Truncated Boolean Representable Simplicial Complexes
aa r X i v : . [ m a t h . C O ] A p r Truncated boolean representable simplicial complexes
Stuart Margolis, John Rhodes and Pedro V. Silva
April 9, 2019
In this paper, we extend the theory of TBRSC (truncated boolean representable simplicial complexes)created in [14, Sec. 8.2]. The paper is reasonably self-contained, but familiarity with [14] will bevery useful.Matroids [12], BRSC (boolean representable simplicial complexes) and TBRSC, as models of discrete geometry , are all concerned with the generalized notion of independence . All matroids admita boolean representation (usually many), so do BRSC, but not conversely, so BRSC are beyondmatroids [2]. The set of independent subsets of a finite set of points V form a simplicial complex( V, H ) (in the sense of elementary algebraic topology [15]), H being a nonempty collection of subsetsof V closed under taking subsets.We are interested in simplicial complexes S = ( V, H ) arising from a geometry , as to be explainedbelow. In this introduction we only concern ourselves with simple simplicial complexes, i.e. thosesuch that all pairs v v of (distinct) elements of V are in H . But non-simple simplicial complexesare also considered in [14] and several papers.Simple matroids arise through transversals of the partial differences for chains in geometric lat-tices, where we identify the vertices with the atoms of the lattice. More generally, a BRSC may beobtained through transversals of the partial differences for chains in an arbitrary lattice, the role ofatoms being played by any join-generating set [14, Chapter 5].Given a simplicial complex S = ( V, H ), the combinatorial and algebraic fields use the term rank r as the cardinality of the largest set in H . The topological and geometric fields use d = r − dimension of S . We use both as will be explained. We say that S is paving (of dimension d ) andwrite S ∈ Pav( d ) if S has dimension d and H contains all subsets of V of cardinality d . Thus wemay identify the class of (finite) graphs with Pav(1).For (T)BRSC the geometry comes in similarly to matroids as will be explained. Let us restrict tothe paving case for simplicity. Let M = ( V, H ) be a matroid in Pav( d ). For simplicity of explanationwe assume that d = 2, but all generalizes to arbitrary d .1he lattice of flats L ( M ) of M induces a closure operator on 2 V (see [14, Section 4.2] for themost general version). Let X denote the closure of X ⊆ V . Let L = { F ∈ L ( M ) | ≤ | F | < | V |} . If M is a matroid, then each pair of points of V is contained in a unique block in L. This makes La PBD ( pairwise balanced design ) (with λ = 1) in design theory . Conversely, and except for trivialcases, every such PBD determines a matroid in Pav(2) (see e.g [10, Proposition 4.2]).This generalizes to BPav(2) (the BRSC in Pav(2)) as follows (see [14, Section 6.3]). Let S =( V, H ) ∈ BPav(2). Then L = { F ∈ L ( S ) | ≤ | F | < | V |} is a partial geometry since | F ∩ F | ≤ F , F ∈ L. This implies that each pair of points of V is contained in at most one blockin L (pretty much the central notion). If S is not a matroid, this is not enough to produce a PBD,but we get a PBD by adding to L all pairs of points of V contained in no element of L.If S = ( V, H ) is a simplicial complex and k ≥
1, then T k ( S ) = ( V, H ∩ P ≤ k ( V )) is the truncation of S to rank k . Note that we use rank here, not dimension.In general, the truncation of a BRSC is not a BRSC (so we get a wider concept denoted byTBRSC), but it is easily characterized as follows.Given a a simplicial complex S = ( V, H ) of dimension d , we define ε ( S ) = ε ( H ) = { X ⊆ V | ∀ Y ∈ H ∩ P ≤ d ( X ) ∀ p ∈ V \ X Y ∪ { p } ∈ H} . Then transversals of the partial differences for chains in ε ( S ) define a BRSC denoted by S ε .In the paving case, we actually have ε ( S ) = L ( S ε ). Moreover, S ε truncated to the rank of S isthe unique largest TBRSC S contained in S . We mean largest with respect to inclusion of faces(for the same vertex set). This relation is called the weak order in matroid theory [12, Section 7.3].Now S is a TBRSC if and only if T d +1 ( S ε ) = S . This provides a useful criterion to recognize aTBRSC.We note that even if M is a matroid, M ε is not necessarily a matroid (see [10, Examples 5.5 and5.6]). Indeed, ε ( M ) constitutes a lattice under intersection, but not necessarily a geometric lattice.See [10], where this is further developed.We note that, if M is a matroid of rank r , then ε ( M ) consists precisely of the ( r − closed subsets of V in the terminology introduced by Crapo [3]. But TBRSC also shed a new light onCrapo’s concept of erection. An erection of a matroid M = ( V, H ) of rank r is a matroid M ′ of rank ≤ r + 1 such that M = T r ( M ′ ). In [3], Crapo proved that a collection { B , . . . , B k } ⊂ V form acollection of maximal flats of an erection of M if and only if:(1) the closure of each B i in L ( M ) is V ;(2) each B i is ( r − ε ( M ));(3) every facet of M (i.e. maximal subset in H ) is contained in a unique B i .He then proves that the collection of all erections of M (including the trivial erection M ) form alattice for the weak order, and the maximal erection is called the free erection.These conditions were designed to remain in the matroid context. We go beyond matroids to thewider class of TBRSC and consider arbitrary differences of rank, generalizing the work of Crapo andothers [3, 11, 13]. For a TBRSC S , these operations can be viewed as strong maps ( ∨ -maps) of S ε (see [9] and Chapter 5, especially Sections 5.4 - 5.5 and 8.2 of [14]).2he outline of this paper is as follows:In Sections 2 and 3 we provide the basic theory of BRSC and TBRSC, respectively. In Section4 we discuss low dimensions.Section 5 deals with the join operator: ( V, H ) ∪ ( V, H ′ ) = ( V, H ∪ H ′ ). In this paper we use joinreferring to the lattice of all simplicial complexes on a fixed vertex set V , ordered by inclusion. Thisis the same as the lattice of semigroup ideals of the monoid (2 V , ∩ ).In general, (T)BRSC are not closed under join, but the class TBPav( d ) (the TBRSC in Pav( d ))is. A key resource are the complexes B d ( V, L ) (where 2 ≤ d ≤ | L | < | V | ), containing all subsets of V with at most d points and all subsets of V with d + 1 points which intersect L in exactly d points.Note that, given S = ( V, H ) ∈ TBPav( d ), we have B d ( V, L ) ⊆ S if and only if L ∈ ε ( S ).Moreover, S is a TBRSC if and only if S = [ L ∈ L B d ( V, L ) = T d +1 ( S ε )where L ⊆ P ≥ d ( V ) \ { V } . More generally, S = T d +1 ( S ε ) is the largest paving TBRSC containedin S , and is therefore the largest subcomplex of S allowing some geometrical features. This strictlyincludes all paving matroids.In Section 6, we show that the maximum number of vertices for a minimal S ∈ TBPav( d ) \ BPav( d )is ( d + 1)( d + 2). On the other hand, the prevariety T BP (consisting of all paving TBRSC ofdimension ≤
2) is not finitely based.In Section 7, we discuss three questions involving the largest pure subcomplex of a BRSC or ofone of its truncations. We answer them negatively in the general case, but we show them to hold forlow dimensions,Finally, we discuss in Section 8 some of the topological properties of the geometric realization ofa TBRSC, generalizing previous results for BRSC.
For the material presented in this section, the reader is referred to [14]. All the results mentionedhere will be used throughout the paper without further reference.All lattices and simplicial complexes in this paper are assumed to be finite. Given a set V and n ≥
0, we denote by P n ( V ) (respectively P ≤ n ( V ) , P ≥ n ( V )) the set of all subsets of V with precisely(respectively at most, at least) n elements.A (finite) simplicial complex is a structure of the form S = ( V, H ), where V is a finite nonempty setand H ⊆ V contains P ( V ) and is closed under taking subsets. The elements of V and H are calledrespectively vertices and faces . To simplify notation, we shall often denote a face { x , x , . . . , x n } by x x . . . x n .A face of S which is maximal with respect to inclusion is called a facet . We denote by fct( S ) theset of facets of S . The rank and dimension of S are defined respectively byrk( S ) = max {| I | : I ∈ H} , dim( S ) = rk( S ) − . We say that S = ( V, H ) is: • simple if P ( V ) ⊆ H ; 3 paving if P dim ( S ) ( V ) ⊆ V .We denote by Pav( d ) the class of all paving simplicial complexes of dimension d .Two simplicial complexes ( V, H ) and ( V ′ , H ′ ) are isomorphic if there exists a bijection ϕ : V → V ′ such that X ∈ H if and only if Xϕ ∈ H ′ holds for every X ⊆ V .If S = ( V, H ) is a simplicial complex and W ⊆ V is nonempty, we call S | W = ( W, H ∩ W )the restriction of S to W . It is obvious that S | W is still a simplicial complex.A simplicial complex M = ( V, H ) is called a matroid if it satisfies the exchange property :(EP) For all I, J ∈ H with | I | = | J | + 1, there exists some i ∈ I \ J such that J ∪ { i } ∈ H .An important example of matroids are the uniform matroids U k,n : for all 1 ≤ k ≤ n , we write U k,n = ( V, P ≤ k ( V )) with | V | = n .Given an R × V matrix M and Y ⊆ R , X ⊆ V , we denote by M [ Y, X ] the submatrix of M obtained by deleting all rows (respectively columns) of M which are not in Y (respectively X ).A boolean matrix M is lower unitriangular if it is of the form . . .
0? 1 0 . . .
