M of genus 0. We consider this connection of independent interest; it is natural to expect that similar ideas can be used for other matrix valued OPs, as long as the corresponding Riemann surface M is of genus 0. The rest of the method consists of two parts, and mainly follows the lines of a previous work of Charlier, Duits, Kuijlaars and Lenells. First, we perform a Deift–Zhou steepest descent analysis to obtain asymptotics for the scalar valued OPs. The main difficulty is the study of an equilibrium problem in the complex plane. Second, the asymptotics for the OPs are substituted in the double contour integral and the latter is analyzed using the saddle point method. Our main results are the limiting densities of the lozenges in the disordered flower‐shaped region. However, we stress that the method allows in principle to rigorously compute other meaningful probabilistic quantities in the model.We analyze a random lozenge tiling model of a large regular hexagon, whose underlying weight structure is periodic of period 2 in both the horizontal and vertical directions. This is a determinantal point process whose correlation kernel is expressed in terms of non‐Hermitian matrix valued orthogonal polynomials (OPs). This model belongs to a class of models for which the existing techniques for studying asymptotics cannot be applied. The novel part of our method consists of establishing a connection between matrix valued and scalar valued OPs. This allows to simplify the double contour formula for the kernel obtained by Duits and Kuijlaars by reducing the size of a Riemann–Hilbert problem. The proof relies on the fact that the matrix valued weight possesses eigenvalues that live on an underlying Riemann surface  As the size of the hexagon tends to infinity (while the size of the lozenges is kept fixed), the local statistical properties of this model are described by universal processes.A lozenge tiling of a hexagon is a collection of three different types of lozenges =∏p∈Twp. Provided that at least one tiling has a positive weight, this defines a probability measure on the set of tilings by  size nHn:={ it suffices to analyze the asymptotic behavior of the correlation kernel. However, until recently,By putting points on the paths as shown in , each tiY) to a 2p×2p Riemann–Hilbert (RH) problem. This RH problem is related to certain orthogonal polynomials (OPs), which are nonstandard in two aspects:p×p matrix valued,the OPs and the weight are the orthogonality conditions are non‐Hermitian. The size of the RH problem, the size of the weight, and the size of the OPs all depend on p, but quite interestingly not on q.The main result of Ref. p×q periodic weightings are rather unexplored up to now. To the best of our knowledge, the model considered in Ref. Y (of size 2 × 2) with a nonstandard saddle point analysis of the double contour integral. However, because p=1, the associated OPs are scalar (this fact was extensively used in the proof) and it is not clear how to generalize these techniques to the case p≥2.Lozenge tiling models of the hexagon with 1.U) to 2 × 2 RH problem related to scalar OPs. This formula allows for a simpler analysis than the original formula from Ref.  First, we establish a connection between matrix valued and scalar valued OPs. In particular, in Theorem 2.U via the Deift–Zhou steepest descent method. The construction of the equilibrium measure and the associated g‐function is the main difficulty. The remaining part of the RH analysis is rather standard. Second, we perform an asymptotic analysis of the RH problem for 3. Third, the asymptotics for the OPs are substituted in the double contour integral and the latter is analyzed using the saddle point method. The first step is the main novel part of the paper. The remaining two steps were first developed in Ref. The aim of this paper is precisely to develop a method to handle a situation involving matrix valued OPs. We will implement this method on a particular lozenge tiling model with 2 × 2 periodic weightings, which presents one simply connected liquid region , six frozen regions, and six staircase regions ; it will be presented in more detail in Section Our main results, which are stated in Theorem 1.1M, which turns out to be of genus 0. This fact is crucial to obtain the new formula for the kernel in terms of scalar OPs. We expect that ideas similar to the ones presented here can be applied to other tiling models with periodic weightings, as long as the corresponding Riemann surface M is of genus 0.The eigenvalues and eigenvectors of the 2 × 2 orthogonality weight play an important role in the first step of the analysis. They are naturally defined on a two‐sheeted Riemann surface p=q=1). In fact, it is expected that the smallest periods that lead to the presence of a gas phase are either p=2,q=3 or p=3,q=2. For models that present gas phases, we expect M to have genus at least 1, and then new techniques are required. This is left for future works.Lozenge tiling models of large hexagons with periodic weightings can feature all of the three possible types of phases known in random tiling models: the solid, liquid, and gas phases. A solid region  is filled with one type of lozenges. In the liquid and gas phases, all three types of lozenges coexist. The difference between these two phases is reflected in the correlations between two points: in the liquid region, the correlation decay is polynomial with the distance between the points, while in the gas region the decay is exponential. It is known that there is no gas phase for the uniform measure :|x|+|y|≤n+1} with 2 × 1 or 1 × 2 rectangles , where n>0 is an integer which parameterizes the size of the covered region. Uniform domino tilings of the Aztec diamond features four solid regions and one liquid region. The associated discrete point process is determinantal, and turns out to belong to the class of Schur processes  do not belong to the Schur class, but have been studied using other techniques based on some connections with Hahn polynomials:The Aztec diamond is a well‐studied tiling model.n→+∞. This same model was analyzed soon afterward in Ref. n. The analysis of Refs. 2×k periodic Aztec diamond, for an arbitrary k. The class of models for which the formula from Ref. The doubly periodic Aztec diamond exhibits all three phases. It still defines a determinantal point process, but it falls outside of the Schur process class. However, Chhita and Young found in Ref. However, lozenge tiling models of the finite hexagon cannot be represented as models with infinitely many paths (as opposed to the Aztec diamond and the infinite hexagon). In particular, they do not belong to the class of models studied in Ref. 1.3dimer coverings, which are perfect matchings of a certain bipartite graph. We refer to Ref. In addition to being in bijection with nonintersecting paths, lozenge tilings of the hexagon are also in bijection with 2In this section, we present a lozenge tiling model with 2 × 2 periodic weightings. We also introduce the necessary material to invoke the double contour formula from Ref. 2.1For the presentation of the model and the results, it is convenient to define the hexagon and the lozenges as in  and 3)..3). HoweHn is mapped by this transformation to a hexagon whose six sides are of equal length. Above the definition . In the figures, we will assign the colors red, green, and yellow for the three lozenges in (4), from left to right, respectively.so that finition  of Hn, w2.2Hn has corners located at , , , , , and . We normalize the lozenges such that they cover each a surface of area 1, and the vertices of the lozenges have integer coordinates. We recall that each lozenge tiling of Hn gives rise, through , and the bottom left vertex has coordinates . We denote the paths bypj(0)=j+12 and ending positions pj(2n)=n+j+12. The particular 2 × 2 periodic lozenge tiling model that we consider depends on a parameter α∈,) is an edge of Gn, thenThe regular hexagon  through , to a sy0, n), n,n, >0 for all T, and thus we have a well‐defined probability measure via =0 for all T. So in this case,  and let n≥1 be an integer. There exists a unique tiling Tmax of Hn such that W(T)≤αW(Tmax) for all T≠Tmax. Furthermore,Let For any values of ightings  are suchis case,  does notGn, and we omit the details.▪The proof of Proposition α→0, the randomness disappears because the tiling Tmax becomes significantly more likely than any other tiling. Therefore, our model interpolates between the uniform measure over the tilings (for α=1) and a particular totally frozen tiling Tmax (as α→0), see Figures It follows from Proposition W(T) of a tiling T can alternatively be defined asSeveral tiling models in the literature   α∈, and whose edges are of the form e=,) with x2=x1+1 and y2−y1∈{0,1}. The weighting  for the weight associated to the edge e=,) of G∞. This weight can be obtained from =Tx,x+1 for all y1,y2∈Z. The weightings Tx,x+1 can be represented as two 2 × 2 block Toeplitz matrices  that are infinite in both directions. These two infinite matrices can be encoded in two 2 × 2 matrix symbols Ax,x+1(z), whose entries )i+1,j+1, 0≤i,j≤1, are given byIt will be convenient for us to define eighting  was defined from  and onlyTx,x+1 from its symbol byGn is then obtained by taking the following product 2Nz2N2.4Y(z)=Y defined byY is characterized as the unique solution to the following RH problem.RH problems for scalar OPs have been introduced by Fokas, Its, and Kitaev in Ref. (a)Y:C∖γ→C4×4 is analytic.(b)Y(z) as z approaches γ from inside and outside exist, are continuous on γ and are denoted by Y+ and Y−, respectively. Furthermore, they are related byThe limits of (c)z→∞, we have Y(z)=(I4+O(z−1))zNI20202z−NI2.As 2.5K denote the associated kernel. By definition of determinantal point processes, for integers k≥1, and x1,…,xk,y1,…,yk with ≠ if i≠j we haveProposition 2(from Ref. 20) Let α∈ is the unique bivariate polynomial of degree ≤N−1 in both variables w and z, which satisfies the following reproducing property:P of degree ≤N−1. Because it satisfies  is called a reproducing kernel.From Ref. roperty:12πi∫γP=0 or Ψ′(ζ)=0. In the liquid region, Proposition s, lying in the upper half plane. This saddle plays an important role in our analysis, and some of its properties are stated in Propositions s in Theorem Remark 1α=1, our model reduces to the uniform measure and the kernel can be expressed in terms of scalar‐valued OPs. However, our approach is based on the formulas . Because the limiting densities for the lozenges in this case are already well‐known,α∈ to avoid unnecessary discussions.If formulas  and 22)α=1, our The new double contour formula for the kernel in terms of scalar OPs is stated in Theorem 3.1W byWe define the scalar weight (a)U:C∖γC→C2×2 is analytic, where γC is a closed curve surrounding c and c−1 once in the positive direction, but not surrounding 0.