Triangulations of polygons and stacked simplicial complexes: separating their Stanley–Reisner ideals

A triangulation of a polygon has an associated Stanley–Reisner ideal. We obtain a full algebraic and combinatorial understanding of these ideals and describe their separated models. More generally, we do this for stacked simplicial complexes, in particular for stacked polytopes.


Introduction
Triangulations of polygons constitute a basic yet rich topic going into many directions.The most classical fact about these is perhaps that they are counted by the Catalan numbers, [11,Chap.23].Their Stanley-Reisner ideals seem hitherto not to have been systematically studied.Here we get a full understanding of their algebraic and combinatorial nature.Considerably more generally, we do this for the Stanley-Reisner ideals of stacked simplicial complexes.
Example 1.1.Consider the triangulation of the heptagon in Figure 1.This may be built up step by step from triangles, by successively attaching the triangles 127, 257, 567, 245, 234.
Each triangle after the first is attached to a single edge of some earlier triangle.This is a type of shelling called a stacking, and every triangulation of a polygon Date: August 30, 2022.2010 Mathematics Subject Classification.Primary: 13F55; Secondary: 05C69, 05C70.Key words and phrases.triangulation of polygon, stacked simplicial complex, separation of ideal, regular sequence, independent vertices.is a stacking.Moreover to a triangulation of the polygon we may associate a tree (drawn in red in Figure 1), showing how the triangles are attached to each other.This gives our two fundamental notions: That of stacking and the associated (hyper)tree.
Let X be a simplicial complex on a set A, i.e., a family of subsets of A such that if F ∈ X and G ⊆ F , then G ∈ X.Let F 1 , F 2 , . . ., F k be an ordering of the facets (the maximal faces) of X.We assume that the F i 's all have the same cardinality.Let X p be the simplicial complex generated by F 1 , . . ., F p .
The sequence F 1 , . . ., F k is a stacking of X if each F p is attached to X p−1 along a single codimension-one face of X p−1 .So we may write F p = G p ∪ {v p } where G p is a face of X p−1 and v p is not a vertex of X p−1 .This is a shelling, but a particularly simple kind of shelling, since each F p is attached to a single codimension-one face, in contrast to a union of one or more such faces.A simplicial complex that has a stacking as above is called a stacked simplicial complex.Such simplicial complexes have appeared in the literature also as "facet constructible complexes" (see [6,10]).Although they have been previously studied from the point of view of commutative algebra, to the best of our knowledge it was with a different persepective than in this paper.For instance, in [6] the focus is more on the homological invariants and Cohen-Macaulayness of such complexes.
To a stacked simplicial complex X we associate a (hyper)tree as in the example above.Let V be an index set for the facets of X.For a codimension-one face G of X which is on at least two facets, let e G = {v ∈ V | F v ⊇ G}.This gives a hypergraph on V whose edges are the sets e G .In fact this hypergraph is a hypertree T : it is connected, each pair of edges intersects in at most one vertex, and there are no cycles.The hypertree T is an ordinary tree, like in Figure 1, when each codimension-one face is on at most two facets.Then X is a triangulated ball.In fact, X may then be realized as a stacked polytope (see [12]), and every stacked polytope is of this kind.For the relationship between stacked simplicial complexes and stacked polytopes, we refer to Section 4.5 of [10].
Given an (ordinary) tree T , let V be the vertices of T , and E the edge set of T .Let C ⊆ E × V be the incidence relation consisting of pairs (e, v) such that v is a vertex on the edge e.Let k[x C ] be the polynomial ring in the variables x e,v for (e, v) ∈ C. We associate a squarefree monomial ideal I(T ) in the polynomial ring k[x C ] as follows.Given a pair of vertices v, w of T , there is a unique path between v and w in the tree T : v v ′ e f w w ′ Associate to the pair of vertices {v, w} the monomial m v,w = x e,v • x f,w .
The ideal I(T ) is the monomial ideal generated by the m v,w as v and w run through all distinct pairs of vertices of V .
We show that the Stanley-Reisner ring of any triangulation of a polygon is obtained from k[x C ]/I(T ) by dividing out by a suitable regular sequence of variable  differences x e,v ′ − x f,w ′ .More generally any Stanley-Reisner ring of a stacked simplicial complex is obtained this way.The rings k[x C ]/I(T ) for trees T are thus the "initial objects" or "free objects" for Stanley-Reisner rings of stacked simplicial complexes.Formulated otherwise, let I be the Stanley-Reisner ring of a stacked simplicial complex.The separated models of I are one or more of the I(T ).Then I(T ) is the Stanley-Reisner ideal of a stacked simplicial complex of dimension 3 (with facets of cardinality 4) with eight vertices and five facets.Dividing out by the variable difference x a,2 − x d,4 , we get the Stanley-Reisner ring of the triangulation of the heptagon k[x C ]/(I(T ) + (x a,2 − x d,4 )). Figure 2, on the right, shows the triangulation with our new labelings of the vertices.
The ideals I(T ) are introduced in [2] where they are shown to be all possible polarizations of the square of the graded maximal ideal (x e ) 2 e∈E in k[x e ] e∈E .If I X is the Stanley-Reisner ring of a stacked simplicial complex we therefore have processes: Each of the arrows above preserves the graded Betti numbers.Hence every I X has the same graded Betti numbers as a second power of a graded maximal ideal (x e ) e∈E .Figure 3.
)∈C be the linear subspace of one-dimensional forms in the polynomial ring k[x C ].A subspace L of this linear space is a regular linear space if it has a basis consisting of a regular sequence of variable differences for k[x C ]/I(T ).The quotient ring by the space L of linear forms will still be a polynomial ring divided by a monomial ideal.We show the following.Theorem 5.16.There is a one-to-one correspondence between regular linear spaces for k[x C ]/I(T ) and partitions of the vertex set V .
In particular for the partition with one part, the whole of V , the regular sequence consists of all variable differences x e,v − x e,w for e = {v, w} ∈ E, and the quotient ring is k[x e ] e∈E /m 2 , where m = (x e ) e∈E is the irrelevant maximal ideal.Theorem 6.2.There is a one-to-one correspondence between regular linear spaces for k[x C ]/I(T ) giving squarefree quotient rings, and partitions of the vertex set V into sets of independent vertices.These quotient rings give the Stanley-Reisner rings of stacked simplicial complexes.
In [9] the first author gives a one-to-one correspondence between partitions of the vertex set of a tree T into (r + 1) independent sets, and partitions of the edge set of T into r sets.We recall this in the appendix.The above may then be reformulated as: Theorem 6.3.There is a one-to-one correspondence between regular linear spaces for k[x C ]/I(T ) giving squarefree quotient rings, and partitions of the edge set E(T ).Moreover, the dimension of the simplicial complex associated to this quotient Stanley-Reisner ring is one less than the number of parts in the partition.
