Total positivity of a Cauchy kernel Thomas Simon
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Thomas Simon. Total positivity of a Cauchy kernel. 2013. hal-00820576
HAL Id: hal-00820576 https://hal.archives-ouvertes.fr/hal-00820576 Preprint submitted on 6 May 2013
HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. TOTAL POSITIVITY OF A CAUCHY KERNEL
THOMAS SIMON
Abstract. We study the total positivity of the kernel 1 (x2 +2 cos(πα)xy+y2) The case of infinite order is characterized by an application of Schoenberg’s theorem. We then give necessary conditions for the cases of any given finite order with the help of Chebyshev polynomials of the second kind. Sufficient conditions for the finite order cases are also obtained, thanks to Propp’s formula for the Izergin-Korepin determinant. As a by-product, we give a partial answer to a question of Karlin on positive stable semi-groups.
1. Introduction
Let I be some real interval and K some real kernel defined on I I. The kernel K is called × totally positive of order n (TPn) if
det [K(x y )] 0 i j 1≤i j≤m ≥ for every m 1 n x < < x and y < < y If these inequalities hold for all ∈ { } 1 m 1 m n one says that K is TP∞. The kernel K is called sign-regular of order n (SRn) if there exists ε 1 1 such that { m}1≤m≤n ∈ {− } ε det [K(x y )] 0 m i j 1≤i j≤m ≥ for every m 1 n x < < x and y < < y If these inequalities hold for all n one ∈ { } 1 m 1 m says that K is SR∞. The above four properties are called strict, with corresponding notations STP and SSR, when all involved inequalities are strict. We refer to [7] for the classic account on this field and its various connections with analysis, especially Descartes’ rule of signs. We also mention the recent monograph [11] for a more linear algebraic point of view and updated references. A function f : R R+ is called P´olya frequency of order n (PF ) if the kernel K(x y) = → ≤ ∞ n f(x y) is TP on R R When this kernel is STP we will use the notation f SPF Probability − n × n ∈ n densities belonging to the class PF∞ have been characterized by Schoenberg - see e.g. Theorem 7.3.2 (i) p. 345 in [7] - through the meromorphic extension of their Laplace transform, whose
2000 Mathematics Subject Classification. 15A15, 15B48, 33C45, 60E07. Key words and phrases. Alternating sign matrix - Cauchy kernel - Chebyshev polynomial - Izergin-Korepin deter- minant - Positive stable semi-group - Total positivity. 1 2 THOMAS SIMON reciprocal is an entire function having the following Hadamard factorization: ∞ 1 2 (1) = e−γs +δs (1 + a s)e−ans E[esX ] n n=0 with X the associated random variable, γ 0 δ R and a2 < The classical example is ≥ ∈ n ∞ the Gaussian density. PF2 functions are easily characterized by the log-concavity on their support - see Theorems 4.1.8 and 4.1.9 in [7]. But except for the cases n = 2 or n = there is no handy ∞ criterion for testing the PFn character of a given function resp. the TPn character of a given kernel, and such questions might be difficult. In this paper we consider the kernel 1 K (x y) = α x2 + 2 cos(πα)xy + y2 over (0 + ) (0 + ) with α [0 1) Because of its proximity with the standard Cauchy density ∞ × ∞ ∈ we may call Kα a Cauchy kernel, eventhough the denomination Poisson kernel would also be justified. Occurrences of the kernel Kα in the literature are many and varied. We show the following
Theorem. (a) One has
K STP K SR α 1 2 1 3 1 n 0 α ∈ ∞ ⇔ α ∈ ∞ ⇔ ∈ { } (b) For every n 2 one has ≥ K TP K SR α 1 2 1 3 1 n or α < 1 n α ∈ n ⇔ α ∈ n ⇒ ∈ { } (c) For every n 2 one has α 1 n 1 (n2 n 6) K STP ≥ ≤ ∧ − − + ⇒ α ∈ n
The well-known fact that K0 and K1 2 are STP∞ is a direct consequence of explicit classical formulæ for the involved determinants, due respectively to Cauchy and Borchardt - see e.g. (2.7) and (3.9) in [8] - and which will be recalled thereafter. The characterization obtained in Part (a) follows without difficulty from Schoenberg’s theorem and Gauss’ multiplication formula, and will be given in Section 2. The more involved proofs of Part (b) and Part (c) rely both on an analysis of the derivative determinant ∂i+j−2K ∆n = det α α n ∂xi−1∂yj−1 where, here and throughout, we set detn for the determinant of a matrix whose rows and columns are indexed by (i j) 1 n 2 To obtain Part (b) we establish a closed expression for ∆n(1 0+) ∈ { } α in terms of Chebyshev polynomials of the second kind, an expression which is negative whenever α 1 2 1 3 1 n or α > 1 n The computations are performed in Section 3, in four manners ∈ { } involving respectively Wronskians, Schur functions, rectangular matrices and alternating sign ma- trices. The observation that these four very different approaches all lead to the same formula was TOTAL POSITIVITY OF A CAUCHY KERNEL 3 interesting to the author. On the other hand only the last approach, which is based on Propp’s formula for the Izergin-Korepin determinant, seems successful to get the more difficult Part (c). This latter result is partial because our upper condition on α is probably not optimal. We raise the
