Homology, Homotopy and Applications, vol.4(2), 2002, pp.409–426

SOME CONJECTURES ABOUT THE HILBERT SERIES OF GENERIC IDEALS IN THE EXTERIOR ALGEBRA

GUILLERMO MORENO-SOCÍAS and JAN SNELLMAN

(communicated by Larry Lambe)

Abstract We calculate the Hilbert series of a quotient of the exterior algebra by a generic form of even degree, and give conjectures about the Hilbert series of other generic quotients.

To Jan–Erik Roos on his sixty–fifth birthday

1. Introduction

In the symmetric algebra K[x_{1}, …, x_{n}], the set of Hilbert series coming from homogeneous quotients are classified by Macaulays theorem [16, 6, 3]. There is an infinite number of possible series, but if we fix positive integers d_{1}, …, d_{r}, and restrict our study to quotients by homogeneous ideals I of “type” or “numerical character” (d_{1}, …, d_{r}), ie generated by forms of those prescribed degrees, then there are only finitely many Hilbert series. Furthermore, in the affine space parametrising these homogeneous ideals, there is a Zariski-open subset of ideals with the same Hilbert series, and the Hilbert series obtained on this open set is minimal [9, 7].

Unfortunately, even though we know the set of all Hilbert series, we do not know what Hilbert series arise from ideals of numerical character (d_{1}, …, d_{r}). In fact, we do not even know the “generic” series, but it is conjectured [17, 7] that it is ⟨(1 − t)^{−n} ∏_{i=1}^{r}(1 − t^{d_{i}})⟩; the brackets mean “truncate before the first non-positive coefficient”.

In the exterior algebra ⋀ V_{n}, we also know the set of all Hilbert series of homogeneous quotients, by the so-called Kruskal-Katona theorem [15, 14, 5, 2]. Here, this set is finite, so one would think that it should be easy to find the subset of Hilbert series coming from quotients by ideals having a prescribed numerical character. In particular, it should be easy to find the generic value. However, very little is known.

In this article, we give one new result (the series for a quotient by one form of even degree) and several conjectures, supported by extensive computer calculations.

It is worthwhile to point out that the problem of determining the Hilbert series of quotients by generic quadratic forms is especially interesting, since it determines

Snellman was supported grants from Svenska institutet and by grant n. 231801F from Centre International des Etudiants et Stagiaires while visiting cole Polytechnique, and by grants from Svenska Institutet and Kungliga Vetenskapsakademin while visiting University of Wales, Bangor Received December 15, 2000, revised March 16, 2001; published on July 12, 2002. 2000 Mathematics Subject Classification: 15A75, 13D40. Key words and phrases: exterior algebra, generic ideals, Hilbert series. \begin{matrix}c \\ ◯ \end{matrix} 2002, Guillermo Moreno-Socías and Jan Snellman. Permission to copy for private use granted.

the Koszulness of the quadratic algebras in question. We refer to the recent article by Fröberg and Löfwall [10].

2. Notation

Let K be a field of characteristic 0. Then ℚ is the prime subfield of K. For any positive integer n, let V = V_{n} be an n-dimensional vector space over K, with a distinguished basis X_{n} = {x_{1}, …, x_{n}}. Let K[x_{1}, …, x_{n}] denote the symmetric algebra on V_{n}, and let ⋀ V_{n} denote the exterior algebra on V_{n}. We define 𝔖(V_{n}), the square-free algebra on V_{n}, to be the commutative K-algebra generated by X_{n}, with the relations x_{i}^{2} = 0; in other words, 𝔖(V_{n}) = \frac{K[x_{1},...,x_{n}]}{(x_{1}^{2},...,x_{n}^{2})}. There is an isomorphism of graded vector spaces between ⋀ V_{n} and 𝔖(V_{n}), but they are not isomorphic as Kalgebras, since the exterior algebra is skew-commutative and 𝔖(V_{n}) is commutative.

We shall need the following operations for formal power series.

Definition 2.1. Let f(t) = ∑_{i=0}^{∞} a_{i}t^{i} ∈ ℤ[[t]], g(t) = ∑_{i=0}^{∞} b_{i}t^{i} ∈ ℤ[[t]]. We say that f ⩾ g if a_{i} ⩾ b_{i} for all i. We define

max(f(t), g(t)) = ∑_{i=0}^{∞} max(a_{i}, b_{i})

⟨f(t)⟩ = ∑_{i=0}^{ℓ} a_{i}t^{i},  ℓ = max({ i a_{j} > 0 for j ⩽ i })

⟩f(t)⟨ = ∑_{i=ℓ}^{∞} a_{i}t^{i},  ℓ = min({ i a_{j} > 0 for j ⩾ i })

We use the conventions max(∅) = −1 = min(ℕ), min(∅) = +∞ = max(ℕ).

Let [X_{n}] denote the free abelian monoid on X_{n}, and denote by Y_{n} the subset of square-free monomials. Then Y_{n} is a K-basis for both ⋀ V_{n} and 𝔖(V_{n}). We define the degree of a monomial in [X_{n}] (and in Y_{n}) in the usual way, and denote by [X_{n}]^{d} and Y_{n}^{d} the subset of monomials (square-free monomials) of degree d.

A form ⋀ K[x_{1}, …, x_{n}] ∋ f = ∑_{m∈[X_{n}]^{d}} c_{m}m is said to be generic if the coefficients c_{m} ∈ K fulfil the following conditions:

1. c_{m} ∉ ℚ,

2. m ≠ m′ ⟹ c_{m} ≠ c_{m′},

3. The set of all c_{m}’s is algebraically independent over ℚ.

A homogeneous ideal I ⊂ K[x_{1}, …, x_{n}] is called generic if it can be minimally generated by a finite set of generic forms, so that all of the occuring coefficients of the forms are different, and so that the set of all occuring coefficients is algebraically independent over ℚ. If the forms have degrees d_{1}, …, d_{r}, then we say that I has “numerical character” (d_{1}, …, d_{r}). It is an important fact that any two generic ideals of the same numerical character have the same initial ideal and the same Hilbert series.

