Enumerations deciding the weak Lefschetz property
aa r X i v : . [ m a t h . A C ] D ec Draft:
ENUMERATIONS DECIDING THE WEAK LEFSCHETZ PROPERTY
DAVID COOK II ⋆ , UWE NAGEL Abstract.
We introduce a natural correspondence between artinian monomial almost com-plete intersections in three variables and punctured hexagonal regions. We use this corre-spondence to investigate the algebras for the presence of the weak Lefschetz property. Inparticular, we relate the field characteristics in which such an algebra fails to have the weakLefschetz property to the prime divisors of the enumeration of signed lozenge tilings of theassociated punctured hexagonal region. On the one side this allows us to establish the weakLefschetz property in many new cases. On the other side we can determine some of theprime divisors of the enumerations by means of an algebraic argument.For numerous classes of punctured hexagonal regions we find closed formulae for theenumerations of signed lozenge tilings, and thus the field characteristics in which the asso-ciated algebras fail to the have the weak Lefschetz property. Further, we offer a conjecturefor a closed formula for the enumerations of signed lozenge tilings of symmetric puncturedhexagonal regions. These formulae are exploited to lend further evidence to a conjectureby Migliore, Mir´o-Roig, and the second author that classifies the level artinian monomialalmost complete intersections in three variables that have the weak Lefschetz property incharacteristic zero. Moreover, the formulae are used to generate families of algebras whichnever, or always, have the weak Lefschetz property, regardless of field characteristic. Finally,we determine (in one case, depending on the presence of the weak Lefschetz property) thesplitting type of the syzygy bundle of an artinian monomial almost complete intersection inthree variables, when the characteristic of the base field is zero.Our results convey an intriguing interplay between problems in algebra, combinatorics,and algebraic geometry, which raises new questions and deserves further investigation.
Contents
1. Introduction 22. Compiling the tool-chain 43. Almost complete intersections 64. Punctured hexagons and friends 115. Interlude of signs 176. Determinants 217. Centralising the puncture 308. Interesting families and examples 339. Splitting type and regularity 36Appendix A. Hyperfactorial calculus 41References 45
Mathematics Subject Classification.
Key words and phrases.
Monomial ideals, weak Lefschetz property, determinants, lozenge tilings, non-intersecting lattice paths, perfect matchings.Part of the work for this paper was done while the authors were partially supported by the NationalSecurity Agency under Grant Number H98230-09-1-0032. ⋆ Corresponding author. Introduction
The starting point of this paper has been an intriguing conjecture in [25] on the weakLefschetz property of certain algebras. Though the presence of this property implies con-siderable restrictions on invariants of the algebra, many algebras are expected to have theweak Lefschetz property. However, establishing this property is often rather difficult. In thispaper we make progress on the above conjecture and illustrate the depth of the problem byconsidering a larger class of algebras and relating the problem to a priori seemingly unrelatedquestions in combinatorics and algebraic geometry. This builds on the work of many authors(e.g., [4], [6], [8], [22], and [25]).Throughout this work we consider in particular the question of how the weak Lefschetzproperty of a certain K -algebra A depends on the characteristic of the field K . We begin byrelating the algebra A to two square integer matrices, N and Z , where the entries of N arebinomial coefficients and Z is a zero-one matrix. We show that A has the weak Lefschetzproperty if and only if the determinant of either of these matrices does not vanish modulothe characteristic of K . Next, we establish that the determinant of N enumerates signedlozenge tilings of a punctured hexagonal region and that the determinant of Z enumeratessigned perfect matchings of a bipartite graph associated to the same punctured hexagonalregion. The relation to the weak Lefschetz property implies that both determinants havethe same prime divisors; in fact, we show that their absolute values are the same by usingcombinatorial arguments. Finally, we show that in certain cases deciding the presence of theweak Lefschetz property is equivalent to determining the splitting type of some semistablerank three vector bundles on the projective plane.We now describe the contents of this paper in more detail. Let R = K [ x , . . . , x n ] be thestandard graded n -variate polynomial ring over the infinite field K , and let A be a standardgraded K -algebra over R . We say A is artinian if A is finite dimensional as a vector spaceover K . Further, an artinian K -algebra A is said to have the weak Lefschetz property ifthere exists a linear form ℓ ∈ [ A ] such that, for all integers d , the multiplication map × ℓ : [ A ] d → [ A ] d +1 has maximal rank, that is, the map is injective or surjective. Such alinear form is called a Lefschetz element of A .The weak Lefschetz property has been studied extensively for many reasons, especiallyfor the relation to the Hilbert function (see, e.g., [1], [16], [26], and [29]). A convenient wayto encode the Hilbert function of an artinian K -algebra A is the h -vector , a finite sequence h ( A ) = ( h , . . . , h e ) of positive integers h i = dim K [ A ] i . Using this notation, one immediateconsequence ([16, Remark 3.3]) of A having the weak Lefschetz property is that the h -vectorof A is strictly unimodal . Further, the positive part of the first difference of h ( A ) is h ( A/ℓA ),where ℓ is any Lefschetz element of A .The weak Lefschetz property is known to be subtle to both deformations (see, e.g., [9], [24],and [25]) but also to field characteristic. The latter, considering the weak Lefschetz propertyin positive characteristic, is an exciting and active direction of research. Migliore, Mir´o-Roig,and the second author [25], as well as Zanello and Zylinski [29], began explorations into theconnection between the weak Lefschetz property and positive characteristic, and also posedseveral interesting questions.In [8], the authors found a connection between certain families of level artinian mono-mial almost complete intersections and lozenge tilings of hexagons; independently, Li andZanello [22] found a similar connection for artinian monomial complete intersections (seealso Corollary 6.5). However, both were without combinatorial bijection until one was found NUMERATIONS DECIDING THE WLP 3 by Chen, Guo, Jin, and Liu [6]; Boyle, Migliore, and Zanello [2] have pushed this connectionfurther. Brenner and Kaid [5] also consider artinian monomial complete intersections inthree variables with generators all of the same degree. We also note that in their study ofpure O -sequences Boij, Migliore, Mir´o-Roig, the second author, and Zanello [1] have exploredthe relation between the weak Lefschetz property and pure O -sequences.In this paper we extend the connection found by Chen, Guo, Jin, and Liu to a connectionbetween artinian monomial almost complete intersections in three variables and lozengetilings of more general regions that we call punctured hexagons. In Section 2 we gather afew useful tools for dealing with the weak Lefschetz property. In Section 3 we introducethe algebras we are interested in: artinian monomial almost complete intersections in threevariables. If the syzygy bundle is not semistable, then the algebra has the weak Lefschetzproperty in characteristic zero ([4]). Thus we focus on the algebras that have semistablesyzygy bundles, which we classify numerically (Proposition 3.3). Then we prove that suchan algebra has the weak Lefschetz property if and only if a particular map between the peakhomogeneous components of the algebra is a bijection (Corollary 3.7). Using this, we showthat to each of the studied algebras A we can associate a zero-one matrix Z A such that A hasthe weak Lefschetz property in positive characteristic p if and only if p is not a prime divisorof the determinant of Z A (Proposition 3.8). We also describe a matrix N A with binomialentries that also has the analogous property (Proposition 3.9). Moreover, we demonstratethat a rather simple algebraic argument can be used to determine some of the prime divisorsof the determinants for both Z A and N A (Proposition 3.10).In Section 4 we organise the monomials generating the peak homogeneous componentsof such an algebra in a plane. It turns out that the monomials fill a punctured hexagon(Theorem 4.1). Using the well-known bijection between lozenge tilings and non-intersectinglattice paths, and the Lindstr¨om-Gessel-Viennot theorem ([13], [14], [23]) on non-intersectinglattice paths, we show that the determinant of the binomial matrix N A is the enumeration ofthe signed lozenge tilings of the punctured hexagon, up to sign (Theorem 4.5). Furthermore,using another well-known bijection between lozenge tilings and perfect matchings (see, e.g.,[21]), we argue that the determinant of the zero-one matrix Z A is an enumeration of thesigned perfect matchings of the associated bipartite graph, up to sign (Theorem 4.8).In Section 5 we use the aforementioned connections to show that the determinant of thezero-one matrix Z A and the binomial matrix N A are the same, up to sign (Theorem 5.3).Moreover, in a special case of Kasteleyn’s theorem [17] about enumerating perfect matchings,when the puncture has an even side-length, then the determinant and the permanent of Z A are also the same, up to sign (Corollary 5.4).In Section 6 we prove first that the determinant of N A is non-zero when the puncture is ofeven side-length (Theorem 6.3), thus establishing the weak Lefschetz property in many newcases. We then find closed formulae for the determinants when the puncture is trivial (Propo-sition 6.4), when any one side of the hexagonal region has length zero (Proposition 6.6), whena vertex of the puncture touches one of the sides of the region (Proposition 6.9), and when aside of the puncture touches one of the sides of the region (Proposition 6.12). We close witha complete description of when the region is symmetric. In particular, we show that whencertain parity conditions hold the determinant is zero (Proposition 6.14) and we providea conjecture for a closed formula of the determinant when the same parity conditions fail(Conjecture 6.15). D. COOK II, U. NAGEL
In Section 7 we explore two different ways to centralise the puncture. We call the puncture axis-central when it is central along each of the three axes, independently. Using veryinvolved computations, Ciucu, Eisenk¨olbl, Krattenthaler, and Zare [7] found closed formulaefor the enumerations and signed enumerations of regions with an axis-central puncture;therein axis-central is called simply “central”. We use these closed formulae to describethe permanents of the zero-one matrices Z A (Corollary 7.2) and the determinants of bothmatrices (Corollary 7.3), Z A and N A , when the puncture is axis-central. We call the puncture gravity-central when its vertices are equidistant from the sides of the containing hexagon; thiscondition is equivalent to the associated algebra being level, that is, its socle is concentratedin one degree. Using this observation we provide further evidence for a conjecture by Migliore,Mir´o-Roig, and the second author [25] about the presence of the weak Lefschetz property forlevel artinian monomial almost complete intersections in characteristic zero (Proposition 7.7).In Section 8 we describe a method, for any positive integer n , to generate a subfamilyof algebras whose associated matrices have determinant n (Proposition 8.2). From this wegenerate a subfamily of algebras which always have the weak Lefschetz property, regardlessof the field characteristic (Corollary 8.3); we also describe a different subfamily of algebraswhich always have the weak Lefschetz property (Proposition 8.4). Moreover, we describe theunique algebras which retain certain properties yet have minimal multiplicity (Example 8.6).In Section 9 we explicitly determine (in one case, depending on the presence of the weakLefschetz property) the splitting type of all artinian monomial almost complete intersections.In particular, we consider separately the cases when the syzygy bundle is non-semistable(Proposition 9.3) and semistable (Propositions 9.6 and 9.7). Moreover, in the case of idealsassociated to punctured hexagons, we relate the weak Lefschetz property to a number ofother problems in algebra, combinatorics, and algebraic geometry (Theorem 9.9).Finally, in Appendix A we provide a technique, a “picture-calculus”, for working withhyperfactorials, a basic unit for the aforementioned closed formulae. We demonstrate thatseveral nice polynomials can be written as ratios of products of hyperfactorials (Proposi-tion A.1 and Corollary A.5). Further, this shows that MacMahon’s formula for the numberof lozenge tilings of a (non-punctured) hexagon is a polynomial in one of the side-lengthswhen the other two are fixed (Corollary A.3).2. Compiling the tool-chain
Let R = K [ x , . . . , x n ] be the standard graded n -variate polynomial ring over the infinitefield K , and let A be an artinian standard graded K -algebra over R . Then the minimal freeresolution of A ends with the free module L mi =1 R ( − t i ) r i , where 0 < t < · · · < t m and 0 < r i for all i . In this case, A is called level if m = 1, the socle degrees of A are t i − n , for all i ,and the socle type of A is the sum P mi =1 t i .We recall that once multiplication by a general linear form is surjective, then it remainssurjective. Proposition 2.1. [25, Proposition 2.1(a)]
Let A = R/I be an artinian standard graded K -algebra, and let ℓ be a general linear form. If the map × ℓ : [ A ] d → [ A ] d +1 is surjective, then × ℓ : [ A ] d +1 → [ A ] d +2 is surjective. This generalises to modules generated in degrees that are sufficiently small.
NUMERATIONS DECIDING THE WLP 5
Lemma 2.2.
Let M be an R -module generated in degrees bounded by e , and let ℓ be ageneral linear form. If the map × ℓ : [ M ] d → [ M ] d +1 is surjective, and d ≥ e , then the map × ℓ : [ M ] d +1 → [ M ] d +2 is surjective.Proof. Consider the sequence[ M ] d × ℓ −→ [ M ] d +1 → [ M/ℓM ] d +1 → . Notice the first map is surjective if and only if [
M/ℓM ] d +1 = 0 . By assumption the mapis surjective, so [
M/ℓM ] d +1 = 0. Hence [ M/ℓM ] d +2 is zero unless there is a generator of M with degree beyond d . However, the assumption is that no generators exist with degreebeyond d. (cid:3) From this we get a result analogous to [25, Proposition 2.1(b)] for non-level algebras.
Proposition 2.3.
Let A = R/I be an artinian standard graded K -algebra, and let ℓ be ageneral linear form. If the map × ℓ : [ A ] d − → [ A ] d is injective, and d is no greater than thesmallest socle degree of A , then × ℓ : [ A ] d − → [ A ] d − is injective.Proof. The K -dual of A , M , is a shift of the canonical module of A and is generated in degreesthat are a linear shift of the socle degrees of A . Consider now the map × ℓ : [ M ] i → [ M ] i +1 .Using Lemma 2.2 we see that once i is at least as large as the largest degree in which M isgenerated, and the map is surjective, then the map is surjective thereafter. The result thenfollows by duality. (cid:3) Further recall that a monomial algebra has the weak Lefschetz property exactly when thesum of the variables is a Lefschetz element.
Proposition 2.4. [25, Proposition 2.2]
Let A = R/I be an artinian standard graded K -algebra with I generated by monomials. Then A has the weak Lefschetz property if and onlyif x + · · · + x n is a Lefschetz element of A . Hence, the weak Lefschetz property can be decided for monomial ideals, in a small numberof cases, by simple invariants. The following lemma is a generalisation of [22, Proposition 3.7].