Conjecture. For every n 2 one has ≥ (2) α < 1 n K STP ⇒ α ∈ n This would show that the inclusion in Part (b) is actually an equivalence, which is true for n = 2 3 by Part (c) of the theorem because then n (n2 n 6) In Section 5 we show (2) ≥ − − + for n = 4 5, and we also give some heuristics reasons supporting the validity of this inclusion for all n. The latter seems however to require wild computations. In Section 5, we state three other conjectures of combinatorial nature whose fulfilment would entail (2). The most natural one is a criterion for the positivity of the generating function
ν(A) ν(AQ) fn k(z) = z z¯
A ∈ An ( A) = k evaluated at a certain complex number z, where is the set of alternating sign matrices of size n, An AQ is the anticlockwise quarter-turn rotation of A (A) is the number of negative entries and ν(A) the inversion number (see the precise notations below). Whereas we can show this criterion for k = 0 or k = max (see below Propositions 3 and 4), unfortunately we cannot do so for all k since there is no sufficiently explicit general formula for fn k. Let us stress that the single evaluation of f (1) = ♯ A (A) = k is a difficult open problem, solved only for certain values of k - see n k { ∈ An } [10] and the references therein. The present paper was initially motivated by a question of S. Karlin - see Section 6 below for its precise statement - on the total positivity in space-time of the positive stable semi-group (t x) p (t x) on (0 + ) (0 + ) which we recall to be defined by the Laplace transform → α ∞ × ∞ ∞ α (3) p (t x)e−λx dx = e−tλ λ 0 α ≥ 0 As a simple consequence of Part (b), in Section 6 we show the
Corollary. For every n 2 one has ≥ p SR α 1 2 1 3 1 n or α < 1 n α ∈ n ⇒ ∈ { } The inclusion α 1 n p TP was proved in [12] for n = 2 and we believe that it is ≤ ⇒ α ∈ n true in general. This would give a complete answer to Karlin’s question and also entail the above conjecture - see Section 6 for an explanation. This will be the matter of further research. 4 THOMAS SIMON
2. Proof of part (a)
This easy part of the theorem is a consequence of the following Proposition 1 and Corollary 1.
Proposition 1. One has
K STP K TP α 1 2 1 3 1 n 0 α ∈ ∞ ⇔ α ∈ ∞ ⇔ ∈ { } Proof. We begin with the second equivalence. Consider the generalized logistic distribution with density sin(πα) g (x) = α 2πα(cosh(x) + cos(πα)) over R Simple transformations - see Theorem 1.2.1. in [7] - entail that the TP∞ character of Kα amounts to the fact that g PF For every s ( 1 1) compute α ∈ ∞ ∈ − ∞ s sx sin(πα) u sin(παs) e gα(x) dx = du = R πα u2 + 2u cos(πα) + 1 α sin(πs) 0 where the right-hand side follows from the residue theorem and is meant as a limit for α = 0 If α 1 2 1 3 1 n 0 the function ∈ { } α sin(πs) s → sin(παs) has a pole at 1 α so that g PF by the aforementioned Theorem 7.3.2 (i) in [7]. If α = 1 n for α ∈ ∞ some n 2 writing ≥ sin(πs) Γ(1 s)Γ(1 + s) (4) = − n sin(πs n) Γ(1 s n)Γ(1 + s n) − and applying Gauss’ multiplication formula and Weierstrass formula for the Gamma function - see e.g. 1.2(11) p.4 and 1.1(3) p.1 in [5] - shows that this function is of the type (1), in other words that g PF If α = 0 the same conclusion holds true thanks to the Eulerian formula α ∈ ∞ sin(πs) s2 = 1 πs − n2 n ≥1 This finishes the proof of the second equivalence. To show the first one, it remains to prove that K STP whenever α 1 2 1 3 1 n 0 Cauchy’s double alternant formula (see e.g. α ∈ ∞ ∈ { } (2.7) in [8] or Example 4.3 in [11]) 1 (y2 y2)(x2 x2) 1≤i (6) log(Γ ) + + log(Γ ) log(Γ ) log(Γ ) 1 n (n−1) n − 1 n − − (n−1) n where Γt is for every t > 0 the random variable with density xt−1e−x 1 Γ(t) {x>0} Second, it is straightforward that the density of log(Γt) is SPF∞ This property conveys to f1 n thanks to the above factorization and the composition formula. Remark 1. (a) The above