THE HOLONOMIC RANK FORMULA FOR A-HYPERGEOMETRIC SYSTEM ^{1}

Go Okuyama

1. Introduction

Given a finite set A of d-dimensional integral vectors which belong to one hyperplane off the origin in ℚA and a parameter vector β ∈ ℂ^{d}, Gel’fand, Kapranov and Zelevinskii [5] defined a system of differential equations, called an A-hypergeometric system M_{A}(β). They proved that the holonomic rank of M_{A}(β) equals the normalized volume of the convex hull of A and the origin (denote by vol(A)) for any β when the semigroup ring ℂ[ℕA] determined by A is Cohen-Macaulay. In general, the rank is not less than the volume (see [1], [13], Theorem 3.5.1). Meanwhile Adolphson [1] showed that even when ℂ[ℕA] is not Cohen-Macaulay, the holonomic rank equals vol(A), as long as β is generic in a certain sense. After Strumfels and Takayama showed that the holonomic rank can actually be greater than vol(A) for non-generic parametes β, Cattani, D’Andrea and Dickenstein showed that if the covex hull of A is a segment, then there exists a rank-jumping parameter whenever ℂ[ℕA] is not a Cohen-Macaulay ring. Saito, who generalized this result by using different methods, showed that there exist rank-jumping parameters for any non-Cohen-Macaulay simplicial semigroup ring ℂ[ℕA]. Matusevich [6] showed that, if the toric ideal defined by A is generic in a certain sense and non-Cohen-Macaulay, then there exists a rank-jumping parameter. However, when we fix a parameter β, it is not well-known how the holonomic rank is described explicitly except when the covex hull of A is simplicial (see [10], Theorem 6.3). In this paper, using combinatorial notion, we provide a rank formula in the case where the rank of A is three.

1.1. Definition of A-hypergeometric system. Let A = (a_{1}, …, a_{n}) = (a_{ij}) be a d × n-matrix of rank d with coefficients in ℤ. Let k be a field of characteristic zero and ℕ the set of nonnegative integers. We denote the set {a_{1}, …, a_{n}} by A as well. Let F_{A} denote the face lattice of the cone

ℚ_{≥0}A := {∑_{j=1}^{n} c_{j}a_{j}| c_{j} ∈ ℚ_{≥0}}.

Let ℕA denote the semigroup generated by A and by k[ℕA] its semigroup ring contained in the Laurant polynomial ring k[t_{1}^{±}, …, t_{d}^{±}]. For a face σ in F_{A}, we denote by ℕ(A ∩ σ) the semigroup generated by A ∩ σ, and by ℤ(A ∩ σ) the group generated by A ∩ σ. When A ∩ σ = ∅, we agree that ℕ(A ∩ σ) = ℤ(A ∩ σ) = 0. We consider the k-algebra homomorphism ϕ_{A} : k[∂_{1}, …, ∂_{n}] → k[ℕA] defined by

1The detailed version of this paper has been submitted for publication elsewhere.

ϕ_{A}( ∑_{u∈ℕ^{n}} c_{u}∂^{u}) := ∑_{u∈ℕ^{n}} c_{u}t^{Au},

where c_{u} ∈ k, ∂^{u} := ∂_{1}^{u_{1}} ⋯ ∂_{n}^{u_{n}}, and t^{Au} := t_{1}^{(Au)_{1}} ⋯ t_{d}^{(Au)_{d}}. We denote by I_{A}(∂) the kernel of ϕ_{A} and call it the toric ideal of A. Since ϕ_{A} is an epimorphism, we have

k[∂]/I_{A}(∂) ≃^{ϕ̄_{A}} k[ℕA].

Given a column parameter vector β = ^{t}(β_{1}, …, β_{d}) ∈ k^{d}, let H_{A}(β) denote the left ideal of the n-th Weyl algebra

D = k⟨x_{1}, …, x_{n}, ∂_{1}, …, ∂_{n}⟩

generated by I_{A}(∂) and ∑_{j=1}^{n} a_{ij}θ_{j} − β_{i} (i = 1, …, d), where θ_{j} := x_{j}∂_{j}. We call the quotient D-module M_{A}(β) := D/H_{A}(β) the A-hypergeometric system with parameter β. This system was introduced in the late eighties by Gel’fand, Graev, and Zelevinski (see [4]); its systematic study was started by Gel’fand, Zelevinski, and Kapranov (see, e.g. [5]).