0? ? 1 . . . . . . Two matrices are congruent if we can transform one into the other by independently permutingrows/columns. A boolean matrix is nonsingular if it is congruent to a lower unitriangular matrix.Equivalently, nonsingular matrices can be characterized through the concept of permanent . Thepermanent of a square matrix M = ( m ij ) (a positive version of the determinant) is defined byPer M = X π ∈ S n n Y i =1 m i,iπ . But, even though our matrix is boolean, we compute its permanent in the superboolean semiring SB ,which can be described as the quotient of the usual semiring ( N , + , · ) by the congruence with classes { } , { } , { , . . . } . Then a square boolean matrix is nonsingular if and only if its permanent is 1 in SB .We note that the classical results on determinants involving only a rearrangement of the permu-tations extend naturally to permanents.Given an R × V boolean matrix M , we say that the subset of columns X ⊆ V is M - independent if there exists some Y ⊆ R such that the submatrix M [ Y, X ] is nonsingular.A simplicial complex S = ( V, H ) is boolean representable (BRSC) if there exists some booleanmatrix M such that H is the set of all M -independent subsets of V . Since P ( V ) ⊆ H , this implies4hat all the columns of M are nonzero. Moreover, for all distinct p, q ∈ V , the columns M [ R, p ] and M [ R, q ] are different if and only if pq ∈ H .By restricting the set of columns, it is easy to see that a restriction of a BRSC is still a BRSC[14, Proposition 8.3.1(i)].All matroids are boolean representable [14, Theorem 5.2.10], but the converse is not true.A subset F of 2 V is called a Moore family if V ∈ F and F is closed under intersection (thatis, a Moore family is a submonoid of the monoid of all subsets of V under intersection). EveryMoore family, under inclusion, constitutes a lattice (with intersection as meet and the determinedjoin F ∨ F = ∩{ F ∈ F | F ∪ F ⊆ F } ). We say that X ⊆ V is a transversal of the successivedifferences for a chain F ⊂ F ⊂ . . . ⊂ F k in F if X admits an enumeration x , . . . , x k such that x i ∈ F i \ F i − for i = 1 , . . . , k .If Tr( F ) is the set of transversals of the successive differences for chains in F , then ( V, Tr( F ))constitutes a BRSC. Moreover, every BRSC can be obtained this way by taking as Moore family its lattice of flats (see [14, Chapters 5 and 6]):We say that X ⊆ V is a flat of S = ( V, H ) if ∀ I ∈ H ∩ X ∀ p ∈ V \ X I ∪ { p } ∈ H . The set of all flats of S is denoted by L ( S ). Note that V, ∅ ∈ L ( S ) in all cases, and L ( S ) is indeed aMoore family (and therefore a lattice). Note also that P ≤ d − ( V ) ⊆ L ( S ) for every S ∈ Pav( d ).It follows from [14, Corollary 5.2.7] that a simplicial complex S = ( V, H ) is boolean representableIf and only if the transversals of the successive differences for chains in L ( S ) are precisely the elementsof H .By [14, Proposition 8.3.3(i)], the flats of a BRSC determine flats on any restriction: if F is a flatof a BRSC S = ( V, H ) and W ⊆ V , then F ∩ W ∈ L ( S | W ).The lattice L ( S ) induces a closure operator on 2 V defined by X = ∩{ F ∈ L ( S ) | X ⊆ F } for every X ⊆ V . It follows from the definitions that X = V when X contains a facet of S .By [14, Corollary 5.2.7], S = ( V, H ) is boolean representable if and only if every X ∈ H admitsan enumeration x , . . . , x k satisfying x ⊂ x x ⊂ . . . ⊂ x . . . x k . (1)It is well known that in the case of matroids, this enumeration can be chosen arbitrarily [12]. In this section, we exposit the basic facts about TBRSCs. The proofs of the results can be found in[14, Section 8.2], but we include them in the Appendix for the sake of completeness and conveniencefor the reader.Given a simplicial complex S = ( V, H ) and k ≥
1, the k - truncation of S is the simplicial complex T k ( S ) = ( V, T k ( H )), where T k ( H ) = H ∩ P ≤ k ( V ).5e say that a simplicial complex S = ( V, H ) is a TBRSC if S = T k ( S ′ ) for some BRSC S ′ and k ≥
1. For every d ≥
1, we denote by TBPav( d ) the class of all paving TBRSCs of dimension d .To recognize a TBRSC, it is convenient to develop an alternative characterization. The key isbuilding the flats of a canonical BRSC having our TBRSC as a truncation. The following resultcharacterizes the flats of a truncation with respect to the flats of the original complex. Proposition 3.1 [14, Proposition 8.2.2]
Let S = ( V, H ) be a simplicial complex and let k ≥ . Then L ( T k ( S )) = { X ∈ L ( S ) | fct( T k ( S )) ∩ X = ∅} ∪ { V } . Proof . In the Appendix. (cid:3)
It follows that the lattice of flats of T k ( S ) is obtained from the lattice of flats of S by identifyingthe elements of an up set (namely the subset of flats containing some facet of T k ( S )). In semigroup-theoretic terms, this makes L ( T k ( S )) a Rees quotient of the ∨ -semilattice of L ( S ).For any simplicial complex S = ( V, H ) of dimension d , we define ε ( S ) = ε ( H ) = { X ⊆ V | ∀ Y ∈ H ∩ P ≤ d ( X ) ∀ p ∈ V \ X Y ∪ { p } ∈ H} . Note that ε ( S ) generalizes to arbitrary simplicial complexes what Crapo calls d-closed sets of amonoid in his fundamental paper from 1970 [3].The following lemma is clear from the definition. Lemma 3.2 [14, Lemma 8.2.3]
Let S = ( V, H ) be a simplicial complex. Then:(i) ε ( S ) is a Moore family;(ii) L ( S ) ⊆ ε ( S ) . Given S = ( V, H ), write H ε = Tr( ε ( S )) and let S ε = ( V, H ε ) denote the BRSC defined by thelattice ε ( S ). Lemma 3.3 [14, Lemma 8.2.4]
Let S = ( V, H ) be a simplicial complex of dimension d . Then:(i) T d +1 ( H ε ) ⊆ H ;(ii) ε ( S ) ⊆ L ( S ε ) ;(iii) S ε is a BRSC. Proof . In the Appendix. (cid:3)
Now we can state the main result of this section:
Theorem 3.4 [14, Theorem 8.2.5]
Let S = ( V, H ) be a simplicial complex of dimension d . Then thefollowing conditions are equivalent:(i) S = T d +1 ( S ′ ) for some boolean representable simplicial complex S ′ ;(ii) S = T d +1 ( S ε ) .Furthermore, in this case we have L ( S ε ) = ε ( S ) . Proof . In the Appendix. (cid:3)
6e present now two examples which show that the class of TBRSCs is intermediate between theclass of BRSCs and the class of simplicial complexes. We analyze these examples in the Appendix.The first example shows that a TBRSC is not necessarily a BRSC, even in the paving case.
Example 3.5
Let V = { , . . . , } , H = ( P ≤ ( V ) \ { , , , , , } ) and S = ( V, H ) .Then S ∈ TBPav(2) but is not boolean representable.
The second complex shows that a (paving) simplicial complex is not necessarily a TBRSC.
Example 3.6
Let V = { , . . . , } , H = P ≤ ( V ) ∪ { } and S = ( V, H ) . Then S is not a TBRSC. With respect to the equality L ( S ε ) = ε ( S ) in Theorem 3.4, we can show it holds for all pavingsimplicial complexes: Proposition 3.7
Let d ≥ and S = ( V, H ) ∈ Pav( d ) . Then L ( S ε ) = ε ( S ) . Proof . In the Appendix. (cid:3)
However, the next example, also analyzed in the Appendix, shows that equality may not hold.
Example 3.8
Let S = ( V, H ) be the simplicial complex defined by V = { , . . . , } and H = ( P ≤ ( V ) \ { , , } ) ∪ { } . Then ε ( S ) ⊂ L ( S ε ) . Proposition 4.1
Every TBRSC of dimension 1 is boolean representable.
Proof . Let S = ( V, H ) be a TBRSC of dimension 1. For every a ∈ V , let F a = { a } ∪ { b ∈ V \ { a } | ab / ∈ H} . Every a ∈ V is a transversal of the successive differences for the chain ∅ ⊂ V in L ( S ).Suppose now that a, b ∈ V are distinct and ab ∈ H . Then ab is a transversal of the successivedifferences for the chain ∅ ⊂ F a ⊂ V , so it suffices to show that F a ∈ L ( S ).Since S is a TBRSC, there exists a boolean matrix M with column space V such that, for every X ∈ P ≤ ( V ), we have X ∈ H if and only if X is M -independent. Since P ( V ) ⊆ H , all the columnsof M are nonzero, so X ∈ H if and only if the columns of X are different. Thus F a is the set of all b ∈ V having columns in M equal to the column of a .Let X ∈ H ∩ F a and p ∈ V \ F a . Then | X | ≤ p is different, so X ∪ { p } ∈ H and so F a ∈ L ( S ) as required. (cid:3) Example 3.5 shows that Proposition 4.1 fails for dimension 2, even in the paving case.The next lemma features a class of matroids which is useful to build counterexamples.
Lemma 4.2
Let V be a finite nonempty set and let F ⊆ P ( V ) ∪ P ( V ) be such that F ∩ F ′ = ∅ forall distinct F, F ′ ∈ F . Let H consist of all X ∈ P ≤ ( V ) containing no element of F . Then ( V, H ) isa matroid. Proof . In the Appendix. (cid:3)
The next result shows that, when it comes to separate BRSCs from TBRSCs, Example 3.5 hasthe minimum number of vertices.
Proposition 4.3
Every TBRSC with at most 5 vertices is boolean representable. roof . Let S = ( V, H ) be a TBRSC with | V | ≤
5. In view of Proposition 4.1, we may assume thatdim( S ) ≥ | V | = 3. Then S is the uniform matroid U , , hence a BRSC.Suppose next that | V | = 4. We may assume that dim( S ) = 2, otherwise S = U , . If S := T ( S ′ )for some BRSC S ′ of dimension 3, then S = U , and is therefore a BRSC.Thus we may assume that | V | = 5. If dim( S ) = 4, then S = U , is a BRSC. If dim( S ) = 3 and S = T ( S ′ ) for some BRSC S ′ of dimension 4, then S = U , and is also a BRSC. Hence we mayassume that dim( S ) = 2.Suppose that S is not a BRSC. Then L ( S ) ⊂ ε ( S ). Let Z ∈ ε ( S ) \ L ( S ). Comparing thedefinitions of ε ( S ) and L ( S ), we see that H ∩ P ( Z ) = ∅ , hence we may take a a a ∈ H ∩ P ( Z ).Since a a a ∈ H ⊆ H ε , we may assume that there exists a chain ∅ = Z ⊂ Z ⊂ Z ⊂ Z in ε ( S )such that a i ∈ Z i \ Z i − for i = 1 , ,
3. By Lemma 3.2(i), a a a is also a transversal of the successivedifferences for the chain ∅ ⊂ Z ∩ Z ⊂ Z ∩ Z ⊂ Z ∩ Z in ε ( S ), hence there exists a chain ∅ = Z ′ ⊂ Z ′ ⊂ Z ′ ⊂ Z ′ ⊂ Z ′ = V (2)in ε ( S ). Since | V | = 5, there exists some j ∈ | Z ′ j \ Z ′ j − | = 2. Without loss ofgenerality, we may assume that Z ′ j = Z ′ j − ∪ X ∈ P ( V ) does not contain 12, then X is a transversal of the successive differences for thechain (2), hence X ∈ T ( H ε ) = H . Thus the only possible elements of P ( V ) \ H = P ( V ) \ H ε are123 , , / ∈ H , we have necessarily H = { X ∈ P ≤ ( V ) | X } , because any other 2-subset is contained in some element of H∩ P ( V ). By Lemma 4.2, S is a matroid,hence boolean representable.Thus we may assume that we have one of the following four cases:(C1) H = P ≤ ( V ) \ { , , } ;(C2) H = P ≤ ( V ) \ { , } ;(C3) H = P ≤ ( V ) \ { } ;(C4) H = P ≤ ( V ).Now (C3) and (C4) are clearly both matroids (hence BRSCs). We can show that (C1) is a BRSC bychecking that 34 , ,
45 are flats. Similarly, (C2) is a BRSC because 15 , , ,
45 are flats. Thereforeevery TBRSC with 5 vertices is a BRSC. (cid:3) Join
Given two simplicial complexes S = ( V, H ) and S ′ = ( V, H ′ ) we define the join of S and S ′ as thesimplicial complex S ∨ S ′ = ( V, H ∪ H ′ ) . Notice that given a simplicial complex ( V, H ), then H is just a down set of 2 V under inclusion. Thedown sets of 2 V form a lattice equal to the lattice of semigroup ideals of the monoid (2 V , ∩ ), andthis construction is precisely the join in this lattice. Proposition 5.1