Example 1.3.Consider Example 1.2 above.The regular linear space L is the space L = x a,2 − x d,4 .It corresponds to the partitions of vertices and partitions of edges of the tree in Figure 3.These partitions are respectively There are three parts in the edge partition and so the dimension of the associated simplicial complex is one less, the dimension of the triangulated polygon.
Finally we show the following.
Theorem 8.7.There is a one-to-one correspondence between regular linear spaces for k[x C ]/I(T ) giving squarefree quotient rings whose associated simplicial complex is a triangulated ball, and partitions of the edge set E(T ) into sets of independent edges.
In particular the last two theorems above give that triangulations of simplicial polygons correspond to partitions of the edges of trees T into three parts, each part being a set of independent edges.In particular only trees T whose maximal vertex degree is 3 arise in this context (which is easy to see directly like in Example 1.1).
The organization of this article is as follows.In Section 2 we recall the notions of separating and joining variables in monomial ideals.We develop basic auxiliary results for doing this.We also recall the notion of separated model.In Sections 3 and 4 we recall basic notions for simplicial complexes.We define stacked simplicial complexes and hypertrees.We show that the separated models of stacked simplicial complexes are the ideals I(T ).
Section 5 is the main technical part and gives the combinatorial description of which linear spaces of variable differences are regular linear spaces for k[x C ]/I(T ).Section 6 describes the regular linear spaces that give squarefree quotient rings.Section 7 describes the ordering relation between partitions of vertices that corresponds to inclusion of regular linear spaces.Lastly in Section 8 we describe those regular linear spaces where the quotient ring is associated to a triangulation of a ball, or equivalently of a stacked polytope.We also describe the Stanley-Reisner ring of the boundary of these polytopes, which are simplicial spheres.
The appendix recalls the correspondence between partitions of vertices of V (T ) into independent sets and partitions of edges of E(T ).
Acknowledgements.We thank Lars Hällström, Veronica Crispin Quinonez, Russ Woodroofe and the anonymous referee for all their comments, which improved this paper.In particular Lars Hällström suggested the conceptual gain of indexing the variables in k[x C ] by pairs of edges and vertices (e, v) such that v ∈ e, instead of letting the variables be indexed by E × {0, 1}, as we did in a preliminary version of this article.The second author was supported by the Finnish Academy of Science and Letters, with the Vilho, Yrjö and Kalle Väisälä Fund.
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Separations and joins for Stanley-Reisner ideals
We recall the notion of separation for monomial ideals I.It is a converse to the notion of dividing a quotient ring S/J out by a variable difference which is a non-zero divisor.When I = I X or J = J X is a Stanley-Reisner ideal, we descibe how the simplicial complex X transforms under these processes.
For a set V denote by k[x V ] the polynomal ring in the variables x v for v ∈ V .With some abuse of notation, for R ⊆ V let x R denote the monomial r∈R x r .(This should not cause confusion since in the polynomial ring we always surround x V with square brackets.)2.1.Separating a variable.The following definition is from [8, Section 2].Definition 2.1.Let V ′ p −→ V be a surjection of finite sets with the cardinality of V ′ one more than that of V .Let v 1 and v 2 be the two distinct elements of V ′ which map to a single element v in V .Let I be a monomial ideal in the polynomial ring k[x V ] and J a monomial ideal in k[x V ′ ].We say J is a simple separation of I if the following hold: i.The monomial ideal I is the image of J by the map k ii.Both the variables x v 1 and x v 2 occur in some minimal generators of J (usually in distinct generators).iii.The variable difference are monomial ideals such that J is obtained by a succession of simple separations of I, J is a separation of I.If J has no further separation, we call J a separated model (of I).
Let X be a simplicial complex on the set V .This is a family of subsets of V such that F ∈ X and G ⊆ F implies G ⊆ F .The set of v ∈ V with {v} ∈ X is the support of X.For R ⊆ V the restriction X R if the simplicial complex on R consisting of all F ∈ X such that F ⊆ R. Denote by X −R the restriction X R c where R c is the complement of R in V .The link lk X R is the simplicial complex on R c consisting of all F ⊆ R c such that F ∪ R ∈ X.If Y ⊆ X are simplicial complexes, denote by X \ Y the relative simplicial complex, consisting of those F ∈ X which are not in Y .
Let I X be the Stanley-Reisner ideal of X, the monomial ideal in k[x V ] whose generators are the monomials x R for R ∈ X. Suppose we use v to separate I X to an ideal I X ′ in the polynomial ring k[x V ′ ].Write the minimal set of monomial generators of I X as M 0 ∪ M v , where M 0 consists of those that do not contain x v and M v of those of the form x v • x R .The separated ideal I X ′ will then have minimal generators ).Here M v,i consists of those minimal generators that contain x v i .There is a bijection between M v,1 ∪ M v,2 and M v by sending 2.2.Criteria for separating a variable.Here is a general description of how I X can be separated using the variable x v .Proposition 2.2.We may separate I X using the variable x v iff the following holds: there is a partition of the faces into two non-empty parts where each F i is closed under taking smaller sets in the sense that if G ⊆ F and F ∈ F i , then either G ∈ F i or G ∈ lk X v.The facets of the simplicial complex X ′ in the separated ideal I X ′ are then obtained from the facets F of X as follows: • and so in I X .But then F could not have been in X.
Suppose conversely we have the partition F 1 ⊔ F 2 .Let the ideal I X ′ be constructed as just before this Subsection 2.2 so it is generated by For v ∈ V and X a simplicial complex, let the neigborhood of v be Corollary 2.3.Let X be a flag simplicial complex on V , i.e., I X is generated by quadratic monomials.Let v be in the support of X. Suppose X −N (v) = X 1 ∪ X 2 where X 1 and X 2 are simplicial complexes supported on disjoint vertex sets V 1 and V 2 .Then using x v the ideal I X ⊆ k[x V ] may be separated to an ideal The facets of X ′ correspond to the facets F of X as follows: • We then let F 1 be the set of those F such that F 0 is a subset of V 1 and similarly for F 2 .These will then be disjoint and closed under taking smaller sets.
Example 2.4.In Figure 1 in the introduction one may apply the above corollary to the vertex v = 5.This gives the separated ideal I(T ) of Example 1.2.The vertex v = 5 is the only vertex we may use to get a separated ideal.These things may also be seen by Proposition 2.2.

