Total positivity of a Cauchy kernel

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Total positivity of a Cauchy kernel

Thomas Simon

To cite this version:

Thomas Simon. Total positivity of a Cauchy kernel. 2013. �hal-00820576�

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 ordern (TPn) if

det [K(x_i, y_j)]_1≤i,j≤m ≥ 0

for every m ∈ {1, . . . , n}, x₁ < . . . < x_m and y₁ < . . . < y_m. If these inequalities hold for all n one says that K is TP_∞. The kernel K is called sign-regular of order n (SR_n) if there exists {ε_m}1≤m≤n∈ {−1,1} such that

ε_mdet [K(x_i, y_j)]_1≤i,j≤m ≥ 0

for every m∈ {1, . . . , n}, x₁ < . . . < x_m and y₁< . . . < y_m. If these inequalities hold for alln one says thatK 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≤ ∞(PFn) if the kernel K(x, y) = f(x−y) is TP_nonR×R.When this kernel is STP_n,we will use the notationf ∈SPF_n.Probability 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

reciprocal is an entire function having the following Hadamard factorization:

(1) 1

E[e^sX] = e^−γs2+δs Y∞ n=0

(1 +a_ns)e^−ans with X the associated random variable, γ ≥ 0, δ ∈ R, and P

a²_n < ∞. The classical example is 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 PF_ncharacter of a given function resp. the TP_ncharacter of a given kernel, and such questions might be difficult. In this paper we consider the kernel

K_α(x, y) = 1

x²+ 2 cos(πα)xy+y²

over (0,+∞)×(0,+∞),withα∈[0,1).Because of its proximity with the standard Cauchy density we may call K_α 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_α ∈STP_∞ ⇔ K_α ∈SR_∞ ⇔ α∈ {1/2,1/3, . . . ,1/n, . . . ,0}. (b) For every n≥2, one has

K_α ∈TP_n ⇔ K_α ∈SR_n ⇒ α∈ {1/2,1/3, . . . ,1/n} or α <1/n.

(c) For every n≥2, one has α≤1/n∧1/(n²−n−6)+ ⇒ Kα ∈STPn.

The well-known fact that K0 and K_1/2 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

∆ⁿ_α = det_n

∂^i+j−2K_α

∂xⁱ⁻¹∂y^j−1

where, here and throughout, we set det_n 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_α(1,0+) in terms of Chebyshev polynomials of the second kind, an expression which is negative whenever α6∈ {1/2,1/3, . . . ,1/n} orα >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

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 onαis probably not optimal. We raise the Conjecture. For every n≥2,one has

(2) α <1/n ⇒ K_α ∈STP_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

f_n,k(z) = X

A∈ An

µ(A) =k

z^ν(A)z¯^ν(AQ)

evaluated at a certain complex numberz, whereAnis the set of alternating sign matrices of sizen, A^Q is the anticlockwise quarter-turn rotation ofA, µ(A) is the number of negative entries andν(A) the inversion number (see the precise notations below). Whereas we can show this criterion for k= 0 ork=µ_max (see below Propositions 3 and 4), unfortunately we cannot do so for all ksince there is no sufficiently explicit general formula forf_n,k. Let us stress that the single evaluation of f_n,k(1) =♯{A∈ Aⁿ, µ(A) =k}is a difficult open problem, solved only for certain values of k- see [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)7→p_α(t, x) on (0,+∞)×(0,+∞),which we recall to be defined by the Laplace transform (3)

Z ∞ 0

p_α(t, x)e^−λxdx = e^−tλα, λ≥0.

As a simple consequence of Part (b), in Section 6 we show the Corollary. For every n≥2,one has

p_α ∈SR_n ⇒ α∈ {1/2,1/3, . . . ,1/n} or α <1/n.

The inclusion α ≤ 1/n ⇒ p_α ∈ TP_n was proved in [12] for n = 2 and we believe that it is 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.

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_α ∈STP_∞ ⇔ K_α ∈TP_∞ ⇔ α ∈ {1/2,1/3, . . . ,1/n, . . . ,0}.

