aa r X i v : . [ m a t h . C O ] J un HILBERT BASES AND LECTURE HALL PARTITIONS
MCCABE OLSEN
Abstract.
In the interest of finding the minimum additive generating set for the set of s -lecture hall partitions, we compute the Hilbert bases for the s -lecture hall cones in cer-tain cases. In particular, we determine the Hilbert bases for two well-studied families ofsequences, namely the 1 mod k sequences and the ℓ -sequences. Additionally, we provide acharacterization of the Hilbert bases for u -generated Gorenstein s -lecture hall cones in lowdimensions. Introduction
For any n ∈ Z ≥ , let [ n ] := { , , · · · , n } . Let s = ( s , s , · · · , s n ) be a sequence such that s i ∈ Z ≥ for each i . Given any s -sequence, define the s -lecture hall partitions to be the set L ( s ) n := (cid:26) λ ∈ Z n : 0 ≤ λ s ≤ λ s ≤ · · · ≤ λ n s n (cid:27) . In the case when s is weakly (or strictly) increasing, s -lecture hall partitions are a refinementof the set of all partitions. Bousqet-M´elou and Eriksson first introduced the notion of s -lecture hall partitions in two seminal papers [4, 5], and since then these objects have beenvastly studied in various contexts. For example, lecture hall partitions give rise to variationsand generalizations of classical partition identities, which are of interest in combinatorialnumber theory. Lecture hall partitions also give rise to various discrete geometric objects,namely rational cones, lattice polytopes, and rational polytopes. These objects have givenrise to interesting Hilbert series and Ehrhart theoretic results leading to generalizations ofEulerian polynomials. See the excellent survey of Savage [12] for an overview of much of thiswork.One question that remains open in general is the following: Question 1.1.
Can we determine the unique minimal additive generating set for L ( s ) n for anarbitrary s ? Are there nontrivial bounds on the cardinality of this set?While this is in general a difficult question to answer, one method is to employ tools frompolyhedral geometry. Given a sequence s = ( s , · · · , s n ), we define the s -lecture hall cone tobe the rational, pointed, simplical polyhedral cone given by C ( s ) n := (cid:26) λ ∈ R n : 0 ≤ λ s ≤ λ s ≤ · · · ≤ λ n s n (cid:27) . Date : September 24, 2018.2010
Mathematics Subject Classification.
Key words and phrases. lecture hall partitions, Hilbert bases, Gorenstein cones.The author thanks the American Institute of Mathematics, as this work began at the November 2016workshop on polyhedral geometry and partition theory. The author thanks his advisor, Benjamin Braun, forhelpful comments and suggestions throughout this project. The author also thanks the anonymous refereesfor reading the manuscript carefully and providing helpful suggestions and comments.
For a rational, pointed cone C ∈ R n , the Hilbert basis of C is the unique minimal additivegenerating set of C ∩ Z n . Noting that C ( s ) n ∩ Z n = L ( s ) n , we can now reformulate Question 1.1in terms of polyhedral geometry. Question 1.2.
Can we determine the Hilbert basis of C ( s ) n for arbitrary s ? Can we givenontrivial bounds on the cardinality of this set?The reformulation in this question seems fruitful. Determining the Hilbert basis for apolyhedral cone allows for the study of the Hilbert series of cone, as well as other algebraicinterests such as free resolutions of the defining ideal of the cone. This extension of possiblequestions and problems indicates that it may be a worthwhile pursuit. Additionally, someresults on Hilbert bases of lecture hall cones are known. Specifically, Beck, Braun, K¨oppe,Savage, and Zafeirakopoulos [2] show that the elements of the Hilbert basis of C ( s ) n for s =(1 , , · · · , n ) are naturally indexed by subsets A ⊆ [ n − C ( s ) n and they show that the numeratorof the Hilbert series with respect to this grading is an Eulerian polynomial. This motivateslooking for a general form for arbitrary s .Unfortunately, it is unlikely that there is a general structure for the Hilbert bases of s -lecture hall cones, and it is almost a certainty that no nontrivial bounds on the cardinalityexist. This can be seen in the simplest case n = 2. Let s = ( s , s ) and notice that we haveupper and lower bounds; namely s + 1 forms an upper bound given by enumerating latticepoints in the fundamental parallelepiped of C ( s ) n and 3 is a lower bound provided s ≥ s = 1). These bounds are in fact sharp, as the sequence s = ( s , k · s + 1)for any k ∈ Z ≥ gives a cone whose Hilbert basis has cardinality s + 1, whereas the cone forthe sequence s = ( s , k · s −
1) for any k ∈ Z ≥ has a Hilbert basis of cardinality 3.Subsequently, in order to obtain meaningful results, we must place some additional re-strictions. Motivated by recent work on lecture hall cones [1, 2], we restrict to the case of u -generated Gorenstein s -lecture hall cones (defined in Section 2). We pose the followingquestion. Question 1.3.
Can we determine the Hilbert basis of C ( s ) n where s is an arbitrary u -generated Gorenstein sequence? Can we give the cardinality of the set of Hilbert basiselements, or find nontrivial bounds to this set?In this paper, we make progress towards answering Question 1.3. Section 2 is devoted toproviding necessary definitions and terminology. In Sections 3 and 4, we consider well-studiedfamilies of sequences, namely the 1 mod k sequences and the ℓ -sequences. In particular, wehaving the following descriptions of the Hilbert bases: Theorem 1.4 (Theorem 3.1) . For all k ≥ , the Hilbert basis for the k cones in R n ,denoted C k,n , consist of the following elements: • The element v A := (0 , , · · · , , a , a , · · · , a k , a k + 1) for each A ⊆ [ n − where A = { a < a < · · · < a k } ; • Element w ∈ L k,n , where w n − = ( n − k + 1 and w n = ( n − k + 1 ;where L n,k denotes the set of k lecture hall partitions. Theorem 1.5 (Theorem 4.1) . Let s = ( s , s , · · · , s n ) be an ℓ -sequence for some ℓ ≥ . TheHilbert basis H ℓn for the ℓ -sequence cone C ℓn is ILBERT BASES AND LECTURE HALL PARTITIONS 3 H ℓn = n [ i =0 (cid:8) λ ∈ L ℓn : λ n − = s i , λ n = s i +1 (cid:9) where L ℓn denotes the set of ℓ -sequence lecture hall partitions. The necessary definitions and terminology used in these theorems appear in greater de-tail in Sections 2, 3, and 4. These main results provide two different generalizations ofknown Hilbert basis results, as both the 1 mod k sequences and ℓ -sequences specialize to thesequence s = (1 , , · · · , n ) for k = 1 and ℓ = 2.In Sections 5, 6, and 7, we provide a characterization for the Hilbert bases of u -generatedGorenstein s -lecture hall cones in R n for n ≤
4, noting that the complexity of the Hilbertbases grows rapidly as the dimension increases. We conclude the paper in Section 8 byproviding some direction for future work in the context of commutative algebra, particularlythe study of toric ideals and free resolutions.2.
