On the structure of the tight-span of a totally split-decomposable metric

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Onthestructureofthetight-spanofatotallysplit-decomposablemetric

ArticleinEuropeanJournalofCombinatorics·April2006

DOI:10.1016/j.ejc.2004.05.007·Source:DBLP

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European Journal of Combinatorics 27 (2006) 461–479

www.elsevier.com/locate/ejc

On the structure of the tight-span of a totallysplit-decomposable metric

K.T. Hubera, J.H. Koolenb,∗, V. Moultona

aSchool of Computing Sciences, University of East Anglia, Norwich, NR4 7TJ, UKbDivision of Applied Mathematics, KAIST, 373–1 Kusongdong, Yusongku, Daejon 305 701, Republic of Korea

Received 22 January 2004; accepted 31 May 2004Available online 8 December 2004

Abstract

The tight-span of a finite metric space is a polytopal complex with a structure that reflectsproperties of the metric. In this paper we consider the tight-span of a totally split-decomposablemetric. Such metrics are used in the field of phylogenetic analysis, and a better knowledge of thestructure of their tight-spans should ultimately provide improved phylogenetic techniques. Here weprove that a totally split-decomposable metric is cell-decomposable. This allows us to break up thetight-span of a totally split-decomposable metric into smaller, easier to understand tight-spans. Asa consequence we prove that the cells in the tight-span of a totally split-decomposable metric arezonotopes that are polytope isomorphic to either hypercubes or rhombic dodecahedra.© 2004 Elsevier Ltd. All rights reserved.

1. Introduction

In this paperX will denote a finite set with|X| ≥ 2. Given a metricd on X, i.e. asymmetric mapd : X × X → R that vanishes precisely on the diagonal and satisfiesthe usual triangle inequality, associate a polytopal complexT(d) to d as follows. LetRX

∗ Corresponding author.E-mail addresses:[email protected] (K.T. Huber), [email protected] (J.H. Koolen),

[email protected] (V. Moulton).

0195-6698/$ - see front matter © 2004 Elsevier Ltd. All rights reserved.doi:10.1016/j.ejc.2004.05.007

462 K.T. Huber et al. / European Journal of Combinatorics 27 (2006) 461–479

denote the set of functions that mapX to R. Associate the polyhedron

P(d) = { f ∈ RX : f (x)+ f (y) ≥ d(x, y) for all x, y ∈ X},to d, and letT(d) consist of the bounded faces ofP(d). The complexT(d) is known asthe tight-spanof d.

The tight-span of a metric was first introduced by Isbell [18], and has been subsequentlyrediscovered and studied in e.g. [3,7]. More recently, tight spans have been seen to arisenaturally in the context oftropical geometry, where they have been shown to be related totropical polytopes[4,5]. The structure ofT(d) can be used to deduce properties ofd. Forexample, a study of metrics having “tree-like” tight-spans in [7] led in part to the conceptof a totally split-decomposablemetric [2], which, besides having applications to the theoryof finite metric spaces [6], is now regularly used within phylogenetic analysis (cf. e.g. [1,13,17,20]).

In this paper we are interested in better understanding the structure of the tight-spanof a totally split-decomposable metric. Building upon results on this structure presentedin [9,10,12], in our main result (Theorem 7.1) we prove that the tight-span of a totallysplit-decomposable metric iscell-decomposable, a property that implies that the tight-span consists of smaller, easier to understand tight-spans. As a consequence, using resultsin [15,16], we prove inCorollary 7.3that the cells in the tight-span of a totally split-decomposable metric are zonotopes that are polytope isomorphic either to hypercubes orrhombic dodecahedra. We expect that this improved understanding of the tight-span willultimately lead to better tools for phylogenetic analysis.

The rest of the paper is organised as follows. InSection 2, we present somepreliminaries including a definition for theBuneman complex. This is a polyhedral complexthat can be associated to a totally split-decomposable metric. InSection 3we present somenew results concerning the structure of the Buneman complex. We then consider a mapκ

introduced in [9], that relates the Buneman complex of a totally split-decomposable metricto its tight-span. In particular, inTheorem 4.3we characterise whenκ maps the Bunemancomplex into the tight-span. As a corollary, inSection 5we give conditions for whenκinduces an injection from the set of maximal cells of the Buneman complex into the set ofthe maximal cells of the tight-span, and subsequently, inSection 6, we characterise whenκ induces a bijection. Using these results together with some from [15,16], in Section 7,we conclude with the proofs ofTheorem 7.1andCorollary 7.3.

2. Preliminaries

In this section, we review some properties of the Buneman complex and the tight-span. We begin by recalling some basic definitions concerning polytopes and polytopalcomplexes.

2.1. Polytopal complexes

We follow [19] and [21]. A polyhedronin Rn, n ∈ N, is the intersection of a finitecollection of halfspaces inRn and apolytope is a bounded polyhedron. Aface of apolyhedronP is the empty-set,P itself, or the intersection ofP with a supporting

K.T. Huber et al. / European Journal of Combinatorics 27 (2006) 461–479 463

hyperplane and, if dim(P) = d, i.e. P is d-dimensional, then its 0-dimensional faces arecalled itsvertices. The collection of all faces of a polytope forms a lattice with respect to theordering given by set inclusion, and we say that two polytopes arepolytope isomorphiciftheir face-lattices are isomorphic. Apolyhedral complexC is a finite collection of polyhedra(which we callcells) such that each face of a member ofC is itself a member ofC, and theintersection of two members ofC is a face of each. If all of the cells inC are polytopes, wecall C a polytopal complex. Given a polyhedral complexC, we will not usually distinguishbetweenC and itsunderlying set

⋃C∈C C. For anyc in the underlying set ofC, we let[c]

denote the minimal cell inC (under inclusion of cells), that containsc. Also, if c is in theunderlying set ofC andC is a cell inC with C = [c], then we say thatc is a generatorof C.

2.2. Totally split-decomposable metrics

A split of X is a bipartition ofX, and a setS of splits of X is a split system(on X).Denote the split system consisting of all possible splits ofX by S(X). For everyx ∈ Xand any splitS of X, we denote byS(x) the element ofS that containsx, and byS(x)the complement ofS(x). To avoid certain non-essential technicalities, in this paper we willassume that all split systems are non-empty and, forS a split system onX, that, for allx �= y in X, there exists some splitS ∈ S with S(x) �= S(y). A weightingon a split systemS ⊆ S(X) is a mapα : S → R>0 : S �→ αS = α(S), and such a pair(S, α) is called aweighted split system(on X). We call a split systemS ⊆ S(X) weakly compatibleif thereexist no three distinct splitsS1, S2, S3 ∈ S and four elementsx0, x1, x2, x3 ∈ X such that

Sj (xi ) = Sj (x0) if and only if i = j . (1)

Now, a metricd on X is calledtotally split-decomposableif there exists a weighted weaklycompatible split system(S, α) on X with

d = dS,α =∑S∈S

αSδS,

where, for any splitS ∈ S(X) and allx, y ∈ X,

δS(x, y) ={

1 if S(x) �= S(y),0 else.

