Hypergeometric series with algebro-geometric dressing Alicia - PowerPoint PPT Presentation

Solutions to hypergeometric recurrences A n := ( 훼 ) n ( 훽 ) n A n x n . ∑ F ( 훼, 훽, 훾 ; x ) = ( 훾 ) n n ! , 훾 / ∈ ℤ < 0 , n ≥ 0 Key equivalence If we define A n = 0 for all n ∈ ℤ < 0 , the coefficients A n satisfy the recurrence: ( 1 + n )( 훾 + n ) A n + 1 − ( 훼 + n )( 훽 + n ) A n = 0 , for all n ∈ ℤ (2) (2) is equivalent to the fact that F ( 훼, 훽, 훾 ; x ) satisfies Gauss differential equation: Θ = x d [Θ(Θ + 훾 − 1 ) − x (Θ + 훼 )(Θ + 훽 )]( F ) = 0 , dx A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 5 / 46

Solutions to hypergeometric recurrences B n := ( 훼 ) n ( 훽 ) n B n x n . G ( 훼, 훽, 훾, 훿 ; x ) = ∑ , 훾, 훿 / ∈ ℤ < 0 , ( 훾 ) n ( 훿 ) n n ≥ 0 Caveat ( 훿 + n )( 훾 + n ) B n + 1 − ( 훼 + n )( 훽 + n ) B n = 0 , for all n ∈ ℕ . (3) but G ( 훼, 훽, 훾 ; x ) does not satisfy the differential equation: [(Θ + 훿 − 1 )(Θ + 훾 − 1 ) − x (Θ + 훼 )(Θ + 훽 )]( G ) = 0 . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 6 / 46

Solutions to hypergeometric recurrences B n := ( 훼 ) n ( 훽 ) n B n x n . G ( 훼, 훽, 훾, 훿 ; x ) = ∑ , 훾, 훿 / ∈ ℤ < 0 , ( 훾 ) n ( 훿 ) n n ≥ 0 Caveat ( 훿 + n )( 훾 + n ) B n + 1 − ( 훼 + n )( 훽 + n ) B n = 0 , for all n ∈ ℕ . (3) but G ( 훼, 훽, 훾 ; x ) does not satisfy the differential equation: [(Θ + 훿 − 1 )(Θ + 훾 − 1 ) − x (Θ + 훼 )(Θ + 훽 )]( G ) = 0 . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 6 / 46

Solutions to hypergeometric recurrences B n := ( 훼 ) n ( 훽 ) n B n x n . G ( 훼, 훽, 훾, 훿 ; x ) = ∑ , 훾, 훿 / ∈ ℤ < 0 , ( 훾 ) n ( 훿 ) n n ≥ 0 Caveat ( 훿 + n )( 훾 + n ) B n + 1 − ( 훼 + n )( 훽 + n ) B n = 0 , for all n ∈ ℕ . (3) but G ( 훼, 훽, 훾 ; x ) does not satisfy the differential equation: [(Θ + 훿 − 1 )(Θ + 훾 − 1 ) − x (Θ + 훼 )(Θ + 훽 )]( G ) = 0 . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 6 / 46

Solutions to hypergeometric recurrences B n := ( 훼 ) n ( 훽 ) n B n x n . G ( 훼, 훽, 훾, 훿 ; x ) = ∑ , 훾, 훿 / ∈ ℤ < 0 , ( 훾 ) n ( 훿 ) n n ≥ 0 The normalization hides the initial condition If we define B n = 0 for all n ∈ ℤ < 0 , then (n+1) ( 훿 + n )( 훾 + n ) B n + 1 − (n+1) ( 훼 + n )( 훽 + n ) B n = 0 , for all n ∈ ℤ . (4) G ( 훼, 훽, 훾 ; x ) does satisfy the differential equation: [Θ(Θ + 훿 − 1 )(Θ + 훾 − 1 ) − x (Θ + 1 )(Θ + 훼 )(Θ + 훽 )]( G ) = 0 . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 7 / 46

Solutions to hypergeometric recurrences B n := ( 훼 ) n ( 훽 ) n B n x n . G ( 훼, 훽, 훾, 훿 ; x ) = ∑ , 훾, 훿 / ∈ ℤ < 0 , ( 훾 ) n ( 훿 ) n n ≥ 0 The normalization hides the initial condition If we define B n = 0 for all n ∈ ℤ < 0 , then (n+1) ( 훿 + n )( 훾 + n ) B n + 1 − (n+1) ( 훼 + n )( 훽 + n ) B n = 0 , for all n ∈ ℤ . (4) G ( 훼, 훽, 훾 ; x ) does satisfy the differential equation: [Θ(Θ + 훿 − 1 )(Θ + 훾 − 1 ) − x (Θ + 1 )(Θ + 훼 )(Θ + 훽 )]( G ) = 0 . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 7 / 46

Solutions to hypergeometric recurrences B n := ( 훼 ) n ( 훽 ) n B n x n . G ( 훼, 훽, 훾, 훿 ; x ) = ∑ , 훾, 훿 / ∈ ℤ < 0 , ( 훾 ) n ( 훿 ) n n ≥ 0 The normalization hides the initial condition If we define B n = 0 for all n ∈ ℤ < 0 , then (n+1) ( 훿 + n )( 훾 + n ) B n + 1 − (n+1) ( 훼 + n )( 훽 + n ) B n = 0 , for all n ∈ ℤ . (4) G ( 훼, 훽, 훾 ; x ) does satisfy the differential equation: [Θ(Θ + 훿 − 1 )(Θ + 훾 − 1 ) − x (Θ + 1 )(Θ + 훼 )(Θ + 훽 )]( G ) = 0 . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 7 / 46

Hypergeometric recurrences in two variables Naive generalization Let a mn , m , n ∈ ℕ such that there exist two rational functions R 1 ( m , n ) , R 2 ( m , n ) expressible as products of (affine) linear functions in ( m , n ) , such that a m + 1 , n a m , n + 1 = R 1 ( m , n ) , = R 2 ( m , n ) (5) a mn a mn (with obvious compatibility conditions). Write P 1 ( m , n ) P 2 ( m , n ) R 1 ( m , n ) = R 2 ( m , n ) = Q 1 ( m + 1 , n ) , Q 2 ( m , n + 1 ) . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 8 / 46

Hypergeometric recurrences in two variables Naive generalization Let a mn , m , n ∈ ℕ such that there exist two rational functions R 1 ( m , n ) , R 2 ( m , n ) expressible as products of (affine) linear functions in ( m , n ) , such that a m + 1 , n a m , n + 1 = R 1 ( m , n ) , = R 2 ( m , n ) (5) a mn a mn (with obvious compatibility conditions). Write P 1 ( m , n ) P 2 ( m , n ) R 1 ( m , n ) = R 2 ( m , n ) = Q 1 ( m + 1 , n ) , Q 2 ( m , n + 1 ) . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 8 / 46

Hypergeometric recurrences in two variables Naive generalization Let a mn , m , n ∈ ℕ such that there exist two rational functions R 1 ( m , n ) , R 2 ( m , n ) expressible as products of (affine) linear functions in ( m , n ) , such that a m + 1 , n a m , n + 1 = R 1 ( m , n ) , = R 2 ( m , n ) (5) a mn a mn (with obvious compatibility conditions). Write P 1 ( m , n ) P 2 ( m , n ) R 1 ( m , n ) = R 2 ( m , n ) = Q 1 ( m + 1 , n ) , Q 2 ( m , n + 1 ) . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 8 / 46

Hypergeometric recurrences in two variables Naive generalization, suite m , n ∈ ℕ a mn x m 1 x n Consider the generating function F ( x 1 , x 2 ) = ∑ 2 and the differential operators ( 휃 i = x i ∂ ∂ x i ) : Δ 1 = Q 1 ( 휃 1 , 휃 2 ) − x 1 P 1 ( 휃 1 , 휃 2 ) Δ 2 = Q 2 ( 휃 1 , 휃 2 ) − x 2 P 2 ( 휃 1 , 휃 2 ) . Then, the recurrences (5) in the coefficients a mn are equivalent to Δ 1 ( F ) = Δ 2 ( F ) = 0 if Q 1 ( 0 , n ) = Q 2 ( m , 0 ) = 0 and in this case, if we extend the definition of a mn by 0 , the recurrences Q 1 ( m + 1 , n ) a m + 1 , n − P 1 ( m , n ) = Q 2 ( m , n + 1 ) a m , n + 1 − P 2 ( m , n ) = 0 hold for all ( m , n ) ∈ ℤ 2 . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 9 / 46

Hypergeometric recurrences in two variables Naive generalization, suite m , n ∈ ℕ a mn x m 1 x n Consider the generating function F ( x 1 , x 2 ) = ∑ 2 and the differential operators ( 휃 i = x i ∂ ∂ x i ) : Δ 1 = Q 1 ( 휃 1 , 휃 2 ) − x 1 P 1 ( 휃 1 , 휃 2 ) Δ 2 = Q 2 ( 휃 1 , 휃 2 ) − x 2 P 2 ( 휃 1 , 휃 2 ) . Then, the recurrences (5) in the coefficients a mn are equivalent to Δ 1 ( F ) = Δ 2 ( F ) = 0 if Q 1 ( 0 , n ) = Q 2 ( m , 0 ) = 0 and in this case, if we extend the definition of a mn by 0 , the recurrences Q 1 ( m + 1 , n ) a m + 1 , n − P 1 ( m , n ) = Q 2 ( m , n + 1 ) a m , n + 1 − P 2 ( m , n ) = 0 hold for all ( m , n ) ∈ ℤ 2 . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 9 / 46