Preliminaries
We recall a few definitions from polyhedral geometry. A polyhedral cone C in R n is thesolution set to a finite collection of linear inequalities Ax ≥ A , orequivalently for some elements w , w , · · · , w j ∈ R n , C = span R ≥ { w , w , · · · , w j } . The elements w i are called ray generators . The cone C is said to be rational if the matrix A contains only rational entries (equivalently if each w i ∈ Q n ), it is said to be simplicial ifit is defined by n independent inequalities (equivalently if j = n and { w i } ni =1 are linearlyindependent), and it is said to be pointed if it does not contain a linear subspace of R n . Let C ◦ denote the interior of C .Given any pointed rational cone C ⊂ R n , a proper grading of C is a function g : C ∩ Z n → Z r ≥ , for some r , satisfying (i) g ( λ + µ ) = g ( λ ) + g ( µ ); (ii) g ( λ ) = 0 implies λ = 0; and(iii) for any v ∈ N r , g − ( v ) is finite. Moreover, the integer points C ∩ Z n form a semigroup.Semigroups of this type have unique minimal generating sets known as the Hilbert basis of C . Additionally, pointed rational cones give rise to a semigroup algebra structure C [ C ] := C [ C ∩ Z n ]. For background and details see [3, 9].We say that a pointed, rational cone C ⊂ R n is Gorenstein if there exists a point c ∈ C ◦ ∩ Z n such that C ◦ ∩ Z n = c + ( C ∩ Z n ). This point is known as the Goren-stein point of C . Due to theorems of Stanley [15], this notion of Gorenstein is equivalent tothe commutative algebra notion of Gorenstein, as C is Gorenstein if and only if the algebra C [ C ] is Gorenstein. For reference and commutative algebra details, see [7, 16].It will also be useful to recall several definitions for convex polytopes and Ehrhart Theory.Let P ⊂ R n be a n -dimensional convex polytope with vertex set { v , v , · · · , v d } . We say P is a lattice polytope if v i ∈ Z n for each i . Likewise, we say that P is a rational polytope if v i ∈ Q n for each i . The lattice point enumerator of P is the function i ( P , t ) = t · P ∩ Z n )where t · P = { t · α : α ∈ P} is the t th dilate of P with t ∈ Z ≥ . By theorems of Ehrhart[8], if P is lattice, i ( P , t ) is a polynomial in the variable t of degree n and if P is rational, i ( P , t ) is a quasipolynomial in the variable t of degree d . Subsequently, we will call i ( P , t ) MCCABE OLSEN the
Ehrhart polynomial of P or the Ehrhart quasipolynomial of P in each respective case.For reference and background on Ehrhart Theory, see [3, 17].Given a sequence s = ( s , · · · , s n ), the s -lecture hall cone is the rational, pointed, simplicalpolyhedral cone defined as C ( s ) n := (cid:26) λ ∈ R n : 0 ≤ λ s ≤ λ s ≤ · · · ≤ λ n s n (cid:27) . Alternatively, one may consider a ray generator description with integral generators C ( s ) n = span R ≥ { (0 , · · · , , , (0 , · · · , , s i , s i +1 , · · · , s n − , s n ) : 1 ≤ i ≤ n − } . It is easy to see that C ( s ) n ∩ Z n = L ( s ) n . There are many choices for properly grading the L ( s ) n ,though three useful notions are as follows: • λ ( λ , λ , · · · , λ n ); • λ λ n ; • λ ( λ n − λ n − ).In a similar manner, we define the s -lecture hall polytope to be P ( s ) n := (cid:26) λ ∈ R n : 0 ≤ λ s ≤ λ s ≤ · · · ≤ λ n s n ≤ (cid:27) . A related geometric structure is the rational s -lecture hall polytope , which is defined similarly: R ( s ) n := (cid:26) λ ∈ R n : 0 ≤ λ s ≤ λ s ≤ · · · ≤ λ n s n ≤ s n (cid:27) . Remark . For a given lecture hall cone C ( s ) n , we may assume that gcd( s , s , · · · , s n ) = 1.If we have gcd( s , s , · · · , s n ) = m >
1, we could consider the sequence t = ( t , · · · , t n )defined by t i = s i /m and notice that it is clear by definition that C ( s ) n = C ( t ) n . However,when considering the lecture hall polytope P ( s ) n or the rational lecture hall polytope R ( s ) n , itis not permissible to make this assumption. Given two rational polytopes P , Q ∈ R n , wesay P ∼ = Q if Q = f U ( P ) + v where f U is the linear transformation defined by a unimodularmatrix U and v ∈ R n . Note that P ( s ) n = P ( t ) n and R ( s ) n = R ( t ) n . In fact, we have P ( s ) n = m · P ( t ) n and R ( s ) n = m · R ( t ) n .There has been much study of these three polyhedral geometric objects (see, e.g., [1, 2, 10,11, 12, 13]). In particular, a characterization of which s -sequences yield Gorenstein coneswas implicitly given by Bousquet-M´elou and Eriksson in [5] and explicitly stated by Beck,Braun, K¨oppe, Savage, and Zafeirakopoulos in [1] as follows: Theorem 2.2 (Beck et al [1, Corollary 2.6], Bousquet-M´elou, Eriksson [5, Proposition 5.4]) . For a positive integer sequence s , the s -lecture hall cone C ( s ) n is Gorenstein if and only ifthere exists some c ∈ Z n satisfying c j s j − = c j − s j + gcd( s j , s j +1 ) for j > , with c = 1 . Moreover, in the case of s -sequences where gcd( s i , s i +1 ) = 1 holds for all i , we havea refinement to this theorem. We say that s is u -generated by a sequence u of positiveintegers if s = u s − s i +1 = u i s i − s i − for i > ILBERT BASES AND LECTURE HALL PARTITIONS 5
Theorem 2.3 (Beck et al [1, Theorem 2.8], Bousquet-M´elou, Eriksson [5, Proposition 5.5]) . Let s = ( s , · · · , s n ) be a sequence of positive integers such that gcd( s i , s i +1 ) = 1 for ≤ i < n . Then C ( s ) n is Gorenstein if and only if s is u -generated by some sequence u =( u , u , · · · , u n − ) of positive integers. When such a sequence exists, the Gorenstein point c for C ( s ) n is defined by c = 1 , c = u , and for ≤ i < n , c i +1 = u i c i − c i − . It is a natural question to consider the Hilbert basis of a given polyhedral cone. While thequestion of characterizing the Hilbert bases for C ( s ) n for arbitrary s is intractible, a naturalredirection is to restrict to the case of u -generated Gorenstein s -sequences. To providefurther motiviation, Beck, Braun, K¨oppe, Savage, and Zafeirakopoulos in [2] give an explicitdescription of the Hilbert basis in the case of s = (1 , , · · · , n ), which is u -generated by u = (3 , , , · · · , Theorem 2.4 (Beck et al [2, Theorem 5.1]) . For each A = { a < a < · · · < a k } ⊆ [ n − ,define the element v A to be v A = (0 , · · · , , a , a , · · · , a l , a k + 1) . The Hilbert basis for L (1 , , ··· ,n ) n is H (1 , , ··· ,n ) n := { v A : A ⊆ [ n − } . As a corollary, the semigroup algebra C [ C (1 , , ··· ,n ) n ] is generated entirely by elements in degree with respect to the grading given by λ ( λ n − λ n − ) . The k sequences For any k ∈ Z ≥ , we define the 1 mod k sequence to be s = (1 , k + 1 , k + 1 , · · · , ( n − k + 1) . For convenience of notation, let L k,n := L ( s ) n , let C k,n := C ( s ) n , and let P k,n := P ( s ) n . Thissequence is u -generated by u = ( k + 2 , , , · · · , k = 1, we obtain the sequence s = (1 , , · · · , n ). This generalization has been well studied,most notably by Savage and Viswanathan [13] using a discrete geometric point of view. Wenow give a concise description for the Hilbert basis of C k,n . Theorem 3.1.
