Truncated Boolean Representable Simplicial Complexes
Stuart W. Margolis, John Rhodes, Pedro Silva

TL;DR
This paper extends the theory of truncated boolean representable simplicial complexes, broadening its scope to include all matroids and enhancing the applicability of combinatorial geometry to finite simplicial complexes.
Contribution
It significantly advances the theory of truncated boolean representable simplicial complexes, including all matroids, and expands their role in combinatorial geometry.
Findings
Includes all matroids within the class of complexes
Extends the theoretical framework of simplicial complexes
Enhances applications of combinatorial geometry
Abstract
We extend, in significant ways, the theory of truncated boolean representable simplicial complexes introduced in 2015. This theory, which includes all matroids, represents the largest class of finite simplicial complexes for which combinatorial geometry can be meaningfully applied
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Truncated boolean representable simplicial complexes
Stuart Margolis, John Rhodes and Pedro V. Silva
Abstract
We extend, in significant ways, the brief theory of truncated boolean representable simplicial complexes introduced in 2015. This theory, which includes all matroids, represents the largest class of finite simplicial complexes for which combinatorial geometry can be meaningfully applied.
2010 Mathematics Subject Classification: 05B35, 05E45, 14F35, 55P15, 55U10
Keywords: boolean representable simplicial complex, truncation, matroid, join, prevariety, topology, erection
1 Introduction
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 be very 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 admit a boolean representation (usually many), so do BRSC, but not conversely, so BRSC are beyond matroids [2]. The set of independent subsets of a finite set of points form a simplicial complex (in the sense of elementary algebraic topology [15]), being a nonempty collection of subsets of closed under taking subsets.
We are interested in simplicial complexes arising from a geometry, as to be explained below. In this introduction we only concern ourselves with simple simplicial complexes, i.e. those such that all pairs of (distinct) elements of are in . But non-simple simplicial complexes are also considered in [14] and several papers.
Simple matroids arise through transversals of the partial differences for chains in geometric lattices, where we identify the vertices with the atoms of the lattice. More generally, a BRSC may be obtained through transversals of the partial differences for chains in an arbitrary lattice, the role of atoms being played by any join-generating set [14, Chapter 5].
Given a simplicial complex , the combinatorial and algebraic fields use the term rank as the cardinality of the largest set in . The topological and geometric fields use as the dimension of . We use both as will be explained. We say that is paving (of dimension ) and write if has dimension and contains all subsets of of cardinality . Thus we may identify the class of (finite) graphs with .
For (T)BRSC the geometry comes in similarly to matroids as will be explained. Let us restrict to the paving case for simplicity. Let be a matroid in Pav. For simplicity of explanation we assume that , but all generalizes to arbitrary .
The lattice of flats of induces a closure operator on (see [14, Section 4.2] for the most general version). Let denote the closure of . Let
[TABLE]
If is a matroid, then each pair of points of is contained in a unique block in . This makes a PBD (pairwise balanced design) (with ) in design theory. Conversely, and except for trivial cases, every such PBD determines a matroid in Pav (see e.g [10, Proposition 4.2]).
This generalizes to (the BRSC in ) as follows (see [14, Section 6.3]). Let . Then is a partial geometry since for all distinct . This implies that each pair of points of is contained in at most one block in (pretty much the central notion). If is not a matroid, this is not enough to produce a PBD, but we get a PBD by adding to all pairs of points of contained in no element of .
If is a simplicial complex and , then is the truncation of to rank . 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 by TBRSC), but it is easily characterized as follows.
Given a a simplicial complex of dimension , we define
[TABLE]
Then transversals of the partial differences for chains in define a BRSC denoted by .
In the paving case, we actually have . Moreover, truncated to the rank of is the unique largest TBRSC contained in . 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 is a TBRSC if and only if . This provides a useful criterion to recognize a TBRSC.
We note that even if is a matroid, is not necessarily a matroid (see [10, Examples 5.5 and 5.6]). Indeed, constitutes a lattice under intersection, but not necessarily a geometric lattice. See [10], where this is further developed.
We note that, if is a matroid of rank , then consists precisely of the -closed subsets of in the terminology introduced by Crapo [3]. But TBRSC also shed a new light on Crapo’s concept of erection. An erection of a matroid of rank is a matroid of rank such that . In [3], Crapo proved that a collection form a collection of maximal flats of an erection of if and only if:
- (1)
the closure of each in is ;
- (2)
each is -closed (i.e. belongs to );
- (3)
every facet of (i.e. maximal subset in ) is contained in a unique .
He then proves that the collection of all erections of (including the trivial erection ) form a lattice 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 the wider class of TBRSC and consider arbitrary differences of rank, generalizing the work of Crapo and others [3, 11, 13]. For a TBRSC , these operations can be viewed as strong maps (-maps) of (see [9] and Chapter 5, especially Sections 5.4 - 5.5 and 8.2 of [14]).
The outline of this paper is as follows:
In Sections 2 and 3 we provide the basic theory of BRSC and TBRSC, respectively. In Section 4 we discuss low dimensions.
Section 5 deals with the join operator: . In this paper we use join referring to the lattice of all simplicial complexes on a fixed vertex set , ordered by inclusion. This is the same as the lattice of semigroup ideals of the monoid .
In general, (T)BRSC are not closed under join, but the class (the TBRSC in ) is. A key resource are the complexes (where ), containing all subsets of with at most points and all subsets of with points which intersect in exactly points.
Note that, given , we have if and only if . Moreover, is a TBRSC if and only if
[TABLE]
where . More generally, is the largest paving TBRSC contained in , and is therefore the largest subcomplex of allowing some geometrical features. This strictly includes all paving matroids.
In Section 6, we show that the maximum number of vertices for a minimal is . On the other hand, the prevariety (consisting of all paving TBRSC of dimension ) is not finitely based.
In Section 7, we discuss three questions involving the largest pure subcomplex of a BRSC or of one of its truncations. We answer them negatively in the general case, but we show them to hold for low dimensions,
Finally, we discuss in Section 8 some of the topological properties of the geometric realization of a TBRSC, generalizing previous results for BRSC.
