradiction, for ρ sufficiently small with respect to k, since, due to the fact that κ which is a contradiction, for ρ sufficiently small with respect to k, since, as above, Lemma 12.2] applied at the branch point E j we deduce that the angle between the oriented segments σ A j E j and σ E j F j tends to 0 as k → ∞.By the same argument applied at the branch point F j we deduce the same property for the angle between the oriented segments σ E j F j and σ F j D j .As a consequence, the angle between the oriented segments σ A j D j and σ E j F j tends to 0 as k → ∞.Again by [2, Lemma 12.2], the angles ∠C j E j F j and ∠E j F j B j are equal to π 2 + C(k) where C(k) tends to 0 as k → ∞.  17.Representation of (3a), (3b), (3c).
Next, using that the angle ∠E j F j D j differs from π by a positive constant which depends only on k, we observe that the plane containing A j , C j , F j (and therefore also E j ) is obtained from the plane containing D j , B j , E j (and therefore also F j ) by a rotation O around the line containing σ E j F j such that, for any fixed k, O − Id < f (ρ), where f (ρ) tends to 0 as ρ → 0. This implies that the angle between the oriented segments σ B j C j and σ A j E j is larger than π 2 − c(k), where c(k) tends to 0 as k → ∞.This is a contradiction for ρ sufficiently small and k sufficiently large, since for any k the angle between the oriented segments σ B j C j and σ A j D j tends to 0 as ρ → 0 and for any ρ the angle between the oriented segment σ A j E j and the oriented segment σ A j D j tends to 0 (independently of ρ) as k → ∞.
(3c) We exclude this case as the corresponding set is not the support of any current with boundary ∂(S n j Q ′ ), because both the segments σ A j E j and σ E j D j should have multiplicity θ, thus the multiplicity of σ E j F j would be zero.

Conclusion.
In order to conclude the proof of Lemma 4.3, for j and k sufficiently large we now exclude the case in which S n j coincides with Z j close to at least one point p i .Indeed, should that happen, we could build a better competitor than T for b by adding 1 k h i=1 T B n j −1 (p i ) to S n j .
We claim that, for j sufficiently large and for k ≥ k 0 (α), we have the inequality being strict unless S passes through all the p i 's and around every p i the current S n j has the shape W j , i.e. (remember that Q ′ depends on i) {i : p i / ∈ supp(S)} = ∅ = {i : S n j Q ′ = Z j }. which is supported in exactly two points.Since necessarily Sn j ∈ OTP(∂ Sn j ), we deduce that for j sufficiently large.Combining (4.30), (4.18) and (4.4), we obtain that, for j sufficiently large and for k ≥ k 0 (α)  Observe that equality holds if and only if the negative terms above vanish which yields the validity of the claim in the case S = T .Moreover (4.31) trivially holds also in the case S = T , which concludes the validity of the claim in the general case, and of Lemma 4.3.
Proof of Proposition 4.1.Since the conclusion of Lemma 4.3 holds for every converging subsequence S n j , we deduce that S n = T n and therefore OTP(b n ) = {T n }, for n sufficiently large.
otherwise.If µ − and µ + are elements of M + (K) such that M(µ − ) = M(µ + ), the optimal branched transportation problem with boundary b = µ + − µ − seeks a normal current T ∈ N 1 (K) which minimizes the α-mass M α among all currents S with boundary ∂S = b.Hence, we denote by TP(b) the set of transport paths with boundary b as TP(b) := {T ∈ N 1 (K) : ∂T = b}, and the least transport energy associated to b as E α (b) := inf{M α (T ) : T ∈ TP(b)}.

3. 1 .Lemma 3 . 1 .
Reduction to integral boundaries and integer polyhedral minimizers.For any b ∈ A C and ε > 0, there exist δ > 0 and a boundary b ′′ ∈ A C−δ with F K (b − b ′′ ) < ε and b ′′ = ηb I for some η > 0 and b I ∈ I 0 (K).Proof.Without loss of generality and up to rescaling, we can assume C = 1 and write A instead of A C .Let b ∈ A and T ∈ OTP(b) and define T and denote b ′′ = ∂T ′′ .Observe that by(3.4) and (3.3) we have

.Lemma 3 . 6 .
The validity of (a) follows from the absence of loops.On the other hand, if a point p as in (a) violated (b), one could restrict the currents T and T ′ respectively to the complementary of σ ij and σ ′ ij , thus contradicting the minimality of M .The validity of (a) and (b) is a contradiction because the multiplicities a ij and a ′ ij correspond to the multiplicity of p as point in the support of b.Let b ∈ I 0 (K) and S, T ∈ OTP(b) with supp(S) = supp(T ).Then S = T .Proof.Assume by contradiction S = T .By Lemma 3.2, S − T ∈ P 1 (K) ∩ I 1 (K) is a nontrivial current with ∂(S − T ) = 0 and by assumption supp(S − T ) ⊂ supp(S).As in the proof of Lemma 3.2 we deduce that supp(S − T ) contains a loop.In particular, so does supp(S), which contradicts [2, Proposition 7.8].Lemma 3.7.Let b ∈ I 0 (K) be a boundary.Then OTP(b) is finite.Proof.By Lemma 3.4 the range of the integer M of Definition 3.3 among all T ∈ OTP(b) is finite.In turn this implies that the set of possible topologies of currents T ∈ OTP(b) is finite.Indeed the topology of a polyhedral current T as in Definition 3.3, up to choosing the order of the points {x 1 , . . ., x M }, is uniquely determined by the M × M matrix A := (|sign(a ij )|) ij .Hence it is sufficient to prove that if T and T ′ are in OTP(b) and have the same topology, then T = T ′ , and by Lemma 3.6 it suffices to prove that supp(T ) = supp(T ′ ).

Proposition 4 . 1 .
There exists k 0 = k 0 (α) such that for (b n ) n as in (4.1) with k ≥ k 0 and for n sufficiently large, OTP(b n ) = {T n }.In the next lemma, for any set A and ρ > 0 we denote B ρ (A) := a∈A B ρ (a).

Figure 3 .
Figure 3. Representation of parts of S n j Q.