For all k ≥ , the Hilbert basis H k,n of L k,n consists of the following elements: • The element v A := (0 , , · · · , , a , a , · · · , a k , a k + 1) for each A ⊆ [ n − where A = { a < a < · · · < a k } . • Element w ∈ L k,n , where w n − = ( n − k + 1 and w n = ( n − k + 1 .Proof. The Hilbert basis for the case of k = 1 is known by Theorem 2.4 and the descriptioncan be translated to be written in this language with ease. Subsequently, we will prove theresult assuming k ≥ v A are all possible elements of degree one with respect to the gradinggiven by deg( λ ) = λ n − λ n − . Let a = ( a , a , · · · , a n − , a n ) ∈ L k,n such that a n − a n − = 1and a n − < n −
1. We can see that a = v A for some set A for the following reasons:(i) For each 1 ≤ i ≤ n − a n − < n − a n − i < n − i because we have theinequlities n − ik ( n − i ) + 1 ≤ n − i + 1 k ( n − i + 1) + 1 ≤ · · · ≤ n − k ( n −
3) + 1 ≤ n − k ( n −
2) + 1
MCCABE OLSEN but we also clearly have n − i + 1 k ( n − i ) + 1 n − i + 1 k ( n − i + 1) + 1 ;(ii) We must have a i < a i +1 for all i ≤ n − a i +1 − k ( i −
1) + 1 < a i +1 ki + 1are equivalent to a i +1 ≤ i , but we also clearly have a i +1 k ( i −
1) + 1 a i +1 ki + 1 . Hence, we have a = (0 , · · · , , a j , a j +1 , · · · , a n − , a n − + 1) which means a = v A for the set A = { a j < a j +1 < · · · < a n − } ⊂ [ n − a ∈ L k,n and suppose that a n − = j ≥ n −
1. Notice that a n ≥ j + 2, because if we suppose that a n = j + 1, then wearrive at a contradiction as jk ( n −
2) + 1 ≤ j + 1 k ( n −
1) + 1holds if and only if j < ( n − a must be of degree 2or higher.Second, note that w ∈ L k,n , with w n − = ( n − k + 1 and w n = ( n − k + 1 cannot bewritten as a combination of elements of the type v A . This follows from a grading argument as w has degree k . If we consider a = P ki =1 v A i , it is clear that a n − ≤ k ( n − < k ( n −
2) + 1 = w n − and we have the result.Now, suppose that a ∈ L k,n . There are three possible cases:(1) a n − < k ( n −
2) + 1 and a n < k ( n −
1) + 1;(2) a n − < k ( n −
2) + 1 and a n ≥ k ( n −
1) + 1;(3) a n − ≥ k ( n −
2) + 1 and a n ≥ k ( n −
1) + 1.Case 1: Suppose that a n − < k ( n −
2) + 1 and a n < k ( n −
1) + 1. Given that s = 1, thiscondition forces a = 0, because a ≤ a n k ( n −
1) + 1 < ≤ i ≤ n − a i < k ( i −
1) + 1 because a i k ( i −
1) + 1 < . Moreover, we note that for all such i , we have a i k ( i −
1) + 1 < a i +1 ki + 1because equality would force a i +1 = a i + k · a i k ( i −
1) + 1which cannot be an integer by our previous observation and the fact that gcd( k, k ( i − j be the largest index such that a j < j −
1. We now write a = b + c where b = (0 , a , · · · , a j , j, j + 1 , · · · , n − ILBERT BASES AND LECTURE HALL PARTITIONS 7 and c = (0 , , · · · , , a j +1 − j, a j +2 − ( j + 1) , · · · , a n − ( n − . It is clear that b = v A for some A ⊆ [ n − c ∈ L k,n , notice that for all i ≥ ja i − i + 1 k ( i −
1) + 1 ≤ a i +1 − ki + 1is equivalent to a i ( ki + 1) + 1 ≤ a i +1 ( k ( i −
1) + 1)which is equivalent to a i k ( i −
1) + 1 < a i +1 ki + 1and thus we have the desired result. So by induction, a of this form can be written as thesum of elements of the type v A .Case 2: Suppose that a n − < k ( n −
2) + 1 and a n ≥ k ( n −
1) + 1. We claim that a − v ∅ = a − (0 , , · · · , , ∈ L k,n . If a n > k ( n −
1) + 1, this is immediate. So, suppose that a n = k ( n −
1) + 1, then a n − k ( n −
2) + 1 ≤ k ( n − k ( n −
2) + 1 < k ( n − k ( n −
1) + 1 = a n − k ( n −
1) + 1holds because k >
0. Thus, for a of this form we can reduce to Case 1.Case 3: Suppose that a n − ≥ k ( n −
2) + 1 and a n ≥ k ( n −
1) + 1. Let j be the largestindex such that a j < k ( j −
1) + 1. We write a = b + c , where b = ( a , a , · · · , a j , kj + 1 , k ( j + 1) + 1 , · · · , k ( n −
2) + 1 , k ( n −
1) + 1)and c = (0 , · · · , , a j +1 − ( kj +1) , a j +2 − ( k ( j +1)+1) , · · · , a n − − ( k ( n − , a n − ( k ( n − . It is clear that b ∈ L k,n with b n − = k ( n −
2) + 1 and b n = k ( n −
1) + 1, which is an elementof our proposed Hilbert basis. Moreover, because for all i ≥ j we have a i ≥ k ( i −
1) + 1 byassumption, it is immediate that c ∈ L k,n . Thus, by induction, this case will reduce to eitherCase 1 or Case 2 showing the result. (cid:3) In addition to the description of the Hilbert basis, we can also give the cardinality of theHilbert basis by using Ehrhart theoretic methods.
