Orthogonal Systems in Finite Graphs
aa r X i v : . [ m a t h . G R ] J u l Orthogonal systems in finite graphs
Andrew J. Duncan Ilya V. KazachkovVladimir N. Remeslennikov
November 13, 2018
Abstract
Let Γ be a finite graph and G Γ be the corresponding free par-tially commutative group. In this paper we construct orthogonalitytheory for graphs and free partially commutative groups. The theorydeveloped here provides tools for the study of the structure of thecentraliser lattice of partially commutative groups. Contents
Glossary of Notation
Γ — a finite undirected graph with vertex set X Γ ⊕ Γ — the join of graphs Γ and Γ C G ( S ) — the centraliser of a a subset S of G C ( G ) — the set of centralisers of a group GG or G (Γ) — the (free) partially commutative group with underly-ing graph Γlg( w ) — the length of a geodesic word w ′ such that w = G w ′ d ( x, y ) — the distance from x to y , x, y ∈ Γ O Z ( Y ) — the orthogonal complement of Y in Z , i.e. { u ∈ Z | d ( u, y ) ≤ , for all y ∈ Y } Y ⊥ — the orthogonal complement of Y in X , O X ( Y )cl Z ( Y ) — the closure of Y in Z with respect to O Z ( Y ), i.e.cl Z ( Y ) = O Z ( O Z ( Y ))cl( Y ) — the closure of Y in X , i.e. cl( Y ) = Y ⊥⊥ L (Γ) or L — the lattice of closed sets of Γ X — the set X ∪ { t } Γ — the graph (
X, E (Γ) ∪ E t ), E t = { ( t, x ) | x ∈ J t } , J t ⊆ XL — the lattice of closed sets, L (Γ), of Γ L t — { Y ⊆ X | Y = C ∩ J t , where C ∈ L } ˜ L — the set L ∪ L t h ( L ) — the height of a lattice LY ∼ ⊥ Z — Y, Z ⊆ X are ⊥ -equivalent in X , that is Y ⊥ = Z ⊥ acl( S ) — the Abelian closure of a simplex S , that is the unionof T ⊆ X such that S ∼ ⊥ TY ∼ o Z — subsets Y, Z ⊆ X are o -equivalent, i.e. Y ⊥ r Y = Z ⊥ r Z fcl( A ) — the free-closure of a free co-simplex A , that is theunion of all free co-simplexes B such that A ∼ o B x ] ⊥ — the ⊥ -equivalence class of x , that is { y ∈ X | x ∼ ⊥ y } [ x ] o — the o -equivalence class of x , that is { y ∈ X | x ∼ o y } x ∼ y — x, y ∈ X are equivalent, i.e. either x ∼ ⊥ y or x ∼ o y [ x ] — the equivalence class of x with respect to ∼ Γ c — the compression of the graph Γ Introduction
This paper is a continuation of a series of papers [4, 5] where the authorsdevelop the theory of free partially commutative groups.Free partially commutative groups arise in many branches of mathematicsand computer science and consequently are known by a variety of names: semifree groups , graph groups , right-angled Artin groups , trace groups , locallyfree groups . We refer the reader to [2], [9] and references there for a survey ofthese groups, which we shall refer to here as partially commutative groups.The analysis of proofs of results on partially commutative groups showsthat these rely heavily upon two main ideas: divisibility and orthogonality.The divisibility theory of partially commutative groups has been formalised in[9] and is a convenient tool for solving major algorithmic problems. The ideaof considering orthogonal complements of subsets of vertices of the underlyinggraph of a partially commutative implicitly occurs in many papers, see forinstance, [14, 12] and also [11] pp. 650-651. In this paper we formalise thisidea and establish the main results of orthogonality theory for graphs. Definition.
Let G (Γ ) be the partially commutative group with underlyinggraph Γ = (
X, E ) . For a vertex x ∈ X we define x ⊥ to be the set of allvertices of Γ connected with x . For a subset Y ⊆ X we define Y ⊥ = \ y ∈ Y y ⊥ . Let L (Γ) be the set of all subsets Z of X of the form Y ⊥ for some Y ⊆ X .We call L (Γ) the lattice of closed sets of Γ . The importance of the lattice of closed sets L (Γ) for the theory of par-tially commutative groups is a consequence of the the fact that the lattice L (Γ) is isomorphic to the lattice of parabolic centralisers (see Section 1) of3 (Γ) which, in turn, is crucial for study of the group G (Γ) itself and itsautomorphism group Aut( G (Γ)).The main problem that we consider in this paper is how the lattice ofclosed set behaves when one joins a vertex v to the graph Γ to form a newgraph ¯Γ. Naturally this depends on which vertices of Γ are joined to v . Inparticular, we prove that the lattices L = L (Γ) and L = L (Γ) are isomorphicif and only if v is joined to the orthogonal complement of a simplex S ⊂ X ;see Theorem 2.38.Moreover, we prove that the height h ( L ) of the extended lattice L is h ( L ) = h ( L ) + m , where m = 0 , L behaves nicely under these operations. We then introduce the notion ofcompression of a graph Γ which plays an important role in the study ofpartially commutative groups and prove that the lattices of closed sets forthe graph Γ and its compression are closely related. The compression of agraph allows us to give a decomposition of the automorphism group of thegraph as a semi-direct product of the automorphism group of the compressionwith a direct sum of symmetric groups.The results of the current paper play a key role in two papers of au-thors which are currently under preparation: one on the structure of latticesof centralisers of a given partially commutative group G , the other on thestructure of the automorphism group Aut( G ), [7, 8].A major part of our research on partially commutative groups, [4, 5, 6, 7,8] was carried out while the second and the third authors were visiting theUniversity of Newcastle Upon Tyne, thanks to the support of the EPSRCgrants EP/D065275/1 and GR/S61900/01. Graph will mean undirected, finite graph throughout this paper. If x and y are vertices of a graph then we define the distance d ( x, y ) from x to y to bethe minimum of the lengths of all paths from x to y in Γ. A subgraph S ofa graph Γ is called a full subgraph if vertices a and b of S are joined by anedge of S whenever they are joined by an edge of Γ.4et Γ be a graph with V (Γ) = X . A subset Y of X is called a simplex ifthe full subgraph of Γ with vertices Y is isomorphic to a complete graph. Amaximal simplex is called a clique . A subset Y of X is called a free co-simplex if the full subgraph of Γ with vertices Y is isomorphic to the null graph. Thereason why the word “free” is necessary here will become apparent later (seeSection 2.10).Let Γ i be a graph with vertex set X i , for i = 1 ,
2. The join Γ ⊕ Γ of Γ and Γ is the graph with vertex set the disjoint union X ⊔ X and edge setconsisting of all the edges of Γ i , for i = 1 and 2 and an edge joining x to x for all x ∈ X and x ∈ X . Let P be a partially ordered set with order relation ≤ . Then P is said to bea lattice if every pair of elements of P has a unique infimum and a uniquesupremum. We usually write s ∧ t and s ∨ t for the infimum and supremum,respectively, of s and t .A lattice is said to be bounded if it has both a minimum and a maximumelement. An ascending chain in a lattice is a sequence of elements a , a , . . . such that a i < a i +1 . The length of a finite chain a < · · · < a k is said to be k . Descending chains are defined analogously. A lattice may be boundedand have infinite ascending or descending chains (or both). The height of alattice L is defined to be the maximum of the lengths of all chains in L , if itexists, and ∞ otherwise.A homomorphism of partially ordered sets is a map from one partiallyordered set to another which preserves the order relation. If P and Q arelattices then a homomorphism of partially ordered sets f : P → Q is calleda homomorphism of lattices if f ( s ∨ t ) = f ( s ) ∨ f ( t ) and s ∧ t = f ( s ) ∧ f ( t ),for all s, t ∈ P . For further details on lattices we refer the reader to [1]. If S is a subset of a group G then the centraliser of S in G is C G ( S ) = { g ∈ G : gs = sg, for all s ∈ S } . We write C ( S ) instead of C G ( S ) when themeaning is clear. Let C ( G ) denote the set of centralisers of a group G . Therelation of inclusion then defines a partial order ‘ ≤ ’ on C ( G ). We define the5nfimum of a pair of elements of C ( G ) in the obvious way as: C ( M ) ∧ C ( M ) = C ( M ) ∩ C ( M ) = C ( M ∪ M ) . Moreover the supremum C ( M ) ∨ C ( M ) of elements C ( M ) and C ( M )of C ( G ) may be defined to be the intersection of all centralisers containing C ( M ) and C ( M ). Then C ( M ) ∨ C ( M ) is minimal among centraliserscontaining C ( M ) and C ( M ). These definitions make C ( G ) into a lattice,called the centraliser lattice of G . This lattice is bounded as it has a greatestelement, G = C (1), and a least element, Z ( G ), the centre of G . Lattices ofcentralisers have been extensively studied; a brief survey of results can befound in [4].The centraliser dimension of a group G is defined to be the height ofthe centraliser lattice of G and is denoted cdim( G ). Centralisers have theproperties that, for all subsets S and T of G , if S ⊆ T then C ( S ) ≥ C ( T )and C ( C ( C ( S ))) = C ( S ). Therefore if C < C < · · · is an ascending chainthen · · · > C ( C ) > C ( C ) is a descending chain and both chains are eitherinfinite or of the same length. Thus cdim( G ) is the maximum of the lengthsof descending chains of centralisers in G , if such a maximum exists, and isinfinite otherwise. Let Γ be a finite, undirected, simple graph. Let X = V (Γ) be the set ofvertices of Γ and let F ( X ) be the free group on X . For elements g, h of agroup we denote the commutator g − h − gh of g and h by [ g, h ]. Let R = { [ x i , x j ] ∈ F ( X ) | x i , x j ∈ X and there is an edge from x i to x j in Γ } . We define the partially commutative group with ( commutation ) graph Γ to bethe group G (Γ) with presentation h X | R i . (Note that these are the groupswhich are called finitely generated free partially commutative groups in [3].)Let Γ be a simple graph, G = G (Γ) and let w ∈ G . Denote by lg( w ) thelength of a geodesic word in X ∪ X − that represents the element w ∈ G :that is a word of minimal length amongst those representing w . We say that w ∈ G is cyclically minimal if and only iflg( g − wg ) ≥ lg( w )for every g ∈ G . 