Hypergeometric recurrences in two variables Naive generalization, suite m , n ∈ ℕ a mn x m 1 x n Consider the generating function F ( x 1 , x 2 ) = ∑ 2 and the differential operators ( 휃 i = x i ∂ ∂ x i ) : Δ 1 = Q 1 ( 휃 1 , 휃 2 ) − x 1 P 1 ( 휃 1 , 휃 2 ) Δ 2 = Q 2 ( 휃 1 , 휃 2 ) − x 2 P 2 ( 휃 1 , 휃 2 ) . Then, the recurrences (5) in the coefficients a mn are equivalent to Δ 1 ( F ) = Δ 2 ( F ) = 0 if Q 1 ( 0 , n ) = Q 2 ( m , 0 ) = 0 and in this case, if we extend the definition of a mn by 0 , the recurrences Q 1 ( m + 1 , n ) a m + 1 , n − P 1 ( m , n ) = Q 2 ( m , n + 1 ) a m , n + 1 − P 2 ( m , n ) = 0 hold for all ( m , n ) ∈ ℤ 2 . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 9 / 46

Hypergeometric recurrences in two variables Naive generalization, suite m , n ∈ ℕ a mn x m 1 x n Consider the generating function F ( x 1 , x 2 ) = ∑ 2 and the differential operators ( 휃 i = x i ∂ ∂ x i ) : Δ 1 = Q 1 ( 휃 1 , 휃 2 ) − x 1 P 1 ( 휃 1 , 휃 2 ) Δ 2 = Q 2 ( 휃 1 , 휃 2 ) − x 2 P 2 ( 휃 1 , 휃 2 ) . Then, the recurrences (5) in the coefficients a mn are equivalent to Δ 1 ( F ) = Δ 2 ( F ) = 0 if Q 1 ( 0 , n ) = Q 2 ( m , 0 ) = 0 and in this case, if we extend the definition of a mn by 0 , the recurrences Q 1 ( m + 1 , n ) a m + 1 , n − P 1 ( m , n ) = Q 2 ( m , n + 1 ) a m , n + 1 − P 2 ( m , n ) = 0 hold for all ( m , n ) ∈ ℤ 2 . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 9 / 46

Hypergeometric recurrences in two variables Naive generalization, suite m , n ∈ ℕ a mn x m 1 x n Consider the generating function F ( x 1 , x 2 ) = ∑ 2 and the differential operators ( 휃 i = x i ∂ ∂ x i ) : Δ 1 = Q 1 ( 휃 1 , 휃 2 ) − x 1 P 1 ( 휃 1 , 휃 2 ) Δ 2 = Q 2 ( 휃 1 , 휃 2 ) − x 2 P 2 ( 휃 1 , 휃 2 ) . Then, the recurrences (5) in the coefficients a mn are equivalent to Δ 1 ( F ) = Δ 2 ( F ) = 0 if Q 1 ( 0 , n ) = Q 2 ( m , 0 ) = 0 and in this case, if we extend the definition of a mn by 0 , the recurrences Q 1 ( m + 1 , n ) a m + 1 , n − P 1 ( m , n ) = Q 2 ( m , n + 1 ) a m , n + 1 − P 2 ( m , n ) = 0 hold for all ( m , n ) ∈ ℤ 2 . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 9 / 46

Two examples from combinatorics Dissections A subdivision of a regular n -gon into ( m + 1 ) cells by means of nonintersecting diagonals is called a dissection . How many dissections are there? 1 ( n − 3 ) ( m + n − 1 ) 0 ≤ m ≤ n − 3 . d m , n = ; m + 1 m m So, the generating function is naturally defined for ( m , n ) belonging to the lattice points in the rational cone { ( a , b ) / 0 ≤ a ≤ b − 3 } (and 0 outside). A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 10 / 46

Two examples from combinatorics Dissections A subdivision of a regular n -gon into ( m + 1 ) cells by means of nonintersecting diagonals is called a dissection . How many dissections are there? 1 ( n − 3 ) ( m + n − 1 ) 0 ≤ m ≤ n − 3 . d m , n = ; m + 1 m m So, the generating function is naturally defined for ( m , n ) belonging to the lattice points in the rational cone { ( a , b ) / 0 ≤ a ≤ b − 3 } (and 0 outside). A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 10 / 46

Two examples from combinatorics Dissections A subdivision of a regular n -gon into ( m + 1 ) cells by means of nonintersecting diagonals is called a dissection . How many dissections are there? 1 ( n − 3 ) ( m + n − 1 ) 0 ≤ m ≤ n − 3 . d m , n = ; m + 1 m m So, the generating function is naturally defined for ( m , n ) belonging to the lattice points in the rational cone { ( a , b ) / 0 ≤ a ≤ b − 3 } (and 0 outside). A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 10 / 46

Two examples from combinatorics Dissections A subdivision of a regular n -gon into ( m + 1 ) cells by means of nonintersecting diagonals is called a dissection . How many dissections are there? 1 ( n − 3 ) ( m + n − 1 ) 0 ≤ m ≤ n − 3 . d m , n = ; m + 1 m m So, the generating function is naturally defined for ( m , n ) belonging to the lattice points in the rational cone { ( a , b ) / 0 ≤ a ≤ b − 3 } (and 0 outside). A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 10 / 46

Two examples from combinatorics [Example 9.2, Gessell and Xin, The generating function of ternary trees and continued fractions , EJC ’06] ( m + n 1 − xy ) x m y n , ∑ GX ( x , y ) = 1 − xy 2 − 3 xy − x 2 y = 2 m − n m , n ≥ 0 ( a is defined as 0 if b < 0 or a − b < 0 . ) where b So we are summing over the lattice points in the convex rational cone { ( a , b ) ∈ ℝ 2 : 2 a − b ≥ 0 , 2 b − a ≥ 0 } = ℝ ≥ 0 ( 1 , 2 ) + ℝ ≥ 0 ( 2 , 1 ) . Or: the terms are defined over ℤ 2 extending by 0 outside the cone. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 11 / 46

Two examples from combinatorics [Example 9.2, Gessell and Xin, The generating function of ternary trees and continued fractions , EJC ’06] ( m + n 1 − xy ) x m y n , ∑ GX ( x , y ) = 1 − xy 2 − 3 xy − x 2 y = 2 m − n m , n ≥ 0 ( a is defined as 0 if b < 0 or a − b < 0 . ) where b So we are summing over the lattice points in the convex rational cone { ( a , b ) ∈ ℝ 2 : 2 a − b ≥ 0 , 2 b − a ≥ 0 } = ℝ ≥ 0 ( 1 , 2 ) + ℝ ≥ 0 ( 2 , 1 ) . Or: the terms are defined over ℤ 2 extending by 0 outside the cone. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 11 / 46

Two examples from combinatorics [Example 9.2, Gessell and Xin, The generating function of ternary trees and continued fractions , EJC ’06] ( m + n 1 − xy ) x m y n , ∑ GX ( x , y ) = 1 − xy 2 − 3 xy − x 2 y = 2 m − n m , n ≥ 0 ( a is defined as 0 if b < 0 or a − b < 0 . ) where b So we are summing over the lattice points in the convex rational cone { ( a , b ) ∈ ℝ 2 : 2 a − b ≥ 0 , 2 b − a ≥ 0 } = ℝ ≥ 0 ( 1 , 2 ) + ℝ ≥ 0 ( 2 , 1 ) . Or: the terms are defined over ℤ 2 extending by 0 outside the cone. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 11 / 46

Our results through an example Data ( 2 m − n + 2 )! Consider the hypergeometric terms a m , n = ( − 1 ) n for ( m , n ) n ! m ! ( m − 2 n )! integers with m − 2 n ≥ 0 , n ≥ 0 , which satisfy the recurrences: = ( 2 m − n + 4 ) ( 2 m − n + 3 ) P 1 ( m , n ) a m + 1 , n = ( m + 1 ) ( m + 1 − 2 n ) Q 1 ( m + 1 , n ) a m , n P 1 ( m , n ) = ( 2 m − n + 4 ) ( 2 m − n + 3 ) , Q 1 ( m , n ) = m ( m − 2 n ) = − ( m − 2 n ) ( m − 2 n − 1 ) P 2 ( m , n ) a m , n + 1 ( 2 m − n + 2 ) ( n + 1 ) = Q 2 ( m , n + 1 ) a m , n P 2 ( m , n ) = − ( m − 2 n ) ( m − 2 n − 1 ) , Q 2 ( m , n ) = ( 2 m − n + 3 ) n A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 12 / 46