Corollary 3.2. |H k,n | = ( k + 1) n − + ( k − k + 2 n − . Proof.
Given that we have an element v A for all A ⊆ [ n − n − elements. Toenumerate the remaining Hilbert basis elements, note that there is a clear bijection between w ∈ L k,n with w n − = ( n − k + 1 and w n = ( n − k + 1, and elements w ′ ∈ L k,n − suchthat w ′ n − ≤ ( n − k + 1. However, for any such w ′ , one can identify w ′ as a lattice pointin the polytope P k,n − . Savage and Viswanathan [13, Theorem 2] prove that the Ehrhartpolynomial of P k,n is given by i ( P k,n , t ) = ( − t t X p =0 (cid:18) k − t − p (cid:19)(cid:18) − k p (cid:19) ( kp + 1) n . MCCABE OLSEN
Evaluating i ( P k,n − , t ) at t = 1 yields i ( P k,n − ,
1) = ( − (cid:18) k − (cid:19) + ( − (cid:18) − k ( k + 1) n − (cid:19) = ( k + 1) n − + ( k − k . Thus, the proof is complete. (cid:3) The ℓ -sequences For any ℓ ∈ Z ≥ , define the ℓ -sequence to be s = ( s , s , · · · , s n ) recursively as follows: s i +1 = ℓs i − s i − with s = 0 and s = 1. For convenience of notation let L ℓn := L ( s ) n , C ℓn := C ( s ) n , and R ℓn := R ( s ) n . Note that it is easy to see that any ℓ -sequence is strictlyincreasing. Moreover, ℓ -sequences are u -generated by the sequence u = ( ℓ + 1 , ℓ, ℓ, · · · , ℓ ),and hence C ℓn is Gorenstein. If we let ℓ = 2, we reduce to the known case of s = (1 , , · · · , n ).The ℓ -sequences have appeared from a number theoretic point of view by way of the ℓ -lecture hall theorem and ℓ -Euler theorems studied in [5] and [14]. We now give an explicitdescription of the Hilbert basis for any ℓ -sequence lecture hall cone. Theorem 4.1.
Let s = ( s , s , · · · , s n ) be an ℓ -sequence for some ℓ ≥ . The Hilbert basisof C ℓn is H ℓn = n [ i =0 (cid:8) λ ∈ L ℓn : λ n − = s i , λ n = s i +1 (cid:9) Proof.
Note that the Hilbert basis for ℓ = 2 is given by Theorem 2.4 and can be translatedinto this notation with ease. We will use the convention that s i = 0 if i ≤
0. We first claimthat there are no redundancies in H ℓn . First note that w ∈ L ℓn with w n − = s = ℓ and w n = s = ℓ − w = ℓ · v ′ + c · u where v n − = 1, v n = ℓ , u n − = 0, and u n = 1, but this is contradiction as w n = ℓ + c for some positive integer c .Now, suppose that for some i ≥ w ∈ L ℓn such that w n − = s i and w n = s i +1 with w = P v j where each v j is an element of the proposed Hilbert basis as well. This wouldimply that s i = m X k =1 a k · s k where a k ∈ Z ≥ and m < i and that s i +1 = m X k =1 a k · s k +1 . However, since s i +1 = ℓ · s i − s i − , combining these two gives us that s i − = m X k =1 a k · s k − . We can now use this equality along with s i = ℓ · s i − − s i − to deduce that s i − = m X k =1 a k · s k − . ILBERT BASES AND LECTURE HALL PARTITIONS 9
In fact, we can continue this iteration so that s i − j = m X k =1 a k · s k − j . In the case j = i − s = m X k =1 a k · s k − i +2 = a i − · s = a i − which implies that m = i − a m = a i − = ℓ as s = ℓ . However, this implies that s i = ℓ · s i − + i − X k =1 a k · s k with a k ∈ Z ≥ , which is a contradiction to s i = ℓ · s i − − s i − . Thus, we have no redundancy.Let λ = ( λ , λ , · · · , λ n − , λ n ) ∈ L ℓn . First note that if λ n − ≥ s i then λ n ≥ s i +1 . Noticethat the inequality s i s n − < s i +1 s n is equivalent to s i s n < s i +1 s n − , and making the substitutions s n = ℓs n − − s n − and s i +1 = ℓs i − s i − and simplifying leadsto the new equivalent statement s i − s n − < s i s n − . Repeating this process similarly shows that the above inequalities are equivalent to s i − j s n − j < s i − j +1 s n − j − for any 1 ≤ j ≤ i . So, if j = i , note that s = 0, s = 1, and we have that 0 < s n − j − whichis necessarily true. Moreover, if λ n < s i +1 then we have the inequality s i s n − ≤ s i +1 − s n which is equivalent to s i s n ≤ s i +1 s n − − s n − . Making similar reductions as above for any 1 ≤ j ≤ i this is equivalent to one of the following: ( s i − j s n − j ≤ s i − j +1 s n − j − − s n − , if j is even s i − j s n − j + s n − ≤ s i − j +1 s n − j − , if j is odd . If we consider j = i , either of the preceding is equivalent to s n − ≤ s n − j − which is a contradiction because s is a strictly increasing sequence for any ℓ . Ergo, λ n − ≥ s i implies λ n ≥ s i +1 .Consider λ n − . If λ n − ≥ s n − , we have λ n ≥ s n . Let j be the smallest integer such that λ j ≥ s j . Notice that ( λ , · · · , λ j − , s j , · · · , s n − , s n ) ∈ L ℓn and λ − ( λ , · · · , λ j − , s j , · · · , s n − , s n ) ∈ L ℓn follows immediately.Now suppose that s i ≤ λ n − < s i +1 . Notice, since s = 1, that we can write the element λ n − = k · s i + P a p ∈ A s a p where 1 ≤ k < ℓ , A is a multiset of elements of [ i −