1.2. Known results on the holonomic rank of M_{A}(β). In this note, we define the holonomik rank of the A-hypergeometric system rank(M_{A}(β)) as follows:

rank(M_{A}(β)) := dim_{k(x)}(k(x) ⊗_{k[x]} M_{A}(β)).

Here k(x) = k(x_{1}, …, x_{n}) is the field of rational functions. One of the results shown in [5] about the holonomic rank of A-hypergeometric system is that rank(M_{A}(β)) = vol(A) for any β ∈ k^{d} when the semigroup ring k[ℕA] is a Cohen-Macaulay ring. Here vol(A) means the normalized volume of the convex hull in ℚ^{d} of A and the origin. This equality can fail if k[ℕA] is not a Cohen-Macaulay ring. However, even if we drop the assumption that k[ℕA] is a Cohen-Macaulay ring, we have

rank(M_{A}(β)) ≥ vol(A)

for any β ∈ k^{d}, and the equality holds for generic β. So we write j_{A}(β) for the gap between the holonomic rank and the volume in this talk.

Moreover, in fact, Matusevich, Miller and Walther [7] completely showed that the rank of M_{A}(β) is indepent of β, that is, j_{A}(β) = 0 for any β if and only if ℂ[ℕA] is Cohen-Macaulay. However, given a parameter β, we do not know the formula of the rank of M_{A}(β) except when the convex hull of A is simplicial.

2. Main Result

2.1. Combinatorial term F_{A}(β). As in the previous section, in order to compute the gap j_{A}(β), we introduce a combinatorial term as follows. First, for λ ∈ ℤA and β, we define the subset J (λ; β) of F_{A} by

J (λ; β) := {σ ∈ F_{A}| λ ∉ ℕA + ℤ(A ∩ σ), β − λ ∈ k(A ∩ σ)}.

Second, we define a preorder on ℤA as follows:

λ < μ ⟺_{def} for any σ ∈ J (λ; β), λ + ℤ(A ∩ σ) = μ + ℤ(A ∩ σ).

Then we have the following proposition on this set.

Proposition 2.1. (1) The set J (λ; β) does not contain ℚ_{≧0}A.

(2) If λ ∈ ℕA, then we have J (λ; β) = ∅.

(3) If λ < μ, then we have J (λ; β) ⊂ J (μ; β).

Now, we consider the subset of ℤA \ ℕA:

E_{A}(β) = {λ ∈ ℤA \ ℕA| J (λ; β) ≠ ∅}.

We denote by F_{A}(β) the inductive limit of the set (E_{A}(β), <) which we regard as an inductive system. In other words, F_{A}(β) coincides with the set of maximal elements in ((ℤA \ ℕA)/ ∼, <), where ∼ means the equivalence relation defined by

λ ∼ μ ⟺_{def} λ < μ and λ > μ.

Since λ ∈ β + ⋃_{τ∈F_{A}} k(A ∩ τ) for any λ ∈ F_{A}(β) and [ℤA ∩ ℚτ : ℤ(A ∩ τ)] < ∞ for any face τ, we see that F_{A}(β) is a finite set.

2.2. Main result. Let d = 3 to the end of this note. First assume that the cardinality of F_{A}(β) is one. Let F_{A}(β) = {λ} and J̃(λ; β) denote the set of maximal elements in J (λ; β). Then the sets J̃(λ; β) can be classified into four cases:

(1): J̃(λ; β) = ∅,

(2): J̃(λ; β) consists of one proper face σ, (3): J̃(λ; β) consists of all facets, (4): none of the above.

For each case, we have the following theorem:

Theorem 2.2. Let d = 3. Assume that the cardinality of F_{A}(β) is one. Then we have the following.

(1) J̃(λ; β) satisfies the case (1) ⇒ j_{A}(β) = 0,

(2) J̃(λ; β) satisfies the case (2) ⇒

j_{A}(β) = \begin{cases}0 & if σ is a facet, \\ vol(A ∩ σ) & if σ is an edge, \\ 2 & if σ = \{0\}, \end{cases}

(3) J̃(λ; β) satisfies the case (3) ⇒ j_{A}(β) = 0,

(4) J̃(λ; β) satisfies the case (4) ⇒ j_{A}(β) = ∑_{σ∈J̃(λ;β):edges}(vol(A ∩ σ)) + m − 1. Here m means the number of connected components of the finite graph G_{λ} = {σ ∈ F_{A}| {0} ≠ σ ⊂ τ for some τ ∈ J̃(λ; β)} with respect to the inclusion relation.