Our results through an example Data ( 2 m − n + 2 )! Consider the hypergeometric terms a m , n = ( − 1 ) n for ( m , n ) n ! m ! ( m − 2 n )! integers with m − 2 n ≥ 0 , n ≥ 0 , which satisfy the recurrences: = ( 2 m − n + 4 ) ( 2 m − n + 3 ) P 1 ( m , n ) a m + 1 , n = ( m + 1 ) ( m + 1 − 2 n ) Q 1 ( m + 1 , n ) a m , n P 1 ( m , n ) = ( 2 m − n + 4 ) ( 2 m − n + 3 ) , Q 1 ( m , n ) = m ( m − 2 n ) = − ( m − 2 n ) ( m − 2 n − 1 ) P 2 ( m , n ) a m , n + 1 ( 2 m − n + 2 ) ( n + 1 ) = Q 2 ( m , n + 1 ) a m , n P 2 ( m , n ) = − ( m − 2 n ) ( m − 2 n − 1 ) , Q 2 ( m , n ) = ( 2 m − n + 3 ) n A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 12 / 46

Our results through an example Data ( 2 m − n + 2 )! Consider the hypergeometric terms a m , n = ( − 1 ) n for ( m , n ) n ! m ! ( m − 2 n )! integers with m − 2 n ≥ 0 , n ≥ 0 , which satisfy the recurrences: = ( 2 m − n + 4 ) ( 2 m − n + 3 ) P 1 ( m , n ) a m + 1 , n = ( m + 1 ) ( m + 1 − 2 n ) Q 1 ( m + 1 , n ) a m , n P 1 ( m , n ) = ( 2 m − n + 4 ) ( 2 m − n + 3 ) , Q 1 ( m , n ) = m ( m − 2 n ) = − ( m − 2 n ) ( m − 2 n − 1 ) P 2 ( m , n ) a m , n + 1 ( 2 m − n + 2 ) ( n + 1 ) = Q 2 ( m , n + 1 ) a m , n P 2 ( m , n ) = − ( m − 2 n ) ( m − 2 n − 1 ) , Q 2 ( m , n ) = ( 2 m − n + 3 ) n A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 12 / 46

Our results through an example Data ( 2 m − n + 2 )! Consider the hypergeometric terms a m , n = ( − 1 ) n for ( m , n ) n ! m ! ( m − 2 n )! integers with m − 2 n ≥ 0 , n ≥ 0 , which satisfy the recurrences: = ( 2 m − n + 4 ) ( 2 m − n + 3 ) P 1 ( m , n ) a m + 1 , n = ( m + 1 ) ( m + 1 − 2 n ) Q 1 ( m + 1 , n ) a m , n P 1 ( m , n ) = ( 2 m − n + 4 ) ( 2 m − n + 3 ) , Q 1 ( m , n ) = m ( m − 2 n ) = − ( m − 2 n ) ( m − 2 n − 1 ) P 2 ( m , n ) a m , n + 1 ( 2 m − n + 2 ) ( n + 1 ) = Q 2 ( m , n + 1 ) a m , n P 2 ( m , n ) = − ( m − 2 n ) ( m − 2 n − 1 ) , Q 2 ( m , n ) = ( 2 m − n + 3 ) n A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 12 / 46

Our results through an example Data ( 2 m − n + 2 )! Consider the hypergeometric terms a m , n = ( − 1 ) n for ( m , n ) n ! m ! ( m − 2 n )! integers with m − 2 n ≥ 0 , n ≥ 0 , which satisfy the recurrences: = ( 2 m − n + 4 ) ( 2 m − n + 3 ) P 1 ( m , n ) a m + 1 , n = ( m + 1 ) ( m + 1 − 2 n ) Q 1 ( m + 1 , n ) a m , n P 1 ( m , n ) = ( 2 m − n + 4 ) ( 2 m − n + 3 ) , Q 1 ( m , n ) = m ( m − 2 n ) = − ( m − 2 n ) ( m − 2 n − 1 ) P 2 ( m , n ) a m , n + 1 ( 2 m − n + 2 ) ( n + 1 ) = Q 2 ( m , n + 1 ) a m , n P 2 ( m , n ) = − ( m − 2 n ) ( m − 2 n − 1 ) , Q 2 ( m , n ) = ( 2 m − n + 3 ) n A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 12 / 46

Our results through an example We have that the terms t m , n = a mn for m − 2 n ≥ 0 , n ≥ 0 and t ( m , n ) = 0 for any other ( m , n ) ∈ ℤ 2 , satisfy the recurrences: Q 1 ( m + 1 , n ) t m + 1 , n − P 1 ( m , n ) t m , n = Q 2 ( m , n + 1 ) t ( m , n + 1 ) − P 2 ( m , n ) t m , n = 0 . (6) Question Which other terms t m , n , ( m , n ) ∈ ℤ 2 satisfy (6) ? Remark When the linear forms in the polynomials P i , Q i defining the recurrences have generic constant terms, the solution is given by the Ore-Sato coefficients. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 13 / 46

Our results through an example We have that the terms t m , n = a mn for m − 2 n ≥ 0 , n ≥ 0 and t ( m , n ) = 0 for any other ( m , n ) ∈ ℤ 2 , satisfy the recurrences: Q 1 ( m + 1 , n ) t m + 1 , n − P 1 ( m , n ) t m , n = Q 2 ( m , n + 1 ) t ( m , n + 1 ) − P 2 ( m , n ) t m , n = 0 . (6) Question Which other terms t m , n , ( m , n ) ∈ ℤ 2 satisfy (6) ? Remark When the linear forms in the polynomials P i , Q i defining the recurrences have generic constant terms, the solution is given by the Ore-Sato coefficients. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 13 / 46

Our results through an example We have that the terms t m , n = a mn for m − 2 n ≥ 0 , n ≥ 0 and t ( m , n ) = 0 for any other ( m , n ) ∈ ℤ 2 , satisfy the recurrences: Q 1 ( m + 1 , n ) t m + 1 , n − P 1 ( m , n ) t m , n = Q 2 ( m , n + 1 ) t ( m , n + 1 ) − P 2 ( m , n ) t m , n = 0 . (6) Question Which other terms t m , n , ( m , n ) ∈ ℤ 2 satisfy (6) ? Remark When the linear forms in the polynomials P i , Q i defining the recurrences have generic constant terms, the solution is given by the Ore-Sato coefficients. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 13 / 46

Our results through an example Question Which other terms t m , n , ( m , n ) ∈ ℤ 2 satisfy (6) ? Answer There are three other solutions b mn , c mn , d mn (up to linear combinations) A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 14 / 46

Our results through an example Question Which other terms t m , n , ( m , n ) ∈ ℤ 2 satisfy (6) ? Answer There are three other solutions b mn , c mn , d mn (up to linear combinations) A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 14 / 46

Our results through an example Answer There are four solutions a mn , b mn , c mn , d mn (up to linear combinations), with generating series F 1 , . . . , F 4 : ( 2 m − n + 2 )! a m , n = ( − 1 ) n a m , n x m 1 x n F 1 = ∑ n ! m ! ( m − 2 n )! , 2 , m − 2 n ≥ 0 n ≥ 0 ( 2 m − n − 1 )! b m , n = ( − 1 ) m b m , n x m 1 x n F 2 = ∑ n ! m ! ( − 2 m + n + 3 )! , − 2 m + n ≥ 3 2 m ≥ 0 ( − m − 1 )! ( − n − 1 )! c m , n = ( − 1 ) m + n c m , n x m 1 x n F 3 = ∑ ( m − 2 n )! ( − 2 m + n − 3 )! , m − 2 n ≥ 0 2 − 2 m + n ≥ 3 F 4 = x − 2 x − 1 d − 2 , − 1 = 1 , 2 . 1 In all cases, t mn = 0 outside the support of the series. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 15 / 46

Pictorially 6 m 1 = 0 -2m 1 +m 2 =3 5 4 3 m 1 -2m 2 = 0 2 1 m 2 = 0 0 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 0 0 -1 0 -2 0 -3 -4 0 -5 A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 16 / 46 -6

Explanations The generating functions F i satisfy the differential equations: [Θ 1 (Θ 1 − 2 Θ 2 ) − x 1 ( 2 Θ 1 − Θ 2 + 4 ) ( 2 Θ 1 − Θ 2 + 3 )]( F ) = 0 , [Θ 2 ( − 2 Θ 1 + Θ 2 − 3 ) − x 2 ( 2 Θ 2 − Θ 1 ) ( 2 Θ 2 − Θ 1 + 1 )]( F ) = 0 . Consider the system of binomial equations: q 1 = ∂ 11 ∂ 31 − ∂ 22 , q 2 = ∂ 21 ∂ 41 − ∂ 32 in the commutative polynomial ring ℂ [ ∂ 1 , . . . , ∂ 4 ] . The zero set q 1 = q 2 = 0 has two irreducible components, one of degree 3 and mutiplicity 1 , which intersects ( ℂ ∗ ) 4 (it is the twisted cubic), and another component “at infinity”: { ∂ 2 = ∂ 3 = 0 } , of degree 1 and multiplicity 1 = min { 2 × 2 , 1 × 1 } . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 17 / 46