Now consider the affine space V = A^{\binom{n+d_{1}−1}{d_{1}}} × ⋯ × A^{\binom{n+dr −1}{dr}} parametrising the set of homogeneous ideals of numerical character (d_{1}, …, d_{r}). Since there are countably many conditions to be fulfilled for an ideal to be generic, the subset of the parameter space corresponding to generic ideals is not open, but a countable intersection of open sets, hence dense. However, in V there is a Zariski-open subset corresponding to ideals with the same Hilbert function, and the generic ideals are contained in this subset [9].

We make similar definitions for the square-free algebra, and for the exterior algebra. Here, a generic form is a generic linear combination of square-free monomials of a certain degree. It is still true that the generic Hilbert series is attained on an open component of the parameter space, and that the generic ideals are contained in this component.

3. Hilbert series for generic principal ideals in the symmetric and square-free algebra

3.1. Principal ideals in the symmetric algebra

If f ∈ K[x_{1}, …, x_{n}] is a non-zero form of degree d, not necessarily homogeneous, then clearly the Hilbert series of the quotient \frac{K[x_{1},...,x_{n}]}{(f)} is (1 − t)^{−n}(1 − t^{d}).

3.2. Principal ideals in the square-free algebra

If f ∈ 𝔖(V_{n}) is a generic form of degree d, then there is a similar simple formula for \frac{𝔖(V_{n})}{(f)}(t) (the Hilbert series of the quotient). To state the formula, we need some additional notation.

Definition 3.1. We denote the zero series by 0, and define

Δ_{n,d}(t) = ⟩(t^{d} − 1)(1 + t)^{n}⟨

= ∑_{v=⌈(n−d)/2⌉}^{n} (\binom{n}{v} − \binom{n}{v + d}) t^{v}

δ_{n,d}(t) = ⟨(1 + t)^{n}(1 − t^{d})⟩

= ∑_{v=0}^{⌊(n−d)/2⌋}(\binom{n}{v + d} − \binom{n}{v})t^{v}

The following result is due to Frberg [8].

Theorem 3.2. Let f ⊂ 𝔖(V_{n}) be a generic form of degree d. Then

\frac{𝔖(V_{n})}{(f)}(t) = δ_{n,d}(t)  (1)

Proof. By considering the graded exact sequence

0 ⟶ ann(f)(−d) ⟶ 𝔖(V_{n})(−d) ⟶^{·f} 𝔖(V_{n}) ⟶ \frac{𝔖(V_{n})}{(f)} ⟶ 0  (2)

in each degree r, we see that (1) holds if and only if multiplication by f, regarded as a linear map ϕ_{r} from 𝔖(V_{n})_{r} to 𝔖(V_{n})_{r+d}, is injective when \binom{n}{r} ⩽ \binom{n}{r+d}, and surjective when \binom{n}{r} ⩾ \binom{n}{r+d}.

Write f = ∑_{m∈Y_{n}^{d}} c_{m}m. For 0 ⩽ r ⩽ n − d, Y_{n}^{r} is a basis of 𝔖(V_{n})_{r}, and Y_{n}^{r+d} is a basis of 𝔖(V_{n})_{r+d}. Thus, we must show that for each r, the matrix of ϕ_{r} in this basis has maximal rank. This matrix has rows indexed by Y_{n}^{r+d} and columns indexed by Y_{n}^{r}. The entry at row R, column C is

\begin{cases}0 & C |̸R \\ c_{m} & R = mC \end{cases}

If we specialise this matrix, the rank can only decrease, so if we can prove that some specialised matrix has full rank, then we are done. Putting all c_{m} = 1, we obtain the incidence matrix of r-subsets of [n] into r + d-subsets of [n], that is, the rows are indexed by r-subsets and the columns by r + d-subsets, with a 1 at the a, b’th position iff a ⊂ b, and 0 otherwise. It has been shown by combinatorialists that this matrix has full rank [18, 13, 11].

4. Principal ideals in the exterior algebra — the difference between even and odd degree

Let f ∈ ⋀ V_{n} be a generic form of degree d. Denote the Hilbert series of \frac{⋀ V_{n}}{f} by q_{n,d}(t), that of the annihilator of f by a_{n,d}(t), and that of the principal ideal (f) by p_{n,d}(t). From the the graded exact sequence

0 ⟶ ann(f)(−d) ⟶ ⋀ V_{n}(−d) ⟶^{·f} ⋀ V_{n} ⟶ \frac{⋀ V_{n}}{(f)} ⟶ 0  (3)

we get that

\begin{aligned}q_{n,d}(t) & = t^{d}a_{n,d}(t) − t^{d}(1 + t)^{n} + (1 + t)^{n} \\ & = t^{d}a_{n,d}(t) + (1 + t)^{n}(1 − t^{d}) \\ a_{n,d}(t) & = t^{−d} (q_{n,d}(t) − (1 + t)^{n}(1 − t^{d})) \end{aligned}  (4)

If d is even, we shall prove that the vector space map

⋀^{v} V_{n} ⟶^{·f} ⋀^{v+d} V_{n}  (5)

is injective “when it can be”, ie when \binom{n}{v} ⩽ \binom{n}{v+d}, and surjective “when it can be”, ie when \binom{n}{v} ⩾ \binom{n}{v+d}. This leads immediately to the formulæ

\begin{aligned}q_{n,d}(t) & = ⟨(1 + t)^{n}(1 − t^{d})⟩ = δ_{n,d}(t) \\ a_{n,d}(t) & = t^{−d} (q_{n,d}(t) − (1 − t^{d})(1 + t)^{n}) \\ & = t^{−d} (δ_{n,d}(t) − (1 − t^{d})(1 + t)^{n}) \\ & = t^{−d} ∑_{r=0}^{n}[max (0, \binom{n}{r + d} − \binom{n}{r}) − (\binom{n}{r + d} − \binom{n}{r})]t^{r} \\ & = t^{−d} ∑_{r=0}^{n} max (0, −\binom{n}{r + d} + \binom{n}{r}) t^{r} \\ & = t^{−d}Δ_{n,d}(t) \end{aligned} (6)

In particular, as n → ∞, (1 + t)^{−n}q_{n,d}(t) → (1 − t^{d}), and a_{n,d}(t) → 0, with respect to the t-adic norm on ℤ[[t]].