Lemma 2.5.
Let A = R/I be an artinian standard graded K -algebra with I generated bymonomials. Suppose that a is the least positive integer such that x ai ∈ I , for ≤ i ≤ n ,and suppose that the Hilbert function of R/I weakly increases to degree s + 1 . Then, for anypositive prime p such that a ≤ p m ≤ s + 1 for some positive integer m , A fails to have theweak Lefschetz property in characteristic p .Proof. By Proposition 2.4, we need only consider ℓ = x + · · · + x n . Suppose the characteristicof K is p , then by the Frobenius endomorphism ℓ · ℓ p m − = ℓ p m = x p m + · · · + x p m n . Moreover,as a ≤ p m , then ℓ p m = 0 in A while ℓ = 0 in A . Hence × ℓ p m − : [ A ] → [ A ] p m is not injectiveand thus A does not have the weak Lefschetz property. (cid:3) Further, for monomial ideals, if the weak Lefschetz property holds in characteristic zero,then it holds for almost every characteristic.
Lemma 2.6.
Let I be an artinian monomial ideal in R . If R/I has the weak Lefschetzproperty when char K = 0 , then R/I has the weak Lefschetz property for char K sufficientlylarge. D. COOK II, U. NAGEL
Proof.
By Proposition 2.4, we need only consider ℓ = x + · · · + x n . As R/I is artinian, thenthere are finitely many maps that need to be checked for the maximal rank property, andthis in turn implies finitely many determinants that need to be computed. Further, becauseof the form of ℓ , the matrices in question are all zero-one matrices. Thus, the determinantsto be checked are integers. Simply let p be the smallest prime larger than all prime divisorsof the determinants, then the determinants are all non-zero modulo p and so R/I has theweak Lefschetz property if char K ≥ p . (cid:3) And (pseudo-)conversely, again for monomial ideals, if the weak Lefschetz property holdsin some positive characteristic, then it holds for characteristic zero.
Lemma 2.7.
Let I be an artinian monomial ideal in R . If R/I has the weak Lefschetzproperty when char K = p > , then R/I has the weak Lefschetz property for char K = 0 .Proof. The proof is the same as that of Lemma 2.6 except we notice that if an integer d isnon-zero modulo a prime p , then d is not zero. (cid:3) Last, we note that any artinian ideal in two variables has the weak Lefschetz property.This was proven for characteristic zero in [16, Proposition 4.4] and then for arbitrary char-acteristic in [26, Corollary 7], though it was not specifically stated therein, as noted in [22,Remark 2.6]. We provide a brief, direct proof of this fact to illustrate the weak Lefschetzproperty. Unfortunately, the simplicity of this proof fails in three variables, even for mono-mial ideals.
Proposition 2.8.
Let R = K [ x, y ] , where K is an infinite field with arbitrary characteristic.Every artinian algebra in R has the weak Lefschetz property.Proof. Assume I = ( g , . . . , g t ) and the given generators are minimal. Let s = min { deg g i | ≤ i ≤ t } . Then h ( R/I ), the h -vector of R/I , strictly increases by one from h to h s − and h s − ≥ h s , thus the positive part of the first difference of h ( R/I ), ∆ + h ( R/I ), is s ones. More-over, for a general linear form ℓ ∈ R , R/ ( I, ℓ ) ∼ = K [ x ] /J where J = ( x s ) so h ( R/ ( I, ℓ )) is s ones, that is, ∆ + h ( R/I ) = h ( R/ ( I, ℓ )). Hence
R/I has the weak Lefschetz property withLefschetz element ℓ . (cid:3) Almost complete intersections
Here we restrict to artinian monomial almost complete intersections in three variables.These are the ideals discussed in [4, Corollary 7.3] and [25, Section 6].Let K be an infinite field, and consider the ideal I a,b,c,α,β,γ = ( x a , y b , z c , x α y β z γ )in R = K [ x, y, z ], where 0 ≤ α < a, ≤ β < b, and 0 ≤ γ < c . If α = β = γ = 0, then wedefine I a,b,c, , , to be ( x a , y b , z c ) which is a complete intersection and is studied extensivelyin [22] and [6]. Assume at most one of α, β, and γ is zero. Proposition 3.1. [25, Proposition 6.1]
Let I = I a,b,c,α,β,γ be defined as above. Assume,without loss of generality, that ≤ α ≤ β ≤ γ . (i) If α = 0 , then R/I has socle type with socle degrees a + β + c − and a + b + γ − ;thus R/I is level if and only if b − β = c − γ . (ii) If α > , then R/I has socle type with socle degrees α + b + c − , a + β + c − ,and a + b + γ − ; thus R/I is level if and only if a − α = b − β = c − γ . NUMERATIONS DECIDING THE WLP 7 (iii)
Moreover, the minimal free resolution of
R/I has the form (3.1) 0 → R ( − a − b − γ ) ⊕ R ( − a − β − c ) ⊕ R n ( − α − b − c ) → R ( − a − β − γ ) ⊕ R ( − α − b − γ ) ⊕ R ( − α − β − c ) ⊕ R ( − a − b ) ⊕ R ( − a − c ) ⊕ R n ( − b − c ) → R ( − α − β − γ ) ⊕ R ( − a ) ⊕ R ( − b ) ⊕ R ( − c ) → R → R/I → where n = 1 if α > and n = 0 if α = 0 . Moreover, we see that in characteristic zero the weak Lefschetz property follows for certainchoices of the parameters.
Proposition 3.2. [25, Theorem 6.2]
Let K be an algebraically closed field of characteristiczero. Then R/I a,b,c,α,β,γ has the weak Lefschetz property if a + b + c + α + β + γ . Semi-stability.
The syzygy module syz I of I = I a,b,c,α,β,γ fits into the exact sequence0 −→ syz I −→ R ( − α − β − γ ) ⊕ R ( − a ) ⊕ R ( − b ) ⊕ R ( − c ) −→ I a,b,c,α,β,γ −→ . The sheafification ] syz I is a rank 3 bundle on P , and it is called the syzygy bundle of I .Recall that a vector bundle E on projective space is said to be semistable if, for everycoherent subsheaf F ⊂ E , the following inequality holds: c ( F ) rk ( F ) ≤ c ( E ) rk ( E ) . We analyse when I a,b,c,α,β,γ has a semistable syzygy bundle. (Note, the slightly awkwarddefinition of s in the following is kept for consistency with [25, Section 7], the starting pointof this work.) Proposition 3.3.
Let K be an algebraically closed field of characteristic zero. Further, let I = I a,b,c,α,β,γ , and define the following rational numbers s := 13 ( a + b + c + α + β + γ ) − ,A := s + 2 − a,B := s + 2 − b,C := s + 2 − c, and M := s + 2 − ( α + β + γ ) . Then I has a semistable syzygy bundle if and only if the following conditions all hold: (i) 0 ≤ M , (ii) 0 ≤ A ≤ β + γ , (iii) 0 ≤ B ≤ α + γ , and D. COOK II, U. NAGEL (iv) 0 ≤ C ≤ α + β .Proof. Using [3, Corollary 7.3] we have that I has a semistable syzygy bundle if and only if(a) max { a, b, c, α + β + γ } ≤ s + 2,(b) min { α + β + c, α + b + γ, a + β + γ } ≥ s + 2, and(c) min { a + b, a + c, b + c } ≥ s + 2.Notice that condition (a) is equivalent to A, B, C, and M being non-negative. Moreover,condition (b) is equivalent to the upper bounds on A, B, and C . We claim that condition(c) follows directly from condition (a).Indeed, by condition (a) we have that C + M ≥ A + B + C + M = s + 2 ≥ A + B = 2( s + 2) − a − b , thus a + b ≥ s + 2. Similarly, we have a + c ≥ s + 2 and b + c ≥ s + 2.Thus condition (c) holds if condition (a) holds. (cid:3) This gives further conditions on the parameters that force the weak Lefschetz property incharacteristic zero (see [4, Theorem 3.3]). This extends [25, Lemma 6.7].
Corollary 3.4.
Let K be an algebraically closed field of characteristic zero, and let I = I a,b,c,α,β,γ . If any of the conditions (i)-(iv) in Proposition 3.3 fail, then R/I has the weakLefschetz property.
The above definitions of s, A, B, C, and M are not without purpose. Before going further,we make a few comments about the given parameters. Remark 3.5.
Suppose s, A, B, C, and M are defined as in Proposition 3.3. Then clearly s is an integer if and only if a + b + c + α + β + γ ≡ s is an integer, then so are A, B, C, and M . Further, A + B + C + M = s + 2 and A + B + C = α + β + γ .3.2. Associated matrices.
Given the minimal free resolution of
R/I (see (3.1)), we caneasily compute the h -vector of R/I as a weighted sum of binomial coefficients dependentonly on the parameters a, b, c, α, β, and γ .We say h ( A ) has twin peaks if there exists an integer s such that h s = h s +1 . When I a,b,c,α,β,γ has parameters as in Proposition 3.3 and s is an integer, then the algebras R/I a,b,c,α,β,γ alwayshave twin peaks and the peaks are bounded by the socle degrees. This extends the resultsin [25, Lemma 7.1] wherein the level algebras
R/I a,b,c,α,β,γ with twin peaks are identified.
Lemma 3.6.
Assume the parameters of I = I a,b,c,α,β,γ satisfy the conditions in Proposi-tion 3.3 and suppose a + b + c + α + β + γ ≡ . Then R/I has twin peaks in degrees s and s + 1 . Moreover, s + 1 is bounded above by the socle degrees of R/I .Proof.
The upper bounds on
A, B, and C are exactly those required to force the ultimateand penultimate terms in the minimal free resolution of R/I , given in Proposition 3.1(iii),to not contribute to the computation of the h -vector for degrees up to s + 1. Moreover, as A, B, C, and M are non-negative, and using (cid:0) n +12 (cid:1) − (cid:0) n (cid:1) = n for n ≥
0, then h s +1 − h s = (cid:18)(cid:18) s + 32 (cid:19) − (cid:18) A + 12 (cid:19) − (cid:18) B + 12 (cid:19) − (cid:18) C + 12 (cid:19) − (cid:18) M + 12 (cid:19)(cid:19) − (cid:18)(cid:18) s + 22 (cid:19) − (cid:18) A (cid:19) − (cid:18) B (cid:19) − (cid:18) C (cid:19) − (cid:18) M (cid:19)(cid:19) = s + 2 − ( A + B + C + M )=0 . NUMERATIONS DECIDING THE WLP 9
Suppose, without loss of generality, that α ≤ β ≤ γ . The socle degrees of R/I are α + b + c − a + β + c −
3, and a + b + γ −
3, with the first removed if α = 0. The followingargument shows that α + b + c − s + 1, however, with a simple changing of namesit can be used to show that each of the socle degrees is at least s + 1.As we are considering the socle degree α + b + c −
3, we may assume α ≥
1. Notice that α + b + c − A + B + C + 2 M + α −
3, which is at least s + 1 = A + B + C + M − A + M + α ≥
2. If A + M ≥
1, then we are done. Suppose A + M = 0, then A = M = 0and b + c = α + β + γ . Moreover, since b > β and c > γ , then α + β + γ = b + c ≥ β + γ + 2.Thus α ≥ (cid:3) An immediate consequence of the previous lemma is that exactly one map need be con-sidered for each algebra in order to determine the presence of the weak Lefschetz property.
Corollary 3.7.
Assume the parameters of I = I a,b,c,α,β,γ satisfy the conditions in Proposi-tion 3.3 and suppose a + b + c + α + β + γ ≡ . Then R/I has the weak Lefschetzproperty if and only if the map × ( x + y + z ) : [ R/I ] s → [ R/I ] s +1 is injective (or surjective).Proof. This follows immediately from Lemma 3.6 by using Propositions 2.1–2.4. (cid:3)
This leads to the definition of two matrices with determinants that determine the weakLefschetz property. The first is a zero-one matrix and the second is a matrix of binomialcoefficients.
Proposition 3.8.
Assume the parameters of I = I a,b,c,α,β,γ satisfy the conditions in Propo-sition 3.3 and suppose a + b + c + α + β + γ ≡ .Then there exists a matrix Z = Z a,b,c,α,β,γ such that (i) Z is a square integer matrix of size h s , (ii) R/I has the weak Lefschetz property if and only if det Z K ) , and (iii) the entries of Z are given by ( Z ) i,j = (cid:26) n j is a multiple of m i , otherwise , where { m , . . . , m h s } and { n , . . . , n h s } are the monomial bases of [ R/I ] s and [ R/I ] s +1 ,respectively, and are given in lexicographic order.Proof. We notice that the map × ( x + y + z ) : [ R/I ] s → [ R/I ] s +1 can be represented as amatrix Z with rows and columns indexed by fixed monomial bases of [ R/I ] s and [ R/I ] s +1 ,respectively. This follows immediately from viewing [ R/I ] d as a vector space over K .Claim (i) follows from Lemma 3.6 wherein it is shown that h s = h s +1 . Since Z is square,then the injectivity of × ( x + y + z ) : [ R/I ] s → [ R/I ] s +1 is equivalent to Z being invertible,that is, equivalent to det Z being non-zero in K . Thus, claim (ii) follows from Corollary 3.7wherein it is shown that the injectivity of the map × ( x + y + z ) : [ R/I ] s → [ R/I ] s +1 exactly determines the presence of the weak Lefschetz property for R/I . Claim (iii) followsimmediately from the construction of the map. (cid:3)
The following generalises the results in [25, Theorem 7.2 and Corollary 7.3].
Proposition 3.9.
Assume the parameters of I = I a,b,c,α,β,γ satisfy the conditions in Propo-sition 3.3, and suppose a + b + c + α + β + γ ≡ .Then there exists a matrix N = N a,b,c,α,β,γ such that (i) N is a square integer matrix of size C + M , (ii) R/I has the weak Lefschetz property if and only if det N K ) , and (iii) the entries of N are given by ( N ) i,j = (cid:18) cA + j − i (cid:19) if ≤ i ≤ C, (cid:18) γA + C − β + j − i (cid:19) if C + 1 ≤ i ≤ C + M. Proof.