Explanations The generating functions F i satisfy the differential equations: [Θ 1 (Θ 1 − 2 Θ 2 ) − x 1 ( 2 Θ 1 − Θ 2 + 4 ) ( 2 Θ 1 − Θ 2 + 3 )]( F ) = 0 , [Θ 2 ( − 2 Θ 1 + Θ 2 − 3 ) − x 2 ( 2 Θ 2 − Θ 1 ) ( 2 Θ 2 − Θ 1 + 1 )]( F ) = 0 . Consider the system of binomial equations: q 1 = ∂ 11 ∂ 31 − ∂ 22 , q 2 = ∂ 21 ∂ 41 − ∂ 32 in the commutative polynomial ring ℂ [ ∂ 1 , . . . , ∂ 4 ] . The zero set q 1 = q 2 = 0 has two irreducible components, one of degree 3 and mutiplicity 1 , which intersects ( ℂ ∗ ) 4 (it is the twisted cubic), and another component “at infinity”: { ∂ 2 = ∂ 3 = 0 } , of degree 1 and multiplicity 1 = min { 2 × 2 , 1 × 1 } . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 17 / 46

Explanations The generating functions F i satisfy the differential equations: [Θ 1 (Θ 1 − 2 Θ 2 ) − x 1 ( 2 Θ 1 − Θ 2 + 4 ) ( 2 Θ 1 − Θ 2 + 3 )]( F ) = 0 , [Θ 2 ( − 2 Θ 1 + Θ 2 − 3 ) − x 2 ( 2 Θ 2 − Θ 1 ) ( 2 Θ 2 − Θ 1 + 1 )]( F ) = 0 . Consider the system of binomial equations: q 1 = ∂ 11 ∂ 31 − ∂ 22 , q 2 = ∂ 21 ∂ 41 − ∂ 32 in the commutative polynomial ring ℂ [ ∂ 1 , . . . , ∂ 4 ] . The zero set q 1 = q 2 = 0 has two irreducible components, one of degree 3 and mutiplicity 1 , which intersects ( ℂ ∗ ) 4 (it is the twisted cubic), and another component “at infinity”: { ∂ 2 = ∂ 3 = 0 } , of degree 1 and multiplicity 1 = min { 2 × 2 , 1 × 1 } . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 17 / 46

Explanations The generating functions F i satisfy the differential equations: [Θ 1 (Θ 1 − 2 Θ 2 ) − x 1 ( 2 Θ 1 − Θ 2 + 4 ) ( 2 Θ 1 − Θ 2 + 3 )]( F ) = 0 , [Θ 2 ( − 2 Θ 1 + Θ 2 − 3 ) − x 2 ( 2 Θ 2 − Θ 1 ) ( 2 Θ 2 − Θ 1 + 1 )]( F ) = 0 . Consider the system of binomial equations: q 1 = ∂ 11 ∂ 31 − ∂ 22 , q 2 = ∂ 21 ∂ 41 − ∂ 32 in the commutative polynomial ring ℂ [ ∂ 1 , . . . , ∂ 4 ] . The zero set q 1 = q 2 = 0 has two irreducible components, one of degree 3 and mutiplicity 1 , which intersects ( ℂ ∗ ) 4 (it is the twisted cubic), and another component “at infinity”: { ∂ 2 = ∂ 3 = 0 } , of degree 1 and multiplicity 1 = min { 2 × 2 , 1 × 1 } . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 17 / 46

Explanations The generating functions F i satisfy the differential equations: [Θ 1 (Θ 1 − 2 Θ 2 ) − x 1 ( 2 Θ 1 − Θ 2 + 4 ) ( 2 Θ 1 − Θ 2 + 3 )]( F ) = 0 , [Θ 2 ( − 2 Θ 1 + Θ 2 − 3 ) − x 2 ( 2 Θ 2 − Θ 1 ) ( 2 Θ 2 − Θ 1 + 1 )]( F ) = 0 . Consider the system of binomial equations: q 1 = ∂ 11 ∂ 31 − ∂ 22 , q 2 = ∂ 21 ∂ 41 − ∂ 32 in the commutative polynomial ring ℂ [ ∂ 1 , . . . , ∂ 4 ] . The zero set q 1 = q 2 = 0 has two irreducible components, one of degree 3 and mutiplicity 1 , which intersects ( ℂ ∗ ) 4 (it is the twisted cubic), and another component “at infinity”: { ∂ 2 = ∂ 3 = 0 } , of degree 1 and multiplicity 1 = min { 2 × 2 , 1 × 1 } . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 17 / 46

Explanations The generating functions F i satisfy the differential equations: [Θ 1 (Θ 1 − 2 Θ 2 ) − x 1 ( 2 Θ 1 − Θ 2 + 4 ) ( 2 Θ 1 − Θ 2 + 3 )]( F ) = 0 , [Θ 2 ( − 2 Θ 1 + Θ 2 − 3 ) − x 2 ( 2 Θ 2 − Θ 1 ) ( 2 Θ 2 − Θ 1 + 1 )]( F ) = 0 . Consider the system of binomial equations: q 1 = ∂ 11 ∂ 31 − ∂ 22 , q 2 = ∂ 21 ∂ 41 − ∂ 32 in the commutative polynomial ring ℂ [ ∂ 1 , . . . , ∂ 4 ] . The zero set q 1 = q 2 = 0 has two irreducible components, one of degree 3 and mutiplicity 1 , which intersects ( ℂ ∗ ) 4 (it is the twisted cubic), and another component “at infinity”: { ∂ 2 = ∂ 3 = 0 } , of degree 1 and multiplicity 1 = min { 2 × 2 , 1 × 1 } . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 17 / 46

Explanations Consider the system of binomial equations: q 1 = ∂ 1 1 ∂ 1 3 − ∂ 2 2 , q 2 = ∂ 1 2 ∂ 1 4 − ∂ 2 3 in the commutative polynomial ring ℂ [ ∂ 1 , . . . , ∂ 4 ] . The zero set q 1 = q 2 = 0 has two irreducible components, one of degree 3 and mutiplicity 1 , which intersects ( ℂ ∗ ) 4 , and another component “at infinity”: { ∂ 2 = ∂ 3 = 0 } , of degree 1 and multiplicity 1 = min { 2 × 2 , 1 × 1 } . This multiplicity equals the intersection multiplicity at ( 0 , 0 ) of the system of two binomials in two variables: p 1 = ∂ a 3 − ∂ b 2 , p 2 = ∂ c 2 − ∂ d a = 1 , b = 2 , c = 1 , d = 2 3 , . The multiplicity of this only ( non homogeneous ) component at infinity is equal to the dimension of the space of solutions of the recurrences with finite support. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 18 / 46

Explanations Consider the system of binomial equations: q 1 = ∂ 1 1 ∂ 1 3 − ∂ 2 2 , q 2 = ∂ 1 2 ∂ 1 4 − ∂ 2 3 in the commutative polynomial ring ℂ [ ∂ 1 , . . . , ∂ 4 ] . The zero set q 1 = q 2 = 0 has two irreducible components, one of degree 3 and mutiplicity 1 , which intersects ( ℂ ∗ ) 4 , and another component “at infinity”: { ∂ 2 = ∂ 3 = 0 } , of degree 1 and multiplicity 1 = min { 2 × 2 , 1 × 1 } . This multiplicity equals the intersection multiplicity at ( 0 , 0 ) of the system of two binomials in two variables: p 1 = ∂ a 3 − ∂ b 2 , p 2 = ∂ c 2 − ∂ d a = 1 , b = 2 , c = 1 , d = 2 3 , . The multiplicity of this only ( non homogeneous ) component at infinity is equal to the dimension of the space of solutions of the recurrences with finite support. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 18 / 46

Explanations Consider the system of binomial equations: q 1 = ∂ 1 1 ∂ 1 3 − ∂ 2 2 , q 2 = ∂ 1 2 ∂ 1 4 − ∂ 2 3 in the commutative polynomial ring ℂ [ ∂ 1 , . . . , ∂ 4 ] . The zero set q 1 = q 2 = 0 has two irreducible components, one of degree 3 and mutiplicity 1 , which intersects ( ℂ ∗ ) 4 , and another component “at infinity”: { ∂ 2 = ∂ 3 = 0 } , of degree 1 and multiplicity 1 = min { 2 × 2 , 1 × 1 } . This multiplicity equals the intersection multiplicity at ( 0 , 0 ) of the system of two binomials in two variables: p 1 = ∂ a 3 − ∂ b 2 , p 2 = ∂ c 2 − ∂ d a = 1 , b = 2 , c = 1 , d = 2 3 , . The multiplicity of this only ( non homogeneous ) component at infinity is equal to the dimension of the space of solutions of the recurrences with finite support. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 18 / 46