Note that ifd is such a metric, then it follows by results in [2] that if d = dS ′,α′ for someweakly compatible split systemS ′ and weightingα′ on S ′, thenS ′ = S andα′ = α.Totally split-decomposable metrics were introduced in [2]. Besides having mathematicalinterest, such metrics play a useful role in phylogenetic analysis (cf. e.g. [13,17]).

2.3. The Buneman complex

We begin by recalling some further definitions concerning splits and split systems.Recall thatX is a finite set. For every proper non-empty subsetA ⊆ X, we denote thesplit {A, A} by SA. Given a split systemS ⊆ S(X), we define its underlying setU(S) by

U(S) =⋃S∈S

S= {A ⊆ X | there existsS∈ S with A ∈ S}.

464 K.T. Huber et al. / European Journal of Combinatorics 27 (2006) 461–479

We call two distinct splitsS, S′ ∈ S(X) compatibleif there exists someA ∈ S and someA′ ∈ S′ with A ∩ A′ = ∅, otherwise we callS andS′ incompatible. We call a split systemS ⊆ S(X) incompatibleif every pair of distinct splits inS is incompatible. We also defineany split system with cardinality one to be incompatible.

Now, given any mapφ : U(S) → R, we define

supp(φ) = {A ∈ U(S) | φ(A) �= 0},and put

S(φ) = {S∈ S : S ⊆ supp(φ)}.Given a weighted split system(S, α) on X, put

H (S, α) ={φ ∈ RU(S) : φ(A) ≥ 0 andφ(A)+ φ(A) = αSA

2for all A ∈ U(S)

}.

It is straight-forward to check that this is a polytope inRU(S) that is polytope isomorphicto an|S|-dimensional hypercube. The subsetB(S, α) of H (S, α) defined by

B(S, α) = {φ ∈ H (S, α) : A1, A2 ∈ supp(φ) andA1 ∪ A2 = X ⇒ A1 ∩ A2 = ∅}is a polytopal complex called theBuneman complexassociated to(S, α). This complexwas introduced in [8]—see also [9] (note in the definition that we present forH (S, α), wehave introduced a factor of12 for scaling purposes).

It can be shown that the mapd1 : RU(S)×RU(S) → R≥0 defined, for allφ, φ′ ∈ RU(S),by

d1(φ, φ′) =

∑A∈U(S)

|φ(A)− φ′(A)|

restricts to give a metric on bothH (S, α) and B(S, α), and that the map fromX intoB(S, α) defined by taking an elementx ∈ X to the function

φx : U(S) → R≥0 : A �→{αSA

2if x �∈ A,

0 else,

is an embedding of(X,dS,α) into (B(S, α),d1) [8, Section 2]. We will make use of thefollowing results:

(B1) [8, Section 2] Ifφ ∈ B(S, α), then

[φ] = {ψ ∈ H (S, α) | supp(ψ) ⊆ supp(φ)}.(B2) [8, Lemma 5.2] For allS ′ ⊆ S the (restriction) map

B(S, α) → B(S ′, α) : φ �→ φ |S ′

is surjective.(B3) S ′ ⊆ S is a maximal incompatible split system, then there exists a unique maximal

cell C in B(S, α) with S(φ) = S ′, for any generatorφ of C. (This followsby [8, Proposition 3.3], which states that a split systemS is incompatible if and onlyif H (S, α) = B(S, α), and (B2).)

K.T. Huber et al. / European Journal of Combinatorics 27 (2006) 461–479 465

(B4) If φ ∈ B(S, α) with [φ] a maximal cell of B(S, α), then S(φ) is a maximalincompatible split system inS. (This follows from the definition ofB(S, α), (B1),(B2), and [8, Proposition 3.3].)

(B5) [8, Section 2] IfC is a cell inB(S, α) andφ is any generator ofC, then dim(C) =|S(φ)|.

2.4. The tight-span

Suppose thatd is a metric onX. It can be shown (cf. [7,18]) that the mapd∞ :RX × RX → R≥0 defined, for f, g ∈ RX, by

d∞( f, g) = maxx∈X

| f (x)− g(x)|,restricts to give a metric onP(d) andT(d), and that the map

Ψ : X → T(d) : y �→ (hy : X → R : x �→ d(x, y))

is an embedding of the metric space(X,d) into (T(d),d∞), i.e. Ψ is an injection withd∞(Ψ (x),Ψ (y)) = d(x, y) holding for allx, y ∈ X.

Now, given f ∈ P(d), define a graphK ( f ) with vertex setX and edge set consisting ofthose subsets{x, y} of X with f (x)+ f (y) = d(x, y). Proofs for the following statementscan be found in [7]:

(TS1) If f ∈ T(d), then

[ f ] = {g ∈ T(d) : K ( f ) ⊆ K (g)}.(TS2) If f ∈ P(d), then f ∈ T(d) if and only if for all x ∈ X there is somey ∈ X distinct

from x with {x, y} an edge ofK ( f ).(TS3) If f ∈ T(d) and f (y) = 0 for somey ∈ X, then f = hy.

3. Gates in the Buneman complex

In this section, we prove some results concerning the Buneman complex. Suppose that(S, α) is any weighted split system onX. Note that every cellC in the Buneman complexB(S, α) is X-gated, that is, for everyx ∈ X there is an elementγ in C, called thegate forx in C, with

d1(φx, ψ) = d1(φx, γ )+ d1(γ,ψ)

holding for allψ ∈ C. Now, for anyx ∈ X and any generatorφ of C, define

γ x = γ xC : U(S) → R≥0 : A �→

{φx(A) if A ∈ U(S(φ)),φ(A) else.

We first prove thatγ x is a gate forx in C.

Lemma 3.1. Suppose(S, α) is a weighted split system on X, C is any cell in B(S, α), andφ ∈ B(S, α) is any generator of C. Then the following statements hold.

(i) If A ∈ U(S − S(φ)) andψ ∈ C, thenφ(A) ∈ {0, αSA2 } andψ(A) = φ(A).

466 K.T. Huber et al. / European Journal of Combinatorics 27 (2006) 461–479

(ii) For any x∈ X, the mapγ x defined above is a gate for x in C.(iii) For all x, y ∈ X,

d1(γx, γ y) =

∑S∈S(φ)

αSδS(x, y).

Proof. Supposex, y ∈ X, C is a cell in B(S, α), andφ ∈ B(S, α) is a generator ofC.It is straight-forward to see that (i)–(iii) all hold in caseC is a vertex. So, without loss ofgenerality, we will assume dim(C) > 0. In particular, by (B5)S(φ) �= ∅.

(i): SupposeA ∈ U(S − S(φ)). Then|{A, A} ∩ supp(φ)| = 1, by the definition ofS(φ).Henceφ(A) ∈ {0, αSA

2 }, asφ ∈ H (S, α). Now supposeψ ∈ C. By (B1) it follows that|{A, A} ∩ supp(ψ)| ≤ |{A, A} ∩ supp(φ)| = 1 and so, again by (B1),ψ(A) = φ(A).