Notice that R/ ( I, x + y + z ) ∼ = S/J , where S = K [ x, y ] and J = ( x a , y b , ( x + y ) c , x α y β ( x + y ) γ ) . Thus the sequence [
R/I ] d × ( x + y + z ) −−−−−−→ [ R/I ] d +1 → [ R/ ( I, x + y + z )] d +1 → × ( x + y + z ) : [ R/I ] s → [ R/I ] s +1 is injective exactly when [ S/J ] s +1 = 0. Henceit suffices to show that all s + 2 monomials of the form x i y j where i + j = s + 1 are in J .Clearly if i ≥ a or j ≥ b , then x i y j is in J . This leaves s + 2 − ( s + 2 − a ) − ( s + 2 − b ) = s + 2 − A − B = C + M monomials that are not trivially in J . Thus there are C + M equations and unknowns, all of which only involve the non-monomial terms (after reductionby the monomial terms). Associated to this system of equations is a square integer matrixof size C + M , call it N . Then N is invertible if and only if det N is non-zero in K . Thus,claims (i) and (ii) hold.There are s + 2 − c = C ways to scale ( x + y ) c and s + 2 − ( α + β + γ ) = M ways to scale x α y β ( x + y ) γ to be degree s +1. In both cases consider the binomial coefficient indexed by thedegree of y . Then ( N ) i,j is the coefficient on x a − j y A + j − in the scaling x C − i y i − ( x + y ) c for1 ≤ i ≤ C , i.e., (cid:0) cA + j − i (cid:1) , and in the scaling x C + M − i y i − C − x α y β ( x + y ) γ for C +1 ≤ i ≤ C + M ,i.e., (cid:0) γA + C − β + j − i (cid:1) . Thus claim (iii) holds. (cid:3) Clearly det Z a,b,c,α,β,γ and det N a,b,c,α,β,γ must both be either zero or have the same set ofprime divisors. We can determine a few of the prime divisors from the known failure of theweak Lefschetz property. Proposition 3.10.
Assume the parameters of I = I a,b,c,α,β,γ satisfy the conditions in Propo-sition 3.3, and suppose a + b + c + α + β + γ ≡ . If K has positive characteristic p and their exists a positive integer m such that max { a, b, c } ≤ p m ≤ s + 1 = 13 ( a + b + c + α + β + γ ) − , then (i) R/I fails to have the weak Lefschetz property, (ii) p is a prime divisor of the determinant of Z a,b,c,α,β,γ , and (iii) p is a prime divisor of the determinant of N a,b,c,α,β,γ .Proof. By Lemma 3.6, the Hilbert function of
R/I weakly increases to degree s + 1, hencepart (i) follows by Lemma 2.5. Parts (ii) and (iii) then follow from Propositions 3.8 and 3.9,respectively. (cid:3) In the next section we will see a nice combinatorial interpretation for both matrices aswell as the defined values s, A, B, C, and M . NUMERATIONS DECIDING THE WLP 11 Punctured hexagons and friends
Recall the definition of s, A, B, C, and M , and the conditions thereon, from Proposition 3.3.In this section we assume, without exception, that I = I a,b,c,α,β,γ has parameters matchingthese conditions and further that a + b + c + α + β + γ ≡ Punctured hexagons.
Notice that every monomial in [ R ] d is of the form x i y j z k where i, j, and k are non-negativeintegers such that i + j + k = d . Hence we can organise the monomials in [ R ] d into a triangleof side-length d +1 with x d at the lower-center, y d at the upper-right, and z d at the upper-left.(See Figure 4.1.) Figure 4.1.
The monomial triangle for [ R ] Notice that we can interlace the monomials of [ R ] d − within the monomials of [ R ] d . Ifwe stay consistent with our orientation (i.e., largest power of x at the lower-center, largestpower of y at the upper-right, and largest power of z at the upper-left), then two monomialsare adjacent if and only if one divides the other. (See Figure 4.2.) We call such a figure the interlaced basis region of [ R ] d − and [ R ] d . Figure 4.2.
The interlaced basis region of [ R ] and [ R ] If we compute the interlaced basis region of [
R/I a,b,c,α,β,γ ] s and [ R/I a,b,c,α,β,γ ] s +1 , then weget a punctured hexagonal region. Theorem 4.1.
Let I = I a,b,c,α,β,γ satisfy the conditions in Proposition 3.3, and suppose a + b + c + α + β + γ ≡ . Then the interlaced basis region H a,b,c,α,β,γ of [ R/I ] s and [ R/I ] s +1 is in the shape of a hexagon with side-lengths (in clockwise cyclic order, starting atthe bottom) ( A, B + M, C, A + M, B, C + M ) and with a puncture in the shape of an equilateral triangle of side-length M . The puncturehas sides parallel to the sides of the hexagon of lengths A + M, B + M, and C + M . Moreover, the puncture is located α, β, and γ units from the sides of length A + M, B + M, and C + M ,respectively. (See Figure 4.3.) Figure 4.3. H a,b,c,α,β,γ , the interlaced basis region of [ R/I ] s and [ R/I ] s +1 Proof.
The interlaced basis region of [
R/I ] s and [ R/I ] s +1 corresponds to a spatial placementof the monomials of the associated components of R/I . As I is a monomial ideal, we caneasily get restrictions on the monomials x i y j z k in the region:(i) The generator x a forces 0 ≤ i < a ; this corresponds to the lower-center missingtriangle which has side-length s + 2 − a = A .(ii) The generator y b forces 0 ≤ j < b ; this corresponds to the upper-right missingtriangle which has side-length s + 2 − b = B .(iii) The generator z c forces 0 ≤ k < c ; this corresponds to the upper-left missing trianglewhich has side-length s + 2 − c = C .(iv) The generator x α y β z γ forces one of i < α, j < β , or k < γ to also hold; thiscorresponds to the center missing triangle, which has side-length s +2 − α − β − γ = M .This further forces the particular placement of the puncture.Moreover, the conditions in Proposition 3.3 force the regions to have non-negative side-lengths and to not overlap. (cid:3) Remark 4.2.
The ideals I = I a,b,c,α,β,γ in Theorem 4.1 are in bijection with their hexagonalregions (assuming a fixed orientation and assuming a puncture of side-length zero is stillconsidered to be in a particular position). Suppose we have a punctured hexagonal region,as in Figure 4.3, with parameters A, B, C, M, α, and β . Then a = B + C + M , b = A + C + M , c = A + B + M , and γ = A + B + C − ( α + β ).Moreover, we notice that, in characteristic zero, these ideals are exactly the artinianmonomial almost complete intersections which do not immediately have the weak Lefschetzproperty from Proposition 3.2 or Proposition 3.3.Notice that by Lemma 3.6 we have h s = h s +1 , so the region H a,b,c,α,β,γ has the same numberof upward pointing triangles as it has downward pointing triangles. In particular, it may NUMERATIONS DECIDING THE WLP 13 then be possible to tile the region by lozenges (i.e., rhombi with unit side-lengths and anglesof 60 ◦ and 120 ◦ ; we also note a pair of alternate names used in the literature: calissons anddiamonds).4.2. Non-intersecting lattice paths.
We follow [7, Section 5] (similarly, [12, Section 2]) to translate lozenge tilings of H a,b,c,α,β,γ to families of non-intersecting lattice paths. An example of a lozenge tiling and its associatedfamily of non-intersecting lattice paths is given in Figure 4.4. Hexagon tiling by lozenges Family of non-intersecting lattice paths
Figure 4.4.
Example of a lozenge tiling and its associated family of non-intersecting lattice pathsIn order to transform a lozenge tiling of a punctured hexagon H a,b,c,α,β,γ into a family ofnon-intersecting lattice paths, we follow three simple steps (see Figure 4.5):(i) Mark the midpoints of the triangle edges parallel to the sides of length C and C + M with vertices. Further, label the midpoints, always moving lower-left to upper-right,(a) along the hexagon side of length C as A , . . . , A C ,(b) along the puncture as A C +1 , . . . , A C + M , and(c) along the hexagon side of length C + M as E , . . . , E C + M .(ii) Using the lozenges as a guide, we connect any pair of vertices that occur on a singlelozenge.(iii) Thinking of motion parallel to the side of length A as horizontal and motion parallelto the side of length B as vertical, we orthogonalise the lattice (and paths) andconsider the lower-left vertex as the origin.Given the above transformation of H a,b,c,α,β,γ to the integer lattice, we see that A i and E j have easy to compute coordinates: A i = (cid:26) ( i − , B + M + i −
1) if 1 ≤ i ≤ C, ( β + i − C − , B − α + i −
1) if C + 1 ≤ i ≤ C + M, and E j = ( A + j − , j −
1) for 1 ≤ j ≤ C + M. Now we associate to each family of non-intersecting lattices paths a permutation and useit to assign a sign to the family of paths. (i) Mark midpoints with vertices and labelparticular vertices (ii) Connect vertices using the tiling(iii) Orthogonalise the path family The family by itself
Figure 4.5.
Example of converting lozenge tilings to families of non-intersecting lattice paths
Definition 4.3.
Let L be a family of non-intersecting lattice paths as above, and let λ ∈ S C + M be the permutation so that A i is connected to E λ ( i ) . We define the sign of L to bethe signature (or sign) of the permutation λ . That is, sgn L := sgn λ .Now we are ready to use a beautiful theorem relating (signed) enumerations of families ofnon-intersecting lattice paths with determinants. In particular, we use a theorem first givenby Lindstr¨om in [23, Lemma 1] and stated independently in [14, Theorem 1] by Gessel andViennot. Stanley gives a very nice exposition of the topic in [28, Section 2.7].Here we give a specialisation of the theorem to the case when all edges have the sameweight—one. In particular, this result is given in [7, Lemma 14]. Theorem 4.4.
Let A , . . . , A n , E , . . . , E n be distinct lattice points on N where each A i isabove and to the left of every E j . Then det ≤ i,j ≤ n ( P ( A i → E j )) = X λ ∈ S n sgn( λ ) P + λ ( A → E ) , NUMERATIONS DECIDING THE WLP 15 where P ( A i → E j ) is the number of lattice paths from A i to E j and, for each permutation λ ∈ S n , P + λ ( A → E ) is the number of families of non-intersecting lattice paths with pathsgoing from A i to E λ ( i ) . Thus, we have an enumeration of the signed lozenge tilings of a punctured hexagon withsigns given by the non-intersecting lattice paths.
Theorem 4.5.
The enumeration of signed lozenge tilings of H a,b,c,α,β,γ , with signs given bythe signs of the associated families of non-intersecting lattice paths (Definition 4.3), is givenby det N a,b,c,α,β,γ , where the matrix N a,b,c,α,β,γ is defined in Proposition 3.9.Proof. Notice that the number of lattice paths from ( u, v ) to ( x, y ), where u ≤ x and v ≥ y ,is given by (cid:0) x − u + v − yx − u (cid:1) as there are x − u + v − y steps and x − u must be horizontal steps(equivalently, v − y must be vertical steps). Thus the claim follows immediately from thesteps above. (cid:3) However, we need not consider all ( C + M )! permutations λ ∈ S C + M as the vast majoritywill always have P + λ ( A → E ) = 0. Given our choice of A i and E j the only possible choicesof λ are given by λ k = (cid:18) · · · k k + 1 k + 2 · · · C C + 1 C + 2 · · · C + M · · · k M + k + 1 M + k + 2 · · · C + M k + 1 k + 2 · · · M + k (cid:19) , where 0 ≤ k ≤ C and k corresponds to the number of lattice paths that go below thepuncture. In particular, the three parts of λ k correspond to the paths going below, above,and starting from the puncture. We call these permutations the admissible permutations of H a,b,c,α,β,γ .We will use this connection to compute determinants in Section 6, but first we look at analternate combinatorial connection.4.3. Perfect matchings.
Lozenge tilings of a punctured hexagon can be associated to perfect matchings on a bi-partite graph. This connection was first used by Kuperberg in [21] to study symmetries onplane partitions. An example of a lozenge tiling and its associated perfect matching of edgesis given in Figure 4.6. (i) Hexagon tiling by lozenges (ii) Perfect matching of edges
Figure 4.6.
Example of a lozenge tiling and its associated perfect matching of edges
In order to transform a lozenge tiling of a punctured hexagon H a,b,c,α,β,γ into a perfectmatching of edges, we follow three simple steps (see Figure 4.7):(i) Put a vertex at the center of each triangle.(ii) Connect the vertices whose triangles are adjacent.(iii) Select the edges which the lozenges cover–this set is the perfect matching. (i) Put vertices in triangle centers (ii) Connect vertices of adjacent triangles(iii) Select edges covered by lozenges The perfect matching by itself Figure 4.7.
Example of converting lozenge tilings to perfect matchings of edgesNotice that the graph associated to the punctured hexagon H a,b,c,α,β,γ is a bipartite graphwith colour classes given by monomials in [ R/I ] s and [ R/I ] s +1 . Thus we can represent thisbipartite graph by a bi-adjacency matrix with rows enumerated by the monomials in [ R/I ] s and columns enumerated by the monomials in [ R/I ] s +1 . We fix the order on the monomialsto be the lexicographic order. Clearly then the matrix Z a,b,c,α,β,γ from Proposition 3.8 is thebi-adjacency matrix described here.Consider the permanent of Z = Z a,b,c,α,β,γ , that is,perm Z = X π ∈ S hs h s Y i =1 ( Z ) i,π ( i ) . As Z has entries which are either zero or one, we see that all summands in perm Z are eitherzero or one. Moreover, each non-zero summand corresponds to a perfect matching, as it cor-responds to an isomorphism between the two colours classes of the bipartite graph, namely, NUMERATIONS DECIDING THE WLP 17 the monomials in [
R/I ] s and [ R/I ] s +1 . Thus, perm Z enumerates the perfect matchings ofthe bipartite graph associated to H a,b,c,α,β,γ , and hence perm Z also enumerates the lozengetilings of H a,b,c,α,β,γ . Proposition 4.6.
The number of lozenge tilings of H a,b,c,α,β,γ is perm Z a,b,c,α,β,γ . Since each perfect matching is an isomorphism between the two colour classes, it can beseen as a permutation π ∈ S h s . As with Definition 4.3, it is thus natural to assign a sign toa given perfect matching. Definition 4.7.