argument based on the composition formula yields the STP∞ character of all kernels (ax + by + c)−d for a b d > 0 and c 0 (this is the main result of [4] - see Theorem ≥ 3.1 therein), in writing 1 ∞ = e−axue−byuud−1e−cu du (ax + by + c)d 0 (b) It is easy to see that x K (√x √y) is a completely monotone function for every y > 0 and → α that there exists a certain positive finite kernel L (x y) on (0 + ) (0 + ) such that α ∞ × ∞ ∞ −x2u Kα(x y) = e Lα(u y)du 0 When α is the reciprocal of an integer, by (6) it is possible to express Lα as a convolution of weighted Laplace transformation kernels and show that it is RR∞ The kernel Lα is less explicit in the other cases but we feel that its sign-regularity index matches the total positivity index of Kα in other terms that K TP L RR α ∈ n ⇔ α ∈ n for all n Proposition 2. The function g is positive-definite for every α [0 1) α ∈ 6 THOMAS SIMON Proof. The beginning of the proof of Proposition 1 entails by analytic continuation that isx sinh(παs) e gα(x) dx = R α sinh(πs) for every s R Since α [0 1) we can apply the Fourier inversion formula to obtain ∈ ∈ isx sinh(παs) gα(x) = e ds x R R 2πα sinh(πs) ∈ The conclusion follows from Bochner’s theorem. Remark 2. The above proposition is very well-known, but we gave a proof for the reader’s comfort. It is also true that gr is positive-definite for every α [0 1) and r > 0 - see e.g. Exercise 5.6.22 (ii) α ∈ in [2]. Corollary 1. For every α [0 1) and n 2 one has ∈ ≥ K TP K SR α ∈ n ⇔ α ∈ n Proof. Proposition 2 entails that for every α [0 1) the function ∈ 1 + cos(πα) x → cosh(x) + cos(πα) is the characteristic function of the random variable Xα with density (1 + cos(πα)) sinh(παt) t → sin(πα) sinh(πt) For every n 2 s s R and 0 < x < < x this yields ≥ 1 n ∈ 1 n n 2 1 iykXα sisjKα(xi xj) = E tke > 0 2(1 + cos(πα)) 1≤i j≤n k=1 with the notation ti = si xi and yi = log(xi) Hence, the quadratic form [Kα(xi xj)]1≤i j≤n is positive definite for every n 2 and 0 < x < < x In particular, one has ≥ 1 n det [Kα(xi xj)]1≤i j≤n > 0 for every n 2 and 0 < x < < x This shows the required equivalence. ≥ 1 n 3. Proof of part (b) Our argument for this part is the same as the one we have used in [13], in the framework of confluent hypergeometric functions. Let X = x < < x and Y = y < < y be two { 1 n} { 1 n} TOTAL POSITIVITY OF A CAUCHY KERNEL 7 sets of positive variables and V = (x x ) and V = (y y ) X j − i Y j − i 1≤i 3.1. Wronskians. We fix α [0 1) set c = cos(πα) and s = sin(πα) We use the notations ∈ α α Qα(x y) = y + x cos(πα) Qα(x y) = Qα(y x) Rα(x y) = Qα(x y)Qα(x y) 2 2 and Pα(x y) = x + 2xy cos(πα) + y In the following, we will often skip the dependence in (x y) for concision. Observe that (9) R c P = s2 xy and Q2 P = s2 y2 α − α α α α − α − α Introduce the kernels p T = (2p)! ( 1)p−kCp−k(4Q2 )kKp+k+1 α 2p − p+k α α k=0 and p T = 2(2p + 1)! Q ( 1)p+1−kCp−k (4Q2 )kKp+k+2 α 2p+1 α − p+1+k α α k=0 for any integer p Lemma 1. For every n 0 one has ≥ ∂nK α = T ∂yn α n Proof. The formula is obviously true for n = 0 1 By induction, it suffices to show that ∂T ∂T α 2p = T and α 2p+1 = T ∂y α 2p+1 ∂y α 2p+2 for any integer p The latter are elementary computations, whose details are left to the reader. Lemma 1 allows to express ∆n as the Wronskian of (T T ) built on x derivatives, in α α 0 α n−1 − other words one has ∂i−1T (10) ∆n = det α j−1 α n ∂xi−1 Observe that p sin((2p + 1)πα) T (1 0+) = (2p)! ( 1)kCk (2 cos(πα))2p−2k = (2p)! α 2p − 2p−k sin(πα) k=0 where the second equality comes after identifying two standard definitions of the Chebyshev poly- nomial of the second kind U2p Making the same identification for Tα 2p+1(1 0+) and putting the two together entail sin((r + 1)πα) (11) T (1 0+) = ( 1)rr! α r − sin(πα) for every r 0 We next compute the successive x derivatives of T at (1 0+) The outline of ≥ − α r the proof is analogous to Lemma 1, but it is more involved and so we give the details. TOTAL POSITIVITY OF A CAUCHY KERNEL 9 Lemma 2. For every j 1 and r 0 one has ≥ ≥ ∂jT (r + j + 1)! (12) α r (1 0+) = ( 1)j T (1 0+) ∂xj − (r + 1)! α r Proof. We first consider the case r = 2p Setting (r + j + 1)! ( 1)j ∂jT a = Eα = − α 2p F α = ( 1)p−kCp−k4kQ2k−1 Gα = Q F α j r (r + 1)! j p (2p)! ∂xj k p − p+k α k p α k p we will show by induction that for every j 1 one has ≥p α j α p+k+j+1 α (13) Ej p = aj rQα Gk pKα + yHj p k=0 for some smooth function Hα on (0 ) [0 ) Specifying to (x y) = (1 0+) clearly