(ii): By (B1) and the definition ofγ x, φx, andS(φ), it follows thatγ x is contained inC.Now supposeψ is any element ofC. By (i) and the definition ofγ x,

d1(φx, ψ) =∑

A∈U(S)|φx(A)− ψ(A)|

=∑

A∈U(S−S(φ))|φx(A)− ψ(A)| +

∑A∈U(S(φ))

|φx(A)− ψ(A)|

=∑

A∈U(S−S(φ))|φx(A)− γ x(A)| +

∑A∈U(S(φ))

|γ x(A)− ψ(A)|

= d1(φx, γx)+ d1(γ

x, ψ).

Henceγ x is a gate forx in C.

(iii): By (i), (ii), and the definition ofφz andγ z, z ∈ X,

d1(γx, γ y) =

∑A∈U(S(φ))

|φx(A)− φy(A)|

=∑

S∈S(φ)φy(S(x))+

∑S∈S(φ)

∣∣∣αS

2− φy(S(x))

∣∣∣=

∑S∈S(φ)

αSδS(x, y). �

Note that ifC is a cell ofB(S, α) with dim(C) > 0, thend1 restricted to the set

Γ (C) = {γ xC : x ∈ X}

is a metric. Later we shall be interested in the case where the metricd1 |Γ (C) is antipodal(recall that a metricd on a finite setY is antipodal if there is an involutionσ : Y → Y,mapping each elementy in Y to an elementy distinct fromy, called theantipodeof y, withd(y, z) + d(z, y) = d(y, y) holding for allz ∈ Y). In this situation, we callC antipodalX-gatedand, forx, y ∈ X, we callγ x theantipodeof γ y in C if γ x is the antipode ofγ y

in (Γ (C),d1 |Γ (C)).Proposition 3.2. Suppose(S, α) is a weighted split system on X, C is a cell in B(S, α)with dim(C) > 0, and φ is any generator of C. Suppose in addition that d1 |Γ (C) is

K.T. Huber et al. / European Journal of Combinatorics 27 (2006) 461–479 467

an antipodal metric onΓ (C), and x, y ∈ X distinct. Then the following statements areequivalent.

(i) γ x is the antipode ofγ y in C.

(ii) S(x) �= S(y) for all S ∈ S(φ).(iii) d1(γ

x, γ y) = d1(γx, ψ) + d1(ψ, γ

y) for all ψ ∈ C.

(iv) For all S ∈ S(φ) and allψ ∈ C,

αS =∑A∈S

|φx(A)− ψ(A)| + |ψ(A)− φy(A)|.

(v) d1(γx, γ y) = ∑

S∈S(φ) αS.

Proof. (i) ⇒ (ii): Supposeγ x is the antipode ofγ y in C, and that there is someS0 ∈ S(φ)with S0(x) = S0(y). Let z ∈ S0(x). Sinceγ x is the antipode ofγ y,

d1(γx, γ y) = d1(γ

x, γ z)+ d1(γz, γ y),

and so, usingLemma 3.1, αSδS(x, y) = ∑A∈S |φx(A) − φz(A)| + |φz(A) − φy(A)|, for

all S∈ S(φ). Thus,

0 = αS0δS0(x, y)

=∑A∈S0

|φx(A)− φz(A)| + |φz(A)− φy(A)|

= 2(|φx(S0(x))− φz(S0(x))| + |φz(S0(x))− φy(S0(x))|)= 4φz(S0(x)),

and soφz(S0(x)) = 0. Thusz ∈ S0(x), a contradiction.

(ii) ⇒ (iv): SupposeS ∈ S(φ) andψ ∈ C. Then

αS = ψ(S(x))+ αS

2− ψ(S(x))+ αS

2− ψ(S(x))+ ψ(S(x))

=∑A∈S

|φx(A)− ψ(A)| + |φy(A)− ψ(A)|.

(iv) ⇒ (iii): Supposeψ ∈ C. By Lemma 3.1(i), γ x(A) = γ y(A) = φ(A) = ψ(A) holdsfor all A ∈ U(S − S(φ)). (iii) now follows.(iii) ⇒ (i): This is clear sinceΓ (C) ⊆ C.(ii) ⇒ (v): This follows byLemma 3.1(iii).(v) ⇒ (ii): Suppose (v) holds and there exists someS′ ∈ S(φ) with S′(x) = S′(y). Thensince

∑A∈U(S(φ)) |φx(A) − φy(A)| = d1(γ

x, γ y) = ∑S∈S(φ) αS, there must exist some

S′′ ∈ S(φ) with

αS′′ <∑A∈S′′

|φx(A)− φy(A)| = 2|φx(S′′(x))− φy(S

′′(x))| = αS′′,

which is impossible. �

468 K.T. Huber et al. / European Journal of Combinatorics 27 (2006) 461–479

Corollary 3.3. Suppose that the conditions stated in the last proposition all hold and thatin addition the cell C is maximal. Then the following statements hold.

(i) If S(x) �= S(y) for all S ∈ S(φ), thenφ(S′(x)) = 0 for all S′ ∈ S − S(φ) withS′(x) = S′(y).

(ii) φx, γx, φ, γ y, φy is a geodesic in B(S, α) if and only ifγ x is the antipode ofγ y in

C.(iii) Suppose x1, x2, y1, y2 ∈ X and that the antipode ofγ yi in C is γ x j for all i , j ∈

{1,2}. Then dS,α(x1, y1)+ dS,α(x2, y2) = dS,α(x1, y2)+ dS,α(x2, y1).

Proof. (i): SupposeS′ ∈ S − S(φ) with S′(x) = S′(y). Since S(φ) is maximalincompatible by (B4), there must exist someS ∈ S(φ) which is compatible withS′. Asy ∈ S′(x) andS(x) �= S(y) by assumption, eitherS(x)∪ S′(x) = X or S(y)∪ S′(x) = X.SinceS(x), S(y) ∈ supp(φ) and S(x) ∩ S′(x) �= ∅ �= S(y) ∩ S′(y) = S(y) ∩ S′(x), itfollows thatφ(S′(x)) = 0.(ii): Supposeφx, γ

x, φ, γ y, φy is a geodesic inB(S, α). Then clearlyd1(γx, γ y) =

d1(γx, φ)+ d1(φ, γ

y). Hence, byProposition 3.2, γ x is the antipode ofγ y in C.Conversely, suppose thatγ x is the antipode ofγ y in C. By (i)

αSδS(x, y) = 2(φ(S(x))+ |φy(S(x))− φ(S(x))|)for all S ∈ S − S(φ). Now usingLemma 3.1andProposition 3.2, it is straight-forward tocheck that

d1(φx, φy) =∑S∈S

αSδS(x, y)

=∑

S∈S(φ)αS +

∑S∈S−S(φ)

αSδS(x, y)

= d1(γx, γ y)+

∑S∈S−S(φ)

2(φ(S(x))+ |φy(S(x))− φ(S(x))|)

= d1(γx, γ y)+

∑A∈U(S−S(φ))

(|φx(A)− φ(A)| + |φy(A)− φ(A)|)

= d1(γx, γ y)+ d1(φx, γ

x)+ d1(γy, φy)

holds. But byProposition 3.2, d1(γx, γ y) = d1(γ

x, φ)+d1(φ, γy). It immediately follows

thatφx, γx, φ, γ y, φy is a geodesic inB(S, α).