Let P be a perfect matching of the bipartite graph associated to H a,b,c,α,β,γ ,and let π ∈ S h s be the associated permutation (as described above). We define th sign of P to be the signature of the permutation π . That is, sgn P := sgn π .Since the sign is the sign that is used in computing the determinant of the matrix Z a,b,c,α,β,γ ,we get an enumeration of the signed lozenge tilings of a punctured hexagon with signs givenby the perfect matchings. Theorem 4.8.
The enumeration of signed perfect matchings of the bipartite graph associatedto H a,b,c,α,β,γ , with signs given by the signs of the related perfect matchings, is given by det Z a,b,c,α,β,γ , where the matrix Z a,b,c,α,β,γ is defined in Proposition 3.8. Remark 4.9.
Kasteleyn [17] provided, in 1967, a general method for computing the numberof perfect matchings of a planar graph as a determinant. Moreover, he provided a classicalreview of methods and applications of enumerating perfect matchings. Planar graphs, suchas the “honeycomb graphs” described here, are studied for their connections to physics; inparticular, honeycomb graphs model the bonds in dimers (polymers with only two structuralunits) and perfect matchings correspond to so-called dimer coverings . Kenyon [18] gives amodern recount of explorations on dimer models, including random dimer coverings andtheir limiting shapes.
Remark 4.10.
Recall that Proposition 3.10 provides a numerical constraint that determinessome of the prime divisors of the determinants of the matrices Z a,b,c,α,β,γ and N a,b,c,α,β,γ bymeans of some algebra deciding the weak Lefschetz property for the algebra R/I a,b,c,α,β,γ .Hence, by Theorems 4.5 and 4.8, we see that information from algebra can indeed be usedto determine some of the prime divisors of the enumerations of signed lozenge tilings and ofsigned perfect matchings.Finally, we note that in [27], Propp gives a history of the connections between lozengetilings (of non-punctured hexagons), perfect matchings, plane partitions, non-intersectinglattice paths. 5.
Interlude of signs
In the preceding section we discussed three related combinatorial structures from which wecan extract the primes p for which the algebras R/I a,b,c,α,β,γ fail to have the weak Lefschetzproperty. Therein we discussed two different ways to assign a sign to a lozenge tiling: bythe associated family of non-intersecting lattice paths (Definition 4.3) and by the associatedperfect matching (Definition 4.7). We now show that the two signs indeed agree.Fix a hexagonal region H = H a,b,c,α,β,γ , and fix a lozenge tiling T of H . As discussed inSection 4, we associate to the tiling T a family of non-intersecting lattice paths L T and a perfect matching P T . Moreover, we introduced a permutation λ T ∈ S C + M associated to L T (see Definition 4.3) and a permutation π T ∈ S h s associated to P T via Z a,b,c,α,β,γ (seeDefinition 4.7).We first notice that “rotating” particular lozenge groups of T do not change the permu-tation associated to the non-intersecting lattice paths. Lemma 5.1.
Let T be a lozenge tiling of H a,b,c,α,β,γ . Pick any triplet of lozenges in T which iseither an up or a down grouping, as in Figure 5.1, and let U be T with the triplet exchanged Figure 5.1. up and down lozenge groups with lattice path pieces superimposed for the other possibility (i.e., rotated ◦ ). Then U is a lozenge tiling of H a,b,c,α,β,γ and λ T = λ U . Moreover, π U = τ π T , for some three-cycle π ∈ S h s .Proof. First, we note that if T is a lozenge tiling of H a,b,c,α,β,γ then clearly so is U as thechange does not modify any tiles besides the three in the triplet.Next, notice that exchanging the triplet in T for its rotation only modifies the associatedfamily of non-intersecting lattice paths in one path. Moreover, it does not change the startingor ending points of the path, merely the order in which it gets there, that is, either rightthen down or down then right. Thus, λ T = λ U .Last, suppose, without loss of generality, that our chosen triplet is an up lozenge group.Label the three upward pointing triangles in the triplet i, j, k as in Figure 5.2. Thus we see Figure 5.2. An up lozenge group with labelingthat π U ( i ) = π T ( k ), π U ( j ) = π T ( i ), π U ( k ) = π T ( j ), and π U ( m ) = π T ( m ) for m not i, j, or k .Hence π U = τ π T where τ is the three-cycle ( π T ( k ) , π T ( j ) , π T ( i )). (cid:3) It follows that two lozenge tilings that have the same λ permutation have π permutationswith the same sign. Proposition 5.2.
For each H a,b,c,α,β,γ there exists a constant i ∈ { , − } such that for alllozenge tilings T of H a,b,c,α,β,γ the expression sgn L T = i · sgn P T holds, where L T is the familyof non-intersecting lattice paths associated to T and P T is the family of perfect matchingsassociated to T .Proof. Step 1 :Let T and U be two lozenge tilings of H a,b,c,α,β,γ with λ T = λ U . As λ T = λ U , then thefamilies of non-intersecting lattice paths associated to T and U start and end at the same NUMERATIONS DECIDING THE WLP 19 places. Hence T can be modified by a series of, say n , rotations, as in Lemma 5.1, to U .Thus π U = τ n τ n − · · · τ π T , where τ , . . . , τ n ∈ S h s are three cycles by Lemma 5.1. As sgn τ i = 1 for 1 ≤ i ≤ n , and sgnis a group homomorphism, we see that sgn π T = sgn π U . Step 2 :By the comments following Theorem 4.5 we only need to consider the admissible permu-tations λ , . . . , λ C . Moreover, sgn λ k = ( − M ( C − k ) so sgn λ k = ( − M sgn λ k +1 .Let T and U be two lozenge tilings of H = H a,b,c,α,β,γ with λ T = λ k and λ U = λ k +1 . Thatis, sgn λ T = ( − M sgn λ U . First, α ≥ C − k by the existence of T as C − k paths go abovethe puncture and so must go through a gap of size α , and similarly β ≥ k + 1 by the existenceof U .By Step 1 , we may pick T and U however we wish, as long as λ T = λ k and λ U = λ k +1 . Inparticular, let T , and similarly U , be defined as follows (see Figure 5.3): (i) The tiling T (ii) The tiling U Figure 5.3.
An example of tilings T and U of H , , , , , , for k = 1, which are“minimal” below the puncture and “maximal” everywhere else; both tilingshave the regions of similarity highlighted.(i) The C − k paths above the puncture ( C − k − U ) always move right beforemoving down.(ii) The M paths from the puncture always move right before moving down.(iii) The k paths below the puncture ( k + 1 for U ) always move down before movingright.With the idea of up and down triplets from Lemma 5.1, we can say a path is “minimal” ifit contains no up triplets and a path is “maximal” if it contains no down triplets. Thus, T and U are “minimal” below the puncture and “maximal” everywhere else.Given this choice, T and U have exactly the same paths for the top C − k − k paths below the puncture. Hence we can trim off thesepaths to make two new tilings, T ′ and U ′ , of H ′ = H B + M +1 ,A + M +1 ,c,α − ( C − k − ,β − k,γ . Noticethat H and H ′ have the same A, B, M, and γ , only C, α, and β have changed; in particular, C ′ = 1. See Figure 5.4 parts (i) and (ii) for an example of the tilings T ′ and U ′ with theirregion-of-difference highlighted in bold. (i) The tiling T ′ (ii) The tiling U ′ Figure 5.4.
The punctured hexagon H , , , , , ; both tilings have the region-of-difference highlighted.Clearly then T ′ and U ′ differ in four ways: (i) the upper path in T ′ except the small overlapnear the end, (ii) the lower path in U ′ , (iii) the position of the bend in the puncture-paths,and (iv) the part past the bend of the bottom puncture-path in T ′ . The difference between T ′ and U ′ is exactly 2( A + B + M ) + M − A + B + M ) + M − σ , such that π T ′ = σπ U ′ . Wethen have sgn π T ′ = ( − A + B + M )+ M − − sgn π U ′ = ( − M sgn π U ′ . That is, sgn π T = ( − M sgn π U . Since sgn λ T = ( − M sgn λ U , the claim follows. (cid:3) We conclude that Z a,b,c,α,β,γ and N a,b,c,α,β,γ have the same determinant, up to sign. Theorem 5.3.
Consider the punctured hexagon H a,b,c,α,β,γ . Then | det Z a,b,c,α,β,γ | = | det N a,b,c,α,β,γ | . Proof.
Combine Theorems 4.5 and 4.8 via Proposition 5.2. (cid:3)
Moreover, when the puncture is of even length, the determinant and permanent of Z a,b,c,α,β,γ are the same. Corollary 5.4.
Consider the punctured hexagon H a,b,c,α,β,γ . If M is even, then perm Z a,b,c,α,β,γ = | det Z a,b,c,α,β,γ | . Proof.
A simple analysis of the proof of Proposition 5.2 implies that when M is even thensgn π T = sgn π U for all tilings T and U of H a,b,c,α,β,γ . Thus, the enumeration of signed lozengetilings of H a,b,c,α,β,γ is, up to sign, the enumeration of (unsigned) lozenge tilings of H a,b,c,α,β,γ .Thus, the claim follows from Proposition 4.6 and Theorem 4.8. (cid:3) Remark 5.5.
We make a pair of remarks regarding the preceding corollary.(i) The corollary can be viewed as a special case of Kasteleyn’s theorem on enumeratingperfect matchings [17]. To see this, notice that when M is even, then all “faces” ofthe bipartite graph have size congruent to 2 (mod 4).(ii) The corollary extends [6, Theorem 1.2], where punctured hexagons with trivial punc-ture (i.e., M = 0) are considered. We further note that [18, Section 3.4] provides,independently, essentially the same proof as [6], and the proof of Lemma 5.1 buildson this technique. NUMERATIONS DECIDING THE WLP 21
We conclude this section with some observations on the signs introduced here.Let T be a lozenge tiling of H a,b,c,α,β,γ , and let L T and P T be the associated family ofnon-intersecting lattice paths and perfect matching, respectively. By Proposition 5.2, wemay assume that sgn L T = sgn P T . Thus we may assign to T the sign sgn T = sgn L T .Recall that there are C admissible permutations λ , . . . , λ C (see the discussion after The-orem 4.5) associated to H a,b,c,α,β,γ . Further, sgn λ k = ( − M ( C − k ) and so if M is even thensgn λ k = 1 for all k . Hence, we need only consider M odd. In this case, sgn λ k = 1 if andonly if C − k is even. Thus, the sign of T is ( − C − k . (i) The sign of a family ofnon-intersecting lattice paths (ii) The sign of a lozengetiling (iii) The sign of a perfectmatching Figure 5.5.
Example of interpreting the signBy definition of λ k , C − k is the number of lattice paths in the family that go above thepuncture; see Figure 5.5(i). For the lozenge tiling T , C − k is the number of edges of lozengesof T that touch the line formed by extending the edge of the puncture parallel to the side oflength C to the side of length A + M ; see Figure 5.5(ii). Note that this interpretation is inline with the definition of the statistic n ( · ) in [7, Section 2]. Last, for the perfect matching, C − k is the number of non -selected edges that correspond to those on the edge describedfor lozenge tilings; see Figure 5.5(iii).6. Determinants
We continue to use the notation introduced in Proposition 3.3 and Theorem 4.1. Through-out this section we assume that
A, B, C, and M meet conditions (i)-(iv) of Proposition 3.3and a + b + c + α + β + γ ≡ N a,b,c,α,β,γ given in Proposi-tion 3.9 using Theorem 4.5. In particular, we are chiefly interested in whether the determi-nant is zero and if we can compute an upper bound on the prime divisors. In some cases wecan explicitly compute the determinant.6.1. A few properties.
First, a brief remark about the polynomial nature of the determinants.
Remark 6.1.
The argument in [7, Section 6] demonstrates that for fixed
A, B, and C and α, β, and γ satisfying certain restraints, then the determinant of N a,b,c,α,β,γ is polynomial in M , the side-length of the puncture of H a,b,c,α,β,γ , for M of a fixed parity. This argumentcenters around an alternate bijection between the lozenge tilings and non-intersecting latticepaths. We note that the argument is completely independent of the restrictions on α, β, and γ .Thus, their argument can be easily seen to generalise to show that, for fixed A, B, C, α, β, and γ , the determinant of N a,b,c,α,β,γ is polynomial in M , for M of a fixed parity.We demonstrate that every punctured hexagonal region H a,b,c,α,β,γ has at least one tiling. Lemma 6.2.
Every region H a,b,c,α,β,γ has at least one lozenge tiling.Proof. In this case, it is easier to show there exists a family L of non-intersecting latticepaths. In particular, it is sufficient to show that the sum of the maximum number ofpaths that can go above and below the puncture is at least C . By analysis of H a,b,c,α,β,γ ,we see that at most min { C, β, B + C − α } paths can go below the puncture and at mostmin { C, α, A + C − β } paths can go above the puncture. However, as 0 ≤ A, B, C and C ≤ α + β , then min { C, β, B + C − α } + min { C, α, A + C − β } ≥ C . (cid:3) Thus when M is even, the determinant is always positive. Theorem 6.3. If a + b + c is even, then M is even and det N a,b,c,α,β,γ > . Thus I a,b,c,α,β,γ = ( x a , y b , z c , x α y β z γ ) has the weak Lefschetz property in characteristic zero and when the characteristic is suffi-ciently large.Proof. Recall the definition of the admissible partitions λ k , for 0 ≤ k ≤ C (see the discussionfollowing Theorem 4.5). Since M is even, then sgn λ k = 1 for 0 ≤ k ≤ C and hencedet N a,b,c,α,β,γ is the number of tilings of H a,b,c,α,β,γ . Thus, by Lemma 6.2, det N a,b,c,α,β,γ > (cid:3) Mahonian determinants.
MacMahon computed the number of plane partitions (finite two-dimensional arrays thatweakly decrease in all columns and rows) in an A × B × C box as (see, e.g., [27, Page 261])Mac( A, B, C ) := H ( A ) H ( B ) H ( C ) H ( A + B + C ) H ( A + B ) H ( A + C ) H ( B + C ) , where A, B, and C are non-negative integers and H ( n ) := Q n − i =0 i ! is the hyperfactorial of n .David and Tomei proved in [10] that plane partitions in an A × B × C box are in bijectionwith lozenge tilings in an non-punctured hexagon of side-lengths ( A, B, C, A, B, C ). We notethat Propp states on [27, Page 258] that Klarner was likely the first to have observed this.See Figure 6.1 for an illustration of the connection.