2.3.
Criteria for joining variables.We present here basic results on dividing out a Stanley-Reisner ring by variable differences.
Let X be a simplicial complex on a set V , and F a facet of X.Then for the algebraic set A(X) in the affine space A V k defined by the Stanley-Reisner ideal k where all coordinates x v = 0 for v ∈ F , while the x v for v ∈ F may take arbitrary values. For Let the ideal I be the image of I X .Then I may or may not be squarefree.If I is squarefree we say that x v 1 − x v 2 cuts squarefree.Then let I = I X 1 where X 1 is the associated simplicial complex.We then have a commutative diagram of algebraic sets: . Let e v be the point in affine space A V k where x v takes value 1 and the other variables value 0. The map φ above sends (1) divisor and cuts squarefree, the facets of X 1 are G 1 , . . ., G r where: - Proof.a.The associated primes of I X are the ideals generated by variables (x v ) v ∈F , one such ideal for each facet F .The variable difference is a nonzero divisor iff it is not in any of these ideals.This means that never both v 1 and v 2 are in such an ideal, or equivalently never both v 1 and v 2 are outside of a facet F . b.The ideal I is squarefree iff there is no minimal generator c.This follows by (1) above.
A sequence of linear forms ℓ 1 , . . ., ℓ r is a regular sequence for S/I X iff for every facet F , it cuts down A(F ) successively by one dimension for every ℓ k .
Corollary 2.6.Let X be a simplicial complex on a set V .Let B be a forest on V , and denote B 1 , . . ., B m the trees in B and V i the support of B i for each i.Then {x v − x w | {v, w} edge of B} is a regular sequence for S/I X iff for each facet F and each V i , at most one of the vertices of V i is not in F .
Proof.A facet F of X gives the irreducible component A(F ) of the algebraic set associated to X.When cutting down A(F ) by the sequence of variable differences associated to the edges of B i we have: Hence using the edges of the forest B the linear space A(F ) is cut down to a linear space whose dimension is i (|V i | − 1) less than A(F ) for each facet F of X.But then the set of variable differences is a regular sequence.
Conversely assume the sequence is regular.If there are V i and F with {v ′ , v ′′ } ⊆ V i \ F , then the variable differences x v − x w associated to edges {v, w} in B i would only give the restrictions x v = 0 for v ∈ V i ∩ F .This only cuts down dimension at most by |V i | − 2, contrary to the sequence being regular.

Stacked simplicial complexes
Let X be a simplicial complex on a set A. In the previous section we used V for the vertex set of X, but in the sequel we reserve V for the vertex set of the hypertree T associated to the stacked simplicial complex X.In other words V is an index set for the facets of X.
We show that I X may be successively separated to an ideal I X ′ , where X ′ is a stacked simplicial complex of dimension two less than the number of facets |V |. (See Figure 4 for two examples of such X ′ .)3.1.Stacked simplicial complexes and associated hypertree.A facet F of a simplicial complex X is a leaf if there is a vertex v of F such that F is the only facet containing v.Such a vertex is a free vertex of X.If v is the only free vertex of F we say F is stacked on X −{v} .(In general, a free face of a simplicial complex is a face that is not a facet and that lies on exactly one facet, see for instance [15].The term "leaf" is quite standard in graph theory, and less common in the setting of simplicial complexes, but see for instance [7].)Definition 3.1.A pure simplicial complex (i.e., where all the facets have the same dimension) is stacked if there is an ordering of its facets F 0 , F 1 , . . ., F k such that if X p−1 is the simplicial complex generated by F 0 , . . ., F p−1 , then F p is stacked on X p−1 .
Remark 3.2.This is a special case of shellable simplicial complexes, see [14, Subsection 8.2].It is not the same as the notion of simplicial complex being a tree as in [7], even if the tree is pure.Rather the notion of stacked simplicial complex is more general.For instance the triangulation of the heptagon given in Example 1.1, is not a tree in the sense of [7], since removing the triangles 234 and 257 one has no facet which is a leaf.
} is contained in an edge.If T ′ and T are hypertrees on the same vertex set, T ′ is a refinement of T if (i) every edge of T ′ is contained in an edge of T , and (ii) every edge of T is a union of edges of T ′ .
Definition 3.4.Let X be a stacked simplicial complex with facets F v indexed by a set V .We associate a hypertree to X on the vertex set V .For each codimensionone face , the set of those e G containing at least two facets.
The simplicial complex X is a triangulated ball iff its associated hypertree T is an ordinary tree, [4,Theorem 11.4].It can then be realized as a stacked polytope.Such polytopes are extremal in the following sense: they have the minimal number of faces, given the number of vertices (see [3]).Observation 3.5.Let X be a stacked simplicial complex which is a cone with p vertices in the cone apex.Thus X is a join X 1 * ∆ p where ∆ p is a simplex on p elements and X 1 is not a cone.Then X 1 is also stacked and both X and X 1 have the same associated hypertree.

3.2.
Separating stacked simplicial complexes.Lemma 3.6.Let X be a stacked simplicial complex of dimension d with hypertree T .If T has ≥ d + 3 vertices, then using the procedure of Corollary 2.3, X may be separated to a simplicial complex X ′ which is also stacked, and whose hypertree T ′ is a refinement of T .
Proof.Note that if d = 0, then X is a collection of ≥ 3 vertices and the hypertree T has one edge, the set of all facet indices V .By Corollary 2.3, X may be separated.
Let d ≥ 1 and F 0 , . . ., F k be a stacking order for X.The facet F k is stacked on some previous facet F p , with p < k.Let v be the vertex of be the partition given in Corollary 2.3.
has two components, supported on respectively V 1 and V 2 , and we may apply Corollary 2.3.2. Suppose N(v ′ ) contains G and not w.Then X −N (v ′ ) has components {w} together with at least one other component and we may again apply Corollary 2.3.