Proof. We begin with the second equivalence. Consider the generalized logistic distribution with density

g_α(x) = sin(πα)

2πα(cosh(x) + cos(πα))

over R.Simple transformations - see Theorem 1.2.1. in [7] - entail that the TP_∞ character ofK_α amounts to the fact that g_α∈PF_∞.For everys∈(−1,1),compute

Z

R

e^sxgα(x)dx = sin(πα) πα

Z ∞ 0

u^s

u²+ 2ucos(πα) + 1du = sin(παs) αsin(πs)

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

s 7→ αsin(πs) sin(παs)

has a pole at 1/αso thatg_α 6∈PF_∞by the aforementioned Theorem 7.3.2 (i) in [7]. If α= 1/nfor somen≥2,writing

(4) sin(πs)

nsin(πs/n) = Γ(1−s)Γ(1 +s) 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 thatg_α ∈PF_∞.If α= 0,the same conclusion holds true thanks to the Eulerian formula

sin(πs)

πs = Y

n≥1

1− s²

n²

·

This finishes the proof of the second equivalence. To show the first one, it remains to prove that K_α ∈STP_∞ whenever α ∈ {1/2,1/3, . . . ,1/n, . . . ,0}. Cauchy’s double alternant formula (see e.g.

(2.7) in [8] or Example 4.3 in [11]) det_n

"

1 x²_i +y_j²

#

= Q

1≤i<j≤n(y_j²−y_i²)(x²_j−x²_i) Q

1≤i,j≤n(x²_i +y_j²)

entails immediately thatK_1/2∈STP∞.Analogously, Borchardt’s formula (see e.g. (3.9) in [8])

(5) det_n

1 (x_i+y_j)²

= det_n 1

x_i+y_j

× perm_n 1

x_i+y_j

yields K₀∈STP_∞. An alternative way to prove these two latter facts consists in writing 1

x²+y² = Z ∞

0

e^−x2ue^−uy2du and 1 (x+y)² =

Z ∞ 0

e^−xue^−uyudu.

Indeed, the composition formula - see Lemma 3.1.1 in [7] - entails that K₀ and K_1/2 are STP_∞ because the kernele^−uxis SRR_∞ on (0,+∞)×(0,+∞) - see [7] p. 18 for the latter fact and p. 12 for an explanation of the notation SRR∞.

The remaining cases α 6∈ {0,1/2} are slightly more involved. First, it follows from (4) and a fractional moment identification that f_1/n is the density of the independent sum

(6) log(Γ_1/n) + · · · + log(Γ_(n−1)/n) − log(Γ_1/n) − · · · − log(Γ_(n−1)/n) where Γ_tis for every t >0 the random variable with density

x^t−1e^−x

Γ(t) 1_{x>0}.

Second, it is straightforward that the density of log(Γ_t) is SPF_∞. This property conveys to f_1/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 andc≥0 (this is the main result of [4] - see Theorem 3.1 therein), in writing

1

(ax+by+c)^d = Z _∞

0

e^−axue^−byuu^d−1e^−cudu.

(b) It is easy to see thatx7→K_α(√

x,√y) is a completely monotone function for everyy >0,and that there exists a certain positive finite kernelL_α(x, y) on (0,+∞)×(0,+∞) such that

K_α(x, y) = Z ∞

0

e^−x2uL_α(u, y)du.

When α is the reciprocal of an integer, by (6) it is possible to express L_α as a convolution of weighted Laplace transformation kernels and show that it is RR_∞.The kernelL_α is less explicit in the other cases but we feel that its sign-regularity index matches the total positivity index of K_α, in other terms that

K_α ∈TP_n⇔L_α ∈RR_n for all n.

Proposition 2. The function g_α is positive-definite for every α∈[0,1).

Proof. The beginning of the proof of Proposition 1 entails by analytic continuation that Z

R

e^isxg_α(x)dx = sinh(παs) αsinh(πs)

for everys∈R.Since α∈[0,1), we can apply the Fourier inversion formula to obtain g_α(x) =

Z

R

e^isx sinh(παs)

2παsinh(πs)ds, x∈R. 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 thatg_α^r is positive-definite for everyα ∈[0,1) andr >0 - see e.g. Exercise 5.6.22 (ii) in [2].

Corollary 1. For every α∈[0,1) and n≥2, one has

K_α ∈TP_n ⇔ K_α ∈SR_n. Proof. Proposition 2 entails that for everyα ∈[0,1) the function

x 7→ 1 + cos(πα) cosh(x) + cos(πα)

is the characteristic function of the random variable X_α with density t 7→ (1 + cos(πα)) sinh(παt)

sin(πα) sinh(πt) · For every n≥2, s1, . . . , sn∈Rand 0< x1 < . . . < xn,this yields

X

1≤i,j≤n

s_is_jK_α(x_i, x_j) = 1

2(1 + cos(πα)) E





Xn k=1

t_ke^iykXα

2

 > 0,

with the notation t_i = s_i/x_i and y_i = log(x_i). Hence, the quadratic form [K_α(x_i, x_j)]_1≤i,j≤n is positive definite for every n≥2 and 0< x₁ < . . . < x_n.In particular, one has

det [K_α(x_i, x_j)]_1≤i,j≤n > 0,

for everyn≥2 and 0< x₁< . . . < x_n.This shows the required equivalence.