(b)U(ζ) as ζ approaches γC from inside and outside exist, are continuous on γC and are denoted by U+ and U−, respectively. Furthermore, they are related byThe limits of (c)ζ→∞, we have U(ζ)=(I2+O(ζ−1))ζ2N00ζ−2N.As U to the above RH problem is unique (provided it exists), and can be expressed in terms of scalar‐valued OPs as follows:p2N and q2N−1 are polynomials of degree 2N and 2N−1, respectively, satisfying the following conditions:RU is defined byTheorem 1x∈{1,…,2N−1}, y∈Z, and εx∈{0,1}, we haveγC is a closed curve surrounding c and c−1 once in the positive direction that does not go around 0, and where HK and HK are given byFor It is knownRemark 2RY from , they are ordered as follows:The function given by , r1, r3.3, x,y∈{0,1,…,4N}, are denoted byThe densities for the three types of lozenges at a point ∑j=13Pj=1. Because our model is 2 × 2 periodic, P1, P2, and P3 depend crucially on the parity of x and y, and it is convenient to consider the following matrices:x,y∈{0,1,…,2N−1}. Let {}N≥1 be a sequence satisfying lies in the hexagon belongs to the liquid region Lα⊂H.and satisfy 3.4∈H, there are in total eight saddles for the double contour integral (Q(ζ) is given by . This saddle plays a particular role in the analysis of Section Proposition 3∈Ho (the interior set of H). Then, there exists at most one solution ζ=s to  to  in C+={ζFor each integral , which agiven by . FollowiDefinition 1Lα⊂H bys:Lα→C+ by ↦s.We define the liquid region ∈Lα and s=r+ for all α∈. We now describe some properties of ↦s. Consider the following three circles:R0=α, Rα=(1−α)α, and R1=1−αα , θα∈, and θ0∈. We also defineRemark 3A, the notation A¯ stands for the closure of A.For a given set It is clear from  and 41)41) that π]}⊂γα,4Σ0={ζ∈C:|Proposition 4↦s satisfies s=s, and(a){ξ=0}∩Lα onto Σ1¯∩C+,it maps (b){η=ξ2}∩Lα onto (γ1∖Σ1)∩C+,it maps (c){η=ξ}∩Lα onto Σ0¯∩C+,it maps (d){η=−ξ}∩Lα onto (γ0∖Σ0)∩C+,it maps (e){η=0}∩Lα onto Σα¯∩C+,it maps (f){η=2ξ}∩Lα onto (γα∖Σα)∩C+.it maps The map Q(ζ)1/2 with a branch cut joining r− to r+ along Σ1, such that Q(ζ)1/2∼12ζ as ζ→∞, and denote the associated Riemann surface by Rα:1, and the sheets are ordered such that w=Q(ζ)1/2 on the first sheet and w=−Q(ζ)1/2 on the second sheet. For each solution ζ to (w satisfying w2=Q(ζ), and such thatDefinition 2↦w is defined by w2=Q), such that  and w=w.The map uch that  holds wiBy definition, the saddles lie in the complex plane. We show here that they can be naturally projected on a Riemann surface. Define ion ζ to , there eProposition 5↦,w) is a diffeomorphism from Lα toLαl={∈Lα∣ξ<0} to the upper half‐plane of the first sheet of Rα, and it maps Lαr={∈Lα∣ξ>0} to the upper half‐plane of the second sheet. Moreover, its inverse ↦=,η) is explicitly given byThe map ↦, we conclude that Lα is symmetric with respect to the origin. Also, this equation has real coefficients, so s and s¯ are both solutions whenever ∈Lα. At the boundary ∂Lα of the liquid region, s and s¯ coalesce in the real line, so ∂Lα is part of the zero set of the discriminant of ↦. As we will see, these regions are frozen (or semifrozen).From Propositions After clearing the denominator in , we get(ζ−r1)2(ζ(ζ−r1)2. Denote0(ζ−r1)2∈Lα. We obtain the following limits:ϕk,ij, 1≤i,j,k≤2, and ϕ3,11(l), ϕ3,11(r), ϕ3, 12, ϕ3, 21, ϕ3,22(l), and ϕ3,22(r) are the angles represented in Figure Let tisfying  with ∈Lα. We havej=1,2,3.Let tisfying  with ∈{0,…,2N−1} be such that =∈Fj,α, j∈{1,…,6}. In Figure From Figure ∈Fj,α, j=1,…,6, respectively. Corollary depending on whether 3.6M. The proof of Theorem M is of genus 0. The proof of Theorem In Section U. This analysis goes via a series of transformations U↦T↦S↦R. The first transformation U↦T uses a so‐called g‐function, which is obtained in Section In Section Rα and play a central role in the large N analysis of the kernel. In Section NΦ={ζ∈C:ReΦ(ζ)=ReΦ(s)}, which is of crucial importance to find the contour deformations that we need to consider for the saddle point analysis.The functions Φ and Ψ denote the restrictions of the phase function Ξ to the first and second sheets of The proofs of Propositions α∈, even though it will not be written explicitly.As mentioned in Remark 44.1M given byM are real, Proposition M has at least six zeros on the real line. This can be proved by a direct inspection of the values of M(ζ) at ζ=−∞,r1,0,αc,r2,αc,c,r3,c−1,+∞:∈Ho, whereM are 1−(ξ−η)2>0. We conclude the following:η≠−ξ, M has at least one simple root on  and at least one simple root on ,if η≠ξ2, M has at least one simple root on  and at least one simple root on ,if η≠2ξ, M has at least one simple root on  and at least one simple root on .if  Finally, other computations show that M′(r1)=0 if η=−ξ, that M′(r2)=0 if η=ξ2 and that M′(r3)=0 if η=2ξ. So M has at least 6 real zeros (counting multiplicities) for each ∈Ho.By , the sadζ−r2)2ζ−r2(ζ−r+)(ζ4.2w satisfies w2=Q(ζ). This can be rewritten asξ,η∈R, we geta1,a2,a3,a4>0, we have↦,w) is a bijection from Lα to Rα+. This mapping is clearly differentiable, and therefore it is a diffeomorphism. Replacing ↦ in the right‐hand side of ↦. This implies the symmetry s=s. It remains to prove that ∈Lαl is mapped to a point ,w) lying in the upper half plane of the first sheet. The proof of this claim is splitted in the next two lemmas.Lemma 1Im(2ζ(ζ−c)(ζ−c−1)(ζ−1)(ζ+1)Q(ζ)1/2)=0 if and only if ζ∈R∪Σ1¯.We have We start with the proof of Proposition itten as−(ζ−α), there are eight solutions ζ∈C to f(ζ)=x. The claim follows if we show that all these solutions lie on R∪Σ1. First, note that the function f is positive on the real line, has poles at −1,αc,αc−1,1, and zeros at r1,r2,r3. Because −1<r1<αc<r2<αc−1<r3<1, the equation f(ζ)=x has at least six real solutions (counting multiplicities) for each x∈. Writing ζ=c−1+R1eit∈γ1, t∈, some computations show thatt↦f(c−1+R1eit) is even, positive, and decreases from f(c−1+R1) to 0 as t increases from 0 to θ1, which finishes the proof.▪Consider the function Lemma 2∈Rα+ such that w=Q(s)1/2  is in the first sheet). Then, ξ=ξ<0.Let s∈Σ1¯, which implies that the sign of s−c−(s−1)↦ that lead to simpler proofs of (b)–(f). We only sketch the proof of (e). First, we rewrite  has the same sign asζ∈R∪Σα, which proves part (e). We omit the proofs of parts (b), (c), (d), and (f).This finishes the proof of Proposition  rewrite  asξ+−(ζte (ξ+−ζ−(ζ+1)2Nz−2N. To do this, we follow an idea of Duits and KuijlaarsA. Then, we use this factorization to perform a first transformation Y↦X on the RH problem.To describe the behavior of 5.1A(z) defined in (z−z−) such that it is analytic in C∖, with an asymptotic behavior at ∞ given byA are in the columns of the following matrix:Λ(z)=diag(λ1(z),λ2(z)) is the matrix of eigenvalues. The matrix E(z) is analytic for z∈C∖, and satisfiesσ1=0110.The matrix fined in  has the 5.2Remark 4detY≡1. Note however that the Y↦X transformation does not preserve the unit determinant. Indeed, because detE(z)=−(1−α)(z−z+)(z−z−), we have detX(z)=(1−α)2(z−z+)(z−z−).By standard arguments,The first transformation of the RH problem diagonalizes the 2 × 2 upper right block of the jumps, and is defined byE given by X:C∖→C4×4 is analytic, where we recall that γ is a closed contour surrounding 0 once in the positive direction.(b)X are given byZ:=γ∩. Depending on γ, Z can be the empty set, a finite set, or an infinite set. If Z contains one or several intervals, then on these intervals the jumps areThe jumps for (c)z→∞, we have X(z)=(I4+O(z−1))zNE(z)0202z−NE(z). As z→z− or as z→z+, X(z)=O(1)E(z)0202E(z).As 6A obtained in A obtaineproperty  of RY imM associated to the eigenvalues and eigenvectors of A. This Riemann surface is of genus 0 and therefore there is a one‐to‐one map between it and the Riemann sphere .Now, we introduce the Riemann surface 6.1M is defined byz‐plane glued along . On the first sheet we require y=z+O(1) as z→∞, and on the second sheet we require y=−z+O(1) as z→∞. To shorten the notations, a point  lying on the Riemann surface will simply be denoted by z when there is no confusion, that is, we will omit the y‐coordinate. If we want to emphasize that the point  is on the j‐th sheet, j∈{1,2}, then we will use the notation z(j). With this convention, the two points at infinity are denoted by ∞(1) and ∞(2). The function y satisfiesλ1(z) and λ2(z) are defined on the z‐plane (see (M as follows:M with two simple poles at ∞(1) and ∞(2) and no other poles. Using ((2) and 1α(2), and because M has genus 0, λ has no other zeros. From , defined byM to the Riemann sphere. The first sheet of M is mapped by  is given byz is on the first sheet if |ζ|>1 and on the second sheet if |ζ|<1. By definition, the above function z(ζ) vanishes at ζ(0(1)) and ζ(0(2)). Because it has simple poles at ζ=0 and ζ=∞, and because z(ζ)=z+−z−4ζ+O(1) as ζ→∞, (z(ζ) and ζ(z) satisfyz∈M, Imz=0, z∉, follows the straight line segments , , , , the function ζ(z) increases from −∞ to +∞. In particular, we havec byThe Riemann surface ane (see ), and toatisfiesy(1α(2))=os. From , the matfined byζ=2z−(z++given byz=z++z−2+fined byζ=2z−(z++6.2w(j) on the j‐th sheet of M and z(k) on the k‐th sheet, we define RM(w(j),z(k)) bye1=10 and e2=01. Note that RM:M∗×M∗→C is scalar valued, with M∗=M∖{∞(1),∞(2)}. It is convenient for us to consider formal sums of points on M, which are called divisors in the literature. More precisely, a divisor D can be written in the formD≥0 if n1,…,nk≥0. The divisor of a nonzero meromorphic function f on M is defined byz1,…,zk1 are the zeros of f of multiplicities n1,…,nk1, respectively, and zk1+1,…,zk2 are the poles of f of order nk1+1,…,nk2, respectively. Given a divisor D, we define L(−D) as the vector space of meromorphic functions on M given byLN:=L(−DN) is the vector space of meromorphic functions on M, with poles at ∞(1) and ∞(2) only, such that the pole at ∞(1) is of order at most N−1, and the pole at ∞(2) is of order at most N. Similarly we define LN∗=L(−DN∗). From the Riemann–Roch theorem, we haveLemma 3(a)z↦RM∈LN for every w∈M∗,(b)w↦RM∈LN∗ for every z∈M∗,(c)RM is a reproducing kernel for LN in the sense thatf∈LN, where γM is a closed contour surrounding once 0(1) and 0(2) on the Riemann surface M in the positive direction (in particular γM visits both sheets).We haveFor RM and RX given by   and at most N+1 at ∞(2). Therefore, we have shown thatw∈M∗ a linear combination of the functions fj, so belong to L(−(DN+∞(1)+∞(2))) as a function of z. By definitions of RM and RX, the numerator vanishes for z=w(1) and for z=w(2). Thus, the division by z−w in ((1) and ∞(2) by one, and therefore z↦RM∈LN as claimed in part (a). Now, we turn to the proof of part (b). First, we note thatdetY≡1, by using condition (c) of the RH problem for X, we havegj are also analytic in M∗. On the other hand, by using the asymptotics Y(w)=I4+O(w−1) as w→∞ together with the fact that detY≡1, we can obtain asymptotics for X−1(w) as w→∞ using  in (z=w (on any sheet). Finally, let us take P(w)=p(w)e1T=p(w)10 in =11=e1T+e2T, it givesek, we obtain(1) and γ(2) for the projections of γ on the first and second sheets of M, respectively. Using  and 0(2) on M in the positive direction, and thusz∈M∗. Let us now take P(w)=p(w)e2T=p(w)01 in (w∈M↦11+α(λ(w)−α2−w) on M. By proceeding in a similar way as for =p1(z)+p2(z)(λ(z)−α2−z)withp1,p2twopolynomialsofdegree≤N−1}. Because z↦λ−α2−z has a simple pole at ∞(2) (and no other poles), we conclude that L⊆LN. Note also that dimL=dimLN=2N, and thus we have L=LN. This finishes the proof.