Suppose
) may be written as a disjoint union X 1 ∪ X 2 , with X 2 supported on V 2 and X 1 supported on V 1 ∪ {w}.Again we may apply Corollary 2.3.
Let us now show that T ′ is a refinement of T .Let G be a codimension-one face of X contained in two or more facets F i for i ∈ D, so D is an edge in T .Denote by x v the variable used in the separation.
G and similarly define D 2 .There might also be a facet F = G ∪ {v} containing G, in which case we extend both D 1 and D 2 with the index of this facet.Then D 1 and D 2 are edges of T ′ and they have at most one vertex in common.
Proposition 3.7.Let X be a stacked simplicial complex of dimension d which is not a cone, and let T be the associated hypertree.d.If T is an (ordinary) tree with d + 2 vertices, then X is inseparable and the isomorphism class of X is uniquely determind by T .
Example 3.8.For d = 2, Figure 4 shows the two stacked simplicial complexes of dimension 2 with four facets.The corresponding trees are also drawn in red.
Proof of Proposition 3.7.a,b.Let F 0 , . . ., F k be a stacking order of facets.Let X p be the complex generated by F 0 , . . ., F p .Let C p = ∩ p i=0 F i and G p be the codimension-one face of F p which attaches it to X p−1 .Then for p ≥ 1, If T has en edge of cardinality ≥ 3, some G p equals some G r for r < p. Then C p−1 ⊆ G r ⊆ G p and we get C p−1 = C p .Thus |C q | ≥ d + 2 − q for q ≥ p, and so if X is not a cone, k ≥ d + 2. c.This is shown in Lemma 3.6.d.Let X have associated tree T .Label the vertices of T with {0, 1, . . ., d + 1}.We assume the labeling is such that the induced subgraph on [0, p] is always a tree for p = 0, . . ., d + 1.Then the corresponding ordering F 0 , F 1 , . . ., F d+1 of the facets of X is a stacking order.
Let Y be another stacked simplicial complex with tree S isomorphic to T .Transferring the labeling from T , we get a stacking order G 0 , G 1 , . . ., G d+1 of the facets of Y .Let where ℓ < d + 1 is such that F d+1 is stacked on F ℓ .The following restrictions are cones by part a, since they have ≤ d + 1 vertices and X ′ and Y ′ are not cones (since X and Y are not cones).Their trees are obtained from T and S by removing the vertices labeled d + 1.The F ′ i = lk F i v ′ for i = 0, 1, . . ., d form a stacking order for X ′ and similarly the G ′ i = lk G i w ′ form a stacking order for Y ′ .
By induction there is a bijection between V \ {v, v ′ } and W \ {w, w ′ } sending the facet F ′ i of X ′ to the facet G ′ i of Y ′ .Extend this to a bijection between V and W by v → w, v ′ → w ′ .Then the facet F i is sent to the facet G i for i = 0, . . ., d.
So consider the facets F d+1 and G d+1 .Let the vertex (d + 1) of T be attached to vertex p ≤ d.So F d+1 is attached by the codimension-one face F d+1 ∩ F p .But this is F d+1 \ {v} and does not contain v ′ (F d+1 does not contain v ′ since X is not a cone).So this codimension-one face is

Trees and the associated separated model
Given a tree T we define the ideal I(T ).These ideals are the separated models of stacked simplicial complexes.
Let T be a tree whose set of vertices is V .Let E = E(T ) be its set of edges.The incidence relation C ⊆ E × V is the set of pairs (e, v) such that v ∈ e.It comes with a natural involution τ : C → C sending (e, v) → (e, w) where e = {v, w}.
For v, w ∈ V , denote by vT w the unique path from v to w v e f w and let e, f be the edges incident to respectively v, w on this path.For a set A denote by (A) 2 the set of subsets {a 1 , a 2 } of cardinality 2. From the directed tree T on V , we get a map For a graph G on V those vertices that are incident to an edge of G are called the vertices of G.The edges Ψ(E(G)) give a graph ΨG whose vertices are C.
• If G 1 and G 2 have disjoint vertex sets, the same holds for ΨG 1 and ΨG 2 .
• If G is a forest, then ΨG is a forest, since a cycle in ΨG must come from a cycle in G.The following is a basic object in this article.this variation is defined as {v, w} → {(e, v ′ ), (f, w ′ )}.
We will divide the ring k[x C ]/I(T ) by the following variable differences: Definition 4.4.For each pair {v, w} in (V ) 2 let h v,w be the variable difference associated to the edge Ψ{v, w}.So Note that h w,v = −h v,w .Sometimes we write this as h e where e = {v, w} when this sign plays no role.