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 = {x₁ < . . . < x_n} and Y = {y₁ < . . . < y_n} be two

sets of positive variables and

V_X = Y

1≤i<j≤n

(x_j−x_i) and V_Y = Y

1≤i<j≤n

(y_j−y_i)

be the usual Vandermonde determinants. Consider Dⁿ_α(X, Y) = det_n[K_α(x_i, y_j)] and the afore- mentioned derivative determinant

∆ⁿ_α(x, y) = det_n

∂^i+j−2K_α

∂xⁱ⁻¹∂y^j−1(x, y)

, x, y >0.

Using repeatedly the formula Xp k=0

(−1)^p−kC_p^kf(z+kε) ∼ f^(p)(z)ε^p, ε→0+

which is valid for any smooth real function f, elementary operations on rows and columns show that

(7) ∆ⁿ_α(x, y) = sf(n−1)² lim

ε→0+

Dⁿ_α(X_ε, Y_ε) V_XεV_Yε

where we have set x^ε_i = x+ (i−1)ε, y^ε_i = y+ (i−1)ε for i = 1. . . n, X_ε = (x^ε₁, . . . , x^ε_n), Y_ε = (y₁^ε, . . . , y_n^ε),and

sf(k) = Yk i=0

i!

for the superfactorial number. By Proposition 2, this entails that ∆ⁿ_α(x, x) ≥0 for any x >0 and below it will be established that the inequality is actually everywhere strict - see Remark 5.

On the other hand, if ∆^k_α(1,0+) < 0 for some k ∈ {2, . . . , n}, then (7) entails that K_α is not TP_nand hence not SR_n by Corollary 1. We will prove that this is the case as soon asα≥1/nand α /∈ {1/2,1/3, . . . ,1/n}.More precisely, setting

U_k^α = sinkπα

sinπα and V_n^α = Yn k=1

U_k^α

for everyk, n≥1,we will show that

(8) ∆ⁿ_α(1,0+) = sf(n−1)²V_n^α.

We give four different proofs of (8), each corresponding to a specific approach to evaluate the deriv- ative determinant ∆ⁿ_α(x, y) in closed form. The first two proofs involve classical tools, respectively Wronskians and Schur functions. The last two proofs rely on more elaborate and recent results on the Izergin-Korepin determinant, using combinatorial formulæ due respectively to Lascoux and Propp. The latter formula, involving alternating sign matrices, will be also used to obtain Part (c).

3.1. Wronskians. We fix α∈[0,1), setc_α = cos(πα) ands_α = sin(πα).We use the notations Q_α(x, y) =y+xcos(πα), Qc_α(x, y) =Q_α(y, x), R_α(x, y) =Q_α(x, y)Qc_α(x, y), and Pα(x, y) =x²+ 2xycos(πα) +y².In the following, we will often skip the dependence in (x, y) for concision. Observe that

(9) Rα−cαPα=s²_αxy and Qb²_α−Pα =−s²_αy². Introduce the kernels

T_α,2p = (2p)!

Xp k=0

(−1)^p−kC_p+k^p−k(4Q²_α)^kK_α^p+k+1 and

T_α,2p+1 = 2(2p+ 1)!Q_α Xp k=0

(−1)^p+1−kC_p+1+k^p−k (4Q²_α)^kK_α^p+k+2 for any integer p.

Lemma 1. For every n≥0, one has

∂ⁿK_α

∂yⁿ = T_α,n.

Proof. The formula is obviously true forn= 0,1.By induction, it suffices to show that

∂T_α,2p

∂y = T_α,2p+1 and ∂T_α,2p+1

∂y = T_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_α as the Wronskian of (Tα,0, . . . , Tα,n−1) built onx−derivatives, in other words one has

(10) ∆ⁿ_α = det_n

∂ⁱ⁻¹T_α,j−1

∂xⁱ⁻¹

. Observe that

T_α,2p(1,0+) = (2p)!

Xp k=0

(−1)^kC_2p−k^k (2 cos(πα))^2p−2k = (2p)!sin((2p+ 1)πα) sin(πα)

where the second equality comes after identifying two standard definitions of the Chebyshev poly- nomial of the second kind U_2p. Making the same identification for T_α,2p+1(1,0+) and putting the two together entail

(11) T_α,r(1,0+) = (−1)^rr!sin((r+ 1)πα) sin(πα)

for every r ≥0. We next compute the successive x−derivatives of T_α,r at (1,0+). The outline of the proof is analogous to Lemma 1, but it is more involved and so we give the details.

Lemma 2. For every j≥1 and r≥0,one has

(12) ∂^jT_α,r

∂x^j (1,0+) = (−1)^j(r+j+ 1)!