2 Boolean representable simplicial complexes
For the material presented in this section, the reader is referred to [14]. All the results mentioned here 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 and , we denote by (respectively ) the set of all subsets of with precisely (respectively at most, at least) elements.
A (finite) simplicial complex is a structure of the form , where is a finite nonempty set and contains and is closed under taking subsets. The elements of and are called respectively vertices and faces. To simplify notation, we shall often denote a face by .
A face of which is maximal with respect to inclusion is called a facet. We denote by the set of facets of . The rank and dimension of are defined respectively by
[TABLE]
We say that is:
- •
simple if ;
- •
paving if .
We denote by the class of all paving simplicial complexes of dimension .
Two simplicial complexes and are isomorphic if there exists a bijection such that
[TABLE]
holds for every .
If is a simplicial complex and is nonempty, we call
[TABLE]
the restriction of to . It is obvious that is still a simplicial complex.
A simplicial complex is called a matroid if it satisfies the exchange property:
- (EP)
For all with , there exists some such that .
An important example of matroids are the uniform matroids : for all , we write with .
Given an matrix and , , we denote by the submatrix of obtained by deleting all rows (respectively columns) of which are not in (respectively ).
A boolean matrix is lower unitriangular if it is of the form
[TABLE]
Two matrices are congruent if we can transform one into the other by independently permuting rows/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. The permanent of a square matrix (a positive version of the determinant) is defined by
[TABLE]
But, even though our matrix is boolean, we compute its permanent in the superboolean semiring , which can be described as the quotient of the usual semiring by the congruence with classes . Then a square boolean matrix is nonsingular if and only if its permanent is 1 in .
We note that the classical results on determinants involving only a rearrangement of the permutations extend naturally to permanents.
Given an boolean matrix , we say that the subset of columns is -independent if there exists some such that the submatrix is nonsingular.
A simplicial complex is boolean representable (BRSC) if there exists some boolean matrix such that is the set of all -independent subsets of . Since , this implies that all the columns of are nonzero. Moreover, for all distinct , the columns and are different if and only if .
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 of is called a Moore family if and is closed under intersection (that is, a Moore family is a submonoid of the monoid of all subsets of under intersection). Every Moore family, under inclusion, constitutes a lattice (with intersection as meet and the determined join ). We say that is a transversal of the successive differences for a chain
[TABLE]
in if admits an enumeration such that for .
If is the set of transversals of the successive differences for chains in , then 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 is a flat of if
[TABLE]
The set of all flats of is denoted by . Note that in all cases, and is indeed a Moore family (and therefore a lattice). Note also that for every .
It follows from [14, Corollary 5.2.7] that a simplicial complex is boolean representable If and only if the transversals of the successive differences for chains in are precisely the elements of .
By [14, Proposition 8.3.3(i)], the flats of a BRSC determine flats on any restriction: if is a flat of a BRSC and , then .
The lattice induces a closure operator on defined by
[TABLE]
for every . It follows from the definitions that when contains a facet of .
By [14, Corollary 5.2.7], is boolean representable if and only if every admits an enumeration satisfying
[TABLE]
It is well known that in the case of matroids, this enumeration can be chosen arbitrarily [12].
3 Truncation
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 convenience for the reader.
Given a simplicial complex and , the -truncation of is the simplicial complex , where .
We say that a simplicial complex is a TBRSC if for some BRSC and . For every , we denote by the class of all paving TBRSCs of dimension .
To recognize a TBRSC, it is convenient to develop an alternative characterization. The key is building the flats of a canonical BRSC having our TBRSC as a truncation. The following result characterizes the flats of a truncation with respect to the flats of the original complex.
Proposition 3.1
[14, Proposition 8.2.2]* Let be a simplicial complex and let . Then*
[TABLE]
Proof. In the Appendix.
It follows that the lattice of flats of is obtained from the lattice of flats of by identifying the elements of an up set (namely the subset of flats containing some facet of ). In semigroup-theoretic terms, this makes a Rees quotient of the -semilattice of .
For any simplicial complex of dimension , we define
[TABLE]
Note that generalizes to arbitrary simplicial complexes what Crapo calls d-closed sets of a monoid in his fundamental paper from 1970 [3].
The following lemma is clear from the definition.
Lemma 3.2
[14, Lemma 8.2.3]* Let be a simplicial complex. Then:*
- (i)
* is a Moore family;*
- (ii)
.
Given , write and let denote the BRSC defined by the lattice .
Lemma 3.3
[14, Lemma 8.2.4]* Let be a simplicial complex of dimension . Then:*
- (i)
;
- (ii)
;
- (iii)
* is a BRSC.*
Proof. In the Appendix.
Now we can state the main result of this section:
Theorem 3.4
[14, Theorem 8.2.5]* Let be a simplicial complex of dimension . Then the following conditions are equivalent:*
- (i)
* for some boolean representable simplicial complex ;*
- (ii)
.
Furthermore, in this case we have .
Proof. In the Appendix.
We present now two examples which show that the class of TBRSCs is intermediate between the class 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 , and . Then but is not boolean representable.
The second complex shows that a (paving) simplicial complex is not necessarily a TBRSC.
Example 3.6
Let , and . Then is not a TBRSC.
With respect to the equality in Theorem 3.4, we can show it holds for all paving simplicial complexes:
Proposition 3.7
Let and . Then .
Proof. In the Appendix.
However, the next example, also analyzed in the Appendix, shows that equality may not hold.
Example 3.8
Let be the simplicial complex defined by and
[TABLE]
Then .
4 Low dimensions
Proposition 4.1
Every TBRSC of dimension 1 is boolean representable.
Proof. Let be a TBRSC of dimension 1. For every , let . Every is a transversal of the successive differences for the chain in . Suppose now that are distinct and . Then is a transversal of the successive differences for the chain , so it suffices to show that .