▪Using the definitions of given by  and 68)RM and RXgiven by ) with pry =∑y =∑z(2),andgj(w)=y=∑(w)10 in , with p y. Using , the abo(w)01 in , and notand thusp(z)=12πi6.3z=z(ζ) and w=w(ω) byz (resp. w) is on the first sheet if |ζ|>1 (resp. |ω|>1), and on the second sheet if |ζ|<1 (resp. |ω|<1). We define RU in terms of RM as follows:Proposition 6W and c be defined as in  is meromorphic on M, with a pole of order at most N−1 at ∞(1) and a pole of order at most N at ∞(2). Because z(0)=∞(2) and z(∞)=∞(1), we conclude that for each ω∈C, the function ζ↦RM(w(ω),z(ζ)) is meromorphic on C∪{∞}, with a pole of order at most N−1 at ∞ and another pole of order at most N at 0. Therefore, for each ω∈C, the function ζ↦RU is a polynomial of degree at most 2N−1. From part (b) of Lemma ζ∈C, the function ω↦RU is a polynomial of degree at most 2N−1. So we have proved that RU is a bivariate polynomial of degree ≤2N−1 in both ω and ζ.From part (a) of Lemma ω(0(1))=c−1, ω(0(2))=c, (∂ωw)(c−1)>0, and (∂ωw)(c)<0. In particular, the map w↦ω(w) is conformal in small neighborhoods of 0(1) and 0(2). Because conformal maps preserve orientation, the curve γM which surrounds both 0(1) and 0(2) once in the positive direction, is mapped by w↦ω(w) onto a curve γC on the complex plane, which surrounds c and c−1 once in the positive direction. Furthermore, because ω(∞(2))=0, the curve γC does not surround 0. By changing variables ↦ in ) is meromorphic on the Riemann sphere, with a pole of degree at most N at ζ=0 and a pole of degree at most N−1 at ζ=∞. In other words, ζ↦ζNf(z(ζ))=:p(ζ) is a polynomial of degree at most 2N−1. By multiplying the above equality by ζN, we thus have▪Now, we turn to the proof of . It can  in , and by  in , 76), a, aω(0(1) in , we obtatituting  in the aRU in terms of the solution U to the 2 × 2 RH problem presented in Section Proposition 7RU defined by  and E(z)−1 asRX↦RM transformation given by (γM is a closed contour surrounding once 0(1) and 0(2) on M in the positive direction. By performing the change of variables w=w(ω) and z=z(ζ) as in  is defined for ω,ζ∈C and εx∈{0,1} byNow, using the results of Sections given in  to rewri we haveK−1=11(z)−1 asE(w)=11λ−1=11iven in  as in  and 87)x∈{1,…,2Nζ) as in  and 77)x∈{1,…,2Nζ) as in , and alsgiven by , we getng again  and (77)finition  of W, weiven by x∈{1,…,2N7Pj, j=1,2,3, defined in , j=1,2,3, in terms of RU. In the rest of this section, we follow Ref. η≤ξ2≤0 of the liquid region for the proof of Theorem This section is about the lozenge probabilities fined in . In Sect7.1q and q∼ are given byPj, j=1,2,3, are obtained via a series of lemmas. Let us first recall that the paths pj:{0,1,…,4N}→Z+12, j=0,…,2N−1 are defined in ..(98)K|>1 whenever ζ∈γ∼C and ω∈γC.For The next lemma establishes a double integral formula for the expectation value of the height function.X be the random variable that counts the number of paths going through the point , x∼,y∼∈{0,1,…,4N}. Because X∈{0,1}, we have P=1)=E). Also, note that the identity =1)=K. Thus, by definition ∪C, where C denotes a circle oriented positively centered at a of radius r. We see from |→+∞ as ζ tends to c or c−1. Thus, by choosing r sufficiently small, we can make sure that γ∼C lies in the interior region of γC, and thatε>0. Therefore, uniformly for ζ∈γ∼C and ω∈γC, we haveεy=0 follows after combining , , X. By . By , for εx,,formula  follows ,formula  with E be a smooth parameterization of a curve σ, satisfying ζ′(t)≠0 for all t∈. σ is a trajectory of the quadratic differential Q(ζ)dζ2 if Q(ζ(t))ζ′(t)2<0 for every t∈, and an orthogonal trajectory if Q(ζ(t))ζ′(t)2>0 for every t∈. σ is critical if it contains a zero or a pole of Q. Note that these definitions are independent of the choice of the parameterization.Let r+ and r− are simple zeros of Q, there are three critical trajectories  emanating from each of the points r±. Recall the definitions of γ0,γα,γ1,Σ0,Σα, and Σ1 given in Section Lemma 8Σ0¯, Σα¯, and Σ1¯ are three critical trajectories of Q(ζ)dζ2 joining r− with r+, and γ0∖Σ0, γα∖Σα, and γ1∖Σ1 are each the union of two critical orthogonal trajectories of Q(ζ)dζ2. An illustration is shown in Figure The arcs Because t↦ζ=ζ(t)=c−1+R1eit, t∈, be a parameterization of γ1. Writing r±=c−1+R1e±iθ1 with θ1∈, and noting that ζ′=iReit, we have(ζ−r2)=2R1eit2cost2, andt∈, positive for t∈∪, and zero for t=−π,−θ1,θ1,π. We conclude that Σ1¯ is a critical trajectory and that γ1∖Σ1 is the union of two orthogonal critical trajectories. The statement about Σα¯, γα∖Σα, Σ0¯, γ0∖Σ0 can be proved a similar way, and we provide less details. For ζ=ζ(t)=R0eit, t∈, after long but straightforward computations, we obtainΣ0¯ is a critical trajectory and that γ0∖Σ0 is the union of two orthogonal critical trajectories. For ζ=ζ(t)=αc−1+Rαeit, t∈, we obtainΣα¯ is a critical trajectory and that γα∖Σα is the union of two orthogonal critical trajectories. This finishes the proof.▪Let )2.Using , we show, we getQ(ζ)(ζ′)2ions in (Q(ζ)(ζ′)2e obtainQ(ζ)(ζ′)28.3g′ can be expressed asQ(ζ)1/2. To obtain a g‐function with the desired properties (a), (b), and (c), it turns out that the branch cut needs to be taken along the critical trajectory Σ1 (as in Section Definition 3Q1/2 as(ζ−r+)(ζ−r−) is taken on Σ1 such thatWe define As can be seen in , g′ can Q1/2.Definition 4ϕ:C∖→C by∈iR for ζ∈Σ1. Therefore, Reϕ is single‐valued and continuous in C∖{0,αc,αc−1,c,c−1}, and Reϕ is also harmonic in C∖. Finally, by combining Definition We first state some basic properties of ϕ. By –154), Q, Q154), Nϕ of Reϕ. This will be useful in Section g‐function. Let us defineLemma 9Nϕ divides the complex plane in three regions. The sign of Reϕ in these regions is as shown in Figure We haveIn the rest of this subsection, we determine the zero set ϕ′=Qα1/2 changes sign when it crosses each of the nine points r1, 0, αc, r2, αc−1, c, r3, c−1, c−1+R1. Because ϕ′(ζ)=2−1ζ−1+O(ζ−2) as ζ→∞, we have ϕ′>0 on the intervalsϕ′<0 on the intervalsReϕ admits no other zeros on ∪∪. On the intervals  and , Reϕ admits a local minimum at r1 and r3, respectively, and on the interval , it admits a local maximum at r2. Thus, (Reϕ is strictly monotone on each of the curves (γ0∖Σ0)∩C+, (γα∖Σα)∩C+, and (γ1∖Σ1)∩C+. The expressions (Reϕ is strictly increasing on (γ0∖Σ0)∩C+ oriented from r+ to r1, strictly increasing on (γα∖Σα)∩C+ oriented from r+ to r3, and strictly decreasing on (γ1∖Σ1)∩C+ oriented from r+ to r2. In particular this proves , , (167)Nϕressions , togethe. Thus, (Reϕ(r1)>0how thatNϕ∩R=≠0 for ζ∈Σ1, we must have σ∩Σ1=∅. Also, in view of (C∖(R∪Σ1), and the max/min principle for harmonic functions would then imply that Reϕ in constant on the whole bounded region delimited by σ. By .▪8.4Definition 5Σ1=supp(μ) is given by 1/2 denotes the limit of Q(ξ)1/2 as ξ→ζ∈Σ1 with ξ in the exterior of the circle γ1.We define the measure μ bygiven by , and is Proposition 13The measure μ defined in  is a pro∫Σ1dμ by residue calculation. Because Q+=−Q−, we haveC is a closed curve surrounding Σ1 once in the positive direction, but not surrounding any of the poles of Q. By deforming C into another contour C∼ surrounding 0, αc, αc−1, c, and c−1, we pick up some residues:P={0,αc,αc−1,c,c−1}. By combining Definition Q(ζ)1/2=12ζ+O(ζ−2) as ζ→∞, we find1. Let ζ(t)=c−1+R1eit, −θ1<t<θ1, be a parameterization of Σ1. Consider the functionQ(ζ(t))(ζ′(t))2<0 for t∈ by Lemma t=−θ1 and equals 1 for t=θ1 by (▪We compute given by1πiQ−(ζ(tfunctiont↦∫r−ζ(t) we find∫Σ1dμ(ζ)=given by1πiQ− has a branch cut along =log|ζ|+O(ζ−1), as ζ→+∞.The ℓ∈C byWe define the variational constant g is consistent with (g satisfies (Proposition 14g and ϕ are related byg‐function satisfies the properties (a), (b), and (c) listed at the beginning of Section γC=γ1.The functions we have10g+(ζ)+g−The next proposition shows, among other things, that Definition ent with , and thaatisfies .Proposiζ∈C∖Σ1, we haveC is a closed curve surrounding Σ1 once in the positive direction, but not surrounding any of the poles of Q, and not surrounding ζ. By deforming C into another contour C∼ surrounding 0, αc, αc−1, c, c−1, and ζ, we obtainP={0,αc,αc−1,c,c−1}. By deforming C∼ to ∞, noting that Q(ξ)1/2=O(ξ−1) as ξ→∞, the integral on the right‐hand side of =0. Then,  is fulfilled. For ζ∈ is also fulfilled. For ζ∈γ1∖Σ1¯, by (ζ∈Σ1 we haver− to r+. So (▪We first prove . For a fe obtain∫Σ1dμ(ξ)ζ side of , and we /2.Using  and ϕ′=Q0. Then,  follows 0. Then, . Becausen Σ1, by  we haveom which  and T:C∖γ1→C2×2 is analytic.(b)T are given byBy using , 140), , 140), a(c)ζ→∞, we have T(ζ)=I+O(ζ−1). As ζ tends to r+ or r−, T(ζ) remains bounded.As T will be important for the saddle point analysis of Section Proposition 15T(ζ)=O(N1/6) and T−1(ζ)=O(N1/6) as N→∞, uniformly for ζ∈C∖γ1. In addition, for every δ>0 fixed, we have T(ζ)=O(1) and T−1(ζ)=O(1) as N→∞ uniformly forWe have The following estimates for The rest of this section is devoted to the proof of Proposition 9.2ζ∈Σ1, the jumps for T can be factorized as follows:ϕ+(ζ)+ϕ−(ζ)=0 for ζ∈Σ1. We define the lenses γ+ and γ− byT↦S transformation is given by S(ζ)=T(ζ)W(ζ), whereS satisfies the following RH problem.Note that for (a)S:C∖(γ1∪γ+∪γ−)→C2×2 is analytic.(b)S are given byThe jumps for (c)ζ→∞, we have S(ζ)=I+O(ζ−1). As ζ tends to r+ or r−, S(ζ) remains bounded.As 9.3S in different regions of the complex plane. By Lemma Reϕ(ζ)>0 for ζ∈γ+∪γ−, Reϕ(ζ)<0 for ζ∈γ1∖Σ1¯, and Reϕ(ζ)=0 for ζ∈Σ1. So the jumps for S on γ+∪γ−∪(γ1∖Σ1¯) are exponentially close to the identity matrix matrix as N→∞, uniformly outside fixed neighborhoods of r− and r+. By ignoring these jumps, we are left with the following RH problem, whose solution is denoted as P(∞). We will show in Section P(∞) is a good approximation to S away from r+ and r−.In this subsection, we find good approximations to (a)P(∞):C∖Σ1¯→C2×2 is analytic.