If d is odd, then we have that f^{2} = 0, hence fg = 0 whenever g ∈ (f), hence ann(f) ⊇ (f), hence a_{n,d}(t) ⩾ p_{n,d}(t). In other words, there is a graded complex

(⋀ V ) (−d) ⟶^{·f} ⋀ V ⟶^{·f} (⋀ V ) (d)  (7)

the graded homology of which determines a_{n,d}(t) − p_{n,d}(t). In the (not very interesting) case d = 1, then we know from [1] that this homology vanishes. For odd d > 1, we guess that for a fixed degree r, and n very large, this homology vanishes. Hence, in degree r, the “obstruction to injectivity” in (5) is as small as possible. An equivalent formulation: consider the start of a minimal free graded ⋀ V_{n}-resolution of \frac{⋀ V_{n}}{(f)} ,

\frac{⋀ V_{n}}{(f)} ← ⋀ V_{n} ⟵^{·f} ⋀ V_{n} ← ⨁_{j=1}^{r}(⋀ V_{n})(−β_{2,i}),

where β_{2,i} are the graded Betti numbers. Then we guess that as n increases, and for a fixed i ≠ 2d, β_{2,i} = 0. On the other hand, for sufficiently large n, we guess that β_{2,2d} = 1. Since β_{2,i} is the dimension of the degree i − d part of a certain Tor group, this conjecture can also be stated in terms of Cartan homology (see [2]).

We show the order (ie the smallest ℓ for which t^{ℓ} occurs with non-zero coefficient) of a_{n,d}(t) − p_{n,d}(t) for small n, d in Table 1. It would seem that the order of the difference grows linearly in n, so that a_{n,d}(t) − p_{n,d}(t) → 0 rather rapidly.

Let us turn to the consequences of this conjecture. We get that a_{n,d}(t) ∼ p_{n,d}(t) with respect to the (t)-adic filtration. It then follows from (4) that

q_{n,d}(t) ∼ t^{d}p_{n,d}(t) + (1 + t)^{n}(1 − t^{d})  (8)

Substituting p_{n,d}(t) = (1 + t)^{n} − q_{n,d}(t) and solving for q_{n,d}(t) we get that

q_{n,d}(t) ∼ \frac{(1 + t)^{n}t^{d} + (1 + t)^{n}(1 − t^{d})}{(1 + t^{d})} = \frac{(1 + t)^{n}}{(1 + t^{d})} ,  (9)

d 3 5 7 9 11 13 15 17

n

3 1
4 1
5 1 1
6 2 1
7 3 1 1
8 3 2 1
9 3 3 1 1
10 4 3 2 1
11 5 3 3 1 1
12 5 4 3 2 1
13 5 5 4 3 1 1
14 6 5 4 3 2 1
15 7 6 5 3 3 1 1
16 7 6 5 4 3 2 1
17 - - - 5 4 3 1 1
18 - - - - 4 3 2 1
19 - - - - - 3 3 1
20 - - - - -  - 3 2
Table 1: Order of a_{n,d}(t) − p_{n,d}(t) for small n, d

hence

\frac{\frac{⋀ V_{n}}{(f)} (t)}{⋀(V_{n})(t)} = \frac{q_{n,d}(t)}{(1 + t)^{n}} → \frac{1}{1 + t^{d}}  as n → ∞.  (10)

5. Principal ideals on generic forms of even degree in the exterior algebra

If d = 2 then we can change coordinates on V and replace f with the form x_{1}x_{2} + x_{3}x_{4} + ⋯, as is demonstrated in [4]. The Hilbert series of the quotient can now be easily calculated. We get that \frac{⋀(V_{n})}{(f)}(t) = ⟨(1 + t)^{n}(1 − t^{2})⟩, which is the same as the Hilbert series for the corresponding quotient in the square-free algebra.

Remark 5.1. It is not true that if f_{e} = ∑_{1⩽i<j⩽n} α_{ij}x_{i}x_{j} is a non-generic quadratic form in ⋀ V_{n}, and f_{s} = ∑_{1⩽i<j⩽n} α_{ij}x_{i}x_{j} is the corresponding form in 𝔖(V_{n}), then \frac{⋀ V_{n}}{(f_{e})} and \frac{𝔖(V_{n})}{(f_{s})} have the same Hilbert series. For an example, consider the form x_{1}x_{2} + x_{1}x_{3} + x_{1}x_{4} + x_{3}x_{4}. The quotient of ⋀ V_{4} by this form has Hilbert series 5t^{2} + 4t + 1, but the corresponding quotient of 𝔖(V_{4}) has series t^{3} + 5t^{2} + 4t + 1.

We next show that if the degree d of f is even, then the Hilbert series of the quotient \frac{⋀ V_{n}}{(f)} is the same as for the square-free algebra. To this end, we need some combinatorial results, which we have collected in the appendix. With the aid of these, we can prove:

Theorem 5.2. Let f ∈ ∧^{d}V, with d even, be a generic form. Then the linear transformation

∧^{r}V ⟶^{f·} ∧^{r+d}V  (11)

is injective for 2r + d ⩽ n, and surjective for 2r + d ⩾ n.