(r+ 1)! T_α,r(1,0+).

Proof. We first consider the caser= 2p.Setting a_j,r = (r+j+ 1)!

(r+ 1)! , E_j,p^α = (−1)^j (2p)!

∂^jT_α,2p

∂x^j , F_k,p^α = (−1)^p−kC_p+k^p−k4^kQ^2k−1_α , G^α_k,p = Q_αF_k,p^α , we will show by induction that for every j≥1,one has

(13) E_j,p^α = a_j,rQb^j_α Xp k=0

G^α_k,pK_α^p+k+j+1 + yH_j,p^α

for some smooth functionH_j,p^α on (0,∞)×[0,∞).Specifying to (x, y) = (1,0+) clearly entails (12).

We begin with the casej= 1,in computing E_1,p^α = 2

Xp k=0

(p+k+ 1)F_k,p^α R_αK_α^p+k+2 − c_α Xp k=1

kF_k,p^α P_αK_α^p+k+2

!

= (2p+ 2)Qb_α Xp k=0

G^α_k,pK_α^p+k+2 + 2xys²_α Xp k=1

F_k,p^α K_α^p+k+2

= (2p+ 2)Qb_α Xp k=0

G^α_k,pK_α^p+k+2 + yH_1,p^α

for some smooth function H_1,p^α on (0,∞)×[0,∞), where in the second equality we used the first identity in (9). Suppose now that (13) holds for somej ≥1.Differentiating, we obtain

E_j+1,p^α = a_j,rQb^j−1_α Xp k=0

(2(p+k+j+ 1)G^α_k,pQb²_α−2kc_αP_αF_k,p^α Qb_α−jG^α_k,p)K_α^p+k+j+2 − y∂H_j,p^α

∂x

= (2p+j+ 2)a_j,rQb^j+1_α Xp k=0

G^α_k,pK_α^p+k+j+2 + 2(R_α−c_αP_α)Qb^j_α Xp k=0

kF_k,p^α K_α^p+k+j+2

+ (Qb²_α−P_α)Qb^j−1_α Xp k=0

G^α_k,pK_α^p+k+j+2 − y∂H_j,p^α

∂x

= a_j+1,rQb^j+1_α Xp k=0

G^α_k,pK_α^p+k+j+2 + yH_j+1,p^α

for some smooth functionH_j+1,p^α on (0,∞)×[0,∞),where in the second equality we used the two identities in (9). This completes the proof of (13) forr = 2p.An analogous induction shows that

(−1)^j (2p+ 1)!

∂^jT_α,2p+1

∂x^j = 2a_j,2p+1Q_αQb^j_α Xp k=0

(−1)^p+1−kC_p+1+k^p−k (4Q²_α)^kK_α^p+k+j+2 + yI_j,p^α for every j ≥ 1, p ≥ 0, where I_j,p^α is a smooth function on (0,∞)×[0,∞). This yields (12) for r= 2p+ 1,and finishes the proof.

We finally deduce from (10), (11), Lemma 2, and Leibniz’s formula for determinants, that

∆ⁿ_α(1,0+) = sf(n−1)V_n^αdet_n

(i+j−1)!

j!

= sf(n−1)²V_n^αdet_nh

C_i+j−1^j i

= sf(n−1)²V_n^α,

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)7→∆ⁿ_α(x, y).

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:

X

k≥0

U_k+1^α z^k = K_α(1,−z), |z|<1.

This entails

Dⁿ_α(X, Y) = 1 Yn j=1

x²_j det_n



X

k≥0

U_k+1^α (y_iz_j)^k





with the above notation forXandY,having writtenzj =−x⁻¹_j and assuming, here and throughout,

|y_iz_j|<1 for alli, j= 1, . . . , n.The determinant on the right-hand side, denoted by F_αⁿ(X, Y),can be further evaluated by multilinear expansion:

F_αⁿ(X, Y) = X∞ k1...kn=0

det_nh

z^k_jⁱi Yⁿ

l=1

U_k^α

l+1y^k_l^l

! .

Notice that in the multiple sum, all indicesk_i can be chosen distinct since otherwise the summand is zero. Setting Sn for the symmetric group of sizen and

U_K^α = Yn l=1

U_k^α_l+1

for any n−tuple K = (k1, . . . , kn),one obtains F_αⁿ(X, Y) = X

k1>···>kn≥0

U_K^α X

σ∈Sn

det_nh

z_j^kσ(i)i Yⁿ

l=1

y^k_l^σ(l)

!!