Figure 6.1.
An example of a 3 × × NUMERATIONS DECIDING THE WLP 23
We can use MacMahon’s formula to compute the determinant of N a,b,c,α,β,γ in many cases.Also, note that the prime divisors of Mac( A, B, C ) are sharply bounded above by A + B + C −
1. A first case is when the puncture is trivial. This extends [8, Theorem 4.5] where thelevel algebras of this family are considered.
Proposition 6.4. If a + b + c = 2( α + β + γ ) , then M = 0 and det N a,b,c,α,β,γ is Mac(
A, B, C ) . Thus, I a,b,c,α,β,γ has the weak Lefschetz property if the characteristic of K is zero or at least A + B + C = α + β + γ = ( a + b + c ) . Figure 6.2.
When the puncture has side-length zero, the region is a simple hexagon.
Proof.
When M = 0 then there is no puncture in the region H a,b,c,α,β,γ . Hence the region is asimple hexagon with side-lengths ( A, B, C, A, B, C ), exactly the region to which MacMahon’sformula applies. (cid:3)
This result allows us to recover some earlier results about complete intersections.
Corollary 6.5. If a + b + c is even, then the complete intersection J = ( x a , y b , z c ) hasthe weak Lefschetz property if and only if the characteristic of K is not a prime divisor of Mac(
A, B, C ) . That is, the algebra R/J has the weak Lefschetz property if the characteristicof K is zero or at least A + B + C = α + β + γ = ( a + b + c ) .Proof. Set α = ( − a + b + c ) , β = ( a − b + c ) , γ = ( a + b − c ) , and consider I =( x a , y b , z c , x α y β z γ ). Then Proposition 6.4 applies to I and the mixed term, x α y β z γ , has totaldegree s + 2. Thus we have that [ R/I ] i ∼ = [ R/J ] i for i ≤ s + 1. That is, the twin peaks of R/I are isomorphic to the twin peaks of the complete intersection
R/J . Hence
R/J has the weakLefschetz property if and only if
R/I has the weak Lefschetz property, and Proposition 6.4gives the claim. (cid:3)
In particular, the corollary recovers [22, Theorem 3.2(1)] when combined with Proposi-tion 3.9 and [6, Theorem 1.2] when combined with Corollary 5.4. Further, the special casein [22, Theorem 4.2] can be recovered if we set a = β + γ, b = α + γ, and c = α + β .MacMahon’s formula can be used again in another special case, when C = 0. (Notice if A or B is zero, then we can simply relabel the sides to ensure C is zero.) We notice thisextends [8, Theorem 4.3] where the level algebras of this family are considered. Proposition 6.6. If c = ( a + b + α + β + γ ) , then C = 0 and det N a,b,c,α,β,γ is Mac(
M, A − β, B − α ) . Thus, I a,b,c,α,β,γ has the weak Lefschetz property if the characteristic of K is zero or at least A + B + M − α − β = c − α − β . Figure 6.3.
When C is zero, the lightly shaded region has tiles that are fixed,leaving the only variation in the darkly shaded region. Proof.
In this case, it is easier to consider families of non-intersecting lattice paths. Inparticular, since C = 0, then the only starting points, the A i , are those on the puncture.Further, since lattice paths must move only right and down, then we can focus on the isolatedregion between the puncture and the bottom-right edge. If we convert this region back intoa punctured hexagon, then it is just a hexagon without a puncture and with side-lengths( M, A + C − β, B + C − α, M, A + C − β, B + C − α ). (cid:3) Remark 6.7.
Notice that in the preceding proof, we show that the only possible latticepaths come from the puncture to the opposite edge. Converting this back to the languageof lozenge tilings, we see this means that a large region of the figure has fixed tiles leavingonly a small region in which variation can occur. See Figure 6.3 for an illustration of this.Further, given the condition in Proposition 6.6, we see that the pure power of z , z c , hastotal degree c = s + 2. Thus, if we let I = I a,b,c,α,β,γ , then we have that [ R/I ] i ∼ = [ R/J ] i for i ≤ s + 1, where J = ( x a , y b , x α y β z γ ). Thus, the twin peaks of R/I are isomorphic to thetwin peaks of the non-artinian algebra
R/J . Corollary 6.8.
Let J = ( x a , y b , x α y β z γ ) and c = ( a + b + α + β + γ ) , with parameters stillsuitably restricted. Then the map [ R/J ] i × ( x + y + z ) −→ [ R/J ] i +1 is injective for i ≤ c . Further, MacMahon’s formula can be used when C is maximal, that is, C = α + β . Proposition 6.9. If c = ( a + b + γ ) − α − β , then C = α + β and det N a,b,c,α,β,γ is Mac(
A, B, C + M ) . Thus, I a,b,c,α,β,γ has the weak Lefschetz property if the characteristic of K is zero or at least A + B + C + M = s + 2 = c + α + β . NUMERATIONS DECIDING THE WLP 25
Proof.
In this case, it is easier to consider families of non-intersecting lattice paths. Inparticular, since C = α + β , then γ = A + B and so the puncture has a point touchingthe side labeled C ; see Figure 6.4. Thus the lattice paths starting from A , . . . , A β have Figure 6.4.
When C is maximal, the lightly shaded region has tiles whichare fixed, leaving the only variation in the darkly shaded region.the first M moves being down and the lattice paths starting from A β +1 , . . . , A C have thefirst M moves being right. However, we then see that each A i “starts” on the same line,the line running through the lower-right side of the puncture. If we convert the region-of-interest back into a punctured hexagon, then it is a simple hexagon with side-lengths( A, B, C + M, A, B, C + M ). (cid:3) The next case considered, when the mixed term is in two variables, needs a special deter-minant calculation which may be of independent interest.
Lemma 6.10.
Let T be an n -by- n matrix defined as follows ( T ) i,j = (cid:18) pq + j − i (cid:19) if ≤ j ≤ m, (cid:18) pq + r + j − i (cid:19) if m + 1 ≤ j ≤ n, where p, q, r, and m are non-negative integers and ≤ m ≤ n . Then det T = Mac( m, q, r ) Mac( n − m, p − q − r, r ) H ( q + r ) H ( p − q ) H ( n + r ) H ( n + p ) H ( n + p − q ) H ( n + q + r ) H ( p ) H ( r ) . Proof.
In this case, we can use [7, Equation (12.5)] to evaluate det T to be Y ≤ i Lemma 6.10 generalises the result of [22, Lemma 2.2] where the case r = 1 isdiscussed. Further, when r = 0, then det T = Mac( n, p − q, q ), as expected (see the runningexample, det (cid:0) a + ba − i + j (cid:1) , in [19]).The case when the mixed term has only two variables follows immediately. Proposition 6.12. If γ = 0 , then | det N a,b,c,α,β,γ | is Mac( β − A, A, M ) Mac( α − B, B, M ) H ( A + M ) H ( B + M ) H ( C + M ) H ( A + B + C + M ) H ( a ) H ( b ) H ( c ) H ( M ) . Thus, the type ideal I a,b,c,α,β, = ( x a , y b , z c , x α y β ) has the weak Lefschetz property if the characteristic of K is zero or at least A + B + C + M . Figure 6.5. When γ is zero, the starting points A C +1 , . . . , A C + M coincidewith the M consecutive ending points E A − β +1 , . . . , E A − β + M . Proof. As γ = 0, N = N a,b,c,α,β,γ has entries given by( N ) i,j = (cid:0) cA + j − i (cid:1) if 1 ≤ i ≤ C, (cid:26) j = i + β − A − C j = i + β − A − C (cid:27) if C + 1 ≤ i ≤ C + M. . NUMERATIONS DECIDING THE WLP 27 Further, if we define the matrix T by( T ) i,j = ( (cid:0) cA + j − i (cid:1) if 1 ≤ j ≤ β − A, (cid:0) cA + M + j − i (cid:1) if β − A + 1 ≤ j ≤ C , then | det N | = | det T | due to the structure of the lower-part of N . Thus, if we let p = c, q = A, r = M, m = β − A, and n = C , then by Lemma 6.10 we have the desired determinantevaluation.Moreover, α + M and β + M are smaller than A + B + C + M , so the prime divisors ofdet N are strictly bounded above by A + B + C + M . (cid:3) Remark 6.13. Proposition 6.12 deserves a pair of comments:(i) The evaluation of the determinant includes two Mahonian terms and a third non-Mahonian term. It should be noted that both hexagons associated to the Mahonianterms actually show up in the punctured hexagon. See Figure 6.6 where the darkly Figure 6.6. The darkly shaded hexagons correspond to the two Mahonianterms in the determinantal evaluation.shaded hexagons correspond to the Mahonian terms. It is not clear (to us) where thethird term comes from, though it may be of interest to note that if one subtracts M from each hyperfactorial, before the evaluation, then what remains is Mac( A, B, C ).(ii) We notice the proposition also extends [25, Lemma 6.6] where it was shown thatthe associated almost complete intersection always has the weak Lefschetz propertyin characteristic zero (i.e., the determinant is non-zero). That is, all level type2 artinian monomial almost complete intersections in R have the weak Lefschetzproperty in characteristic zero.6.3. Exploring symmetry. When a = b (equivalently, A = B ) and α = β , then H a,a,c,α,α,γ is symmetric; see Figure 6.7.In this case, c is even exactly when M = (2 a + c − α − γ ) is even; similarly, γ is evenexactly when C = (2 a − c + 2 α + γ ) is even. Moreover, α = A + ( C − γ ).When C and M are odd, we can exploit symmetry to show det N a,a,c,α,α,γ is 0. This resultextends the evaluation in [25, Corollary 7.4] and offers a (more) direct combinatorial proof,rather than one based on linear algebra. Proposition 6.14. If c and γ are odd, a = b , and α = β , then H a,a,c,α,α,γ is symmetric withan odd puncture (i.e., M odd; see Figure 6.7) and det N a,a,c,α,α,γ is 0. Thus, I a,a,c,α,α,γ = ( x a , y a , z c , x α y α z γ ) never has the weak Lefschetz property, regardless of characteristic. Figure 6.7. When a = b and α = β , then H a,a,c,α,α,γ is symmetric. Proof. Recall the admissible partitions of H a,a,c,α,α,γ are λ , . . . , λ C . For 0 ≤ i ≤ C − we seethat P + λ i ( A → E ) = P + λ C − i ( A → E ) by symmetry, and further that sgn λ i = − sgn λ C − i , assgn λ k = ( − M ( C − k ) and C is odd. Hence, det N a,b,c,α,β,γ = P Ci =0 sgn λ i P + λ i ( A → E ) = 0 . (cid:3) From the preceding proof we see that if we consider c even instead of c odd (i.e., M eveninstead of M odd), then det N a,a,c,α,α,γ is even, when γ is odd (i.e., C is odd).Recall the definitions of A, B, C, and M from Proposition 3.3, H a,a,c,α,α,γ from Theorem 4.1,and N a,b,c,α,β,γ from Proposition 3.9. If C or M is even, then the region H a,a,c,α,α,γ is sym-metric and we offer the following conjecture for a closed form for det N a,a,c,α,α,γ . Notice thatin this case α = A + ( C − γ ). Conjecture 6.15. Suppose a = b and α = β so H a,a,c,α,α,γ is symmetric. If c or γ is even,then det N a,b,c,α,β,γ is ( − M ⌈ C ⌉ × H ( M + C ) H ( M + γ ) H ( M + A + (cid:4) C (cid:5) ) H ( M + A + (cid:6) C (cid:7) ) H ( M + 2 A + C ) H ( M + 2 A ) H ( M + A + C ) H ( M + C + γ ) × H ( (cid:4) M (cid:5) ) H ( (cid:4) M (cid:5) + A ) H ( (cid:4) M (cid:5) + C + γ ) H ( (cid:4) M (cid:5) + A + C − γ ) H ( (cid:4) M + C (cid:5) ) H ( (cid:4) M + γ (cid:5) ) H ( (cid:4) M + C (cid:5) + A ) H ( (cid:4) M − γ (cid:5) + A ) × H ( (cid:6) M (cid:7) ) H ( (cid:6) M (cid:7) + A ) H ( (cid:6) M (cid:7) + C + γ ) H ( (cid:6) M (cid:7) + A + C − γ ) H ( (cid:6) M + C (cid:7) ) H ( (cid:6) M + γ (cid:7) ) H ( (cid:6) M + C (cid:7) + A ) H ( (cid:6) M − γ (cid:7) + A ) × H ( A − (cid:4) γ (cid:5) ) H ( (cid:4) C (cid:5) ) H ( (cid:4) γ (cid:5) ) H ( A − (cid:6) γ (cid:7) ) H ( (cid:6) C (cid:7) ) H ( (cid:6) γ (cid:7) ) H ( γ ) H ( A + C − γ ) . Further, the ideal I a,a,c,α,α,γ = ( x a , y a , z c , x α y α z γ ) has the weak Lefschetz property when the characteristic of K is zero or at least A + C + M . Remark 6.16. The above symmetry conjecture deserves a few remarks.(i) Note that by Remark 6.1, det N a,b,c,α,β,γ is polynomial in M . Further, the conjecturedform of the determinant would imply that the polynomial factors completely intolinear terms and has degree AC + (cid:4) γ ( C − γ ) (cid:5) .(ii) If Conjecture 6.15 were shown to hold, then it would complete the ( − NUMERATIONS DECIDING THE WLP 29 (iii) As expected, the conjecture corresponds to Proposition 6.6 when C = 0, to Propo-sition 6.9 when A = γ (this implies α = C and so C = 2 α , which is maximal),and to Proposition 6.12 when γ = 0. Moreover, when A = C = γ , then H a,a,c,α,α,γ has an axis-central puncture (see Section 7.1) and the conjecture corresponds toCorollary 7.3.