Let S = ( V, H ) and S ′ = ( V, H ′ ) be BRSCs with | V | ≤ . Then S ∨ S ′ is a BRSC. Proof . If dim( S ∨ S ′ ) = 1, we may use Proposition 4.1 and Theorem 5.3. The only other nontrivialcase is dim( S ∨ S ′ ) = 2. But it is easy to check [14, Example 5.2.11] that if | V | = 4 and dim( S ) = 2,then S is a BRSC if and only if |H ∩ P ( V ) | 6 = 1. It follows that if S ∨ S ′ is not a BRSC, then S or S ′ is not a BRSC. (cid:3) The next example shows that neither BRSCs nor TBRSCs are closed under join when we consider5 vertices (even at dimension ≤ Example 5.2
Let V = 12345 . Let S = ( V, P ≤ ( V )) and S = ( V, H ) be defined by H = { X ∈ P ≤ ( V ) | , X } . But things work out better in the paving case:
Theorem 5.3
Let d ≥ and let ( V, H ) , ( V, H ′ ) ∈ TBPav( d ) . Then ( V, H ∪ H ′ ) ∈ TBPav( d ) . Proof . Let R = { Z ∩ Z ′ | Z ∈ ε ( H ) , Z ′ ∈ ε ( H ′ ) } . In view of Lemma 3.2(i), R is a Moore family. Hence ( V, Tr( R )) is a BRSC. We claim that H ∪ H ′ = T d +1 (Tr( R )) . (3)Let X ∈ H . By Theorem 3.4, there exists a chain Z ⊂ Z ⊂ . . . ⊂ Z n (4)in ε ( H ) and an enumeration x , . . . , x n of the elements of X such that x i ∈ Z i \ Z i − for every i . Since V ∈ ε ( H ′ ), then (4) is also a chain in R , hence X ∈ Tr( R ). But | X | ≤ d + 1, thus H ⊆ T d +1 (Tr( R ))and also H ′ ⊆ T d +1 (Tr( R )) by symmetry.Conversely, let X ∈ T d +1 (Tr( R )). Since H , H ′ ∈ Pav( d ), we may assume that | X | = d + 1. Thenthere exists a chain R ⊂ R ⊂ . . . ⊂ R d +1 (5)in Tr( R ) and an enumeration x , . . . , x d +1 of the elements of X such that x i ∈ R i \ R i − for every i .Write R d = Z ∩ Z ′ with Z ∈ ε ( H ) and Z ′ ∈ ε ( H ′ ). Since x d +1 / ∈ R d , we may assume that x d +1 / ∈ Z . Since ( V, H ) ∈ TBPav( d ) yields P ≤ d − ( V ) ⊆ ε ( H ), then ∅ ⊂ x ⊂ x x ⊂ . . . ⊂ x . . . x d − ⊂ Z ⊂ V is a chain in ε ( H ) having X as a transversal of the successive differences. Thus X ∈ H by Theorem3.4 and so (5.3) holds. Note also that P ≤ d ( V ) ⊆ H ⊆ H ∪ H ′ .Therefore ( V, H ∪ H ′ ) = T d +1 ( V, Tr( R )) ∈ TBPav( d ). (cid:3) d ) by BPav( d )in Theorem 5.3. Example 5.4
Let V = 123456 , H = P ≤ ( V ) ∪ { , , , } and H ′ = P ≤ ( V ) ∪ { X ∈ P ( V ) (cid:12)(cid:12) | X ∩ | = 1 } . Then ( V, H ) , ( V, H ′ ) ∈ BPav( d ) but ( V, H ∪ H ′ ) / ∈ BPav( d ) . Let V be a finite nonempty set and let L ⊆ V be such that 2 ≤ d ≤ | L | < | V | . We write B d ( V, L ) = ( V, B d ( V, L )) = ( V, Tr( P ≤ d − ( V ) ∪ { V, L } )) . This is easily seen to be equivalent to the following condition: B d ( V, L ) = P ≤ d ( V ) ∪ { X ∈ P d +1 ( V ) (cid:12)(cid:12) | X ∩ L | = d } . If V is clear from the context, we may omit V from B d ( V, L ) and B d ( V, L ). Lemma 5.5
Let V be a finite nonempty set and let L ⊆ V be such that ≤ d ≤ | L | < | V | . Then B d ( V, L ) ∈ BPav( d ) . Proof . It is immediate that P ≤ d − ( V ) ∪ { V, L } ⊆ L ( B d ( L )), hence every X ∈ B d ( L ) is a transversalof the successive differences for some chain in L ( B d ( L )). Thus B d ( L ) ∈ BPav( d ). (cid:3) We can now prove the following result, characterizing TBPav( d ). Theorem 5.6
Let d ≥ and S = ( V, H ) ∈ Pav( d ) . Then the following conditions are equivalent:(i) S ∈ TBPav( d ) ;(ii) S = ∨{ B d ( V, L ) | L ∈ L } for some nonempty L ⊆ P ≥ d ( V ) \ { V } . Proof . (i) ⇒ (ii). Let L = ( P ≥ d ( V ) \ { V } ) ∩ ε ( S ). Since dim( S ) = d , we have L = ∅ .Let X ∈ H . Since P ≤ d ( V ) ⊆ B d ( L ) for every L ∈ L, we may assume that | X | = d + 1. ByTheorem 3.4, there exists a chain Z ⊂ Z ⊂ . . . ⊂ Z d +1 in ε ( S ) and an enumeration a , . . . a d +1 of the elements of X so that a i ∈ Z i \ Z i − for i = 1 , . . . , d + 1.Now a . . . a i ∈ L ( S ) ⊆ ε ( S ) for i = 0 , . . . , d −
1, hence X is a transversal of the chain ∅ ⊂ a ⊂ a a ⊂ . . . a . . . a d − ⊂ Z d ⊂ V and so X ∈ B d ( T d ). Since Z d ∈ L, we get
H ⊆ ∨{ B d ( L ) | L ∈ L } .The opposite inclusion is immediate.(ii) ⇒ (i). By Lemma 5.5 and Theorem 5.3. (cid:3) In such a decomposition ( S = ∨{ B d ( V, L ) | L ∈ L } ), we may refer to the elements of L as lines .The following lemma shows that the decomposition provided by Theorem 5.6 is not unique. Lemma 5.7
Let d ≥ and let V be a finite set with | V | ≥ d + 1 . For every a ∈ V , we have B d ( V, V \ { a } ) = [ L ∈ L a B d ( V, L ) , (6) where L a = { L ∈ P d ( V ) | a ∈ L } . roof . It suffices to show that both sides of (6) contain the same X ∈ P d +1 ( V ). So let X ∈ P d +1 ( V ).Suppose that X ∈ B d ( V, V \ { a } ). Then | X ∩ ( V \ { a } ) | = d , hence a ∈ X . Take b ∈ X \ { a } .Then X \ { b } ∈ L a and so X ∈ B d ( V, X \ { b } ) ⊆ [ L ∈ L a B d ( V, L ) . Conversely, suppose that X ∈ B d ( V, L ) with L ∈ L a . Since | X | = d + 1 and | L | = d , we must have X = L ∪ { c } for some c ∈ V \ L . Hence a ∈ L ⊂ X yields | X ∩ ( V \ { a } ) | = d and X ∈ B d ( V, V \ { a } ).Therefore (6) holds as required. (cid:3) We prove next a version of Theorem 5.6 for BPav( d ). Theorem 5.8
Let d ≥ and S = ( V, H ) ∈ Pav( d ) . Then the following conditions are equivalent:(i) S ∈ BPav( d ) ;(ii) S = ∨{ B d ( V, L ) | L ∈ L ( S ) \ ( P ≤ d − ( V ) ∪ { V } ) } ;(iii) S = ∨{ B d ( V, L ) | L ∈ L } for some nonempty L ⊆ P ≥ d ( V ) \ { V } satisfying | L ∩ L ′ | ≤ d − for all distinct L, L ′ ∈ L . (7) Proof . (iii) ⇒ (i). Since P ≤ d ( V ) ⊆ H , we have P ≤ d − ( V ) ⊆ L ( S ). Let K ∈ L and suppose that X ∈ H ∩ K and p ∈ V \ K . Since P ≤ d ( V ) ⊆ H ⊆ P ≤ d +1 ( V ), we may assume that | X | = d or d + 1.Suppose that | X | = d + 1. Since X ∈ H = ∪ L ∈ L B d ( L ), we have X ∈ B d ( L ) for some L ∈ L . Thus | X ∩ L | = d and so | K ∩ L | ≥ d . In view of (7), we get K = L , hence X ⊆ L , a contradiction since | X | = d + 1 and | X ∩ L | = d . Therefore | X | = d , hence X ∪ { p } ∈ B d ( K ) ⊆ H and so K ∈ L ( S ).Let a , . . . , a d − ∈ V be distinct. Then ∅ ⊂ a ⊂ a a ⊂ . . . ⊂ a . . . a d − ⊂ V (8)is a chain in L ( S ). If a , . . . , a d − ∈ L ∈ L, then (8) can be refined to ∅ ⊂ a ⊂ a a ⊂ . . . ⊂ a . . . a d − ⊂ L ⊂ V, (9)also a chain in L ( S ). It is easy to check that every X ∈ H is a partial transversal of the successivedifferences for some chain of type (8) or (9), hence S is boolean representable.(i) ⇒ (ii). Since P ≤ d − ( V ) ∪ { V } ⊆ L ( S ) and the maximum length of a chain in L ( S ) is d + 1, it follows easily that the maximal chains in L ( S ) must be of the form (8) or (9), with L ∈ L ( S ) \ ( P ≤ d − ( V ) ∪ { V } ). Thus (ii) holds.(ii) ⇒ (iii). Suppose that L, L ′ ∈ L ( S ) are distinct and satisfy | L ∩ L ′ | ≥ d . We may assume that L ∩ L ′ ⊂ L . Let a , . . . , a d − ∈ L be distinct. Since P ≤ d − ( V ) ⊆ L ( S ), we get a chain of length d + 2 ∅ ⊂ a ⊂ a a ⊂ . . . ⊂ a . . . a d − ⊂ L ∩ L ′ ⊂ L ⊂ V in L ( S ), contradicting dim H = d . Therefore | L ∩ L ′ | ≤ d − (cid:3) orollary 5.9 Let d ≥ and and let V be a finite set with | V | ≥ d + 1 . Let ∅ 6 = L ⊆ V be suchthat | L | ∈ { d, d + 1 , | V | − } for every L ∈ L . Then ∨{ B d ( V, L ) | L ∈ L } is boolean representable. Proof . In view of Lemma 5.7, we may assume that | L | ∈ { d, d + 1 } for every L ∈ L. Let
L, L ′ ∈ Lbe distinct. It is easy to check that B d ( L ) ∪ { L } = ∪{B d ( L \ { a } ) | a ∈ L } (10)holds for every L ∈ P d +1 ( V ). Now we may use (10) for replacing L by some equivalent L ′ satisfying(7):(1) If L, L ′ ∈ L ∩ P d +1 ( V ) are such that | L ∩ L ′ | = d , we replace B d ( L ) ∪ B d ( L ′ ) by( ∪{B d ( L \ { a } ) | a ∈ L } ) ∪ ( ∪{B d ( L ′ \ { a } ) | a ∈ L ′ } ) . (2) If L ∈ L ∩ P d +1 ( V ) and L ′ ∈ P d ( L ), we replace B d ( L ) ∪ B d ( L ′ ) by ∪{B d ( L \ { a } ) | a ∈ L } .Indeed, these replacements are legitimate in view of (10), and each such replacement decreases thenumber of L ∈ L ∩ P d +1 ( V ). Eventually, we end up with some L ′ satisfing (7). By Theorem 5.8, ourcomplex is boolean representable. (cid:3) Another way of ensuring closure under join is by restricting the type of complexes in BPav( d ).We define, for every d ≥ V with at least d + 2 elements, Y ( V ) = { ( V, H ) ∈ BPav( d ) | ( V, H ) has no restriction isomorphic to U d,d +2 } . Proposition 5.10
Let d ≥ and let V be a finite set with at least d + 2 elements. Then ( V, H ) , ( V, H ) ∈ Y ( V ) implies ( V, H ∪ H ) ∈ Y ( V ) . Proof . We show that Fl( V, H i ) ⊆ P ≤ d +1 ( V ) ∪ { V } . (11)Let i ∈ { , } and F ∈ L ( V, H i ). Suppose that d + 1 < | F | < | V | . Since P d ( V ) ⊆ H i and norestriction of ( V, H i ) to a d -subset of F is isomorphic to U d,d +2 , there exists some X ∈ P d +1 ( F ) ∩ H i .But then F contains a facet of ( V, H i ) and so F = V , a contradiction. Therefore (11) holds.Now let L i = { F ∈ L ( V, H i ) | d ≤ | F | < | V |} . By (11), we have L i = { F ∈ L ( V, H i ) (cid:12)(cid:12) | F | ∈ { d, d + 1 } } . (12)Since ( V, H i ) ∈ BPav( d ), we have H i = [ L ∈ L i B d ( V, L ) by Theorem 5.8. Thus H ∪ H = ∪{B d ( V, L ) | L ∈ L ∪ L } and so ( V, H ∪ H ) ∈ BPav( d ) by (12) and Proposition 5.9.Suppose that there exists some W ∈ P d +2 ( V ) such that P d +1 ( W ) ∩ ( H ∪ H ) = ∅ . Then P d +1 ( W ) ∩ H = ∅ , contradicting ( V, H ) ∈ Y ( V ). Therefore ( V, H ∪ H ) ∈ Y ( V ). (cid:3) S ∈ Pav( d ), there is a largest pavingTBRSC contained in S , and it is precisely T d +1 ( S ε ). Theorem 5.11
Let S = ( V, H ) ∈ Pav( d ) and τ ( S ) = { ( V, H ′ ) ∈ TBPav( d ) | H ′ ⊆ H} ∪ { ( V, P ≤ d ( V )) } . Then:(i) there exists some (unique) S = ( V, H ) ∈ τ ( S ) such that H ′ ⊆ H for every ( V, H ′ ) ∈ τ ( S ) ;(ii) S = T d +1 ( S ε ) . Proof . (i) Let H = ∪{H ′ | ( V, H ′ ) ∈ τ ( S ) } . Clearly, P ≤ d ( V ) ⊆ H ′ for every ( V, H ′ ) ∈ τ ( S ). Inview of Theorem 5.3, it follows that ( V, H ) ∈ τ ( S ) and we are done.(ii) By Lemma 3.3, T d +1 ( S ε ) is a TBRSC and T d +1 ( H ε ) ⊆ H . Since S ∈ Pav( d ), we have P ≤ d − ( V ) ∪ { V } ⊆ ε ( S ) and so P ≤ d ( V ) ⊆ H ε . Hence T d +1 ( S ε ) is paving and T d +1 ( S ε ) ∈ τ ( S ).Therefore T d +1 ( H ε ) ⊆ H .To prove the opposite inclusion, we may assume that dim( S ) = d (otherwise H = P ≤ d ( V ) ⊆ H ε and we are done). It follows from Theorem 3.4 that S = T d +1 (( S ) ε ), so it suffices to show that( H ) ε ⊆ H ε , which follows itself from ε ( S ) ⊆ ε ( S ). We prove the latter inclusion.Let Z ∈ ε ( S ). Then ∀ X ∈ H ∩ P ≤ d ( Z ) ∀ p ∈ V \ Z X ∪ { p } ∈ H . Suppose that X ∈ H ∩ P ≤ d ( Z ) and p ∈ V \ Z . Since S ∈ Pav( d ), we have X ∈ H ∩ P ≤ d ( Z ) and Z ∈ ε ( S ) yields X ∪ { p } ∈ H ⊆ H . Thus Z ∈ ε ( S ) and so ε ( S ) ⊆ ε ( S ) as required. (cid:3) The next example, analyzed in the Appendix, shows that the paving requirement for subcom-plexes cannot be removed from the definition of τ ( S ). Example 5.12
Let S = ( V, H ) ∈ Pav(3) be defined by V = { , . . . , } and H = P ≤ ( V ) ∪ { abc | a ∈ , b ∈ , c ∈ } . Then S has no largest truncated boolean representable subcomplex. ( d ) \ BPav ( d ) We proved in Proposition 4.3 that we need at least 6 points to separate TBPav(2) from BPav(2).This section starts with a full account of the 6 point case.