6he centraliser dimension of partially commutative groups is finite be-cause all partially commutative groups are linear [10] and all linear groupshave finite centraliser dimension, [13]. In [5] it is shown that the centraliserdimension of a partially commutative group is easy to calculate and dependsonly on the centralisers of subsets of X . If Y ⊆ X then we call C ( Y ) a canonical parabolic centraliser. It is not hard to prove that the intersec-tion of two canonical parabolic centralisers is again a canonical paraboliccentraliser and, as shown in [5], the supremum, in C ( G ), of two canonicalparabolic centralisers is also a canonical parabolic centraliser. Hence the set C ( X ; G ) of canonical parabolic centralisers forms a sublattice of C ( G ). In [5,Theorem 3.3] it is shown that the centraliser dimension of G is equal to theheight of the lattice C ( X ; G ). In [6] we give a short proof of this fact using themethods developed in this paper and give a characterisation of centralisersof arbitrary subsets of a partially commutative group. Moreover in [7, 8] weuse these tools to give a description of the automorphism group of a partiallycommutative group. As before let Γ be a finite, undirected, simple graph, with vertices X , and let G = G (Γ) be the partially commutative group defined by Γ. Given vertices x, y in the same connected component of Γ we define the distance d ( x, y )from x to y to be the minimum of the lengths of paths from x to y . If x and y are in distinct connected components then we define d ( x, y ) = ∞ .Let Y and Z be subsets of X . We define the orthogonal complement of Y in Z to be O Z ( Y ) = { u ∈ Z | d ( u, y ) ≤ , for all y ∈ Y } . By convention we set O Z ( ∅ ) = Z . If Z = X we call O X ( Y ) the orthogonalcomplement of Y , and if no ambiguity arises then we shall sometimes write Y ⊥ instead of O X ( Y ). Also, if every vertex of Z is either in Y or is joined byan edge of Γ to every vertex of Y then we write [ Y, Z ] = 1. Thus [
Y, Z ] = 1if and only if Z ⊆ O X ( Y ) if and only if every element of Y commutes withevery element of Z in the group G . For future reference we record some ofthe basic properties of orthogonal complements in the next lemma.7 emma 2.1. Let Y , Y , Y and Z be subsets of X .1. If Y ⊆ Z then Y ⊆ O Z ( O Z ( Y )) .2. If Y ⊆ Z then O Z ( Y ) = O Z ( O Z ( O Z ( Y ))) .3. If Y ⊆ Y then O Z ( Y ) ⊆ O Z ( Y ) .4. O Z ( Y ∩ Y ) ⊇ O Z ( Y ) ∪ O Z ( Y ) .5. O Z ( Y ∪ Y ) = O Z ( Y ) ∩ O Z ( Y ) .6. Y is a simplex if and only if Y ⊆ Y ⊥ .7. Y is a clique if and only if Y = Y ⊥ .In particular from 1 and 2 we have Y ⊆ Y ⊥⊥ and Y ⊥ = Y ⊥⊥⊥ , where wewrite Y ⊥⊥ for ( Y ⊥ ) ⊥ .Proof. If y ∈ Y ⊆ Z then, for all u ∈ O Z ( Y ), we have d ( u, y ) ≤ y ∈ O Z ( O Z ( Y )) and 1 follows. Statement 3 follows directly fromthe definition of orthogonal complement. Statement 2 follows from 1 and3. Statement 4 follows from 3. To see 5 suppose first that Z = X . It fol-lows from 3 that O X ( Y ∪ Y ) ⊆ O X ( Y ) ∩ O X ( Y ). From 4 and 1 we have O X ( O X ( Y ) ∩ O X ( Y )) ⊇ Y ∪ Y . Hence, from 1 and 3, O X ( Y ) ∩ O X ( Y ) ⊆O X ( O X ( O X ( Y ) ∩O X ( Y ))) ⊆ O X ( Y ∪ Y ), so 5 holds in this case. In general, O Z ( Y ∪ Y ) = O X ( Y ∪ Y ) ∩ Z = O X ( Y ) ∩ O X ( Y ) ∩ Z = O Z ( Y ) ∩ O Z ( Y ),as required. If Y is a simplex and y ∈ Y then d ( y, z ) = 1, for all z ∈ Y , z = y . Hence Y ⊂ Y ⊥ . Conversely, if Y ⊆ Y ⊥ and y, z ∈ Y then d ( y, z ) ≤ Y is a simplex. Therefore 6 holds. If Y is a clique and x ∈ Y ⊥ \ Y then Y ∪ { x } is a simplex, contrary to maximality of Y . Hence, using 6, Y = Y ⊥ .Conversely, if Y = Y ⊥ , then Y is a simplex and, by a similar argument, thereis no simplex strictly containing Y . Hence 7 holds. Example 2.2.
1. In general the inclusions of Lemma 2.1 are strict. Forinstance, take Γ to be the graph of Figure 2.1, let Y = { a, c } and Y = { b, c, d } . Then Y ⊥ = { b } , Y ⊥ = { c } and ( Y ∩ Y ) ⊥ = { b, c, d } :so ( Y ∩ Y ) ⊥ = Y ⊥ ∪ Y ⊥ . Moreover Y ⊥⊥ = { a, b, c } 6 = Y .2. The subgroup G ( X ⊥ ) is the centre of the group G = G (Γ).8Sfrag replacements a b c d Figure 2.1: A path graph3. If X = X ⊔ X is a disjoint union of X and X and Γ is the direct sumof graphs Γ( X ) and Γ( X ) then G = G ( X ) × G ( X ). If O X ( X ) = O X ( X ) = ∅ then the groups G ( X i ), i = 1 , O X ( X ) = X and O X ( X ) = X .The connection between orthogonal complements and centralisers is madeexplicit in the following lemma. Lemma 2.3.
Let G = G (Γ) and Y ⊆ X . Then C G ( Y ) = G ( Y ⊥ ) .Proof. If x ∈ X then C G ( x ) ⊇ G ( x ⊥ ). From [9, Lemma 2.4] we also have C G ( x ) ⊆ G ( x ⊥ ). Hence C G ( Y ) = ∩ y ∈ Y C G ( y ) = ∩ y ∈ Y G ( y ⊥ ) = G ( ∩ y ∈ Y y ⊥ ) = G ( Y ⊥ ).For subsets Y and Z of X we define the closure of Y in Z to be cl Z ( Y ) = O Z O Z ( Y ). When Z = X we write cl( Y ) for cl X ( Y ). The closure operatorin Γ satisfies the following properties. Lemma 2.4.
Let Y , Y , Y and Z be subsets of X .1. Y ⊆ cl( Y ) .2. cl( Y ⊥ ) = Y ⊥ .3. cl(cl( Y )) = cl( Y ) .4. If Y ⊆ Y then cl( Y ) ⊆ cl( Y ) .5. cl( Y ∩ Y ) ⊆ cl( Y ) ∩ cl( Y ) and cl( Y ) ∪ cl( Y ) ⊆ cl( Y ∪ Y ) .6. If Z = cl( Y ) then Z = U ⊥ , for some U ⊂ X , and then cl( U ) = Z ⊥ = Y ⊥ .7. If cl( Y ) = cl( Y ) then Y ⊥ = Y ⊥ .8. Y is a simplex if and only if cl( Y ) is a simplex if and only if cl( Y ) ⊆ Y ⊥ . . If Y ⊆ Y then cl(cl( Y ) ∩ Y ) = cl( Y ) .10. cl(cl( Y ) ∪ cl( Y )) = cl( Y ∪ Y ) .11. cl(cl( Y ) ∩ Y ) ∩ Y = cl( Y ) ∩ Y .Proof. Statements 1 and 2 are restatements of Lemma 2.1, 1 and 2, respec-tively. To see 3 apply the operator O X to both sides of 2. Statement 4 isa consequence of Lemma 2.1.3. Statement 5 follows from 4. If Z = cl( Y )then Z = U ⊥ , where U = Y ⊥ . If Z = U ⊥ then cl( U ) = U ⊥⊥ = Z ⊥ =(cl( Y )) ⊥ = Y ⊥ , using Lemma 2.1.2. Hence 6 holds. To see 7 apply theoperator O X to both cl( Y ) and cl( Y ) and use Lemma 2.1.2. For 8, if cl( Y )is a simplex then cl( Y ) ⊆ cl( Y ) ⊥ , from Lemma 2.1.6, so from 1 and Lemma2.1.2 cl( Y ) ⊆ Y ⊥ . If cl( Y ) ⊆ Y ⊥ then, from 1 and Lemma 2.1.6, Y ⊆ Y ⊥ so Y is a simplex, and cl( Y ) ⊆ cl( Y ⊥ ) = cl( Y ) ⊥ , so cl( Y ) is a simplex. Y ⊥⊥ ⊆ ( Y ⊥⊥ ) ⊥ = Y ⊥ so Y ⊆ Y ⊥⊥ ⊆ Y ⊥ ; and Y is a simplex. In the settingof 9 we have, from 1, Y ⊆ cl( Y ) ∩ Y , so cl( Y ) ⊆ cl(cl( Y ) ∩ Y ). On the otherhand cl( Y ) ∩ Y ⊆ cl( Y ) so, from 3 and 4, cl(cl( Y ) ∩ Y ) ⊆ cl( Y ). To see 10use the second part of 5 and then 3 to obtain cl(cl( Y ) ∪ cl( Y )) ⊆ cl( Y ∪ Y ).For the opposite inclusion use 1 to obtain Y ∪ Y ⊆ cl( Y ) ∪ cl( Y ) and then4 implies that cl( Y ∪ Y ) ⊆ cl(cl( Y ) ∪ cl( Y )), as required. For 11 first notethat 1 implies that cl( Y ) ∩ Y ⊆ cl(cl( Y ) ∩ Y ) ∩ Y . Also cl( Y ) ∩ Y ⊆ cl( Y )so 4 and 3 imply that cl(cl( Y ) ∩ Y ) ⊆ cl( Y ). On intersection with Y thisgives the inclusion required to complete the proof. Example 2.5.
1. If x ∈ X and Y = cl( x ) = x ⊥⊥ then Y is a simplex.2. In terms of the group G the subset Y of X is a simplex if and only if G ( Y ) is Abelian. As C G ( Z ) = G ( Z ⊥ ), for any subset Z of X , Lemma2.1.6 states that G ( Y ) is Abelian if and only if G ( Y ) ⊆ C G ( Y ). Thecontent of Lemma 2.4.8 is that G ( Y ) is Abelian if and only if C G ( Y ) isAbelian if and only if C G ( Y ) ⊆ C G ( Y ). Definition 2.6.
A subset Y of X is called closed ( with respect to Γ) if Y =cl( Y ) . Denote by L (Γ) the set of all closed subsets of X . We list some basic properties of L (Γ). Lemma 2.7.
Let Y be a subset of X . The following hold.1. cl( Y ) ∈ L (Γ) . . X is the unique maximal element of L (Γ) .3. Y is closed if and only if Y = O X ( U ) , for some U ∈ L (Γ) .4. O X ( X ) is the unique minimal element of L (Γ) .5. If Y , Y ∈ L (Γ) then Y ∩ Y ∈ L (Γ) .Proof.