1] ofcardinality r < ∞ , and each a p is chosen to be as large as possible. Then we have that λ n > k · s i +1 + P a p ∈ A s a p +1 . This is an elementary exercise akin to the previous proof that λ n − ≥ s i implies λ n > s i +1 . To see that λ − ( λ , · · · , λ n − , s i , s i +1 ) ∈ L ℓn , first suppose thatwe write λ n − = k · s i + i − X t =0 b t · s t where b t ∈ Z ≥ are the multiplicities of the elements of themultiset described above. Now, we have k · s i + i − X t =0 b t · s t ! − s i s n − ≤ k · s i +1 + i − X t =0 b t · s t +1 ! − s i +1 s n ≤ λ n − s i +1 s n . The second equality is immediate by previous observation and the first inequality is equiva-lent to ( k − s i s n + s n · i − X t =0 b t · s t ≤ ( k − s i +1 s n − + s n − · i − X t =0 b t · s t +1 . By expanding using s n = ℓ · s n − − s n − on the right hand side and s i +1 = ℓ · s i − s i − and s t +1 = ℓ · s t − s t − on the left hand side, we have after simplification( k − s n − s i + i − X t =0 b t · s t ≥ ( k − s n − s i − + i − X t =0 b t · s t − . In a similar manner to the above, we arrive at the equivalent statement( k − s n − j s i − j + i − X t =0 b t · s t − j ≤ ( k − s n − j − s i − j +1 + i − X t =0 b t · s t − j +1 for any 0 ≤ j ≤ i . When j = i ,0 = ( k − s n − i s ≤ ( k − s n − i − s = ( k − s n − i − which is necessarily true. Therefore, by induction, we have a complete Hilbert basis. (cid:3) We now provide a method for computing the cardinality of the Hilbert basis for any ℓ -sequence. Though not given by an explicit algebraic expression, this formula gives a com-binatorial interpretation for the cardinality of the Hilbert basis elements of ℓ -sequences. Corollary 4.2. |H ℓn | = 2 + n − X j =1 i ( R ℓn − , s j ) where i ( R ℓn − , t ) denotes the Ehrhart quasipolynomial of the rational lecture hall polytope R ℓn − .Proof. Suppose that λ ∈ L ℓn such that λ n − = s i +1 and λ n = s i +2 for some 1 ≤ i ≤ n − λ n − ≤ s i by similar applying arguments used in the proof of Theorem4.1. Therefore, we can bijectively associate λ with a lattice point λ ′ in the s i th dilate of therational lecture hall polytope R ℓn − , so λ ′ ∈ ( s i · R ℓn − ∩ Z n − ). Therefore, all such Hilbertbasis elements are enumerated by i ( R ℓn − , s i ). All Hilbert basis elements are counted in thisway with the exception of two, namely (0 , · · · , , ,
1) and (0 , · · · , , , ℓ ), as s = 1 and s = ℓ . Thus, we have the desired. (cid:3) ILBERT BASES AND LECTURE HALL PARTITIONS 11
As an aside, note that the ( n − i ( R ℓn − , s n − ) = i ( P ℓn − , P ℓn − , which one may havesuspected from the results in the 1 mod k cones. This phenomenon occurs in later cases aswell. 5. Two-dimensional Gorenstein sequences
We begin our low-dimensional characterization for the Hilbert bases of u -generated Goren-stein lecture cones by considering the two-dimensional case. Notice that when n = 2, Remark2.1 implies that there is no distinction between Gorenstein and u -generated Gorenstein. Ap-plying Theorems 2.2 and 2.3 provides the following description for the Gorenstein condition. Lemma 5.1.
Suppose that s = ( s , s ) such that C ( s )2 is Gorenstein. Then s = ( s , ks − for k ≥ . Using this description, we will now classify the Hilbert bases for all two-dimensional Goren-stein lecture hall cones as follows.
Theorem 5.2.
Let C ( s )2 be a Gorenstein lecture hall cone with s = ( s, ks − for some k ≥ .The Hilbert basis of C ( s )2 is H ( s )2 = { (0 , , ( s, ks − , (1 , k ) } .Proof. Let ( a, b ) ∈ L ( s )2 . First, suppose that a ≥ s and note that this immediately impliesthat b ≥ ks −
1. We have that ( a, b ) − ( s, ks − ∈ L ( s )2 because a − ss ≤ b − ( ks − ks − as ≤ bks − a ≥ s and b ≥ ks − ≤ a ≤ s −
1. If a ≥
1, then b ≥ k because s < kks − , but s > k − ks − .Observe that as < bks − b = ak − as which by the assumption 1 ≤ a ≤ s − a, b ) − (1 , k ) ∈ L ( s )2 , as a − s ≤ b − kks − a ( ks − < bs, which is equivalent to our observation above.Finally, note that if a = 0 and b ≥
1, ( a, b ) − (0 , ∈ L ( s )2 is immediate. Thus, by induction,we have a complete Hilbert basis. (cid:3) We note that when n = 2, the Gorenstein condition ensures that the Hilbert basis is of thesmallest possible cardinality, |H ( s )2 | = 3 when s ≥ |H ( s )2 | = 2 if s = 1. This furthermotivates the restriction to u -generated Gorenstein cones. Three-dimensional u -generated Gorenstein sequences We continue our low dimensional characterization for u -generated Gorenstein lecture hallcones by considering the three-dimensional case. When n = 3, a direct application of Theo-rem 2.3 yields the following description. Lemma 6.1.
Suppose that s = ( s , s , s ) such that C ( s )3 is Gorenstein with gcd( s i , s i +1 ) = 1 for all i . Then s = ( s, ks − , ℓ ( ks − − s ) for integers s ≥ , k ≥ and ℓ ≥ . Using the above lemma, we now completely characterize the Hilbert bases for all u -generated Gorenstein lecture hall cones for n = 3. Theorem 6.2.
Suppose that s = ( s, ks − , ℓ ( ks − − s ) . Then • If s ≥ , then the Hilbert basis is H ( s )3 = { (0 , , , (0 , , ℓ ) , (0 , k, ℓk − , (1 , k, ℓk − , ( j, ks − , ℓ ( ks − − s ) ∀ ≤ j ≤ s } . • If s = 1 , then the Hilbert basis is H ( s )3 = { (0 , , , (0 , , ℓ ) , (0 , k − , ℓ ( k − − , (1 , k − , ℓ ( k − − } . Proof.