Since is a TBRSC, there exists a boolean matrix with column space such that, for every , we have if and only if is -independent. Since , all the columns of are nonzero, so if and only if the columns of are different. Thus is the set of all having columns in equal to the column of .
Let and . Then and the column of is different, so and so as required.
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 be a finite nonempty set and let be such that for all distinct . Let consist of all containing no element of . Then is a matroid.
Proof. In the Appendix.
The next result shows that, when it comes to separate BRSCs from TBRSCs, Example 3.5 has the minimum number of vertices.
Proposition 4.3
Every TBRSC with at most 5 vertices is boolean representable.
Proof. Let be a TBRSC with . In view of Proposition 4.1, we may assume that .
Suppose first that . Then is the uniform matroid , hence a BRSC.
Suppose next that . We may assume that , otherwise . If for some BRSC of dimension 3, then and is therefore a BRSC.
Thus we may assume that . If , then is a BRSC. If and for some BRSC of dimension 4, then and is also a BRSC. Hence we may assume that .
Suppose that is not a BRSC. Then . Let . Comparing the definitions of and , we see that , hence we may take . Since , we may assume that there exists a chain in such that for . By Lemma 3.2(i), is also a transversal of the successive differences for the chain
[TABLE]
in , hence there exists a chain
[TABLE]
in . Since , there exists some such that . Without loss of generality, we may assume that .
If does not contain , then is a transversal of the successive differences for the chain (2), hence . Thus the only possible elements of are .
If , we have necessarily
[TABLE]
because any other 2-subset is contained in some element of . By Lemma 4.2, is a matroid, hence boolean representable.
Thus we may assume that we have one of the following four cases:
- (C1)
;
- (C2)
;
- (C3)
;
- (C4)
.
Now (C3) and (C4) are clearly both matroids (hence BRSCs). We can show that (C1) is a BRSC by checking that are flats. Similarly, (C2) is a BRSC because are flats. Therefore every TBRSC with 5 vertices is a BRSC.
5 Join
Given two simplicial complexes and we define the join of and as the simplicial complex
[TABLE]
Notice that given a simplicial complex , then is just a down set of under inclusion. The down sets of form a lattice equal to the lattice of semigroup ideals of the monoid , and this construction is precisely the join in this lattice.
Proposition 5.1
Let and be BRSCs with . Then is a BRSC.
Proof. If , we may use Proposition 4.1 and Theorem 5.3. The only other nontrivial case is . But it is easy to check [14, Example 5.2.11] that if and , then is a BRSC if and only if . It follows that if is not a BRSC, then or is not a BRSC.
The next example shows that neither BRSCs nor TBRSCs are closed under join when we consider 5 vertices (even at dimension ). We analyze this example in the Appendix.
Example 5.2
Let . Let and be defined by
[TABLE]
But things work out better in the paving case:
Theorem 5.3
Let and let . Then .
Proof. Let
[TABLE]
In view of Lemma 3.2(i), is a Moore family. Hence is a BRSC. We claim that
[TABLE]
Let . By Theorem 3.4, there exists a chain
[TABLE]
in and an enumeration of the elements of such that for every . Since , then (4) is also a chain in , hence . But , thus and also by symmetry.
Conversely, let . Since , we may assume that . Then there exists a chain
[TABLE]
in and an enumeration of the elements of such that for every .
Write with and . Since , we may assume that . Since yields , then
[TABLE]
is a chain in having as a transversal of the successive differences. Thus by Theorem 3.4 and so (5.3) holds. Note also that .
Therefore .
The next example, analyzed in the Appendix, shows that we cannot replace by in Theorem 5.3.
Example 5.4
Let , and
[TABLE]
Then but .
Let be a finite nonempty set and let be such that . We write
[TABLE]
This is easily seen to be equivalent to the following condition:
[TABLE]
If is clear from the context, we may omit from and .
Lemma 5.5
Let be a finite nonempty set and let be such that . Then .
Proof. It is immediate that , hence every is a transversal of the successive differences for some chain in . Thus .
We can now prove the following result, characterizing .
Theorem 5.6
Let and . Then the following conditions are equivalent:
- (i)
;
- (ii)
* for some nonempty .*
Proof. (i) (ii). Let . Since , we have .
Let . Since for every , we may assume that . By Theorem 3.4, there exists a chain
[TABLE]
in and an enumeration of the elements of so that for . Now for , hence is a transversal of the chain
[TABLE]
and so . Since , we get .
The opposite inclusion is immediate.
(ii) (i). By Lemma 5.5 and Theorem 5.3.
In such a decomposition (), we may refer to the elements of as lines.
The following lemma shows that the decomposition provided by Theorem 5.6 is not unique.
Lemma 5.7
Let and let be a finite set with . For every , we have
[TABLE]
where .
Proof. It suffices to show that both sides of (6) contain the same . So let .
Suppose that . Then , hence . Take . Then and so
[TABLE]
Conversely, suppose that with . Since and , we must have for some . Hence yields and . Therefore (6) holds as required.
We prove next a version of Theorem 5.6 for .
Theorem 5.8
Let and . Then the following conditions are equivalent:
- (i)
;
- (ii)
;
- (iii)
* for some nonempty satisfying*
[TABLE]
Proof. (iii) (i). Since , we have . Let and suppose that and . Since , we may assume that or .
Suppose that . Since , we have for some . Thus and so . In view of (7), we get , hence , a contradiction since and . Therefore , hence and so .
Let be distinct. Then
[TABLE]
is a chain in . If , then (8) can be refined to
[TABLE]
also a chain in . It is easy to check that every is a partial transversal of the successive differences for some chain of type (8) or (9), hence is boolean representable.
(i) (ii). Since and the maximum length of a chain in is , it follows easily that the maximal chains in must be of the form (8) or (9), with . Thus (ii) holds.
(ii) (iii). Suppose that are distinct and satisfy . We may assume that . Let be distinct. Since , we get a chain of length
[TABLE]
in , contradicting . Therefore .
Corollary 5.9
Let and and let be a finite set with . Let be such that for every . Then is boolean representable.
