Lecture 1. Frits Beukers. October 23, Lecture 1 October 23, / 34

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1 Lecture 1 Frits Beukers October 23, 2018 Lecture 1 October 23, / 34

2 Gauss hypergeometric function Euler and Gauss defined 2F 1 ( α β γ ) z = n=0 (α) n (β) n n!(γ) n z n where (x) n = x(x + 1) (x + n 1) (Pochhammer symbol). Examples ( ) F 1 2 z = 1 z log(1 z) ( ) 2 1/2 1 2F 1 1 z = (1 z) 1/2 ( ) 3 1/2 1/2 2F 1 1 z = 2 π 1 0 dt (1 t 2 )(1 zt 2 ) We will use the notation F (α, β, γ z). Introduction Lecture 1 October 23, / 34

3 Differential equation z(z 1)F + ((α + β + 1)z γ)f + αβf = 0 This is a Fuchsian differential equation of order 2 with singularities at 0, 1,. Local solutions at z = 0: F (α, β, γ z) z 1 γ F (α + 1 γ, β + 1 γ, 2 γ z) At z = (1/z) α F (α, α + 1 γ, α + 1 β 1/z) (1/z) β F (β, β + 1 γ, β + 1 α 1/z) Introduction Lecture 1 October 23, / 34

4 Monodromy Let V be solution space of hypergeometic equation. Analytic continuation gives us the monodromy representation ρ : π 1 (C \ {0, 1}) GL(V ) Y z0 0 1 X Monodromy matrices: M i := ρ(g i ), i = 0, 1, with relation M M 1 M 0 = 1. Introduction Lecture 1 October 23, / 34

5 Monodromy properties Denote e(x) = exp(2πix). Eigenvalues M 0 : 1, e( γ) M 1 : 1, e(γ α β) M : e(α), e(β) Proposition Let A, B GL(2, C) with eigenvalues a 1, a 2 resp b 1, b 2 and such that A 1 B has eigenvalue 1. Let G = A, B. Then G irreducible {a 1, a 2 } {b 1, b 2 } =. In that case A, B are uniquely determined up to common conjugation. Application: A = M 1 0, B = M. So, monodromy irreducible {α, β}(mod Z) and {0, γ}(mod Z) disjoint. Introduction Lecture 1 October 23, / 34

6 Explicit matrices Characteristic polynomial of M 1 0 is x 2 (1 + e(γ))x + e(γ) of M is x 2 (e(α) + e(β))x + e(α + β). Up to common conjugation: ( ) M0 1 0 e(γ) = e(γ) M = ( ) 0 e(α + β). 1 e(α) + e(β) Theorem Suppose α, β, γ (0, 1]. Then there exists a Hermitian form F on C 2 such that F (gx, gy) = F (x, y) for all g M 0, M. This form is definite if and only if γ lies between α and β. Introduction Lecture 1 October 23, / 34

7 Schwarz s list In 1873 H.A. Schwarz ( ) gave a list of all parameter triples α, β, γ such that 2 F α β 1 γ z is algebraic in z. All triples are in Q. ( ) An example, 2 F 19/60 49/60 1 4/5 z is algebraic of degree 720. Its Galois group is a central extension of the alternating group A 5 by a cyclic group of order 60. Such functions were used in F.Klein s Vorlesungen über das Ikosaeder. Introduction Lecture 1 October 23, / 34

8 Clausen-Thomae functions Let α 1,..., α d and β 1,..., β d 1 be any parameters and β d = 1. Define ( ) α1,..., α d df d 1 β 1,..., β d 1 z (α 1 ) k (α d ) k = (β 1 ) k (β d 1 ) k k! zk k=0 where (x) n = x(x + 1) (x + n 1) is the Pochhammer symbol. Hypergeometric equation z(d+α 1 ) (D+α d )F = (D+β 1 1) (D+β d 1)F, D = z d dz This is a Fuchsian differential equation of order d with singularities at 0, 1,. Introduction Lecture 1 October 23, / 34

9 Monodromy Theorem Monodromy irreducible {α 1,..., α d } and {β 1,..., β d } disjoint modulo Z. Levelt s theorem (1960) Write d i=1 (x e(β i)) = x d + B 1 x d B d and d i=1 (x e(α i)) = x d + A 1 x d A d. Then up to common conjugation M and M0 1 equal A d A d A d A B d B d B d B 1 Introduction Lecture 1 October 23, / 34