1. This follows from Lemma 2.4.3.2. This is clear, given the previous statement and the fact that X ⊆ cl( X ).3. It follows, from Lemma 2.4, 2 and 6, that Y ∈ L (Γ) if and only if Y = O X ( U ), for some subset U of X . If Y is closed and Y = O X ( U )then Y = cl( Y ) = O X (cl( U )) and, as cl( U ) is closed, the result follows.4. From the previous statement it follows that O X ( X ) ∈ L (Γ). If Y ∈ L (Γ) then Y = O X ( U ), for some U ⊆ X . From Lemma 2.1 then O X ( X ) ⊆ O X ( U ) = Y , as required.5. From Lemma 2.4, 1 and 5, we have Y ∩ Y ⊆ cl( Y ∩ Y ) ⊆ cl( Y ) ∩ cl( Y ) = Y ∩ Y , the last equality holding by definition of closed set. Therefore Y ∩ Y =cl( Y ∩ Y ).The relation Y ⊆ Y defines a partial order on the set L (Γ). As theclosure operator cl is inclusion preserving and maps arbitrary subsets of X into closed sets we can make L (Γ) into a lattice by defining the the infimum Y ∧ Y of Y and Y by Y ∧ Y = cl( Y ∩ Y ) = Y ∩ Y and the supremum Y ∨ Y = cl( Y ∪ Y ). Proposition 2.8.
The set L (Γ) with operations ∧ and ∨ above is a completelattice.Proof. As we have seen L (Γ) is a lattice. From Lemma 2.7 it has maximumelement X and minimum element O X ( X ), so is complete. Proposition 2.9.
The operator O X maps L (Γ) to itself and is a lattice anti-automorphism. roof. If Y ∈ L (Γ) then, from Lemma 2.7, O X ( Y ) ∈ L (Γ); so O X maps L (Γ) to itself. From Lemma 2.1 O X is inclusion reversing. Moreover, for Y ∈ L (Γ) we have O X ( O X ( Y )) = Y ; so the restriction of O X to L (Γ) is abijection. Hence this restriction is a lattice anti-automorphism.If Z ⊆ X and Γ Z is the full subgraph of Γ with vertex set Z then, byabuse of notation, we write L ( Z ) for L (Γ Z ). As long as it is clear that Γ isfixed this should cause no confusion. We have O Z ( Y ) = O X ( Y ) ∩ Z so L ( Z )consists of subsets Y of Z such that Y = cl Z ( Y ) = O X ( O X ( Y ) ∩ Z ) ∩ Z . Now suppose that X is a disjoint union X = X ⊔ X , where X and X arenon-empty, and Γ = Γ( X ) ⊔ Γ( X ) (that is no edge of Γ joins a vertex of X to a vertex of X ). Write Γ i for Γ( X i ), i = 1 ,
2. We wish to describe L (Γ) interms of the lattices L (Γ i ). First of all we note the following lemma. Lemma 2.10.
With the hypotheses above, if U is a non-empty subset of X i then O X i ( U ) = O X ( U ) .Proof. By definition O X i ( U ) ⊆ O X ( U ). We have O X ( U ) = { x ∈ X | d ( u, x ) ≤ , ∀ u ∈ U } . If x / ∈ X i then, as U = ∅ , there is some u ∈ U such that d ( x, u ) = ∞ . Hence x ∈ O X ( U ) implies x ∈ X i , so x ∈ O X i ( U ).The relationship between L (Γ) and the L (Γ i ) is specified by the followingproposition. Proposition 2.11.
Let
Γ = Γ ⊔ Γ , as above.1. ∅ ∈ L (Γ) .2. A non-empty set Y is in L (Γ) \{ X, X , X } if and only if Y is in pre-cisely one of L (Γ i ) \{ X i } , i = 1 or .3. If O X i ( X i ) = ∅ then ∅ ∈ CS (Γ i ) and X i / ∈ CS (Γ) .4. If O X i ( X i ) = ∅ then ∅ / ∈ CS (Γ i ) and X i ∈ CS (Γ) .Proof.
1. As X i is non-empty it follows that ∅ = O X ( X ), so ∅ ∈ L (Γ).12. Let Y be a non-empty element of L (Γ) \{ X, X , X } . Then Y = O X ( U ), for some subset U of X . If U ∩ X i = ∅ , for i = 1 and 2,then O X ( U ) = ∅ . Hence U ⊆ X i , for i = 1 or 2. If U = ∅ then Y = X ,so U = ∅ and, from Lemma 2.10, Y = O X i ( U ) so is in L (Γ i ). Notethat in this case Y ⊆ X i and is non-empty; so cannot be in L (Γ j ), j = i . Conversely if Y is a non-empty element of L (Γ i ) \{ X i } then Y = O X i ( U ), for some U ⊆ X i . As Y = X i we have U = ∅ and so,from Lemma 2.10 again, Y ∈ CS (Γ).3. From Lemma 2.7, ∅ ∈ L (Γ i ). From Lemma 2.10 we have ∅ = O X i ( X i ) = O X ( X i ). If X i ∈ L (Γ) then X i = O X ( U ), for some U ∈ L (Γ). Hence ∅ = O X ( X i ) = U which implies X i = O X ( U ) = X , a contradiction.4. As O X i ( X i ) is the minimal element of L (Γ i ), in this case ∅ / ∈ L (Γ i ).We have X i = O X i ( U ), for some U ∈ L (Γ i ), so U = ∅ and U ⊆ X i .That X i ∈ L (Γ) now follows from Lemma 2.10.Let L = L (Γ), L i = L (Γ i ) and L ′ i = L (Γ i ) \{ X i } . Then Figure 2.2illustrates the composition of L (Γ) in terms of the L (Γ i ). Now supposePSfrag replacements XL L ′ X X ⊥ L L ′ X X ⊥ ∅O X i ( X i ) = ∅ , i = 1 , XL L ′ X X ⊥ L L ′ X X ⊥ ∅O X ( X ) = ∅ , O X ( X ) = ∅ ,PSfrag replacements XL L ′ X X ⊥ L L ′ X X ⊥ ∅O X i ( X i ) = ∅ , i = 1 , L of closed sets in a disconnected graphthat Γ has connected components Γ , . . . , Γ m , where V (Γ i ) = X i . Assumethat O X i ( X i ) = ∅ , for i = 1 , . . . , r and that O X i ( X i ) = ∅ , for i > r . A13traightforward induction using Proposition 2.11 shows that the lattice L (Γ)takes the form shown in Figure 2.3: where we use the obvious extension ofthe notation introduced above for the lattices L (Γ i ). We may often therefore ... ... PSfrag replacements XL L r L ′ r +1 X X ⊥ L r L ′ L ′ m X r X ⊥ r ∅ Figure 2.3: The lattice L of the graph with connected components Γ , . . . , Γ m .reduce to the study of L (Γ) where Γ is a connected graph.Now suppose that X ⊥ = ∅ and set X ∗ = X \ X ⊥ . Let Γ( X ∗ ) = Γ ∗ the fullsubgraph of Γ with vertex set X ∗ . Proposition 2.12.
The set O X ∗ ( X ∗ ) = ∅ and the lattice L (Γ) is isomorphicto the lattice L (Γ ∗ ) .Proof. From the definitions it follows that O X ( X ∗ ) = O X ( X ). Therefore O X ∗ ( X ∗ ) = O X ( X ∗ ) ∩ X ∗ = O X ( X ) ∩ X ∗ = ∅ . If Y = O X ( U ), where U ∈ L (Γ) then Y \ X ⊥ = O X ( U ) \O X ( X ) = O X ∗ ( U \O X ( X )) . Hence the map φ : Y → Y \ X ⊥ maps L (Γ) to L (Γ ∗ ).Clearly φ is inclusion preserving. To see that φ is surjective, note thatif V ⊆ X ∗ then O X ∗ ( V ) = C \ X ⊥ , where C = O X ( V \O X ( X )). Therefore φ is a surjective homomorphism of partially ordered sets. Since Y ∈ L (Γ)implies X ⊥ ⊆ Y it follows that φ is also injective; so φ is an isomorphism oflattices.The set O X ( X ) is called the kernel of the graph Γ. Given the propositionabove we may restrict to the study of lattices with the trivial kernel.14ow suppose that Γ = Γ( X ) ⊕ Γ( X ), for some partition X = X ∪ X of X (see Section 1.1). Let Γ i = Γ( X i ) and let G i = G (Γ i ), i = 1 ,
2; so G = G × G . Proposition 2.13.
In the above notation, if
Γ = Γ ⊕ Γ then L (Γ) = L (Γ ) × L (Γ ) . In this case the study of the lattice L (Γ) reduces to the study of L (Γ )and L (Γ ). We now consider the effect on the lattice of closed sets of the addition to Γ,or removal from Γ, of a vertex. In particular we shall see how the heights ofthese lattices are related and how to make restrictions on the way in whichthe new vertex is added to obtain isomorphism of the two lattices.We shall see below that if we adjoin a single vertex to Γ then the heightof the lattice of closed sets of the new graph is equal to h ( L (Γ)) + k , where k = 0 , V (Γ) = X and edges E (Γ). Let t be anelement not in X and define X = X ∪ { t } . Let J t be a subset of X . DefineΓ to be the graph with vertices X and edges E (Γ) ∪ E t , where E t is the set E t = { ( t, x ) | x ∈ J t } . Let L = L (Γ) and L = L (Γ).In order to understand how L and L are related we introduce a latticeintermediate between L and L . This will help us to give a simple descriptionof the structure of the lattice L in terms of the lattice L . Let L t = { Y ⊆ X | Y = C ∩ J t , where C ∈ L } . Now define the set of subsets ˜ L of X to be ˜ L = L ∪ L t . We shall see that ˜ L isa lattice which embeds in the lattice L . Note that if Y ∈ L t then Y = C ∩ J t ,for some C ∈ L , so Y = Y ∩ J t ⊆ cl( Y ) ∩ J t ⊆ cl( C ) ∩ J t = C ∩ J t = Y. Hence Y = cl( Y ) ∩ J t and it follows that cl( Y ) is the minimal element of L which intersects with J t to give Y , for all Y ∈ L t . Setting Z = cl( Y )this gives Z = cl( Y ) = cl(cl( Y ) ∩ J t ) = cl( Z ∩ J t ). Also if Y ∈ L t \ L thencl( Y ) = Y = cl( Y ) ∩ J t , so Z = cl( Y ) * J t . Conversely, given Z ∈ L such15hat Z * J t and Z = cl( Z ∩ J t ) then cl( Z ∩ J t ) = Z ∩ J t , so Z ∩ J t ∈ L t \ L .Therefore L t \ L = { Y = Z ∩ J t | Z ∈ L, Z * J t and Z = cl( Z ∩ J t ) } . (2.1)We define a closure operation icl X = icl on subsets of X byicl( U ) = (cid:26) cl X ( U ) , if U * J t cl X ( U ) ∩ J t , if U ⊆ J t , for U ⊆ X . Then icl( U ) ∈ ˜ L , for all U ⊆ X .Now assume that Y and Y are in ˜ L and Y ⊆ Y . If Y * J t thenicl( Y ) = cl X ( Y ) and icl( Y ) = cl X ( Y ) so icl( Y ) ⊆ icl( Y ). If Y ⊆ J t thenicl( Y ) = cl X ( Y ) ∩ J t ⊆ cl X ( Y ) ∩ J t ⊆ icl( Y ). Therefore icl is an inclusionpreserving map from subsets of X to ˜ L . It also follows from the definitionand 2.1 that icl( U ) = U , for all U ∈ ˜ L , so ˜ L is a retract of X . We maytherefore make ˜ L into a lattice by defining Y ∧ Y = icl( Y ∩ Y ) and Y ∨ Y = icl( Y ∪ Y ) , for Y , Y ∈ ˜ L . Lemma 2.14. If U, V ∈ ˜ L then U ∧ V = U ∩ V and U ∨ V = (cid:26) cl( U ∪ V ) ∩ J t , if U ∪ V ⊆ J t cl( U ∪ V ) , otherwise . Proof.