Proof. We put k = r + d. Suppose that

f = ∑_{K∈\binom{[n]}{d}} c_{K}x_{K}  (12)

The matrix of the map (11) is an \binom{n}{r+d} × \binom{n}{d} matrix, { \atop M_{r,r+d,n}}, where the rows are indexed by (r + d)-subsets K, and the columns by d-subsets T. The entry at position (K, T) is

\begin{cases}0 & if T ⊈ K \\ σ(T, K)c_{T} & if T ⊆ K \end{cases}  (13)

We must prove that this matrix has maximal rank. Clearly, the rank can not increase under specialisation, so if we prove that the matrix obtained by replacing each c_{T} with 1 has maximal rank, then so does { \atop M_{r,r+d,n}}. However, the specialised matrix is nothing but the matrix M_{r,r+d,n} of Theorem Appendix A.6, so it has full rank.

Theorem 5.3. Let f ∈ ⋀ V_{n} be a generic form of degree d, with d even. Then

\frac{⋀ V_{n}}{(f)} (t) = ⟨(1 + t)^{n}(1 − t^{d})⟩ = δ_{n,d}(t)  (14)

Proof. This follows from Theorem 5.2, together with (3).

6. Principal ideals on generic forms of odd degree in the exterior algebra

Let d be an odd integer. Recall that we’ve conjectured that a_{n,d}(t) − p_{n,d}(t) → 0 as n → ∞, and that this conjecture leads to the conclusions that p_{n,d}(t) ∼ (1 + t)^{n}(1 + t^{d})^{−1}. In this section, we shall try to guess the exact value of q_{n,d}(t).

Since a_{n,d}(t) ⩾ p_{n,d}(t), a_{n,d}(t) ⩾ Δ_{n,d}(t), it follows that a_{n,d}(t) ⩾ max(p_{n,d}(t), Δ_{n,d}(t)). We tabulate the difference a_{n,d}(t) − max(p_{n,d}(t), Δ_{n,d}(t)) in Table 2 and Table 3.

Using the data of Table 3, we make the following conjecture:

Conjecture 6.1. Let d be an odd integer > 3. Then, putting τ_{n,d}(t) = a_{n,d}(t) − max (p_{n,d}(t), Δ_{n,d}(t)),

τ_{n,d}(t) = \begin{cases}t^{v(v−1)/2} & ∃v, s ∈ ℕ : v > 0, n − d = −1 + \frac{5}{2}v + \frac{1}{2}v^{2}, d = 5 + 2vs \\ 0 & otherwise \end{cases}

Table 2: Difference between true and predicted Hilbert series of the annihilator of a generic form of odd degree

n deg=3  5 7 9 11 13 15 17 19
3 0
4 0
5 t  0
6 0  0
7 0  t 0
8 0  0 0
9 3t^{3}  0 t 0
10 t^{4}  0 0 0
11 t^{5}  t^{3} 0 t 0
12 t^{6} + 12t^{5}  0 0 0 0
13 t^{7} + 13t^{6} + t^{5}  0 0 0 t 0
14 t^{8} + 14t^{7} + 91t^{6}  0 0 0 0 0
15 15t^{8} + 105t^{7}  0 0 t^{3} 0 t 0
16 16t^{9} + 120t^{8} + 559t^{7} t^{6} 0 0 0 0 0
17  0 0 0 t 0
18  0 0 0 0
19  t^{3} 0 t 0
20  0 0 0
21  0 t

Table 3: Difference between true and predicted Hilbert series of the annihilator of a generic form of odd degree > 3

n − d deg=5 7 9 11 13 15 17 19
0  0  0 0 0 0 0 0 0
1  0  0 0 0 0 0 0 0
2  t  t t t t t t t
3  0  0 0 0 0 0 0
4  0  0 0 0 0 0 0
5  0  0 0 0 0 0 0
6  t^{3}  0 t^{3} 0 t^{3}
7  0  0 0 0
8  0  0 0
9  0  0
10 0
11 t^{6}

This conjecture yields a formula for the Hilbert series, but since said formula is very complicated, we do not write it down; instead we show how to derive q_{n,d}(t). From

\begin{aligned}a_{n,d}(t) & = τ_{n,d}(t) + max (p_{n,d}(t), Δ_{n,d}(t)) \\ q_{n,d}(t) & = a_{n,d}(t)t^{d} + (1 + t)^{n}(1 − t^{d}) \\ p_{n,d}(t) & = (1 + t)^{n} − q_{n,d}(t) \end{aligned}  (16)

we get

p_{n,d}(t) = (1 + t)^{n} − q_{n,d}(t)
= (1 + t)^{n} − a_{n,d}(t)t^{d} − (1 + t)^{n}(1 − t^{d})
= (1 + t)^{n} − t^{d}τ_{n,d}(t) − t^{d} max (p_{n,d}(t), Δ_{n,d}(t)) − (1 + t)^{n}(1 − t^{d})
= t^{d} ((1 + t)^{n} − τ_{n,d}(t) − max(p_{n,d}(t), Δ_{n,d}(t)))

(17)

Hence, writing p_{n,d}(t) = ∑_{i=0}^{n} a_{i}t^{i}, with the a_{i}’s as undetermined coefficients, and denoting the t^{i}-coefficient of τ_{n,d}(t) by b_{i}, we get the equation

a_{ℓ} = \binom{n}{ℓ − d} − b_{i−ℓ} − max(a_{ℓ−d}, \binom{n}{ℓ − d} − \binom{n}{ℓ})  (18)

which we can solve recursively, using the initial values

a_{0} = ⋯ = a_{d−1} = 0, a_{d} = a_{n} = 1.