10 Invariant Hermitean form Suppose that α i, β j R for all i, j. Theorem Then there exists a unique (up to scalars) monodromy invariant Hermitean form F. That is, F (gx, gy) = F (x, y) for all monodromy matrices g. Theorem The Hermitian form F is definite if and only if the sets {α 1,..., α d } and {β 1,..., β d } interlace modulo Z. Introduction Lecture 1 October 23, / 34

11 Interlacing Interlacing sets in [0, 1) when d = 4, 0 1 Two sets {α 1,..., α d } and {β 1,..., β d } are said to interlace modulo Z if the sets {α i α i } i=1,...,d and {β i β i } i=1,...,d interlace in [0, 1). Introduction Lecture 1 October 23, / 34

12 Finite monodromy Suppose {α 1,..., α d } and {β 1,..., β d } are sets of rational numbers disjoint modulo Z. Let N be a common denominator. Suppose the monodromy group is finite. Then there is an invariant definite Hermitian form. Hence the parameter sets interlace mod Z. Monodromy matrices have elements in Z[e(1/N)]. Apply Galois element ζ N ζ p N, gcd(p, N) = 1 to monodromy matrices. Get monodromy with parameter sets {pα i } and {pβ i }. Hence they interlace modulo Z. Introduction Lecture 1 October 23, / 34

13 Algebraic hypergeometric functions Converse also holds. Theorem (Beukers-Heckman, 1986) A hypergeometric group is finite if and only if the sets {pα 1,..., pα d } and {pβ 1,..., pβ d } interlace mod Z for every integer p with gcd(p, N) = 1. Example: F (x) = 8 F 7 ( 1/30 7/30 11/30 13/30 17/30 19/30 23/30 29/30 1/5 1/3 2/5 1/2 3/5 2/3 4/5 ) x which equals n 0 (30n)!n! ( z ) n. (15n)!(10n)!(6n)! Introduction Lecture 1 October 23, / 34

14 Appell s functions Consider F 1 (α, β, β, γ, x, y) = m,n 0 (α) m+n (β) m (β ) n m!n!(γ) m+n x m y n F 2 (α, β, β, γ, γ, x, y) = m,n 0 F 3 (α, α, β, β, γ, x, y) = m,n 0 F 4 (α, β, γ, γ, x, y) = m,n 0 (α) m+n (β) m (β ) n m!n!(γ) m (γ ) n (α) m (α) n (β) m (β ) n m!n!(γ) m+n (α) m+n (β) m+n m!n!(γ) m (γ ) n x m y n x m y n x m y n These are the Appell hypergeometric functions in two variables, introduced in Bivariate hypergeometric functions Lecture 1 October 23, / 34

15 Appell differential equation The Appell functions satisfy a system of partial linear differential equations of order 2. For example, F 4 (α, β, γ, γ, x, y) satisfies x(1 x)f xx y 2 F yy 2xyF xy + γf x (α + β + 1)(xF x + yf y ) = αβf y(1 y)f yy x 2 F xx 2xyF xy + γ F y (α + β + 1)(xF x + yf y ) Studied by Picard and Goursat. = αβf Bivariate hypergeometric functions Lecture 1 October 23, / 34

16 Lauricella functions Further generalisation by Lauricella (1893), F A (a, b, c x) = m 0 F B (a, b, c x) = m 0 F C (a, b, c x) = m 0 F D (a, b, c x) = m 0 (a) m (b) m x m x x n < 1 (c) m m! (a) m (b) m (c) m m! xm i : x i < 1 (a) m (b) m x m x x n < 1 (c) m m! (a) m (b) m (c) m m! xm i : x i < 1 When n = 2 these functions coincide with Appell s F 2, F 3, F 4, F 1 respectively. When n = 1, they all coincide with Gauss 2 F 1. Bivariate hypergeometric functions Lecture 1 October 23, / 34

17 The A-polytope Start with a finite subset A Z r R r. We assume The Z-span of A is Z r There is a linear form h such that h(a) = 1 for all a A. Define a vector of parameters α = (α 1,..., α r ) R r Remember: The set A and the vector α will completely characterise a so-called A-hypergeometric system of differential equations. A-hypergeometric functions Lecture 1 October 23, / 34

18 Lattice of relations Write A = {a 1,..., a N }. The lattice of relations L Z N is formed by all l = (l 1,..., l N ) Z N such that l 1 a 1 + l 2 a l N a N = 0. Let h be the form such that h(a i ) = 1 for i = 1,..., r. Apply h to any relation l 1 a l N a N = 0. Then we get N i=1 l i = 0 for all l L. A-hypergeometric functions Lecture 1 October 23, / 34