Proof. In view of Lemma 5.7, we may assume that for every . Let be distinct. It is easy to check that
[TABLE]
holds for every . Now we may use (10) for replacing by some equivalent satisfying (7):
- (1)
If are such that , we replace by
[TABLE]
- (2)
If and , we replace by .
Indeed, these replacements are legitimate in view of (10), and each such replacement decreases the number of . Eventually, we end up with some satisfing (7). By Theorem 5.8, our complex is boolean representable.
Another way of ensuring closure under join is by restricting the type of complexes in . We define, for every and every finite set with at least elements,
[TABLE]
Proposition 5.10
Let and let be a finite set with at least elements. Then
[TABLE]
Proof. We show that
[TABLE]
Let and . Suppose that . Since and no restriction of to a -subset of is isomorphic to , there exists some . But then contains a facet of and so , a contradiction. Therefore (11) holds.
Now let
[TABLE]
By (11), we have
[TABLE]
Since , we have by Theorem 5.8. Thus and so by (12) and Proposition 5.9.
Suppose that there exists some such that . Then , contradicting . Therefore .
Finally, we prove that, when we start with a complex , there is a largest paving TBRSC contained in , and it is precisely .
Theorem 5.11
Let and
[TABLE]
Then:
- (i)
there exists some (unique) such that for every ;
- (ii)
.
Proof. (i) Let . Clearly, for every . In view of Theorem 5.3, it follows that and we are done.
(ii) By Lemma 3.3, is a TBRSC and . Since , we have and so . Hence is paving and . Therefore .
To prove the opposite inclusion, we may assume that (otherwise and we are done). It follows from Theorem 3.4 that , so it suffices to show that , which follows itself from . We prove the latter inclusion.
Let . Then
[TABLE]
Suppose that and . Since , we have and yields . Thus and so as required.
The next example, analyzed in the Appendix, shows that the paving requirement for subcomplexes cannot be removed from the definition of .
Example 5.12
Let be defined by and . Then has no largest truncated boolean representable subcomplex.
6 On
We proved in Proposition 4.3 that we need at least 6 points to separate from . This section starts with a full account of the 6 point case.
Proposition 6.1
Up to isomorphism, the complexes with 6 points in are of the form for:
- (1)
;
- (2)
;
- (3)
;
- (4)
;
- (5)
.
Moreover, all the above 5 cases are nonisomorphic.
Proof. In the Appendix.
Ordering the sets of faces through inclusion, we can build the following diagram
[TABLE]
The missing triangles in the three lowest elements are respectively
[TABLE]
[TABLE]
hence all the edges correspond to covering relations (check the enumeration of the missing triangles for (1)–(5) in the proof of Proposition 6.1).
We note that:
- •
by Lemma 5.5.
- •
. Indeed, suppose that there exist such that . Since , we successively get and , a contradiction. In view of Theorem 3.4, this implies .
- •
. Similar to the preceding case.
- •
No simplicial complex isomorphic to (4) embeds in (3). To prove this, recall the missing triangles in (3) and (4). We can check that is contained in a missing triangle of (3) for every . Similarly, is contained in a missing triangle of (3) for every . Suppose that is such that the isomorphic image of (3) through (call it (3”)) has (4) as subcomplex. Then the missing triangles of (3”) are a proper subset of the missing triangles of (4). Hence is contained in a missing triangle of (4) for every , and is contained in a missing triangle of (4) for every . However, only satisfies this property, yielding , a contradiction.
Note also that an arbitrary needs not having a restriction isomorphic to . The complexes featuring Proposition 6.1 constitute all counterexamples for .
We intend now to show that is in some sense finitely generated. We start with a couple of lemmas.
Let (respectively ) denote the class of all finite truncated boolean representable simplicial 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].
Lemma 6.2
The classes and are prevarieties of simplicial complexes.
Proof. In the Appendix.
Let . By Lemma 6.2, every restriction of is in (with possibly lower dimension). We say that is minimal if every proper restriction of is boolean representable.
Lemma 6.3
Let . Then the maximum number of vertices for a minimal is .
Proof. Let be minimal. Hence but every proper restriction of is boolean representable. By [14, Theorem 8.5.2(ii)], we get .
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 the Swirl is not. The Swirl is defined as follows:
Let and for . Write also and
[TABLE]
We define
[TABLE]
It is easy to check that all the fall into four cases (not necessarily disjoint):
- (a)
there exist with ;
- (b)
there exist ;
- (c)
there exist with ;
- (d)
.
Define
[TABLE]
It is straightforward to check that , hence by Theorem 5.6. Since , we have found some minimal with vertices as required.
Let be a prevariety of simplicial complexes. We say that is finitely based if there exists some such that every simplicial complex not in admits a restriction not in with at most vertices.
Given a prevariety of simplicial complexes and , we define the prevariety
[TABLE]
Let denote the class of all finite paving boolean representable simplicial complexes. By [14, Theorem 8.5.2], is finitely based for every . Since by Proposition 4.1, it follows that is finitely based.
Theorem 6.4
* is not finitely based.*
Proof. It suffices to build arbitrary large simplicial complexes not in with all proper restrictions in .
Suppose that . Then , and so
[TABLE]
Let and take as vertex set
[TABLE]
where we identify
[TABLE]
Let
[TABLE]
and . We show next that:
[TABLE]
Indeed, such an contains necessarily some element of . Without loss of generality, we may assume that this element is among (the other cases follow by symmetry).
Suppose that , say with . Since , then are not consecutive integers. If , then and , hence and we are done. Thus we may assume that . If , then , hence and we are done. Thus we may assume that . Since , this implies . Since , we get either (yielding ) or (yielding ).
Hence we may assume that at least one of the other elements of (say ) is not of the form . Let be such that . It is easy to check that . Therefore (14) holds.
Next we show that
[TABLE]
Let and assume that . Assume first that . Since for , we get successively . Since , we get .
Suppose now that . Since , we use the same argument to deduce that . Since , we get .
Finally, suppose that . Since and , we use the same argument to deduce that . Since , we get and (15) is proved.