= X

k1>···>kn≥0

U_K^α det_nh z_j^kii

det_nh y^k_jⁱi

,

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. Settingki=n−i+λi,one has

F_αⁿ(X, Y) = X

λ1≥···≥λn≥0

U_K^αdet_nh

z_j^n−i+λii det_nh

y^n−i+λ_j ⁱi

= V_YV_Z X

λ

U_K^α s_λ(Y)s_λ(Z) where in the second equality the sum is meant on all partitions of size nand 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

∆ⁿ_α(x, y) = sf(n−1)² x^n(n+1)/2

X

λ

Yn i=1

U_n−i+λ^α _i+1

!

s_λ(1, . . . ,1)²(−yx⁻¹)^|λ|

for all 0< y < x,with the notation|λ|=λ₁+· · ·+λ_n.By the change of variableµ_i =λ_n+1−i,the classical formula

s_λ(1, . . . ,1) = Q

1≤i<j≤n(λ_i−λ_j+j−i) sf(n−1)

which can be recovered from the Jacobi-Trudi identity - see e.g. Proposition 4.2 in [3] - and a standard binomial determinant evaluation, entails finally

(14) ∆ⁿ_α(x, y) = 1 x^n(n+1)/2

X∞ k=0

(−yx⁻¹)^k X

µ₁≤ · · · ≤µn

µ₁+· · ·+µn=k

Yn i=1

U_i+µ^α _i

! Y

1≤i<j≤n

(µ_j −µ_i+j−i)².

We can now compute explicitly ∆ⁿ_α(1,0+) because the summation is made on the single partition (0, . . . ,0) : we get

∆ⁿ_α(1,0+) = Yn i=1

U_i^α × Y

1≤i<j≤n

(j−i)² = sf(n−1)² V_n^α, as required by (8).

Remark 4. Because of the alternate signs, the above expression (14) does not seem very helpful either to study the everywhere positivity of ∆ⁿ_α(x, y).

3.3. Rectangular matrices. This third approach hinges upon a certain closed expression for the determinant

D_q(X, Y) = det_n

1

(x_i+y_j)(qx_i+y_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 Theoremq in [9]. It is given by

D_q(X, Y) = V_XV_Y

P_q(X, Y)det_n

H_X×E_Y^q

where P_q(X, Y) = Q

1≤i,j≤n(x_i+y_j)(qx_i+y_j),and H_X = [h_k−i(X)]1≤i≤n,1≤k≤2n−1, E_Y^q =

q^k−j+1−q^j−1

q−1 e_n−k+j−1(Y)

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

(15) X

k≥0

h_k(X)t^k = Yn r=1

(1−x_rt)⁻¹ and X

k≥0

e_k(Y)t^k = Yn r=1

(1 +y_rt).

Settingq =e^2iπα, X_x^q = (xq^−1/2, . . . , xq^−1/2),and Y_y = (y, . . . , y),Lascoux’s formula and (7) yield

∆ⁿ_α(x, y) = sf(n−1)²q^{−n(n−1)/4}K_α(x, y)ⁿ²det_nh H_X^q

x×E_Y^q

y

i.

By (15), we have hr(Xx^q) =C_n+r−1^r x^rq^−r/2,whence H_X^q

x = h

C_n+k−1−i^k−i q^(i−k)/2x^k−ii

1≤i≤n,1≤k≤2n−1.

On the other hand, (15) entails e_r(Y_y) =C_n^ry^r,so that after some simplifications E_Yy = h

C_n^n−k+j−1q^(k−1)/2U_k+2−2j^α y^n−k+j−1i

1≤k≤2n−1,1≤j≤n, with our above notation for U_r^α. The i-th row of the productH_Xq

x ×E_Y

y having a factor q^(i−1)/2, we finally obtain

(16) ∆ⁿ_α(x, y) = sf(n−1)²K_α(x, y)ⁿ²det_n[A_n(x)×B^α_n(y)]

with the notations

An(x) = h

C_n+k−1−i^k−i x^k−ii

1≤i≤n,1≤k≤2n−1

and

B_n^α(y) = h

C_n^n−k+j−1U_k+2−2j^α y^n−k+j−1i

1≤k≤2n−1,1≤j≤n. SettingL^α_n(x, y) = det_n[A_n(x)×B_n^α(y)],the Cauchy-Binet formula entails

(17) L^α_n(x, y) = X

1≤σ1<...<σn≤2n−1

Aσ(x)B^α_σ(y)

where A_σ(x) is the n×n minor obtained from the columns σ₁, . . . , σ_n inA_n(x) and B^α_σ(y) is the n×nminor obtained from the rowsσ₁, . . . , σ_n inB_n^α(y).By Leibniz’s formula, one has

Aσ(x) = Aσ(1)x^nσ and B^α_σ(y) = B^α_σ(1)y^{n(n−1)−nσ} with the notation

n_σ = Xn

i=1

(σ_i−i).