Proposition 6.1
Up to isomorphism, the complexes with 6 points in TBPav (2) \ BPav (2) are of theform (123456 , H ) for:(1) H = B (1234) ∪ B (12) ;(2) H = B (1234) ∪ B (12) ∪ B (15) ;(3) H = B (1234) ∪ B (12) ∪ B (15) ∪ B (25) ;(4) H = B (1234) ∪ B (12) ∪ B (35) ;(5) H = B (1234) ∪ B (12) ∪ B (15) ∪ B (35) .Moreover, all the above 5 cases are nonisomorphic. Proof . In the Appendix. (cid:3) ❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡❡ (2) ❙❙❙❙❙❙❙❙❙❙❙❙❙❙❙❙❙❙❙❙ (4) ❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦ (1) ❙❙❙❙❙❙❙❙❙❙❙❙❙❙❙❙❙❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦ ( V, B (1234) ∪ { } ) ❙❙❙❙❙❙❙❙❙❙❙❙❙❙❙ ( V, B (1234) ∪ { } ) ❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦ ( V, B (1234))The missing triangles in the three lowest elements are respectively124 , , , , , , , , , , , , , , , , , , , , • ( V, B (1234)) ∈ BPav(2) by Lemma 5.5. • ( V, B (1234) ∪ { } ) / ∈ TBPav2. Indeed, suppose that there exist Z ∈ ε ( B (1234) ∪ { } )such that | Z ∩ | = 2. Since 124 , , / ∈ H , we successively get 4 ∈ Z and 1234 ⊆ Z , acontradiction. In view of Theorem 3.4, this implies ( V, B (1234) ∪ { } ) / ∈ TBPav2. • ( V, B (1234) ∪ { } ) / ∈ TBPav2. Similar to the preceding case. • No simplicial complex isomorphic to (4) embeds in (3). To prove this, recall the missingtriangles in (3) and (4). We can check that 3 x is contained in a missing triangle of (3) for every x = 3. Similarly, 4 y is contained in a missing triangle of (3) for every y = 4. Suppose that ϕ ∈ S is such that the isomorphic image of (3) through ϕ (call it (3”)) has (4) as subcomplex.Then the missing triangles of (3”) are a proper subset of the missing triangles of (4). Hence(3 ϕ ) x is contained in a missing triangle of (4) for every x = 3 ϕ , and (4 ϕ ) y is contained ina missing triangle of (4) for every y = 4 ϕ . However, only 4 satisfies this property, yielding3 ϕ = 4 = 4 ϕ , a contradiction.Note also that an arbitrary S ∈ TBPav( d ) \ BPav( d ) needs not having a restriction isomorphicto U d,d +2 . The complexes featuring Proposition 6.1 constitute all counterexamples for d = 2.We intend now to show that TBPav( d ) \ BPav( d ) is in some sense finitely generated. We startwith a couple of lemmas.Let T BR (respectively
T BP ) denote the class of all finite truncated boolean representable sim-plicial complexes (respectively finite paving truncated boolean representable simplicial complexes).A class of simplicial complexes closed under isomorphism and restriction is called a prevariety .For details on prevarieties, see [14, Sections 8.4 and 8.5].14 emma 6.2
The classes
T BR and
T BP are prevarieties of simplicial complexes.
Proof . In the Appendix. (cid:3)
Let S ∈ TBPav( d ) \ BPav( d ). By Lemma 6.2, every restriction of S is in T BP (with possiblylower dimension). We say that S is minimal if every proper restriction of S is boolean representable. Lemma 6.3
Let d ≥ . Then the maximum number of vertices for a minimal S ∈ TBPav( d ) \ BPav( d ) is ( d + 1)( d + 2) . Proof . Let S = ( V, H ) ∈ TBPav( d ) \ BPav( d ) be minimal. Hence S / ∈ BPav( d ) but every properrestriction of S is boolean representable. By [14, Theorem 8.5.2(ii)], we get | V | ≤ ( d + 1)( d + 2).Now we consider the Swirl , the simplicial complex defined in the proof of [14, Theorem 8.5.2(ii)],where it is proved that every proper restriction of this complex is boolean representable, but theSwirl is not. The Swirl is defined as follows:Let A = { a , . . . , a d } and B i = { b i , . . . , b id } for i = 0 , . . . , d . Write also A i = A \ { a i } and C i = P d +1 ( A i ∪ ( B i \ { b i } )) ∪ { B i } . We define V = A ∪ d [ i =0 B i , H = P ≤ d +1 ( V ) \ d [ i =0 C i . It is easy to check that all the X ∈ H ∩ P d +1 ( V ) fall into four cases (not necessarily disjoint):(a) there exist b ij , b kℓ ∈ X with i = k ;(b) there exist b i , a j ∈ X ;(c) there exist b ij , a i ∈ X with j > X = a . . . a d .Define L = { L ∈ P d ( V ) | there exist some b ij , b kℓ ∈ L with i = k }∪ { L ∈ P d ( V ) | there exist some b i , a j ∈ L }∪ { L ∈ P d ( V ) | there exist some b ij , a i ∈ L with j > }∪ { A i ∪ B i | i = 0 , . . . , d } . It is straightforward to check that H = ∪{B d ( L ) | L ∈ L } , hence S ∈ TBPav( d ) by Theorem 5.6.Since | V | = ( d + 1)( d + 2)), we have found some minimal S ∈ TBPav( d ) \ BPav( d ) with ( d + 1)( d + 2)vertices as required. (cid:3) Let V be a prevariety of simplicial complexes. We say that V is finitely based if there exists some m ≥ V admits a restriction not in V with at most m vertices.Given a prevariety V of simplicial complexes and d ∈ N , we define the prevariety V d = { S ∈ V | dim( S ) ≤ d } . Let BP denote the class of all finite paving boolean representable simplicial complexes. By [14,Theorem 8.5.2], BP d is finitely based for every d ≥
1. Since
T BP = BP by Proposition 4.1, itfollows that T BP is finitely based. Theorem 6.4
T BP is not finitely based. roof . It suffices to build arbitrary large simplicial complexes not in T BP with all proper restric-tions in T BP .Suppose that S = ( V, H ) ∈ Pav(2). Then P ≤ ( V ) ⊆ L ( S ) ⊆ ε ( S ), and so S ∈ T BP if and only if, for every X ∈ H ∩ P ( V ),there exists some Z ∈ ε ( S ) such that | X ∩ Z | = 2 . (13)Let n ≥ V = { x , . . . , x n , y , . . . , y , z , . . . , z } , where we identify x = y = z , x = z = y , y = z = x n . Let T = { x i x i +1 x i +2 | i = 0 , . . . , n − } ∪ { y i y i +1 y i +2 | ≤ i ≤ } ∪ { z i z i +1 z i +2 | ≤ i ≤ } , H = P ≤ ( V ) \ T and S = ( V, H ). We show next that:for every X ∈ ( H ∩ P ( V )) \ { x x y } , there exists some F ∈ L ( S ) such that | X ∩ F | = 2 . (14)Indeed, such an X contains necessarily some element of V \ x x y . Without loss of generality, wemay assume that this element is among x , . . . , x n − (the other cases follow by symmetry).Suppose that X ⊂ x . . . x n , say X = x i x j x k with i < j < k . Since X ∈ H , then i, j, k are notconsecutive integers. If k < n , then k > k − i >
2, hence x i x k ∈ L ( S ) and we are done. Thuswe may assume that k = n . If i >
1, then k − i >
2, hence x i x k ∈ L ( S ) and we are done. Thus wemay assume that i ≤
1. Since k = n , this implies 2 ≤ j ≤ n −
1. Since n ≥
6, we get either k − j > x j x k ∈ L ( S )) or j − i > x i x j ∈ L ( S )).Hence we may assume that at least one of the other elements of X (say a ) is not of the form x j .Let i ∈ { , . . . , n − } be such that x i ∈ X . It is easy to check that x i a ∈ L ( S ). Therefore (14) holds.Next we show that for every Z ∈ ε ( S ), | Z ∩ { x x y }| 6 = 2. (15)Let Z ∈ ε ( S ) and assume that | Z ∩ { x x y }| ≥
2. Assume first that x , x ∈ Z . Since x i x i +1 x i +2 / ∈ H for i = 0 , . . . , n −
2, we get successively x ∈ Z, . . . , x n ∈ Z . Since x n = y , we get x x y ⊆ Z .Suppose now that x , y ∈ Z . Since x = y , we use the same argument to deduce that y ∈ Z, . . . , y ∈ Z . Since y = x , we get x x y ⊆ Z .Finally, suppose that x , y ∈ Z . Since x = z and y = z , we use the same argument to deducethat z ∈ Z, . . . , z ∈ Z . Since z = x , we get x x y ⊆ Z and (15) is proved.In view of (13), it follows from (15) that S / ∈ T BP .Fix now v ∈ V and write W = V \ { v } . We must show that S | W ∈ T BP (since T BP is closedunder restrictions, this implies that S | W ′ ∈ T BP for any W ′ ⊂ V ).Since S ∈ Pav(2), we only need to show that the righthand side of (13) holds when we replace S by S | W . Let X ∈ H ∩ P ( W ). Suppose first that X = x x y . By (14), there exists some F ∈ L ( S )such that | X ∩ F | = 2. It follows that F ∩ W ∈ L ( S | W ). Since | X ∩ ( F ∩ W ) | = 2, the desiredcondition is satisfied if X = x x y . 16hus we may assume that X = x x y . It follows that either v = x i with 2 ≤ i ≤ n − v = y j or v = z j with 2 ≤ j ≤ v = x i . Let Z = x . . . x i − . It is immediate that Z ∈ ε ( S | W ) and | Z ∩ x x y | = 2.If v = y j (respectively v = z j ), we take Z = y . . . y j − (respectively Z = z . . . z j − ). Therefore, inview of (13), we get S | W ∈ T BP as required. (cid:3) Let S = ( V, H ) be a simplicial complex of dimension d . We say that S is pure if all the facets of S have dimension d .We define pure( S ) = ( V ′ , H ′ ) by V ′ = ∪ ( H ∩ P d +1 ( V )) , H ′ = ∪{ X | X ∈ H ∩ P d +1 ( V ) } . It is immediate that pure( S ) is the largest pure subcomplex of S , also called the pure core of S .This section is devoted to the following questions, all related to the concept of pure core: Problem 7.1
Let S be a BRSC. Is pure( S ) a BRSC? Problem 7.2
Let S be a BRSC and let k ≥ . Is pure( T k ( S )) a BRSC? Problem 7.3
Let H be a BRSC and let k ≥ . Is pure( T k ( S )) a TBRSC? Note that Problem 7.3 admits the equivalent statement:If S is a TBRSC and k ≥
1, is pure( T k ( S )) a TBRSC? (16)For the nontrivial implication, let S be a TBRSC of rank r and assume that Problem 7.3 has apositive answer. Since S is a TBRSC, we have S = T r ( S ′ ) for some BRSC S ′ .Suppose first that k ≥ r . Then T k ( S ) = S , hence we must show that pure( S ) is a TBRSC. Since S = T r ( S ′ ), our goal follows from the answer of Problem 7.3 for S ′ and r .Assume now that k < r . It is easy to check that T k ( S ) = T k ( S ′ ). Since the answer of Problem 7.3implies that pure( T k ( S ′ )) is a TBRSC, then pure( T k ( S )) is a TBRSC and so (16) has also a positiveanswer.Since matroids are pure and closed under restriction [12], all questions have positive answers formatroids. We show that none of them admits a positive answer in general, but we establish particularcases.The following example, analyzed in the Appendix, answers Problem 7.1 negatively for dimension3. It also answers Problems 7.2 and 7.3 for dimension 3 and k = 4. Example 7.4
Let V = { , . . . , } and R = {∅ , , , , , , , , V } . Then S = ( V, Tr( R )) isa BRSC but pure( S ) is not a TBRSC. However, we can find positive answers for all the problems in particular cases as we shall see.The following lemma will prove useful:
Lemma 7.5
Let S = ( V, H ) be a simplicial complex and let I, J ∈ H be such that I ⊆ J . Then thereexists some I ′ ∈ H such that I ⊆ I ′ and I ′ = J . Proof . In the Appendix. (cid:3) M = ( V, H ) is a matroid if and only iffor all X, Y ∈ H , X = Y implies | X | = | Y | . (17)Indeed, if X = Y and | X | > | Y | , it follows easily from the exchange property that Y ∪ { x } ∈ H forsome x ∈ X \ Y , hence Y ∪ { x } = Y . In the case of matroids, the enumeration in (1) can be chosenarbitrarily (see [12]). By taking x as last, we reach a contradiction. Thus | X | = | Y | .Conversely, suppose that (17) holds. Let I, J ∈ H be such that | I | = | J | + 1. Suppose that J ∪ { i } / ∈ H for every i ∈ I \ J . Then I ⊆ J and so by Lemma 7.5 this contradicts (17). Thus M satisfies the exchange property and is therefore a matroid.We define a simplicial complex S = ( V, H ) to be a near-matroid if X = Y ⊂ V implies | X | = | Y | for all X, Y ∈ H . The rank function ρ : L ( S ) \ { V } → N is defined by F ρ = | X | , where X ∈ H is such that X = F. Note that such an X exists by [14, Proposition 4.2.4].It follows from (17) that every matroid is a near-matroid. The following result shows that theconverse fails, in fact a near matroid needs not be boolean representable. Proposition 7.6
Let S be a simplicial complex of dimension d ≥ .(i) If S is paving, then S is a near-matroid.(ii) If S is boolean representable and d ≤ , then S is a near-matroid. Proof . (i) Write S = ( V, H ) and suppose that X, Y ∈ H are such that X = Y ⊂ V . By [14,Proposition 4.2.3], X and Y are not facets (it is easy to check that the closure of a facet must be V ). Suppose that | X | < d . Since P d ( V ) ⊆ H , it follows that X = X , so in this case we get indeed Y = X . Thus we may assume by symmetry that | X | , | Y | ≥ d . Since X and Y are not facets, then | X | = d = | Y | and so S is a near-matroid.(ii) Let X, Y ∈ H be such that X = Y ⊂ V . Since ∅ = ∅ and the closure of a facet is V , we mayassume that X, Y / ∈ fct( S ) ∪ {∅} .Assume first that X = { a } . Let M be an R × V boolean matrix representing S . If C a = { b ∈ V | M [ R, a ] = M [ R, b ] } , then C a ⊆ a . Moreover, J ∈ H ∩ C a implies | J | ≤
1, and since all thecolumns of M are nonzero (we have P ( V ) ⊆ H ), it follows that a = C a . Thus Y = X = C a yields | Y | = 1 = | X | .Assume now that | X | >
1. By the previous case, we also have | Y | >
1. Since dim( S ) ≤ X, Y / ∈ fct( S ), we have necessarily | X | = 2 = | Y | . Therefore S is a near-matroid. (cid:3) Before discussing boolean representable near-matroids, we present two lemmas.