19 Formal A-hypergeometric series Choose γ 1,..., γ N such that α = γ 1 a γ N a N. Note that γ = (γ 1,..., γ N ) is determined modulo L R. Let v 1,..., v N be variables and consider Φ = l L v l 1+γ 1 1 v l N+γ N N Γ(l 1 + γ 1 + 1) Γ(l N + γ N + 1). A-hypergeometric functions Lecture 1 October 23, / 34

20 Homogeneity equations For any j = 1,..., N write a j = (a 1j,..., a rj ) t. Note that a i1 l a in l N = 0 for every l L and every i. For i = 1,..., r define the differential operator Note that Z i = a i1 v a in v N v 1 v N Z i (v l 1+γ 1 1 v l N+γ N N ) = (a i1 (l 1 + γ 1 ) + + a in (l N + γ N ))v l+γ = α i v l+γ Hence (Z i α i )Φ = 0. These equations reflect the homogeneity property Ψ(t a 1 v 1,, t a N v N ) = t α Ψ(v 1,..., v N ) for any solution Ψ and any t (C ) r. Here t a denotes t a 1 1 tar r. A-hypergeometric functions Lecture 1 October 23, / 34

21 Box equations Let (λ 1,..., λ N ) L. Define the operator ) λi λ = λ i >0 ( v i ) λi λ i <0 ( v i Let λ + be the vector with components max(0, λ i ) and λ with components min(0, λ i ). Then λ = λ + λ. Notice that λ v l+γ Γ(l + γ + 1) = v l+γ λ+ Γ(l + γ λ + + 1) v l+γ+λ Γ(l + γ + λ + 1). Since λ + λ = λ L summation over L gives equal sums that cancel. A-hypergeometric functions Lecture 1 October 23, / 34

22 A-hypergeometric system of equations The system of differential equations λ Φ = 0, λ L and (Z i α i )Φ = 0, i = 1, 2,..., r was first explicitly described by Gel fand, Kapranov and Zelevinsky around They called these equations A-hypergeometric equations and their analytic solutions A-hypergeometric functions. We denote the system by H A (α). In his book on Generalised hypergeometric equations, which appeared in 1990, B.Dwork independently arrives at the same equations, but in the language of differential modules. A-hypergeometric functions Lecture 1 October 23, / 34

23 Example 1, Gauss 2 F 1 Gauss F (α, β, γ z) is proportional to n 0 Application of Γ-identities gives n 0 Γ(n + α)γ(n + β) Γ(n + γ)γ(n + 1) zn. z n Γ( n + 1 α)γ( n + 1 β)γ(n + γ)γ(n + 1). The lattice L is spanned by ( 1, 1, 1, 1). A set A is given by A-hypergeometric functions Lecture 1 October 23, / 34

24 A-hypergeometric equations for 2 F 1 Recall that L = ( 1, 1, 1, 1) and F (α, β, γ z) is proportional to z n Γ( n + 1 α)γ( n + 1 β)γ(n + γ)γ(n + 1). n 0 Formal A-hypergeometric solution: n 0 v n α 1 v n β 2 v n+γ 1 3 v n 4 Γ( n + 1 α)γ( n + 1 β)γ(n + γ)γ(n + 1). The A-hypergeometric equations read ( )Φ = 0 (v v α)φ = 0 (v v β)φ = 0 ( v v γ 1)Φ = 0 A-hypergeometric functions Lecture 1 October 23, / 34

25 Classical equations for 2 F 1 Reduction of the A-hypergeometric system gives, after setting v 1 = v 2 = 1, v 3 = 1, v 4 = z, z(z 1)F + ((α + β + 1)z γ)f + αβf = 0 A-hypergeometric functions Lecture 1 October 23, / 34

26 Example 2, Appell F 1 Appell F 1 (α, β, β, γ x, y) is proportional to m,n 0 Application of Γ-identities gives Γ(m + n + α)γ(m + β)γ(n + β ) Γ(m + n + γ)γ(m + 1)Γ(n + 1) x m y n. x m y n m,n 0 Γ( m n+1 α)γ( m+1 β)γ( n+1 β )Γ(m+n+γ)Γ(m+1)Γ(n+1) The lattice L is spanned by ( 1, 1, 0, 1, 1, 0) and ( 1, 0, 1, 1, 0, 1). A corresponding set A, e 1, e 2, e 3, e 4, e 2 + e 1 e 4, e 3 + e 1 e 4 R 4 A-hypergeometric functions Lecture 1 October 23, / 34