The expression for U ∨ V is merely a restatement of the definitions.If U ∈ L then icl( U ) = cl X ( U ). Therefore, for U and V in L we have (in thelattice ˜ L ) U ∧ V = U ∩ V . If either U or V belongs to L t then U ∩ V ⊆ J t so U ∧ V = cl( U ∩ V ) ∩ J t ⊆ cl( U ) ∩ cl( V ) ∩ J t = U ∩ V ⊆ cl( U ∩ V ) ∩ J t and the Lemma follows. Definition 2.15.
Define ˜ β to be the inclusion map of L into ˜ L and ˜ γ to bethe map from ˜ L to L given by ˜ γ ( Y ) = cl X ( Y ) , for Y ∈ ˜ L . Lemma 2.16.
The maps ˜ β and ˜ γ are homomorphisms of partially orderedsets and ˜ γ ˜ β = id L . We have ˜ β ( Y ∧ Z ) = ˜ β ( Y ) ∧ ˜ β ( Z ) , for all Y, Z ∈ L , and ˜ γ ( U ∨ V ) = ˜ γ ( U ) ∨ ˜ γ ( V ) , for all U, V ∈ ˜ L . If U, V ∈ ˜ L such that U = V and ˜ γ ( U ) = ˜ γ ( V ) then (after interchanging U and V if necessary) U ∈ L \ L t and V ∈ L t \ L and U = cl X ( V ) . a bce fx y Figure 2.4: Example 2.17
Proof.
The first statement is a direct consequence of the definitions, as is thefact that ˜ β respects the lattice infimum operation. For all U, V ∈ ˜ L we have˜ γ ( U ) ∨ ˜ γ ( V ) = cl(cl( U ) ∪ cl( V )) = cl( U ∪ V ) , from Lemma 2.4.10. If U ∪ V * J t then ˜ γ ( U ∨ V ) = cl cl( U ∪ V ) = cl( U ∪ V ).On the other hand, if U ∪ V ⊆ J t then ˜ γ ( U ∨ V ) = cl(cl( U ∪ V ) ∩ J t ) =cl( U ∪ V ), using Lemma 2.4.9. Hence ˜ γ ( U ) ∨ ˜ γ ( V ) = ˜ γ ( U ∨ V ), for all U, V ∈ ˜ L .Let U, V ∈ ˜ L . If U, V ∈ L then ˜ γ ( U ) = ˜ γ ( V ) implies U = V . If U, V ∈ L t then U = cl( U ) ∩ J t and V = cl( V ) ∩ J t and cl( U ) = cl(cl( U ) ∩ J t ) = ˜ γ ( U )and similarly cl( V ) = ˜ γ ( V ). Therefore ˜ γ ( U ) = ˜ γ ( V ) implies that U =cl( U ) ∩ J t = cl( V ) ∩ J t = V . Therefore, if U = V and ˜ γ ( U ) = ˜ γ ( V ) then oneof U, V is in L \ L t and the other in L t \ L . Assume then that U ∈ L \ L t and V ∈ L t \ L . In this case U = ˜ γ ( U ) = ˜ γ ( V ) = cl(cl( V ) ∩ J t ) = cl( V ).In general ˜ β does not preserve supremums and ˜ γ does not preserve infi-mums. Example 2.17.
In the graph of Figure 2.4 the sets B = { b } and C = { c } areclosed. The supremum B ∨ C = cl( B ∪ C ) = { b, c, y } and setting J t = { b, c } wehave ˜ β ( B ∨ C ) = { b, c, y } and ˜ β ( B ) ∨ ˜ β ( C ) = cl( { b, c } ) ∩{ b, c } = { b, c } . In thesame graph cl( x ) = { a, x, c } and cl( y ) = { b, y, c } . Set J t = { x, y } and then U = cl( x ) ∩ J t = { x } and V = cl( y ) ∩ J t = { y } are both elements of L t . Now U ∧ V = ∅ so ˜ γ ( U ∧ V ) = cl( ∅ ) = ∅ . However ˜ γ ( U ) ∧ ˜ γ ( V ) = cl( x ) ∩ cl( y ) = { c } .Next we show that the lattice ˜ L is embedded, as a partially ordered set,in L . Definition 2.18.
Let β : ˜ L → L and γ : L → ˜ L be the maps given by β ( Y ) = cl X ( Y ) , for Y ∈ ˜ L and γ ( Z ) = icl( Z \{ t } ) , for Z ∈ L . emma 2.19. The maps β and γ are homomorphisms of partially orderedsets and γβ = id ˜ L ; so β is injective and γ is surjective. We have γ ( Z ) = Z \{ t } , for all Z ∈ L , and β ( Y ) = (cid:26) Y if O X ( Y ) * J t ,Y ∪ { t } if O X ( Y ) ⊆ J t , (2.2) for all Y ∈ ˜ L .If Z and Z are elements of L such that Z = Z then γ ( Z ) = γ ( Z ) if and only if (after interchanging Z and Z if necessary) t ∈ Z and Z = Z \{ t } ∈ L .Proof. Since the closure operations in ˜ L and L preserve inclusion of sets itfollows from the definitions that β and γ are homomorphisms of partiallyordered sets.Now let U ∈ ˜ L . If U * J t then O X ( U ) = O X ( U ). On the other hand if U ⊆ J t then O X ( U ) = O X ( U ) ∪ { t } . Therefore, if U * J t then β ( U ) = O X O X ( U ) = (cid:26) O X O X ( U ) , if O X ( U ) * J t O X O X ( U ) ∪ { t } , if O X ( U ) ⊆ J t and (2.2) holds as U * J t implies that U ∈ L . If U ⊆ J t then β ( U ) = O X ( O X ( U ) ∪ { t } ) = O X ( O X ( U )) ∩ ( J t ∪ { t } )so β ( U ) = (cid:26) O X O X ( U ) ∩ J t , if O X ( U ) * J t ( O X O X ( U ) ∩ J t ) ∪ { t } , if O X ( U ) ⊆ J t . In this case, as U ⊆ J t we have O X O X ( U ) ∩ J t = cl X ( U ) ∩ J t = U . Thus, inall cases, (2.2) holds.Now suppose that Z ∈ L and let Y ∈ L such that Z = O X ( Y ). If t ∈ Y then Z = O X ( Y ) ⊆ O X ( t ) = J t ∪ { t } . Conversely if Z ⊆ J t ∪ { t } then t ∈ Y = O X ( Z ). Hence Z \{ t } ⊆ J t if and only if t ∈ Y . Similarly Y \{ t } ⊆ J t if and only if t ∈ Z . To show that γ ( Z ) = Z \{ t } we considervarious cases.1. Suppose that t ∈ Z and that t / ∈ Y . Then Y ⊆ J t and Z = O X ( Y ) = O X ( Y ) ∪ { t } . Therefore Z \{ t } = O X ( Y ) ∈ L and, since Z \{ t } * J t ,it follows that γ ( Z ) = icl( Z \{ t } ) = Z \{ t } .18. Assume that t ∈ Z and t ∈ Y . Then Y ⊆ J t ∪ { t } and Z = O X ( Y ) = O X (( Y \{ t } ) ∪ { t } )= O X ( Y \{ t } ) ∩ O X ( t )= ( O X ( Y \{ t } ) ∪ { t } ) ∩ ( J t ∪ { t } )= ( O X ( Y \{ t } ) ∩ J t ) ∪ { t } . Therefore Z \{ t } = O X ( Y \{ t } ) ∩ J t and, since Z \{ t } ⊆ J t , wehave, using Lemma 2.4.11, icl( Z \{ t } ) = cl( O X ( Y \{ t } ) ∩ J t ) ∩ J t = O X ( Y \{ t } ) ∩ J t . Therefore γ ( Z ) = icl( Z \{ t } ) = Z \{ t } .3. Assume that t / ∈ Z and t / ∈ Y . In this case Z = O X ( Y ) ∈ L and, since Z * J t , it follows that γ ( Z ) = cl( Z ) = Z = Z \{ t } .4. Assume that t / ∈ Z and t ∈ Y . Since t / ∈ Z this means that Z ⊆ J t and γ ( Z ) = cl( Z ) ∩ J t . Now Z = O X ( Y ) = O X ( Y \{ t } ) ∩ ( J t ∪ { t } )= O X ( Y \{ t } ) ∩ J t ∈ L t , as Y \{ t } * J t . Hence Z = cl( Z ) ∩ J t and so γ ( Z ) = Z = Z \{ t } .Thus γ ( Z ) = Z \{ t } , for all Z ∈ L .Now suppose that Z , Z ∈ L such that Z = Z . Suppose that γ ( Z ) = γ ( Z ). As γ ( Z i ) = Z i \{ t } we must have, after interchanging Z and Z ifnecessary, Z = Z ∪ { t } ; so t ∈ Z ∈ L and Z \{ t } ∈ L . Definition 2.20.
Let β : L → L be the map given by β ( Y ) = cl X ( Y ) , for Y ∈ L . Let γ : L → L be the map given by γ ( Z ) = cl X ( X ∩ Z ) , for Z ∈ L . Corollary 2.21.
We have β = β ˜ β and γ = ˜ γγ . The maps β and γ arehomomorphisms of partially ordered sets. For Y ∈ Lβ ( Y ) = (cid:26) Y if O X ( Y ) * J t ,Y ∪ { t } if O X ( Y ) ⊆ J t . Moreover γβ = id L , β is injective and γ is surjective. .4 The height of the extended lattice In this section we determine the possible differences in height between thelattices L and L . By a strong ascending chain in a partially ordered set L is meant a sequence C , C . . . of elements of L such that C i < C i +1 , forall i ≥ Strong descending chains are defined analogously, replacing < by > . The length of a finite strong chain C , . . . , C d is d . If C , C . . . isa sequence of elements of L such that C i ≤ C i +1 , for all i ≥
0, then wecall C , C . . . a weak ascending chain. Weak descending chains are definedanalogously. The length of a weak chain C is the maximum of the lengths ofstrong chains obtained by taking subsequences of C . We shall from now onuse chain to mean either weak or strong chain, if the meaning is clear. Wedenote the length of a chain C by l ( C ). Let L and L ′ be partially ordered setsand let φ : L → L ′ be a homomorphism or anti-homomorphism of partiallyordered sets. If C is a chain C , . . . , C d in L then we denote by φ ( C ) the chain φ ( C ) , . . . , φ ( C d ), in L ′ . Clearly the length of C is greater than or equal tothe length of φ ( C ). Definition 2.22.