Regular quotients of tree ideals
We describe precisely what sequences of variable differences are regular for k[x C ]/I(T ).The combinatorial description is in terms of partitions of the vertex set of T , Theorem 5.16.Definition 5.1.Let T be an (undirected) tree with vertex set V .
• The sequence of vertices v, u, w is T -aligned if u is on the path in T linking v and w. • The set {v, u, w} is non-aligned for T , if no ordering of them makes a T -aligned sequence.Proof.This is by Lemma 2.5 and the description in Lemma 4.2 of the facets of the simplicial complex associated to I(T ).Given any edge outside of im Ψ, one may find a facet F v disjoint from this edge.
The following is the basic obstruction for a sequence of h v,w 's to be regular.Lemma 5.4.Let v, u, w be T -aligned.Then h v,u and h v,w do not form a regular sequence.
Proof.Let the path vT w be: ))-regular, by showing that x g,w is in the colon ideal (I(T ) + (h v,u )) : h v,w .Indeed x g,w h v,w = x g,w (x e,v ′ − x g,w ′ ) = −x g,w x g,w ′ + x g,w x e,v ′ = −x g,w x g,w ′ + x g,w x f,u ′ + x g,w h v,u is an element of I(T ) + (h v,u ).
The following is straight-forward.
Definition 5.6.Let T be a tree with vertex set V .Let U ⊆ V and let S be a tree on U (S is a priori unrelated to T ).The tree S flows with T if whenever v, u, w are T -aligned vertices with v, u, w ∈ U, then v, u, w are S-aligned.
Example 5.7.The tree T in Figure 5 has black edges and seven vertices.The trees S are drawn in red.In the first case U = {2, 3, 4, 6}.The sequence of vertices 2, 3, 4 is T -aligned but not S-aligned, so S does not flow with T .In the second case U = {2, 4, 5, 7} and 4, 5, 7 is a T -aligned and S-aligned sequence.This tree S flows with T .
Lemma 5.8.Let U ⊆ V and let S and T be trees with vertex sets U and V , respectively.Then S flows with T iff whenever {v, w} is an edge in S, there is no u ∈ U \ {v, w} such that v, u, w are T -aligned.
Proof.Let S flow with T and let {v, w} be an S-edge.If there is u such that v, u, w are T -aligned, then v, u, w would be S-aligned, which is not the case since {v, w} is an edge in S. Conversely suppose the condition holds for edges in S. Let v, u, w be vertices in U which are T -aligned, so {v, w} is not an edge of S. Suppose the path vSw does not contain u.We argue by induction on the length ℓ S (v, w) of vSw that this is not possible.Since ℓ S (v, w) ≥ 2 let r ∈ U on vSw be distinct from v, w (note that r = u).Then ℓ S (v, w) > ℓ S (v, r) and ℓ S (v, w) > ℓ S (r, w).
Consider in T a path p from r to a vertex on the path vT w.We may assume only the end vertex of p is on vT w.If p first hits vT w in the path segment vT u, then r, u, w are T -aligned and with the path rSw being such that ℓ S (r, w) < ℓ S (v, w).By induction this situation is not possible.The case when p first hits vT w in uT w is similar.Corollary 5.9.For any U ⊆ V , there is a tree S with vertices U flowing with T .
Proof.Let v ∈ V .Consider v as a center from which the tree T branches out.Let U 0 be the subset of U consisting of w ∈ U such that the path vT w contains no other vertex in U than w (in particular if v ∈ U then U 0 = {v}).
Now define S to be the tree whose edges are: and ii) the path uT w intersects U only in {u, w}.• Give the vertices in U 0 a total order.If u, w ∈ U 0 are successive let {u, w} be an edge in S. The tree S fulfills the criterion of the lemma above, and hence flows with T .Definition 5.10.If S is a tree on the vertex set U ⊆ V , let L(S) be the linear space with basis the h e = h v,w where e = {v, w} are the edges of S. If L(S) has a basis that is a regular sequence of variable differences for k[x C ]/I(T ), we say that L(S) is a regular linear space.(Equivalently some basis or any basis of L(S) is a regular sequence.)Lemma 5.11.Let S be a tree with vertex set U, and assume that only the end vertices v and w of the path vT w are contained in U.
• If S flows with T , then h v,w ∈ L(S).
• If L(S) is a regular linear space, then h v,w ∈ L(S).
Proof.Let v = v 0 , v 1 , . . ., v n = w be a path in S of length n ≥ 2. Then by assumption v 1 , . . ., v n−1 are not in vT w.If the v i -incident edges on v i−1 T v i and v i T v i+1 are always distinct, the paths would splice to give the unique path from v to w.This cannot be the case since this path vT w only has the end vertices in U. Hence for at least one v p , 1 ≤ p ≤ n − 1, these two v p -incident edges are equal.We have three possibilities: } is non-aligned for T .For case i), if S flows with T , this would give that v p , v p−1 , v p+1 are S-aligned, which is not the case since v p−1 , v p , v p+1 are S-aligned.Similarly the second case ii) is excluded.If L(S) is regular the first and second cases are aslo excluded by Lemma 5.3.Hence only the last possibility iii) is left.
If S flows with T , then if v p−1 T v p+1 contains an element of U, such an element would be either on v p−1 T v p or on v p T v p+1 .But this is not the case by Lemma 5.8 since v p−1 , v p and v p , v p+1 are edges in S. Then we take out the edge {v p−1 , v p } from S and take in the edge {v p−1 , v p+1 } to get a new tree S ′ which still flows with T by Lemma 5.8.By induction on the length ℓ S ′ (v, w), we have h v,w ∈ L(S).
If L(S) is a regular linear space, then we again replace S with S ′ .Due to Lemma 5.5 we have L(S) = L(S ′ ) and again we get h v,w ∈ L(S).Proposition 5.12.Let S be a tree on U ⊆ V .If L(S) is a regular linear space, then S flows with T .
Proof.Let {v, w} be an edge in S, so h v,w ∈ S. Suppose v, u, w are T -aligned vertices in U.If we show this is not possible, then S flows with T by Lemma 5.8.Choose u as close as possible to v, so vT u only contains v and u from U. By Lemma 5.11, h v,u is in L(S).So both h v,w and h v,u are in L(S).By Lemma 5.4, these two elements do not form a regular sequence, contradicting the fact that L(S) is regular.Hence there can be no u such that v, u, w are T -aligned.So S flows with T .Lemma 5.13.For any two trees R and S on U flowing with T , one has L(R) = L(S).Thus U determines a unique regular linear space, denoted L(U).
Proof.Let {v, w} be an R-edge.We show h v,w ∈ L(S).Since R flows with T , the path vT w does not contain any elements of U save the end vertices.Since S flows with T , Lemma 5.11 gives h v,w ∈ L(S).We now use induction on the length n of the cycle to show that L(O) cannot be regular linear space.Not every sequence then h vp,v p−1 and h vp,v p+1 do not form a regular sequence, against the assumption.By the same reason v p v p+1 v p−1 are not T -aligned.Hence . Take the edges {v p−1 , v p } and {v p , v p+1 } out from the cycle O and take in the edge v p−1 , v p+1 to make a new cycle L(O ′ ) ⊆ L(O).By induction L(O ′ ) is not a regular linear space and so neither is L(O).b.Suppose each S i gives a regular sequence.This sequence is determined by the edges of ΨS i , and this is a forest.By Corollary 2.6 this is equivalent to each tree in ΨS i giving a regular sequence.But the disjoint union of the trees in the ΨS i are precisely the trees in ΨS.Hence Corollary 2.6 gives the result.
The following is the converse of Proposition 5.12: Proposition 5.15.Let S be a tree on U ⊆ V .If S flows with T , then L(S) is a regular linear space.
Proof.By Lemma 5.13 above, if U is the vertex set of S, we may choose S to be any tree on U that flows with T .
Let v ∈ V .Consider the face Let U ⊆ V and define the tree S flowing with T with vertices in U as in Corollary 5.9.This tree comes with two types of edges: • Edges {u, w} where u and w are two successive elements in the ordering of U 0 .Then Ψ{u, w} = {(f, u ′ ), (g, w ′ )} where f = {u, u ′ } is the edge on uT v going out from u and similarly g = {w, w ′ } the edge on vT w going out form w. These {(f, u ′ ), (g, w ′ )} give a tree T 0 with vertices from the incidence relation C (actually a line graph).• Edges {u ′ , w} where u ′ is the element in U on the path vT w closest to w.
If f = {w ′ , w} and g = {u ′ , u} are the edges on u ′ T w, one has Ψ{u ′ , w} = {(g, u), (f, w ′ )}.Each such w gives a unique u ′ , but one u ′ may correspond to several g's and w's.For each pair u ′ , g these edges form a tree T u ′ ,g , a star, with vertices from C. The trees T 0 and T u ′ ,g (with vertices from C) are all disjoint.Together the edges of these trees give all variable differences h v,w for {v, w} an S-edge.The vertices of T 0 are contained in F v .Each T u ′ ,g has all its vertices save (g, u) contained in F v .By Corollary 2.6, the linear space L(S) is regular.Theorem 5.16.There is a one-to-one correspondence between regular linear spaces for k[x C ]/I(T ) and partitions Q of the vertex set V .If the partition of Proof.By Lemma 5.13, each U i determines a unique linear space L(U i ).If S i is a tree on U i flowing with T , then L(S i ) = L(U i ).Let S = ∪ r i=0 S i .By Lemma 5.14, the edges of ΨS give a regular sequence.This regular sequence is a basis for L(Q).
Conversely if L is a regular linear space generated by the regular elements {h e } e∈G for some graph G on V , by Lemma 5.14 the graph G decomposes into a forest and we get a partition of V where each U i is the vertex set of each tree in the forest.(The vertices v of V not incident to any edge of G give singletons {v} in the partition.) Corollary 5.17.The length of the longest regular sequence of variable differences for k[x C ]/I(T ) is |E|.Such a sequence corresponds to the trivial partition of V with only one part, the set V itself.The corresponding tree that flows with T is just T itself.Hence this regular sequence is given by {h e } e∈E and the quotient ring is k[x E ]/(x e | e ∈ E) 2 .