entails (12). j p ∞ × ∞ We begin with the case j = 1 in computing p p Eα = 2 (p + k + 1)F α R Kp+k+2 c kF α P Kp+k+2 1 p k p α α − α k p α α k=0 k=1 p p α p+k+2 2 α p+k+2 = (2p + 2)Qα Gk pKα + 2xysα Fk pKα k=0 k=1 p α p+k+2 α = (2p + 2)Qα Gk pKα + yH1 p k=0 for some smooth function Hα on (0 ) [0 ) where in the second equality we used the first 1 p ∞ × ∞ identity in (9). Suppose now that (13) holds for some j 1 Differentiating, we obtain ≥ p ∂Hα Eα = a Qj−1 (2(p + k + j + 1)Gα Q2 2kc P F α Q jGα )Kp+k+j+2 y j p j+1 p j r α k p α − α α k p α − k p α − ∂x k=0 p p = (2p + j + 2)a Qj+1 Gα Kp+k+j+2 + 2(R c P )Qj kF α Kp+k+j+2 j r α k p α α − α α α k p α k=0 k=0 p ∂Hα + (Q2 P )Qj−1 Gα Kp+k+j+2 y j p α − α α k p α − ∂x k=0 p j+1 α p+k+j+2 α = aj+1 rQα Gk pKα + yHj+1 p k=0 for some smooth function Hα on (0 ) [0 ) where in the second equality we used the two j+1 p ∞ × ∞ identities in (9). This completes the proof of (13) for r = 2p An analogous induction shows that p ( 1)j ∂jT − α 2p+1 = 2 a Q Qj ( 1)p+1−kCp−k (4Q2 )kKp+k+j+2 + yIα (2p + 1)! ∂xj j 2p+1 α α − p+1+k α α j p k=0 for every j 1 p 0 where Iα is a smooth function on (0 ) [0 ) This yields (12) for ≥ ≥ j p ∞ × ∞ r = 2p + 1 and finishes the proof. 10 THOMAS SIMON We finally deduce from (10), (11), Lemma 2, and Leibniz’s formula for determinants, that (i + j 1)! ∆n(1 0+) = sf(n 1) V α det − α − n n j! = sf(n 1)2 V α det Cj = sf(n 1)2 V α − n n i+j−1 − n where the last equality follows from the standard evaluation of a binomial determinant which is left to the reader. This completes the proof of (8). Remark 3. It does not seem that (10) and standard Wronskian transformations can provide any information which is enough explicit to characterize the everywhere positivity of (x y) ∆n(x y) → α The latter is crucial for (2) - see Section 4 below. 3.2. Schur functions. This second approach relies on the following well-known expression for the generating function of Chebyshev polynomials of the second kind: U α zk = K (1 z) z < 1 k+1 α − | | k≥0 This entails 1 Dn(X Y ) = det U α (y z )k α n n k+1 i j 2 k≥0 xj j =1 with the above notation for X and Y having written z = x−1 and assuming, here and throughout, j − j y z < 1 for all i j = 1 n The determinant on the right-hand side, denoted by F n(X Y ) can | i j| α be further evaluated by multilinear expansion: ∞ n n ki α kl Fα (X Y ) = detn zj Ukl+1yl k1 k n=0 l=1 Notice that in the multiple sum, all indices ki can be chosen distinct since otherwise the summand is zero. Setting for the symmetric group of size n and Sn n α α UK = Ukl+1 l=1 for any n tuple K = (k k ) one obtains − 1 n n n α kσ(i) kσ(l) Fα (X Y ) = UK detn zj yl k1>··· >kn≥0 σ ∈Sn l=1 α ki ki = UK detn zj detn yj k1>··· >kn≥0 TOTAL POSITIVITY OF A CAUCHY KERNEL 11 where in the second equality we used twice Leibniz’s formula. The above can now be rewritten in terms of Schur functions and Vandermonde determinants. Setting k = n i + λ one has i − i n α n−i+λi n−i+λi α Fα (X Y ) = UK detn zj detn yj = VY VZ UK sλ(Y )sλ(Z) λ1≥···≥ λn≥0 λ where in the second equality the sum is meant on all partitions of size n and we use the standard notation for Schur functions, displayed e.g. in Chapter 4 p.124 of [3]. Putting everything together and differentiating with (7) yield after some simplifications 2 n n sf(n 1) α 2 −1 |λ| ∆α(x y) = − Un−i+λi+1 sλ(1 1) ( yx ) xn(n+1) 2 − λ i=1 for all 0 < y < x with the notation λ = λ + + λ By the change of variable = λ the | | 1 n i n+1−i classical formula (λi λj + j i) s (1 1) = 1≤i Remark 4. Because of the alternate signs, the above expression (14) does not seem very helpful n either to study the everywhere positivity of ∆α(x y) 3.3. Rectangular matrices. This third approach hinges upon a certain closed expression for the determinant 1 D (X Y ) = det q n (x + y )(qx + y ) i j i j The latter is called the Izergin-Korepin determinant in the literature, and appears in the context of the six-vertex model - see Chapter 7 in [3]. We use its evaluation in terms of rectangular matrices separating the variables, which is due to Lascoux - see Theorem q in [9]. It is given by VX VY q Dq(X Y ) = detn HX EY Pq(X Y ) × 12 THOMAS SIMON where Pq(X Y ) = 1≤i j≤n(xi + yj)(qxi + yj) and qk−j+1 qj−1 H = [h (X)] Eq = − e (Y ) X k−i 1≤i≤n 1≤k≤2n−1 Y q 1 n−k+j−1 − 1≤k≤2n−1 1≤j≤n are two rectangular matrices involving the complete resp. elementary symmetric functions, which we recall to be defined through