The height h ( L ) of a lattice L is the length of its maximalchain, if this exists, and is infinite otherwise. The following is a corollary of Lemmas 2.16 and 2.19
Corollary 2.23. h ( L ) ≤ h ( ˜ L ) ≤ h ( L ) .Proof. If C is a maximal chain in L then ˜ β ( C ) is a chain in ˜ L . As ˜ β is injective˜ β ( C ) has the same length as C and the result follows. The second inequalityfollows similarly. Example 2.24.
Let Γ be the graph of Figure 2.1 and let J t = { a, c } . Then L consists of X , the orthogonal complements (in X ) of a , b , c and d , andalso { b, c } = O X { b, c } , { b } = O X { a, c } , { c } = O X { b, d } and ∅ . Therefore h ( L ) = 4. ˜ L contains in addition the set J t and the set { a } = J t ∩ O X ( a ). Itfollows that h ( ˜ L ) = 4 as well. Finally, the maximal proper subsets of L arethe orthogonal complements (in X ) of a , b , c and t (as O X ( d ) ⊆ O X ( c )). Theonly one of these sets with 4 elements is O X ( c ). However, the intersection of O X ( c ) with any other proper maximal subset has at most 2 elements. Hence L can have height at most 4. As h ( ˜ L ) = 4 it now follows that h ( L ) = h ( ˜ L ) = h ( L ) = 4. 20 xample 2.25. Let Γ be the graph of Figure 2.1 and Γ be the graph obtainedby removing vertex c . Then, with t = c we have X = { a, b, d } and J t = { b, d } .In this case L consists of the sets X , O X ( a ), O X ( d ) and ∅ , so h ( L ) = 2. L t contains in addition the sets J t and O X ( a ) ∩ J t = { b } . Thus h ( ˜ L ) = 3.Moreover, from the previous example h ( L ) = 4. (The semibraid group on n generators is the partially commutative group G n with presentation h x , . . . , x n | [ x i , x j ] = 1 , if | i − j | ≥ i . The graphs of this example are those of G and G , see [5] for further details)In fact these two examples illustrate the two extremes in differences ofheight between L and ˜ L and between ˜ L and L : as the following propositionsshow. Proposition 2.26. h ( ˜ L ) = h ( L ) + m , where m = 0 or .Proof. Let C = Z < · · · < Z k be a strictly ascending chain in ˜ L , with k = h ( ˜ L ). Then ˜ γ ( C ) is an ascendingchain in L . If Z i ∈ L for all i then ˜ γ ( C ) = C , so Lemma 2.23 implies that h ( ˜ L ) = h ( L ). Assume then that Z i / ∈ L , for some i , and let r be the smallestinteger such that Z r ∈ L , for all i ≥ r . Then Z i ⊆ J t , so Z i ∈ L t , for all i ≤ r −
1. Using Lemma 2.16, ˜ γ ( Z r ) < · · · < ˜ γ ( Z k ) and ˜ γ ( Z ) < · · · < ˜ γ ( Z r − )are strictly ascending chains in L . The length of ˜ γ ( C ) is therefore at least k − h ( ˜ L ) −
1; so h ( L ) ≥ h ( ˜ L ) −
1, and the lemma follows from Lemma2.23.
Proposition 2.27. h ( L ) = h ( ˜ L ) + m , where m = 0 or .Proof. Let C = Z < · · · < Z k be a strictly ascending chain in L . As γ isinclusion preserving the sequence γ ( C ) is ascending. Let r be the least integersuch that t ∈ Z i for r ≥ i . Then, from Lemma 2.19, γ ( Z ) < · · · < γ ( Z r − ) ≤ γ ( Z r ) < γ ( Z r +1 ) < · · · < γ ( Z k ) , so γ ( C ) has length at least k − Theorem 2.28. h ( L ) = h ( L ) + m , where m = 0 , or . a b cd efghijkl t Γ PSfrag replacements a b cd e f g hijk lt Γ Figure 2.5: Examples 2.29 and 2.30The next two examples show that a difference of one between the heightsof L and L may occur and may be due either to a difference in height between L and ˜ L or between ˜ L and L . Example 2.29.
Let Γ be the graph obtained by removing vertex t fromthe graph Γ = Γ of Figure 2.5 and let J t = { a, b, c } . Then h ( L ) = 4 and h ( L ) = 5. In this case the height of the lattice ˜ L is 5, with a maximal chain X > O X ( d ) > J t > O X ( f ) ∩ J t > O X ( f ) ∩ J t ∩ O X ( a ) > ∅ . Example 2.30.
Let Γ be the graph obtained by removing vertex t from thegraph Γ = Γ of Figure 2.5 and again let J t = { a, b, c } . Then h ( L ) = 5 and h ( ˜ L ) = 5. In this case the maximal chains in the lattice L involve only thevertices g, h, i, j, k, l and the sets of L t involve only vertices a, b, c . Thereforethe lattice ˜ L has some new chains of length 5 but none of length 6. Howevercomputation shows (see [5]) that h ( L ) = 6. Next we use the results of the Section 2.3 to describe the lattice L in terms ofthe lattice L . We make the following definition. Suppose that L is a latticewhich is a subset of a lattice L ′ and that the partial ordering in L is therestriction of the partial ordering in L ′ . Assume that L contains a subset22 such that there is an isomorphism of partially ordered sets, ρ , from S to L ′ \ L . Then we say that L ′ is obtained from L by doubling S along ρ .Recall from Section 2.3 that if Z ∈ ˜ L and Z * J t then Z ∈ L . This,together with (2.1), prompts the following definition. Definition 2.31.
Let R = { Z ∈ ˜ L | Z * J t and cl( Z ∩ J t ) = Z } and let ρ be the map from R to ˜ L given by ρ ( Z ) = Z ∩ J t . If Z ∈ R then Z ∈ L and Z / ∈ L t , as Z * J t . Furthermore, from (2.1), ρ ( Z ) ∈ L t \ L = ˜ L \ L . Proposition 2.32. ˜ L is obtained from L ⊆ ˜ L by doubling R along ρ .Proof. As ρ clearly preserves inclusion it suffices to show that ρ is a bijection.If ρ ( Y ) = ρ ( Z ), with Y, Z ∈ R then Z = cl( Z ∩ J t ) = cl( Y ∩ J t ) = Y , so ρ isinjective. From (2.1) if follows that ρ is also surjective.The lattice L is obtained from ˜ L by a doubling on an appropriate subsetof ˜ L . To see this we use the following strengthening of the final part ofLemma 2.19. We remark that condition (2.3) of the lemma can be expressedmore succinctly in terms of complements by noting that1. O X ( O X ( Y ) ∩ J t ) = O X ( O J t ( Y )) and2. if Y ⊆ J t then ( O X ( O X ( Y ) ∩ J t )) ∩ J t = cl J t ( Y ) ∈ L ( J t ). Lemma 2.33.
Let Y ⊂ X . Then Y and Y ∪ { t } belong to L if and only if O X ( Y ) * J t and Y = (cid:26) O X ( O X ( Y ) ∩ J t ) , if Y * J t O X ( O X ( Y ) ∩ J t ) ∩ J t , if Y ⊆ J t . (2.3) Proof.
Suppose that O X ( Y ) * J t . If Y * J t thencl X ( Y ∪ { t } ) = O X ( O X ( Y ) ∩ J t )= O X ( O X ( Y ) ∩ J t ) ∪ { t } . (2.4)23f, on the other hand, Y ⊆ J t then O X ( Y ∪ { t } ) = ( O X ( Y ) ∩ J t ) ∪ { t } socl X ( Y ∪ { t } ) = O X (cid:0) ( O X ( Y ) ∩ J t ) ∪ { t } (cid:1) = (cid:0) O X ( O X ( Y ) ∩ J t ) ∪ { t } (cid:1) ∩ ( J t ∪ { t } )= (cid:0) O X ( O X ( Y ) ∩ J t ) ∩ J t (cid:1) ∪ { t } . (2.5)In both cases, if in addition (2.3) holds then cl X ( Y ∪ { t } ) = Y ∪ { t } and Y ∪ { t } ∈ L .Now, given that O X ( Y ) * J t and (2.3) holds, choose x ∈ O X ( Y ) suchthat x / ∈ J t . Then O X ( x ) = O X ( x ) ⊇ cl X ( Y ) ⊇ Y and t / ∈ O X ( x ). Fromthe above Y ∪ { t } ∈ L , so Y ∪ { t } = O X ( Z ), for some Z ∈ L . Then O X ( Z ∪ { x } ) = O X ( Z ) ∩ O X ( x ) = Y ; and Y ∈ L .Conversely suppose that Y and Y ∪ { t } belong to L . In this case if O X ( Y ) ⊆ J t then O X ( Y ) ⊆ O X ( Y ) ∪{ t } so cl X ( Y ) ⊇ O X ( O X ( Y )) ∩ ( J t ∪{ t } ).Thus t ∈ cl X ( Y ) and Y / ∈ L , a contradiction. Thus O X ( Y ) * J t . If Y * J t then, from (2.4), Y = cl X ( Y ∪ { t } ) \{ t } = O X ( O X ( Y ) ∩ J t ). If, on theother hand, Y ⊆ J t then (2.5) implies that Y = ( O X ( O X ( Y ) ∩ J t )) ∩ J t , asclaimed.The lemma prompts the following definition. Definition 2.34.
Let S = { Y ⊂ X | Y * J t , O X ( Y ) * J t , and Y = O X ( O X ( Y ) ∩ J t ) } and S = { Y ⊂ X | Y ⊆ J t , O X ( Y ) * J t , and Y = ( O X ( O X ( Y ) ∩ J t )) ∩ J t } . Let S = S ∪ S and let T = { Y ∪ { t }| Y ∈ S } . Let σ be the map from S to T given by σ ( Y ) = Y ∪ { t } . From Lemma 2.33 it follows that S ∪ T ⊆ L and by definition S ⊆ ˜ L .Moreover, from Lemma 2.19, β ( Y ) = Y , for all Y ∈ S , so S = β ( S ) ⊆ β ( ˜ L ) ⊆ L . Proposition 2.35.