This shows that L^α_n(x, y) is a homogeneous polynomial of degree n(n−1) with coefficient

(18) X

nσ=k

A_σ(1)B^α_σ(1)

for the termx^ky^n(n−1)−k.Besides, by (16) and symmetry, we haveL^α_n(x, y) =L^α_n(y, x) so that these coefficients are palindromic. We compute

L^α_n(1,0+) = L^α_n(0+,1) = AId˜(1)B^α_Id˜(1) with the notation ˜Id = (n, . . . ,2n−1). One finds immediately B^α_˜

Id(1) =U₁^α× · · · ×U_n^α and some elementary linear transformations yield A^α_˜

Id(1) = 1.Since Kα(0+,1) = 1,we finally deduce (8).

3.4. Alternating sign matrices. In this last approach we use Izergin-Korepin’s original formula for the determinant D_q(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

(19) D_q(X, Y) = V_XV_Y P_q(X, Y)

X

A∈An

(−1)^µ(A)(1−q)^2µ(A)q^C2n−I(A) Yn i=1

x^µ_i^i(A)y_i^µi(A) Y

1≤i, j≤n aij= 0

(α_ijx_i+y_j)

where An stands for the set of n×n alternating sign matrices (ASM) viz. those matrices made 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: µ_i(A) resp. µⁱ(A) is the number of−1s in the i-th row resp. i-th column of A,µ(A) the total number of−1s inA,I(A) the generalized inversion number of A viz.

I(A) = X

i<k,l<j

a_ija_kl, and

α_ij =

q if P

k≤ia_kj = P

l≤ja_il, 1 otherwise.

Notice that ASM matrices without −1 are permutation matrices. In particular, one recovers Bor- chardt’s formula (5) from (19) in setting q = 1 - see Exercise 7.2.14 in [3]. In the following we will set J(A) = ♯{(i, j), P

k≤ia_kj = P

l≤ja_il}, which is the number of southwest or northeast molecules in the terminology of the six-vertex model, and satisfies the formula

(20) J(A) = 2I(A) − 2µ(A)

(see Exercise 7.1.8 in [3]). Setting q = e^2iπα, we see that (7) entails after some simplifications hinging upon (20) the following expression for the derivative determinant ∆ⁿ_α(x, y):

(21) sf(n−1)²K_α(x, y)ⁿ² X

A∈An

(P_α(x, y))^µ(A)(Q_α(x, y))^J(A)( ¯Q_α(x, y))n(n−1)−2I(A)

with the notations

P_α(x, y) = 4 sin²(πα)xy and Q_α(x, y) = e^iπα/2x+e^−iπα/2y.

This yields

∆ⁿ_α(1,0+) = sf(n−1)²e−iπn(n−1)α/2 X

A∈Sn

e^2iπαI(A),

whereSnstands for the set of permutation matrices of sizen.Using the generating function ofI(A) for permutation matrices given e.g. in Corollary 3.5 of [3], further trigonometric simplifications entail

∆ⁿ_α(1,0+) = sf(n−1)²V_n^α, as required by (8).

Remark 5. The formula (21) also shows that

∆ⁿ_α(x, x) = sf(n−1)² (2xcos(πα/2))ⁿ⁽ⁿ⁺¹⁾

X

A∈An

(4 sin²(πα/2))^µ(A) > 0 for all x >0 andα ∈[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_α(x, y)>0 for every k∈ {2, . . . , n}and x, y >0,then Kα is STPn.By an induction, we hence need to show that

(22) α≤1/n∧1/(n²−n−6)₊ ⇒ ∆ⁿ_α(x, y)>0 for allx, y >0.

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 anyA∈ An,introduce the statistics

ν(A) = X

1≤i < i^′≤n 1≤j < j^′≤n

AijA_i^′_j^′ = J(A) 2

(see pp. 5-6 in [1] for an explanation of the second equality). Set µ_n= max{µ(A), A ∈ An} and notice that µ_n= (n−1)²/4 for nodd and thatµ_n=n(n−2)/4 for neven (see again p. 6 in [1]).

From (21), it is clear that if

(23) F_n,k^α = X

A∈ An

µ(A) =k

Q^2ν(A)_α Q¯n(n−1)−2ν(A)−2k

α > 0

for all k= 0. . . µ_n,then ∆ⁿ_α is everywhere positive. Notice first that F_n,k^α is real since it writes 1

2 X

A∈ An

µ(A) =k

(Q^2ν(A)_α Q¯n(n−1)−2ν(A)−2k

α + ¯Q^2ν(A)_α Qn(n−1)−2ν(A)−2k

α ).