(b)P(∞) are given byThe jumps for (c)ζ→∞, we have P(∞)(ζ)=I+O(ζ−1). As ζ→ζ∗∈{r+,r−}, P(∞)(ζ)=O((ζ−ζ∗)−1/4).As P(∞)(ζ) as ζ→ζ∗∈{r+,r−} has been added to ensure existence of a solution. This RH problem is independent of N, and its unique solution is given bya(ζ):=(ζ−r+ζ−r−)1/4 is analytic in C∖Σ1 and such that a(ζ)∼1 as ζ→∞.The condition on the behavior of P(∞) is not a good approximation to S in small neighborhoods of r+,r−; this can be seen from the behaviorsδ>0 in Proposition Dr+ and Dr− be small open disks of radius δ/2 centered at r+ and r−, respectively. We now construct local approximations P(r+) and P(r−)  to S in Dr+ and Dr−, respectively. We require P(r±) to satisfy the same jumps as S inside Dr±, to remain bounded as ζ→r±, and to satisfy the matching conditionζ∈∂Dr±. The density of μ vanishes like a square root at the endpoints r+ and r−, and therefore P(r±) can be built in terms of Airy functions.z∈Dr±.Note that 9.4S↦R of the steepest descent is defined byS and P(r±) satisfy the same jumps inside Dr±, R is analytic inside (Dr+∖{r+})∪(Dr−∖{r−}). Furthermore, S and P(r±) remain bounded near r±, so the singularities of R at r± are removable. We conclude that R is analytic inR−−1R+ are O(N−1) on ∂Dr+∪∂Dr+, and by Lemma R−−1R+=O(e−cN) on (γ1∪γ+∪γ−)∖(Dr+∪Dr−) for a certain c>0. It follows by standard theoryR and R−1 remain bounded as N→∞.The final transformation ∪∂Dr−.By , the jumined by16R(ζ)=S(ζlytic inC∖(γ1∪γ+∪W(ζ) and W(ζ)−1 are bounded as N→+∞, uniformly for ζ∈C. Proposition Inverting the transformations  and 196196, we g10N→+∞ will be in the form e2N−Φ), for a certain function Φ which is described below. The analytic continuation of Φ to the second sheet of Rα is denoted Ψ—it will also play a role in the saddle point analysis and is presented below.In Section The content of this section is a preparation for the saddle point analysis of Section 10.1Definition 7∈H and ζ∈C∖, we define Φ and Ψ byFor ave used  and (179and Ψ byΦ(ζ)=Φ and ζ↦e±2NΨ have no jumps along ∈H, ReΦ and ReΨ are harmonic on C∖, and well defined and continuous on C∖{0,αc,αc−1,c,c−1}. For ∈Ho, we note the following basic properties of Φ:In the formulas that will be used in Section Φ′ and Ψ′. For the saddle point analysis, it will be important to know: (1) the sign of |s−c−1|−R1 and (2) whether Φ′(s)=0 or Ψ′(s)=0. We summarize the different cases in the next lemma.Lemma 10∈Lα and s=s. Then, we have(a)Φ′(s)=0 and |s−c−1|<R1 if and only if ξ<0 and η<ξ2,(b)Φ′(s)=0 and |s−c−1|>R1 if and only if ξ<0 and η>ξ2,(c)Ψ′(s)=0 and |s−c−1|<R1 if and only if ξ>0 and η>ξ2,(d)Ψ′(s)=0 and |s−c−1|>R1 if and only if ξ>0 and η<ξ2,(e)|s−c−1|=R1 if and only if ξ=0 or η=ξ2.Let Because the saddle points are the solutions to , it foll▪This is an immediate consequence of Propositions 10.2η≤ξ2<0. We have represented NΦ for different values of  in Figures Φ′ and Ψ′. From , , and . This determines the location of six saddles. The remaining two are s and s¯, and we already know from Lemma Φ′(s)=0=Φ′(s¯). Therefore, Φ′≠0 on ∪. Because Φ′(ζ)∈R for ζ∈R∖{0,αc,αc−1,c,c−1}, this implies by ∩C+ is either the empty set or a singleton.We show with the next two lemmas that the set ζ∈C∖{0,αc,αc−1,c,c−1}, we define the following functions:Lemma 11(Σα∪Σ1¯)∩C+ from left to right, then(1)Ref1 is strictly decreasing on Σα∩C+ and constant on Σ1∩C+,(2)Ref2 is constant on Σα∩C+ and strictly decreasing on Σ1∩C+,(3)Ref3 is strictly decreasing.If ζ moves along For ddtRef1(αc−1+Rαe−it) has the same sign as sint. In particular, Ref1(ζ) is strictly decreasing along Σα∩C+ as ζ moves from left to right. Another (and simpler) computation givesRef1 is constant on Σ1. The proofs for f2 and f3 are similar, so we omit them.▪A long and tedious computation shows that Corollary 3η≤ξ2<0, the function ζ↦ReΦ(ζ) is strictly decreasing as ζ moves along (Σα∪Σ1¯)∩C+ from left to right.For Reϕ=0 on Σα∪Σ1. Therefore, from the expression  for the open and bounded region delimited by σ.Φ′(s)=0, there are four curves {Γj}j=14 emanating from s that belongs to NΦ. By Corollary NΦ∩(Σα∪Σ1¯)∩C+ is either the empty set or a singleton, so at least three of the Γj's, say Γ1,Γ2,Γ3, do not intersect (Σα∪Σ1¯)∩C+. The curves Γj, j=1,2,3 cannot lie entirely in C+; otherwise the max/min principle for harmonic functions would imply that ReΦ is constant within the region int(Γj). Therefore, Γj, j=1,2,3 have to intersect R. Note that Φ(ζ)¯=Φ(ζ¯) implies that NΦ is symmetric with respect to R. In particular, the curves Γj, j=1,2,3 join s with s¯. The next lemma states that Γ4 is not contained in the region int(Σα∪Σ1¯).Lemma 12NΦ∩(Σα∪Σ1¯)∩C+ is a singleton.Because 4 lies entirely in int(Σα∪Σ1¯), and denote pj for the intersection point of Γj with R. We assume without loss of generality that p1<p2<p3<p4. There is at most one pj inside each of the intervalspj's, and each of them leads to a contradiction. Let us treat the caseRe(Φ(ζ)−Φ(s)) changes sign as ζ crosses NΦ∖{s,s¯}, by ∩NΦ=∅, and σ2 is a closed curve surrounding either c or c−1, such that int(σ2)∩NΦ=∅. Because NΦ intersects both  and  exactly once, σ1 surrounds αc and σ2 surrounds c−1. Then, the max/min principle implies that ReΦ is constant on int(Γ3∪Γ4¯)∖int(σ2), which is a contradiction. The four other cases than →+∞ as ζ→∞, so Γ4 intersects the real line, and then by symmetry ends at s¯. So each of the Γj's intersects R. We denote pj for the intersection point of Γj with R, and choose the ordering such that p1<p2<p3. We recall that Re(Φ(ζ)−Φ(s)) is harmonic for ζ∈C∖ and changes sign as ζ crosses NΦ∖{s,s¯}. Therefore, by (int(Γ1∪Γ2¯) must contain at least one of the singularities αc and αc−1, and int(Γ2∪Γ3¯) must contain at least one of the singularities c and c−1. There are still quite a few cases that can occur. The figures provide a fairly good overview (though not complete) of what can happen:1.αc,αc−1,c∈int(Γ1∪Γ2¯), c−1∈int(Γ2∪Γ3¯). In Figure 2.αc,αc−1∈int(Γ1∪Γ2¯) and c,c−1∈int(Γ2∪Γ3¯). In Figures 3.αc−1∈int(Γ1∪Γ2¯) and c∈int(Γ2∪Γ3¯). In Figures  Furthermore, Γ4 intersects both Σ1 and  in Figure Σα and  in Figure Σα and  in Figure γζ∗ and γω∗ as described in the following proposition. These contours are illustrated for two different situations in Figures Proposition 16∈Lα with η<ξ2<0. There exist contours γζ∗ and γω∗ such thatγω∗⊂int(Σα∪Σ1¯), it surrounds αc and αc−1, and it goes through s and s¯ in such a way thatγζ∗⊂int(γ1), surrounds c and c−1, and it goes through s and s¯ in such a way thatLet Lemma now from  that ReΦfore, by , the regη=ξ2, we know from Proposition s lies on γ1∖Σ1¯. For the saddle point analysis, we will need γζ∗ lying inside γ1 (not necessarily strictly inside). To prove existence of such a contour γζ∗, we need to know that ReΦ(ζ)−ReΦ(s) is strictly negative for ζ∈γ1∖Σ1¯ .Lemma 13η=ξ2<0. For ζ∈γ1∖(Σ1¯∪{s})∩C+, we have ReΦ(ζ)<ReΦ(s).Let If ζ=c−1+R1eit. For t∈, we havea1,a2 are given by a1=α2+(1−α)1−α+α22(1−α) and a2=2−3α+2α2+α32(1−α)1−α+α2 and satisfy a1>a2>1. The expression , such that Re(Φ′(ζ)dζ)=0, and this must be s. This implies that ReΦ(ζ)−ReΦ(s) is of constant sign on γ1∖(Σ1¯∪{s})∩C+. By (Re(Φ′(ζ)dζ)>0 at t=θ1 , so the claim is proved.▪Let  we haveRe(Φ′(ζ)d we haveRe(Φ′(ζ)dγζ∗ and γω∗ as described in Proposition Proposition 17∈Lα with η=ξ2<0. There exist contours γζ∗ and γω∗ such thatγω∗⊂int(Σα∪Σ1¯), it surrounds αc and αc−1, and it goes through s and s¯ in such a way thatγζ∗⊂int(γ1)¯, surrounds c and c−1, and it goes through s and s¯ in such a way thatLet Therefore, we can find contours 11 in the lower left part of the liquid region, that is for ∈Lα∩{η≤ξ2≤0}. We divide the proof in three subcases: η≤ξ2<0, η<ξ2=0, and η=ξ=0.Remark 6 lies in the other quadrants of the liquid region. Note however that this is not needed, thanks to the symmetries of Section By adapting the analysis of this section and of Section In this section, we prove Proposition 11.1I is defined in  another formula for I in terms of the second column of U. Therefore, the choice of γC will matter. To be able to use the steepest descent of Section γC=γ1. Recall that T is expressed in terms of U via  is uniformly bounded as ζ and ω stay bounded away from r+ and r−. We will need the analytic continuation in ω of R∼T from the interior of γ1 to the bounded region delimited by Σ1∪Σα¯ , and by (Reϕ(ω)<0 for ω∈int((γ1∖Σ1)∪Σα), so R∼T,a remains bounded as N→+∞, uniformly for ζ and ω bounded away from r+ and r−, as long as ω∈int(Σ1¯∪Σα). Our next goal is to prove the following.Proposition 18 be coordinates inside the hexagon, such that ξ:=xN−1 and η:=yN−1 satisfy ∈Lα with η≤ξ2<0. Take γζ∗ and γω∗ as in Proposition η<ξ2, and as in Proposition η=ξ2 I is defiof U via , and def, and by  it is giRemark 7γζ∗ intersects γ1∖Σ1¯ whenever η=ξ2. We do not indicate whether we take the + or − boundary values in the integrand of  and R∼−T denote the limits of R∼T as ω′→ω from the interior and exterior of γ1, respectively.where Remark 8x,y∈{1,2,…,2N−1}, we definem appear in the integrand of (q and q∼ are defined in (H satisfies the conditions stated in Proposition g(ω) is bounded for ω in compact subsets and satisfies g(ω)∼log(ω) as ω→∞. Therefore, the following properties hold:(i)ζ↦m is analytic in C∖{ω,0,c,c−1},The function (ii)ω↦m is analytic in (C∪{∞})∖.The function  The statement that ω↦m is analytic at ∞ deserves a little computation: because x,y∈{1,2,…,2N−1}, we have m=O(ω−1−2N+N−y+x)=O(ω−2) as ω→∞.For grand of . We recafined in , that H η<ξ2, Proposition γζ lies strictly inside γ1, so in this case we can (and do) take γζ=γζ∗ in ). After a small computation, we find that this residue is the first term on the right‐hand side of ∈Lα∩{η≤ξ2<0}, and define ξN:=xN/N−1 and ηN:=yN/N−1. By ∈Lα∩{η≤ξ2<0} for all large enough N  in , we getsN=s, ΦN(ζ):=Φ, and the contours γζ∗ and γω∗ also depend on N, even though this is not indicated in the notation. Because γζ∗ and γω∗ do not pass through r+ and r−, Proposition C1>0, and where Dε is the union of two small disks of radii ε>0 surrounding s and s¯. Because sN and sN¯ are simple zeros of ΦN′, we have the estimatesC2>0. Therefore, the left‐most term in  in  by 2∫γich give .▪11.2∈L, such that η′<ξ′2<0. In this case, the set NΦ contains four curves emanating from s: three of these curves, namely, Γ1, Γ2, and Γ3, lie in int(Σ1∪Σα¯), the other curve Γ4 intersects once Σ1∪Σα¯. Denote pj for the intersection of Γj with R, and recall that the ordering for Γ1, Γ2, and Γ3 is such that p1<p2<p3.Let us briefly recall first the situation for → with η<0  tends to a point s=s lying on Σ1. In this limit, both Γ3 and Γ4 tend to the arc1 tends to Σ1∖Σs. Thus, the case ξ=0 gives less freedom for the contour deformations and the saddle point analysis is more involved. To handle this case, we need information about both NΦ and NΨ, whereNΦ and NΨ are represented in Figure Lemma 14ξ=0, we have Σ1⊂NΦ and Σ1⊂NΨ.For As Reϕ(ζ)=0 for ζ∈Σ1, by Definition ζ∈Σ1.