Suppose that s ≥
2. We claim that the proposed Hilbert basis has no redundancy.It is sufficient to show that ( j, ks − , ℓ ( ks − − s ) cannot be written as a sum of otherproposed elements. Suppose this is possible, then there exist positive integers α, β , and γ such that α ( k ) + β = ks −
1. This has solutions α = s − i and β = ki − ≤ i < s .However, we must also have α ( ℓk −
1) + β ( ℓ ) + γ = ℓ ( ks − − s and evaluating at the abovesolution implies that γ = − i . This is a contradiction.Let ( a, b, c ) ∈ L ( s )3 . First note that if a ≥ s , this implies that b ≥ ks − c ≥ ℓ ( ks − − s .It is clear then that ( a, b, c ) − ( s, ks − , ℓ ( ks − − s ) ∈ L ( s )3 . If 0 ≤ a < s and b ≥ ks − c ≥ c ≥ ℓ ( ks − − s and ( a, b, c ) − ( a, ks − , ℓ ( ks − − s ) ∈ L ( s )3 .Next suppose that 1 ≤ a < s and b < ks −
1. Notice that a ≥ b ≥ k and c ≥ ℓk − s < kks − < ℓk − ℓ ( ks − − s but s > k − ks − and kks − > ℓk − ℓ ( ks − − s . Additionally, wecan see that the inequalities must be strict: as < bks − < cℓ ( ks − − s . This follows because equality of the first and second fractions implies that b = ak − as whichis not an integer by the assumption 1 ≤ a < s −
1, and equality of second and third fractionsimplies that c = bℓ − bsks − which is not an integer by the assumption b < ks −
1. Now, weclaim that ( a, b, c ) − (1 , k, ℓk − ∈ L ( s )3 , as a − s ≤ b − kks − ≤ c − ℓk + 1 ℓ ( ks − − s is equivalent to a ( ks −
1) + 1 ≤ bs and b ( ℓ ( ks − − s ) ≤ c ( ks − − a ( ks − < bs and b ( ℓ ( ks − − s ) < c ( ks − a = 0. If b ≥ k , we have (0 , b, c ) − (0 , k, ℓk − ∈ L ( s )3 immediatelyby the previous argument. So, suppose that 1 ≤ b < k , and notice that this implies that ILBERT BASES AND LECTURE HALL PARTITIONS 13 c ≥ bℓ as bks − < bℓℓ ( ks − − s . However, we also have bks − > bℓ − ℓ ( ks − − s as this is equivalent to sb < ks − b ≤ k −
1. We now claim that (0 , b, c ) − (0 , ℓ ) ∈ L ( s )3 as wehave the following inequalities b − ks − ≤ bℓ − ℓℓ ( ks − − s ≤ c − ℓℓ ( ks − − s . The second inequality is immediate by c ≥ bℓ and the first inequality is equivalent to b ≥ L ( s )3 can be written as a sum of these elements and wehave the Hilbert basis.Now, we suppose that s = 1. It is clear that there is no redundancy in the proposedHilbert basis. Note that we must have k ≥
2. Let ( a, b, c ) ∈ L ( s )3 . Consider b . If b ≥ k − c ≥ ℓ ( k − −
1. If a ≥
1, then ( a, b, c ) − (1 , k − , ℓ ( k − − ∈ L ( s )3 is immediate. If a = 0, then ( a, b, c ) − (0 , k − , ℓ ( k − − ∈ L ( s )3 is also immediate. Now, if 1 ≤ b < k − a = 0 and c ≥ bℓ , which follows from the same argument given in the previouscase. Moreover, we also have ( a, b, c ) − (0 , , ℓ ) ∈ L ( s )3 immediately from work of the previouscase. Thus, by induction, we have the Hilbert basis. (cid:3) We note that in this case, the cardinality of the Hilbert basis is directly dependent on thestarting value s , with |H ( s )3 | = s + 5 when s ≥ |H ( s )3 | = 4 when s = 1.7. Four-dimensional u -generated Gorenstein sequences We conclude our low-dimensional characterization of u -generated Goresntein lecture hallcones in the case of four dimensions. We have the following description for the Hilbert bases. Theorem 7.1.
Suppose that s = ( s , s , s , s ) is u -generated by u = ( u , u , u ) such that C ( s )4 is a Gorenstein lecture hall cone. Recall that c = ( c , c , c , c ) is the Gorenstein pointof C ( s )4 , with c = 1 , c = u , and c i +1 = u i c i − c i − for i ≥ . Then(a) If s = 1 and u = 2 and the Hilbert basis is H ( s )4 = { (0 , , , , (0 , , , u ) , (0 , , s , s ) , (0 , , s , s ) , (1 , , s , s ) } . (b) If s = 1 and u ≥ and the Hilbert basis is H ( s )4 = ( (0 , j, s , s ) for all ≤ j ≤ s (1 , s , s , s ) , (0 , , , , (0 , , , u ) , (0 , , u , u u − , (0 , , u , u u − ) . (c) If s = 2 and u = 1 , then the Hilbert basis is H ( s )4 = { (2 , , s , s ) , (1 , , s , s ) , (0 , , s , s ) , (0 , , s , s ) , (0 , , , u ) , (0 , , , } . (d) If s ≥ and u = 1 , then the Hilbert basis is H ( s )4 = ( λ ∈ L ( s )4 with λ = s and λ = s (0 , , , , (0 , , , u ) , (0 , , c , c ) , (0 , , c , c ) , (1 , , c , c ) ) . (e) If s ≥ and u ≥ , then the Hilbert basis is H ( s )4 = λ ∈ L ( s )4 with λ = s and λ = s (0 , j, c , c ) for all ≤ j ≤ c ( c , c , c , c ) , (0 , , , , (0 , , , u ) , (0 , , u , u u − , (0 , , u , u u − . Proof.