Indeed, setting A^Q for the anticlockwise quarter-turn rotation of A, one has A^Q ∈ Aⁿ with µ(A^Q) =µ(A) and 2ν(A^Q) =n(n−1)−2ν(A)−2µ(A) (see p. 7 in [1]).

Suppose now that k6= 0. Then necessarilyn≥3, ν(A)≥1 andν(A^Q)≥1 (see p. 7 in [1]) and one gets the factorization

F_n,k^α = |Q_α|⁴ X

A∈ An

µ(A) =k

ℜ(Q^2(ν(A)−1)_α Q¯^2(ν_α ^(AQ)−1))

= |Qα|⁴

νn,k

X

i=1

aⁱ_n,kℜ(Q²⁽ⁱ⁻¹⁾_α Q¯n(n−1)−2k−2(i+1)

α )

where aⁱ_n,k are non-negative integer coefficients and ν_n,k= max{ν(A), A∈ An, µ(A) = k}.Notice in passing that no closed expression foraⁱ_n,k or evenν_n,kseem available in the literature. Forn= 3, necessarily k= 1 =ν_3,1 and there is only one summand, so that F_3,1^α =|Q_α|⁴ >0. Forn≥4, it is sufficient to check that

ℜ(Q²⁽ⁱ⁻¹⁾_α Q¯n(n−1)−2k−2(i+1)

α ) > 0

for all i= 1. . . ν_n,k.Noticing thatn(n−1)−2k−2(i+ 1)≤n²−n−6 and recalling that Q_α(x, y) = e^iπα/2x+e^−iπα/2y,

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_n,0^α >0 wheneverα≤1/n(n−1).The following proposition yields the optimal conditionα≤1/n.

Together with the above discussion, it concludes the proof of Part (c).

Proposition 3. For every n≥2, one has F_n,0^α >0 whenever α≤1/n.

Proof. Using again the generating function of the inversion numbers of permutation matrices, we obtain the explicit formula

F_n,0^α = Yn i=1

Q¯²ⁱ_α −Q²ⁱ_α Q¯²_α−Q²_α

.

By an induction argument it is hence sufficient to prove that α≤1/n ⇒ Q¯²ⁿ_α −Q²ⁿ_α

Q¯²_α−Q²_α > 0

for everyn≥2.Settinga=e^2iπα and recalling the definition of Qα,we write Q¯²ⁿ_α −Q²ⁿ_α

Q¯²_α−Q²_α = X

p+q=n−1

X2p j=0

X2q k=0

(a^1/4x)^j(a^−1/4y)^2p−j(a^−1/4x)^k(a^1/4y)^2p−k

=

2n−2X

i=0

c_ixⁱy²ⁿ⁻²⁻ⁱ

for some palindromic sequence {c_i, i= 0, . . . ,2n−2} which is given by

c_i = X

p+q=n−1 j+k=i

j≤2p k≤2q

a(j−k+q−p)/2.

Summing appropriately and using some standard trigonometry, one can show that c_i = cos(π(n−1−i)α)−cos(πnα)

sin²(πα)

for everyi= 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 {ci} is unimodal.

5. Open questions

In this section we give some heuristic reasons supporting the validity of the inclusion (24) α <1/n ⇒ ∆ⁿ_α(x, y)>0 for allx, y >0

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).

5.1. Alternating sign matrices. By (21), the positivity of ∆ⁿ_α amounts to that of X

A∈An

P_α^µ(A)Q^2ν(A)_α Q¯^2ν(A_α ^Q),

with the notations of Section 4. The above sum can be expressed with the bivariate generating function

Zn(x, y) = X

A∈An

x^ν(A)y^µ(A),

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 ofAⁿ, was the matter of a whole story which is told in the book [3]. As in the proof of Part (c) let us now write

X

A∈An

P_α^µ(A)Q^2ν(A)_α Q¯^2ν(A_α ^Q) =

µn

X

k=0

P_α^µ(A)F_n,k^α ,

recalling the notations µ_n= max{µ(A), A∈ An}and F_n,k^α = X

A∈ An

µ(A) =k

Q^2ν(A)_α Q¯^2ν(A_α ^Q).

In view of Proposition 3, it is natural to raise the following question.

Conjecture 1. For every n≥2 and every k= 0, . . . , µ_n,one has F_n,k^α >0 whenever α≤1/n.

The kernel F_n,k^α can be expressed in terms of the generating function Z_n,k(x) = 1

k!

