On the nature of the generating series of walks in the quarter plane - PowerPoint PPT Presentation

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 1/27

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

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 . � f D , i , j , k x i y j t k . • Generating series: F D ( x , y , t ) := 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

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

The kernel of the walk Identify directions in D by ( i , j ) , i , j ∈ {− 1 , 0 , 1 } . Consider � x i y j , S D ( x , y ) = ( i , j ) ∈D and the kernel of the walk is K D ( x , y , t ) := xy ( 1 − tS D ( x , y )) . Example D = {← , ↑ , ց} = { ( − 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 ) . 5/27

The functional equation of the walk 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 ) . 6/27

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

Reduction to an elliptic case. Over the 2 8 possible walks, only 79 need to be studied. • ∀ t , # G t < ∞ for 23 walks. → A. Bostan, M. Bousquet-Mélou, M. Kauers, M. Mishna • ∃ t , # G t = ∞ for 56 walks. • E t has genus zero for 5 walks. • E t has genus one for 51 walks. → 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 8/27

Rough statement of the main result. 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. 9/27

Elliptic functions • M er ( E t ) = meromorphic function on E t . • ∃ ω 1 , t ∈ i R > 0 , ω 2 , t ∈ R > 0 , such that M er ( E t ) = { f ( ω ) ∈ M er ( C ) | f ( ω ) = f ( ω + ω 1 , t ) = f ( ω + ω 2 , t ) } . • We define the Weierstrass function: � ℘ t ( ω ) = 1 1 1 ω 2 + ( ω + p ω 1 , t + q ω 2 , t ) 2 − ( p ω 1 , t + q ω 2 , t ) 2 . p , q ∈ Z 2 \ ( 0 , 0 ) • M er ( E t ) = C ( ℘ t ( ω ) , ∂ ω ℘ t ( ω )) . 10/27

Analytic continuation 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. 11/27

Functional equation evaluated on E t The meromorphic continuation satisfy �� � � , � x ( ω + ω 3 , t ) − x ( ω ) � τ F x , D ( ω ) = F x , D ( ω ) + y ( − 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. 12/27

Some consequences of difference Galois theory 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 ∈ M er ( E t ) such that ∂ n ω ( b ) + c n − 1 ∂ n − 1 ( b ) + · · · + c 1 ∂ ω ( b ) + c 0 b = τ ( h ) − h . ω 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. 13/27

Poles of b 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 . 14/27

The base field 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 . 15/27

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

Sketch of proof in the case • The poles of b are { ( ∞ , ± i ) , ( ± i , ∞ ) , ( ± i , ± i t + 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 2 k ( ∞ , i ) = ( ∞ , i ) . No fixed point by τ implies k = 0. Contradiction. 17/27

Sketch of proof in the case • The poles of b are { ( ∞ , ± i ) , ( ± i , ∞ ) , ( ± i , ± i t + 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 , ± i t + t ) } . 17/27

Triple pole case ( , ∈ D ) / 18/27

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

Double pole case ( ∈ D ) / 19/27

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 � � . Then, � F x , D , � Assume that D ∈ F y , D are diff. tr. 19/27

A symmetric case: 20/27

A symmetric case: 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. 20/27

Algebraic cases 21/27

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

A criteria of algebraicity 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 + Z ω 3 , t . 23/27

Uni-orbit, simple pole case 24/27

Uni-orbit, simple pole case Lemma b ∈ M er ( 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. 24/27

Uni-orbit, double pole case 25/27