Lemma 7.7
Let S = ( V, H ) be a near-matroid and let F, F ′ ∈ L ( S ) be such that F ⊂ F ′ ⊂ V . Then F ρ < F ′ ρ . Proof . In the Appendix. (cid:3) emma 7.8 Let S = ( V, H ) be a near-matroid and let F, F ′ ∈ L ( S ) be such that F ⊂ F ′ ⊂ V . Let a ∈ F ′ \ F and k = F ′ ρ − F ρ . Then there exist a , . . . a k ∈ V such that F ⊂ F ∪ a ⊂ F ∪ a a ⊂ . . . ⊂ F ∪ a . . . a k = F ′ . Proof . In the Appendix. (cid:3)
Theorem 7.9
Let S be a boolean representable near-matroid and let k ≥ . Then:(i) T k ( S ) is a BRSC;(ii) pure( T k ( S )) is a BRSC. Proof . (i) Write S = ( V, H ) and let X denote the closure of X ⊆ V in L ( S ). We define F k = { F ∈ L ( S ) | F ρ < k } ∪ { V } . Since L ( S ) is closed under intersection, it follows from Lemma 7.7 that F k is a Moore family. Hence( V, Tr( F k )) is a BRSC. We show that T k ( S ) = ( V, Tr( F k )).Let X ∈ T k ( H ) and let s = | X | . Then there exists an enumeration a , . . . , a s of the elements of X such that a ⊂ a a ⊂ . . . ⊂ a . . . a s . Hence X is a transversal of the successive differences for ∅ ⊂ a ⊂ a a ⊂ . . . ⊂ a . . . a s − ⊂ V, which is a chain in F k . Thus X ∈ Tr( F k ).Conversely, assume that X ∈ Tr( F k ). Since F k ⊆ L ( S ), it follows that X ∈ H .Suppose that | X | > k . Since X ∈ Tr( F k ), there exist some F ∈ F k and x ∈ X such that F ∩ X = X \ { x } . But F = Y for some Y ∈ T k − ( H ). Hence X \ { x } ⊆ Y and by Lemma 7.5 thereexists some Z ∈ H such that X \ { x } ⊆ Z and Z = Y = F . But then | Z | ≥ | X \ { x }| ≥ k > | Y | , acontradiction since S is a near-matroid. Thus | X | ≤ k and so T k ( H ) = Tr( F k ) as claimed.(ii) Let F ′ k denote the set of all flats of S occurring in chains of the form F ⊂ F ⊂ . . . ⊂ F k in F k . We claim that F ′ k is a Moore family.Let F, F ′ ∈ F ′ k . We may assume that F, F ′ = V . We have F ∩ F ′ ∈ F k since F k is a Moorefamily. Since F ∈ F ′ k \ { V } , there exists some F ′′ ∈ L ( S ) such that F ⊆ F ′′ and F ′′ ρ = k −
1. Nowwe apply Lemma 7.8 to both inclusions ∅ ⊆ F ∩ F ′ ⊆ F ′′ . This ensures that F ∩ F ′ will appear insome chain of flats of length k in L ( S ) of the form ∅ ⊂ . . . ⊂ F ′′ ⊂ V. Since F ′′ ρ = k −
1, it follows from Lemma 7.7 that this is in fact a chain in F k and therefore in F ′ k .Thus F ∩ F ′ ∈ F ′ k . Since V ∈ F ′ k , then F ′ k is a Moore family. Writing V ′ = ∪F ′ k , it follows that( V ′ , Tr( F ′ k )) is a BRSC. We claim that pure( T k ( S )) = ( V ′ , Tr( F ′ k )).19et X ∈ H ∩ P k ( V ). Then there exists an enumeration a , . . . , a k of the elements of X such that a ⊂ a a ⊂ . . . ⊂ a . . . a k . Hence X is a transversal of the successive differences for ∅ ⊂ a ⊂ a a ⊂ . . . ⊂ a . . . a k − ⊂ V, which is a chain of length k in F k . Thus X ∈ Tr( F ′ k ).Conversely, assume that X ∈ Tr( F ′ k ). We may assume that X is a facet of ( V ′ , Tr( F ′ k )). Thenthere exists some chain F ⊂ F ⊂ . . . ⊂ F s (18)in F ′ r and some enumeration a , . . . , a s of the elements of X such that a i ∈ F i \ F i − for i = 1 , . . . , s .Since X is a facet, we must have F = ∅ and F s = V . Suppose that F s − ρ = r < k −
1. Since F s − ∈ F ′ k , then it must occur in some chain of length k in F ′ k , hence we have some chain F s − = F ′ ⊂ F ′ ⊂ . . . ⊂ F ′ t ⊂ F ′ t +1 = V in F ′ k for some t ≥
1. Since a s ∈ F ′ t +1 \ F ′ , we have a s ∈ F ′ j \ F ′ j − for some j ∈ { , . . . , t + 1 } , hencethere exists some Y ∈ Tr( F ′ k ) ∩ P s + t ( V ) containing (strictly) X , contradicting X ∈ fct( V ′ , Tr( F ′ k )).Thus F s − ρ = k − a i ∈ F i \ F i − for i = 1 , . . . , s − s − k in L ( S ) of the form F ⊂ F ∪ a ⊆ . . . ⊆ F ⊂ F ∪ a ⊆ . . . ⊆ F s − ⊂ F s , which admits a transversal of the successive differences containing X . Since X ∈ fct( V ′ , Tr( F ′ k )),it follows that s = k and so in view of Lemma 7.7 we have X ∈ H ∩ P k ( V ), hence X is a facet ofpure( T k ( S )). Therefore pure( T k ( S )) = ( V ′ , Tr( F ′ k )) as claimed. (cid:3) Together with Proposition 7.6, this yields:
Corollary 7.10
Problems 7.1, 7.2 and 7.3 have positive answers for boolean representable near-matroids. In particular, they hold for:(i) paving BRSC;(ii) BRSC of dimension ≤ . As remarked earlier, Example 7.4 answers negatively Problem 7.2 for dimension 3 and k = 4. Onthe other hand, Problem 7.2 has a positive answer for k ≤
2: if S is a BRSC, then T k ( S ) is a BRSCby Proposition 4.1 and pure( T k ( S )) is a BRSC by Corollary 7.10(ii).The next example (discussed in the Appendix) answers negatively Problem 7.2 for dim( S ) = 3and k = 3. Example 7.11
Let S = ( V, H ) with V = ∪{ i, i ′ , i ′′ | i ∈ Z } , Z = ∪{ i ( i + 1)( i + 1) ′ , i ′′ ( i + 1)( i + 1) ′ | i ∈ Z } and H = ( P ≤ ( V ) \ Z ) ∪ { ii ′′ ( i + 1) p | i ∈ Z , p ∈ V \ ii ′′ ( i + 1)( i + 1) ′ }∪ { ii ′′ ( i + 1) ′ p | i ∈ Z , p ∈ V \ ii ′′ ( i + 1)( i + 1) ′ } . Then S is a BRSC but pure( T ( S )) is not. boolean module B (4) : asimplicial complex of dimension 3 admitting a 4 × (2 −
1) boolean matrix representation where allcolumns are distinct and nonzero (so we have all possible nonzero columns).