∂^kZ_n

∂y^k (x,0)

but unfortunately no tractable closed formula is known for the latter. Even its value at x = 1, which counts the number of elements of An with k negative entries, is known in closed form only 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_3,1^α >0 whenever α ≤1/3. Let us now show the validity of Conjecture 1 for n≥ 4 andk =µ_n. Recall that µ₂= 0 and µ₃= 1 so that there is no loss of generality in considering n≥4.

Proposition 4. For every n≥4 one has F_n,µ^α _n >0 whenever α≤1/n.

Proof. First, suppose that n = 2p+ 1 is odd. Then µ_n = p² and this concerns one single matrix with inversion number ν=p(p+ 1)/2 (see p. 6 in [1]). We deduce

Z_n,µn(x) = x^p(p+1)/2 and F_n,µ^α _n = |Q_α|^2p(p+1) > 0.

Second, suppose that n = 2p is even. Then µ_n = p(p−1) and this concerns two matrices with inversion numbersν=p(p+ 1)/2 resp. ν =p(p−1)/2 (see again p. 6 in [1]). We deduce

Z_n,µn(x) = x^p(p−1)/2(1 +x^p) and F_n,µ^α _n = |Q_α|^2p(p−1)(Qⁿ_α+ ¯Qⁿ_α) > 0 if α≤1/n.

Since there is no closed formula for F_n,k^α if k 6∈ {0, µ_n}, it seems quite difficult to establish Conjecture 1 for all n, k. Let us conclude this paragraph in checking its validity for n = 4 and n= 5.Since µ4 = 2 and µ5= 4 we just have to consider the cases k= 1 resp. k= 1,2,3.

•Z_4,1(x) = 2x(1 +x)³ and F_4,1^α = 2|Q_α|⁴(Q²_α+ ¯Q²_α)³ > 0 if α≤1/2.

•Z_5,1(x) = x(3 + 14x+ 35x²+ 48x³+ 48x⁴+ 35x⁵+ 14x⁶+ 3x⁷),which rewrites 3x

x⁵−1 x³−1

+ 8x²

x⁴−1 x²−1

+ 10x³

x⁴−1 x³−1

+ 2x⁴

x²−1 x−1

and yields

F_5,1^α = 3|Q_α|⁴F_5,0^α

F_3,0^α + 8|Q_α|⁸F_4,0^α

F_2,0^α + 10|Q_α|¹²F_4,0^α

F_3,0^α + 2|Q_α|¹⁶F_2,0^α . The latter is positive forα≤1/5,by Proposition 3.

•Z_5,2(x) = x(2 + 12x+ 21x²+ 24x³+ 21x⁴+ 12x⁵+ 2x⁶),which rewrites 2x(1 +x)(1 +x+x²)(1 +x+x²+x³) + 6x²(1 +x²)² + 11x³(1 +x²) and yields

F_5,2^α = 2|Q_α|⁴F_4,0^α + 6|Q_α|⁸(Q⁴_α+ ¯Q⁴_α)² + 11|Q_α|¹²(Q⁴_α+ ¯Q⁴_α).

The latter is positive forα≤1/4,again by Proposition 3.

•Z_5,3(x) = x(1 + 6x²+ 6x³+x⁵) = x(1 +x)(3x²+ (1 +x²)²−x(1−x)²),so that F_5,3^α = |Q_α|⁴(Q²_α+ ¯Q²_α)(3|Q_α|⁸+ (Q⁴_α+ ¯Q⁴_α)²− |Q_α|⁴(Q²_α−Q¯²_α)²).

Since (Q²_α−Q¯²_α)² =−4(x−y)²sin²(πα),we see thatF_5,3^α >0 if α≤1/2.

Observe that the generating function Z_5,3 has vanishing terms so that F_5,3^α cannot be suitably written in terms of F_5,0^α , F_4,0^α , F_3,0^α , F_2,0^α ,contrary toF_5,2^α andF_5,1^α .In general, testing the positivity of F_n,k^α seems to depend on bizarre rearrangements.

5.2. Rectangular matrices. In this paragraph we study in more details the minors A_σ(1) and B^α_σ(1), with the notations of Section 3.3. Indeed, formulæ (16), (17) and (18) show that the positivity of ∆ⁿ_α is ensured by that of A_σ(1) and B^α_σ(1).The analysis for A_σ(1) is easy.

Proposition 5. For everyσ :{1, . . . , n} → {1, . . . ,2n−1}increasing, one hasA_σ(1) = A_σ˜(1)>0.

Proof. The generating function

X

r≥0

C_n−1+r^r x^r = 1 (1−x)ⁿ

and Edrei’s criterion - see Theorem 1.2 p. 394 in [7] - show that the sequence {C_n−1+r^r , r ≥0} is TP_∞.In other words, the matrixA_n(1) is TP_∞viz. all its minors are non-negative. In particular,