Explanations Consider the system of binomial equations: q 1 = ∂ 1 1 ∂ 1 3 − ∂ 2 2 , q 2 = ∂ 1 2 ∂ 1 4 − ∂ 2 3 in the commutative polynomial ring ℂ [ ∂ 1 , . . . , ∂ 4 ] . The zero set q 1 = q 2 = 0 has two irreducible components, one of degree 3 and mutiplicity 1 , which intersects ( ℂ ∗ ) 4 , and another component “at infinity”: { ∂ 2 = ∂ 3 = 0 } , of degree 1 and multiplicity 1 = min { 2 × 2 , 1 × 1 } . This multiplicity equals the intersection multiplicity at ( 0 , 0 ) of the system of two binomials in two variables: p 1 = ∂ a 3 − ∂ b 2 , p 2 = ∂ c 2 − ∂ d a = 1 , b = 2 , c = 1 , d = 2 3 , . The multiplicity of this only ( non homogeneous ) component at infinity is equal to the dimension of the space of solutions of the recurrences with finite support. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 18 / 46

Explanations Consider the system of binomial equations: q 1 = ∂ 1 1 ∂ 1 3 − ∂ 2 2 , q 2 = ∂ 1 2 ∂ 1 4 − ∂ 2 3 in the commutative polynomial ring ℂ [ ∂ 1 , . . . , ∂ 4 ] . The zero set q 1 = q 2 = 0 has two irreducible components, one of degree 3 and mutiplicity 1 , which intersects ( ℂ ∗ ) 4 , and another component “at infinity”: { ∂ 2 = ∂ 3 = 0 } , of degree 1 and multiplicity 1 = min { 2 × 2 , 1 × 1 } . This multiplicity equals the intersection multiplicity at ( 0 , 0 ) of the system of two binomials in two variables: p 1 = ∂ a 3 − ∂ b 2 , p 2 = ∂ c 2 − ∂ d a = 1 , b = 2 , c = 1 , d = 2 3 , . The multiplicity of this only ( non homogeneous ) component at infinity is equal to the dimension of the space of solutions of the recurrences with finite support. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 18 / 46

Finite recurrences and polynomial solutions A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 19 / 46

Finite recurrences and polynomial solutions A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 19 / 46

Finite recurrences and polynomial solutions A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 19 / 46

Finite recurrences and polynomial solutions A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 19 / 46

General picture Let B ∈ ℤ n × 2 with rows b 1 , . . . , b n satisfying b 1 + ⋅ ⋅ ⋅ + b n = 0 . ∣ b ji ∣− 1 P i = ∏ ∏ ( b j ⋅ 휃 + c j − l ) , (7) b ji < 0 l = 0 b ji − 1 ∏ ∏ Q i = ( b j ⋅ 휃 + c j − l ) , and (8) b ji > 0 l = 0 H i = Q i − x i P i , (9) where b j ⋅ 휃 = ∑ 2 k = 1 b jk 휃 x k . The operators H i are called Horn operators and generate the left ideal Horn ( ℬ , c ) in the Weyl algebra D 2 . Call d i = ∑ b ij > 0 b ij = − ∑ b ij < 0 b ij the order of the operator H i . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 20 / 46

General picture Let B ∈ ℤ n × 2 with rows b 1 , . . . , b n satisfying b 1 + ⋅ ⋅ ⋅ + b n = 0 . ∣ b ji ∣− 1 P i = ∏ ∏ ( b j ⋅ 휃 + c j − l ) , (7) b ji < 0 l = 0 b ji − 1 ∏ ∏ Q i = ( b j ⋅ 휃 + c j − l ) , and (8) b ji > 0 l = 0 H i = Q i − x i P i , (9) where b j ⋅ 휃 = ∑ 2 k = 1 b jk 휃 x k . The operators H i are called Horn operators and generate the left ideal Horn ( ℬ , c ) in the Weyl algebra D 2 . Call d i = ∑ b ij > 0 b ij = − ∑ b ij < 0 b ij the order of the operator H i . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 20 / 46

General picture Let B ∈ ℤ n × 2 as above and let A ∈ ℤ ( n − 2 ) × n such that the columns b ( 1 ) , b ( 2 ) of B span ker ℚ ( A ) . Write any vector u ∈ ℝ n as u = u + − u − , where ( u + ) i = max ( u i , 0 ) , and ( u − ) i = − min ( u i , 0 ) . Definition T i = ∂ b ( i ) + − ∂ b ( i ) i = 1 , 2 . − , The left D n -ideal H ℬ ( c ) is defined by: H ℬ ( c ) = ⟨ T 1 , T 2 ⟩ + ⟨ A ⋅ 휃 − A ⋅ c ⟩ ⊆ D n . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 21 / 46

General picture Let B ∈ ℤ n × 2 as above and let A ∈ ℤ ( n − 2 ) × n such that the columns b ( 1 ) , b ( 2 ) of B span ker ℚ ( A ) . Write any vector u ∈ ℝ n as u = u + − u − , where ( u + ) i = max ( u i , 0 ) , and ( u − ) i = − min ( u i , 0 ) . Definition T i = ∂ b ( i ) + − ∂ b ( i ) i = 1 , 2 . − , The left D n -ideal H ℬ ( c ) is defined by: H ℬ ( c ) = ⟨ T 1 , T 2 ⟩ + ⟨ A ⋅ 휃 − A ⋅ c ⟩ ⊆ D n . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 21 / 46

General picture Let B ∈ ℤ n × 2 as above and let A ∈ ℤ ( n − 2 ) × n such that the columns b ( 1 ) , b ( 2 ) of B span ker ℚ ( A ) . Write any vector u ∈ ℝ n as u = u + − u − , where ( u + ) i = max ( u i , 0 ) , and ( u − ) i = − min ( u i , 0 ) . Definition T i = ∂ b ( i ) + − ∂ b ( i ) i = 1 , 2 . − , The left D n -ideal H ℬ ( c ) is defined by: H ℬ ( c ) = ⟨ T 1 , T 2 ⟩ + ⟨ A ⋅ 휃 − A ⋅ c ⟩ ⊆ D n . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 21 / 46

General picture Theorem [D.- Matusevich - Sadykov ’05] For generic complex parameters c 1 , . . . , c n , the ideals Horn ( ℬ , c ) and H ℬ ( c ) are holonomic. Moreover, rank ( H ℬ ( c )) = rank ( Horn ( ℬ , c )) = d 1 d 2 − ∑ 휈 ij = g ⋅ vol ( A ) + ∑ 휈 ij , b i , b j b i , b j depdt indepdt where the the pairs b i , b j of rows lie in opposite open quadrants of ℤ 2 . Remarks Solutions to recurrences with finite support correspond to (Laurent) polynomial solutions. These solutions come from (non homogeneous) primary components at infinity of the binomial ideal ⟨ T 1 , T 2 ⟩ . There are ∑ 휈 ij many linearly independent. For special parameters a special study is needed, along the lines in [D. - Matusevich and Miller ’10]. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 22 / 46

General picture Theorem [D.- Matusevich - Sadykov ’05] For generic complex parameters c 1 , . . . , c n , the ideals Horn ( ℬ , c ) and H ℬ ( c ) are holonomic. Moreover, rank ( H ℬ ( c )) = rank ( Horn ( ℬ , c )) = d 1 d 2 − ∑ 휈 ij = g ⋅ vol ( A ) + ∑ 휈 ij , b i , b j b i , b j depdt indepdt where the the pairs b i , b j of rows lie in opposite open quadrants of ℤ 2 . Remarks Solutions to recurrences with finite support correspond to (Laurent) polynomial solutions. These solutions come from (non homogeneous) primary components at infinity of the binomial ideal ⟨ T 1 , T 2 ⟩ . There are ∑ 휈 ij many linearly independent. For special parameters a special study is needed, along the lines in [D. - Matusevich and Miller ’10]. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 22 / 46

General picture Theorem [D.- Matusevich - Sadykov ’05] For generic complex parameters c 1 , . . . , c n , the ideals Horn ( ℬ , c ) and H ℬ ( c ) are holonomic. Moreover, rank ( H ℬ ( c )) = rank ( Horn ( ℬ , c )) = d 1 d 2 − ∑ 휈 ij = g ⋅ vol ( A ) + ∑ 휈 ij , b i , b j b i , b j depdt indepdt where the the pairs b i , b j of rows lie in opposite open quadrants of ℤ 2 . Remarks Solutions to recurrences with finite support correspond to (Laurent) polynomial solutions. These solutions come from (non homogeneous) primary components at infinity of the binomial ideal ⟨ T 1 , T 2 ⟩ . There are ∑ 휈 ij many linearly independent. For special parameters a special study is needed, along the lines in [D. - Matusevich and Miller ’10]. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 22 / 46

General picture Theorem [D.- Matusevich - Sadykov ’05] For generic complex parameters c 1 , . . . , c n , the ideals Horn ( ℬ , c ) and H ℬ ( c ) are holonomic. Moreover, rank ( H ℬ ( c )) = rank ( Horn ( ℬ , c )) = d 1 d 2 − ∑ 휈 ij = g ⋅ vol ( A ) + ∑ 휈 ij , b i , b j b i , b j depdt indepdt where the the pairs b i , b j of rows lie in opposite open quadrants of ℤ 2 . Remarks Solutions to recurrences with finite support correspond to (Laurent) polynomial solutions. These solutions come from (non homogeneous) primary components at infinity of the binomial ideal ⟨ T 1 , T 2 ⟩ . There are ∑ 휈 ij many linearly independent. For special parameters a special study is needed, along the lines in [D. - Matusevich and Miller ’10]. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 22 / 46