Second not assume that the cardinality of F_{A}(β) is one. In this case, it suffices that for each λ ∈ F_{A}(β) we compute the number determind by the previous theorem, that is, we compute the right hand side of the equality in the theorem, regarding F_{A}(β) as the singleton set {λ}. For each λ ∈ F_{A}(β), let l_{λ} denote the number in this meaning. Then we have the rank formula as desired:

Theorem 2.3. Let d = 3. Then we have j_{A}(β) = ∑_{λ∈F_{A}(β)} l_{λ}.

3. Examples

Example 1. Let A_{1} = (\begin{matrix}1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 2 & 3 & 5 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 \end{matrix}) . Then we have vol(A_{1}) = 7.

First we consider the case where β = ^{t}(1, 2, 0). Then we have F_{A_{1}}(β) = {β} and J̃(β; β) = {ℚ_{≧0}a_{1}, ℚ_{≧0}a_{4}}. Hence we have j_{A_{1}}(β) = 1.

Second we consider the case where β = ^{t}(2/5, 1, 0). Then we have F_{A_{1}}(β) = {^{t}(1, 1, 0),^{t}(1, 4, 0)} and J̃(^{t}(1, 1, 0); β) = {ℚ_{≧0}a_{1}} and J̃(^{t}(1, 4, 0); β) = {ℚ_{≧0}a_{4}}. Hence F_{A_{1}}(β) is semisimple. Since ℚ_{≧0}a_{1} and ℚ_{≧0}a_{4} are both edges, we have j_{A_{1}}(β) = 1 + 1 = 2.

[figure: s_{3}; s_{2}; s_{1} = 1]

Figure 1. The set A_{1}

Example 2. Let A_{2} = (\begin{matrix}1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 0 & 2 & 0 & 1 \\ 0 & 0 & 0 & 1 & 1 & 2 & 2 \end{matrix}) . Then we have vol(A_{2}) = 6.

Let β = ^{t}(1, 1, 1). Then we have F_{A_{2}}(β) = {β} and J̃(β; β) = {{0}}. Hence we have j_{A_{2}}(β) = 2.

[figure: s_{3}; s_{2}; s_{1} = 1]

Figure 2. The set A_{2}

Example 3. Let A_{3} = (\begin{matrix}1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 2 & 3 & 0 & 2 & 3 \\ 0 & 0 & 0 & 1 & 1 & 1 \end{matrix}) . Then we have vol(A_{3}) = 6.

Let β = ^{t}(0, 1, 0). Then we have F_{A_{3}}(β) = {β} and J̃(β; β) = {ℚ_{≧0}a_{1}+ℚ_{≧0}a_{4}, ℚ_{≧0}a_{3}+ ℚ_{≧0}a_{6}}. Hence we have j_{A_{3}}(β) = 1.

[figure: s_{3}; s_{2}; s_{1} = 1]

Figure 3. The set A_{3} Example 4. Let A_{4} = (\begin{matrix}3 & 2 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 1 & 0 & 0 \end{matrix}); not homogeneous. Then we have vol(A_{4}) =

12.

First we consider the case where β = ^{t}(1, 1, 0). Then we have F_{A_{4}}(β) = {β} and J̃(β; β) = {ℚ_{≧0}a_{1}}. Hence we have j_{A_{4}}(β) = vol(A_{4} ∩ ℚ_{≧0}a_{1}) = 3.

Second we consider the case where β = ^{t}(2, 2, 0). Then F_{A_{4}}(β) = {β} and J̃(β; β) = {{0}}. Hence we have j_{A_{4}}(β) = 2.

[figure: s_{3}; s_{2}; s_{1} = 1; s_{3}; s_{2}; s_{1} = 2; s_{3}; s_{2}; s_{1} = 3]

Figure 4. The set A_{4}

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Department of Mathematics Hokkaido University Sapporo, 060-0810 JAPAN

E-mail address: go zoku@lapis.plala.or.jp