(iii): Using (i) andProposition 3.2it is straight-forward to show that

dS,α(xi , yj ) = d1(φxi , γxi )+ d1(γ

xi , γ yj )+ d1(γyj , φyj )

holds for alli , j ∈ {1,2}. But by uniqueness of gates,γ x1 = γ x2 andγ y1 = γ y2 and (iii)now easily follows. �

4. Teutoburgan split systems

Given a weighted split system(S, α) on X, define a map

κ : RU(S) → RX : φ �→ (X → R : x �→ d1(φ, φx)).

K.T. Huber et al. / European Journal of Combinatorics 27 (2006) 461–479 469

The mapκ was originally introduced in [9].1 Note that it immediately follows from thisdefinition thatκ(B(S, α)) ⊆ P(dS,α) and that, by (TS3),κ(φx) = hx, for all x ∈ X.Moreover, using the fact that, fory a fixed element inX, and for all x ∈ X and allφ ∈ B(S, α),

κ(φ)(x) = 2∑

{A∈U(S):y∈A}|φ(A)− φx(A)|, (2)

it is straight-forward to check thatκ induces a non-expanding map fromB(S, α) toP(dS,α), i.e.

d∞(κ(φ), κ(ψ)) ≤ d1(φ,ψ)

holds for allφ,ψ ∈ B(S, α). In this section we will characterise those split systemsS ofX for whichκ(B(S, α)) ⊆ T(dS,α) holds for any weightingα onS.

We begin by proving two useful lemmas. Abusing notation, to anyφ ∈ B(S, α)associate the graphK (φ) which has vertex setX and edge set consisting of those subsets{x, y} of X with d1(φx, φy) = d1(φx, φ) + d1(φ, φy). It is straight-forward to check that{x, y} is an edge ofK (φ) if and only if {x, y} is an edge ofK (κ(φ)).

Lemma 4.1. Let (S, α) be a weighted split system on X. Suppose C is a maximal cell inB(S, α), φ is any generator of C, and d1 |Γ (C) is an antipodal metric onΓ (C). Then, forx, y ∈ X distinct, the following statements are equivalent.

(i) γ x is the antipode ofγ y in C.(ii) {x, y} ⊆ X is an edge of K(κ(φ)) or – equivalently – of K(φ).(iii) κ(φx), κ(γ

x), κ(φ), κ(γ y), κ(φy) is a geodesic in P(dS,α).

Proof. (i) ⇒ (iii): Supposeγ x is the antipode ofγ y in C. By Corollary 3.3(ii), φx, γx,

φ, γ y, φy is a geodesic inB(S, α). Sinceκ is non-expanding, it immediately follows thatκ(φx), κ(γ

x), κ(φ), κ(γ y), κ(φy) is a geodesic inP(dS,α).(iii) ⇒ (ii): Supposeκ(φx), κ(γ

x), κ(φ), κ(γ y), κ(φy) is a geodesic inP(dS,α). Thenclearly

d∞(κ(φx), κ(φ))+ d∞(κ(φ), κ(φy)) = d∞(κ(φx), κ(φy)).

But, for anyz ∈ X, κ(φz) = hz and sod∞(κ(φz), κ(φ)) = d∞(hz, κ(φ)) = κ(φ)(z). (ii)now follows immediately.(ii) ⇒ (i): Suppose{x, y} is an edge ofK (κ(φ)). Thend1(φx, φy) = d1(φx, φ)+d1(φ, φy).Sinceγ x and γ y are gates inC for x and y, respectively, it immediately follows thatφx, γ

x, φ, γ y, φy is a geodesic inB(S, α). Hence, byCorollary 3.3(ii), γ x is the antipodeof γ y in C. �

A split systemS ⊆ S(X) is calledantipodalif for all x ∈ X there exists somey ∈ Xsuch thatS(x) �= S(y) holds for allS ∈ S. Such split systems were studied in [11]. Wenow relate them to antipodalX-gated cells in the Buneman complex.

1 In [9] this map is denoted byλ. Since our definition ofκ is slightly different from the mapλ presented in [9],we useκ as opposed toλ to prevent confusion. It can be easily checked that the results stated in [9] concerningλalso hold forκ.

470 K.T. Huber et al. / European Journal of Combinatorics 27 (2006) 461–479

Lemma 4.2. Suppose C⊆ B(S, α) is a cell withdim(C) > 0 andφ is a generator of C.Then the following statements are equivalent.

(i) C is antipodal X-gated.(ii) S(φ) is antipodal.

Proof. (i) ⇒ (ii): This follows immediately fromProposition 3.2.

(ii) ⇒ (i): Supposex ∈ X. SinceS(φ) is antipodal by assumption, there is somey ∈ Xwith S(x) �= S(y) holding for all S ∈ S(φ). Note that if y′ ∈ X distinct from y withS(x) �= S(y′) for all S ∈ S(φ) then S(y) = S(y′) and soγ y = γ y′

follows by thedefinition ofγ y andγ y′

. Hence, the map which takes, for anyu ∈ X, the gateγ u to γ v

with v ∈ ⋂S∈S(φ) S(u) is a well-defined involution onΓ (C). Moreover, for allz ∈ X and

all A ∈ U(S(φ)),|φx(A)− φy(A)| = |φx(A)− φz(A)| − |φz(A)− φy(A)|

and hence, byLemma 3.1(i),

d1(γx, γ y) = d1(γ

x, γ z)+ d1(γz, γ y).

Thusd1 |Γ (C) is an antipodal metric onΓ (C), and, therefore,C is antipodalX-gated. �

We now give the characterisation promised above.

Theorem 4.3. Suppose thatS is a split system on X. Then the following statements areequivalent:

(i) Every maximal incompatible split system contained inS is antipodal.(ii) For every weightingα : S → R>0, every maximal cell in B(S, α) is antipodal

X-gated.(ii ′) For some weightingα : S → R>0, every maximal cell in B(S, α) is antipodal

X-gated.(iii) For every weightingα : S → R>0, every cell in B(S, α) with non-zero dimension is

antipodal X-gated.(iii ′) For some weightingα : S → R>0, every cell in B(S, α) with non-zero dimension is

antipodal X-gated.(iv) For every weightingα : S → R>0, κ(B(S, α)) ⊆ T(dS,α).(iv ′) For some weightingα : S → R>0, κ(B(S, α)) ⊆ T(dS,α).