In view of (6), it follows from (15) that .
Fix now and write . We must show that (since is closed under restrictions, this implies that for any ).
Since , we only need to show that the righthand side of (6) holds when we replace by . Let . Suppose first that . By (14), there exists some such that . It follows that . Since , the desired condition is satisfied if .
Thus we may assume that . It follows that either with or or with .
Suppose that . Let . It is immediate that and . If (respectively ), we take (respectively ). Therefore, in view of (6), we get as required.
7 The pure core
Let be a simplicial complex of dimension . We say that is pure if all the facets of have dimension .
We define by
[TABLE]
It is immediate that is the largest pure subcomplex of , also called the pure core of .
This section is devoted to the following questions, all related to the concept of pure core:
Problem 7.1
Let be a BRSC. Is pure a BRSC?
Problem 7.2
Let be a BRSC and let . Is pure a BRSC?
Problem 7.3
Let be a BRSC and let . Is pure a TBRSC?
Note that Problem 7.3 admits the equivalent statement:
[TABLE]
For the nontrivial implication, let be a TBRSC of rank and assume that Problem 7.3 has a positive answer. Since is a TBRSC, we have for some BRSC .
Suppose first that . Then , hence we must show that pure is a TBRSC. Since , our goal follows from the answer of Problem 7.3 for and .
Assume now that . It is easy to check that . Since the answer of Problem 7.3 implies that pure is a TBRSC, then pure is a TBRSC and so (16) has also a positive answer.
Since matroids are pure and closed under restriction [12], all questions have positive answers for matroids. We show that none of them admits a positive answer in general, but we establish particular cases.
The following example, analyzed in the Appendix, answers Problem 7.1 negatively for dimension 3. It also answers Problems 7.2 and 7.3 for dimension 3 and .
Example 7.4
Let and . Then is a BRSC but 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 be a simplicial complex and let be such that . Then there exists some such that and .
Proof. In the Appendix.
Matroids admit a wide variety of characterizations. One of them is the following: a simplicial complex is a matroid if and only if
[TABLE]
Indeed, if and , it follows easily from the exchange property that for some , hence . In the case of matroids, the enumeration in (1) can be chosen arbitrarily (see [12]). By taking as last, we reach a contradiction. Thus .
Conversely, suppose that (17) holds. Let be such that . Suppose that for every . Then and so by Lemma 7.5 this contradicts (17). Thus satisfies the exchange property and is therefore a matroid.
We define a simplicial complex to be a near-matroid if
[TABLE]
for all . The rank function is defined by
[TABLE]
Note that such an exists by [14, Proposition 4.2.4].
It follows from (17) that every matroid is a near-matroid. The following result shows that the converse fails, in fact a near matroid needs not be boolean representable.
Proposition 7.6
Let be a simplicial complex of dimension .
- (i)
If is paving, then is a near-matroid.
- (ii)
If is boolean representable and , then is a near-matroid.
Proof. (i) Write and suppose that are such that . By [14, Proposition 4.2.3], and are not facets (it is easy to check that the closure of a facet must be ). Suppose that . Since , it follows that , so in this case we get indeed . Thus we may assume by symmetry that . Since and are not facets, then and so is a near-matroid.
(ii) Let be such that . Since and the closure of a facet is , we may assume that .
Assume first that . Let be an boolean matrix representing . If , then . Moreover, implies , and since all the columns of are nonzero (we have ), it follows that . Thus yields .
Assume now that . By the previous case, we also have . Since and , we have necessarily . Therefore is a near-matroid.
Before discussing boolean representable near-matroids, we present two lemmas.
Lemma 7.7
Let be a near-matroid and let be such that . Then .
Proof. In the Appendix.
Lemma 7.8
Let be a near-matroid and let be such that . Let and . Then there exist such that
[TABLE]
Proof. In the Appendix.
Theorem 7.9
Let be a boolean representable near-matroid and let . Then:
- (i)
* is a BRSC;*
- (ii)
* is a BRSC.*
Proof. (i) Write and let denote the closure of in . We define
[TABLE]
Since is closed under intersection, it follows from Lemma 7.7 that is a Moore family. Hence is a BRSC. We show that .
Let and let . Then there exists an enumeration of the elements of such that
[TABLE]
Hence is a transversal of the successive differences for
[TABLE]
which is a chain in . Thus .
Conversely, assume that . Since , it follows that .
Suppose that . Since , there exist some and such that . But for some . Hence and by Lemma 7.5 there exists some such that and . But then , a contradiction since is a near-matroid. Thus and so as claimed.
(ii) Let denote the set of all flats of occurring in chains of the form
[TABLE]
in . We claim that is a Moore family.
Let . We may assume that . We have since is a Moore family. Since , there exists some such that and . Now we apply Lemma 7.8 to both inclusions . This ensures that will appear in some chain of flats of length in of the form
[TABLE]
Since , it follows from Lemma 7.7 that this is in fact a chain in and therefore in . Thus . Since , then is a Moore family. Writing , it follows that is a BRSC. We claim that .
Let . Then there exists an enumeration of the elements of such that
[TABLE]
Hence is a transversal of the successive differences for
[TABLE]
which is a chain of length in . Thus .
Conversely, assume that . We may assume that is a facet of . Then there exists some chain
[TABLE]
in and some enumeration of the elements of such that for . Since is a facet, we must have and . Suppose that . Since , then it must occur in some chain of length in , hence we have some chain
[TABLE]
in for some . Since , we have for some , hence there exists some containing (strictly) , contradicting . Thus .
Now for and so we can apply Lemma 7.8 times to refine (18) to a chain of length in of the form
[TABLE]
which admits a transversal of the successive differences containing . Since , it follows that and so in view of Lemma 7.7 we have , hence is a facet of . Therefore as claimed.
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 . On the other hand, Problem 7.2 has a positive answer for : if is a BRSC, then is a BRSC by Proposition 4.1 and is a BRSC by Corollary 7.10(ii).