Squarefree quotients
We determine what regular linear spaces give quotient rings of k[x C ]/I(T ) whose associated ideals are squarefree.These are the Stanley-Reisner rings of stacked simplicial complexes.Let T be a tree with vertices V .Lemma 6.1.Let U ⊆ V and let S be a tree on U. If S flows with T , then the regular quotient of k[x C ]/I(T ) by {h e } e∈S is a squarefree monomial ideal iff the vertex set U is an independent vertex set in V for the tree T .
Proof.The following is essential to note: The variables in the quotient ring modulo the sequence {h e } e∈S correspond precisely to the connected components of the graph ΨS with vertex set the incidence relation C.
If the vertex set U is dependent, say contains end vertices of an edge e = {v, w}, then we divide out by x e,v − x e,w and the ideal of the quotient ring will contain x 2 e as a generator and so is not squarefree.Suppose then that U is independent.Let {v, w} be a pair of vertices in U. Suppose the associated monomial x e,v x f,w becomes a square after dividing out by the regular sequence.This means that (e, v) and (f, w) are in the same connected component of ΨS.Let the edge e have vertices v, v ′ and the edge f vertices w ′ , w.So v ′ and w ′ are on the path vT w.Removing the edge e from T we get a component T v containing v, and similarly removing f from T we get a component T w containing w.
Any edge in ΨS containing (e, v) is the image of an edge {v, v ′ } in S where v ∈ T v .Similarly we have an edge {w ′ , w} in S where w ∈ T w .But since (e, v) and (f, w) are in the same connected component of ΨS, there must in S be an edge {ṽ, w} where ṽ is in T v and w is in T w .Then either v ′ or w ′ from U is in the interior of the path ṽT w.Since S flows with T this cannot be the case by Lemma 5.8.Theorem 6.2.There is a one-to-one correspondence between regular linear spaces for k[x C ]/I(T ) giving squarefree quotient rings, and partitions of V into sets of independent vertices.
Proof.Suppose we have a squarefree quotient ring.Each part U i of the partition gives a regular linear space L(U i ).By Lemma 6.1, U i is independent.Conversely, if we have a partition of V into independent sets U i , let S i be a tree on U i flowing with T .The images ΨS i have disjoint vertex sets as i varies.Lemma 6.1 above shows that the quotient is squarefree.
Using Theorem A.1, the above may equivalently be formulated as follows: Theorem 6.3.There is a one-to-one correspondence between regular linear spaces for k[x C ]/I(T ) giving squarefree quotient rings, and partitions of the edge set E.
If P is a partition of the edge set corresponding to the partition Q into independent vertex sets, write L(P ) = L(Q).
Corollary 6.4.The length of the longest regular sequence of variable differences giving a squarefree quotient of k[x C ]/I(T ) is |E| − 1.It corresponds to the unique partition of V into two independent sets of V for the tree T .Thus the associated regular linear space is also unique.

Partial order on partitions
If Q and Q ′ are partitions of the vertex set V of a tree T , we get the linear spaces L(Q) and L(Q ′ ).What does the inclusion relation on linear spaces correspond to on partitions?Since the linear spaces depend on additional structure coming from the tree T , this is not simply refinement of partitions.Note that such a U ′ may be convex in some U while not being convex in V .Lemma 7.2.Let U ′ and U be subsets of Proof.Suppose L(U ′ ) is a nonzero subspace of L(U) and there exists v ∈ U ′ \ U.There is another w ∈ U ′ such that h v,w ∈ L(U ′ ).Consider the path vT w in T : If the edge e occured in some h a,b generating L(U), since v ∈ U, one of a or b would have to be v ′ , and v ′ ∈ U.But then this h a,b would contain x e,v instead of x e,v ′ .Hence U ′ ⊆ U.
Let us show that U ′ is convex in U. Let v, w ∈ U ′ be such that vT w contains some u ∈ U \ U ′ .By possibly moving v and w closer to u, and u closer to v, we may assume on vT w that v and w are the only vertices in U ′ , and on vT u that v and u are the only vertices in U.But then h v,w ∈ L(U ′ ) and h v,u ∈ L(U) by Lemma 5.11.If L(U ′ ) ⊆ L(U) this could not be the case by Lemma 5.4, since L(U) is regular.Hence, if we have inclusion, U ′ must be convex in U.
Conversely if U ′ is convex in U, then letting S ′ be a tree on U ′ flowing with T , by Lemma 5.11, for each edge e in S ′ we have h e ∈ L(U).
The following is immediate from the above.
In a partition Q of V , if U i and U j are parts such that either U i or U j is not convex in U i ∪ U j , we say that U i and U j are intertwined.
Corollary 7.4.The maximal partitions for the partial order are the partitions Q such that any two parts U i and U j in the partition are intertwined.
Example 7.5.In the introduction, looking at Figure 3, the partition of vertices in Example 1.3 is not maximal.We may join The latter vertex partition is maximal since it is intertwined.Also note that the first partition is not {1, 2, 3, 4, 5}, since {1, 5} and {2} (as well as {4}) are intertwined.
The partition {1, 2, 3, 4, 5} corresponds to the quotient ring k[x E ]/(x e ) 2 e∈E , which is k[x E ] divided by the square of the maximal graded ideal.Hence this ring is not a quotient ring of k[x C ]/(I(T ) + (x a,2 − x d,4 )) of Example 1.2, by a regular linear space.(But it is of course a quotient taking a suitable general linear space.)Corollary 7.6.Let V be the partition of V into singletons.Then for any partition Q of V the interval [V , Q] with respect to the partial order is a Boolean lattice.
Proof.Given a subset U of V , we must show that the lattice of partitions of U into convex parts is a Boolean lattice.Let v be extremal in U in the sense that every other vertex of U is on the same side of v, i.e., there is an edge e = {v, w} from v such that the path from v to any other vertex of U starts with the edge e.Let U ′ = U \ {v}.By induction the lattice of partitions of U ′ into convex subsets is a Boolean lattice B. The partitions Q of U into convex subsets are now of two types: either {v} is a singleton class, or v and w are in the same class.This gives that the lattice of partitions of U identifies as B × {0, 1} and so is Boolean.Proposition 7.7.Let P ′ and P be partitions of E. Then L(P ′ ) ⊆ L(P ) iff each part E i of P is a union of parts of P ′ which are convex for E i .Write then P ′ P .
Corollary 7.8.The maximal partitions for the partial order are the partitions P such that any two parts E i and E j in the partition are intertwined.Example 7.9.In the introduction, looking at Figure 3, the partition of edges in Example 1.3 is not maximal.We may join {a, d} ∪ {b} ∪ {c} {a, d} ∪ {b, c}.
In the latter partition the parts are intertwined and so it is maximal.It correspond to the vertex partition {1, 5} ∪ {3, 4} ∪ {2}.This vertex partition is also maximal (but that does not necessarily follow from the edge partition being maximal).
Corollary 7.10.Let E be the partition of E into singletons.Then for any partition P of E the interval [E, P ] with respect to the partial order is a Boolean lattice.