Proof. We will prove (i)⇒ (ii) ⇒ (ii ′) ⇒ (i), (ii) ⇒ (iii) ⇒ (iii ′) ⇒ (ii ′), and (ii)⇒ (iv)⇒ (iv ′) ⇒ (ii ′).

The implications (ii)⇒ (ii ′), (iii) ⇒ (iii ′), (iv) ⇒ (iv′), and(iii ′) ⇒ (ii ′) clearly allhold.

(i) ⇒ (ii): Suppose thatα : S → R>0 is a weighting, andC is a maximal cell inB(S, α)with generatorφ. By (B4),S(φ) is maximal incompatible and soS(φ) must be antipodal,by assumption. Thus, byLemma 4.2, C is antipodalX-gated.(ii) ⇒ (iii): Supposeα : S → R>0 is a weighting,C is a cell inB(S, α) with dim(C) > 0,and D is any maximal cell containingC. Let φ and ψ be generators ofC and D,respectively. By assumption,D is antipodalX-gated, and so for anyx ∈ X there exists

K.T. Huber et al. / European Journal of Combinatorics 27 (2006) 461–479 471

somey ∈ X with γ yD the antipode ofγ x

D in D. By Proposition 3.2, S(x) �= S(y) forall S ∈ S(ψ). By (B1), S(φ) ⊆ S(ψ) and soS(φ) is antipodal. (iii) now follows byLemma 4.2.(ii ′) ⇒ (i): Supposeα : S → R>0 is a weighting so that every maximal cell inB(S, α)is antipodalX-gated. SupposeS ′ ⊆ S is a maximal incompatible split system. Then, by(B3),S ′ = S(φ) whereφ ∈ B(S, α) is a generator of some maximal cell inB(S, α). Since[φ] is antipodalX-gated by assumption,S ′ is antipodal byLemma 4.2.(ii) ⇒ (iv): Supposeα : S → R>0 is a weighting,C is a maximal cell inB(S, α), φ is agenerator ofC, andx ∈ X. Then there must exist somey ∈ X distinct fromx with γ y theantipode ofγ x in C. By Lemma 4.1, {x, y} is an edge ofK (κ(φ)). Sinceκ(φ) ∈ P(dS,α),(TS2) impliesκ(φ) ∈ T(dS,α). By (TS1), it follows thatκ(B(S, α)) ⊆ T(dS,α).(iv ′) ⇒ (ii ′): Supposeα : S → R>0 is a weighting withκ(B(S, α)) ⊆ T(dS,α). Let Cbe a maximal cell inB(S, α) with generatorφ, and letx ∈ X. Sinceκ(φ) ∈ T(dS,α), by(TS2) there is somey ∈ X distinct fromx with {x, y} an edge ofK (κ(φ)). Hence, for allS ∈ S(φ), we must haveS(x) �= S(y) since, otherwise, if there were someS ∈ S(φ) withS(x) = S(y) then

0 = αSδS(x, y) =∑A∈S

|φx(A)− φ(A)| + |φy(A)− φ(A)| = 4φ(S(x)),

which is impossible sinceS ∈ S(φ). ThusS(φ) is antipodal, and so, byLemma 4.2, C isantipodalX-gated. �

Motivated by this last theorem, we call a split systemS ⊆ S(X) Teutoburganif everymaximal incompatible subset of splits inS is antipodal. Since every weakly compatible,yet incompatible split system is antipodal [11], it immediately follows that every weaklycompatible split system is Teutoburgan. Note, however, that a Teutoburgan split system isnot necessarily weakly compatible (e.g. take the split system of cardinality 3 on the set ofvertices of a 3-cube induced by removing collections of parallel edges).

Remark 4.4. If (S, α) is a weighted split system for which the mapΦ : X → B(S, α)mapsX surjectively onto the set of vertices ofB(S, α), then it is straight-forward to checkthatS is Teutoburgan. Moreover, it can be shown that such a split system can be associatedto anyBuneman graph[8] (by taking X to be the vertex set of the graph, andS to bethe split system induced by the “parallel classes” of edges of the Buneman graph). Thisprovides a large additional class of Teutoburgan split systems.

5. Maximal cells of the tight-span

In this section we shall show that ifS is a Teutoburgan split system then, for anyweightingα on S, κ induces an injective map from the set of maximal cells ofB(S, α)into the set of maximal cells ofT(dS,α). This is essentially a consequence of the followingresult.

Theorem 5.1. Let (S, α) be a weighted split system on X withS Teutoburgan. Suppose Cis a maximal cell of B(S, α), andφ is any generator of C. Then the following statementshold.

472 K.T. Huber et al. / European Journal of Combinatorics 27 (2006) 461–479

(i) For all x ∈ X, κ(γ x) ∈ [κ(φ)].(ii) κ(φ) is a generator of a maximal cell of T(dS,α).(iii) Supposeψ ∈ B(S, α). Thenψ ∈ C if and only ifκ(ψ) ∈ [κ(φ)].Proof. (i): Suppose{u, v} is an edge ofK (κ(φ)) with u �= v, which exists by (TS2). SinceS is Teutoburgan, byLemma 4.1γ u is the antipode ofγ v in C. By Proposition 3.2andCorollary 3.3(ii), φu, γ

u, γ z, γ v, φv is a geodesic inB(S, α). Hence, sinceγ u andγ v aregates inC for u andv, respectively,

d1(φu, φv) = d1(φu, γz)+ d1(γ

z, φv) = κ(γ z)(u)+ κ(γ z)(v).

Hence{u, v} is an edge ofK (κ(γ z)). Thus, K (κ(φ)) ⊆ K (κ(γ z)), and so by (TS1)κ(γ z) ∈ [κ(φ)].(ii): By Theorem 4.3[κ(φ)] is a cell ofT(dS,α). Suppose that[κ(φ)] is not maximal. Thenthere exists somef ∈ T(dS,α) with [κ(φ)] � [ f ]. By (TS1), K ( f ) � K (κ(φ)) andso there existx1, y1 ∈ X with {x1, y1} an edge ofK (κ(φ)) but not of K ( f ). Note thatx1 �= y1. For, if not, thenκ(φ) = hx1 by (TS3), and, takingγ z to be the antipode ofγ x1

in C, for z ∈ X (which exists byTheorem 4.3), by (i) we obtainκ(γ z) = κ(φ) = hx1. Soκ(γ z)(x1) = 0, which is impossible because, sinceγ z is the antipode ofγ x1 in C,

d1(γz, φx1) ≥

∑A∈U(S(φ))

|γ z(A)− φx1(A)| =∑

A∈U(S(φ))|γ z(A)− γ x1(A)| > 0.