▪Because γζ∗ and γω∗ as follows ¯, is such that (Σ1∖Σs)⊂γω∗, surrounds αc and αc−1, and it satisfiesγζ∗⊂int(γ1)¯, is such that Σs⊂γζ∗, surrounds c and c−1, and it satisfiesWe choose {∈Lα∩{η<ξ2=0}, and define ξN:=xN/N−1 and ηN:=yN/N−1. By Lemma ∈Lα satisfies ηN<0 and ξN=0 for all N. The proof of Proposition γζ∗ and γω∗, and as in , Φ(ζ)=Φ, and the contours γζ∗ and γω∗ depend on N. We also take the + boundary value in (ω∈γ1. Because ReΦ(ζ)=ReΦ(s) for all ζ∈Σs and ReΦ(ω)=ReΦ(s) for all ω∈Σ1∖Σs¯, we need additional deformation of contours.Let tisfying  with =Ψ−(ζ) and the jumps for T =Φ∓(ζ) for ζ∈Σ1, and because Σ1⊂NΦ∩Nψ, the signs ofζ∈Σ1, provided ε=ε(ζ)∈R is small enough (ε nonnecessarily positive), see also the signs around Σ1 in Figure ReΨ(ζ)<ReΨ(s) for ζ∈Σs,out. For the first term in (e2N(Φ+(ζ)−2ϕ+(ζ)), and by Definition Ψ=Φ−2ϕ. Therefore, we deform Σs inward to Σs,in, and this contour is chosen such that ReΨ(ζ)<ReΨ(s) for ζ∈Σs,in, see Figure We first treat the contour deformations in the ζ‐variable. Recall the definition  of R∼T,aps for T  to obtais for T (e2NΦ(ζ)TT(s for T (e2NΦ(ζ)T(Σ1∖Σs outward to (γ1∖Σs)ext, see Figure ReΨ(ω)>ReΨ(s) for ω∈(γ1∖Σs)ext. Because Φ+(ω)=Ψ−(ω) on Σ1, the exponential factor of the integrand is e−2NΨ(ω) there. Also, for ω∈Σ1∖Σs, by (e−4Nϕ(ω) remains bounded for ω∈(γ1∖Σs)ext.In the ω‐variable, we simply analytically continue the integrand and deform 1∖Σs, by  we haves,s¯, and that ω stays away from s,s¯,r+,r−. By a similar analysis as the one done in (s or s¯ is O(N−12) as N→+∞. When ω is close to r±, we know by Proposition T−1(ω)=O(N1/6). Because Φ′(r±)≠0≠Ψ′(r±), the contribution to =r+, s¯=r−, and Φ=−Ψ=ϕ =∈Lα with η′<0, part of contour γω∗ lies in the region int(Σα∪Γ1¯), see Figures η′→η=0, Γ1 tends to Σα, so we need additional contour deformations to handle this case. Consider the contours γζ∗:=γ1 and γω∗=γα. By Lemma At the center of the hexagon, we have {=}N≥1, so that ξN:=xN/N−1=0 and ηN:=yN/N−1=0 for all N. In the same way as done in Proposition ω∈Σα.For simplicity, we consider the sequence position I is given by the second line of (Σα outward so that Reϕ(ω)>0, and for the first term, we deform Σα inward so that Reϕ(ω)<0.For  line of , and thur+ and r−. For ζ and ω close to r±, by Proposition T(ζ)=O(N1/6) and T−1(ω)=O(N1/6). The contribution to (r+ and r− is thus bounded byC1,C2,C3>0 and for all large enough N. This finishes the proof of Proposition On the deformed contours, the integrand is uniformly exponentially small, as long as ζ and ω are bounded away from ution to  when ζ a


Given a finite set of European call option prices on a single underlying, we want to know when there is a market model that is consistent with these prices. In contrast to previous studies, we allow models where the underlying trades at a bid–ask spread. The main question then is how large (in terms of a deterministic bound) this spread must be to explain the given prices. We fully solve this problem in the case of a single maturity, and give several partial results for multiple maturities. For the latter, our main mathematical tool is a recent result on approximation by peacocks. Put differently, we want to detect arbitrage in given prices. We do not consider continuous call price surfaces, but restrict to the  case of finitely many strikes and maturities. Therefore, consider a financial asset with finitely many European call options written on it. In a frictionless setting, the consistency problem is well understood: Carr and Madan . To define the payoff of the call options, we use an arbitrary reference price process that evolves within the bid–ask spread. We show =B(t)−1 for the time zero price of a zero‐coupon bond maturing at t, and kt,i=D(t)Kt,i for the discounted strikes. The symbol Ct(K) denotes a call option with maturity t and strike K.Our time index set will be T=1 bid and ask are S_1=$99, respectively, S¯1=$101. Then, the agent might wish to exercise the option to obtain a security for $99 instead of $100, or he may forfeit the option on the grounds that spending $100 would earn him a position whose liquidation value is only $99. The exercise decision cannot be nailed down without making further assumptions. In practice, the quoted ticker price of the underlying is the last price at which an actual transaction has occurred. This price then triggers cash‐settled options. However, this approach is not feasible in our setup, which does not include an order book.In the presence of a bid–ask spread on the underlying, it is not obvious how to define the payoff of an option; this issue seems to have been somewhat neglected in the transaction costs literature. Indeed, suppose that an agent holds a call option with strike $100, and that at maturity SC. This process evolves within the bid–ask spread. It is not a traded asset by itself, but just serves to fix the call option payoff (StC−K)+ for strike K and maturity t. This payoff is immediately transferred to the bank account without any costs.Definition 2.1model consists of a finite probability space  with a discrete filtration (Ft)t∈T and three adapted stochastic processes S_, S¯, and SC, satisfyingA In the literature on option pricing under transaction costs, it is usually assumed that the bid and ask of the underlying are constant multiples of a mid‐price (often assumed to be geometric Brownian motion). This mid‐price is then used as trigger to decide whether an option should be exercised, followed by physical delivery Bichuch, . The assS_t and S¯t denote the bid, respectively, ask price of the underlying at time t. Note that, in our terminology, the initial bid and ask are part of the given prices ), but allow any adapted process inside the bid–ask spread. We now give a definition for consistency of option prices, allowing for (arbitrarily large) bid–ask spreads on both the underlying and the options.Definition 2.2consistent with the absence of arbitrage, if there is a model (in the sense of Definition E∈,1≤i≤Nt,t∈T∗.(S∗)t∈T such that S_t≤St∗≤S¯t for t∈T and such that (D(t)St∗)t∈T is a P‐martingale(Ft)t∈T. The pair  is called a consistent price system.There is a process The prices As for the reference price process S∗ is also called a shadow price. According to Kabanov and Stricker E1{s>0})s∈T as bid price process (and similarly for the ask price), and (B(s)E)s∈T as the process in the second part of Definition The process Stricker   is “too expensive,” and without frictions, buying C2(k)−C1(k) would be an arbitrage opportunity (upon selling one unit of stock if C1(k) expires in the money). In particular, the first condition from Corollary 4.2 in Davis and Hobson −C1(k): the short call −C1(k) finishes in the money with payoff −(c+1). This cannot be compensated by going short in the stock, because its bid price stays at 2. The payoff at time t=2 of this strategy, with shorting the stock at time t=1, isBut with spreads we can choose As mentioned in Section Definition 2.4ε≥0. Then the prices Let Our focus will thus be on a stronger notion of consistency, where the discounted spread on the underlying is bounded. Hence, our goal becomes to determine how large a spread is needed to explain given option prices.SC, made for tractability, and makes sense given the actual size of market prices and spreads . With the same justification, in our main results on ε‐consistency we will assume that all discounted strikes kt,i are larger than ε. If ε=0 and the bid and ask prices in The bound SC. However, it is not hard to show that choosing SC=12(S_+S¯) yields almost the same notion of ε‐consistency.Proposition 2.5ε≥0 and assume that we are interested in arbitrage‐ free models where, in addition to the requirements of Definition arithmetically ε‐consistent. For ε≥0, the prices are arithmetically 2ε‐consistent if and only if they are ε‐consistent.Let As mentioned above, we do not insist on any specific definition of the reference price S_t,S¯t,StC,St∗. We define new bid and ask prices S_t′:=StC∧St∗ and S¯t′:=StC∨St∗. Then S¯t′−S_t′≤B(t)ε. Therefore, the model consisting of S_t′,S¯t′,StC,St∗ is ε‐consistent. Conversely, assume that the given prices are ε‐consistent. Then there exist processes SC and S∗ on a probability space  such that |StC−St∗|≤B(t)ε a.s. We then simply set S_t=StC−B(t)ε and S¯t=StC+B(t)ε, and have thus constructed an arithmetically 2ε‐consistent model.□First, assume that there exists an arithmetically 2ε‐consistent model with corresponding stochastic processes p∈ and p≠12.Note that the statement of Proposition (D(t)StC)t∈T does not have to be a martingale, as SC is not traded on the market. The option prices give us some information about the marginals of the process SC, though. On the other hand, the process (D(t)St∗)t∈T has to be a martingale, but we have no information about its marginals, except that |St∗−StC|≤εB(t). This impliesW∞ denotes the infinity Wasserstein distance, and L the law of a random variable. The distance W∞ is defined on M, the set of probability measures on R with finite mean, by and random pairs  with marginals . See Gerhold and Gülüm ≤ε is equivalent to the existence of a probability space with random variables X′∼LX, Y′∼LY such that |X′−Y′|≤ε a.s. t∈T∗ in M is called a peacock, if μs≤cμt for all s≤t in T∗ . These notions are useful for constructing models for ε‐consistent prices, as made explicit by the following lemma. As is evident from its proof, the sequence (μt) consists of the marginals of a (discounted) reference price, whereas (νt) gives the marginals of a martingale within the bid–ask spread. The proof uses a coupling result from our companion paper t∈T∗ and (νt)t∈T∗ in M such that:(i)Rμt∈ for all t∈T∗ and i∈{1,⋯,Nt}, and μt)=1 for t∈T∗,(ii)(νt)t∈T∗ is a peacock and its mean satisfies EνT∈, and(iii)W∞≤ε for all t∈T∗.For For (μt)t∈T∗ and (νt)t∈T∗ be as above. Recall that Strassen's ((S∼t)t∈T such that νt is the law of S∼t for t∈T∗.Let rassen's  Theorem M^ and S^C such that M^t∼νt,D(t)S^tC∼μt, and |M^t−D(t)S^tC|≤ε for t∈T∗. As in the the proof of Theorem 9.2 in Gerhold and Gülüm (finite probability space with these properties. The sufficiency statement now easily follows from Lemma 9.1 in Gerhold and Gülüm ((Sˇt)t∈T and (StC)t∈T∗ satisfyingSˇ is a martingale,Sˇt∼νt and D(t)StC∼μt for t∈T∗,|Sˇt−D(t)StC|≤ε for t∈T∗. It then suffices to defineS_t≤St∗≤S¯t holds for t∈T and not just T∗.From (iii), and the remark before Definition nd Gülüm , it is end Gülüm . Indeed,t∈T∗, define μt as the law of D(t)StC, and νt as the law of St∗. It is then very easy to see that the stated conditions are satisfied. As for the finite support condition, note that the probability space in Definition □Conversely, assume now that the given prices are ε‐consistent. For tth trading period, for example, becomesDefinition 2.8(i)semistatic portfolio, or semistatic trading strategy, is a tripleφ10∈R, φt0:3t→R are Borel measurable for t∈T∗, analogously for ϕ1, and φt,i∈R for t∈T∗,i∈{1,⋯,Nt}. Here, φt0 denotes the investment in the bank account, φt1 denotes the number of stocks held in the period from t−1 to t, and φt,i∈R is the number of options with maturity t∈T∗ and strike Kt,i which the investor buys at time zero.A (ii)self‐financing, if1≤t<T and s_u,suC,s¯u∈, 1≤u≤t, whereA semistatic portfolio is called (iii)initial portfolio value of a semistatic portfolio Φ is given byFor prices (iv)liquidation value at time T is defined asThe To prepare for the central notions of model‐independent and weak arbitrage, we now define semistatic trading strategies in the bank account, the underlying asset, and the call options. Here, semistatic means that the position in the call options is fixed at time zero. The definition is model‐independent; as soon as a model , 1≤t≤T, that satisfyLΦ(sT)≥0.Let t∈T∗,cf. , we haveHaving defined semistatic portfolios, we can now formulate two useful