(iv) When C is even and M is odd, then using the f a,b ( c ) from Proposition A.1 and f ea,b ( c ) and f oa,b ( c ) from Corollary A.5, we can rewrite the monic (as a polynomial in M ) part of the conjecture as f oC + γ , C + γ ( M ) · f e | C − γ | , | C − γ | ( M +min( C,γ )) · f e min( C,γ ) , max( C,γ ) ( M +2 A − γ ) · f | A − γ | , | A − γ | ( M + C − γ +2 min( A,γ )) f | C + γ − A | , | C + γ − A | ( M +min(2 A − γ,C + γ )) . (Carefully note that the input parameter in each of the polynomials f above is oddas M is odd.)We give an example of using the symmetry conjecture. Example 6.17. Consider A = B = 8, C = 6, γ = 2, and M even. Then α = β = 10, a = b = 14 + M , and c = 16 + M . Moreover, the region H M, M, M, , , is symmetricand does not fall into the case of Remark 6.16(iii).Supposing Conjecture 6.15 holds, then H M, M, M, , , has a ( − − × ( M + 1)( M + 3) ( M + 4) ( M + 5) ( M + 7) × ( M + 12) ( M + 13) ( M + 14) ( M + 15) ( M + 16) ( M + 17) ( M + 18) ( M + 19)( M + 20) . Thus, I M, M, M, , , = ( x M , y M , z M , x y z ) has the weak Lefschetz prop-erty when the characteristic of the ground field is 0 or at least M + 21.So far, in every case where we can bound the prime divisors of det N a,b,c,α,β,γ from above,we can do so linearly in the parameters (actually, always by at most s + 2). This may,however, not always be the case. We provide the following example to demonstrate that thisis true, but also as a contrast to the symmetry conjecture, where some restrictions lead to a(conjectured) closed form. Example 6.18. Consider the level and type 3 algebra given by R/I , where I t, t, t, , , = ( x t , y t , z t , xy z )and t ≥ 4. By Remark 6.1, we have that det N = det N t, t, t, , , is a polynomial in t . Hence we can use interpolation to determine the polynomial in terms of t ; in particular,det N t, t, t, , , is4 H (7) · (cid:26) ( t − t − t − t ( t + 1) ( t + 2)( t + 4)( t + 6)( t + 6 t − 1) if t is odd;( t − ( t − t ( t + 1) ( t + 2)( t + 5)( t + 7)( t + 2 t − 9) if t is even.In 1857, Bouniakowsky conjectured that for every irreducible polynomial f ∈ Z [ t ] of degreeat least 2 with common divisor d = gcd { f ( i ) | i ∈ Z } , there exists infinitely many integers t such that d f ( t ) is prime. We note that the weaker Fifth Hardy-Littlewood conjecture,which states that t + 1 is prime for infinitely many positive integers t , is a special case ofthe Bouniakowsky conjecture.When t is odd, the determinant has the quadratic factor t + 6 t − 1. If we let t = 2 k + 1,then this factor becomes 2(2 k + 8 k + 3), which is an irreducible polynomial over Z [ k ] with common divisor 2 (when k = 4 then the polynomial evaluates to 134 = 2 · t , assuming theBouniakowsky conjecture. Similarly the quadratic factor of the determinant for t even isprime for infinitely many even integers t , again assuming the Bouniakowsky conjecture.Hence, assuming the Bouniakowsky conjecture, for large enough t , the upper bound onthe prime divisors of the determinant grows quadratically in t .The above example falls in to the case of Proposition 7.7(ii)(a) or the second open caseimmediately following the proposition, depending on the parity of t .7. Centralising the puncture In this section we consider two subtlety different ways to centralise the puncture of apunctured hexagon. The first, axis-central , forces the puncture to be centered along eachaxis, individually. The second, gravity-central , forces the puncture to be the same distance,simultaneously, from the three sides of the hexagon that are parallel to the puncture-sides.Throughout this section we assume, in addition to the conditions in Proposition 3.3 and a + b + c + α + β + γ ≡ I a,b,c,α,β,γ has type 3, that is, α , β , and γ are non-zero.7.1. Axis-central. We define a punctured hexagon H a,b,c,α,β,γ to have an axis-central puncture if the punctureis “central” as defined in [7, Section 1]. Specifically, for each side of the puncture, thepuncture-side should be the same distance from the parallel hexagon-side as the puncture-vertex opposite the puncture-side is from the other parallel hexagon-side; see Figure 7.1(i).However, when c has a different parity than both a and b , then an adjustment has to bemade; in particular, translate the puncture parallel to the hexagon-side of length C one-halfunit toward the side of length A ; see Figure 7.1(b). (i) The parity of c agrees with a and b . (ii) The parity of c differs from a and b . Figure 7.1. A punctured hexagon with an axis-central puncture.When H a,b,c,α,β,γ has an axis-central puncture, then the ideal has a nice form. Supposefirst that a , b , and c have the same parity. Then α = a − M − α so a = 2 α + M ; similarly, b = 2 β + M and c = 2 γ + M . Thus, if we set t = M , then I α + t, β + t, γ + t,α,β,γ = ( x α + t , y β + t , z γ + t , x α y β z γ ) . The conditions in Proposition 3.3 simplify to α ≤ β + γ , β ≤ α + γ , γ ≤ α + β , and t ≥ NUMERATIONS DECIDING THE WLP 31 Now, suppose the parity of c differs from that of both a and b . Then α = a − M − α + 1, β = b − M − β − 1, and γ = c − M − γ , so a = 2 α + M − b = 2 β + M + 1, and c = 2 γ + M .Thus, if we set t = M , then I α + t − , β + t +1 , γ + t,α,β,γ = ( x α + t − , y β + t +1 , z γ + t , x α y β z γ ) . The conditions in Proposition 3.3 simplify to α ≤ β + γ + 1, β ≤ α + γ − γ ≤ α + β , and t ≥ N a,b,c,α,β,γ have been calculated for puncturedhexagons with axis-central punctures. We recall the four theorems here, although we forgothe exact statements of the determinant evaluations; the explicit evaluations can be foundin [7]. Theorem 7.1. [7, Theorems 1, 2, 4, & 5] Let A, B, C, and M be non-negative integers andlet H be the associated hexagon with an axis-central puncture. Then (1) The number of lozenge tilings of H is CEKZ ( A, B, C, M ) if A, B, and C share acommon parity. (2) The number of lozenge tilings of H is CEKZ ( A, B, C, M ) if A, B, and C do notshare a common parity. (4) The number of signed lozenge tilings of H is (i) CEKZ ( A, B, C, M ) if A, B, and C are all even, and (ii) 0 if A, B, and C are all odd. (5) The number of signed lozenge tilings of H is CEKZ ( A, B, C, M ) if A, B, and C donot share a common parity.Moreover, the four functions CEKZ i are polynomials in M which factor completely intolinear terms. Further, each can be expressed as a quotient of products of hyperfactorials and,in each case, the largest hyperfactorial term is H ( A + B + C + M ) . Thus, we calculate the permanent of Z a,b,c,α,β,γ . Corollary 7.2. Let H a,b,c,α,β,γ be a hexagon with an axis-central puncture. Then perm Z a,b,c,α,β,γ = (cid:26) CEKZ ( A, B, C, M ) if a, b, and c share a common parity; CEKZ ( A, B, C, M ) otherwise.Proof. This follows from Proposition 4.6 and Theorem 7.1. (cid:3) Moreover, we calculate the determinant of N a,b,c,α,β,γ , and thus can completely classifywhen the algebra R/I a,b,c,α,β,γ has the weak Lefschetz property. Corollary 7.3. Let H a,b,c,α,β,γ be a hexagon with an axis-central puncture. If M is even,then det N a,b,c,α,β,γ = (cid:26) CEKZ ( A, B, C, M ) if a, b, and c share a common parity; CEKZ ( A, B, C, M ) otherwise.If M is odd, then det N a,b,c,α,β,γ = CEKZ ( A, B, C, M ) if a, b, c, and s + 2 share a common parity;0 if a, b, and c share a common paritydifferent from the parity of s + 2 ; CEKZ ( A, B, C, M ) if a, b, and c do not share a common parity. Thus, R/I a,b,c,α,β,γ always fails to have the weak Lefschetz property if a, b, c, and M areodd, regardless of the field characteristic. Otherwise, R/I a,b,c,α,β,γ has the weak Lefschetzproperty if the field characteristic is zero or at least A + B + C + M .Proof. This follows from Theorem 4.5 and Theorem 7.1. (cid:3) As we will see in the following subsection, having a gravity-central puncture is equivalentto the associated algebra being level. Question 7.4. Consider the punctured hexagon H a,b,c,α,β,γ . Is there an algebraic property P of algebras such that H a,b,c,α,β,γ has an axis-central puncture if and only if R/I a,b,c,α,β,γ hasproperty P ?7.2. Gravity-central. We define a punctured hexagon H a,b,c,α,β,γ to have a gravity-central puncture if the verticesof the puncture are each the same distance from the perpendicular side of the hexagon; seeFigure 7.2. That is, we have that B + C − α = A + C − β = A + B − γ , which simplifies to Figure 7.2. A punctured hexagon with a gravity-central puncture.the relation a − α = b − β = c − γ , and this is exactly the condition in Proposition 3.1(ii)for R/I a,b,c,α,β,γ to be level and type 3. Thus, if we let t be this common difference, then wecan rewrite I a,b,c,α,β,γ as I α + t,β + t,γ + t,α,β,γ = ( x α + t , y β + t , z γ + t , x α y β z γ ) . Without loss of generality, assume 0 < α ≤ β ≤ γ . Then the conditions in Proposition 3.3simplify to t ≥ ( α + β + γ ) and γ ≤ α + β ).The ideals I α + t,β + t,γ + t,α,β,γ are studied extensively in [25, Sections 6 & 7]. In particular,[25, Conjecture 6.8] makes a guess as to when R/I α + t,β + t,γ + t,α,β,γ has the weak Lefschetzproperty in characteristic zero. We recall the conjecture here, though we present it in adifferent but equivalent form. Conjecture 7.5. Consider the ideal I α + t,β + t,γ + t,α,β,γ in R where K has characteristic zero, < α ≤ β ≤ γ ≤ α + β ) , t ≥ ( α + β + γ ) , and α + β + γ is divisible by three.If ( α, β, γ, t ) is not (2 , , , or (3 , , , , then R/I α + t,β + t,γ + t,α,β,γ fails to have the weakLefschetz property if and only if t is even, α + β + γ is odd, and α = β or β = γ . Remark 7.6. [25, Conjecture 6.8] is presented in a format that does not elucidate thereasoning behind it. We present the conjecture differently so it says that the weak Lefschetzproperty fails in two exceptional cases and also when a pair of parity conditions and asymmetry condition hold. NUMERATIONS DECIDING THE WLP 33 We add further support to the conjecture. Proposition 7.7. Let I = I α + t,β + t,γ + t,α,β,γ be as in Conjecture 7.5. Then (i) R/I fails to have the weak Lefschetz property when t is even, α + β + γ is odd, and α = β or β = γ ; (ii) R/I has the weak Lefschetz property when (a) t and α + β + γ have the same parity, or (b) t is odd and α = β = γ is even.Proof. Part (i) follows from Proposition 6.14 (also by [25, Corollary 7.4]). Part (ii)(a) implies M is even and so follows by Theorem 6.3. Part (ii)(b) follows from [7, Theorem 4], which isrecalled here in Theorem 7.1(4)(i). (cid:3) We note that Conjecture 7.5 remains open in two cases, both of which are conjectured tohave the weak Lefschetz property:(i) t even, α + β + γ odd, and α < β < γ ;(ii) t odd, α + β + γ even, and α < β or β < γ . Remark 7.8. Notice that the second open case in the above statement is solved for thecases when α = β or β = γ if Conjecture 6.15 is true.7.3. Axis- and gravity-central. Suppose a, b, and c have the same parity. Then the punctured hexagons that are bothaxis- and gravity-central are precisely those with a = b = c = α + t and α = β = γ . In thiscase, we strengthen [25, Corollary 7.6]. Corollary 7.9. Consider the level, type algebra A given by R/I α + t,α + t,α + t,α,α,α = R/ ( x α + t , y α + t , z α + t , x α y α z α ) , where t ≥ α . Then A fails to have the weak Lefschetz property in characteristic zero if andonly if α is odd and t is even. In [20], Krattenthaler described a bijection between cyclically symmetric lozenge tilings ofthe punctured hexagon considered in the previous corollary and descending plane partitionswith specified conditions.If c has a different parity than a and b , then α − β + 1 = γ . Thus for α ≥ M non-negative we have that the ideals of the form I α + M, α + M − , α + M − ,α,α,α = ( x α + M , y α + M − , z α + M − , x α y α − z α − ) , are precisely those that are both axis- and gravity-central.8. Interesting families and examples In this section, we give several interesting families and examples.8.1. Large prime divisors. Throughout the two preceding sections, when we could bound the prime divisors of det N above, we bounded them above by (at most) s + 2. However, this need not always be thecase, as demonstrated in Example 6.18. We provide here a few exceptional-looking thoughsurprisingly common cases. Example 8.1. Recall that s + 2 = ( a + b + c + α + β + γ ). In each case, we specify theparameter set by a sextuple ( a, b, c, α, β, γ ).(i) Consider the parameter set (4 , , , , , s + 2 = 7 and det N = 11. This isthe smallest s + 2 so that det N has a prime divisor greater than s + 2.(ii) For the parameter set (20 , , , , , s + 2 = 28 anddet N = 2 · · · · · · · · . Hence det N is divisible by a prime that is over 700000 times large than s + 2.Moreover, 20554657 is greater than the multiplicity of the associated algebra.(iii) Consider the parameter set (7 , , , , , s +2 = 14 and det N = 13 · · s + 2 so that det N has more than one prime divisor greaterthan s + 2.