Hypertree of quotients and triangulated balls
We describe the squarefree quotients of k[x C ]/I(T ) by regular linear spaces whose associated simplicial complex is a triangulated ball.In particular we describe when we get triangulations of polygons. Let be a partition of the edge set E of the tree T .We may think of the edges of E i as a color class.The partition corresponds by Theorem A.1 to a partition V = U 0 ⊔ U 1 ⊔ • • • ⊔ U r of the vertex set into independent sets of vertices.Let S = S 0 ∪ S 1 ∪ • • • ∪ S r where the S i are trees on U i flowing with T .The image ΨS is a forest and each ΨS j is a collection of connected components (trees) of ΨS.Moreover L(P ) = ⊕ r i=1 L(S i ).In the sequel we also write ΨP for ΨS.Let us describe the variables in the quotient ring k[x C ]/L(P ) (this is a polynomial ring).These variables identifiy as subsets of of the incidence relation C.Those subsets which contain more than one element arise as follows.For each class E i consider maximal sets of edges E ij ⊆ E i such that for every pair of edges f, g in E ij , the only edges in E i on the unique path from f to g are f and g themselves.For given i two such maximal E ij and E ij ′ have at most one edge in common.(In fact the E ij 's form the set of edges in a hypertree on We now describe the facets of the simplicial complex corresponding to the quotient k[x C ]/(I(T ) + L(P )).For each v ∈ V and color class E i , let E v i be the set of edges f in E i such that on the path from f to v the only edge in E i is f itself.Then E v i is a maximal set E ij as above and hence gives a variable x E v i in k[x C ]/L(P ).We have x E v i = x E w j iff i = j and there is no edge from E i on the path from v to w.Let Proof.This is clear.
Lemma 8.5.The facets F v and F w have a codimension-one face G in common if and only if the path from v to w has all edges of the same color.Then for all edges e on this path, the F e are equal, and this is G.In particular G is common to all facets F u for u on this path.
Proof.Suppose the edges on the path are all of the same color red.Let the path be v = u 0 , u 1 , . . ., u m = w with e i the edge {u i−1 , u i }.Then for each edge e i for suitable F i .Since e i and e i+1 are successive red edges we divide out by the variable difference x e i ,u i − x e i+1 ,u i and so (e i , u i ) identifies with (e i+1 , u i ).We also have We must then have F i = F i+1 .Hence all these F i are equal.Suppose the edges on the path are not of the same color.Suppose going from v to w there is first a sequence of red edges, the first one being e = {v = u 0 , u 1 } and then eventually a blue edge f = {u i , u i+1 }.
• The facet F v contains (e, u 0 ) of color red.The facet F w also contains a (class) of a red edge.If this red edge was e it would have to be (e, u 1 ).
Hence (e, u 0 ) is in F v but not in F w .• Similarly the blue (f, u i ) is in F v , and by a similar argument as above, (f, u i ) is not in F w .• The upshot is that F v \ F w contains at least two elements, and so F v and F w do not intersect in codimension one.
Recall that a set of edges in the tree T is independent if no two edges in the set are adjacent.The quotient of k[x C ]/I(T ) by L(P ) is a stacked simplicial complex.It is again a quotient of the polynomial ring k[x C ]/L(P ).Each part E i of E is a subforest of T .Let T ij be the trees of this subforest and V ij ⊆ V the support of T ij .Let T ′ be the hypertree whose edges are the sets V ij .In particular note that if P is a partition whose parts E i consist of independent edges, then each T ij is simply an edge, and so T ′ = T .Proposition 8.6.Let P be a partition of the edge set of T .The quotient of k[x C ]/I(T ) by L(P ) corresponds to a stacked simplicial complex X whose associated hypertree is T ′ .Proof.Consider then the tree T ij .Let v, u, w be three vertices in V ij .If they are T -aligned for some ordering, the facets F v , F u , F w of X have a codimension-one face in common by Lemma 8.5.Suppose {u, v, w} are non-aligned.Consider the path from v to w and let e be its last edge.Then e is also the last edge on the path from u to w. Write F w = F ∪ {(e, w)}.By the argument of Lemma 8.5, all x on these paths have F x containing F .We readily get that F is a codimension-one face of every F x for x ∈ V ij .Thus each V ij form an edge in the hypertree T ′ associated to the simplicial complex X.Theorem 8.7.There is a one-to-one correspondene between: • regular linear spaces giving squarefree quotients of k[x C ]/I(T ) corresponding to triangulated balls, and • partitions P of the edge set E of T into sets of independent edges.The codimension-one faces of this triangulation which are on two facets are precisely the faces F e of Lemma 8.4.Let B(T, P ) be the ideal generated by the x Fe for e ∈ E(T ).Then I(T ) + B(T, P ) is the Stanley-Reisner ideal in k[x C ]/L(P ) definining the boundary of this triangulated ball, a triangulated sphere.
Proof.When the edges are partitioned into independent sets, the hypertree T ′ is an ordinary tree T .And when a stacked simplicial complex gives an ordinary tree T , it is a triangulated ball, and may be realized as a stacked polytope.
The only faces on a stacked simplical complex not on the boundary, are the codimension one faces which are on at least two faces.This gives the statement about the Stanley-Reisner ideal of the boundary.Remark 8.8.In [5] the first author et al. give the construction of large classes of triangulated balls, defined by letterplace ideals of posets.The ideal defining the boundary of triangulated balls is given in a similar way there.
In particular triangulations of simplicial polygons correspond to partitions of trees T into three parts, each part being a set of independent edges.Thus only trees T whose maximal vertex degree is 3 arise in this context.
Corollary 8.9.The length of the longest regular sequence of variable differences giving a squarefree quotient of k[x C ]/I(T ) that corresponds to a triangulated ball is |E| − ∆, where ∆ is the maximal degree of a vertex of T .
Proof.This is because the minimal number of parts in a partition of E(T ) into independent edges, the edge chromatic number of the tree T , is the maximal degree of a vertex in T , [1].