Example 7.12
The boolean module B (4) is pure and its truncation to rank 3 is a pure TBRSC whichis not a BRSC. We turn now our attention to Problem 7.3.As remarked earlier, Example 7.4 also answers negatively Problem 7.3 for dimension 3 and k = 4.The next result shows, that, unlike Problem 7.2, Problem 7.3 admits a positive answer for k ≤ Theorem 7.13
Let S be a BRSC and let ≤ k ≤ . Then pure( T k ( S )) is a TBRSC. Proof . In the Appendix. (cid:3)
In this section, we generalize to TBRSCs results proved in [9] for the topology of BRSCs.Let S = ( V, H ) be a simplicial complex. We say that S is connected if the graph T ( S ) isconnected. The proof for the following result is essentially the proof given for BRSCs in [9, Lemma3.1]. Lemma 8.1
Let S = ( V, H ) be a TBRSC. Then S is connected unless H = P ( V ) and | V | > . Proof . In the Appendix. (cid:3)
It is well known that the geometric realization || S || of a simplicial complex S , a subspace of someeuclidean space R n , is unique up to homeomorphism. For details, see e.g. [14, Appendix A.5].Given a point v ∈ || S || , the fundamental group π ( || S || , v ) is the group having as elements thehomotopy equivalence classes of closed paths v q q the product being determined by the concatenation of paths.If S is connected, then π ( || S || , v ) ∼ = π ( || S || , w ) for all points v , w in || S || , hence we mayuse the notation π ( || S || ) without ambiguity. We produce now a presentation for π ( || S || ). Thiscombinatorial description is also known as the edge-path group of S (for details on the fundamentalgroup of a simplicial complex, see [15]).We fix a spanning tree T of S and we define A = { a pq | pq ∈ H ∩ P ( V ) } ,R T = { a qp a − pq | pq ∈ H ∩ P ( V ) } ∪ { a pq a qr a − pr | pqr ∈ H ∩ P ( V ) } ∪ { a pq | pq ∈ T } . We may view π ( || S || ) as the group defined by the group presentation h A | R T i . (19)We compute next the fundamental group of a connected TBRSC. If it has dimension 1, it is agraph and so it follows easily from the presentation (19) that its fundamental group is free of rank21 − v + 1, where e (respectively v denotes the number of edges (respectively vertices). Note that v − T . Therefore we concentrate our attention in the case ofdimension ≥
2. These TBRSCs are connected by Lemma 8.1.Given a BRSC S = ( V, H ), the graph of flats Γ( L ( S )) has vertex set V and edges p −− q whenever p = q and pq ⊂ V .Let C be a connected component of Γ( L ( S )). If H∩ P ( C ) = ∅ , we shall say that C is H - nontrivial .Otherwise, we say that C is H - trivial . The size of C is its number of vertices.The next result shows that, given a TBRSC S = ( V, H ) of dimension ≥
2, the graph of flatsΓ( L ( S ε )) and the size of its H ε -trivial components determine completely the fundamental group of S . Note that L ( S ε ) = ε ( S ) by Theorem 3.4, hence, for all distinct p, q ∈ V , p −− q is an edge of L ( S ε ) if and only if there exists some Z ∈ ε ( S ) such that pq ⊆ Z ⊂ V . Theorem 8.2
Let S be a TBRSC of dimension ≥ . Assume that Γ( L ( S ε )) has s H ε -nontrivialconnected components and r H ε -trivial connected components of sizes f , . . . , f r . Then π ( ||S|| ) is afree group of rank (cid:18) s + f + . . . + f r − (cid:19) − r X i =1 (cid:18) f i (cid:19) , or equivalently, (cid:18) s − (cid:19) + ( s − f + . . . + f r ) + X ≤ i 2, it follows from Theorem 3.4 that T ( S ) = T ( S ε ). It follows that π ( || S || ) = π ( || S ε || ) as required. (cid:3) Corollary 8.3 Let S be a simple TBRSC of dimension ≥ . Then π ( || S || ) is a free group of rank (cid:0) t − (cid:1) , where t denotes the number of connected components of Γ( L ( S ε )) . Proof . If S = ( V, H ) is simple, then each H ε -trivial connected component of Γ( L ( S ε )) has preciselyone vertex. Hence, by Theorem 8.2, π ( || S || ) is a free group of rank (cid:0) t − (cid:1) . (cid:3) In [9, Example 3.5], it is shown that free groups of rank (cid:0) n (cid:1) ( n ≥ 2) occur effectively as funda-mental groups of simple BRSCs of dimension 2.Let S = ( V, J ) be a simplicial complex. We recall now the definitions of the (reduced) homologygroups of S (see e.g. [4]).If S has c connected components, it is well known that the H ( S ) is isomorphicto the free abelian group of rank c . For dimension k ≥ 1, we proceed as follows.Fix a total ordering of V . Let C k ( S ) denote the free abelian group on J ∩ P k +1 ( V ), that is,all the formal sums of the form P i ∈ I n i X i with n i ∈ Z and X i ∈ J ∩ P k +1 ( V ) (distinct). Given X ∈ J ∩ P k +1 ( V ), write X = x x . . . x k with x < . . . < x k . We define X∂ k = k X i =0 ( − i ( X \ { x i } ) ∈ C k − ( S )22nd extend this by linearity to a homomorphism ∂ k : C k ( S ) → C k − ( S ) (the k th boundary map of S ). Then the k th homology group of S is defined as the quotient H k ( S ) = Ker ∂ k / Im ∂ k +1 . The 0 th reduced homology group of S , denoted by ˜ H ( S ), is isomorphic to the free abelian groupof rank c − 1, where c denotes the number of connected components of S . For k ≥ 1, the k th reducedhomology group of S , denoted by ˜ H k ( S ) coincides with the k th homology group.A wedge of spheres S , . . . , S m (of possibly different dimensions) is a topological space obtainedby identifying m points s i ∈ S i for i = 1 , . . . , m .We say that two topological spaces X and Y have the same homotopy type if there exist continuousmappings α : X → Y and β : Y → X such that: • there exists a homotopy between αβ and 1 X ; • there exists a homotopy between βα and 1 Y .An important theorem of Bj¨orner and Wachs [1] states that shellable simplicial complexes (a classincluding matroids as a particular case) have the homotopy type of a wedge of spheres.Theorem 8.2 also yields the following important consequence, where the proof is essentially theproof given for BRSCs in [9, Theorem 3.6]. Theorem 8.4 Let S be a TBRSC of dimension . Then:(i) the homology groups of S are free abelian;(ii) S has the homotopy type of a wedge of 1-spheres and 2-spheres. Proof . In the Appendix. (cid:3) Acknowledgments The first author acknowledges support from the Binational Science Foundation (BSF) of the UnitedStates and Israel, grant number 2012080. The second author acknowledges support from the SimonsFoundation (Simons Travel Grant Number 313548). The third author was partially supported byCMUP (UID/MAT/00144/2019), which is funded by FCT (Portugal) with national (MCTES) andEuropean structural funds through the programs FEDER, under the partnership agreement PT2020. References [1] A. Bj¨orner and M. L. Wachs, Nonpure shellable complexes and posets I, Trans. Amer. Math.Soc. 348 (1996), 1299–1327.[2] P. Cameron, Beyond matroids? Permutation group bases and boolean representable com-plexes, lecture given at the Conference in honour of Geoff Whittle , Wellington, 14-16/12/2015,http://sms.victoria.ac.nz/foswiki/pub/Events/GeoffWhittleConference/Programme/PC.pdf.[3] H. H. Crapo, Erecting geometries, Ann. New York Acad. Sci. 175 (1970), 89–92.234] A. Hatcher, Algebraic Topology , Cambridge University Press, Cambridge, 2002.[5] Z. Izhakian and J. Rhodes, New representations of matroids and generalizations, preprint,arXiv:1103.0503, 2011.[6] Z. Izhakian and J. Rhodes, Boolean representations of matroids and lattices, preprint,arXiv:1108.1473, 2011.[7] Z. Izhakian and J. Rhodes, C-independence and c-rank of posets and lattices, preprint,arXiv:1110.3553, 2011.[8] S. Margolis, J. Rhodes and P. V. Silva, On the subsemigroup complex of an aperiodic Brandtsemigroup, Semigroup Forum Internat. J. Algebra Comput. J. Comb. Th. B 27 (1979), 216–224.[12] J. G. Oxley, Matroid Theory , Oxford Science Publications, 1992.[13] R. Pendavingh and J. van der Pol, Enumerating matroids of fixed rank, Electron. J. Combin. Boolean Representations of Simplicial Complexes and Matroids ,Springer Monographs in Mathematics, Springer, 2015.[15] E. H. Spanier, Algebraic Topology , Springer, 1966.[16] C. T. C. Wall, Finiteness conditions for CW-complexes, Ann. Math. Appendix We collect in this Appendix several proofs omitted from the main text, and the discussion of severalexamples. Proof of Proposition 3.1 . Let X ∈ L ( T k ( S )). If X contains a facet of T k ( S ), then X = V , so weassume that X contains no facet of T k ( S ). Now let I ∈ H ∩ X and p ∈ V \ X . Since I / ∈ fct( T k ( S )),we have | I | < k and so I ∈ T k ( H ). Now X ∈ L ( T k ( S )) yields I ∪ { p } ∈ T k ( H ) ⊆ H . Therefore X ∈ L ( S ) and the direct inclusion holds.Conversely, assume that X ∈ L ( S ) contains no facet of T k ( S ). Let I ∈ H∩ P ≤ k ( X ) and p ∈ V \ X .Since X ∈ L ( S ), we get I ∪ { p } ∈ H . But I is not a facet of T k ( S ), hence | I | < k and so I ∪ { p } ∈ T k ( H ). Thus X ∈ L ( T k ( S )) as required. (cid:3) roof of Lemma 3.3 . (i) We prove that T k ( H ε ) ⊆ H (20)holds for k = 0 , . . . , d + 1 by induction on k .The case k = 1 being trivial, assume that k ∈ { , . . . , d + 1 } and (20) holds for k − 1. Let X ∈ T k ( H ε ). We may assume that | X | = k . Then there exists an enumeration x , . . . , x k of X and Z , . . . , Z k ∈ ε ( S ) such that Z ⊃ Z ⊃ . . . ⊃ Z k and x i ∈ Z i − \ Z i for i = 1 , . . . , k . Let X ′ = { x , . . . , x k } . Since X ′ ∈ T k − ( H ε ), it follows from the induction hypothesis that X ′ ∈ H . Now | X ′ | ≤ d , X ′ ⊆ Z and x ∈ V \ Z , hence it follows from Z ∈ ε ( S ) that X = X ′ ∪ { x } ∈ H . Thus(20) holds for k = 1 , . . . , d + 1.(ii) Let X ∈ ε ( S ). Let I ∈ H ε ∩ X and p ∈ V \ X . Since I ∈ H ε , there exists an enumeration x , . . . , x k of I and Z , . . . , Z k ∈ ε ( S ) such that Z ⊃ Z ⊃ . . . ⊃ Z k and x i ∈ Z i − \ Z i for i = 1 , . . . , k .Now by Lemma 3.2(i) Z ∩ X ⊃ Z ∩ X ⊃ . . . ⊃ Z k ∩ X is also a chain in ε ( S ) satisfying x i ∈ ( Z i − ∩ X ) \ ( Z i ∩ X ) for i = 1 , . . . , k . Since V ⊃ Z ∩ X isalso a chain in ε ( S ) and p ∈ V \ ( Z ∩ X ), we get I ∪ { p } ∈ H ε and so X ∈ L ( S ε ).