(iv) Last, for the parameter set (8 , , , , , s + 2 = 17 and det N = 2 · · · · N has two prime divisors both greaterthan a + b + c + α + β + γ , the sum of the generating degrees of R/I a,b,c,α,β,γ .Given the previous example and Example 6.18, it seems unlikely that there is a reasonablysimple closed formula for the determinant of N a,b,c,α,β,γ in general, as opposed to the case ofa symmetric region (see Conjecture 6.15).8.2. Fixed determinants. For any positive integer n , there is an infinite family of punctured hexagons with exactly n tilings. Note the algebras are type 2 if β is zero or c = n + β + 1 and type 3 otherwise. Proposition 8.2. Let n be a positive integer. If β ≥ and c ≥ n + β + 1 , then det N c − β − ,β +2 ,c,c − n − β − ,β,n = n. Hence the ideal I c − β − ,β +2 ,c,c − n − β − ,β,n = ( x c − β − , y β +2 , z c , x c − n − β − y β z n ) has the weak Lefschetz property when the characteristic of K is either zero or not a primedivisor of n .Proof. In this case, s = c − A = β + 1, B = c − β − C = 0, and M = 1.Using Proposition 6.6 we have thatdet N c − β − ,β +2 ,c,c − n − β − ,β,n = Mac( M, A − β, B − α ) = Mac(1 , , n − 1) = n. Alternatively, from Proposition 3.9 we have that N c − β − ,β +2 ,c,c − n − β − ,β,n = (cid:18)(cid:18) γA + C − β (cid:19)(cid:19) = (cid:18)(cid:18) n (cid:19)(cid:19) = ( n ) . Clearly then the determinant is n . (cid:3) Thus for any prime p , Proposition 8.2 provides infinitely many monomial almost completeintersections that fail to have the weak Lefschetz property exactly when the field character-istic is p .A result of Proposition 8.2 is an infinite (in fact, two dimensional) family whose membershave a unique tiling. Note that the algebras are type 2 if β is zero or c = β + 2 and type 3otherwise. NUMERATIONS DECIDING THE WLP 35 Corollary 8.3. If β ≥ and c ≥ β + 2 , then det N c − β − ,β +2 ,c,c − β − ,β, is 1. That is, I a,b,c,α,β,γ = ( x c − β − , y β +2 , z c , x c − β − y β z ) has the weak Lefschetz property independent of the field characteristic. Another family whose members have a unique tiling comes from Proposition 6.12. Notethat it is a three dimensional family but also that all of the associated algebras are type 2. Proposition 8.4. If a = b = α + β + c and γ = 0 , then A = B = 0 (see Figure 8.1) and det N a,b,c,α,β,γ is 1. That is, I a,b,c,α,β,γ = ( x α + β + c , y α + β + c , z c , x α y β ) has the weak Lefschetz property independent of the field characteristic. Figure 8.1. When A = B = γ = 0, then H a,b,c,α,β,γ has a unique tiling. Proof. This follows from Proposition 6.12. (cid:3) Several questions were asked in [25], two of which we can answer in the affirmative. Remark 8.5. Question 8.2(2c) asked if there exist non-level almost complete intersectionswhich never have the weak Lefschetz property. The almost complete intersection R/I , , , , , = R/ ( x , y , z , x y z )is non-level and never has the weak Lefschetz property, regardless of field characteristic, asdet N , , , , , = 0 by Proposition 6.14.Further, we notice here that Question 8.2(2b) in [25] is answered in the affirmative bythe comments following Question 7.12 in [25]. In particular, I , , , , , is a level almostcomplete intersection which has odd socle degree (39) and never has the weak Lefschetzproperty, as det N , , , , , = 0.8.3. Minimal multiplicity. The Huneke-Srinivasan Multiplicity Conjecture, which was proven by Eisenbud and Schr-eyer [11, Corollary 0.3], shows that the multiplicity of a Cohen-Macaulay module gives nicebounds on the possible shifts of the Betti numbers. Moreover, as the algebras A can beviewed as finite dimensional vector spaces, then the multiplicity is the dimension of A asa vector space. Thus, algebras that have minimal multiplicity while retaining a particularproperty are the smallest, in the above sense, examples one can generate. Example 8.6. Possibly of interest are a few cases of minimal multiplicity with regard tothe weak Lefschetz property.The following examples never have the weak Lefschetz property, that is, the determinantof their associated matrix N a,b,c,α,β,γ is 0. Note that both examples are type 3. (i) The unique level ideal with minimal multiplicity is I , , , , , = ( x , y , z , xyz ) . Its Hilbert function is (1 , , , , 3) and so it has multiplicity 19. It is worth notingthat this ideal is extensively studied in [4, Example 3.1] and is the basis for anexploration of the subtlety of the Lefschetz properties in [9].(ii) The unique non-level ideal with minimal multiplicity is I , , , , , = ( x , y , z , x y z ) . Its Hilbert function is (1 , , , , , , , , 1) and so it has multiplicity 57. Further,this ideal is the example given in Remark 8.5.Moreover, the following examples always have the weak Lefschetz property, regardlessof the base field characteristic. That is to say, the determinant of their associated matrix N a,b,c,α,β,γ is 1.(i) The two level ideals with minimal multiplicity are I , , , , , = ( x, y , z , yz ) and I , , , , , = ( x, y , z , yz ) . Both ideals have Hilbert function (1 , , 2) and thus multiplicity 5. However, bothideals are isomorphic to ideals in K [ y, z ].(ii) The unique level, type 2 ideal without x as a generator and with minimal multiplicityis I , , , , , = ( x , y , z , xy ) . Its Hilbert function is (1 , , , 2) and so it has multiplicity 9.(iii) The unique level, type 3 ideal with minimal multiplicity is I , , , , , = ( x , y , z , xyz ) . Its Hilbert function is (1 , , , , , , , 3) and so it has multiplicity 46.(iv) The unique non-level, type 2 ideal with minimal multiplicity is I , , , , , = ( x , y , z , yz ) . Its Hilbert function is (1 , , , 1) and so it has multiplicity 8.(v) The unique non-level, type 3 ideal with minimal multiplicity is I , , , , , = ( x , y , z , xyz ) . Its Hilbert function is (1 , , , , 2) and so it has multiplicity 14.Notice that example (ii) and (iv) in the above enumeration differ only slightly in themixed term yet one is level and the other is not. It should also be noted that all of theabove examples were found via an exhaustive search in the finite space of possible idealsusing Macaulay2 [15]. 9. Splitting type and regularity Throughout this section we assume K is an algebraically closed field of characteristic zero.Recall the definition of the ideals given in Section 3; consider I = I a,b,c,α,β,γ = ( x a , y b , z c , x α y β z γ ) , NUMERATIONS DECIDING THE WLP 37 where 0 ≤ α < a, ≤ β < b, ≤ γ < c, and at most one of α, β, and γ is zero. In this sectionwe consider the splitting type of the syzygy bundles of the artinian algebras R/I , regardlessof any extra conditions on the parameters.Recall, also from Section 3, that the syzygy module syz I of I is defined by the exactsequence0 −→ syz I −→ R ( − α − β − γ ) ⊕ R ( − a ) ⊕ R ( − b ) ⊕ R ( − c ) −→ I a,b,c,α,β,γ −→ ] syz I on P of I is the sheafification of syz I . Its restriction to theline H ∼ = P defined by ℓ = x + y + z splits as O H ( − p ) ⊕ O H ( − q ) ⊕ O H ( − r ). The argumentsin [25, Proposition 2.2] (recalled here in Proposition 2.4) imply that ( p, q, r ) is the splittingtype of the restriction of ] syz I to a general line. Thus, we call ( p, q, r ) the generic splittingtype of syz I .In order to compute the generic splitting type of syz I , we use the observation that R/ ( I, ℓ ) ∼ = S/J , where S = K [ x, y ], and J = ( x a , y b , ( x + y ) c , x α y β ( x + y ) γ ). Define syz J bythe exact sequence(9.1) 0 −→ syz J −→ S ( − α − β − γ ) ⊕ S ( − a ) ⊕ S ( − b ) ⊕ S ( − c ) −→ J −→ { x a , y b , ( x + y ) c , x α y β ( x + y ) γ } of J . Thensyz J ∼ = S ( − p ) ⊕ S ( − q ) ⊕ S ( − r ). The Castelnuovo-Mumford regularity of a homogeneousideal I is denoted by reg I . Remark 9.1. For later use, we record the following facts on the generic splitting type ( p, q, r )of syz I a,b,c,α,β,γ .(i) As the sequence in (9.1) is exact, we see that p + q + r = a + b + c + α + β + γ .(ii) Further, if any of the generators of J are extraneous, then the degree of that gener-ator is one of p, q, and r .(iii) As regularity can be read from the Betti numbers of R/J , we get that reg J + 1 =max { p, q, r } .Before moving on, we prove a useful lemma. Lemma 9.2. Let S = K [ x, y ] , where K is a field of characteristic zero, and let a, b, α, β, and γ be non-negative integers with α + β + γ < a + b . Without loss of generality, assumethat < a − α ≤ b − β . Then reg ( x a , y b , x α y β ( x + y ) γ ) is a + β + γ − if α = 0 and < γ ≤ b − β − a ; α + b − if < α, γ ≤ b − β + α − a, and < β or < γ ; and (cid:6) ( a + b + α + β + γ ) (cid:7) − if γ > b − β + α − a. Further still, we always have reg ( x a , y b , x α y β ( x + y ) γ ) ≤ (cid:6) ( a + b + α + β + γ ) (cid:7) − .Proof. We proceed in three steps.First, consider γ = 0, 0 < α , and 0 < β . Then by the form of the minimal free resolutionof the quotient algebra S/ ( x a , y b , x α y β ) we have that reg ( x a , y b , x α y β ) = α + b − γ > α = β = 0. By [16, Proposition 4.4], the algebra S/ ( x a , y b )has the strong Lefschetz property in characteristic zero. Thus the Hilbert function of S/ ( x a , y b , ( x + y ) γ ) isdim K [ S/ ( x a , y b , ( x + y ) γ )] j = max { , dim K [ S/ ( x a , y b )] j − dim K [ S/ ( x a , y b )] j − γ } . By analysing when the difference becomes non-positive, we get that the regularity is a + γ − γ ≤ b − a and (cid:6) ( a + b + γ ) (cid:7) − γ > b − a .Third, consider γ > < α or 0 < β . Notice that( x a , y b , x α y β ( x + y ) γ ) : x α y β = ( x a − α , y b − β , ( x + y ) γ ) . Considering the short exact sequence0 → [ S/ ( x a − α , y b − β , ( x + y ) γ )]( − α − β ) × x α y β −→ S/ ( x a , y b , x α y β ( x + y ) γ ) → S/ ( x a , y b , x α y β ) → , where the first map is multiplication by x α y β , we obtainreg ( x a , y b , x α y β ( x + y ) γ ) = max { α + β + reg ( x a − α , y b − β , ( x + y ) γ ) , reg ( x a , y b , x α y β ) } . The claims then follows by simple case analysis. (cid:3) Recall that the semistability of syz I a,b,c,α,β,γ is completely determined by the parameters a, b, c, α, β, γ in Proposition 3.3.9.1. Non-semistable syzygy bundle. We first consider the case when the syzygy bundle is not semistable. We distinguish threecases. It turns out that in two cases, at least one of the generators of J is extraneous. Proposition 9.3. Let K be a field of characteristic zero and suppose I = I a,b,c,α,β,γ is anideal of R . Let J = ( x a , y b , ( x + y ) c , x α y β ( x + y ) γ ) be an ideal of S . We assume, without lossof generality, that a ≤ b ≤ c so that C ≤ B ≤ A . (i) If M < , then the generator x α y β ( x + y ) γ of J is extraneous. The generic splittingtype of syz I is ( a + c, b, α + β + γ ) if c ≤ b − a and ( (cid:4) ( a + b + c ) (cid:5) , (cid:6) ( a + b + c ) (cid:7) , α + β + γ ) if c > b − a . (ii) If M ≥ and C < , then the generator ( x + y ) c of J is extraneous. Thegeneric splitting type of syz I is ( a + b + α + β + γ − r − , r + 1 , c ) , where r =reg ( x a , y b , x α y β ( x + y ) γ ) (which is given in Lemma 9.2). (iii) If M ≥ , C ≥ , and A > β + γ , then the only destabilising sub-bundle of syz I is syz ( x a , x α y β z γ ) and so the generic splitting type of syz I is ( (cid:4) ( α + b + c ) (cid:5) , (cid:6) ( α + b + c ) (cid:7) , a + β + γ ) .Proof. Assume M < 0, then ( a + b + c ) < α + β + γ and when c ≥ a + b then a + b − ≤ 12 ( a + b + c ) − < α + β + γ. By Lemma 9.2 the regularity of ( x a , y b , ( x + y ) c ) is a + b − c ≥ a + b and ⌈ ( a + b + c ) ⌉− x α y β ( x + y ) γ is contained in ( x a , y b , ( x + y ) c ) and the firstclaim follows.Assume M ≥ C < 0, then 2( α + β + γ ) ≤ a + b + c , c ≥ ( a + b + α + β + γ ), and when α + β + γ ≥ a + b then 2( α + β + γ ) ≤ a + b + c implies c ≥ a + b . By Lemma 9.2, the regularityof ( x a , y b , x α y β ( x + y ) γ ) is a + b − α + β + γ ≥ a + b and at most ⌈ ( a + b + α + β + γ ) ⌉ − x + y ) c is contained in ( x a , y b , x α y β ( x + y ) γ ) and the secondclaim follows.Last, assume M ≥ , C ≥ 0, and A > β + γ . Note that since A + B + C = α + β + γ wethen have that B + C < α and, in particular, B < α + γ and C < α + β . Using Brenner’scombinatorial criterion for the semi-stability of syzygy bundles of monomial ideals (see [3, NUMERATIONS DECIDING THE WLP 39 Corollary 6.4]), we see that that S = syz ( x a , x α y β z γ ) ∼ = R ( − r ), where r = a + β + γ , is theonly destabilising sub-bundle of syz I . Further, (syz I ) / S is a semistable rank two vectorbundle, so by Grauert-M¨ulich theorem, the quotient has generic splitting type ( p, q ) where0 ≤ q − p ≤ 1. Thus, if we consider the short exact sequence0 −→ S −→ syz I −→ (syz I ) / S −→ , then the third claim follows after restricting to ℓ . (cid:3) In the third case, when A > β + γ , the associated ideal J ⊂ S may be minimally generatedby four polynomials, unlike in the other two cases. Example 9.4. Consider the ideals I , , , , , = ( x , y , z , x yz ) and J = ( x , y , ( x + y ) , x y ( x + y ))in R and S , respectively. Notice that in this case, 0 ≤ C ≤ B ≤ A , 0 ≤ M , and A > β + γ sothe syzygy bundle of R/I , , , , , is non-semistable and its generic splitting type is determinedin Proposition 9.3(iii). Further, J