For the case d = 3, we proceed differently: we tabulate q_{n,3}(t) − w_{n,3}(t) in Table 4, and from that, make the following conjecture:

Conjecture 6.2. The Hilbert series of \frac{⋀ V_{n}}{(f)} , where f is a generic cubic form, is given by

\begin{aligned}p_{n,3}(t) & = \frac{t^{d}L_{n}(t) + (1 + t)^{n}}{1 + t^{d}} \\ L_{n}(t) & = \begin{cases}(3t)^{2ℓ−1}(1 + t)^{2} & n = 4ℓ \\ c_{1}(n)t^{2ℓ−1}(1 + t)(1 + (3^{c_{2}(n)} − 1)t + t^{2}) & n = 4ℓ + 1 \\ (3t)^{2ℓ}(1 + t)^{2} & n = 4ℓ + 2 \\ (3t)^{2ℓ+1}(1 + t) & n = 4ℓ + 3 \end{cases} \end{aligned}  (19)

where c_{1}(n), c_{2}(n) are some positive integers.

7. Hilbert series for generic non-principal ideals in the symmetric and square-free algebra

Let I = (f_{1}, …, f_{r}) be a generic ideals in K[x_{1}, …, x_{n}], generated by forms of degree d_{1}, …, d_{r}. There is a famous conjecture [17, 7] for the Hilbert series of the quotient \frac{K[x_{1},...,x_{n}]}{I_{n}}.

\begin{matrix}n & q_{n}(t) − w_{n}(t) \\ 3 & 3t(1 + t) \\ 4 & 3t(1 + t)^{2} \\ 5 & t(1 + t)(t^{2} + 8t + 1) \\ 6 & 9t^{2}(1 + t)^{2} \\ 7 & 27t^{3}(1 + t) \\ 8 & 27t^{3}(1 + t)^{2} \\ 9 & 3t^{3}(1 + t)(t^{2} + 26t + 1) \\ 10 & 81t^{4}(1 + t)^{2} \\ 11 & 243t^{5}(1 + t) \\ 12 & 243t^{5}(1 + t)^{2} \\ 13 & t^{5}(1 + t)(t^{2} + 728t + 1) \\ 14 & 729t^{6}(1 + t)^{2} \\ 15 & 2187t^{7}(1 + t) \\ 16 & 2187t^{7}(1 + t)^{2} \end{matrix}

Table 4: a_{n,d}(t) − p_{n,d}(t) for a cubic generic form

Conjecture 7.1. Let I = (f_{1}, …, d_{r}) ⊂ K[x_{1}, …, x_{n}] be a generic ideal, with |f_{i}| = d_{1} for 1 ⩽ i ⩽ r. Then the Hilbert series of the graded algebra \frac{K[x_{1},...,x_{n}]}{I_{n}} is given by

⟨(1 − t)^{−n} ∏_{i=1}^{r}(1 − t^{d_{i}})⟩  (20)

It is easy to see that if r ⩽ n, the generators form a regular sequence, and hence that

\frac{K[x_{1}, …, x_{n}]}{I_{n}}(t) = (1 − t)^{−n} ∏_{i=1}^{r}(1 − t^{d_{i}}),  for n ⩾ r  (21)

In particular, the conjecture holds for r ⩽ n. The conjecture is also know to be true for r = n + 1.

We note that (21) implies that

lim_{n→∞} \frac{\frac{K[x_{1},...,x_{n}]}{I_{n}}(t)}{K[x_{1}, …, x_{n}](t)} = ∏_{i=1}^{r}(1 − t^{d_{i}})  (22)

Now suppose that I = (f_{1}, …, f_{r}) is a generic ideal in the square-free algebra, and that f_{i} is a generic form of degree d_{i}. Then

\frac{𝔖(V_{n})}{(f_{1}, …, f_{r})} ≃ \frac{K[x_{1}, …, x_{n}]}{(f′_{1}, …, f′_{r}, x_{1}^{2}, …, x_{n}^{2})}

where f′_{i} can be taken to be a generic form in K[x_{1}, …, x_{n}] which maps to f_{i} under the canonical epimorphism K[x_{1}, …, x_{n}] ↠ 𝔖(V_{n}). It seems reasonable to assume that the Hilbert series of the quotient will not change if we replace the squares of variables with generic quadratic forms. Conjecture 7.1 then leads to the following:

Conjecture 7.2. Let r, n, d_{1}, …, d_{r}, and let I_{n} be a generic ideal i 𝔖(V_{n}) with generators of degrees d_{1}, …, d_{r}. Then

\frac{𝔖(V_{n})}{I_{n}}(t) = ⟨(1 + t)^{n} ∏_{i=1}^{r}(1 − t^{d_{i}})⟩  (23)

If this conjecture holds (our computations support this), then it follows that

lim_{n→∞} \frac{\frac{𝔖(V_{n})}{I_{n}}(t)}{𝔖(V_{n})(t)} = ∏_{i=1}^{r}(1 − t^{d_{i}})  (24)

This is analogous to (22).

8. Hilbert series for generic non-principal ideals in the exterior algebra

We now throw all caution to the wind to make some bold conjectures about the Hilbert series of non-principal generic ideals. Let I_{n} = (f_{1}, …, f_{r}) be a generic ideal in ⋀ V_{n}, with |f_{i}| = d_{i}, and consider the exact sequence

0 ⟶ ann(f_{r})(−d_{r}) ⟶ \frac{⋀ V_{n}}{(f_{1}, …, f_{r−1})}(−d_{r}) ⟶^{·f_{r}} \frac{⋀ V_{n}}{(f_{1}, …, f_{r−1})} ⟶ \frac{⋀ V_{n}}{(I)} ⟶ 0

(25)

We denote the Hilbert series of \frac{⋀ V_{n}}{(I)} by q_{n}(t), that of \frac{⋀ V_{n}}{(f_{1},...,f_{r−1})} by u_{n}(t), and that of ann(f_{r}) by a_{n}(t). Then

q_{n}(t) = u_{n}(t) − t^{d_{r}}u_{n}(t) + t^{d}a_{n}(t).  (26)