General phylosophy Moral of this story Key to the answer it the homogenization and translation to the A -side! A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 23 / 46

General phylosophy Moral of this story Key to the answer it the homogenization and translation to the A -side! A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 23 / 46

Examples of rational bivariate hypergeometric series The proof in the talk! Lemma: The series f ( s 1 , s 2 ) ( x ) := ∑ ( s 1 m 1 )!( s 2 m 2 )! x m 1 ( s 1 m 1 + s 2 m 2 )! 1 x m 2 2 . m ∈ ℕ 2 is a rational function for all ( s 1 , s 2 ) ∈ ℕ 2 . m ∈ ℕ 2 x m 1 1 x m 2 1 Proof: f ( 0 , 0 ) ( x 1 , x 2 ) = ∑ = ( 1 − x 1 )( 1 − x 2 ) , 2 ( m 1 + m 2 )! 1 x m 1 1 x m 2 f ( 1 , 1 ) ( x ) = ∑ = 1 − x 1 − x 2 , m ∈ ℕ 2 2 m 1 ! m 2 ! ( 2 m 1 )!( 2 m 2 )! x 2 m 1 ( 2 m 1 + 2 m 2 )! x 2 m 2 f ( 2 , 2 ) ( x 2 1 , x 2 2 ) = ∑ = m ∈ ℕ 2 1 2 1 4 ( f ( 1 , 1 ) ( x 1 , x 2 ) + f ( 1 , 1 ) ( − x 1 , x 2 ) + f ( 1 , 1 ) ( x 1 , − x 2 ) + f ( 1 , 1 ) ( − x 1 , − x 2 )) = 1 − x 2 1 − x 2 2 2 , 1 − 2 x 2 1 − 2 x 2 2 − 2 x 2 1 x 2 2 + x 4 1 + x 4 1 − x 1 − x 2 f ( 2 , 2 ) ( x 1 , x 2 ) = 2 . ⋄ 1 − 2 x 1 − 2 x 2 − 2 x 1 x 2 + x 2 1 + x 2 A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 24 / 46

Examples of rational bivariate hypergeometric series The proof in the talk! Lemma: The series f ( s 1 , s 2 ) ( x ) := ∑ ( s 1 m 1 )!( s 2 m 2 )! x m 1 ( s 1 m 1 + s 2 m 2 )! 1 x m 2 2 . m ∈ ℕ 2 is a rational function for all ( s 1 , s 2 ) ∈ ℕ 2 . m ∈ ℕ 2 x m 1 1 x m 2 1 Proof: f ( 0 , 0 ) ( x 1 , x 2 ) = ∑ = ( 1 − x 1 )( 1 − x 2 ) , 2 ( m 1 + m 2 )! 1 x m 1 1 x m 2 f ( 1 , 1 ) ( x ) = ∑ = 1 − x 1 − x 2 , m ∈ ℕ 2 2 m 1 ! m 2 ! ( 2 m 1 )!( 2 m 2 )! x 2 m 1 ( 2 m 1 + 2 m 2 )! x 2 m 2 f ( 2 , 2 ) ( x 2 1 , x 2 2 ) = ∑ = m ∈ ℕ 2 1 2 1 4 ( f ( 1 , 1 ) ( x 1 , x 2 ) + f ( 1 , 1 ) ( − x 1 , x 2 ) + f ( 1 , 1 ) ( x 1 , − x 2 ) + f ( 1 , 1 ) ( − x 1 , − x 2 )) = 1 − x 2 1 − x 2 2 2 , 1 − 2 x 2 1 − 2 x 2 2 − 2 x 2 1 x 2 2 + x 4 1 + x 4 1 − x 1 − x 2 f ( 2 , 2 ) ( x 1 , x 2 ) = 2 . ⋄ 1 − 2 x 1 − 2 x 2 − 2 x 1 x 2 + x 2 1 + x 2 A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 24 / 46

Examples of rational bivariate hypergeometric series The proof in the talk! Lemma: The series f ( s 1 , s 2 ) ( x ) := ∑ ( s 1 m 1 )!( s 2 m 2 )! x m 1 ( s 1 m 1 + s 2 m 2 )! 1 x m 2 2 . m ∈ ℕ 2 is a rational function for all ( s 1 , s 2 ) ∈ ℕ 2 . m ∈ ℕ 2 x m 1 1 x m 2 1 Proof: f ( 0 , 0 ) ( x 1 , x 2 ) = ∑ = ( 1 − x 1 )( 1 − x 2 ) , 2 ( m 1 + m 2 )! 1 x m 1 1 x m 2 f ( 1 , 1 ) ( x ) = ∑ = 1 − x 1 − x 2 , m ∈ ℕ 2 2 m 1 ! m 2 ! ( 2 m 1 )!( 2 m 2 )! x 2 m 1 ( 2 m 1 + 2 m 2 )! x 2 m 2 f ( 2 , 2 ) ( x 2 1 , x 2 2 ) = ∑ = m ∈ ℕ 2 1 2 1 4 ( f ( 1 , 1 ) ( x 1 , x 2 ) + f ( 1 , 1 ) ( − x 1 , x 2 ) + f ( 1 , 1 ) ( x 1 , − x 2 ) + f ( 1 , 1 ) ( − x 1 , − x 2 )) = 1 − x 2 1 − x 2 2 2 , 1 − 2 x 2 1 − 2 x 2 2 − 2 x 2 1 x 2 2 + x 4 1 + x 4 1 − x 1 − x 2 f ( 2 , 2 ) ( x 1 , x 2 ) = 2 . ⋄ 1 − 2 x 1 − 2 x 2 − 2 x 1 x 2 + x 2 1 + x 2 A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 24 / 46

Examples of rational bivariate hypergeometric series The proof in the talk! Lemma: The series f ( s 1 , s 2 ) ( x ) := ∑ ( s 1 m 1 )!( s 2 m 2 )! x m 1 ( s 1 m 1 + s 2 m 2 )! 1 x m 2 2 . m ∈ ℕ 2 is a rational function for all ( s 1 , s 2 ) ∈ ℕ 2 . m ∈ ℕ 2 x m 1 1 x m 2 1 Proof: f ( 0 , 0 ) ( x 1 , x 2 ) = ∑ = ( 1 − x 1 )( 1 − x 2 ) , 2 ( m 1 + m 2 )! 1 x m 1 1 x m 2 f ( 1 , 1 ) ( x ) = ∑ = 1 − x 1 − x 2 , m ∈ ℕ 2 2 m 1 ! m 2 ! ( 2 m 1 )!( 2 m 2 )! x 2 m 1 ( 2 m 1 + 2 m 2 )! x 2 m 2 f ( 2 , 2 ) ( x 2 1 , x 2 2 ) = ∑ = m ∈ ℕ 2 1 2 1 4 ( f ( 1 , 1 ) ( x 1 , x 2 ) + f ( 1 , 1 ) ( − x 1 , x 2 ) + f ( 1 , 1 ) ( x 1 , − x 2 ) + f ( 1 , 1 ) ( − x 1 , − x 2 )) = 1 − x 2 1 − x 2 2 2 , 1 − 2 x 2 1 − 2 x 2 2 − 2 x 2 1 x 2 2 + x 4 1 + x 4 1 − x 1 − x 2 f ( 2 , 2 ) ( x 1 , x 2 ) = 2 . ⋄ 1 − 2 x 1 − 2 x 2 − 2 x 1 x 2 + x 2 1 + x 2 A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 24 / 46

Examples of rational bivariate hypergeometric series The proof in the talk! Lemma: The series f ( s 1 , s 2 ) ( x ) := ∑ ( s 1 m 1 )!( s 2 m 2 )! x m 1 ( s 1 m 1 + s 2 m 2 )! 1 x m 2 2 . m ∈ ℕ 2 is a rational function for all ( s 1 , s 2 ) ∈ ℕ 2 . m ∈ ℕ 2 x m 1 1 x m 2 1 Proof: f ( 0 , 0 ) ( x 1 , x 2 ) = ∑ = ( 1 − x 1 )( 1 − x 2 ) , 2 ( m 1 + m 2 )! 1 x m 1 1 x m 2 f ( 1 , 1 ) ( x ) = ∑ = 1 − x 1 − x 2 , m ∈ ℕ 2 2 m 1 ! m 2 ! ( 2 m 1 )!( 2 m 2 )! x 2 m 1 ( 2 m 1 + 2 m 2 )! x 2 m 2 f ( 2 , 2 ) ( x 2 1 , x 2 2 ) = ∑ = m ∈ ℕ 2 1 2 1 4 ( f ( 1 , 1 ) ( x 1 , x 2 ) + f ( 1 , 1 ) ( − x 1 , x 2 ) + f ( 1 , 1 ) ( x 1 , − x 2 ) + f ( 1 , 1 ) ( − x 1 , − x 2 )) = 1 − x 2 1 − x 2 2 2 , 1 − 2 x 2 1 − 2 x 2 2 − 2 x 2 1 x 2 2 + x 4 1 + x 4 1 − x 1 − x 2 f ( 2 , 2 ) ( x 1 , x 2 ) = 2 . ⋄ 1 − 2 x 1 − 2 x 2 − 2 x 1 x 2 + x 2 1 + x 2 A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 24 / 46