notions of arbitrage.Definition 2.10ε≥0. The prices weak arbitrage opportunity with respect to spread‐bound ε if there is no model‐independent arbitrage strategy (with respect to spread‐bound ε), but for any model satisfying rΦ is nonpositive,Let ε≥0 and write only model‐independent arbitrage, meaning model‐independent arbitrage with respect to spread‐bound ε, and similarly for weak arbitrage. The notion of weak  arbitrage was first used in Davis and Hobson , but we think of C(εB(1)) as the underlying. The choices for r_0=S_0−2ε and r¯0=S¯0 made in Theorem εB(1) has to satisfyD(1)E[S1C] has to lie in the closed interval , which implies that the price of an option with strike B(1)ε has to lie in the interval . Therefore, in the proof of Theorem Ct(εB(t)) as a reference to the underlying and −Ct(εB(t)) as a reference to a short position in the underlying plus an additional deposit of 2ε in the bank account.We fix 0≤i<j<l≤N, and that its payoff is nonnegative. A call spread is a portfolio of a long and a short call, where the latter has a larger strike.Theorem 3.1ε≥0 and consider prices as at the beginning of Section T=1 and k1>ε All butterfly spreads have nonnegative time‐0 price, that is,(ii)The call prices satisfy(iii)All call spreads have nonnegative time‐0 price, that is,(iv)If a call spread is available for zero cost, then the involved options have zero bid, respectively, ask price, that is,Let Moreover, there is a model‐independent arbitrage, as soon as any of the conditions (i)–(iii) is not satisfied. Finally, if (i)–(iii) hold but (iv) fails, then there is a weak arbitrage opportunity.Before we formulate the main result for a single maturity, we recall that a butterfly contract (with maturity 1) is defined byThis theorem is proved in Appendices ε=0 and r_i=r¯i=ri, the conditions from Theorem Remark 3.2r_i≥r_j or r¯i≥r¯j for i<j, as shown in the following example.Note that in contrast to the frictionless case, it is not required that bid or ask prices decrease as the strike increases, in order to get models that are ε‐consistent with the absence of arbitrage. This means that we do not have to require ε=0 (no spread on the underlying), and the prices are given by S_0=S¯0=5,r¯i=i+5,r_i=1+i2,ki=i for i=1,2. We assume that the bank account is constant until maturity. These prices and a possible choice of shadow prices ei:=D(1)E[(S1C−Ki)+] are shown in Figure μ=δ5, where δ denotes the Dirac delta. This example shows that, in our setting, prices that are admissible from a no‐arbitrage point of view do not necessarily make economic sense: As the payoff of C(K2) at maturity never exceeds the payoff of C(K1), the utility indifference ask‐price of C(K2) should not be higher than the utility indifference ask‐price of C(K1).Consider two call options, where For d Hobson .Remark i,j,l>0, as these conditions depend on ε only for i=0 i∈{1,⋯,Nt} can then be extended to a call function of a measure μt StC is given by μt. Then we have thats∈ and set νt=δs (Dirac delta) for all t∈T∗. Clearly, (νt)t∈T∗ is a peacock, and we set St∗=B(t)s, which implies D(t)St∗∼νt. Finally, we define S_t=St∗∧StC and S¯t=St∗∨StC, and have thus constructed an arbitrage‐free model.□We define random variables (νt) from Lemma W∞,M introduced before Definition ε>0, a measure μ∈M, and m∈, the setRμ of μ (see W∞‐distance ε of a given sequence of measures.Theorem 4.2.(Theorem 3.5 in Gerhold and Gülüm ) Let ε>0 and (μn)n∈N be a sequence in M such that(νn)n∈N such thatm∈I and for all N∈N, x1,⋯,xN∈R, we haveσn=sgn(xn−1−xn) depends on xn−1 and xn. In this case, it is possible to choose Eν1=Eν2=⋯=m.To prepare for our main result on ε‐consistency in the multiperiod model, we now recall the main result of Gerhold and Gülüm , which gof μ see  as follond Gülüm  gives anμt from Lemma DSC) has to be assumed, but the existence of the peacock (νt) can be replaced by fairly explicit conditions, using Theorem Theorem 4.3ε≥0 the prices S¯0−S_0≤ε and there is a sequence of finitely supported measures (μt)t∈T∗ in M such that:(i)Rμt∈ for all t∈T∗ and i∈{1,⋯,Nt}, and μt)=1 for t∈T∗,(ii)N∈{1,⋯,T−1} and x1,⋯,xN∈Rσn is as in Theorem μn:=μT for n>T.There isFor We can now give a partial solution to the multiperiod ε‐consistency problem. The existence of the measures □Immediate from Lemma SC in Definitions Theorem 4.4p∈The prices satisfy Definition (ii)The prices are consistent with the absence of arbitrage.Let As we allow an arbitrary reference price process Definition 4.5p∈ and two probability measures μ,ν on R, we define the modified Prokhorov distance asFor For the proof of Theorem p by h in the right‐hand side.) Note that d0P=W∞. A well‐known result, which was first proved by Strassen  with random variables X∼μ and Y∼ν such thatGiven measures  in Theorem Theorem 4.7.(Theorem 8.3 in Gerhold and Gülüm ) Let (μn)n∈N be a sequence in M, ε>0, and p∈n∈N with mean m such thatThe following result shows that, unlike for Proof of Theorem 4.4(μt)t∈T∗ as in the proof of Proposition Rμt∈ for i∈{1,⋯Nt} and t∈T∗. Now we pick s∈. Then by, Theorem (νt)t∈T∗ with mean s such that dpP≤ε for all t∈T∗. We can now use Proposition (S∼tC)t∈T∗ and (S∼t∗)t∈T∗ whose marginal distributions are given by (μt)t∈T∗ , respectively, (νt)t∈T∗, such that (S∼t∗)t∈T∗ is a martingale and such thatp=0, but the proof trivially extends to p∈. We then simply put□(i) implies (ii) by definition. To show the other implication, we define probability measures nd Gülüm  for the 5t∈T∗ the option prices can be associated to a measure μt, such that Eμt∈ – (6) in Cousot, (μt)t∈T∗ is a peacock.The main result of the preceding section Theorem  gives seε>0), it turns out that we need conditions that involve all maturities simultaneously , x=, I= and J= are given, such that(i)xt∈R for all t∈{1,⋯,u},(ii)σ1∈{−1,1} and σt=sgn(xt−1−xt) for all t∈{2,⋯,u},(iii)jt∈{0,⋯,Nt} and kt,jt=xt+εσt for all t∈{1,⋯,u},(iv)it∈{0,⋯,Nt} and either kt,it≤xt−1+εσt or it=0 for all t∈{2,⋯,u}. Then, we define a CVB with these parameters as the contractCVBu are given byu as the maturity of the CVB.Fix If we consider a bid–ask spread on the underlying and want to check for ε‐consistency according to Definition Lemma 5.2ε≥0. For all parameters u,σ,x,I,J as in Definition rΦ=−r_uCVB, such that for all models satisfying t∈{2,⋯,u+1} one of the following conditions holds:(i)φt0≥0 and φt1=0, or(ii)φt0≥kt,j−εσt and φt1=−1. In particular, all corresponding cash flows are nonnegative.Fix φt0,φt1 are of course the same as in φt0,φt1 inductively. As we are defining a model‐independent strategy, we could also use the deterministic dummy variables (S_u)u≤t,(SuC)u≤t,(S¯u)u≤t, though. We just have to keep in mind that φt0,φt1 have to be constructed as functions of (S_u)u≤t,(SuC)u≤t,(S¯u)u≤t, without using the distribution of these random vectors.The arguments of u<T, therefore we excluded the case u=T.Moreover, note that later on in Theorem Proof of Lemma 5.2r_uCVB. We have to keep in mind that if it=0 for some t∈{2,⋯,u}, then the corresponding expression in jt=0 for some t∈{1,⋯,u}, then the expression −Ct in Kt,i instead of Kt,it and Kt,j instead of Kt,jt.Assume that we buy the contract2−CVBu, we will call this scenario A, or we have one short position in the underlying  and φt+10≥kt,j−εσt; we will refer to this as scenario B. Note that scenarios A and B are not disjoint, but this will not be a problem.We will show inductively that after we have traded at time σ1=−1 and afterward with the case σ1=1. We start with t=1 and first assume that j1>0. If C1 expires out of the money, then we do not trade at time 1 and obtain φ20=2ε≥0, so we are in scenario A. Otherwise we sell one unit of the underlying, and thusD(t)=B(t)−1. If j1=0 then k1,j=ε. We do not close the short position in this case and we get that φ20=4ε≥k1,j−σ1ε, so we also get to scenario B.We will first deal with the case where t−1 we are in scenario A, and in part B, we will assume that at the end of period t−1 we are in scenario B.For the induction step, we split the proof into two parts. In part A, we will assume that after trading time t we can end up either in situation A or B. First we assume that jt,it>0, and so both expressions in t denote options (and not the underlying). Under these assumptions, φt0 satisfiesKt,i≤Kt,j or if both options expire out of the money, then φt+10≥0, and we are in situation A. So, suppose that Kt,i>Kt,j and that StC>Kt,j. This also implies that σt=1. If this is the case, we go short one unit of the underlying, and φt+10 can be bounded from below as follows:jt=0 and it>0. Then we have that kt,j=ε. After trading time t, we end up in scenario B,jt>0 and it=0. As kt,j>ε, we can close the long position in the underlying and end up in scenario A at the end of time t,jt=it=0 is easily handled, because the long and the short position simply cancel out. We are done with part A.Part A: We will show that after we have traded at time t−1 we are in scenario B, and thus φt0=kt−1,j−εσt−1. First we will consider the case where jt,it>0. If at time t the option with strike Kt,j expires in the money, we do not close the short position and havext−1≤xt and xt−1>xt, and always assume that Ct expires out of the money. If xt−1≤xt, then we also have that kt,i≤kt,j and that σt=−1. We close the short position to end up in scenario A,xt−1>xt and σt=1, we do not trade at time t to stay in scenario B,Part B: Assume that after we have traded at time jt=0 and it>0. As before, we have kt,j=ε, and we can close one short position to stay in scenario B,jt>0 and it=0, then we distinguish two cases: either Ct expires out of the money, in which case we cancel out the long and short position in the underlying and haveCt expires in the money. Then we sell one unit of the underlying and hence we end up in scenario B,kt−1,j−εσt−1=xt−1≥kt,i−εσt, and that kt,i=ε.We proceed with the case where jt=it=0 is again easy to handle, because the long and the short position cancel out and we are in scenario B at the end of the (t+1)‐st period.The case where u we are either in scenario A or scenario B, which proves the assertion if σ1=−1.Thus, after we have traded at time σ1=1 is similar. We will first show that after trading at time 1 we can either be in scenario A or scenario B, and the statement of the proposition then follows by induction exactly as in the case σ1=−1.The proof for j1>0. Then, if the option C1 expires out of the money, we are in scenario A; otherwise we go short in the underlying and havej1=0, then we also have that kj,1=ε, and hence we are in scenario B.