the generating functions n n (15) h (X)tk = (1 x t)−1 and e (Y )tk = (1 + y t) k − r k r r=1 r=1 k≥0 k≥0 2iπα q −1 2 −1 2 Setting q = e Xx = (xq xq ) and Yy = (y y) Lascoux’s formula and (7) yield n 2 −n(n−1) 4 n2 q ∆ (x y) = sf(n 1) q Kα(x y) det H q E α − n Xx × Yy q r r −r 2 By (15), we have hr(Xx) = Cn+r−1x q whence k−i (i−k) 2 k−i H q = Cn+k−1−iq x Xx 1≤i≤n 1≤k≤2n−1 r r On the other hand, (15) entails er(Yy) = Cny so that after some simplifications n−k+j−1 (k−1) 2 α n−k+j−1 EY = Cn q Uk+2−2jy y 1≤k≤2n−1 1≤j≤n α (i−1) 2 with our above notation for U The i-th row of the product H q E having a factor q , r Xx × Yy we finally obtain 2 (16) ∆n(x y) = sf(n 1)2 K (x y)n det [A (x) Bα(y)] α − α n n × n with the notations k−i k−i An(x) = Cn+k−1−ix 1≤i≤n 1≤k≤2n−1 and α n−k+j−1 α n−k+j−1 Bn (y) = Cn Uk+2−2jy 1≤k≤2n−1 1≤j≤n Setting Lα(x y) = det [A (x) Bα(y)] the Cauchy-Binet formula entails n n n × n α α (17) Ln(x y) = Aσ(x)Bσ (y) 1≤σ1< <σ n≤2n−1 where A (x) is the n n minor obtained from the columns σ σ in A (x) and Bα(y) is the σ × 1 n n σ n n minor obtained from the rows σ σ in Bα(y) By Leibniz’s formula, one has × 1 n n nσ α α n(n−1)−nσ Aσ(x) = Aσ(1)x and Bσ (y) = Bσ (1)y with the notation n n = (σ i) σ i − i=1 TOTAL POSITIVITY OF A CAUCHY KERNEL 13 This shows that Lα(x y) is a homogeneous polynomial of degree n(n 1) with coefficient n − α (18) Aσ(1)Bσ (1) n σ=k k n(n−1)−k α α for the term x y Besides, by (16) and symmetry, we have Ln(x y) = Ln(y x) so that these coefficients are palindromic. We compute α α α Ln(1 0+) = Ln(0+ 1) = AId˜ (1)BId˜ (1) with the notation Id˜ = (n 2n 1) One finds immediately Bα (1) = U α U α and some − Id˜ 1 × × n elementary linear transformations yield Aα (1) = 1 Since K (0+ 1) = 1 we finally deduce (8). Id˜ α 3.4. Alternating sign matrices. In this last approach we use Izergin-Korepin’s original formula for the determinant Dq(X Y ) and its expression in terms of alternating sign matrices, due to Propp, which is given in Exercise 7.2.13 p. 244 in [3]. It reads n V V 2 i X Y (A) 2 (A) Cn−I(A) i(A) (A) (19) Dq(X Y ) = ( 1) (1 q) q xi yi (αijxi + yj) Pq(X Y ) − − A∈An i=1 1 ≤ i j ≤ n aij = 0 where stands for the set of n n alternating sign matrices (ASM) viz. those matrices made An × out of 0s, 1s and 1s for which the sum of the entries in each row and each column is 1, and the − non-zero entries in each row and column alternate in sign. We refer to [3] for a comprehensive account on this topic, and also to Section 3 in the recent paper [1] for updated results. In (19), the following notations are used: (A) resp. i(A) is the number of 1s in the i-th row resp. i-th i − column of A, (A) the total number of 1s in A, I(A) the generalized inversion number of A viz. − I(A) = aijakl i 2 (21) sf(n 1)2K (x y)n (P (x y)) (A)(Q (x y))J(A)(Q¯ (x y))n(n−1)−2I(A) − α α α α A ∈An 14 THOMAS SIMON with the notations 2 iπα 2 −iπα 2 Pα(x y) = 4 sin (πα)xy and Qα(x y) = e x + e y This yields ∆n(1 0+) = sf(n 1)2e−iπn(n−1)α 2 e2iπαI(A) α − A ∈Sn where stands for the set of permutation matrices of size n Using the generating function of I(A) Sn for permutation matrices given e.g. in Corollary 3.5 of [3], further trigonometric simplifications entail ∆n(1 0+) = sf(n 1)2V α α − n as required by (8). Remark 5. The formula (21) also shows that sf(n 1)2 ∆n(x x) = − (4 sin2(πα 2)) (A) > 0 α (2x cos(πα 2))n(n+1) A ∈An for all x > 0 and α [0 1) ∈ 4. Proof of part (c) This last part of the theorem relies on Karlin’s ETP criterion on the derivative determinant (see Theorem 2.2.6 in [7]) which states that if ∆k (x y) > 0 for every k 2 n and x y > 0 then α ∈ { } Kα is STPn By an induction, we hence need to show that (22) α 1 n 1 (n2 n 6) ∆n(x y) > 0 for all x y > 0 ≤ ∧ − − + ⇒ α This will be obtained from (21) and some considerations on ASM matrices, all to be found in [3] and Section 2.1 of [1]. Again, for concision we will skip the dependence of the involved kernels in (x y). For any A introduce the statistics ∈ An J(A) ν(A) = AijAi′j′ = ′ 2 1 ≤ i< i ≤ n ′ 1 ≤ j < j ≤ n (see pp. 5-6 in [1] for an explanation of the second equality). Set = max (A) A and n { ∈ An} notice that = (n 1)2 4 for n odd and that = n(n 2) 4 for n even (see again p. 6 in [1]). n − n − From (21), it is clear that if α 2ν(A) ¯n(n−1)−2ν(A)−2k (23) Fn k = Qα Qα > 0 A ∈ An ( A) = k TOTAL POSITIVITY OF A CAUCHY KERNEL 15 n α for all k = 0 n then ∆α is everywhere positive. Notice first that Fn k is real