Now define

Z = {z ∈ X | γ zC = γ

y1C }, and

Y = {z ∈ X | γ zC is the antipode ofγ y1

C in C}.Clearly,y1 ∈ Z and, byLemma 4.1, x1 ∈ Y. Since f ∈ T(dS,α) and{x1, y1} is not an edgeof K ( f ), by (TS2) there existx2, y2 ∈ X with x1 �= y2 andx2 �= y1 such that{x1, y2} and{x2, y1} are edges ofK ( f ). SinceK ( f ) ⊆ K (κ(φ)), Lemma 4.1impliesx2 ∈ Y andy2 ∈Z. Hence, byCorollary 3.3(iii), dS,α(x1, y1)+dS,α(x2, y2) = dS,α(x1, y2)+dS,α(x2, y1).But then

f (y1)+ f (x2)+ f (y2)+ f (x1) = dS,α(y1, x2)+ dS,α(y2, x1)

= dS,α(y1, x1)+ dS,α(y2, x2)

≤ f (y1)+ f (x2)+ f (y2)+ f (x1),

and sodS,α(y1, x1) = f (y1)+ f (x1) anddS,α(y2, x2) = f (y2)+ f (x2). Hence,{x1, y1}is an edge ofK ( f ) which is a contradiction.(iii): Supposeψ ∈ [φ] and let {x, y} be an edge ofK (κ(φ)). By Proposition 3.2,Corollary 3.3(ii), andTheorem 4.3φx, γ

x, φ, γ y, φy andφx, γx, ψ, γ y, φy are geodesics

in B(S, α). Thus, sinceγ x andγ y are gates inC, for x andy respectively,

dS,α(x, y) = κ(φ)(x)+ κ(φ)(y) = κ(ψ)(x)+ κ(ψ)(y).

Hence,{x, y} is an edge ofK (κ(ψ)). Thus, by (TS1),κ(ψ) ∈ [κ(φ)].Conversely, supposeκ(ψ) ∈ [κ(φ)]. We can assumeS(φ) �= S since otherwise

supp(ψ) ⊆ U(S) = supp(φ) and soψ ∈ [φ]. We first claim that ifS ∈ S − S(φ),

K.T. Huber et al. / European Journal of Combinatorics 27 (2006) 461–479 473

then there exist elementsx, y ∈ X with S(x) = S(y) andγ x the antipode ofγ y in [φ].Indeed, supposeS ∈ S − S(φ). By (B4),S(φ) is a maximal incompatible split system inS, and so there exists someS′ ∈ S(φ) with S′ andS compatible. Hence there exists somex ∈ X with S(x)∪ S′(x) = X. Since[φ] is antipodalX-gated byTheorem 4.3, there existssomey ∈ X with γ x is the antipode ofγ y in [φ]. By Proposition 3.2, y �∈ S′(x) and soy ∈ S(x). Hence,S(x) = S(y), which completes the proof of the first claim.

We now claim thatφ(A) = ψ(A) holds for all A ∈ U(S − S(φ)). SupposeA ∈U(S − S(φ)). PutS0 = SA. Then, by the claim just above, there exist elementsx, y ∈ Xwith S0(x) = S0(y) and γ x the antipode ofγ y in [φ]. Hence, byProposition 3.2andCorollary 3.3(i), φ(S0(x)) = 0, and, byLemma 4.1, {x, y} is an edge ofK (κ(φ)) ⊆K (κ(ψ)). Thus,d1(φx, ψ)+ d1(ψ, φy) = dS,α(x, y). Since for allS∈ S

αSδS(x, y) ≤∑A∈S

|φx(A)− ψ(A)| + |ψ(A)− φy(A)|,

it follows that 0 = αS0δS0(x, y) = ∑A∈S0

|φx(A) − ψ(A)| + |ψ(A) − φy(A)|. Thus,φx(A) = ψ(A), for all A ∈ S0, and soψ(S0(x)) = 0. In particular, it follows thatφ(A) = ψ(A) holds for all A ∈ U(S − S(φ)) which concludes the proof of the claim.Using (B1), it is now straight-forward to conclude thatψ ∈ [φ]. �

In view of the last theorem it follows that the mapκ ′ = κ ′S,α defined by taking any

maximal cellC in B(S, α) to the cell[κ(φ)], whereφ is any generator ofC, is a well-defined map from the set of maximal cells ofB(S, α) to the set of maximal cells ofT(dS,α). Moreover, we have

Corollary 5.2. If (S, α) is a weighted split system on X withS Teutoburgan, then the mapκ ′ defined above is injective.

Proof. Suppose thatC and C′ are maximal cells inB(S, α) with κ ′(C) = κ ′(C′).Let φ andφ′ be generators forC and C′, respectively. Then[κ(φ)] = [κ(φ′)]. Henceκ(φ) ∈ [κ(φ′)] and soφ ∈ [φ′] by Theorem 5.1(iii). Thus [φ] ⊆ [φ′] by (TS1).Interchanging the roles ofφ andφ′ yields [φ′] ⊆ [φ]. ThereforeC = [φ] = [φ′] = C′.Henceκ ′ is injective. �

6. Totally split-decomposable metrics

For (S, α) a weighted split system onX, by the main result of [9] κ(B(S, α)) =T(dS,α) if and only if S is weakly compatible. We now use this fact to prove that in caseS is a weakly compatible split system, the mapκ ′ defined at the end of the last section is abijection.

Theorem 6.1. Let (S, α) be a weighted split system on X. IfS is weakly compatible, thenthe mapκ ′ is a bijection between the set of maximal cells of B(S, α) and the set of maximalcells of T(dS,α).

Proof. Since any weakly compatible split system is Teutoburgan, byCorollary 5.2 itfollows that the mapκ ′ is injective. Hence it suffices to prove thatκ ′ is surjective.

474 K.T. Huber et al. / European Journal of Combinatorics 27 (2006) 461–479

To this end, suppose thatZ is a maximal cell inT(dS,α). Let h be any generator ofZ.SinceS is weakly compatibleκ mapsB(S, α) onto T(dS,α) [9]. Hence, there must besomeψ ∈ B(S, α) with κ(ψ) = h. SupposeC is a maximal cell inB(S, α) whichcontainsψ, and letφ be a generator ofC. SinceS is Teutoburgan,[κ(φ)] is a maximalcell in T(dS,α) by Theorem 5.1(ii). So h = κ(ψ) ∈ [κ(φ)] by Theorem 5.1(iii), andthus, by (TS1),Z = [h] ⊆ [κ(φ)]. But Z is maximal, and soZ = [κ(φ)]. Thusκ ′ issurjective. �

We conclude this section by giving some new characterisations of weakly compatiblesplit systems (see [9] for some further characterisations). Given a metricd on a finite setY, define theunderlying graph UG(Y,d) to be the graph with vertex setY and edge setconsisting of those subsets{x, y} ⊆ Y for which there is noz ∈ Y distinct fromx andy with d(x, y) = d(x, z) + d(z, y). In addition, define a split systemS ⊆ S(X) to be3-cube-freeif for all 3-subsets{S1, S2, S3} ⊆ S there existsAk ∈ Sk for k = 1,2,3 withA1 ∩ A2 ∩ A3 = ∅.

Theorem 6.2. Suppose thatS ⊆ S(X) is a Teutoburgan split system. Then the followingstatements are equivalent.