If d_{r} is even, we conjecture that a_{n}(t) ∼ 0, hence

q_{n}(t) ∼ (1 − t^{d_{r}})u_{n}(t)  (27)

If d_{r} is odd, we conjecture that the annihilator of f_{r} is “close” to the principal ideal on f_{r}, hence that a_{n}(t) ∼ (u_{n}(t) − q_{n}(t)), which yields

q_{n}(t)(1 + t^{d}) ∼ u_{n}(t)  (28)

By induction, we arrive at the following conjecture:

lim_{n→∞} \frac{q_{n}(t)}{(1 + t)^{n}} = ∏_{i=1}^{r} (1 − (−1)^{d_{i}}t^{d_{i}})^{(−1)^{d_{i}}} ∈ ℤ[[t]],  (29)

with respect to the (t)-adic topology.

One would be tempted to guess that if all d_{i}’s are even, the Hilbert series of \frac{⋀ V_{n}}{(f_{1},...,f_{r})} should be exactly

(1 + t)^{n} ∏_{i=1}^{r}(1 − t^{d_{i}})  (30)

However, this is not true, even for the simplest case r = 2 and d_{1} = d_{2} = 2. In Table 5 we tabulate the difference between the true Hilbert series and (30).

\begin{matrix}n & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \\ Diff & 0 & 0 & 0 & t^{3} & 0 & t^{4} & t^{4} & t^{5} & 10t^{5} & t^{6} + t^{5} & 64t^{6} & t^{7} + 13t^{6} \end{matrix}

Table 5: Difference between the true Hilbert series and the “anticipated Hilbert series” for generic ideals generated by two quadratic forms

Appendix A. The signed incidence matrix has full rank when the difference in cardinality is even

We prove a “signed version” of the well-known theorem that the incidence matrix of r-subsets of [n] = {1, …, n} into d + r-subsets have full rank. Our proof is a modification of the one by Wilson [18].

To begin, let us define the “signs” involved.

Definition Appendix A.1. Let [n] = {1, …, n}, and let C and R be two subsets of [n], with

\begin{aligned}C & = \{t_{1}, …, t_{a}\},  t_{1} < ⋯ < t_{a} \\ R & = \{k_{1}, …, k_{b}\},  k_{1} < ⋯ < k_{b} \end{aligned}

Then define σ(C, R) to be zero if C ⊈ R, and otherwise the sign of the permutation which sorts [C, R \ C] in ascending order. In other words, if C ⊆ R then σ(C, R) is the sign of the uniquely determined permutation γ such that

\begin{aligned}t_{γ(i)} & = k_{i},  1 ⩽ i ⩽ a \\ k_{γ(j)} & = k_{a+j},  1 ⩽ j ⩽ b \end{aligned}

Definition Appendix A.2. Let [n] = {1, …, n}, and let A, B be two subsets of [n], of cardinality a and b, with 0 ⩽ a < b. For a ⩽ r < b, we define

s_{r}(A, B, n) = ∑_{\begin{matrix}C∈\binom{[n]}{r} \\ A⊆C⊆B \end{matrix}} σ(C, B)  (31)

For 0 ⩽ d ⩽ n, we define

s_{d,n} = ∑_{R∈\binom{[n]}{d}} σ(R, [n]) = s_{d}(∅, [n], n)  (32)

Lemma Appendix A.3. With the notations of Definition Appendix A.2, put d = b − r. We have that

s_{r}(A, B, n) = \begin{cases}0 & A ⊈ B \\ (−1)^{d}s_{d,b−a} & A ⊆ B \end{cases}  (33)

Proof. Put d = b − r. If A ⊈ B then clearly s_{r}(A, B, n) = 0. Suppose that A ⊆ B. Then

s_{r}(A, B, n) = ∑_{\begin{matrix}C∈\binom{[n]}{r} \\ A⊆C⊆B \end{matrix}} σ(C, B) = ∑_{\begin{matrix}C∈\binom{B}{r} \\ A⊆C \end{matrix}} σ(C, B),

so the sum is independent of n. Furthermore, we can write A ⊆ C ∈ \binom{B}{r} as a disjoint union C = A ∪ (C \ A), hence the sum can be written

∑_{S∈\binom{B\A}{r−a}} σ(S ∪ A, B) = ∑_{S∈\binom{B\A}{r−a}} σ(S, B \ A).

Now, since S has cardinality r−a, the set (B \A)\S has cardinality b−a−(r−a) = b−r = d, so the permutation which transforms [S, B\A] to [B\A, S] has cardinality (−1)^{d}. Hence, by substituting R = (B \ A) \ S, we get that the sum is equal to

(−1)^{d} ∑_{S∈\binom{B\A}{v−a}} σ((B \ A) \ S, B \ A) = (−1)^{d} ∑_{R∈\binom{B\A}{d}} σ(R, B \ A)

= (−1)^{d} ∑_{R∈\binom{[b−a]}{d}} σ(R, [b − a]),

which is the desired result.

Lemma Appendix A.4. Suppose that 0 < d ⩽ n, and that d is even. Then s_{d,n} > 0.

Proof. The lemma is trivially true for d = n. If d = 2, we note that σ({v, v + 1}, [n]) = 1 for 1 ⩽ v < n, since the permutation transforming [v, v+1, 1, 2, …, v−1, v+2, v+ 3, …, n] to [1, 2, …, n] is even. Furthermore, the signs of σ({v, v + ℓ}, [n]) alternate in sign as ℓ goes from 1 to n − v. Thus, for a fixed v, there are either as many positive as negative σ({v, v + ℓ}, [n]), or 1 more positive than negative, depending on the parity of n − v. By summing over all v, we conclude that there are always strictly more positive than negative signs.