since it writes 1 (Q2ν(A)Q¯n(n−1)−2ν(A)−2k + Q¯2ν(A)Qn(n−1)−2ν(A)−2k) 2 α α α α A ∈ An ( A) = k Indeed, setting AQ for the anticlockwise quarter-turn rotation of A one has AQ with ∈ An (AQ) = (A) and 2ν(AQ) = n(n 1) 2ν(A) 2 (A) (see p. 7 in [1]). − − − Suppose now that k = 0 Then necessarily n 3 ν(A) 1 and ν(AQ) 1 (see p. 7 in [1]) and ≥ ≥ ≥ one gets the factorization Q F α = Q 4 (Q2(ν(A)−1)Q¯2(ν(A )−1)) n k | α| ℜ α α A ∈ An ( A) = k νn k = Q 4 ai (Q2(i−1)Q¯n(n−1)−2k−2(i+1)) | α| n k ℜ α α i=1 where ai are non-negative integer coefficients and ν = max ν(A) A (A) = k Notice n k n k { ∈ An } i in passing that no closed expression for an k or even νn k seem available in the literature. For n = 3 necessarily k = 1 = ν and there is only one summand, so that F α = Q 4 > 0 For n 4, it is 3 1 3 1 | α| ≥ sufficient to check that (Q2(i−1)Q¯n(n−1)−2k−2(i+1)) > 0 ℜ α α for all i = 1 ν Noticing that n(n 1) 2k 2(i + 1) n2 n 6 and recalling that n k − − − ≤ − − iπα 2 −iπα 2 Qα(x y) = e x + e y this inequality becomes clearly true when α 1 (n2 n 6) after expanding the trigonometric ≤ − − polynomials, because all involved cosines are evaluated inside [0 π 2] It remains to consider the case k = 0 Reasoning exactly as above, for every n 2 one obtains ≥ F α > 0 whenever α 1 n(n 1) The following proposition yields the optimal condition α 1 n. n 0 ≤ − ≤ Together with the above discussion, it concludes the proof of Part (c). Proposition 3. For every n 2 one has F α > 0 whenever α 1 n ≥ n 0 ≤ Proof. Using again the generating function of the inversion numbers of permutation matrices, we obtain the explicit formula n Q¯2i Q2i F α = α − α n 0 ¯2 2 Qα Qα i=1 − 16 THOMAS SIMON By an induction argument it is hence sufficient to prove that Q¯2n Q2n α 1 n α − α > 0 ≤ ⇒ Q¯2 Q2 α − α for every n 2 Setting a = e2iπα and recalling the definition of Q we write ≥ α 2p 2q Q¯2n Q2n α − α = (a1 4x)j(a−1 4y)2p−j(a−1 4x)k(a1 4y)2p−k Q¯2 Q2 α α p+q=n−1 − j=0 k=0 2n−2 i 2n−2−i = ci x y i=0 for some palindromic sequence c i = 0 2n 2 which is given by { i − } (j−k+q−p) 2 ci = a p+q = n − 1 j+ k = i j ≤ 2p k ≤ 2q Summing appropriately and using some standard trigonometry, one can show that cos(π(n 1 i)α) cos(πnα) ci = − − − sin2(πα) for every i = 0 2n 2 We omit the details. This completes the proof since the latter expressions − are all positive whenever α 1 n Notice also that these formulæ show that the sequence c is ≤ { i} unimodal. 5. Open questions In this section we give some heuristic reasons supporting the validity of the inclusion (24) α < 1 n ∆n(x y) > 0 for all x y > 0 ⇒ α which, by the ETP criterion and an immediate induction, is enough to show (2). First, we formulate a conjecture on the positivity of certain generating functions of ASM matrices which would entail (24), and we test it on the values n = 3 4 5 Second, we discuss more thoroughly Lascoux’s factorization and settle two problems on the positivity on certain minors of rectangular matrices involving Chebyshev polynomials, whose solution would again entail (24). n 5.1. Alternating sign matrices. By (21), the positivity of ∆α amounts to that of (A) 2ν(A) ¯2ν(AQ) Pα Qα Qα A ∈An TOTAL POSITIVITY OF A CAUCHY KERNEL 17 with the notations of Section 4. The above sum can be expressed with the bivariate generating function ν(A) (A) Zn(x y) = x y A ∈An which is itself given as a certain functional determinant - see Formula (57) p. 21 in [1]. Unfortu- nately the latter determinant is in general very difficult to evaluate, except in certain particular cases. Its value at x = y = 1 counting the cardinal of , was the matter of a whole story which An is told in the book [3]. As in the proof of Part (c) let us now write n (A) 2ν(A) ¯2ν(AQ) (A) α Pα Qα Qα = Pα Fn k A ∈An k=0 recalling the notations = max (A) A and n { ∈ An} α 2ν(A) ¯2ν(AQ) Fn k = Qα Qα A ∈ An ( A) = k In view of Proposition 3, it is natural to raise the following question. Conjecture 1. For every n 2 and every k = 0 one has F α > 0 whenever α 1 n ≥ n n k ≤ α The kernel Fn k can be expressed in terms of the generating function 1 ∂kZ Z (x) = n (x 0) n k k! ∂yk but unfortunately no tractable closed formula is known for the latter. Even its value at x = 1 which counts the number of elements of with k negative entries, is known in closed form only An in some cases - see Chapter 3 in [10] and the references therein. By Proposition 3, Conjecture 1 is true for k = 0 In the preceding section, we