For each of the cases, we will consider C ( s )4 with respect to the grading defined by λ ( λ − λ ). The first two cases (a) and (b) can be reduced to the three dimensional caseand hence follow directly from the proof of Theorem 6.2. Case (c):
First note that to have a valid sequence u ≥ s = 1, s = u −
2, and s = u ( u − −
1. It is clear that there are no redundancies among the elements of theproposed Hilbert basis.We will now show that an arbitrary element of L ( s )4 can be written as a sum of elements ofthis basis by induction. Let λ ∈ L ( s )4 and consider λ . If λ ≥ s , we then consider λ = 0 or λ ≥
1. If λ = 0, it is clear that λ − (0 , , s , s ) ∈ L ( s )4 . If λ ≥
1, we then consider λ = 0, λ = 1, or λ ≥
2. We can see that if λ = 0, then λ − (0 , , s , s ) ∈ L ( s )4 and if λ = 1, then λ − (1 , , s , s ) ∈ L ( s )4 . If λ ≥
2, note that λ ≥ ⌈ λ ⌉ which implies λ − (2 , , s , s ) ∈ L ( s )4 follows from λ − ≤ (cid:24) λ (cid:25) − ≤ λ − ≤ λ < s . This implies that λ = λ = 0 and λ ≥ u λ . The firsttwo are trivial and the latter follows because λ u − < u λ u ( u − − λ u − ≤ u λ − u ( u − − λ ≥ s which is a contradiction. Moreover, we have that λ − (0 , , , u ) ∈ L ( s )4 because λ − u − ≤ u λ − u u ( u − − ≤ λ − u u ( u − − Case (d):
First, we claim that this set contains no redundancy. Note that no element λ ∈ L ( s )4 with λ = c and λ = c can be written as a combination of smaller elements.Given that c = u − c = u u − u −
1, this would imply that c = ( u − u + b where b ∈ Z ≥ which is impossible. Suppose that w ∈ L ( s )4 such that w = s , w = s andthere are additional elements of the proposed Hilbert basis such that P di =1 v i = w . Notethat this would imply there are integers m, n, p ∈ Z ≥ , where m ≤ s − s = m · c + n = m · u − m + n and s = m · c + n · u + p . However, we also have s = u s − s + 1. Combining and simplification yields the result s = m − p −
1, which isa contradiction to m ≤ s −
2, and hence no such sum exists.We will now show that an arbitrary element of L ( s )4 can be written as a sum of elements ofthis basis by induction. Suppose that λ ∈ L ( s )4 and consider λ . If λ ≥ s , then consider λ and λ . One of three cases will hold (i) λ < s , (ii) λ ≥ s with λ < s , or (iii) λ ≥ s . For(i), it is clear that λ − ( λ , λ , s , s ) ∈ L ( s )4 , for (ii) it is clear that λ − ( λ , s , s , s ) ∈ L ( s )4 andfor (iii) it is clear that λ − ( s , s , s , s ) ∈ L ( s )4 all of which are valid lecture hall partitions. ILBERT BASES AND LECTURE HALL PARTITIONS 15
Now, suppose that c ≤ λ < s . We can then write λ = α · c + β where either1 ≤ α < s − ≤ β ≤ c − u − α = s − ≤ β ≤ u −
3, because s = ( s − c + ( u − λ ≥ α · c + β · u . This follows because α · c + βs < α · c + β · u s reduces using c = u − c = u · ( u − − s = u s − u − s , and s = u u s − u s − u u − s + 1 to the inequality − α + β (1 − s ) < , which is obviously true. However, the inequality α · c + βs ≤ α · c + β · u − s reduces in the same manner to α + β ( s − ≥ u s − s − u , which is contradiction because of the conditions 1 that α ≤ s − β ≤ u − α = s − β ≤ u − λ . Suppose that the 1 ≤ λ < s − s , then λ ≥ c λ . This followsbecause λ s < λ ( u − s is equivalent to λ >
0, but the inequality λ s ≤ λ ( u − − s cannot hold because it reduces to s − ≤ λ , which is a contradiction.If λ ≥ λ − s ≤ λ · c − c s ≤ λ − c s = α · c + β − c s ≤ α · c + β · u − c s ≤ λ − c s . Each of these inequalities follows directly from previous observations. It now holds that if λ = 0, we have that λ − (0 , , c , c ) ∈ L ( s )4 in the case λ = 0 and λ − (0 , , c , c ) ∈ L ( s )4 provided λ > ≤ λ < s − λ ≥
1, then λ s < λ s − as equality creates a contradiction.This yields the equivalent inequality λ − s ≤ λ − s − , which implies λ − (1 , , c , c ) ∈ L ( s )4 .Now, suppose that 1 ≤ λ < c . Notice that this implies λ = λ = 0 and λ ≥ u λ . Thefirst two inequalities are immediate, and the latter inequality follows from the fact that λ s < λ · u s is equivalent to s >
1, which is true by assumption, but the inequality λ s ≤ λ · u − s using the observation c = u −
1, reduces to u s − u − s ≤ λ ( s − ≤ ( u − s −
1) = u s − u − s + 2 , which contradicts the assumption that s ≥
3. Subsequently, we have λ − (0 , , , u ) ∈ L ( s )4 by the above observation and applying the arguments used in case (c). Thus, by induction,we have a complete Hilbert basis in this case. Case (e):
We verify that the proposed set contains no redundancy. It is clear that theelements (0 , , , , , , u ), (0 , , u , u u − , , u , u u −
1) cannot be writtenas a combination of one another. So suppose first that an element (0 , j, c , c ) = P ei =1 v i where the v i are elements of smaller degree. This implies that there exist a, b, d ∈ Z ≥ suchthat a · u + b = c and a · ( u u −
1) + b · u + d = c , with the restriction that a ≤ u − c = u u −
1. However, we also have c = u c − c = u c − u , which means that a − d = u , which contradicts a ≤ u −
1. Thus, these elements cannot be written as a sumof elements of lower degree.Now, suppose that w ∈ L ( s )4 with w = s and w = s . If there was some collection ofelements lower degree in the proposed basis such that w = P ei =1 v i , this implies the existenceof integers m, n, p, q ∈ Z ≥ with m ≤ s − n ≤ u − s = m · c + n · u + p , s = m · c + n ( u u −
1) + p · u + q , but also s = u s − s . When we combine and simplify,we have s = m · c + n − q . However, this implies that u s − ≤ ( u − s − − q ,which implies that u + s ≤ − q , which contradicts s ≥ u ≥
2. Hence, there areno redundancies in the proposed Hilbert basis.We will now show that an arbitrary element of L ( s )4 can be written as a sum of elementsof this basis by induction. Let λ ∈ L ( s )4 and consider λ . If λ ≥ s , we can construct anelement of w ∈ L ( s )4 such that w = s and w = s so that λ − w ∈ L ( s )4 by followinganalogous construction to the previous cases.If c ≤ λ < s , note that then we can consider λ . If λ = 0, we have that λ − (0 , λ , c , c ) ∈ L ( s )4 provided that λ ≤ c or λ − (0 , c , c , c ) ∈ L ( s )4 in the case λ > c . If λ ≥