The next example (discussed in the Appendix) answers negatively Problem 7.2 for and .
Example 7.11
Let with ,
[TABLE]
and
[TABLE]
Then is a BRSC but is not.
Another counterexample, also analyzed in the Appendix, is given by the boolean module : a simplicial complex of dimension 3 admitting a boolean matrix representation where all columns are distinct and nonzero (so we have all possible nonzero columns).
Example 7.12
The boolean module is pure and its truncation to rank 3 is a pure TBRSC which is 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 . The next result shows, that, unlike Problem 7.2, Problem 7.3 admits a positive answer for .
Theorem 7.13
Let be a BRSC and let . Then pure is a TBRSC.
Proof. In the Appendix.
8 Topology
In this section, we generalize to TBRSCs results proved in [9] for the topology of BRSCs.
Let be a simplicial complex. We say that is connected if the graph is connected. The proof for the following result is essentially the proof given for BRSCs in [9, Lemma 3.1].
Lemma 8.1
Let be a TBRSC. Then is connected unless and .
Proof. In the Appendix.
It is well known that the geometric realization of a simplicial complex , a subspace of some euclidean space , is unique up to homeomorphism. For details, see e.g. [14, Appendix A.5].
Given a point , the fundamental group is the group having as elements the homotopy equivalence classes of closed paths
[TABLE]
the product being determined by the concatenation of paths.
If is connected, then for all points in , hence we may use the notation without ambiguity. We produce now a presentation for . This combinatorial description is also known as the edge-path group of (for details on the fundamental group of a simplicial complex, see [15]).
We fix a spanning tree of and we define
[TABLE]
[TABLE]
We may view as the group defined by the group presentation
[TABLE]
We compute next the fundamental group of a connected TBRSC. If it has dimension 1, it is a graph and so it follows easily from the presentation (19) that its fundamental group is free of rank , where (respectively denotes the number of edges (respectively vertices). Note that is the number of edges of the spanning tree . Therefore we concentrate our attention in the case of dimension . These TBRSCs are connected by Lemma 8.1.
Given a BRSC , the graph of flats has vertex set and edges whenever and .
Let be a connected component of . If , we shall say that is -nontrivial. Otherwise, we say that is -trivial. The size of is its number of vertices.
The next result shows that, given a TBRSC of dimension , the graph of flats and the size of its -trivial components determine completely the fundamental group of . Note that by Theorem 3.4, hence, for all distinct , is an edge of if and only if there exists some such that .
Theorem 8.2
Let be a TBRSC of dimension . Assume that has -nontrivial connected components and -trivial connected components of sizes . Then is a free group of rank
[TABLE]
or equivalently,
[TABLE]
Proof. This result was proved in [9, Theorem 3.3] for BRSCs (with replaced by ). Therefore it suffices to note that and have the same fundamental group. Indeed, and . Since has dimension , it follows from Theorem 3.4 that . It follows that as required.
Corollary 8.3
Let be a simple TBRSC of dimension . Then is a free group of rank , where denotes the number of connected components of .
Proof. If is simple, then each -trivial connected component of has precisely one vertex. Hence, by Theorem 8.2, is a free group of rank .
In [9, Example 3.5], it is shown that free groups of rank occur effectively as fundamental groups of simple BRSCs of dimension 2.
Let be a simplicial complex. We recall now the definitions of the (reduced) homology groups of (see e.g. [4]).
If has connected components, it is well known that the 0th homology group is isomorphic to the free abelian group of rank . For dimension , we proceed as follows.
Fix a total ordering of . Let denote the free abelian group on , that is, all the formal sums of the form with and (distinct). Given , write with . We define
[TABLE]
and extend this by linearity to a homomorphism (the th boundary map of ). Then the th homology group of is defined as the quotient
[TABLE]
The [math]th reduced homology group of , denoted by , is isomorphic to the free abelian group of rank , where denotes the number of connected components of . For , the th reduced homology group of , denoted by coincides with the th homology group.
A wedge of spheres (of possibly different dimensions) is a topological space obtained by identifying points for .
We say that two topological spaces and have the same homotopy type if there exist continuous mappings and such that:
- •
there exists a homotopy between and ;
- •
there exists a homotopy between and .
An important theorem of Björner and Wachs [1] states that shellable simplicial complexes (a class including 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 the proof given for BRSCs in [9, Theorem 3.6].
Theorem 8.4
Let be a TBRSC of dimension . Then:
- (i)
the homology groups of are free abelian;
- (ii)
* has the homotopy type of a wedge of 1-spheres and 2-spheres.*
Proof. In the Appendix.
Acknowledgments
The first author acknowledges support from the Binational Science Foundation (BSF) of the United States and Israel, grant number 2012080. The second author acknowledges support from the Simons Foundation (Simons Travel Grant Number 313548). The third author was partially supported by CMUP (UID/MAT/00144/2019), which is funded by FCT (Portugal) with national (MCTES) and European structural funds through the programs FEDER, under the partnership agreement PT2020.
Appendix
We collect in this Appendix several proofs omitted from the main text, and the discussion of several examples.
Proof of Proposition 3.1. Let . If contains a facet of , then , so we assume that contains no facet of . Now let and . Since , we have and so . Now yields . Therefore and the direct inclusion holds.
Conversely, assume that contains no facet of . Let and . Since , we get . But is not a facet of , hence and so . Thus as required.
Proof of Lemma 3.3. (i) We prove that
[TABLE]
holds for by induction on .
The case being trivial, assume that and (20) holds for . Let . We may assume that . Then there exists an enumeration of and such that and for . Let . Since , it follows from the induction hypothesis that . Now , and , hence it follows from that . Thus (20) holds for .
(ii) Let . Let and . Since , there exists an enumeration of and such that and for . Now by Lemma 3.2(i)
[TABLE]
is also a chain in satisfying for . Since is also a chain in and , we get and so .
(iii) Let . Then is a transversal of the partition of successive differences for some chain of , and so is any subset of . Thus is a simplicial complex. By (ii), a chain in is also a chain in . Therefore is boolean representable.
