On the nature of the generating series of walks in the quarter plane

Transcription

1 1/27 On the nature of the generating series of walks in the quarter plane Thomas Dreyfus 1 Joint work with Charlotte Hardouin 2 and Julien Roques 3 and Michael Singer 4 1 University Lyon 1, France 2 University Toulouse 3, France 3 University Grenoble 1, France 4 North Carolina State University, USA

2 Abstract 2/27 Consider the walks in the quarter plane starting from (0, 0) with steps in a fixed set D {,,,,,,, }. Example with possible directions D {,,,,, }.

3 Abstract Let f D,i,j,k equals the number of walks in N 2 starting from (0, 0) ending at (i, j) in k steps in D. Generating series: F D (x, y, t) := f D,i,j,k x i y j t k. i,j,k Classification problem: when F D (x, y, t) is algebraic, holonomic, differentially algebraic? Today, we are able to classify in which cases F D is algebraic (resp. holonomic). O. Bernardi, A. Bostan, M. Bousquet-Mélou, F. Chyzak, G. Fayole, M. van Hoeij, R. Iasnogorodski, M. Kauers, I. Kurkova, V. Malyshev, M. Mishna, K. Raschel, B. Salvy... Definition Let f C((x)). We say that f is differentially algebraic if n N, P C(x)[X 0,..., X n ] such that P(f, f,..., f (n) ) = 0. Otherwise we say that f is differentially transcendent. 3/27

4 4/27 1 Classification of the walks 2 Elliptic functions 3 Transcendence of the generating functions 4 Algebraic cases

5 The kernel of the walk 5/27 Identify directions in D by (i, j), i, j { 1, 0, 1}. Consider S D (x, y) = x i y j, (i,j) D and the kernel of the walk is K D (x, y, t) := xy(1 ts D (x, y)). Example D = {,, } = {( 1, 0), (0, 1), (1, 1)}. S D (x, y) = x 1 + y + xy 1, K D (x, y, t) := xy t(y + xy 2 + x 2 ).

6 The functional equation of the walk 6/27 The generating series F D (x, y, t) and the kernel K D (x, y, t) satisfy the following equation K D (x, y, t)f D (x, y, t) = xy K D (x, 0, t)f D (x, 0, t) K D (0, y, t)f D (0, y, t) + K D (0, 0, t)f D (0, 0, t).

7 Group of the walk 7/27 Fix t / Q. Consider the algebraic curve Consider the involutions E t := {(x, y) P 1 (C) 2 K D (x, y, t) = 0}. ι 1 := E t E ( t ) (i, 1) D (x, y) x, x i y (i,1) D x i ι 2 := E t E ( t ) ( 1,j) D (x, y) y j x (1,j) D y, y. j We attach to D the group of the walk G t := ι 1, ι 2.

8 Reduction to an elliptic case. 8/27 Over the 2 8 possible walks, only 79 need to be studied. t, #G t < for 23 walks. t, #G t = for 56 walks. E t has genus zero for 5 walks. E t has genus one for 51 walks. A. Bostan, M. Bousquet-Mélou, M. Kauers, M. Mishna I. Kurkova, K. Raschel From now we fix t / Q such that #G t = and assume that E t has genus one. E t is an elliptic curve

9 Rough statement of the main result. 9/27 Theorem (D-H-R-S 2017) In 42 cases, x F D (x, 0, t), y F D (0, y, t) are diff. tr. In 9 cases, x F D (x, 0, t), y F D (0, y, t) are diff. alg.

10 Elliptic functions 10/27 Mer(E t ) = meromorphic function on E t. ω 1,t ir >0, ω 2,t R >0, such that Mer(E t ) = {f (ω) Mer(C) f (ω) = f (ω+ω 1,t ) = f (ω+ω 2,t )}. We define the Weierstrass function: t (ω) = 1 ω 2 + p,q Z 2 \(0,0) Mer(E t ) = C( t (ω), ω t (ω)). 1 (ω + pω 1,t + qω 2,t ) 2 1 (pω 1,t + qω 2,t ) 2.

11 Analytic continuation 11/27 Proposition (Kurkova, Raschel) The series x F D (x, 0, t), y F D (0, y, t) admit multivalued meromorphic continuation on the elliptic curve E t. Let F x,d (ω) (resp. F y,d (ω)) be the meromorphic continuation of F D (x, 0, t) (resp. F D (0, y, t)), we will see as meromorphic functions on C. explicit f C(X) (resp. g C(X), ω 3,t R >0 ) such that x = f ( t (ω)) (resp. y = g( t (ω ω 3,t /2))). Theorem (Kurkova, Raschel) The function F x,d (ω) (resp. F y,d (ω)) is not holonomic.

12 12/27 Functional equation evaluated on E t The meromorphic continuation satisfy ( τ F x,d (ω)) = F x,d (ω) +y( ω) ( x(ω + ω 3,t ) x(ω) ), ( τ F y,d (ω)) = F y,d (ω) +x(ω)(y( ω) y(ω)), where τ := h(ω) h(ω + ω 3,t ). These are two difference equations and we may use difference Galois theory.

13 Some consequences of difference Galois theory 13/27 Let b := x(ω)(y( ω) y(ω)). Proposition (D-H-R-S 2017) The function F y,d is diff. alg. iff there exist an integer n 1, c 0,..., c n 1 C and h Mer(E t ) such that n ω(b) + c n 1 n 1 ω (b) + + c 1 ω (b) + c 0 b = τ(h) h. Corollary F x,d is diff. alg. F y,d is diff. alg. Corollary Assume that b has a pole ω 0 C, such that, for all 0 k Z, τ k (ω 0 ) not a pole of b. Then, F y,d is diff. tr.