Appendix A. Partitions of the vertices and edges of a tree
We recall the basic result on trees from [9] on the correspondence between partitions of edges and partitons of vertices into independent sets.Let T be a tree with vertex set V and edge set E. We consider partitions of the vertices (2) into disjoint sets such that each V i is an independent set of vertices.(This is almost the same as a coloring of vertices, but not quite: The symmetric group S r acts on colorings by permuting the color labels of the V i .So such a partition is an orbit for the actions of S r .The class of such orbits, or equivalently of partitions (2) are also called non-equivalent vertex colorings, see [13].) We also consider partitions of the edges Here we have no independence requirements.Any partition is good.
Now we make a correspondence as follows.Given such a partition of V , make a partition of E as follows: If v and w are vertices consider the unique path in T linking v and w.Let f , respectively g, be the edge incident to v, respectively w, on this path.If (i) v and w are in the same part V i of V and (ii) no other vertex on this path is in the part V i , then put f and g into the same part of E, and write f ∼ ′ E g.The partition of edges is the equivalence relation ∼ E generated by ∼ ′ E , i.e., the smallest equivalence relation on E containing ∼ ′ E .Note that in general ∼ ′ E alone would not be reflexive nor transitive.Conversely, given a partition of the edge set E, make a partition of V as follows: Let v and w be distinct vertices, and consider again the path from v to w.If (i) the edges f and g are distinct, (ii) f and g are in the same part E j , and (iii) no other edge on this path is in the part E j , then put v and w in the same part of V , and write v ∼ ′ V w.The partition of vertices is the equivalence relation ∼ V generated by ∼ ′ V .Theorem A. 1 ([9]).Let T be a tree with vertex set V and edge set E. The above gives a one-to-one correspondence between partitions of the vertices V into r + 1 independent sets, and partitions of the edges E into r sets.
Example A.2. Any tree has a unique partition of the vertices into two independent sets (two colors modulo S 2 ).This corresponds to the partition of the edges into one part (one color).
a. T has ≥ d + 2 vertices, b.If T has an edge of cardinality ≥ 3, then T has ≥ d + 3 vertices c.If T has ≥ d + 3 vertices, X may be separated to a simplical complex X ′ whose tree T ′ is a refinement of T .

Definition 4 . 1 .Lemma 4 . 2 .Corollary 4 . 3 .
Let k[x C ] be the polynomial ring whose variables are indexed by the incidence relation C. The tree ideal I(T ) in k[x C ] associated to the tree T is the edge ideal of ΨT .It is generated by the monomials m v,w = x e,v x f,w , one monomial for each pair of distinct vertices v, w in V .The edges e and f are incident to v and w, respectively, on the path vT w.These tree ideals are introduced in [2, Section 5] (but in a slightly less conceptual setting by indexing the variables by E × {0, 1}).They are shown to be all the possible separated models for the second power (x e | e ∈ E(T )) 2 of the irrelevant maximal ideal in the polynomial ring k[x e ] e∈E(T ) whose variables are indexed by the edges of T .In particular the ideals I(T ) are Cohen-Macaulay and their graded Betti numbers are precisely those of the graded free resolution of the second power (x e | e ∈ E(T )) 2 of the graded maximal ideal of k[x e ] e∈E(T ) .The following is given in [2, Section 5].The facets of the simplicial complex associated to the Stanley-Reisner ideal I(T ) are F v = {(e, w) ∈ C | w vertex on e closest to v}, one facet for each vertex v ∈ V .The cardinality of these facets is then the number of edges of T .The ideal I(T ) defines the unique non-cone stacked simplicial complex with tree T of dimension |E| − 1 with |E| + 1 vertices, given in Proposition 3.7d.A variation of the map Ψ above is Ψ = (τ ) 2 • Ψ where (τ ) 2 : (C) 2 → (C) 2 is derived from the involution τ of C. Considering the path between v and w

Example 5 . 2 .Lemma 5 . 3 .
Consider the second tree in Figure 7.The sequence of vertices 1, 4, 8 is T -aligned, and the set {1, 5, 8} is non-aligned for T .Recall the variable difference h v,w from Definition 4.4.The variables of the polynomial ring k[x C ] (see Definition 4.1) are indexed by the incidence relation C. The variable differences in k[x C ] which are non-zero divisors for k[x C ]/I(T ) are those coming from the edges of ΨT , i.e., the differences h v,w .

Lemma 5 . 14 .
Let G be a graph on vertex set V (with G a priori unrelated to T ).a.If {h e } e∈G is a regular sequence for k[x C ]/I(T ), then G is a forest.b.If G is a forest consisting of the trees S 1 , . . ., S r , then {h e } e∈G is a regular sequence iff each {h e } e∈S i is a regular sequence.Proof.a.It is enough to show that if G is a cycle O then {h e } e∈O is not a regular sequence.Denote by L(O) their linear span, and let the cycle be v 0 , v 1 , . . ., v n−1 , v n = v 0 of length n.

7. 1 .
Partitions of the vertex set.Definition 7.1.Let U ′ ⊆ U ⊆ V .Then U ′ is convex in U if for every v, w ∈ U ′ ,all vertices on the path vT w that are contained in U are in U ′ .

7. 2 .
Partitions of the edge set.If D ′ ⊆ D ⊆ E are sets of edges of T , we may as above define the notion of D ′ being convex in D. As above we may show: Figure 6.

Figure 6 .
The facet F v of k[x C ]/(I(T ) + L(P )) is of cardinality 3. Its elements are the three maximal sets E v red = {c, e, g}, E v blue = {h, b}, E v green = {f }.Lemma 8.3.The facets of the simplicial complex associated to k[x C ]/(I(T ) + L(P )) are the F v 's, for v ∈ V .In particular the cardinality of each facet is the number of classes in the partition P .Proof.This follows by repeated use of Lemma 2.5.Lemma 8.4.Let e = {v, w} be an edge in T , in the class E k .Then E v i = E w i for i = k.The facets F v and F w have a codimension-one face in common.It is the set Figure 7.
Remark 3.3.Stacked simplicial complexes are flag complexes.Every minimal nonface is an edge.Equivalently the Stanley-Reisner ideal is generated by quadratic monomials.A hypergraph is an ordered pair H = (V, E) where V is a set and E is a collection of subsets of V such that no e ∈ E is contained in another e ′ ∈ E. The elements of V are called the vertices of H and the elements of E are called the edges of H.
A hypergraph H is a hypertree if (i) any two edges intersect in either one or zero elements, (ii) H is connected, i.e., for any two vertices v and w in V there is a sequence e 1 , . . ., e m of edges of H with v ∈ e 1 and w ∈ e m and such that for every i ∈ {1, . . ., m − 1} one has e i ∩ e i+1 = ∅, and (iii) H has no cycle, i.e., no sequence of distinct vertices {w} as a component.If there are other components, we may apply Corollary 2.3.Suppose then {w} is the only component.Then X −{w} must be a cone over v. Let Y be the link lk X v.It is stacked, of dimension d − 1 and has ≥ d + 2 facets.By induction we may use Corollary 2.3 and separate Y , using an element v ′ , to