27 F 1 and F 4 polytope F1 F4 A-hypergeometric functions Lecture 1 October 23, / 34

28 A-hypergeometric equations for F 1 Recall L = ( 1, 1, 0, 1, 1, 0), ( 1, 0, 1, 1, 0, 1) and F 1 proportional to m,n 0 x m y n Γ( m n+1 α)γ( m+1 β)γ( n+1 β )Γ(m+n+γ)Γ(m+1)Γ(n+1) Formal A-hypergeometric solution: m,n Z v m n α 1 v m β 2 v n β 3 v m+n+γ 4 v5 mv 6 n Γ( m n+1 α)γ( m+1 β)γ( n+1 β )Γ(m+n+γ)Γ(m+1)Γ(n+1) Denote i = v i. The A-hypergeometric equations read 1 2 Φ 4 5 Φ = 0, 1 3 Φ 4 6 Φ = 0, 2 6 Φ 3 5 Φ = 0 (v v v α)φ = 0 (v v β)φ = 0 (v v β )Φ = 0 (v 4 4 v 5 5 v 6 6 γ + 1)Φ = 0 A-hypergeometric functions Lecture 1 October 23, / 34

29 Classical equations for F 1 The A-hypergeometric system for F 1 can be reduced to the following system, where we have set v 1 = v 2 = v 3 = v 4 = 1, v 5 = x, v 6 = y, x(1 x)f xx + y(1 x)f xy + [γ (α + β + 1)x]F x βyf y αβf = 0 y(1 y)f yy + x(1 y)f xy + [γ (α + β + 1)y]F y β xf x αβ F = 0 (x y)f xy β F x + βf y = 0 In classical literature the last equation is usually presented as a (non-trivial) consequence of the first two. A-hypergeometric functions Lecture 1 October 23, / 34

30 The rank of an A-hypergeometric system The toric ideal associated to A is the ideal in C[ 1,..., N ] generated by all λ with λ L. Notation: I A. The A-polytope is the convex hull of the set A. Notation: Q A. We assign to Q A a volume normalised such that the volume of a standard simplex is 1. Notation: Vol(Q A ). Theorem (GKZ 1989) The A-hypergeometric system H A (α) has finite rank. Suppose that C[ i ]/I A satisfies the Cohen-Macaulay condition. Then the rank equals Vol(Q A ). By a theorem of Hochster the Cohen-Macaulay condition is satisfied if A is saturated, that is, Z 0 A = Z r R 0 A. A-hypergeometric functions Lecture 1 October 23, / 34

31 Rank jumps A.Adolphson (1994) pointed out that without the Cohen-Macaulay condition the GKZ-theorem on the ranks need not be true. Example, consider A R 2 given by the columns of ( ) Then the rank of H A (α, β) equals 5 if α = 1, β = 2 and 4 otherwise. It is known that the rank of any A-hypergeometric system is finite and Vol(Q A ). L.Matusevich and U.Walther (2005) showed that this difference can be arbitrarily large. A-hypergeometric functions Lecture 1 October 23, / 34

32 Irreducibility Theorem (GKZ 1990) Suppose α + Z r has trivial intersection with the faces of C(A) (non-resonance). Then H A (α) is irreducible. Theorem (F.B, Walther 2011) Suppose that A is saturated and Q A is not a pyramid. If there exists a point of α + Z r contained in a face of C(A), then the A-hypergeometric system is reducible. A-hypergeometric functions Lecture 1 October 23, / 34

33 Reducibility of Gauss equation A-matrix and parameters ( α, β, γ 1). Faces are given by a 1, a 3, equation x 2 = 0 a 1, a 4, equation x 2 + x 3 = 0 a 2, a 3, equation x 1 = 0 a 2, a 4, equation x 1 + x 3 = Non-resonance condition: None of β, β γ, α, α γ is an integer. A-hypergeometric functions Lecture 1 October 23, / 34

34 Reducibility of F 1 The set A associated to F 1 is given by Parameter vector is given by ( α, β, β, γ 1) t. Q A has 5 faces, x 1 = 0, x 2 = 0, x 3 = 0, x 1 + x 4 = 0, x 2 + x 3 + x 4 = 0. Non-resonance: none of the following numbers is an integer, α, β, β, α γ, β + β γ. A-hypergeometric functions Lecture 1 October 23, / 34