Now suppose that we have shown that s_{2k′,n′} > 0 for all k′, n′ such that k′ < k. We want to show that that s_{2k,n} > 0. We have that

s_{2k,n} = ∑_{R∈\binom{[n]}{2k}} σ(R, [n]),

and writing R as a disjoint union of its first two element, and the remaining elements, this becomes

∑_{1⩽k<ℓ⩽n−2} ∑_{R_{2}∈\binom{\{ℓ+1,ℓ+2,...,n\}}{2k−2}} σ(\{k, ℓ\} ∪ R_{2}, [n])

= ∑_{1⩽k<ℓ⩽n−2} ∑_{R_{2}∈\binom{\{ℓ+1,ℓ+2,...,n\}}{2k−2}} σ(R_{2}, \{ℓ + 1, ℓ + 2, …, n\})

= ∑_{1⩽k<ℓ⩽n−2} s_{2k−2,n−ℓ} > 0.

Next, we define the signed incidence matrix.

Definition Appendix A.5. Let 0 < a < b ⩽ n be integers. Then M_{a,b,n} is the \binom{n}{b} × \binom{n}{a} matrix where the rows are indexed by b-subsets of [n], the columns by a-subsets of [n], and where the entry in row B, column A is σ(A, B).

Theorem Appendix A.6. Let 0 < a < b ⩽ n be integers. If d = b − a is even, then M_{a,b,n} has full rank.

Proof. Denote the row indexed by R ∈ \binom{[n]}{b} by τ_{R}, then τ_{R} can be regarded as an element in V_{a}([n]), the free ℚ-vector space on the a-subsets of [n]. If we denote the basis element corresponding to a a-subset C by ϵ_{C}, then

τ_{R} = ∑_{C∈\binom{[n]}{a}} σ(C, R)ϵ_{C}.

The number of rows in M_{a,b,n} is \binom{n}{b}, and the number of columns is \binom{n}{a}. There are less rows than columns if a + b > n, as many rows as columns if a + b = n, and more rows than columns if a + b < n.

1. If a + b ⩾ n, we must prove that the rows are linearly independent. Suppose that there is a linear relation among the τ_{R}’s, so that

∑_{R∈\binom{[n]}{b}} a_{R}τ_{R} = 0  (34)

for some numbers a_{R}. We shall prove that all a_{R} = 0.

Choose an I ⊂ \binom{[n]}{i}, 0 ⩽ i ⩽ a, and define a linear functional H_{I} : V_{a}([n]) → ℚ by

f_{I}(ϵ_{C}) = \begin{cases}1 & I ⊆ C \\ 0 & I ⊈ C \end{cases}  (35)

Then if R ∈ \binom{[n]}{b} we have that

f_{I}(τ_{R}) = f_{I} ( ∑_{C∈\binom{[n]}{a}} σ(C, R)ϵ_{C}) = ∑_{C∈\binom{[n]}{a}} σ(C, R)f_{I}(ϵ_{C})

= ∑_{I⊆C⊆R} σ(C, R) = s_{a}(I, R, n) = \begin{cases}s_{d,b−i} & I ⊆ R \\ 0 & I ⊈ R \end{cases} (36)

The last step follows from Lemma Appendix A.3. Applying f_{I} to (34) we get that

0 = f_{I} ( ∑_{R∈\binom{[n]}{b}} a_{R}τ_{R}) = ∑_{R∈\binom{[n]}{b}} a_{R}f_{I}(τ_{R})

= ∑_{R∈\binom{[n]}{b}} a_{R}s_{a}(I, R) = s_{d,b−i} ∑_{R∈\binom{[n]}{b}R⊇I} a_{R} (37)

Since Lemma Appendix A.4 tells us that s_{d,b−i} ≠ 0, we conclude that

∑_{R⊇I} a_{R} = 0  (38)

Now, for any J ⊂ [n] we have, by exclusion-inclusion, that

∑_{R∩J=∅} a_{R} = ∑_{I⊂J}(−1)^{|I|} ∑_{R⊇I} a_{R}  (39)

Fix R_{0} ∈ \binom{[n]}{b} and put J_{0} = [n] \ R_{0}. Since |J_{0}| = n − b ⩽ a we have, using (38) that

a_{R_{0}} = ∑_{R∩J_{0}=∅} a_{R} = ∑_{I⊆J_{0}}(−1)^{|I|} ∑_{R⊇I} a_{R} = 0  (40)

Since a_{R_{0}} was arbitrary, all a_{R} are zero. This shows that the τ_{R} are linearly independent.

2. If n = a + b, then M is a square matrix. By the previous case, the vectors τ_{R} are linearly independent, but since there are \binom{n}{a} = \binom{n}{b} such vectors, they form a basis of V_{a}([n]); in particular, they span this vector space.

3. Finally, let us consider the remaining case n > a + b, so that there are more rows than columns. We must prove that the rows span V_{a}([n]). We prove this by induction over n − a − b. The case n − a − b = 0 is already proved, and forms the basis of the induction. We assume a, b fixed, and that the assertion has been proved for all a + b ⩽ n′ < n. Let Γ ∈ \binom{[n]}{a} be arbitrary. If we can express α = ϵ_{Γ} as a linear combination of the τ_{R}’s, we are done. To this end, put

α′ = ∑_{\begin{matrix}S∈\binom{[n−1]}{a−1} \\ S∪\{n\}=Γ \end{matrix}} ϵ_{S} ∈ V_{a−1}([n − 1])  (41)

Since a − 1 + b < n − 1, it follows by induction that there are scalars \{ d_{J} J ∈ \binom{[n−1]}{a−1} \} such that

α′ = ∑_{J∈\binom{[n−1]}{a−1}} d_{J}τ′_{J},  τ′_{J} = ∑_{\begin{matrix}S∈\binom{[n−1]}{a−1} \\ S⊆J \end{matrix}} ϵ_{S}  (42)