□First we assume that −CVBu, such that  only depends on σu,ku,j (the investor might have some surplus in the bank account). In the following, we will use this strategy and only write −CVBu, respectively, r_uCVB instead of −CVBu, respectively, r_uCVB. In the case where φu0≥0 and φu1=0, we will say that the CVB expires out of the money; otherwise we will say that it expires in the money.According to Lemma ε≥0.Theorem 5.3ε≥0, s,t,u∈T such that s<t and s<u and i∈{0,⋯,Nt}, j∈{0,⋯,Ns}, l∈{0,⋯,Nu}. Fix prices as at the beginning of Section kt,1>ε for all t∈T. Then, for all CVBs with maturity s∈T and parameters ks,j and σs, the following conditions are necessary for ε‐consistency:(i)(ii) (iii) (iv) Let The next theorem states necessary conditions for the absence of arbitrage in markets with spread‐bound s<t≤u and that i,l>0. Similarly, the other cases can be dealt with. In all four cases (i)–(iv), we will assume that until time s we followed the trading strategy described in Lemma (i)θCt+(1−θ)Cu−CVBs, making an initial profit. If the calendar vertical basket CVBs expires out of the money, then we have model‐independent arbitrage. Otherwise we have a short position in the underlying at time s. To close the short position, we buy θ units of the underlying at time t, and we buy 1−θ units of the underlying at time u. The liquidation value of this strategy at time u is then nonnegative,If (ii)CVBs expires out of the money, then we leave the portfolio as it is. Otherwise we immediately enter a short position and close it at time u. The liquidation value is then nonnegative,Next, assume that (iii)Ct−CVBs for negative cost. Again we can focus on the case where CVBs expires in the money. We sell one unit of the underlying at time s and close the short position at time t. The liquidation value of this strategy at time t is nonnegative,If (iv)CVBs expires in the money is zero, we could simply sell CVBs and follow the trading strategy from Lemma CVBs expires in the money with positive probability, then we can use the same strategy as in the proof of (iii). At time t, the liquidation value of the portfolio is positive with positive probability.□We will show that an ε‐consistent model cannot exist, if We will assume that ε=0, then CVBs has the same payoff as −Cs. Keeping this in mind, it is easy to verify that the conditions from Theorem Note that, if Conjecture 5.4Given the conditions stated in Theorem It remains open whether (i)j1∈{0,⋯,N1} and σ1∈{−1,1} and set x1=k1,j1−εσ1.Pick (ii){x1,⋯,xt−1}, {σ1,⋯,σt−1}, {j1,⋯,jt−1} and {i2,⋯,it−1} first pick jt∈{0,⋯,Nt}.Given (iii)σt distinguishing the following cases:Choose kt,jt≥xt−1+ε set σt=−1;if kt,jt≤xt−1−ε set σt=1;if kt,jt=xt−1 pick σt∈{−1,0,1};if kt,jt∈∖{xt−1} pick σt∈{−1,1}.if (iv)xt=kt,jt−σtε and pick it∈{0,⋯,Nt} such that either kt,it≤xt−1+σtε or it=0.Set (v)Repeat steps (ii)–(iv).Theorem 6We define the notion of ε‐consistent prices, meaning that a set of bid and ask prices for call options and the underlying can be explained by a model with bid–ask spread bounded by ε. For a single maturity, we solve the ε‐consistency problem, recovering the trichotomy consistency/weak arbitrage/model‐independent arbitrage from the frictionless case (Davis & Hobson,


Te author “武风玉" should be “武凤玉".https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5999742/


Scientific Reports 10.1038/s41598-019-39754-6, published online 11 March 2019Correction to: s variable is not scaled by a factor of 104. As a result, in the Results section under the subheading ‘Uplift Rate’,This Article contains a repeated error, where the QQs_min ≈ 6 m3/yr and Qs_max ≈ 60 m3/yr, with a response time of ~ 0.93 Myr for both increased and decreased uplift rate .”“In this case, should read:Qs_min ≈ 6 × 104 m3/yr and Qs_max ≈ 60 × 104 m3/yr, with a response time of ~ 0.93 Myr for both increased and decreased uplift rate .”“In this case, In the Results section under the subheading ‘Precipitation’,Qs_max (19 m3/yr) is attained rapidly with a response time of ~ 0.3 Myr. During a decrease in precipitation, Qs_min (~ 2 m3/yr) is also achieved rapidly but with a more prolonged response time (~ 0.9 Myr).”“In our model, should read:Qs_max (19 × 104 m3/yr) is attained rapidly with a response time of ~ 0.3 Myr. During a decrease in precipitation, Qs_min (~ 2 × 104 m3/yr) is also achieved rapidly but with a more prolonged response time (~ 0.9 Myr).”“In our model, In Figure 1 in panel (b), the y-axis label,s (m3yr-1)”“Qshould read:s (× 104 m3yr−1)”.“QIn Figure 3 in panel (a-b), the y-axis label,s (m3/yr)”“Qshould read:s (× 104 m3yr−1)”.“Qs_min’ and ‘Qs_max’,In Supplementary Table DR2, the units for ‘Magnitude of Change’ for both ‘Q3yr−1)”“(mshould read:4 m3yr-1)”.“”“Qshould read:s (× 104 m3/yr)”.“QIn Supplementary Scenarios 1.1-1.15, Scenarios 2.1-2.15, and Animations, in the graph ‘Change in Qs and provenance (% Source A) over time’, the unit is omitted for the y-axis, wheres”“Qshould read:s (× 104 m3/yr)”.“Q


The combination of light intensity and NSC exhibited significant effects on photosynthetic pigment, nutritional quality, mineral content and antioxidant capacity. That a higher light intensity were readily accessible to higher chlorophyll a/b showed in lettuce of treatment of 350 μmol · m−2 · s−1 × 3/4NSC and 450 μmol · m−2 · s−1 × 1/4NSC. Lower total N contents, higher content of soluble protein, vitamin C, soluble sugar and free amino acid exhibited in lettuce under treatment of 250 and 350 μmol · m−2 · s−1 × 1/4NSC or 3/4NSC. With increasing NSC and LED irradiance, the content of total P and K in lettuce increased and decreased, respectively. The highest and lowest total Ca content were found in treatment of 150 μmol · m−2 · s−1 × 1/4NSC and 450 μmol · m−2 · s−1 × 1/4NSC, respectively, and higher content of total Mg and Zn was observed under 250 μmol m−2 s−1 × 1/4NSC and 150 μmol · m−2 · s−1 × 3/4NSC, respectively. The antioxidant contents generally decreased with increasing NSC level. The higher antioxidant content and capacity occurred in lettuce of 350 μmol · m−2 · s−1 × 1/4NSC treatment. The interaction of 350 μmol · m−2 · s−1 × 1/4NSC might be the optimal condition for lettuce growth in plant factory.Light and nutrient are important factors for vegetable production in plant factory or greenhouse. The total 12 treatments which contained the combination of four light intensity (150, 250, 350 and 450 μmol · m Lactuca sativa L.) is one of the most important vegetable in the world, which is main crop in plant factory. The secondary metabolite in vegetable plays key roles for human body health, such as phenolic compound, vitamin A and C, and carotenoid. These compounds have a function on nutrition and health care1, which could enhance anti-oxidation ability of human body, and the suppression of inflammatory disease and cancer3. Nutritional quality of lettuce is affected by light and nutrient solution in plant factory. Light  plays a crucial role in improvement of plant nutrition quality5. The proper ratio of red and blue light is essential for plant growth and development6. Light intensity not only positively regulates lettuce biomass and morphology, but also nutrition quality and activities of anti-oxidative enzymes8. The content of soluble sugars and ascorbic acid in lettuce increases with increasing light intensity9. Different light intensity is required by different plant for nutritional quality and growth. The highest content of lutein, ß-carotene and chlorophyll in leave shows at 335 μmol · m−2 · s−1 for kale, but 200 μmol · m−2 · s−1 for spinach10. In ten leafy vegetables , higher leaf dry matter, content of protein, K, Ca and Mg, hydrophilic antioxidant activity, and lipophilic antioxidant activity are observed under low light intensity than high intensity11.Lettuce  exhibits the highest weight recovery efficiency in combination of low light intensity and nutrient supply condition14. The combination of high light illumination and low nitrogen contributes to increased vitamin C content and reduced nitrate content15. High CO2 nutrient supply and monochromatic LED synergistically enhance both the lettuce biomass and some amino acids content16. However, it is not yet available research which the interaction between light intensity and nutrient solution regulates vegetable growth and phytonutrients.In hydroponic systems, the arrangement of nutrient solution could modify the quality and yield of vegetable. Nitrate content in lettuce is closely related to solution nitrate concentration−2 · s−1) based on red and blue LED light were investigated. This is aim to provide valuable insights into the optimal combination of light intensity and nutrient solution to improve quality and yield of lettuce in plant factory.In this study, the changes of mineral element contents, nutritional quality, antioxidant activity of lettuce in response to different interaction of nutrient solution concentration  and light intensity  seeds were sowed in sponge block with 1/4 strength nutrient solution concentration (NSC). The full-strength NSC were composed of the following elements: 210.0 mg · L−1 N, 31.0 mg · L−1 P, 234.0 mg · L−1 K, 160.0 mg · L−1 Ca, 48.0 mg · L−1 Mg, 64.0 mg · L−1 S, 5.6 mg · L−1 Fe, 0.5 mg · L−1 B, 0.5 mg · L−1 Mn, 0.05 mg · L−1 Zn, 0.02 mg · L−1 Cu, 0.01 mg · L−1 Mo. The experiment was performed in a growth chamber in South China Agricultural University and the seedlings with three expended true leaves were transplanted in hydroponic system as follow: 22–25/14–18 °C (day/night), 60–80% relative humidity, the nutrient solution aeration rate of 15 min/h, the nutrient solution was renewed every 10 days. Three light intensity  including 150, 250 and 350 μmol · m−2 · s−1, basing on LED: red (660 ± 10 nm): blue (460 ± 10 nm) = 2:1  until to turn white to extract chlorophyll a, b and carotenoid. The absorbance of extract liquor was determined by UV-1200 spectrophotometer  at 645 nm (OD645), 663 nm (OD663) and 440 nm (OD440). The content of pigment was calculated as follow: chlorophyll a concentration (mg/L) = 12.7 × OD663 −2.69 × OD645; chlorophyll b concentration (mg/L) = 22.9 × OD645 −4.86 × OD663; total chlorophyll concentration (mg/L) = 8.02 × OD663 + 20.20 × OD645; carotenoid concentration (mg/L) = 4.7 × OD440 −0.27 × total chlorophyll concentration; chlorophyll a content (mg/g) = (chlorophyll a × 25 mL)/0.5 g; chlorophyll b content (mg/g) = (chlorophyll b × 25 mL)/0.5 g; carotenoid content (mg/g) = (carotenoid × 25 mL)/0.5 g.Pigment content was measured colorimetrically according to Gratani18. A total of 0.5 g fresh lettuces was ground into pulp by liquid nitrogen with 5 mL distilled water. The extract solution was centrifuged at 10000 rpm for 10 min at 4 °C, and 0.05 mL supernatant was combined with 0.95 mL distilled water and 5 mL Coomassie brilliant blue G-250 solution . After 2 min, the soluble protein content was detected at 595 nm by UV-spectrophotometer .Soluble protein content in lettuce was examined by Coomassie brilliant blue G-250 dye method19. 1 g fresh lettuces were heated on the boiling water bath with 10 mL distilled water for 30 min. After the solution cooling, the extracting solution was filtered by volumetric flask. 0.1 mL sample solution was mixed with 0.4 mL 5% salicylic and sulfuric acid and 9.5 mL 8% NaOH. The nitrated content of mixed solution was determined by UV-VIS spectrophotometer  at 410 nm.Nitrate content was determined by ultraviolet spectrophotometry method20. 