14 Poles of b 14/27 We now see b as a function P 1 (C) 2 E t P 1 (C). The set of poles of b is contained in {(, α 1 ), (, α 2 ), (β 1, ), (β 2, ), (β 1, γ 1 ), (β 2, γ 2 )}. }{{}}{{}}{{} Poles of x(ω) Poles of y(ω) Poles of y( ω) Lemma In the poles of x, α 1, α 2 are roots of (1,j) D y j+1. In the poles of y, β 1, β 2 are roots of (i,1) D x i+1.

15 The base field 15/27 Lemma Let Q(t) L C field ext. Let P E t. Then P P 1 (L) 2 τ(p) P 1 (L) 2 ι 1 (P) P 1 (L) 2 ι 2 (P) P 1 (L) 2.

16 Generic case Theorem (D-H-R-S 2017) Assume that {α 1, α 2, β 1, β 2 } (C \ Q(t)). Then, F x,d, F y,d are differentially transcendent. 16/27

17 Sketch of proof in the case 17/27 The poles of b are {(, ±i), (±i, ), (±i, ±it + t)}. Involution σ Gal(Q(i, t) Q(t)). Then σ τ = τ σ. Definition Let P, Q E t. We say that P Q if k Z such that τ k (P) = Q. Lemma (, i) (, i). Proof. Assume that τ k (, i) = (, i). We have τ k (, i) = (, i) and τ 2k (, i) = (, i). No fixed point by τ implies k = 0. Contradiction.

18 Sketch of proof in the case 17/27 The poles of b are {(, ±i), (±i, ), (±i, ±it + t)}. Involution σ Gal(Q(i, t) Q(t)). Then σ τ = τ σ. Definition Let P, Q E t. We say that P Q if k Z such that τ k (P) = Q. Lemma (, i) {(, i), (±i, ), (±i, ±it + t)}.

19 Triple pole case (, / D) 18/27

20 Triple pole case (, / D) 18/27 (, ) double pole of x. (, ) simple pole of y. (, ) only triple pole of b. Corollary Assume that, / D. Then, F x,d, F y,d are diff. tr.

21 Double pole case ( / D) 19/27

22 Double pole case ( / D) (, ) simple pole of x, resp y. (, ) simple pole of x, resp. y( ω). (, ), (, ) are only double poles of b. Lemma If (, ) (, ), then k Z, j {1, 2} s.t. ι j τ k (, ) = τ k (, ). Corollary Assume that D are diff. tr. { }. Then, F x,d, F y,d 19/27

23 A symmetric case: 20/27

24 A symmetric case: 20/27 There are 3 simple poles: (, 0), (0, ), and (0, 1). Lemma If (α, β) (β, α), α, β P 1 (Q(t)), then γ P 1 (Q(t)), s.t. K D (γ, γ, t) = 0. Corollary The series F x,d, F y,d are diff. tr.

25 Algebraic cases 21/27

26 Orbit of the poles, case 22/27 t Polar divisor of b ( 1, t+1 ) +(, 0) +( 1, ) t τ-orbit of one of ( 1, the poles of b t+1 ) τ (0, ) τ (, 0) τ (0, 0) τ ( 1, ) In 8 cases, every poles of b are on the same orbit

27 A criteria of algebraicity 23/27 Proposition (D-H-R-S 2017) The function F y,d is diff. alg. iff for all poles ω 0 of b, we have that s h(ω) = b(ω + n i ω 3,t ) i=1 is analytic at ω 0 where ω 0 + n 1 ω 3,t,..., ω + n s ω 3,t are the poles of b that belong to ω 0 + Zω 3,t.

28 Uni-orbit, simple pole case 24/27

29 Uni-orbit, simple pole case 24/27 Lemma b Mer(E t ) = sum of residues of b is zero. Corollary Assume that D { }. Then, every poles of b are on the same orbit and are simple. Consequently, F x,d, F y,d are diff. alg.

30 Uni-orbit, double pole case 25/27

31 Uni-orbit, double pole case 25/27 Lemma c l If b = (ω ω l k 0 ) l, then b = l k ( 1) l+1 c l (ω + ω 0 ) l. Sketch of proof. We use b( ω) = b(ω). Corollary Assume that D { }. Then, every poles of b are on the same orbit and are at most double. Consequently, F x,d, F y,d are diff. alg.

32 Bi-orbit case 26/27

33 Bi-orbit case Walk t Polar divisor of b ( 1, 2t+1 ), [α] +(, 1), [ α] [Residue] +(, 0), [ α] +( 1, ), [α] t τ-orbit of the poles ( 1, 2t+1 ) (, 1) τ τ (0, ) (0, 0) τ τ (, 0) ( 1, ) Walk Polar divisor of b ( 1, t t+1 ), [α] +(, 1), [ α] [Residue] +( 1, ), [α] +(, 0), [ α] τ-orbit of the poles ( 1, t t+1 ) (, 1) τ τ (0, ) (, 0) τ ( 1, ) 26/27

34 Conclusion and perspectives 27/27 Mix of algebra and analysis allows us to treat every cases. In the differentially algebraic cases, explicit computation of the telescoper should lead to the expression of the differential equations. We now should be able to treat the genus zero case.