Proof of Theorem 3.4. (i) (ii). Write . We start by showing that
[TABLE]
Let . Suppose that and . Since , it follows from that . But now implies and so . Therefore (21) holds.
Now let . Since , there exists an enumeration of and such that and for . By (21), we have and so . Since , then and so . Therefore by Lemma 3.3(i), and so .
(ii) (i). This follows from Lemma 3.3(iii).
It remains to be proved that .
Let . Let and . Then by (ii) and so yields . Since , we get and so . The opposite inclusion follows from Lemma 3.3(ii).
Analysis of Example 3.5. Indeed, it is easy to check that
[TABLE]
are all chains in . Now every is a partial transversal of either chain (if , we use the third chain, if we use the second chain, in the remaining cases we use the first). Hence and so by Lemma 3.3(i). Therefore is a TBRSC by Theorem 3.4.
Consider now .
- •
Since , we get . Since , we get . Since , we get .
- •
Since , we get . Since , we get . Since , we get .
- •
Since , we get . Since , we get . Since , we get .
It follows that is not a BRSC.
Analysis of Example 3.6. Let be an enumeration of . Let contain . Since , we have . Since , we get . Hence and so 123 cannot be a transversal of the successive differences for a chain in . Therefore is not a TBRSC.
Proof of Proposition 3.7. Let . Take and . Since , we have , hence by Lemmas 3.2(ii) and 3.3(ii). Thus and since we get . Thus by Lemma 3.3(i) and so .
Analysis of Example 3.8. Straightforward computation shows that . It follows easily that .
Proof of Lemma 4.2. It is immediate that is a simplicial complex. Let with . We may assume that .
Assume first that . Write and . Suppose that for . Then contains some for . Since , we must have for . But then , a contradiction. Thus for some .
Assume now that . Write and . Suppose that for . Then contains some for . Since , we must have for . But then there exist such that , a contradiction. Thus for some .
Analysis of Example 5.2. Indeed, is a uniform matroid and is a matroid by Lemma 4.2. We may write with
[TABLE]
We have . Let .
If , then yields , and yields .
If , then yields , and yields . Out of symmetry, implies .
It follows that and so is not a TBRSC by Theorem 3.4.
Analysis of Example 5.4. Indeed, it is easy to check that
[TABLE]
and it follows easily that . We have seen in Example 3.5 that .
Analysis of Example 5.12. First, we show that is not a TBRSC. Suppose it is. Then by Theorem 3.4. Then there exists some such that . Out of symmetry, we may assume that .
- •
If , then and yield . Now and yield , a contradiction.
- •
If , then and yield . Now and yield , also a contradiction.
Therefore is not a TBRSC.
Now let be the set of partial transversals of the partial differences for the chain
[TABLE]
Then is a BRSC and . On the other hand, . Since and is not a TBRSC, it follows that admits no largest truncated boolean representable subcomplex.
Proof of Proposition 6.1. We fix as the set of points and we consider . Then there exists some BRSC such that . Given , let (respectively ) denote the closure of in (respectively ).
Since and , there exists some such that
[TABLE]
Without loss of generality, we may assume that . On the other hand, since and , there exists some such that . We may assume that . We claim that
[TABLE]
Indeed, we know already that . Suppose that . Then (and therefore ) for every , yielding , contradicting (22). Without loss of generality, we may assume that for some , say . Hence . Suppose that . Since , this implies , contradicting (22). Thus . Since , we may assume without loss of generality that .
Suppose that . Since , we get . It follows that for every , hence , contradicting (22). Therefore and so (23) holds.
It follows that for all distinct. Since , it follows that . Together with , this implies that the restriction
[TABLE]
misses at least two triangles.
On the one hand, and yield . On the other hand, it follows from (22) that , hence . Since for all distinct, then (if , then ) implies that for some . Therefore has exactly one or two triangles. Since is a restriction of the BRSC , it follows that is a BRSC. On the other hand, it follows from [14, Example 5.2.11] that a paving BRSC with 4 points cannot have exactly one triangle, hence has exactly two triangles, whose intersection has two points, say .
Together with , this implies that . Since , we have . Since we have not distinguished 3 from 4 so far, we may assume that .
In any case, having determines that for all distinct (12 elements), and determines which two elements among the four elements of belong to . Thus we only need to discuss what happens with . If , then (in view of ), implying (and consequently ), contradicting (22). Therefore . Similarly, . It follows that we reduced the discussion to determine whether or not , for each choice of .
If we omit both from , we get the two cases
- (1)
,
- (1’)
,
which are clearly isomorphic.
Now adding (respectively ) is the only consequence of adding (respectively ) as a line, and these additions do not interfere with each other. We are then bound to consider the cases:
- (2)
;
- (2’)
;
- (3)
;
- (4)
;
- (2”)
;
- (5)
.
The cases (2), (2’) and (2”) are clearly isomorphic. Applying the permutations and to in cases (4) and (5), respectively, we have reduced our discussion to the cases
- (1)
;
- (2)
;
- (3)
;
- (4)
;
- (5)
.
We list below the triangles missing in each of the cases:
- (1)
134, 234, 156, 256, 356, 456;
- (2)
134, 234, 256, 356, 456;
- (3)
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). Now 1 appears only once among the missing triangles in (2), and all points appear more often in (4); 1 and 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 . We confirm now that neither of them is a BRSC. For the first three cases, we take .
- (1)
, hence ; , hence ; contains the facet , hence . , hence ; , hence . Similarly, .
- (2)
Same argument as in (1).
- (3)
Same argument as in (1).
For the remaining two cases, we take .
- (4)
, hence ; , hence ; contains the facet , hence . , hence ; , hence . Similarly, .
- (5)
Same argument as in (4).
Proof of Lemma 6.2. Let and let . Since , there exist a BRSC and such that . We claim that
[TABLE]
This is equivalent to the equality
[TABLE]
Now yields and so
[TABLE]
Hence (25) and consequently (24) do hold.