0.5 g fresh leaves were ground into pulp with 3 mL 1% oxalic acid, 1 mL 30% zinc sulfate and 1 mL 15% potassium ferrocyanide. 10 mL extracting solution was mixed with 1 mL phosphate-acetic acid, 2 mL 5% vitriol and 4 mL ammonium molybdate. After 15 min, the mixed solution was determined at 500 nm by UV-VIS spectrophotometer .Vitamin C was performed by using the 2, 6-dichlorophenol indophenol titration method21. 0.5 g fresh leaves were heated on boiling water bath with 10 mL distilled water for 30 min. 0.1 mL supernatant was mixed with 1.9 mL distilled water, 0.5 mL anthrone ethyl acetate and 5 mL vitriol. After shaking, the soluble sugar was detected by UV-VIS spectrophotometer  at 630 nm.Soluble sugar content was performed by anthronesulfuric acid colorimetry method19. 1 g fresh leaves were ground into pulp with 10 mL deionized water, which were heated on a water bath at 80 °C for 30 min. The extract solution was centrifuged at 13000 g for 10 min. The 0.2 mL supernatant was mixed with 0.8 mL 5% (w/v) salicylic acid  and 19 mL 4 mol·L–1 NaOH. The nitrate content was determined by UV-VIS spectrophotometer  at 410 nm.Free amino acid content was determined colorimetrically22. 1.0 g fresh lettuces exacted by 20 mL 60% alcohol (pH = 3.0) were heated on the boiling water bath for 2 h. The exacting solution was filled in volumetric flask. The certain volume of exacting solution using the same extractant to dilute was determined by UV-VIS spectrophotometer  at 535 nm.Anthocyanin content was determined by spectrometric method23. 1 g fresh leaves were pulverized with liquid nitrogen. The sample powder was heated on boiling bath with 8 mL 80% methanol for 60 min, and then the exacting solution was centrifuged with 12000 rpm for 10 min. The supernatant was moved to evaporation flask at 40 °C with 3–4 rpm for 5–6 min. 10 mL distilled water was added to the exacting solution with putting into centrifuge with 8000 rpm for 20 min. 1 mL supernatant was mixed with 7 ml distilled water, 0.5 mL foline-phenol and 11.5 mL 26.7% sodium carbonate. The absorbance of mixed solution was measured at 760 nm by spectrophotometer .Polyphenol content was determined by Folin-Cioealteu methodet al.24. The extracted method of flavonoid identified with polyphenol. 1 mL extract solution was added to 11.5 mL 30% alcohol and 0.7 mL 5% NaNO2. After 5 min, the reaction solution was mixed with 0.7 mL 10% Al(NO3)3, and then 6 min later, the mixture was added 5 mL 5% NaOH. Finally, 10 min later, the mixed solution was determined by spectrophotometer  at 510 nm.Flavonoid content was determined according to Jia et al.23. 0.5 g sample solution which identified with exacting polyphenol was added 2.5 mL 65 μmol · L−1 DPPH solution. After 30 min, the mixed solution was determined by spectrophotometer  at 517 nm.The 2, 2-diphenyl-1-picrylhydrazyl (DPPH) radical scavenging rate was performed by basing on Tadolini 25. 0.4 mL sample solution which identified with exacting polyphenol was mixed with 3.6 mL mixed solution  for 10 min at 37 °C. The mixed solution was determined by spectrophotometer  at 593 nm.The value of ferric-reducing antioxidant power (FRAP) was performed according to Benzie and Strain26, Mo-Sb Colorimetry27, and flame photometry method28, respectively, while Ca, Mg and Zn content were performed by using atomic absorption spectrophotometry method29.Fresh lettuce was heated to de-enzyme at 105 °C for 1 h, then dried at 75 °C. The kiln-dried sample was smashed and stored to measure mineral element content. Total N, P and K was determined by using Ojeda’sp ≤ 0.05 and p ≤ 0.01 level.All the assays were analyzed in triplicates.Variance analysis of one-way in single light intensity or NSC factor, and two-way in the combination of light intensity and NSC was performed by using SPSS17.0 software to determine the significance at −2 · s−1 treatment. A tendency which plant and shoot weight increased at first then decreased with increasing light intensity was observed. Lettuce fresh and dry weight under the combination of 350 μmol · m−2 · s−1 × 3/4NSC was the highest  and dry weight (DW) of plant and shoot Table . There w−2 · s−1 × 1/2NSC, whereas the lowest under 450 μmol · m−2 · s−1 × 1/2NSC. Chlorophyll b content descended with light intensity increasing. A lower chlorophyll a/b was observed under low light intensity treatments (150 and 250 μmol · m−2 · s−1), but no difference under different NSC. Chlorophyll a/b significantly improved with increasing NSC under 350 μmol · m−2 · s−1, along with the highest at 3/4NSC. Hence, 150 μmol · m−2 · s−1 × 1/4NSC was beneficial to accumulation of chlorophyll a and b, and total chlorophyll, while the 350 μmol · m−2 · s−1 × 3/4NSC treatment contributed to the promotion of chlorophyll a/b.As shown in Table p ≤ 0.01), NSC (p ≤ 0.01) and their interaction (p ≤ 0.01)  Table . There went Fig. . These gent Fig. .Figure 1−2 · s−1 × 1/4NSC. Two-way ANOVA analysis confirmed that nitrate content in lettuce was strongly associated with light intensity (p ≤ 0.01), NSC (p ≤ 0.01) and their interaction (p ≤ 0.01)  Table .p ≤ 0.01), NSC level (p ≤ 0.01) and their combination (p ≤ 0.01) exhibited a significant difference on vitamin C content , NSC (p ≤ 0.01) and their interaction (p ≤ 0.01)  Table . It incrNSC Fig. .p ≤ 0.01), NSC (p ≤ 0.01) and their combination (p ≤ 0.01) significantly affected the content of free amino acid in lettuce  with 1/4NSC or 3/4NSC could increase the content of soluble protein and sugar, vitamin C, and free amino acid, and reduce nitrate content in lettuce.In general, 250 μmol · mp ≤ 0.01), NSC level (p ≤ 0.01) and their combination (p ≤ 0.01) exhibited a significant effect on the content of total N, P and K in lettuce   Fig. .Figure 2p ≤ 0.01), NSC level (p ≤ 0.01) and their interactions (p ≤ 0.01)  Table . At 1/4Nely Fig. . The higs−1 Fig. . The lowNSC Fig. .p ≤ 0.01), NSC level (p ≤ 0.01) and their combination (p ≤ 0.01)  Table . AnthocyNSC Fig. . With insed Fig. , the higNSC Fig. .Figure 3p ≤ 0.01), NSC (p ≤ 0.01) and their crosstalk (p ≤ 0.01) (Table −2 · s−1) and the highest flavonoid content, FRAP and DPPH in lettuce were found under 350 μmol · m−2 · s−1 × 1/4NSC. These meant that 350 μmol · m−2 · s−1 × 1/4NSC was the most efficient condition for the improvement of antioxidant component content and capacity in lettuce, including anthocyanin, polyphenol, flavonoid, FRAP and DPPH.Two-way analysis revealed that flavonoid content, FRAP and DPPH in lettuce were significantly related to light intensity (1) Table . Under 31) Table , FRAP (F1) Table  and DPPH1) Table  in lettup ≤ 0.01), flavonoid  and DPPH  were higher than anthocyanin content . Moreover, the content of anthocyanin, polyphenol and flavonoid were greatly relevant to DPPH (p ≤ 0.01), the highest coefficient was found in polyphenol content (r = 0.952), while the minimum in anthocyanin content. Hence, the antioxidant activity mainly derived from polyphenol, flavonoid, anthocyanin in lettuce under light intensity × NSC condition.To obtain a more detailed understanding of antioxidant role in lettuce under light intensity × NSC, the correlation analysis was performed between antioxidant content and DPPH, FRAP Table . The coe−2 s−1 8, while in lettuce increased under 200 μmol · m−2 · s−1 15. Plant biomass and leaf number of lettuce increased  combined with higher nitrogen (15 mmol · L−1 or 23 mmol · L−1) could enhance the content of chlorophyll a and b in lettuce, while the combination of 220 μmo · m−2 · s−1 × 7 mmol · L−1 reduced the content of chlorophyll a and b15. In this study, 150 μmol · m−2 · s−1 × 1/4NSC was conducive to chlorophyll accumulation, whereas the treatment of 450 μmol · m−2 · s−1 × 1/2NSC was unfavourable for chlorophyll accumulation (Table 30. This might be due to that the chlorophyll contents were more affected by light intensity (p ≤ 0.01) than NSC (p ≤ 0.05)  could lead to lower nitrate content but higher content of vitamin C, soluble sugar, soluble protein, and anthocyanin in lettuce31. 220–330 μmol · m−2 · s−1 was the most suitable irradiance level for growth and nutritional quality of Brassica microgreens32. Similarly, lettuce under higher irradiance of 250 or 350 μmol · m−2 · s−1 × 1/4NSC had lower nitrate content, and higher contents of soluble protein, vitamin C, soluble sugar and free amino acid  treatment in comparison with high or low EC33. It was possible that higher light intensity could promote nitrate accumulation through increasing photosynthetic production34. The increase of light intensity could induce accumulation of soluble sugars9. In lettuce, soluble sugar content increased under 150 μmol · m−2 · s−1 to 350 μmol · m−2 · s−1 and decreased under 450 μmol · m−2 · s−1, but was negatively regulated by NSC level  and 1/4NSC or 3/4NSC could be beneficial to improve nutrition quality in lettuce.In plant factory, higher light intensity  than high intensity (800–1200 μmol · m−2 · s−1)11. In lettuce, total N content was the highest under 450 μmol · m−2 · s−1 × 3/4NSC while the lowest under 350 μmol · m−2 · s−1 × 1/2NSC  but the P content decreased10. However, the mineral contents in spinach were significant different under different irradiance levels, content of Ca and Fe decreased at low light levels10. Mineral nutrient played a crucial role in photosynthesis, carbohydrate content in plant36. P, K, Ca, Mg, Zn in lettuce mainly were accumulated under higher light intensity (350 and 450 μmol · m−2 · s−1) × 1/4 or 3/4NSC , NSC level (p ≤ 0.01) and their combination (p ≤ 0.01) (Table −2 · s−1) than higher irradiance (350 and 450 μmol · m−2 · s−1), the highest contents were exhibited under 350 μmol · m−2 · s−1 × 1/4NSC , low light intensity (200–400 μmol · m−2 · s−1) could improve antioxidant activity for ten leafy vegetables15. Brassica microgreens possessed the highest anthocyanin content under 330–440 μmol · m−2 · s−1 irradiance32. Anthocyanin content in lettuce decreased with increasing NSC level , and decreased at first and then increased under higher irradiance (350 and 450 μmol · m−2 · s−1) , flavonoid , DDPH , anthocyanin  and FRAP  Table . These wvel Fig. , while i−1) Fig. . Plant iAP Table . Thus, p−2 · s−1 × 3/4NSC or 1/4NSC was conducive to growth of lettuce, while the irradiance of 250 and 350 μmol · m−2 · s−1 × 1/4 or 3/4NSC contributed to increased content of soluble protein and sugar, vitamin C, nitrate and free acid, and 350 μmol · m−2 · s−1 × 1/4NSC exhibited a dramatically effect on improving antioxidant content and capacity. The 350 μmol · m−2 · s−1 × 1/4NSC treatment was the more suitable condition for lettuce production in plant factory.In plant factory, light and nutrient solution are the most effective factors improving yield and quality of vegetable. This study clearly demonstrated that the combination of light intensity and nutrient solution could significantly affect growth and quality of lettuce. The interaction of 350 μmol · mSupplementary Information.