also proved that F α > 0 whenever α 1 3 Let us now show the validity of Conjecture 1 for n 4 and k = 3 1 ≤ ≥ n Recall that = 0 and = 1 so that there is no loss of generality in considering n 4 2 3 ≥ Proposition 4. For every n 4 one has F α > 0 whenever α 1 n ≥ n n ≤ 2 Proof. First, suppose that n = 2p +1 is odd. Then n = p and this concerns one single matrix with inversion number ν = p(p + 1) 2 (see p. 6 in [1]). We deduce Z (x) = xp(p+1) 2 and F α = Q 2p(p+1) > 0 n n n n | α| Second, suppose that n = 2p is even. Then = p(p 1) and this concerns two matrices with n − inversion numbers ν = p(p + 1) 2 resp. ν = p(p 1) 2 (see again p. 6 in [1]). We deduce − Z (x) = xp(p−1) 2(1 + xp) and F α = Q 2p(p−1)(Qn + Q¯n) > 0 if α 1 n n n n n | α| α α ≤ 18 THOMAS SIMON Since there is no closed formula for F α if k 0 it seems quite difficult to establish n k ∈ { n} Conjecture 1 for all n k Let us conclude this paragraph in checking its validity for n = 4 and n = 5 Since 4 = 2 and 5 = 4 we just have to consider the cases k = 1 resp. k = 1 2 3 Z (x) = 2x(1 + x)3 and F α = 2 Q 4(Q2 + Q¯2 )3 > 0 if α 1 2 • 4 1 4 1 | α| α α ≤ Z (x) = x(3 + 14x + 35x2 + 48x3 + 48x4 + 35x5 + 14x6 + 3x7) which rewrites • 5 1 x5 1 x4 1 x4 1 x2 1 3x − + 8x2 − + 10x3 − + 2x4 − x3 1 x2 1 x3 1 x 1 − − − − and yields α α α α 4 F5 0 8 F4 0 12 F4 0 16 α F5 1 = 3 Qα α + 8 Qα α + 10 Qα α + 2 Qα F2 0 | | F3 0 | | F2 0 | | F3 0 | | The latter is positive for α 1 5 by Proposition 3. ≤ Z (x) = x(2 + 12x + 21x2 + 24x3 + 21x4 + 12x5 + 2x6) which rewrites • 5 2 2x(1 + x)(1 + x + x2)(1 + x + x2 + x3) + 6x2(1 + x2)2 + 11x3(1 + x2) and yields F α = 2 Q 4F α + 6 Q 8(Q4 + Q¯4 )2 + 11 Q 12(Q4 + Q¯4 ) 5 2 | α| 4 0 | α| α α | α| α α The latter is positive for α 1 4 again by Proposition 3. ≤ Z (x) = x(1 + 6x2 + 6x3 + x5) = x(1 + x)(3x2 + (1 + x2)2 x(1 x)2) so that • 5 3 − − F α = Q 4(Q2 + Q¯2 )(3 Q 8 + (Q4 + Q¯4 )2 Q 4(Q2 Q¯2 )2) 5 3 | α| α α | α| α α − | α| α − α Since (Q2 Q¯2 )2 = 4(x y)2 sin2(πα) we see that F α > 0 if α 1 2 α − α − − 5 3 ≤ α Observe that the generating function Z5 3 has vanishing terms so that F5 3 cannot be suitably α α α α α α written in terms of F5 0 F4 0 F3 0 F2 0 contrary to F5 2 and F5 1 In general, testing the positivity α of Fn k seems to depend on bizarre rearrangements. 5.2. Rectangular matrices. In this paragraph we study in more details the minors Aσ(1) and α Bσ (1) with the notations of Section 3.3. Indeed, formulæ (16), (17) and (18) show that the n α positivity of ∆α is ensured by that of Aσ(1) and Bσ (1) The analysis for Aσ(1) is easy. Proposition 5. For every σ : 1 n 1 2n 1 increasing, one has A (1) = A (1) > 0 { } → { − } σ σ˜ Proof. The generating function 1 Cr xr = n−1+r (1 x)n r≥0 − and Edrei’s criterion - see Theorem 1.2 p. 394 in [7] - show that the sequence Cr r 0 is { n−1+r ≥ } TP∞ In other words, the matrix An(1) is TP∞ viz. all its minors are non-negative. In particular, TOTAL POSITIVITY OF A CAUCHY KERNEL 19 all coefficients Aσ(1) are non-negative. We now compute its exact positive value with the help of a formula of Gessel and Viennot on binomial determinants. Substracting the i-th row from the (i 1)-th row successively for i = n 2 and then repeating this operation for i = (n 1) 2 − − p+1 p+1 p i = (n 2) 2 we see indeed from the Pascal relationships Cr = C Cr that the minor − r+1 − Aσ(1) is actually taken from the matrix k−i i Ck−1 = Ck 1≤i≤n 1≤k≤2n−1 0≤i≤n−1 0≤k≤2n−2 The alternative formula mentioned at the bottom of p. 308 in [6] entails (σj σi) A (1) = 1≤i α The analysis for Bσ (1) is however much more delicate. After transposition, we see that it is the n n minor obtained from the columns σ σ in the horizontal matrix × 1 n k−i α Cn U2i−k 1≤i≤n 1≤k≤2n−1 Again, the generating function r r n Cnx = (1 + x) r≥0 k−i α and Edrei’s criterion show that the matrix Cn 1≤i≤n 1≤k≤2n−1 is TP∞ But since some U2i−k are α negative, nothing can be said a priori about the non-negativity of Bσ (1) Transforming the rows (L1 Ln) through the simultaneaous linear operations n L Cj L i → n j j=i multiplies the n n minors by a constant positive factor and yields a matrix whose (i k) coefficient × is given by n j k−j α CnCn U if 1 i n 1 i k n + i j=i 2j−k ≤ ≤ − ≤ ≤ Ck−nU α if i = n n k 2n 1 n 2n−k ≤ ≤ − 0 otherwise. Observe that the (i n + i) coefficient equals n [(n+i) 2] j n+i−j α j n+i−j α α CnCn U2j−n−i = CnCn (U2j−n−i + Un+i−2j) = 0 j=i j=i so that we have a band-matrix of width n whose (i k) coefficient is given by n j k−j α CnCn U if 1 i n i k n 1 + i j=i 2j−k ≤ ≤ ≤ ≤ − 0 otherwise 20 THOMAS SIMON We set Bα n for the above matrix, whose coefficients are all non-negative when α < 1 n We also remark that B is persymmetric, viz. B (i k) = B (n + 1 i 2n k) for any (i k) Indeed, α n α n α n − − n n B (i k) B (n + 1 i 2n k) = Cj Ck−jU α + Cj C2n−k−jU α α n − α n − − n n 2j−k n n 2n−2j−k j=i j= n+1−i n n j j+k−n α j j+k−n α = CnCn U2n−2j−k = CnCn U2n−2j−k j=0 j= n−k and the last sum clearly vanishes. This persymmetry entails Bσ(1) = Bσ˜(1) We now state a natural conjecture which would entail (24). Conjecture 2. For every n 2 all coefficients B (1) are positive whenever α < 1 n ≥ σ The difficulty to prove this conjecture comes from the rectangular shape of the matrices B { α n} which makes it seemingly impossible to use any kind of induction argument. Let us finally display the five first elements of the sequence B whose common shape reminds that of Trinidad and { α n} Tobago’s national flag (the red background being the zeroes, the crucial white diagonal stripes Chebyshev polynomials of the second kind, and the black central parallelogram positive linear combinations thereof) and displays some spatial unimodality around the middle. α α α α α 3U1 3U2 U3 0 0 α 2U1 U2 0 α α α α Bα 1 = (U1 ) Bα 2 = α α Bα 3 = 0 3U2 U3 + 9U1 3U2 0 0 U2 2U1 α α α 0 0 U3 3U2 3U1 α α α α 4U1 6U2 4U3 U4 0 0 0 α α α α α α 0 6U2 4U3 + 24U1 U4 + 16U2 4U3 0 0 Bα 4 = α α α α α α 0 0 4U3 U4 + 16U2 4U3 + 24U1 6U2 0 0 0 0 U α 4U α 6U α 4U α 4 3 2 1 and α α α α α 5U1 10U2 10U3 5U4 U5 0 0 10U α 10U α + 50U α 5U α + 50U α U α + 25U α 0 2 3 1 4 2 5 3 B = 0 0 10U α 5U α + 50U α U α + 25U α + 100U α 0 α 5 3 4 2 5 3 1 0 0 0 5U α U α + 25U α 0 4 5 3 0 0 0 0 U α 5U α 5 1 Using some elementary trigonometry, for these first five values of n we could check that Bα n is TP (viz. all its minors are non-negative) if α < 1 n The following is hence natural Conjecture 3. For every n 1 the matrix B is TP whenever α < 1 n ≥ α n This conjecture is stronger than Conjecture 2 since it involves all minors. Notice that several criteria have appeared in recent years to prove the total positivity of a given matrix without checking every minor. See all the results mentioned in Section 2.5 of [11] and also Theorem 2.16 therein for TOTAL POSITIVITY OF A CAUCHY KERNEL 21 a criterion only on 2 2 minors, interestingly also related to the zeroes of Chebyshev polynomials × of the second kind. Unfortunately, none of this criteria seems particularly helpful in our situation. 6. Proof of the corollary −1 α −1 α Setting fα(x) = pα(1 x) we see from (3) that pα(t x) = t fα(xt ) so that by Theorem 1.2.1. p.18 in [7] the kernel p (t x) has the same sign-regularity over (0 + ) (0 + ) than the α ∞ × ∞ kernel f (ex−y) over R R In Paragraph 7.12.E p.390 of [7] it is shown that α × f (ex−y) STP f (ex−y) TP α 1 2 1 3 1 n α ∈ ∞ ⇔ α ∈ ∞ ⇔ ∈ { } x−y and the question is raised whether fα(e ) should be TP of some finite order when α is not the reciprocal of an integer. In [12], we obtained the equivalence f (ex−y) TP α 1 2 α ∈ 2 ⇔ ≤ Let us now prove the Corollary. We need to show that if α is not the reciprocal of an integer and α < 1 n then f (ex−y) SR If this were true, then the kernels f (ey−x) and α ∈ n α x−y x−u u−y e fα(e )fα(e ) du R would also be SRn by Theorem 1.2.1. and Lemma 3.1.1. in [7]. However, a well-known fractional moment identification - see (3.1) in [12] and the references therein - shows that the latter kernel equals g (x y) with the notations of Section 2. Hence, we get a contradiction to Part (b) of the α − theorem. We finish this paper with a natural conjecture on the total positivity of the positive stable kernel pα which reformulates Karlin’s question in a more precise manner. By Lemma 3.1.1. in [7] and the same argument as in the proof of the corollary, this would also show the conjecture stated in the introduction. But we believe that this last conjecture is harder because the kernel pα is not explicit in general. Conjecture 4. For every n 2 one has ≥ p STP p SR α 1 2 1 3 1 n or α < 1 n α ∈ n ⇔ α ∈ n ⇔ ∈ { } Acknowledgements. I am grateful to Arno Kuijlaars for having lent me his personal copy of [3]. Ce travail a b´en´efici´ed’une aide de l’Agence Nationale de la Recherche portant la r´ef´erence ANR-09-BLAN-0084-01. 22 THOMAS SIMON References [1] R. E. 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