is minimally generated by the four polynomials x , y ,xy (2 x + y ), and x y . Semistable syzygy bundle. Order the entries of the generic splitting type ( p, q, r ) of the semistable syzygy bundle ] syz I such that p ≤ q ≤ r . Then by Grauert-M¨ulich theorem we have that r − q and q − p are both non-negative and at most 1. Moreover, [4, Theorem 2.2] specialises in our case. Theorem 9.5. Let I = I a,b,c,α,β,γ . If R/I has the weak Lefschetz property, then p = q or q = r and r − p ≤ ; otherwise q = p + 1 and r = p + 2 . When a + b + c + α + β + γ I andregularity of J can be computed easily. Proposition 9.6. Let R = K [ x, y, z ] where K is a field of characteristic zero. Suppose I = I a,b,c,α,β,γ is an ideal of R with a semistable syzygy bundle and let J = ( x a , y b , ( x + y ) c , x α y β ( x + y ) γ ) be an ideal of S . Let k = (cid:4) ( a + b + c + α + β + γ ) (cid:5) . Then reg J = k andthe generic splitting type of syz I is (cid:26) ( k, k, k + 1) if a + b + c + α + β + γ = 3 k + 1 , and ( k, k + 1 , k + 1) if a + b + c + α + β + γ = 3 k + 2 . Proof. Let ( p, q, r ) be the generic splitting type of syz I , so a + b + c + α + β + γ = 3( s + 2) = p + q + r . By Proposition 3.2, R/I has the weak Lefschetz property so p = q , q = r ,and r − p ≤ 1. Clearly if p = q = r then p + q + r = 3 p is 0 modulo 3 so cannot be a + b + c + α + β + γ .If p = q < r , then r = p + 1 and p + q + r = 3 p + 1. This matches the case when a + b + c + α + β + γ = 3 k + 1, so p = k and the splitting type of syz I is ( k, k, k + 1).Similarly, if p < q = r , then q = r = p + 1 and p + q + r = 3 p + 2. This matches the casewhen a + b + c + α + β + γ = 3 k + 2, so p = k and the splitting type of syz I is ( k, k + 1 , k + 1).In both cases, we have that k − ≤ reg J ≤ k by Remark 9.1(iii). However, we see thatdim K [ R/I ] k − < dim K [ R/I ] k − so dim K [ R/ ( I, x + y + z )] k − = dim K [ S/J ] k − > J > k − 1. Hence reg J = k . (cid:3) The generic splitting type of I a,b,c,α,β,γ , when the ideal is associated to a punctured hexagon,depends on thew ideal having the weak Lefschetz property. Proposition 9.7. Let R = K [ x, y, z ] where K is a field of characteristic zero. Suppose I = I a,b,c,α,β,γ is an ideal of R with a semistable syzygy bundle (see Proposition 3.3) and a + b + c + α + β + γ ≡ . Let J = ( x a , y b , ( x + y ) c , x α y β ( x + y ) γ ) be an ideal of S and let s + 2 = ( a + b + c + α + β + γ ) . Then (i) If R/I has the weak Lefschetz property, then the generic splitting type of syz I is ( s + 2 , s + 2 , s + 2) and reg J = s + 1 . (ii) If R/I does not have the weak Lefschetz property, then the generic splitting type of syz I is ( s + 1 , s + 2 , s + 3) and reg J = s + 2 .Proof. Let ( p, q, r ) be the generic splitting type of syz I , so a + b + c + α + β + γ = 3( s + 2) = p + q + r .Assume that R/I has the weak Lefschetz property. Suppose p = q , then q = r = p + 1and p + q + r = 3 p + 2, similarly, if q = r , then p = q and r = p + 1 so p + q + r = 3 p + 1;neither case is 0 modulo 3, hence cannot be 3( s + 2). Thus p = q = r = s + 2. Further wethen see that reg J = s + 1 by Remark 9.1(iii).Now assume R/I fails to have the weak Lefschetz property. Then p + q + r = 3 p + 3 =3( s + 2) so p + 1 = s + 2 and p = s + 1. Thus, the generic splitting type of syz I must be( s + 1 , s + 2 , s + 3). As R/I has twin-peaks at s + 1 and s + 2 by Corollary 3.7, we see thatreg J ≤ s + 1 if and only if R/I has the weak Lefschetz property; so reg J ≥ s + 2. However,by Remark 9.1(iii) we have that reg J + 1 ≤ s + 3 so reg J ≤ s + 2, hence reg J = s + 2. (cid:3) This proposition can be combined with the results in the previous sections to compute thegeneric splitting type of many of syzygy bundles of the artinian algebras R/I a,b,c,α,β,γ . Example 9.8. Consider the ideal I , , , , , = ( x , y , z , x y z ) which never has the weakLefschetz property, by Proposition 6.14. The generic splitting type of syz I , , , , , is (9 , , I , , , , , = ( x , y , z , x y z ) has the weak Lefschetz propertyin characteristic zero as det N , , , , , = − I , , , , , is (10 , , I = I a,b,c,α,β,γ is not associated to a punctured hexagon, then we have seen in Proposi-tion 3.2 and Corollary 3.4 that R/I has the weak Lefschetz property in characteristic zero.We summarise part of our results by pointing out that in the case when I is associated to apunctured hexagon then deciding the presence of the weak Lefschetz property is equivalentto determining other invariants of the algebra. Theorem 9.9. Let R = K [ x, y, z ] where K is a field of arbitrary characteristic. Let I = I a,b,c,α,β,γ be associated to a punctured hexagon; in particular, a + b + c + α + β + γ ≡ and syz I is semistable (see Proposition 3.3). Set s = ( a + b + c + α + β + γ ) − .Then the following conditions are equivalent: (i) The algebra R/I has the weak Lefschetz property; (ii) the regularity of S/J is s ; (iii) the determinant of N a,b,c,α,β,γ (i.e., the enumeration of signed lozenge tilings of thepunctured hexagon H a,b,c,α,β,γ ) modulo the characteristic of K is non-zero; and (iv) the determinant of Z a,b,c,α,β,γ (i.e., the enumeration of signed perfect matchings of thebipartite graph associated to H a,b,c,α,β,γ ) modulo the characteristic of K is non-zero. NUMERATIONS DECIDING THE WLP 41 Moreover, if the characteristic of K is zero, then there is one further equivalent condition: (v) The generic splitting type of syz I is ( s + 2 , s + 2 , s + 2) .Proof. Combine Corollary 3.7, Propositions 3.8 and 3.9, Theorems 4.5 and 4.8, and Propo-sition 9.7. (cid:3) This relates the weak Lefschetz property to a number of other problems in algebra, com-binatorics, and algebraic geometry.9.3. Jumping lines. Recall that a jumping line is a line, L = 0, over which the syzygybundle splits differently than in the generic case, x + y + z = 0. Since I = I a,b,c,α,β,γ is amonomial ideal it is sufficient to consider the two cases z = 0 and y + z = 0. Proposition 9.10. Let R = K [ x, y, z ] where K is a field of characteristic zero and let I = I a,b,c,α,β,γ be an ideal of R . The splitting type of syz I on the line z = 0 is ( c, α + b, a + β ) if γ = 0 and ( c, α + β + γ, a + b ) if γ > . And the splitting type of syz I on the line y + z = 0 is ( c, a + β + γ, α + b ) if β + γ < b ≤ c and ( c, α + β + γ, a + b ) if b ≤ min { c, β + γ } .Proof. All four cases follow immediately by analysing the monomial algebra S/J isomor-phic to R/ ( I, L ), where L = 0 is the splitting line, and using Lemma 9.2 to compute theregularities. (cid:3) Appendix A. Hyperfactorial calculus Throughout this manuscript, the hyperfactorial function H on non-negative integers n ,defined by H ( n ) := n − Y i =0 i ! , has been a key ingredient in many of the formulae. In this appendix we will highlight theuses and structure of the hyperfactorial and describe a useful “picture-calculus” approach toworking with hyperfactorials.Notice that, for n ≥ H ( n ) can also be seen as Q n − k =1 k n − k . Thus if we place the numbers1 to n − n − n is the product of all the numbers in the grid. Figure A.1. H (6) = 34560 represented as a triangular grid, both specificallyand as a more generic shapeUsing this pictorial representation of the hyperfactorial, various identities involving hy-perfactorials become more transparent. The first identity is simple, but very useful. Proposition A.1. Let a and b be non-negative integers with a ≤ b . Then the polynomial f a,b ( c ) ∈ Z [ c ] , given by f a,b ( c ) := a Y i =1 ( c + i ) i b − a Y i =1 ( c + a + i ) a a Y i =1 ( c + b + i ) a − i , is equal to H ( a + b + c ) H ( c ) H ( a + c ) H ( b + c ) , for positive integers c .Proof. We proceed with a proof by picture-calculus; note that in each case we choose the − c and that we representthe numbers that are present by a grey shaded region.In Figure A.2(i), we consider H ( a + b + c ) divided by H ( b + c ); note that we align thetriangles at their bottom points. We then multiply by H ( c ), seen in Figure A.2(ii); notethat we align the top edge of the new triangle with the top edge of the triangle associatedto H ( b + c ). Last, we divide by H ( a + c ) creating a parallelogram, seen in Figure A.2(iii). (i) H ( a + b + c ) / H ( b + c ) (ii) Multiply by H ( c ) (iii) Divide by H ( a + c ) Figure A.2. A picture-calculus proof that f a,b ( c ) = H ( a + b + c ) H ( c ) H ( a + c ) H ( b + c ) Notice that the parallelogram is a units tall, b units long, and is shifted to be c units fromthe left edge. Further,(i) The left grey triangular region corresponds to Q ai =1 ( c + i ) i ;(ii) The central grey rectangular region corresponds to Q b − ai =1 ( c + a + i ) a ; and(iii) The right grey triangular region corresponds to Q ai =1 ( c + b + i ) a − i .Thus, this region is exactly the polynomial f a,b ( c ) evaluated at the integer c . (cid:3) Example A.2. For example, notice that f , ( c ) = ( c + 1)( c + 2) ( c + 3) ( c + 4) ( c + 5)and f , ( c ) = ( c + 1)( c + 2) ( c + 3) ( c + 4) ( c + 5) ( c + 6) ( c + 7) . NUMERATIONS DECIDING THE WLP 43 A key example of using Proposition A.1 is with MacMahon’s formula (see, e.g., [7, Equa-tion (1.1)]). MacMahon’s formula for the number of lozenge tilings of a hexagon with side-lengths ( a, b, c, a, b, c ), where a, b, and c are positive integers, is given byMac( a, b, c ) = H ( a ) H ( b ) H ( c ) H ( a + b + c ) H ( a + b ) H ( a + c ) H ( b + c ) , Thus, for fixed a and b , MacMahon’s formula is a polynomial in c . Corollary A.3. Let a and b be non-negative integers with a ≤ b . Then Mac( a, b, c ) is equalto a polynomial in c , when evaluated at positive integers; in particular, Mac( a, b, c ) = H ( a ) H ( b ) H ( a + b ) f a,b ( c ) for positive integers c .Proof. This follows immediately from Proposition A.1 after noticing that H ( a ) H ( b ) H ( a + b ) is inde-pendent of c . (cid:3) When considering polynomials such as f a,b ( c ), we may want only the terms where thefactors are all of the form ( c + i ), where i has a fixed parity. To do this, we define the evenpart of the hyperfactorial of n , a positive integer, to be the even terms in the product H ( n ),that is H e ( n ) := n − Y i =0 ⌊ i ⌋ Y j =1 j, and we define the odd part of the hyperfactorial of n to be H o ( n ) := H ( n ) H e ( n ) . We notice though, that H e ( n ) can be written in terms of hyperfactorials, after an appro-priate scaling. Lemma A.4. For positive integers n , the even part of the hyperfactorial of n , H e ( n ) , is ⌊ n ⌋ ) + ( ⌈ n ⌉ ) H (cid:18)(cid:22) n (cid:23)(cid:19) H (cid:18)(cid:24) n (cid:25)(cid:19) . Proof. By definition, H e ( n ) is the product of the even columns of the pictorial representationof H ( n ). In Figure A.3(i), we see the product of H (cid:0)(cid:4) n (cid:5)(cid:1) and H (cid:0)(cid:6) n (cid:7)(cid:1) and in part (ii) wesee this same product after simplification, that is, condensing the columns to be contiguous.In Figure A.3(iii), we multiply each element of the triangular representation by 2; since thereare (cid:0) ⌊ n ⌋ (cid:1) + (cid:0) ⌈ n ⌉ (cid:1) terms, then we are simply scaling H (cid:0)(cid:4) n (cid:5)(cid:1) H (cid:0)(cid:6) n (cid:7)(cid:1) by 2( ⌊ n ⌋ ) + ( ⌈ n ⌉ ).This is exactly the even columns of H ( n ). (cid:3) Now we can find the polynomials which represent f a,b ( c ) with only factors of the form( c + i ), where i has a fixed parity, present. (i) H (cid:0)(cid:4) n (cid:5)(cid:1) H (cid:0)(cid:6) n (cid:7)(cid:1) (ii) After simplification (iii) After scaling Figure A.3. A picture-calculus proof of an identity of H e ( n ) Corollary A.5. Let a and b be non-negative integers with a ≤ b . Set a = (cid:4) a (cid:5) and b = (cid:4) b (cid:5) .Then the polynomial f ea,b ( c ) ∈ Z [ c ] , given by f ea,b ( c ) := a Y i =1 ( c + 2 i ) i b − a Y i =1 ( c + 2 a + 2 i ) a a Y i =1 ( c + 2 b + 2 i ) b − b + a − i , is equal to H e ( a + b + c ) H e ( c ) H e ( a + c ) H e ( b + c ) if c is even and H o ( a + b + c ) H o ( c ) H o ( a + c ) H o ( b + c ) if c is odd , for positive integers c .Further, the polynomial f oa,b ( c ) ∈ Z [ c ] , given by f oa,b ( c ) := f a,b ( c ) f ea,b ( c ) is equal to H o ( a + b + c ) H o ( c ) H o ( a + c ) H o ( b + c ) if c is even and H e ( a + b + c ) H e ( c ) H e ( a + c ) H e ( b + c ) if c is odd , for positive integers c .Proof. Notice first that f ea,b ( c ) is defined to be the factors of f a,b ( c ) of the form ( c + i ), where i is even. The hyperfactorial representation then follows when considering ( c + i ) would thenbe even exactly when c is even.The second claim follows similarly as the first. (cid:3) Example A.6. 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