The lattice L is obtained from β ( ˜ L ) ⊆ L by doubling S along σ . roof. Using Lemma 2.19, if Y ∈ S and Y ∪ { t } = β ( U ), for some element U ∈ ˜ L , then t ∈ β ( U ) implies that O X ( U ) ∈ J t . However Y ∪ { t } = β ( U ) = U ∪ { t } so U = Y and O X ( Y ) * J t , a contradiction. Hence no element of T belongs to the image of β . If Z ∈ L and Z is not in the image of β then, fromLemma 2.19 again, γ ( Z ) = Z \{ t } ∈ ˜ L and so β ( Z \{ t } ) = Z . Thus either t / ∈ Z and β ( Z \{ t } ) = β ( Z ) = Z ∪{ t } or t ∈ Z and β ( Z \{ t } ) = Z \{ t } . In theformer case Z ∈ L and β ( Z ) = Z ∪{ t } ∈ L so Z ∈ S and Z ∪{ t } ∈ T ∩ Im( β ),a contradiction. Hence β ( Z \{ t } ) = Z \{ t } ∈ L and t ∈ Z . It follows fromLemma 2.33 that Z \{ t } ∈ S so Z ∈ T . That is, T = L \ β ( ˜ L ). As σ is aninclusion preserving bijection the result follows. In those cases where γ is injective it follows, from Corollary 2.21, that γ isa bijection and so an isomorphism of lattices. We now consider under whichconditions this may occur. Let V = O X ( t ) = J t ∪ { t } ∈ L . If γ is injectivethen V = βγ ( V ) = cl X ( J t ) ∪ { t } , so J t = cl X ( J t ) ∈ L . Therefore J t ∈ L is anecessary condition for γ to be injective. We shall show, in Section 2.8, thatif J t is closed then h ( L ) = h ( L ); but we shall also see in Lemma 2.37 that afurther condition is required to ensure that γ is injective. First however weestablish a simple form for γ when J t is closed. Lemma 2.36. If J t ∈ L then ˜ L = L and γ ( Z ) = Z \{ t } , for all Z ∈ L .Moreover, in the notation of Definition 2.34 , S = ∅ so L is obtained from β ( L ) ⊆ L by doubling S along σ .Proof. If J t ∈ L then L t is a subset of L , so ˜ L = L , as claimed. In this case γ = γ and β = β , so the first statement of the Lemma follows from Lemma2.19. If Y ∈ S then Y ∈ L and Y = O X ( W ), where W = O X ( Y ) ∩ J t ∈ L .However this means O X ( Y ) = W ⊆ J t , a contradiction. Lemma 2.37.
The map γ is an isomorphism of lattices if and only if J t = O X ( S ) , where S is a simplex of Γ .Proof. First assume that J t = O X ( A ), where A ⊆ X is a simplex. In thiscase, in the notation of Definition 2.34, Y ∈ S implies Y ∈ L ( J t ), so Y = O J t ( W ), for some W ⊆ J t . Now W ⊆ J t = O X ( A ) = O J t ( A ) which implies O J t ( O J t ( A )) ⊆ O J t ( W ) = Y . As A is a simplex A ⊆ J t so A ⊆ O J t ( O J t (( A ))and thus O X ( Y ) ⊆ J t , contrary to the definition of S . Therefore S = S = ∅ and from Lemma 2.36 L = L . 25n the other hand suppose that J t = O X ( N ), where N is not a simplex.Then, from Lemma 2.1.6, there is s ∈ N such that s / ∈ J t . Therefore t / ∈O X ( N ) and we have J t = O X ( N ) ∈ L . Hence γ ( J t ) = J t = γ ( J t ∪ { t } ) and γ is not injective. From the remarks at the beginning of the Section it followsthat if J t is not the orthogonal complement of a simplex in X then γ is notinjective. (It is not difficult to see that in this case J t ∈ S .)As a consequence of this lemma we obtain the following theorem. Theorem 2.38.
The lattices L and L are isomorphic if and only if J t = O X ( S ) , where S ⊂ X is a simplex, in which case γ is an isomorphism.Proof. From Lemma 2.37, if J t = O X ( S ), where S ⊂ X is a simplex, thenthe lattices are isomorphic and γ is an isomorphism. Now suppose that J t isnot of this form. The map β : L → L is injective so | L | ≤ | L | . If | L | = | L | then, as γβ = id L , it follows that γ is also injective, contrary to Lemma 2.37.Thus | L | < | L | and the lattices are not isomorphic. In this section we consider further the case where the set J t defined above isthe orthogonal complement of a simplex, as in the previous section. First weintroduce some equivalence classes on subsets of vertices Γ. We say that twosubsets S and T of X are ⊥ - equivalent in X and write S ∼ ⊥ T if and only if S ⊥ = T ⊥ ; that is O X ( S ) = O X ( T ). Lemma 2.39.
Let S and T be subsets of X .1. S ∼ ⊥ T if and only if T ⊆ cl X ( S ) and S ⊆ cl X ( T ) .2. If S ∼ ⊥ T and Y ∈ L (Γ) then S ⊆ Y implies that T ⊆ Y .3. If S is a simplex and S ∼ ⊥ T then T is a simplex. In particular, inthis case, G (Γ ′ ) is an Abelian group, where Γ ′ denotes the full subgraphof Γ on S ∪ T .Proof. To see the first statement note that, using Lemma 2.1, S ∼ ⊥ T if andonly if cl X ( S ) = cl X ( T ). It follows that S ∼ ⊥ T implies that S ⊆ cl X ( T ) and T ⊆ cl X ( S ). Conversely if S ⊆ cl X ( T ) then S ⊥ ⊇ T ⊥⊥⊥ = T ⊥ . Similarlyif T ⊆ cl X ( S ) then T ⊥ ⊇ S ⊥ and the result follows. To prove the secondstatement note that by Lemma 2.4, S ⊆ Y and Y closed implies cl X ( S ) ⊆ Y .26hus T ⊆ cl X ( T ) = cl X ( S ) ⊆ Y . For the third statement we have S ⊆O X ( S ) = O X ( T ), since S is a simplex, and so T ⊆ O X ( S ) = O X ( T ). Hence T is a simplex and the result follows.In the light of Lemma 2.39.3 we define the Abelian closure acl( S ) of asimplex S to be the union of subsets T of X such that S ∼ ⊥ T . Then S ⊆ acl( S ) and it is easy to see then that acl( S ) is the unique maximalsimplex such that S ∼ ⊥ acl( S ).Now let ∆ be a graph with vertices V . Let S be a simplex of ∆ and y ∈ V with y / ∈ S and suppose that S ∼ ⊥ { y } in ∆: that is O V ( S ) = O V ( y ). Let∆ y = ∆ \{ y } . Then ∆ y is called an elementary Abelian deflation of ∆ and ∆is called an elementary Abelian inflation of ∆ y . In this case the subgroup of∆ y generated by S is a free Abelian group of rank | S | and the subgroup of∆ generated by S ∪ { y } is free Abelian of rank | S | + 1.If a graph Ω can be obtained from a graph Γ by finitely many elementaryAbelian inflations then Ω is called an Abelian inflation of Γ and Γ is called an
Abelian deflation of Ω. The same terminology carries over to the respectivepartially commutative groups.
Proposition 2.40. If ∆ is an Abelian inflation of Γ then L (∆) ≃ L (Γ) .Proof. It suffices to prove the result in the case where ∆ is an elementaryAbelian inflation of Γ. Suppose then that Γ = ∆ t , for some vertex t of ∆.To be more explicit let V (∆) = X , assume that t ∈ X , S ⊆ X is a simplex, t / ∈ S and S ∼ ⊥ { t } in ∆. Let X = V (Γ). Then, as Γ = ∆ t we have X = X ∪ { t } and O X ( t ) = O X ( S ). Let J t = O X ( t ) \{ t } . Then, as S ⊆ X ,we have O X ( S ) = J t ∈ L (Γ). As ∆ is obtained from Γ by adding the vertex t which is joined to precisely those vertices in J t = O X ( S ), and S is a simplex,it follows from Theorem 2.38 that L (Γ) ≃ L (∆), as claimed. We saw in Section 2.6 that if J t is a closed set then, in the notation ofDefinition 2.34, L is obtained from β ( L ) by doubling S along σ . In thissection we shall show that if J t is closed then h ( L ) = h ( L ). If J t = O X ( S )where S is a simplex then Γ is an Abelian inflation of Γ, so this follows fromProposition 2.40. Therefore we assume that A ⊆ X , such that A is not asimplex, and J t = O X ( A ) ∈ L . As A is not a simplex the set A ′ = A \ J t isnon-empty. Fix a ∈ A ′ . 27ow let Y ∈ L with t ∈ Y . Then Y = O X ( Z ), where Z ⊆ J t ∪ { t } . Thereare two possibilities. Either1. Z ⊆ J t , in which case A ∪ { t } ⊆ O X ( Z ) = Y ; or2. Z * J t , in which case Z = W ∪ { t } , where W ⊆ J t , so a / ∈ Y .In the latter case Y = O X ( W ∪ { t } )= ( O X ( W ) ∪ { t } ) ∩ ( J t ∪ { t } )= ( O X ( W ) ∩ J t ) ∪ { t } whereas O X ( W ∪ { a } ) = O X ( W ) ∩ O X ( a )= ( O X ( W ) ∪ { t } ) ∩ O X ( a )= O X ( W ) ∩ O X ( a ) . This prompts us to define a map α : L → L by α ( Y ) = (cid:26) O X ( W ∪ { a } ) if t ∈ Y, a / ∈ YY otherwise . Note that t / ∈ α ( Y ) and Y \{ t } ∪ { a } ⊆ α ( Y ) , if t ∈ Y and a / ∈ Y (2.6)and that either t / ∈ α ( Y ) or A ∪ { t } ⊆ α ( Y ) for all Y ∈ L. (2.7)Now let C = Z < · · · < Z k be a strong ascending chain in L . Let α ( C ) = α ( Z ) ≤ · · · ≤ α ( Z k ). Lemma 2.41. α ( C ) is a strong ascending chain in L .Proof. Define r = r ( C ) to be the smallest integer such that t ∈ Z r . If nosuch r exists then α ( C ) = C and there is nothing to prove. Suppose thenthat 1 ≤ r ≤ k . Let s be the smallest integer such that A ∪ { t } ⊆ Z s (andset s = k + 1 if A ∪ { t } * Z k ). Then r ≤ s ≤ k + 1. For i such that28 ≤ i ≤ r − s ≤ i ≤ k we have α ( Z i ) = Z i . Therefore we need onlycheck that α ( Z i ) < α ( Z i +1 ) for i such that r − ≤ i ≤ s . If r = s then also α ( Z r ) = Z r and so α ( C ) = C and the Lemma holds.Assume then that r < s . In this case a / ∈ Z r and so a / ∈ Z r − . Therefore a / ∈ α ( Z r − ) = Z r − but a ∈ α ( Z r ). As t / ∈ Z r − we have Z r − ≤ Z r \{ t } ≤ α ( Z r ) , so α ( Z r − ) < α ( Z r ).To see that α ( Z s − ) < α ( Z s ) write Z s = O X ( Y s ), where Y s ⊆ J t and Z s − = O X ( W s − ∪ { t } ), where W s − ⊆ J t . As Z s − < Z s we have W s − ∪{ t } ≥ Y s and, as t / ∈ Y s , W s − ≥ Y s ; so O X ( W s − ) ≤ O X ( Y s ). Therefore α ( Z s − ) = O X ( W s − ) ∩ O X ( a ) ≤ O X ( Y s ) < O X ( Y s ) ∪ { t } = Z s = α ( Z s ) . It remains to check that α ( Z i ) < α ( Z i +1 ), where r ≤ i ≤ s −
2. Givensuch i we have, for j = i and j = i + 1, Z j = O X ( W j ∪ { t } ) = ( O X ( W j ) ∩ J t ) ∪ { t } , where W j ⊆ J t . As Z i < Z i +1 we have W i > W i +1 so O X ( W i ) ≤ O X ( W i +1 ).Therefore α ( Z i ) = O X ( W i ) ∩ O X ( a ) ≤ O X ( W i +1 ) ∩ O X ( a ) = α ( Z i +1 ) . Moreover, as Z i < Z i +1 there is x ∈ O X ( W i +1 ) ∩ J t such that x / ∈ O X ( W i ) ∩ J t .Hence x / ∈ O X ( W i ) and therefore x / ∈ α ( Z i ). However J t ⊆ O X ( a ) so x ∈ O X ( W i +1 ) ∩ J t implies x ∈ α ( Z i +1 ). Thus α ( Z i ) < α ( Z i +1 ).Given a chain C = Z < · · · < Z k in L define γ ( C ) to be the chain γ ( Z ) ≤ · · · ≤ γ ( Z k ). Lemma 2.42. If C is a strictly ascending chain in L such that Z i satisfies (2.7) , for i = 1 , . . . , k , then γ ( C ) is a strictly ascending chain in L .Proof. As before define r = r ( C ) to be the smallest integer such that t ∈ Z r . As J t is closed we have γ ( Z ) = Z \{ t } , for all Z ∈ L . Therefore itsuffices to show that γ ( Z r − ) < γ ( Z r ). We have A ⊆ γ ( Z r ), by (2.7). If A ⊆ Z r − = O X ( Y r − ) then O X ( t ) = J t ∪ { t } = O X ( A ) ⊇ Y r − . In this case t ∈ cl X ( t ) ⊆ Z r − , contrary to the definition of r . Hence A * Z r − and so γ ( Z r − ) < γ ( Z r ). 29ow, given any strictly ascending chain C in L we may, according toLemma 2.41, construct a strictly ascending chain C = α ( C ), satisfying (2.7);as in the hypothesis of Lemma 2.42. Applying γ to C we obtain a strictlyascending chain γ ( C ) in L of the same length as C . Therefore we have thefollowing proposition. Proposition 2.43. If J t is closed then h ( L ) = h ( L ) . A subset A ⊆ X is called a co-simplex if A ∩ O X ( A ) = ∅ . In this section weconsider the case J t = O X ( A ) where A is a co-simplex. In this case if Y ∈ L and Y ⊆ J t then O X ( Y ) ⊇ O X ( J t ) ⊇ A . As A ∩ J t = ∅ we have O X ( Y ) * J t ,for all such Y . Therefore, if A is a co-simplex, S = { Y ∈ L | Y ⊆ J t , Y ∈ L ( J t ) } = L ∩ L ( J t ) = L ( J t ) , as L ( J t ) ⊆ L . Therefore L is obtained from L by doubling L ( J t ) along σ .It is easy to find examples showing that in general there may be elementsof L which are subsets of J t but do not belong to L ( J t ). This motivates thefollowing definition. Definition 2.44.