(iii) Let X ∈ H ε . Then X is a transversal of the partition of successive differences for some chainof ε ( S ), and so is any subset of X . Thus S ε is a simplicial complex. By (ii), a chain in ε ( S ) is alsoa chain in L ( S ε ). Therefore S ε is boolean representable. (cid:3) Proof of Theorem 3.4 . (i) ⇒ (ii). Write S ′ = ( V, H ′ ). We start by showing that L ( S ′ ) ⊆ ε ( S ) . (21)Let F ∈ L ( S ′ ). Suppose that X ∈ H ∩ P ≤ d ( F ) and p ∈ V \ F . Since H ⊆ H ′ , it follows from F ∈ L ( S ′ ) that X ∪ { p } ∈ H ′ . But now | X | ≤ d implies X ∪ { p } ∈ T d +1 ( H ′ ) = H and so F ∈ ε ( S ).Therefore (21) holds.Now let X ∈ H . Since H ⊆ H ′ , there exists an enumeration x , . . . , x k of X and F , . . . , F k ∈ L ( S ′ ) such that F ⊃ F ⊃ . . . ⊃ F k and x i ∈ F i − \ F i for i = 1 , . . . , k . By (21), we have F , . . . , F k ∈ ε ( S ) and so X ∈ H ε . Since dim( S ) = d , then X ∈ T d +1 ( H ε ) and so H ⊆ T d +1 ( H ε ).Therefore H = T d +1 ( H ε ) by Lemma 3.3(i), and so S = T d +1 ( S ε ).(ii) ⇒ (i). This follows from Lemma 3.3(iii).It remains to be proved that L ( S ε ) = ε ( S ).Let X ∈ L ( S ε ). Let I ∈ H ∩ P ≤ d ( X ) and p ∈ V \ X . Then I ∈ H ε by (ii) and so X ∈ L ( S ε )yields I ∪ { p } ∈ H ε . Since | I | ≤ d , we get I ∪ { p } ∈ T d +1 ( H ε ) = H and so X ∈ ε ( S ). The oppositeinclusion follows from Lemma 3.3(ii). (cid:3) Analysis of Example 3.5 . Indeed, it is easy to check that ∅ ⊂ ⊂ ⊂ ⊂ V, ∅ ⊂ ⊂ ⊂ V, ∅ ⊂ ⊂ V are all chains in ε ( S ). Now every X ∈ H is a partial transversal of either chain (if X = 46, we usethe third chain, if X ⊇ 35 we use the second chain, in the remaining cases we use the first). Hence H ⊆ H ε and so S = T ( S ε ) by Lemma 3.3(i). Therefore S is a TBRSC by Theorem 3.4.Consider now 134 ∈ H . • Since 135 / ∈ H , we get 5 ∈ 13. Since 235 / ∈ H , we get 2 ∈ ⊆ 13. Since 123 ∈ fct( S ), we get13 = V . 25 Since 146 / ∈ H , we get 6 ∈ 14. Since 246 / ∈ H , we get 2 ∈ ⊆ 14. Since 124 ∈ fct( S ), we get14 = V . • Since 346 / ∈ H , we get 6 ∈ 34. Since 246 / ∈ H , we get 2 ∈ ⊆ 34. Since 234 ∈ fct( S ), we get34 = V .It follows that S is not a BRSC. (cid:3) Analysis of Example 3.6 . Let a, b, c be an enumeration of 123. Let X ∈ ε ( S ) contain ab . Since ab / ∈ H , we have 4 ∈ X . Since ac / ∈ H , we get c ∈ X . Hence X = V and so 123 cannot be atransversal of the successive differences for a chain in ε ( S ). Therefore S is not a TBRSC. (cid:3) Proof of Proposition 3.7 . Let F ∈ L ( S ε ). Take X ∈ H ∩ P ≤ d ( F ) and p ∈ V \ F . Since P ≤ d ( V ) ⊆ H ,we have P ≤ d − ( V ) ⊆ L ( S ), hence P ≤ d − ( V ) ⊆ ε ( S ) ⊆ L ( S ε ) by Lemmas 3.2(ii) and 3.3(ii). Thus X ∈ H ε and since F ∈ L ( S ε ) we get X ∪ { p } ∈ H ε . Thus X ∪ { p } ∈ T d +1 ( H ε ) ⊆ H by Lemma 3.3(i)and so F ∈ ε ( S ). (cid:3) Analysis of Example 3.8 . Straightforward computation shows that ε ( S ) = {∅ , , V } . It followseasily that 124 ∈ L ( S ε ) \ ε ( S ). (cid:3) Proof of Lemma 4.2 . It is immediate that ( V, H ) is a simplicial complex. Let I, J ∈ H with | I | = | J | + 1. We may assume that | J ∩ I | ≤ J ∩ I = { a } . Write J = ab and I = ac c . Suppose that abc s / ∈ H for s = 1 , abc s contains some F s ∈ F for s = 1 , 2. Since I, J ∈ H , we must have F s = bc s for s = 1 , F ∩ F = { b } , a contradiction. Thus J ∪ { c s } ∈ H for some s .Assume now that J ∩ I = ∅ . Write J = ab and I = c c c . Suppose that abc s / ∈ H for s = 1 , , abc s contains some F s ∈ F for s = 1 , , 3. Since J ∈ H , we must have F s ∈ { ac s , bc s } for s = 1 , , 3. But then there exist i, j ∈ { , , } such that | F i ∩ F j | = 1, a contradiction. Thus J ∪ { c s } ∈ H for some s . (cid:3) Analysis of Example 5.2 . Indeed, S is a uniform matroid and S is a matroid by Lemma 4.2. Wemay write S ∨ S = ( V, H ) with H = P ≤ ( V ) ∪ { X ∈ P ( V ) | , , X } . We have 1235 ∈ H . Let Z ∈ ε ( H ).If 13 ⊆ Z , then 123 / ∈ H yields 2 ∈ Z , and 125 / ∈ H yields 5 ∈ Z .If 15 ⊆ Z , then 125 / ∈ H yields 2 ∈ Z , and 123 / ∈ H yields 3 ∈ Z . Out of symmetry, 35 ⊆ Z implies 1 ∈ Z .It follows that 135 / ∈ H ε and so S ∨ S is not a TBRSC by Theorem 3.4. (cid:3) Analysis of Example 5.4 . Indeed, it is easy to check that L ( V, H ) ⊇ P ≤ ( V ) ∪ { , V } , L ( V, H ′ ) ⊇ P ≤ ( V ) ∪ { , V } , and it follows easily that ( V, H ) , ( V, H ′ ) ∈ BPav( d ). We have seen in Example 3.5 that ( V, H ∪ H ′ ) / ∈ BPav( d ). (cid:3) nalysis of Example 5.12 . First, we show that S is not a TBRSC. Suppose it is. Then 1246 ∈ H ⊆ H ε by Theorem 3.4. Then there exists some Z ∈ ε ( S ) such that | ∩ Z | = 3. Out of symmetry, wemay assume that 24 ⊂ Z . • If 1246 ∩ Z = 124, then 124 ∈ H ∩ Z and 1234 / ∈ H yield 3 ∈ Z . Now 123 ∈ H ∩ Z and1236 / ∈ H yield 6 ∈ Z , a contradiction. • If 1246 ∩ Z = 246, then 246 ∈ H ∩ Z and 2346 / ∈ H yield 3 ∈ Z . Now 234 ∈ H ∩ Z and1234 / ∈ H yield 1 ∈ Z , also a contradiction.Therefore S is not a TBRSC.Now let J ⊆ V be the set of partial transversals of the partial differences for the chain ∅ ⊂ ⊂ ⊂ ⊂ V. Then ( V, J ) is a BRSC and J ⊆ H . On the other hand, P ≤ ( V ) ⊆ H . Since J ∪ P ≤ ( V ) = H and S is not a TBRSC, it follows that S admits no largest truncated boolean representable subcomplex. (cid:3) Proof of Proposition 6.1 . We fix V = { , . . . , } as the set of points and we consider S = ( V, H ) ∈ TBPav(2) \ BPav(2). Then there exists some BRSC S ′ = ( V, H ′ ) such that S = T ( S ′ ). Given X ⊆ V , let X (respectively b X ) denote the closure of X in L ( S ′ ) (respectively L ( S )).Since S / ∈ BPav(2) and P ≤ ( V ) ⊆ L ( S ), there exists some X ∈ P ( V ) ∩ H such that X ⊆ \ X \ { x } for every x ∈ X. (22)Without loss of generality, we may assume that X = 345. On the other hand, since S ′ ∈ BPav(2)and 345 ∈ H ′ , there exists some x ∈ 345 such that x / ∈ \ { x } . We may assume that x = 5. Weclaim that | | = 4 . (23)Indeed, we know already that 5 / ∈ 34. Suppose that 34 = 34. Then 34 y ∈ H ′ (and therefore 34 y ∈ H )for every y ∈ b 34 = 34, contradicting (22). Without loss of generality, we may assumethat 34 y / ∈ H ′ for some y ∈ y = 1. Hence 134 ⊆ 34. Suppose that 34 = 134. Since 134 / ∈ H ,this implies b 34 = 34 = 134, contradicting (22). Thus | | ≥ 4. Since 5 / ∈ 34, we may assume withoutloss of generality that 1234 ⊆ ⊂ 34. Since 5 / ∈ 34, we get 34 = 12346. It follows that 45 z ∈ H ′ for every z ∈ b 45 = 45, contradicting (22). Therefore 34 = 1234 and so (23) holds.It follows that ab , ab ∈ H for all a, b ∈ { , , } 6⊆ H . Together with 134 / ∈ H , this implies that the restriction S ′′ = S ′ | = S | = (1234 , H ′′ )misses at least two triangles.On the one hand, 134 / ∈ H and { , , } 6⊆ H yield 1234 ⊆ b 34. On the other hand, itfollows from (22) that 5 ∈ b 34, hence 12345 ⊆ b 34. Since ab , ab ∈ H ′ for all a, b ∈ / ∈ L ( S ) (if 1234 ∈ L ( S ), then b ⊆ \ { c } ∈ H for some c ∈ S ′′ has exactly one or two triangles. Since S ′′ is a restriction of the BRSC S ′ , it follows27hat S ′′ is a BRSC. On the other hand, it follows from [14, Example 5.2.11] that a paving BRSC with4 points cannot have exactly one triangle, hence S ′′ has exactly two triangles, whose intersection hastwo points, say de .Together with 1234 ∈ L ( S ′ ) ′ , this implies that de ∈ L ( S ′ ). Since 134 / ∈ H , we have de ∈{ , , } . Since we have not distinguished 3 from 4 so far, we may assume that de ∈ { , } .In any case, having 1234 ∈ L ( S ′ ) determines that ab , ab ∈ H for all a, b ∈ de ∈ L ( S ′ ) determines which two elements among the four elements of P (1234)belong to H . Thus we only need to discuss what happens with 156 , , , ∈ H , then35 ∈ L ( S ′ ) (in view of 1234 ∈ L ( S ′ )), implying 35 = 35 (and consequently b 35 = 35), contradicting(22). Therefore 356 / ∈ H . Similarly, 456 / ∈ H . It follows that we reduced the discussion to determinewhether or not 156 , ∈ H , for each choice of de ∈ { , } .If we omit both 156 , 256 from H , we get the two cases(1) H = B (1234) ∪ B (12),(1’) H = B (1234) ∪ B (23),which are clearly isomorphic.Now adding 156 (respectively 256) is the only consequence of adding 15 (respectively 25) as aline, and these additions do not interfere with each other. We are then bound to consider the cases:(2) H = B (1234) ∪ B (12) ∪ B (15);(2’) H = B (1234) ∪ B (12) ∪ B (25);(3) H = B (1234) ∪ B (12) ∪ B (15) ∪ B (25);(4) H = B (1234) ∪ B (23) ∪ B (15);(2”) H = B (1234) ∪ B (23) ∪ B (25);(5) H = B (1234) ∪ B (23) ∪ B (15) ∪ B (25).The cases (2), (2’) and (2”) are clearly isomorphic. Applying the permutations (13) and (132) to12345 in cases (4) and (5), respectively, we have reduced our discussion to the cases(1) H = B (1234) ∪ B (12);(2) H = B (1234) ∪ B (12) ∪ B (15);(3) H = B (1234) ∪ B (12) ∪ B (15) ∪ B (25);(4) H = B (1234) ∪ B (12) ∪ B (35);(5) H = B (1234) ∪ B (12) ∪ B (15) ∪ B (35).We list below the triangles missing in each of the cases:(1) 134, 234, 156, 256, 356, 456;(2) 134, 234, 256, 356, 456; 283) 134, 234, 356, 456;(4) 134, 234, 156, 256, 456;(5) 134, 234, 256, 456.Out of cardinality arguments, we only have to distinguish (2) from (4) and (3) from (5). Now1 appears only once among the missing triangles in (2), and all points appear more often in (4); 1and 2 appear only once among the missing triangles in (3), but only 1 has a single occurrence in (5).Therefore these complexes (1) – (5) are nonisomorphic.By construction, any one of these 5 complexes is in TBPav(2). We confirm now that neither ofthem is a BRSC. For the first three cases, we take 345 ∈ H .(1) 134 / ∈ H , hence 1 ∈ b 34; 234 / ∈ H , hence 2 ∈ b b 34 contains the facet 123, hence b 34 = V .356 / ∈ H , hence 6 ∈ b 35; 456 / ∈ H , hence 4 ∈ b 35. Similarly, 3 ∈ b ∈ H .