(i) S is weakly compatible.(ii) S is 3-cube-free.(iii) for every weightingα : S → R>0, if C is a cell in B(S, α) with dim(C) = 3, then

|Γ (C)| ≤ 6.(iii ′) for some weightingα : S → R>0, if C is a cell in B(S, α) with dim(C) = 3, then

|Γ (C)| ≤ 6.(iv) for every weightingα : S → R>0, if C is a cell in B(S, α) with dim(C) �= 0, then

d1 |Γ (C) is totally split-decomposable.(iv ′) for some weightingα : S → R>0, if C is a cell in B(S, α) with dim(C) �= 0, then

d1 |Γ (C) is totally split-decomposable.

Proof. Clearly (iii) ⇒ (iii ′) and (iv)⇒ (iv′).(i) ⇒ (iv): SupposeS is weakly compatible,α : S → R>0 is a weighting,C is a cell ofB(S, α) with dim(C) > 0, andφ is a generator ofC. Then, byLemma 3.1(iii),

d1(γx, γ y) =

∑S∈S(φ)

αSδS(x, y)

for all x, y ∈ X. SinceS(φ) ⊆ S andS is weakly compatible,S(φ) is weakly compatible,and henced1 |Γ (C) is totally split-decomposable.(iv) ⇒ (iii): Supposeα : S → R>0 is a weighting, and that there is some 3-dimensionalcell C in B(S, α) with |Γ (C)| ≥ 7. By Theorem 4.3C is antipodalX-gated, and, sinceChas eight vertices,|Γ (C)| = 8. Supposeφ is a generator ofC andx0 ∈ X. Then, by (B5),|S(φ)| = 3. PutS(φ) = {S1, S2, S3}. Since each vertex ofC is a gate,

⋂3i=1 Ai �= ∅ for

all Ai ∈ Si , i = 1,2,3, and so we can choose somexi ∈ Si (x0) ∩ Sj (x0) ∩ Sk(x0) with{i , j , k} = {1,2,3}. But then, byLemma 3.1(iii),

d1(γx0, γ xi ) =

∑S∈S(φ)

αSδS(x0, xi ) = αSi

K.T. Huber et al. / European Journal of Combinatorics 27 (2006) 461–479 475

holds fori = 1,2,3. It follows thatγ x0 is a vertex inUG(Γ (C),d1 |Γ (C)) with degree 3.But, since we are assuming thatd1 |Γ (C) is totally split-decomposable, it follows by [16,Theorem 1.2] thatUG(Γ (C),d1 |Γ (C)) is an 8-cycle. This is a contradiction.(iv ′) ⇒ (iii ′): This can be proven using similar arguments to (iv)⇒ (iii).(iii) ⇒ (ii): This follows in a straight-forward manner from the definition ofγ z, z ∈ X.(ii) ⇒ (i): Suppose thatα : S → R>0 is a weighting and thatS is not weakly compatible.Then there exist distinct splitsS1, S2, S3 ∈ S and distinct elementsx0, x1, x2, x3 ∈ Xso that (1) holds. Note thatS ′ = {S1, S2, S3} is incompatible. Hence,B(S ′, α |S ′) =H (S ′, α |S ′) by [8, Proposition 3.3], and so there must exist someφ′ ∈ B(S ′, α |S ′) withS(φ′) = S ′. By (B2) there exists someφ ∈ B(S, α) with φ |S ′ = φ′. Without loss ofgenerality, we may assume thatS(φ) = S(φ′) = S ′. Let C = [φ]. SinceS is Teutoburgan,Theorem 4.3implies thatC is antipodalX-gated and, by (B5), dim(C) = |S(φ)| = 3. Nowby (1), for all i = 1,2,3 and allk, l ∈ {1,2,3} − i distinct,xi ∈ Si (x0) ∩ Sk(x0) ∩ Sl (x0)

and soγ x0, γ x1, γ x2, γ x3 are all distinct gates inC and, by Proposition 3.2, for alli , j ∈ {0,1,2,3} the antipode ofγ xi in C is not γ x j . Hence,|Γ (C)| = 8. But then,by the definition ofγ x j , j = 0,1,2,3,

⋂i=1,2,3 Ai �= ∅ whereAi ∈ Si , i = 1,2,3. It

follows thatS is not 3-cube-free. �

7. Cell-decomposability

Before proving our main result, we recall from [16] the definition of cell-decomposablemetrics and a result about such metrics that will be key in our proof.

Suppose thatd is a metric onX. Given a cellC of T(d) and somex ∈ X, we call a(necessarily unique) elementg ∈ C a gate in C for xif, for all h ∈ C,

d∞(hx,h) = d∞(hx, g)+ d∞(g,h).

We say thatC is X-gatedif there is a gate inC for eachx ∈ X. In case every cell inT(d)is X-gated we calld cell-decomposable.

For d a metric andC a cell of T(d), we let G(C) be the set of gates inC for all theelements inX. We shall use the following restatement of [16, Theorem 1.1]:

• Suppose thatd is a cell-decomposable metric. Then the metricd∞ |G(C) is antipodal,and the mapχ = χC : C → T(d∞ |G(C)) defined, for all f ∈ C and alla ∈ G(C) andx ∈ X with a = f x, byχ( f )(a) = f (x)− f x(x) is a bijective isometry that induces apolytope isomorphism betweenC andT(d∞ |G(C)).In [16] we conjectured that a metric is totally split-decomposable if and only if it is

cell-decomposable. We now prove that the “only if” direction of this conjecture holds.

Theorem 7.1. If d is a totally split-decomposable metric, then d is cell-decomposable.

Proof. Suppose thatd is totally split-decomposable. Let(S, α) be the unique weightedsplit system onX with S weakly compatible andd = dS,α. Note that sinceS is weaklycompatible it is Teutoburgan.

Now, let Z be a maximal cell ofT(d). Suppose thatC is the maximal cell inB(S, α)with κ ′(C) = Z, which exists byTheorem 6.1.

476 K.T. Huber et al. / European Journal of Combinatorics 27 (2006) 461–479

Claim 1: Z is X-gated.Let x be any element ofX. We will show thatκ(γ x

C) is a gate forx in Z.Supposef ∈ Z and letφ be a generator ofC. Then, sinceκ is surjective, there

must exist someψ ∈ B(S, α) with κ(ψ) = f . SinceC is a maximal cell,ψ ∈ Cby Theorem 5.1(iii). Since S is Teutoburgan, byTheorem 4.3and Corollary 3.3(ii)there must exist somey ∈ X with φx, γ

xC, φ, γ

yC, φy a geodesic inB(S, α). By

Proposition 3.2, the fact thatκ is a non-expanding map, and, byTheorem 4.3, it followsthatκ(φx), κ(γ

x), f, κ(γ y), κ(φy) is a geodesic inT(d). But then by (TS4)

d∞(κ(φx), κ(γx))+ d∞(κ(γ x), f ) = d∞(κ(φx), f ) = d∞(hx, f ).

Henceκ(γ xC) is a gate forx in Z. This completes the proof of Claim 1.

Putd′ := d∞ |G(Z). Note that by [16, Theorem 1.1]d′ is antipodal.