Examples of rational bivariate hypergeometric series The proof in the talk! Lemma: The series f ( s 1 , s 2 ) ( x ) := ∑ ( s 1 m 1 )!( s 2 m 2 )! x m 1 ( s 1 m 1 + s 2 m 2 )! 1 x m 2 2 . m ∈ ℕ 2 is a rational function for all ( s 1 , s 2 ) ∈ ℕ 2 . m ∈ ℕ 2 x m 1 1 x m 2 1 Proof: f ( 0 , 0 ) ( x 1 , x 2 ) = ∑ = ( 1 − x 1 )( 1 − x 2 ) , 2 ( m 1 + m 2 )! 1 x m 1 1 x m 2 f ( 1 , 1 ) ( x ) = ∑ = 1 − x 1 − x 2 , m ∈ ℕ 2 2 m 1 ! m 2 ! ( 2 m 1 )!( 2 m 2 )! x 2 m 1 ( 2 m 1 + 2 m 2 )! x 2 m 2 f ( 2 , 2 ) ( x 2 1 , x 2 2 ) = ∑ = m ∈ ℕ 2 1 2 1 4 ( f ( 1 , 1 ) ( x 1 , x 2 ) + f ( 1 , 1 ) ( − x 1 , x 2 ) + f ( 1 , 1 ) ( x 1 , − x 2 ) + f ( 1 , 1 ) ( − x 1 , − x 2 )) = 1 − x 2 1 − x 2 2 2 , 1 − 2 x 2 1 − 2 x 2 2 − 2 x 2 1 x 2 2 + x 4 1 + x 4 1 − x 1 − x 2 f ( 2 , 2 ) ( x 1 , x 2 ) = 2 . ⋄ 1 − 2 x 1 − 2 x 2 − 2 x 1 x 2 + x 2 1 + x 2 A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 24 / 46

Examples of rational bivariate hypergeometric series The proof in the talk! Lemma: The series f ( s 1 , s 2 ) ( x ) := ∑ ( s 1 m 1 )!( s 2 m 2 )! x m 1 ( s 1 m 1 + s 2 m 2 )! 1 x m 2 2 . m ∈ ℕ 2 is a rational function for all ( s 1 , s 2 ) ∈ ℕ 2 . m ∈ ℕ 2 x m 1 1 x m 2 1 Proof: f ( 0 , 0 ) ( x 1 , x 2 ) = ∑ = ( 1 − x 1 )( 1 − x 2 ) , 2 ( m 1 + m 2 )! 1 x m 1 1 x m 2 f ( 1 , 1 ) ( x ) = ∑ = 1 − x 1 − x 2 , m ∈ ℕ 2 2 m 1 ! m 2 ! ( 2 m 1 )!( 2 m 2 )! x 2 m 1 ( 2 m 1 + 2 m 2 )! x 2 m 2 f ( 2 , 2 ) ( x 2 1 , x 2 2 ) = ∑ = m ∈ ℕ 2 1 2 1 4 ( f ( 1 , 1 ) ( x 1 , x 2 ) + f ( 1 , 1 ) ( − x 1 , x 2 ) + f ( 1 , 1 ) ( x 1 , − x 2 ) + f ( 1 , 1 ) ( − x 1 , − x 2 )) = 1 − x 2 1 − x 2 2 2 , 1 − 2 x 2 1 − 2 x 2 2 − 2 x 2 1 x 2 2 + x 4 1 + x 4 1 − x 1 − x 2 f ( 2 , 2 ) ( x 1 , x 2 ) = 2 . ⋄ 1 − 2 x 1 − 2 x 2 − 2 x 1 x 2 + x 2 1 + x 2 A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 24 / 46

Using residues A second proof! m ∈ ℕ 2 ( s 1 m 1 + s 2 m 2 )! ( s 1 m 1 )!( s 2 m 2 )! x m 1 1 x m 2 Proof: The series f ( s 1 , s 2 ) ( x ) := ∑ 2 . defines a rational function for all ( s 1 , s 2 ) ∈ ℕ 2 because it equals the following residue: ( t s 1 1 t s 2 2 / ( t 1 + t 2 + 1 ) dt 1 ∧ dt 2 ) ∑ Res 휉 f ( s 1 , s 2 ) ( x ) = = ( x 1 + t s 1 1 )( x 2 + t s 2 t 1 t 2 2 ) 휉 s 1 1 = − x 1 ,휉 s 2 2 = − x 2 1 1 ∑ = 휉 1 + 휉 2 + 1 . ⋄ s 1 s 2 휉 s 1 1 = − x 1 ,휉 s 2 2 = − x 2 A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 25 / 46

Using residues A second proof! m ∈ ℕ 2 ( s 1 m 1 + s 2 m 2 )! ( s 1 m 1 )!( s 2 m 2 )! x m 1 1 x m 2 Proof: The series f ( s 1 , s 2 ) ( x ) := ∑ 2 . defines a rational function for all ( s 1 , s 2 ) ∈ ℕ 2 because it equals the following residue: ( t s 1 1 t s 2 2 / ( t 1 + t 2 + 1 ) dt 1 ∧ dt 2 ) ∑ Res 휉 f ( s 1 , s 2 ) ( x ) = = ( x 1 + t s 1 1 )( x 2 + t s 2 t 1 t 2 2 ) 휉 s 1 1 = − x 1 ,휉 s 2 2 = − x 2 1 1 ∑ = 휉 1 + 휉 2 + 1 . ⋄ s 1 s 2 휉 s 1 1 = − x 1 ,휉 s 2 2 = − x 2 A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 25 / 46

Using residues A second proof! m ∈ ℕ 2 ( s 1 m 1 + s 2 m 2 )! ( s 1 m 1 )!( s 2 m 2 )! x m 1 1 x m 2 Proof: The series f ( s 1 , s 2 ) ( x ) := ∑ 2 . defines a rational function for all ( s 1 , s 2 ) ∈ ℕ 2 because it equals the following residue: ( t s 1 1 t s 2 2 / ( t 1 + t 2 + 1 ) dt 1 ∧ dt 2 ) ∑ Res 휉 f ( s 1 , s 2 ) ( x ) = = ( x 1 + t s 1 1 )( x 2 + t s 2 t 1 t 2 2 ) 휉 s 1 1 = − x 1 ,휉 s 2 2 = − x 2 1 1 ∑ = 휉 1 + 휉 2 + 1 . ⋄ s 1 s 2 휉 s 1 1 = − x 1 ,휉 s 2 2 = − x 2 A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 25 / 46

Rational bivariate hypergeometric series Question When is a hypergeometric series in 2 variables rational? 2 ) and d j = ( d j 1 , d j Let c i = ( c i 1 , c i 2 ) for i = 1 , . . . , r ; j = 1 , . . . , s be vectors in ℕ 2 . When is the series ∏ r i = 1 ( c i 1 m 1 + c i 2 m 2 )! x m 1 1 x m 2 ∑ 2 ∏ s j = 1 ( d j 1 m 1 + d j 2 m 2 )! m ∈ ℕ 2 the Taylor expansion of a rational function? A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 26 / 46

Rational bivariate hypergeometric series Question When is a hypergeometric series in 2 variables rational? 2 ) and d j = ( d j 1 , d j Let c i = ( c i 1 , c i 2 ) for i = 1 , . . . , r ; j = 1 , . . . , s be vectors in ℕ 2 . When is the series ∏ r i = 1 ( c i 1 m 1 + c i 2 m 2 )! x m 1 1 x m 2 ∑ 2 ∏ s j = 1 ( d j 1 m 1 + d j 2 m 2 )! m ∈ ℕ 2 the Taylor expansion of a rational function? A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 26 / 46

Rational bivariate hypergeometric series Question When is a hypergeometric series in 2 variables rational? 2 ) and d j = ( d j 1 , d j Let c i = ( c i 1 , c i 2 ) for i = 1 , . . . , r ; j = 1 , . . . , s be vectors in ℕ 2 . When is the series ∏ r i = 1 ( c i 1 m 1 + c i 2 m 2 )! x m 1 1 x m 2 ∑ 2 ∏ s j = 1 ( d j 1 m 1 + d j 2 m 2 )! m ∈ ℕ 2 the Taylor expansion of a rational function? A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 26 / 46

Rational bivariate hypergeometric series Answer Theorem: 2 ) and d j = ( d j 1 , d j Let c i = ( c i 1 , c i 2 ) for i = 1 , . . . , r ; j = 1 , . . . , s be vectors in ℕ 2 (with ∑ c i = ∑ d j ). ∏ r i = 1 ( c i 1 m 1 + c i 2 m 2 )! 2 m 2 )! x m 1 1 x m 2 The series ∑ is the Taylor expansion of m ∈ ℕ 2 2 ∏ s j = 1 ( d j 1 m 1 + d j a rational function if and only if it is of the form f ( s 1 , s 2 ) ( x ) . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 27 / 46