(4) 234 / ∈ H , hence 3 ∈ b 24; 134 / ∈ H , hence 1 ∈ b b 24 contains the facet 123, hence b 24 = V .256 / ∈ H , hence 6 ∈ b 25; 456 / ∈ H , hence 4 ∈ b 25. Similarly, 2 ∈ b (cid:3) Proof of Lemma 6.2 . Let S = ( V, H ) ∈ T BR and let ∅ 6 = W ⊆ V . Since S ∈ T BR , there exist aBRSC S ′ = ( V, H ′ ) and m ≥ S = T m ( S ′ ). We claim that S | W = T m ( S ′ | W ) . (24)This is equivalent to the equality H ∩ W = ( H ′ ∩ W ) ∩ P ≤ m ( W ) . (25)Now S = T m ( S ′ ) yields H = H ′ ∩ P ≤ m ( V ) and so H ∩ W = ( H ′ ∩ P ≤ m ( V )) ∩ W = ( H ′ ∩ W ) ∩ P ≤ m ( W ) . Hence (25) and consequently (24) do hold.Since BRSCs are closed under restriction, then S ′ | W is a BRSC and it follows from (24) that S | W ∈ T BR . Thus T BR is closed under restriction. Since it is also closed under isomorphism, then T BR is a prevariety of simplicial complexes.On the other hand, the class of all finite paving simplicial complexes is a prevariety in view of [14,Proposition 8.3.1(ii)]. Since the intersection of two prevarieties is obviously a prevariety, it followsthat T BP is a prevariety itself. (cid:3) nalysis of Example 7.4 . Since R is a Moore family, S is a BRSC. The maximal chains in R are ∅ ⊂ ⊂ ⊂ ⊂ V, ∅ ⊂ ⊂ ⊂ ⊂ V, (26) ∅ ⊂ ⊂ ⊂ V, ∅ ⊂ ⊂ ⊂ V. (27)Hence dim( S ) = 3. Since Tr( R ) is the set of partial transversals of the successive differences for someof these chains, it follows easily thatTr( R ) = ( P ≤ ( V ) \ { , , , , , , , } ) ∪ { a | a ∈ } ∪ { a | a ∈ } ∪ { b | b ∈ } . . Write pure( S ) = ( V, H ′ ). It is routine to check that H ′ = Tr( R ) \ { } . Indeed, it is easy to see that each X ∈ P ( V ) is a partial transversal of the successive differences forsome chain of type (26), and to check which transversals of the successive differences for some chainof type (27) cannot be obtained through chains of type (26).Now we have 1235 ∈ H ′ . Let Z ∈ ε ( H ′ ) be such that | Z ∩ | ≥ 3. We show that 1235 ⊆ Z . • Suppose that 123 ⊆ Z . Since 123 ∈ H ′ and 1234 / ∈ H ′ , we have 4 ∈ Z . Since 34 ∈ H ′ and347 / ∈ H ′ , we have 7 ∈ Z . Since 47 ∈ H ′ and 457 / ∈ H ′ , we get 5 ∈ Z . • Suppose that 125 ⊆ Z . Since 125 ∈ H ′ and 1257 / ∈ H ′ , we have 7 ∈ Z . Since 127 ∈ H ′ and1267 / ∈ H ′ , we have 6 ∈ Z . Since 567 ∈ H ′ and 4567 / ∈ H ′ , we have 4 ∈ Z . Since 24 ∈ H ′ and234 / ∈ H ′ , we get 3 ∈ Z . • Suppose that 135 ⊆ Z . Since 135 ∈ H ′ and 1345 / ∈ H ′ , we have 4 ∈ Z . Since 345 ∈ H ′ and2345 / ∈ H ′ , we get 2 ∈ Z . • Suppose that 235 ⊆ Z . Since 235 ∈ H ′ and 2345 / ∈ H ′ , we have 4 ∈ Z . Since 345 ∈ H ′ and1345 / ∈ H ′ , we get 1 ∈ Z .Thus there exists no Z ∈ ε ( H ′ ) such that | Z ∩ | = 3. By Theorem 3.4, pure( S ) is not a TBRSC. (cid:3) Proof of Lemma 7.5 . Let I ′ ∈ H be maximal with respect to I ⊆ I ′ ⊆ J . If I ′ ⊂ J , we can take p ∈ J \ I ′ and get I ′ ∪ { p } ∈ H ∩ J , contradicting the maximality of I ′ . Thus I ′ = J and we aredone. (cid:3) Proof of Lemma 7.7 . Suppose that F ρ ≥ F ′ ρ . Then there exist I, J ∈ H such that F = I , F ′ = J and | I | ≥ | J | . Hence I ⊆ J and so by Lemma 7.5 there exists some I ′ ∈ H such that I ⊆ I ′ and I ′ = J . But we have then | I ′ | > | I | ≥ | J | , a contradiction since S is a near-matroid. Therefore F ρ < F ′ ρ . (cid:3) roof of Lemma 7.8 . Write F = I with I ∈ H . Since a ∈ F ′ \ F , we have I ∪ a ∈ H . Thus F ⊂ I ∪ a = F ∪ a ⊆ F ′ . Moreover, F ∪ a ρ = | I ∪ a | = | I | + 1 = F ρ + 1 . If F ∪ a = F ′ , we can now iterate this argument to produce a chain F ⊂ F ∪ a ⊂ F ∪ a a ⊂ . . . ⊂ F ∪ a . . . a s = F ′ for some a , . . . a s ∈ V such that F ∪ a . . . a j ρ = F ∪ a . . . a j − ρ + 1 for j = 1 , . . . , s . Thus s = F ′ ρ − F ρ = k and we are done. (cid:3) Analysis of Example 7.11 . It is easy to check that S is indeed a simplicial complex. Clearly, P ≤ ( V ) ⊂ L ( S ). If X ∈ P ( V ) is not contained in any element of Z , then X = X . Hence, if abc ∈ H and ab is not contained in any element of Z , then abc is a transversal of the successive differencesfor the chain ∅ ⊂ a ⊂ ab ⊂ V in L ( S ). On the other hand, it is easy to check that the unique X ∈ P ( V ) ∩ H having all 2-subsetscontained in elements of Z is 123 (see the picture below, where the yellow triangles are the elementsof Z ): 1 ′ ′′ ′ ′′ ′ ′′ Now it is easy to check that ii ′′ ( i + 1)( i + 1) ′ ∈ L ( S ) for every i ∈ Z . It follows that 123 is atransversal of the successive differences for the chain ∅ ⊂ ⊂ ′′ ′ ⊂ V in L ( S ).Finally, each facet of the form ii ′′ ( i + 1) p or ii ′′ ( i + 1) ′ p is a transversal of the successive differencesfor the chain ∅ ⊂ i ⊂ ii ′′ ⊂ ii ′′ ( i + 1)( i + 1) ′ ⊂ V in L ( S ). Since we have now checked all facets, it follows that S is a BRSC.31et Cl( X ) denote the closure of X ⊆ V in L ( T ( S )). For each i ∈ Z , we have i ( i + 1)( i +1) ′ , i ′′ ( i + 1)( i + 1) ′ / ∈ H , so we successively get ( i + 1) ′ ∈ Cl( i ( i + 1)) and i ′′ ∈ Cl( i ( i + 1)). ThusCl( i ( i + 1)) contains ii ′′ ( i + 1) ∈ fct( T ( S )), yielding Cl( i ( i + 1)) = V . But then i ∈ Cl(123 \ { i } ) forevery i ∈ ∈ T ( H ), then there is no chain of the form (1) and so T ( S ) is not booleanrepresentable.We remark that S is not pure since it is straightforward to check that 1 ′ ′ ′′ is a facet. But T ( S ) is pure because there are no facets of dimension 1: given distinct p, q ∈ V , there exists some r ∈ V \ pq such that pqr is not a yellow triangle. (cid:3) Analysis of Example 7.12 . Let M be such a boolean matrix. Since the columns are all distinct andnonzero, every pair of distinct columns is independent. Now let X be a set of independent columnswith | X | = 2 or 3. Let I ⊂ M [ I, X ] is nonsingular. Let j ∈ \ I and let c be the column of M having a 1 at row j and 0 elsewhere. Then the permanentof M [ I ∪ { j } , X ∪ { c } ] equals the permanent of M [ I, X ], hence M [ I ∪ { j } , X ∪ { c } ] is nonsingular.and so X ∪ { c } is independent. Thus B (4) is pure.Since B (4) is by definition a BRSC, then B (4)3 is a TBRSC. Let X denote the closure of X inFl B (4)3 . Consider the columns of M defined by a = , b = , c = . The permanent of the matrix M [134 , abc ] = is 1, hence abc is independent. Define d = , e = , f = g = . We have M [1234 , abd ] = , M [1234 , bde ] = , M [1234 , abe ] = . Since no row of M [1234 , abd ] has precisely two zeroes, abd is dependent. The same occurs with bde .It is immediate that M [123 , abe ] has permanent 1, hence abe is independent. Thus we successivelydeduce d ∈ ab , e ∈ ab and so ab contains the facet abe . Therefore ab = V , where V denotes the fullset of vertices. Out of symmetry, so is ac . 32ow M [1234 , bcf ] = , M [1234 , bcg ] = , M [1234 , bf g ] = . Since no row of M [1234 , bcf ] has precisely two zeroes, bcf is dependent. The same occurs with bcg .It is immediate that M [123 , bf g ] has permanent 1, hence bf g is independent. Thus we successivelydeduce f ∈ bc , g ∈ bc and so bc contains the facet bf g . Therefore bc = V . Together with ab = ac = V ,and abc being independent, this proves that B (4)3 is not a BRSC. (cid:3) Proof of Theorem 7.13 . Suppose first that k ≤ 2. By Proposition 4.1, T k ( S ) is a BRSC, thereforepure( T k ( S )) is a BRSC by Corollary 7.10(ii).Thus we may assume that k = 3. Write pure( T ( S )) = ( V ′ , H ′ ) and consider the restriction S | V ′ .Then pure( T ( S )) = pure( T ( S | V ′ ) ). Since BRSCs are closed under restriction, S | V ′ is also a BRSC.Therefore we may assume that V ′ = V .Let F = F ∈ L( S ) (cid:12)(cid:12) | F ∩ X | 6 = 1 for every X ∈ fct( S ) ∩ P ( V ) } . We claim that F is a Moore family.Clearly, V ∈ F . Let F, F ′ ∈ F . We have F ∩ F ′ ∈ L ( S ). Let X ∈ fct( S ) ∩ P ( V ). Suppose that | ( F ∩ F ′ ) ∩ X | = 1. Then | F ∩ X | = 1 or | F ′ ∩ X | = 1, contradicting F, F ′ ∈ F . Thus F ∩ F ′ ∈ F and so F is a Moore family.Therefore S ′ = ( V, Tr( F )) is a BRSC. We claim that pure( T ( S )) = T ( S ′ ).For every Y ⊆ V , let Y denote its closure in L ( S ). Let X ∈ H ∩ P ( V ). Then there exists anenumeration a, b, c of the elements of X such that ∅ ⊂ a ⊂ ab ⊂ V (28)and c / ∈ ab . Clearly, ∅ , V ∈ F .Suppose that Y ∈ fct( S ) ∩ P ( V ) satisfies | Y ∩ a | = 1. We may write Y = yz with y ∈ a . If z ∈ ab (respectively z / ∈ ab ), then yzc (respectively ybz ) is a transversal of the successive differencesfor (28), contradicting yz ∈ fct( S ). Thus a ∈ F .Suppose now that Y ∈ fct( S ) ∩ P ( V ) satisfies | Y ∩ ab | = 1. We may write Y = yz with y ∈ ab .If y ∈ a (respectively y / ∈ a ), then ybz (respectively ayz ) is a transversal of the successive differencesfor (28), contradicting yz ∈ fct( S ). Thus ab ∈ F .Therefore (28) is a chain in F and so X ∈ Tr( F ). It follows that H ′ ⊆ Tr( F ) ∩ P ≤ ( V ).Conversely, let X ∈ Tr( F ) ∩ P ≤ ( V ). Since F ⊆ L ( S ), we have Tr( F ) ⊆ H and so X ∈ T ( H ).We certainly have X ∈ H ′ if if | X | = 0 or 3, and the case | X | = 1 follows from V ′ = V . Hence wemay assume that | X | = 2. There exists an enumeration a, b of the elements of X and F ∈ F suchthat a ∈ F and b / ∈ F . But then, by definition of F , we get X / ∈ fct( S ). Hence there exists some c ∈ V \ X such that X ∪ { c } ∈ H . Thus X ∪ { c } ∈ H ′ and so X ∈ H ′ . Therefore T (Tr( F )) ⊆ H ′ and so pure( T ( S )) = T ( S ′ ). It follows that pure( T ( S )) is a TBRSC. (cid:3) Proof of Lemma 8.1 . Obviously, S is disconnected if H = P ( V ) and | V | > 1, and connected if | V | = 1. Hence we may assume that pq ∈ H for some distinct p, q ∈ V .33et M be an R × V boolean matrix representing S . It follows from pq ∈ H that M [ R, p ] = M [ R, q ].Thus, for every v ∈ V , we have either M [ R, v ] = M [ R, p ] or M [ R, v ] = M [ R, q ], implying that vp or vq is an edge in H . Therefore S is connected. (cid:3) Proof of Theorem 8.4 . (i) It follows from Lemma 8.1 that S is connected. By Hurewicz Theorem(see [4]), the 1st homology group of S is the abelianization of π ( || S || ), and therefore, in view ofTheorem 8.2, a free abelian group of known rank. The second homology group of any 2-dimensionalsimplicial complex is Ker ∂ ≤ C ( S ), that is, a subgroup of a free abelian group. Therefore H ( S )is itself free abelian.(ii) By [16, Proposition 3.3], any finite 2-dimensional simplicial complex with free fundamentalgroup has the homotopy type of a wedge of 1-spheres and 2-spheres. (cid:3) Stuart Margolis, Department of Mathematics, Bar Ilan University, 52900 RamatGan, Israel E-mail address: [email protected] John Rhodes, Department of Mathematics, University of California, Berkeley,California 94720, U.S.A. E-mail addresses : [email protected], [email protected] Pedro V. Silva, Centro de Matem´atica, Faculdade de Ciˆencias, Universidade doPorto, R. Campo Alegre 687, 4169-007 Porto, Portugal