A closed subset J ∈ L is realisable if L ( J ) = { Y ∈ L | Y ⊆ J } . Lemma 2.45.
An element J ∈ L is realisable if and only if, for all s ∈ X \ J there exists W ⊆ J such that O X ( s ) ∩ J = O X ( W ) ∩ J .Proof. Let J = O X ( A ), where A ∈ L . Suppose that J is realisable and that s ∈ X \ J . Then Y = O X ( s ) ∩ J ∈ L and Y ⊆ J ; so Y ∈ L ( J ). Hence Y = O X ( W ) ∩ J , where W = O X ( U ) ∩ J , for some U ⊆ J , so W ⊆ J asrequired.Now suppose that J satisfies the condition of the Lemma. Let Y ∈ L such that Y ⊆ J . Then Y = O X ( Z ), for some Z ∈ L . Let Z = Z ∩ J and Z = Z \ Z . Fix z ∈ Z . By hypothesis there exists W z ⊆ J such that O X ( z ) ∩ J = O X ( W z ) ∩ J ∈ L ( J ). Therefore O X ( Z ) ∩ J = \ z ∈ Z ( O X ( z ) ∩ J ) ∈ L ( J ) . As Z ⊆ J it is also true that O X ( Z ) ∩ J ∈ L ( J ). We have Y ⊆ J so Y = O X ( Z ) ∩ J = ( O X ( Z ) ∩ J ) ∩ ( O X ( Z ) ∩ J ) ∈ L ( J ), as required.30e now have the following proposition. Proposition 2.46.
Let A be a co-simplex such that O X ( A ) is realisable. If J t = O X ( A ) then L is obtained from L by doubling S = { Y ∈ L | Y ⊆ J t } along σ . We now define another relation on the subsets of X , similar to that of Section2.7: but giving rise to free groups instead of free Abelian groups. If J t is theorthogonal complement of a co-simplex A then O X ( A ) = J t , since A ∩ J t = ∅ ,so O X ( A ) \ A = O X ( t ) \{ t } . This suggests the following definition. If Y and Z are subsets of X we say that Y and Z are o -equivalent and write Y ∼ o Z if O X ( Y ) \ Y = O X ( Z ) \ Z. (2.8)Note that if Y is a co-simplex then Y ∼ o Z implies that O X ( Y, Z ) = O X ( Y )and that G ( O X ( Y )) = C ( Y ∪ Z ) — the centraliser of Y and Z in G (Γ).We call a co-simplex A such that the full subgraph on A is the null grapha free co-simplex. If A is a free co-simplex and B is either a free co-simplexor a single vertex then A ∼ o B implies that the subgroup of G generated by A and B is a free group. We define the free-closure fcl( A ) of a free co-simplex A to be the union of all free co-simplexes B such that A ∼ o B . It is easy tosee that A ⊆ fcl( A ) and that fcl( A ) is the unique maximal free co-simplexsuch that A ∼ o fcl( A ).If J t is the orthogonal complement of a free co-simplex then we say that Γis an elementary free inflation of Γ and that Γ is an elementary free deflation of Γ. We say that ∆ is a free inflation of Γ and Γ is a free deflation of ∆ if∆ is obtained from Γ by a finite sequence of elementary free inflations. We now use the ideas of Sections 2.7 and 2.10 to define an equivalence relationon the vertices of a finite graph Γ; which will give a decomposition of theautomorphism group of Γ. We build this equivalence relation up out of therestrictions to singleton sets of the relations defined in Sections 2.7 and 2.10.The restriction of the relation of ⊥ -equivalence to one-element subsets of X gives and equivalence relation ∼ ⊥ on X such that x ∼ ⊥ y if and only if x ⊥ = y ⊥ . Denote the equivalence class of x under ∼ ⊥ by [ x ] ⊥ .31imilarly, restricting the relation of o -equivalence to one-element sub-sets gives an equivalence relation ∼ o on X such that x ∼ o y if and only if x ⊥ \{ x } = y ⊥ \{ y } . Denote the equivalence class of x under ∼ o by [ x ] o . Lemma 2.47. [ x ] ⊥ is a simplex, for all x ∈ X .2. [ x ] ⊥ ∩ [ x ] o = { x } , for all x ∈ X .3. If | [ x ] ⊥ | ≥ then | [ x ] o | = 1 .4. If | [ x ] o | ≥ then [ x ] o is a free co-simplex and | [ x ] ⊥ | = 1 .Proof. If x ∼ ⊥ y then, as x ∈ x ⊥ we have x ∈ y ⊥ , so [ x ] ⊥ is a simplex. If z ∈ [ x ] ⊥ ∩ [ x ] o then x ∈ x ⊥ = z ⊥ but x / ∈ x ⊥ \{ x } = z ⊥ \{ z } : so it must bethat x = z . If | [ x ] ⊥ | ≥ y = x and y ∼ ⊥ x . If z = x and z ∼ o x then z = y , as [ x ] ⊥ ∩ [ x ] o = { x } from the above. Thus y ∈ y ⊥ = x ⊥ implies y ∈ x ⊥ \{ x } = z ⊥ \{ z } , so z ∈ y ⊥ = x ⊥ , contradicting z ∼ o x . Asimilar argument shows that if | [ x ] o | ≥ | [ x ] ⊥ | = 1. If y = x and y ∼ o x then y / ∈ x ⊥ , as otherwise x ⊥ \{ x } 6 = y ⊥ \{ y } . Hence [ x ] o is a free co-simplexif | [ x ] o | ≥ ∼ on X by x ∼ y if and only if either x ∼ ⊥ y or x ∼ o y . From Lemma 2.47 ∼ is an equivalence relation and we denote theequivalence class of x under ∼ by [ x ]. Define subsets M , M ⊥ and M o of X by M = { x ∈ X : [ x ] = [ x ] o = [ x ] ⊥ = { x }} ,M ⊥ = { x ∈ X : | [ x ] ⊥ | ≥ } and M o = { x ∈ X : | [ x ] o | ≥ } . From Lemma 2.47 it follows that X is the disjoint union X = M ⊔ M ⊥ ⊔ M o .We use the equivalence ∼ to define a quotient graph of Γ. Definition 2.48.
The compression of the graph Γ is the graph Γ c with ver-tices X c = { [ v ] : v ∈ X } and an edge joining [ u ] to [ v ] if and only if ( u ′ , v ′ ) is an edge of Γ for all u ′ ∈ [ u ] and v ′ ∈ [ v ] . c (if there are vertices of Γ such that [ x ] ⊥ has more than two elements). If Γand Γ ′ are graphs without multiple edges, and there is a map f : V (Γ) → V (Γ ′ ) then we say that f induces a graph homomorphism f : Γ → Γ ′ if( f ( u ) , f ( v )) ∈ E (Γ) for all ( u, v ) ∈ Γ. Proposition 2.49.