Rational bivariate hypergeometric series Answer Theorem: 2 ) and d j = ( d j 1 , d j Let c i = ( c i 1 , c i 2 ) for i = 1 , . . . , r ; j = 1 , . . . , s be vectors in ℕ 2 (with ∑ c i = ∑ d j ). ∏ r i = 1 ( c i 1 m 1 + c i 2 m 2 )! 2 m 2 )! x m 1 1 x m 2 The series ∑ is the Taylor expansion of m ∈ ℕ 2 2 ∏ s j = 1 ( d j 1 m 1 + d j a rational function if and only if it is of the form f ( s 1 , s 2 ) ( x ) . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 27 / 46

Rational bivariate hypergeometric series Answer Theorem: 2 ) and d j = ( d j 1 , d j Let c i = ( c i 1 , c i 2 ) for i = 1 , . . . , r ; j = 1 , . . . , s be vectors in ℕ 2 (with ∑ c i = ∑ d j ). ∏ r i = 1 ( c i 1 m 1 + c i 2 m 2 )! 2 m 2 )! x m 1 1 x m 2 The series ∑ is the Taylor expansion of m ∈ ℕ 2 2 ∏ s j = 1 ( d j 1 m 1 + d j a rational function if and only if it is of the form f ( s 1 , s 2 ) ( x ) . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 27 / 46

Rational bivariate hypergeometric series Answer Theorem: 2 ) and d j = ( d j 1 , d j Let c i = ( c i 1 , c i 2 ) for i = 1 , . . . , r ; j = 1 , . . . , s be vectors in ℕ 2 (with ∑ c i = ∑ d j ). ∏ r i = 1 ( c i 1 m 1 + c i 2 m 2 )! 2 m 2 )! x m 1 1 x m 2 The series ∑ is the Taylor expansion of m ∈ ℕ 2 2 ∏ s j = 1 ( d j 1 m 1 + d j a rational function if and only if it is of the form f ( s 1 , s 2 ) ( x ) . A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 27 / 46

Gessell and Xin´s example of a rational bivariate hypergeometric series What if the cone is not the first orthant? We had ∑ ( m + n 1 − xy ) x m y n , GX ( x , y ) = 1 − xy 2 − 3 xy − x 2 y = 2 m − n where we are summing over the lattice points in the (pointed) non unimodular convex cone ℝ ≥ 0 ( 1 , 2 ) + ℝ ≥ 0 ( 2 , 1 ) . Calling m 1 = 2 m − n , m 2 = 2 n − m (so that m = 2 m 1 + m 2 , n = m 1 + 2 m 2 ): 3 3 1 − xy ( m 1 + m 2 )! u m 1 1 u m 2 1 − xy 2 − 3 xy − x 2 y = ∑ 2 , ( m 1 , m 2 ) ∈ L ∩ ℕ 2 m 1 ! m 2 ! where L = ℤ ( 1 , 2 ) + ℤ ( 2 , 1 ) = { ( m 1 , m 2 ) ∈ ℤ 2 : m 1 ≡ m 2 mod 3 } and u 3 1 = x 2 y , u 3 2 = xy 2 . The shape of the non zero coefficients is the expected, but the sum is over a sublattice. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 28 / 46

Gessell and Xin´s example of a rational bivariate hypergeometric series What if the cone is not the first orthant? We had ∑ ( m + n 1 − xy ) x m y n , GX ( x , y ) = 1 − xy 2 − 3 xy − x 2 y = 2 m − n where we are summing over the lattice points in the (pointed) non unimodular convex cone ℝ ≥ 0 ( 1 , 2 ) + ℝ ≥ 0 ( 2 , 1 ) . Calling m 1 = 2 m − n , m 2 = 2 n − m (so that m = 2 m 1 + m 2 , n = m 1 + 2 m 2 ): 3 3 1 − xy ( m 1 + m 2 )! u m 1 1 u m 2 1 − xy 2 − 3 xy − x 2 y = ∑ 2 , ( m 1 , m 2 ) ∈ L ∩ ℕ 2 m 1 ! m 2 ! where L = ℤ ( 1 , 2 ) + ℤ ( 2 , 1 ) = { ( m 1 , m 2 ) ∈ ℤ 2 : m 1 ≡ m 2 mod 3 } and u 3 1 = x 2 y , u 3 2 = xy 2 . The shape of the non zero coefficients is the expected, but the sum is over a sublattice. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 28 / 46

Gessell and Xin´s example of a rational bivariate hypergeometric series What if the cone is not the first orthant? We had ∑ ( m + n 1 − xy ) x m y n , GX ( x , y ) = 1 − xy 2 − 3 xy − x 2 y = 2 m − n where we are summing over the lattice points in the (pointed) non unimodular convex cone ℝ ≥ 0 ( 1 , 2 ) + ℝ ≥ 0 ( 2 , 1 ) . Calling m 1 = 2 m − n , m 2 = 2 n − m (so that m = 2 m 1 + m 2 , n = m 1 + 2 m 2 ): 3 3 1 − xy ( m 1 + m 2 )! u m 1 1 u m 2 1 − xy 2 − 3 xy − x 2 y = ∑ 2 , ( m 1 , m 2 ) ∈ L ∩ ℕ 2 m 1 ! m 2 ! where L = ℤ ( 1 , 2 ) + ℤ ( 2 , 1 ) = { ( m 1 , m 2 ) ∈ ℤ 2 : m 1 ≡ m 2 mod 3 } and u 3 1 = x 2 y , u 3 2 = xy 2 . The shape of the non zero coefficients is the expected, but the sum is over a sublattice. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 28 / 46

Gessell and Xin´s example of a rational bivariate hypergeometric series What if the cone is not the first orthant? We had ∑ ( m + n 1 − xy ) x m y n , GX ( x , y ) = 1 − xy 2 − 3 xy − x 2 y = 2 m − n where we are summing over the lattice points in the (pointed) non unimodular convex cone ℝ ≥ 0 ( 1 , 2 ) + ℝ ≥ 0 ( 2 , 1 ) . Calling m 1 = 2 m − n , m 2 = 2 n − m (so that m = 2 m 1 + m 2 , n = m 1 + 2 m 2 ): 3 3 1 − xy ( m 1 + m 2 )! u m 1 1 u m 2 1 − xy 2 − 3 xy − x 2 y = ∑ 2 , ( m 1 , m 2 ) ∈ L ∩ ℕ 2 m 1 ! m 2 ! where L = ℤ ( 1 , 2 ) + ℤ ( 2 , 1 ) = { ( m 1 , m 2 ) ∈ ℤ 2 : m 1 ≡ m 2 mod 3 } and u 3 1 = x 2 y , u 3 2 = xy 2 . The shape of the non zero coefficients is the expected, but the sum is over a sublattice. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 28 / 46

Gessell and Xin´s example of a rational bivariate hypergeometric series What if the cone is not the first orthant? We had ∑ ( m + n 1 − xy ) x m y n , GX ( x , y ) = 1 − xy 2 − 3 xy − x 2 y = 2 m − n where we are summing over the lattice points in the (pointed) non unimodular convex cone ℝ ≥ 0 ( 1 , 2 ) + ℝ ≥ 0 ( 2 , 1 ) . Calling m 1 = 2 m − n , m 2 = 2 n − m (so that m = 2 m 1 + m 2 , n = m 1 + 2 m 2 ): 3 3 1 − xy ( m 1 + m 2 )! u m 1 1 u m 2 1 − xy 2 − 3 xy − x 2 y = ∑ 2 , ( m 1 , m 2 ) ∈ L ∩ ℕ 2 m 1 ! m 2 ! where L = ℤ ( 1 , 2 ) + ℤ ( 2 , 1 ) = { ( m 1 , m 2 ) ∈ ℤ 2 : m 1 ≡ m 2 mod 3 } and u 3 1 = x 2 y , u 3 2 = xy 2 . The shape of the non zero coefficients is the expected, but the sum is over a sublattice. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 28 / 46

Gessell and Xin´s example of a rational bivariate hypergeometric series What if the cone is not the first orthant? We had ∑ ( m + n 1 − xy ) x m y n , GX ( x , y ) = 1 − xy 2 − 3 xy − x 2 y = 2 m − n where we are summing over the lattice points in the (pointed) non unimodular convex cone ℝ ≥ 0 ( 1 , 2 ) + ℝ ≥ 0 ( 2 , 1 ) . Calling m 1 = 2 m − n , m 2 = 2 n − m (so that m = 2 m 1 + m 2 , n = m 1 + 2 m 2 ): 3 3 1 − xy ( m 1 + m 2 )! u m 1 1 u m 2 1 − xy 2 − 3 xy − x 2 y = ∑ 2 , ( m 1 , m 2 ) ∈ L ∩ ℕ 2 m 1 ! m 2 ! where L = ℤ ( 1 , 2 ) + ℤ ( 2 , 1 ) = { ( m 1 , m 2 ) ∈ ℤ 2 : m 1 ≡ m 2 mod 3 } and u 3 1 = x 2 y , u 3 2 = xy 2 . The shape of the non zero coefficients is the expected, but the sum is over a sublattice. A. Dickenstein (U. Buenos Aires) Hyp. series with AG dressing FPSAC 2010, 08/05/10 28 / 46