Since BRSCs are closed under restriction, then is a BRSC and it follows from (24) that . Thus is closed under restriction. Since it is also closed under isomorphism, then 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 follows that is a prevariety itself.
Analysis of Example 7.4. Since is a Moore family, is a BRSC. The maximal chains in are
[TABLE]
[TABLE]
Hence . Since is the set of partial transversals of the successive differences for some of these chains, it follows easily that
[TABLE]
Write . It is routine to check that
[TABLE]
Indeed, it is easy to see that each is a partial transversal of the successive differences for some chain of type (26), and to check which transversals of the successive differences for some chain of type (27) cannot be obtained through chains of type (26).
Now we have . Let be such that . We show that .
- •
Suppose that . Since and , we have . Since and , we have . Since and , we get .
- •
Suppose that . Since and , we have . Since and , we have . Since and , we have . Since and , we get .
- •
Suppose that . Since and , we have . Since and , we get .
- •
Suppose that . Since and , we have . Since and , we get .
Thus there exists no such that . By Theorem 3.4, is not a TBRSC.
Proof of Lemma 7.5. Let be maximal with respect to . If , we can take and get , contradicting the maximality of . Thus and we are done.
Proof of Lemma 7.7. Suppose that . Then there exist such that , and . Hence and so by Lemma 7.5 there exists some such that and . But we have then , a contradiction since is a near-matroid. Therefore .
Proof of Lemma 7.8. Write with . Since , we have . Thus
[TABLE]
Moreover,
[TABLE]
If , we can now iterate this argument to produce a chain
[TABLE]
for some such that for . Thus and we are done.
Analysis of Example 7.11. It is easy to check that is indeed a simplicial complex. Clearly, . If is not contained in any element of , then . Hence, if and is not contained in any element of , then is a transversal of the successive differences for the chain
[TABLE]
in . On the other hand, it is easy to check that the unique having all 2-subsets contained in elements of is (see the picture below, where the yellow triangles are the elements of ):
1^{\prime}$$2^{\prime\prime}$$3$$3^{\prime}$$1$$3^{\prime\prime}$$2$$2^{\prime}$$1^{\prime\prime}
Now it is easy to check that for every . It follows that is a transversal of the successive differences for the chain
[TABLE]
in .
Finally, each facet of the form or is a transversal of the successive differences for the chain
[TABLE]
in . Since we have now checked all facets, it follows that is a BRSC.
Let denote the closure of in . For each , we have , so we successively get and . Thus contains , yielding . But then for every . Since , then there is no chain of the form (1) and so is not boolean representable.
We remark that is not pure since it is straightforward to check that is a facet. But is pure because there are no facets of dimension 1: given distinct , there exists some such that is not a yellow triangle.
Analysis of Example 7.12. Let be such a boolean matrix. Since the columns are all distinct and nonzero, every pair of distinct columns is independent. Now let be a set of independent columns with or . Let be such that the square matrix is nonsingular. Let and let be the column of having a 1 at row and 0 elsewhere. Then the permanent of equals the permanent of , hence is nonsingular. and so is independent. Thus is pure.
Since is by definition a BRSC, then is a TBRSC. Let denote the closure of in . Consider the columns of defined by
[TABLE]
The permanent of the matrix
[TABLE]
is 1, hence is independent. Define
[TABLE]
We have
[TABLE]
Since no row of has precisely two zeroes, is dependent. The same occurs with . It is immediate that has permanent 1, hence is independent. Thus we successively deduce , and so contains the facet . Therefore , where denotes the full set of vertices. Out of symmetry, so is .
Now
[TABLE]
Since no row of has precisely two zeroes, is dependent. The same occurs with . It is immediate that has permanent 1, hence is independent. Thus we successively deduce , and so contains the facet . Therefore . Together with , and being independent, this proves that is not a BRSC.
Proof of Theorem 7.13. Suppose first that . By Proposition 4.1, is a BRSC, therefore pure is a BRSC by Corollary 7.10(ii).
Thus we may assume that . Write pure and consider the restriction . Then pure. Since BRSCs are closed under restriction, is also a BRSC. Therefore we may assume that .
Let
[TABLE]
We claim that is a Moore family.
Clearly, . Let . We have . Let . Suppose that . Then or , contradicting . Thus and so is a Moore family.
Therefore is a BRSC. We claim that .
For every , let denote its closure in . Let . Then there exists an enumeration of the elements of such that
[TABLE]
and . Clearly, .
Suppose that satisfies . We may write with . If (respectively ), then (respectively ) is a transversal of the successive differences for (28), contradicting . Thus .
Suppose now that satisfies . We may write with . If (respectively ), then (respectively ) is a transversal of the successive differences for (28), contradicting . Thus .
Therefore (28) is a chain in and so . It follows that .
Conversely, let . Since , we have and so . We certainly have if if or 3, and the case follows from . Hence we may assume that . There exists an enumeration of the elements of and such that and . But then, by definition of , we get . Hence there exists some such that . Thus and so . Therefore and so . It follows that is a TBRSC.
Proof of Lemma 8.1. Obviously, is disconnected if and , and connected if . Hence we may assume that for some distinct .
Let be an boolean matrix representing . It follows from that . Thus, for every , we have either or , implying that or is an edge in . Therefore is connected.
Proof of Theorem 8.4. (i) It follows from Lemma 8.1 that is connected. By Hurewicz Theorem (see [4]), the 1st homology group of is the abelianization of , and therefore, in view of Theorem 8.2, a free abelian group of known rank. The second homology group of any 2-dimensional simplicial complex is , that is, a subgroup of a free abelian group. Therefore is itself free abelian.
(ii) By [16, Proposition 3.3], any finite 2-dimensional simplicial complex with free fundamental group has the homotopy type of a wedge of 1-spheres and 2-spheres.
Stuart Margolis, Department of Mathematics, Bar Ilan University, 52900 Ramat Gan, 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 Matemática, Faculdade de Ciências, Universidade do Porto, R. Campo Alegre 687, 4169-007 Porto, Portugal
E-mail address: [email protected]
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