The map c : X → X c given by c ( x ) = [ x ] , for x ∈ X ,induces a surjective graph homomorphism c : Γ → Γ c .Proof. The map c : X → X c is surjective by definition. If c maps edges ofΓ to edges of Γ c then, since neither graph has multiple edges, the inducedmap is a surjective graph homomorphism. Therefore it suffices to show thatif ( u, v ) is an edge of Γ then ([ u ] , [ v ]) is an edge of Γ c .Suppose then that u, v ∈ X , u = v and ( u, v ) is an edge of Γ. If [ u ] = [ v ]and | [ u ] | = 1 or [ u ] = [ u ] o then there are no edges of Γ joining elements of[ u ] to each other. Therefore if [ u ] = [ v ] we may assume that [ u ] = [ v ] = [ u ] ⊥ .In this case [ u ] is a simplex, with more than one element since u = v , and sothere is a loop e in Γ c from [ u ] to itself. Thus ( u, v ) maps to e , as required.Now suppose that [ u ] = [ v ]. If | [ u ] | = | [ v ] | = 1 then ([ u ] , [ v ]) is clearlyan edge of Γ c . Suppose then that | [ v ] | ≥ z ∈ [ v ], z = v . Then( u, v ) ∈ E (Γ) implies u ∈ v ⊥ . As either z ∼ ⊥ v or z ∼ o v and z = v it followsthat z ∈ u ⊥ . If | [ u ] | = 1 this implies that ([ u ] , [ v ]) ∈ E (Γ c ). If | [ u ] | ≥ w ∈ [ u ], w = u . Then w = z (as [ u ] = [ v ]) and z ∈ u ⊥ implies z ∈ w ⊥ .Hence ( w, z ) ∈ E (Γ) and it follows that ([ u ] , [ v ]) is an edge of Γ c .As usual we extend c to a map from subsets of X to subsets of X c bysetting c ( Y ) = ∪ y ∈ Y { c ( y ) } , for Y ⊆ X . If [ y ] ∈ X c then [ y ] ⊥ = { [ u ] ∈ X c : d ([ u ] , [ y ]) ≤ } = { [ u ] ∈ X c : d ( u, y ) ≤ } , by definition of Γ c , so for all y ∈ X , c ( y ) ⊥ = { [ u ] ∈ X c : u ∈ y ⊥ } = [ u ∈ y ⊥ { c ( u ) } = c ( y ⊥ ) . Now suppose that Z = { z , . . . , z n } ⊆ X . Then c ( Z ) ⊥ = ( ∪ ni =1 c ( z i )) ⊥ = ∩ ni =1 c ( z i ) ⊥ = ∩ ni =1 c ( z ⊥ i ) . Clearly ∩ ni =1 c ( z ⊥ i ) ⊇ c ( ∩ ni =1 z ⊥ i ). On the otherhand, if [ u ] ∈ ∩ ni =1 c ( z ⊥ i ) then [ u ] ∈ [ z i ] ⊥ , so d ([ u ] , [ z i ]) ≤ d ( u, z i ) ≤ i = 1 , . . . , n . Therefore u ∈ ∩ ni =1 z ⊥ i from which it follows that [ u ] ∈ c ( ∩ ni =1 z ⊥ i ). Hence c ( Z ) ⊥ = c ( ∩ ni =1 z ⊥ i ) = c ( Z ⊥ ).Now restricting the map c to closed sets we see that if Y ∈ L (Γ) then Y = Z ⊥ , for some Z ⊆ X so c ( Y ) = c ( Z ⊥ ) = c ( Z ) ⊥ ∈ L (Γ c ). Hence c L (Γ) to L (Γ c ), which we denote by c L . Let L denotethe lattice L (Γ) and L c the lattice L (Γ c ). Proposition 2.50.
The map c L : L → L c is a lattice epimorphism whichpreserves the unary relation ⊥ : that is c L ( Y ⊥ ) = c L ( Y ) ⊥ , for all Y ∈ L .Proof. As c is a surjective map it follows that every subset of X c is theimage of a subset of X . If W is a closed subset of X c then W = V ⊥ forsome subset V of X c . Choose Y ⊆ X such that c ( Y ) = V . As we haveseen above we have c ( Y ⊥ ) = V ⊥ = W . As Y ⊥ ∈ L we have c L ( Y ⊥ ) = W ,so c L is a surjective map. It therefore suffices to show that c L is a latticehomomorphism. If S, T ∈ L then S = U ⊥ and T = V ⊥ , for some U, V ∈ L .Then S ∧ T = S ∩ T and c L ( S ) ∧ c L ( T ) = c L ( U ⊥ ) ∩ c L ( V ⊥ )= c L ( U ) ⊥ ∩ c L ( V ) ⊥ = ( c L ( U ) ∪ c L ( V )) ⊥ = ( c L ( U ∪ V )) ⊥ = c L ( U ⊥ ∩ V ⊥ )= c L ( S ∧ T ) . Moreover c L ( S ∨ T ) = c L (( S ∪ T ) ⊥⊥ )= ( c L ( S ∪ T )) ⊥⊥ = ( c L ( S ) ∪ c L ( T )) ⊥⊥ = c L ( S ) ∨ c L ( T ) . Hence c L is a lattice homomorphism as claimed.We make Γ c into a labelled graph as follows. For x ∈ X define µ ( x ) = | [ x ] | and ν ( x ) = 1 , if x ∈ M , ν ( x ) = ⊥ , if x ∈ M ⊥ and ν ( x ) = o , if x ∈ M o . Definea labelling function l : X c → N × { , ⊥ , o } by l ([ y ]) = ( µ ( y ) , ν ( y )), for all y ∈ X c . Example 2.51.
In drawing the compressed graph vertices with labels of theform (1 ,
1) or ( r, ⊥ ) are represented as single circles containing the integer 1or r , respectively, and vertices with labels of the form ( r, o ) are representedas two concentric circles containing the integer r , as in Figure 2.6.34 b cd e 212 [a][c][e] Figure 2.6: A graph Γ and its compression Γ c Let Aut(Γ c ) denote the group of automorphisms of Γ c as a labelled graph:that is φ ∈ Aut(Γ c ) if and only if φ is an automorphism of the graph Γ c suchthat l ( φ ([ v ])) = l ([ v ]), for all [ v ] ∈ X c . Let Aut(Γ) denote the group of graphautomorphisms of Γ and let φ ∈ Aut(Γ). Since φ is an automorphism wehave φ ( u ⊥ ) = φ ( u ) ⊥ , for all u ∈ X . It follows that if u, v ∈ X and [ u ] = [ v ]then [ φ ( u )] = [ φ ( v )]. Applying φ − to the latter equality we see that [ u ] = [ v ]if and only if [ φ ( u )] = [ φ ( v )]. Since c and φ are graph homomorphisms itfollows that φ c = c ◦ φ is an automorphism of Γ c as a labelled graph: thatis φ c ∈ Aut(Γ c ). Denote by Aut( c ) the map which takes φ ∈ Aut(Γ) to φ c ∈ Aut(Γ c ). For [ v ] ∈ X c let S µ ( v ) denote the symmetric group of degree µ ( v ). Proposition 2.52.
The map
Aut( c ) is an epimorphism from Aut(Γ) to Aut(Γ c ) . There is a split short exact sequence → Y [ v ] ∈ X c S µ ( v ) → Aut(Γ)
Aut( c ) −−−→ Aut(Γ c ) → . (2.9) Proof.
We have seen that Aut( c ) is a map from Aut(Γ) to Aut(Γ c ). If φ, φ ′ ∈ Aut(Γ) then ( φ ◦ φ ′ ) c ([ v ]) = [ φ ◦ φ ′ ( v )] = φ c ([ φ ′ ( v )]) = φ c ◦ φ ′ c ([ v ]), for all[ v ] ∈ X c . Hence Aut( c ) is a homomorphism.Let [ v ] ∈ X c and consider the subgraph Γ([ v ]) of Γ. If φ ∈ Aut(Γ([ v ]))then we may extend φ to Γ by setting φ ( u ) = u , for all u / ∈ [ v ]. Hence we mayregard Aut(Γ([ v ])) as a subgroup of Aut(Γ). If u, v ∈ X and [ u ] = [ v ] then φ ◦ φ ′ = φ ′ ◦ φ , for all φ ∈ Aut(Γ([ u ])) and φ ′ ∈ Aut(Γ([ v ])). Moreover, as[ u ] ∩ [ v ] = ∅ we have Aut(Γ([ u ])) ∩ Aut(Γ([ v ])) = 1. Therefore Aut(Γ) contains35he subgroup A = Q [ v ] ∈ X c Aut(Γ([ v ])) . If φ ∈ A then φ ( v ) ∈ [ v ], for all v ∈ [ v ]and for all [ v ] ∈ X c . Therefore φ ∈ ker(Aut( c )) and so A ⊆ ker(Aut( c )).Conversely if φ ∈ ker(Aut( c )) then [ φ ( v )] = [ v ] so φ ( v ) ∈ [ v ], for all v ∈ X .Hence if φ ∈ ker(Aut( c )) then φ | [ v ] ∈ Aut(Γ([ v ])) and so φ ∈ A . Therefore A = ker(Aut( c )). For all [ v ] ∈ X c the graph Γ([ v ]) is either a simplex or afree co-simplex so Aut(Γ([ v ])) is isomorphic to the symmetric group S µ ( v ) ofdegree µ ( v ). Therefore Q [ v ] ∈ X c S µ ( v ) ∼ = A . To show that the sequence (2.9)is exact it remains only to show that Aut( c ) is surjective. However we shallfirst construct an embedding ι : Aut(Γ c ) → Aut(Γ).Fix a transversal V = { v , . . . , v n } for the map c : Γ → Γ c . For i suchthat 1 ≤ i ≤ n choose an ordering ( v i, , . . . v i,µ ( v i ) ) of the class [ v i ], with v i = v i, . Then X = ⊔ ni =1 ⊔ µ ( v i ) j =1 { v i,j } . For i, k such that 1 ≤ i ≤ k ≤ n and l ( v i ) = l ( v k ), define a map τ i,k : [ v i ] → [ v k ] by τ i,k ( v i,j ) = v k,j , j = 1 , . . . , µ ( v i ).Note that, as l ( v i ) = l ( v k ) the map τ i,k is a graph isomorphism from Γ([ v i ])to Γ([ v k ]). If τ i,k is defined and i < k we define τ k,i = τ − i,k . Furthermore if τ i,k and τ k,l are both defined then so is τ i,l and by construction τ i,l = τ k,l ◦ τ i,k .Now let φ c ∈ Aut(Γ c ) and define a map φ of X to itself as follows.Let v ∈ X . Then [ v ] = [ v i ], so v = v i,j , for unique i and j . There is aunique k such that φ c ([ v i ]) = [ v k ] and as l ( φ c ( v i )) = l ( φ c ( v i )) the map τ i,k is defined. Set φ ( v ) = τ i,k ( v i,j ) = v k,j . As all the τ i,k are isomorphisms andas φ c is a graph automorphism it follows that φ is a graph automorphism.Thus ι : φ c → φ is a map from Aut(Γ c ) to Aut(Γ). That ι is an injectivehomomorphism follows directly from the definition.If φ c ∈ Aut(Γ c ) and [ v ] ∈ X c then Aut( c ) ◦ ι ( φ c ) maps [ v ] to [ ιφ c ( v )] = φ c ([ v ]), so Aut( c ) ◦ ι is the identity on Aut(Γ c ). This implies that Aut( c ) issurjective; so the sequence (2.9) is exact. Furthermore ι is a transversal forAut( c ) and so (2.9) splits, as claimed.The compression Γ c of Γ gives rise to a natural decomposition of G (Γ)which we now describe; using the following generalisation of a partially com-mutative group. Let Γ be a graph and to each vertex of Γ associate a group G v . Let F = ∗ v ∈ V (Γ) G v and let N be the normal subgroup of F generatedby all elements of the form [ g u , g v ], where g u ∈ G u , g v ∈ G v and u and v arejoined by an edge of Γ. The group G = F/N is called a partially commuta-tive product of groups . If all the vertex groups G v are infinite cyclic groupsthen G is a partially commutative group. In the case in question take Γ c tobe the underlying graph and associate the the partially commutative groupwith commutation graph Γ([ v ]) to the vertex [ v ]. The vertex groups are all36hen free Abelian groups or free groups. References [1] G. Birkhoff, Lattice Theory,
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