Claim 2: d′ is totally split-decomposable.Since Z is X-gated andκ is a non-expanding map,κ induces an isometry between

(Γ (C),d1 |Γ (C)) and (G(Z),d′). Hence, sinceS is weakly compatible and so, byTheorem 6.2, d1 |Γ (C) is totally split-decomposable we also have thatd′ is totally split-decomposable. This completes the proof of Claim 2.

Sinced′ is antipodal and totally split-decomposable, it immediately follows by [16,Theorem 1.2] thatd′ is cell-decomposable. The theorem now follows directly from:

Claim 3: Every cell inT(d) is X-gated.Let W be any cell ofT(d), andx be any element ofX. Suppose thatZ is any maximal

cell in T(d) containingW.SinceZ is X-gated by Claim 1, there is a gategx for x in Z. Let χZ : Z → T(d′) be

the map given by [16, Theorem 1.1]. Sinced′ = d∞ |G(Z) is cell decomposable, there isa gate forχ(gx) in χ(W). Let p be the inverse image underχ of this gate. We will showthat p is a gate forx in W from which the claim follows.

Let f ∈ W. Sinceχ is a bijective isometry

d∞(gx, f ) = d∞(gx, p)+ d∞(p, f ),

and, sinceZ is X-gated,

d∞(x, f ) = d∞(x, gx)+ d∞(gx, f ).

But, sincep ∈ Z andZ is X-gated,

d∞(x, p) = d∞(x, gx)+ d∞(gx, p).

Hence, in view of these last three equalities, it immediately follows thatd∞(x, f ) =d∞(x, p) + d∞(p, f ). Thus, p is a gate forx in W. This concludes the proof ofClaim 3. �

Before proving our final result, we present a result that we will need concerning thetight-span of an antipodal metric. Recall that a polytope inRn is centrally symmetricifeach cellP contains a pointc called thecentreof P such thatc + x ∈ P if and only ifc − x ∈ P, for x ∈ Rn.

Lemma 7.2. Suppose that d is an antipodal metric on a finite set X. Then T(d) is acentrally symmetric polytope.

K.T. Huber et al. / European Journal of Combinatorics 27 (2006) 461–479 477

Proof. By [15, Theorem 4.2],T(d) is a polytope inRX. To see thatT(d) is centrallysymmetric, we have to show thatT(d) contains a centre, that is, some mapf ∈ T(d) withf + g ∈ T(d) if and only if f − g ∈ T(d), for all g ∈ RX. Consider the map

f : X → R≥0 : y �→ d(y, y)/2.

By [15, Lemma 3.1],d(x, y) = d(y, x) for all x, y ∈ X. Hence, f ∈ P(d). Moreover,by [15, Lemma 4.1], in which it is shown that a maph ∈ P(d) is contained inT(d) if andonly if, for all x ∈ X, h(x)+ h(x) = d(x, x) it immediately follows thatf ∈ T(d).

Now supposeg ∈ RX. We will show that if f + g ∈ T(d), then f − g ∈ T(d), whichwill complete the proof of the lemma. Suppose thatf + g ∈ T(d). We start with showingf − g ∈ P(d). To this end, supposex, y ∈ X. Then, by the definition off , [15, Lemma3.1], and [15, Lemma 4.1],

f (x)− g(x)+ f (y)− g(y)

= f (x)− g(x)+ f (x)+ g(x)+ f (x)+ g(x)− d(x, x)

+ f (y)− g(y)+ f (y)+ g(y)+ f (y)+ g(y)− d(y, y)

= 2( f (x)+ f (y))+ (g + f )(x)+ (g + f )(y)− d(x, x)− d(y, y)

= (g + f )(x)+ (g + f )(y) ≥ d(x, y) = d(x, y),

and so f − g is an element inP(d). Using [15, Lemma 4.1] again and the fact thatf andf + g are elements ofT(d) it is easily seen thatf − g ∈ T(d). �Recall that azonotopeis a polytope inRn all of whose cells are centrally symmetric [21,p. 201]. Note that even though the tight-span of an antipodal metric is centrally symmetric,it is not necessarily a zonotope (for example, the tight-span of the graph metric associatedto the 3-cube contains 2-dimensional cells that are polytope isomorphic to triangles [14]).

We now prove our final result.

Corollary 7.3. Suppose that d is a totally split-decomposable metric. Then every cell inT(d) is a zonotope that is polytope isomorphic either to a hypercube or to a rhombicdodecahedron.

Proof. We first show that every cellZ in T(d) is centrally symmetric from which itimmediately follows thatZ is a zonotope. Suppose thatZ is any cell ofT(d). Denotethe metricd∞ |G(Z) by d′. By Theorem 7.1, d is cell-decomposable. HenceZ is X-gatedand so, by [16, Theorem 1.1],d′ is antipodal. Thus byLemma 7.2, T(d′) is a centrallysymmetric polytope.

Now letχZ : Z → T(d′) be the map given by [16, Theorem 1.1]. Supposef ∈ Z withχ( f ) the centre ofT(d′). We claim thatf is a centre forZ.

Supposeg ∈ RX with f + g ∈ Z. Then, sinceχ is a bijection,χ( f + g) ∈ T(d′) andsoχ( f ) + (χ( f + g) − χ( f )) ∈ T(d′). Sinceχ( f ) is a centre forT(d′), we also haveχ( f )−(χ( f −g)−χ( f )) ∈ T(d′) and so 2χ( f )−χ( f −g) ∈ T(d′). Supposea ∈ G(Z)andx ∈ X with a = f x. Then, by definition ofχ ,

2χ( f )(a)− χ( f − g)(a) = 2 f (x)− 2 f x(x)− f (x)− g(x)+ f x(x)

= χ( f − g)(a).

Hence,χ( f − g) ∈ T(d′) and sof − g ∈ Z. Thus, f is a centre forZ, as claimed.

478 K.T. Huber et al. / European Journal of Combinatorics 27 (2006) 461–479

To complete the proof of the corollary we must show that every cell inT(d) is polytopeisomorphic either to a hypercube or to the rhombic dodecahedron. It suffices to show thatthis holds for every maximal cell ofT(d).

Suppose thatZ is a maximal cell. Then by Claim 2 in the proof ofTheorem 7.1, themetric d∞ |G(Z) is totally split-decomposable and, by [16, Theorem 1.1], it is antipodal.It immediately follows by [16, Theorem 1.2] thatT(d∞ |G(Z)) is polytope isomorphic toeither a hypercube or a rhombic dodecahedron. The proof is completed by applying [16,Theorem 1.1]. �

Acknowledgements

KTH and VM thank The Linnaeus Centre for Bioinformatics, Uppsala University andThe Swedish University of Agricultural Sciences, Sweden, where the results presented inthis paper were established. KTH also thanks The Swedish Research Council (VR) forits support. Part of the work presented in this paper was done while JHK was working atPohang University of Science and Technology. He wishes to thank Com2MaC-KOSEF forits support.

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