For R ∈ \binom{[n]}{a}, n ∈ R, put c′_{R} = d_{R} \ \{n\}. Define

α_{0} = ∑_{\begin{matrix}R∈\binom{[n]}{a} \\ n∈R \end{matrix}} c′_{R}τ_{R} ∈ V_{a}([n])  (43)

If we write

α_{0} = ∑_{C∈\binom{[n]}{a}} a′_{C}ϵ_{C}

we have that for C ∈ \binom{[n]}{a}, n ∈ C, that

a′_{C} = \begin{cases}1 & C = Γ \\ 0 & C ≠ Γ \end{cases}

which implies that

α_{0} = \begin{cases}α & n ∈ Γ \\ 0 & n ∉ Γ \end{cases}

In either case, α − α_{0} has coordinate 0 in component R ∈ \binom{[n]}{a}, unless n ∈ R. Hence, α − α_{0} may be regarded as a vector in V_{a}([n − 1]). By the induction hypothesis, there exist c′′_{R} such that

α − α_{0} = ∑_{R∈\binom{[n−1]}{a}} c′′_{R}τ_{R}  (44)

Defining

c_{R} = \begin{cases}c′_{R} & n ∈ R \\ c′′_{R} & n ∉ R \end{cases}

we get that

α = α_{0} + (α − α_{0}) = ∑_{\begin{matrix}R∈\binom{[n]}{a} \\ n∈R \end{matrix}} c′_{R}τ_{R} + ∑_{\begin{matrix}R∈\binom{[n]}{a} \\ n∉R \end{matrix}} c′′_{R}τ_{R} = ∑_{R∈\binom{[n]}{a}} c_{R}τ_{R}

Appendix B. Calculations

The computer calculations were done on the computers of the UMS Medicis, cole Polytechnique, and on the computers at the Department of Mathematics, Stockholm University. We have used the programme Macaulay 2 [12] to calculate Hilbert series and minimal free resolutions. To save time and memory, the calculations were performed in characteristic 31991. The holes in the tables show that there are limits to what we could calculate, even on a machine with 2 GB of memory.

References

[1] Anetta Aramova, Luchezar L. Avramov, and Jürgen Herzog. Resolutions of monomial ideals and cohomology over exterior algebras. Transactions of the American Mathematical Society, 352(2):579–594, 1999.

[2] Annetta Aramova, Jürgen Herzog, and Takayuki Hibi. Gotzman theorems for exterior algebras and combinatorics. Journal of Algebra, 191:174–211, 1997.

[3] A. M. Bigatti. Aspetti Combinatorici e Computazionali dell’Algebra Commutativa. PhD thesis, Università di Torino, 1995.

[4] N. Bourbaki. Éléments de mathématique. Première partie: Les structures fondamentales de l’analyse. Livre II: Algèbre. Chapitre 9: Formes sesquilinéaires et formes quadratiques. Hermann, Paris, 1959. Actualités Sci. Ind. no. 1272.

[5] G. F. Clements and B. Lindström. A generalization of a combinatorial theorem of Macaulay. J. Combinatorial Theory, 7:230–238, 1969.

[6] David Eisenbud. Commutative Algebra with a View Toward Algebraic Geometry, volume 150 of Graduate Texts in Mathematics. Springer Verlag, 1995.

[7] Ralf Fröberg. An inequality for Hilbert series of graded algebras. Mathematica Scandinavica, 56:117–144, 1985.

[8] Ralf Fröberg and Joachim Hollman. Hilbert series for ideals generated by generic forms. Journal of Symbolic Computation, 17:149–157, 1994.

[9] Ralf Fröberg and Clas Löfwall. On Hilbert series for commutative and noncommutative graded algebras. Journal of Pure and Applied Algebra, 76:33–38, 1991. North Holland.

[10] Ralf Fröberg and Clas Löfwall. Koszul homology and lie algebras with application to generic forms and points. Technical Report 8, Department of Mathematics, Stockholm University, 2000.

[11] J. E. Graver and W. B. Jurkat. The module structure of integral designs. Journal of Combinatorial Theory. Series A, 15:75–90, 1973.

[12] Daniel R. Grayson and Michael E. Stillman. Macaulay 2. Computer algebra

program, available at http://www.math.uiuc.edu/Macaulay2/.

[13] William M. Kantor. On incidence matrices of finite projective and affine spaces. Mathematische Zeitschrift, 124:315–318, 1972.

[14] G. Katona. A theorem of finite sets. In Theory of graphs (Proc. Colloq., Tihany, 1966), pages 187–207. Academic Press, New York, 1968.

[15] Joseph B. Kruskal. The number of simplices in a complex. In Mathematical optimization techniques, pages 251–278. Univ. of California Press, Berkeley, Calif., 1963.

[16] F. S. Macaulay. Some properties of enumeration in the theory of modular systems. Proceedings of the London Mathematical Society, 26:531–555, 1927.

[17] Guillermo Moreno-Socías. Autour de la fonction de Hilbert-Samuel (escaliers d’idéaux polynomiaux). PhD thesis, École Polytechnique, 1991.

[18] Richard M. Wilson. The necessary conditions for t-designs are sufficient for something. Utilitas Mathematica, 4:207–215, 1973.

This article may be accessed via WWW at http://www.rmi.acnet.ge/hha/ or by anonymous ftp at

ftp://ftp.rmi.acnet.ge/pub/hha/volumes/2002/n2a18/v4n2a18.(dvi,ps,pdf)

Guillermo Moreno-Socías moreno@gage.polytechnique.fr

Laboratoire GAGE, cole Polytechnique 91128 Palaiseau Cedex France

Jan Snellman jasne@mai.liu.se

Department of Mathematics Linköping University SE-58183 Linköping SWEDEN