1, we have λ − ( c , c , c , c ) ∈ L ( s )4 . Each of these statements follow identically fromthe arguments for the c ≤ λ < s made in case (d).Now, suppose that u ≤ λ < c = u u −
1. We can write λ = α · u + β whereeither 1 ≤ α ≤ u − ≤ β ≤ u − α = u − ≤ β ≤ u −
2, as c = ( u − u + ( u − λ ≥ α ( u u −
1) + β · u because theinequality α · u + βs < α ( u u −
1) + β · u s reduces to 0 < α · s + β ( u s −
1) which is true by assumption. Additionally, the inequality α · u + βs ≤ α ( u u −
1) + β · u − s reduces to u u s − u − s ≤ αs + β ( u s − α and β implies that either u u s − u − s ≤ ( u − s + ( u − u s −
1) = ( u u s − u − s ) − s + 1 , ILBERT BASES AND LECTURE HALL PARTITIONS 17 which contradicts s ≥
2, or it implies u u s − u − s ≤ ( u − s + ( u − u s −
1) = ( u u s − u − s ) − u s + 2 , which contradicts s ≥ u ≥
2. Moreover, note that this implies that λ < u , whichimplies that λ = 0. Additionally, we have that 1 ≤ λ < u implies that λ ≥ λ u , whichfollows because the inequality λ s = λ s u − < λ u u u s − u − s = λ u s is immediate, but the inequality λ s u − < λ u − u u s − u − s reduces to u s − ≤ λ s which contradicts λ < u . Therefore, we get the inequalities λ − s ≤ λ u − u s ≤ λ − u s and λ − u s = ( α − · u + βs ≤ ( α − · ( u u −
1) + βu s ≤ λ − ( u u − s . Thus, if we have λ − (0 , , u , u u − ∈ L ( s )4 when λ ≥ λ − (0 , , u , u u − ∈ L ( s )4 when λ = 0.If 1 ≤ λ < u , we get λ − (0 , , , u ) ∈ L ( s )4 by repeating analogous arguments to previouscases (see case (c)). Thus, by induction, we have a complete Hilbert basis. (cid:3) Given the explicit Hilbert basis in the case of n = 4, it is additionally of interest toconsider the cardinalities of the set in each case. The following is computation of the thesecardinalities. Corollary 7.2.
For each case of Theorem 7.1, the cardinality of the Hilbert basis is asfollows:(a) If s = 1 and u = 2 , then |H ( s )4 | = 5 .(b) If s = 1 and u ≥ , then |H ( s )4 | = s + 6 .(c) If s = 2 and u = 1 , then |H ( s )4 | = 6 .(d) If s ≥ and u = 1 , then |H ( s )4 | = ( s + 1)( s − .(e) If s ≥ and u ≥ , then |H ( s )4 | = u ( s ( s + 1))2 + u + 6 .Proof. The cases of (a), (b), and (c) are immediate. Consider case (d). It is necessary toenumerate the number of λ ∈ L ( s )4 such that λ = s and λ = s . We should notice thatthis equivalent to determining the number of lattice points in a lecture hall polytope, namely P ( s ,s )2 . Given that this is a lattice triangle, this is an easy task. Note that s = s − P ( s ,s )2 are (0 , , s − s , s − P is a lattice polygon with area A , I interior lattice points, and B boundary lattice points,then A = I + B − must hold (see [3] for details and proof). We can see that there are 2 s lattice points onthe boundary of P ( s ,s )2 as the hypotenuse contains only the vertices as lattice points bygcd( s , s ) = 1. Moreover, since the area is s ( s − , we get that there are s − s − interiorlattice points. Adding the interior points, the boundary points, and the additional fiveHilbert basis elements gives s − s −
22 + 2 s + 5 = ( s + 1)( s − . To show (e), we apply similar methods. We must enumerate the lattice points of P ( s ,s )2 ,where s = u s −
1, which has vertices (0 , , (0 , u s − s , u s − s ( u + 1) boundary points, again noting that the hypotenuse contains only thetwo vertices. Applying Pick’s theorem, yields that there are u s − s ( u +2)+22 interior points.Hence, we have that P ( s ,s )2 contains u ( s ( s +1)2 + 1 lattice points. Additionally, elementsof the form (0 , j, c , c ) account for c + 1 = ( u −
1) + 1 = u elements, and there are 5additional described elements. This gives the cardinality desired. (cid:3) Concluding remarks and future directions
It is possible that one could consider continuing the low-dimensional characterization to n = 5 or greater dimensions. However, there are two observations, which discourage thispursuit. First, as noted by the case of n = 4, as the dimension increases so does thecomplexity and variation of the Hilbert basis. Experimental evidence using Normaliz [6]indicates that there would be many more cases to consider in the case of n = 5, and this willlikely shroud the significance of knowing the Hilbert bases. Secondly, cardinality argumentsare unlikely in general for greater dimension. The cardinality of the Hilbert basis is controlledin large part by the first n − s -sequence. In particular, it appears that toobtain the cardinality of the Hilbert basis, one must always compute the number of latticepoints in P ( s ) n − . In the case of n = 4, n − ≥
3, which makes the task much moredifficult.There are a number of different directions for future research in this vein. To begin, onecould consider the computation of Hilbert bases for more families of s -lecture hall cones.One particular family of well studied sequences which fall under the umbrella of u -generatedGorenstein sequence are the ( k, ℓ ) -sequences (see [12, Section 5] for definition and impor-tance). However, it is certainly possible that some sequences yield lecture hall cones withcombinatorially interesting Hilbert bases that are not u -generated Gorenstein, or even Goren-stein. It may be interesting to consider certain sequences of this type (e.g., the Fibonaccisequence).In addition, knowing the Hilbert bases for s -lecture hall cones opens the door for a numberof questions of an algebraic flavor. It is well known that C [ C ( s ) n ] ∼ = C [ x , · · · , x d ] /I s where d = |H ( s ) n | and I s is a toric ideal. It would be of interest to compute these toric ideals incertain cases to determine if these ideals admit algebraically or combinatorially interestingGr¨obner bases under certain term orders, as well as other algebraic or algebro-geometricproperties. Moreover, one could consider free resolutions of I s to determine the multigradedBetti numbers of the C ( s ) n , either using algebraic or combinatorial methods [18]. These areunknown even in the case s = (1 , , · · · , n ). ILBERT BASES AND LECTURE HALL PARTITIONS 19
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