A COMBINATORIAL NEGATIVE CURVATURE CONDITION IMPLYING GROMOV HYPERBOLICITY 5 1 IOANA-CLAUDIALAZA˘R 0 ’POLITEHNICA’UNIVERSITYOFTIMIS¸OARA,DEPT.OFMATHEMATICS, 2 VICTORIEISQUARE2,300006-TIMIS¸OARA,ROMANIA n E-MAILADDRESS:[email protected] u J Abstract. We explore a local combinatorial condition on a simplicial com- 4 plex,called8–location. Weshowthat8–locationandsimpleconnectivityimply 2 Gromov hyperbolicity. We mention some applications of such combinatorial negative curvaturecondition. ] R 2010 Mathematics Subject Classification: 05C99,05C75. Keywords: 8-location, SD′ property, Gromov hyperbolicity, universal G cover. . h t a m 1. Introduction [ Curvature can be expressed both in metric and combinatorial terms. Metri- 3 cally,one canrefer to nonpositivelycurved(respectively,negativelycurved)metric v 7 spaces in the sense of Aleksandrov, i.e. by comparing small triangles in the space 8 with triangles in the Euclidean plane (hyperbolic plane). These are the CAT(0) 4 (respectively, CAT(-1)) spaces. Combinatorially, one looks for local combinatorial 5 conditions implying some global features typical for nonpositively curved metric 0 spaces. . 1 A very important combinatorial condition of this type was formulated by Gro- 0 mov [Gro87] for cubical complexes, i.e. cellular complexes with cells being cubes. 5 Namely, simply connected cubical complexes with links (that can be thought as 1 : small spheres around vertices) being flag (respectively, 5–large, i.e. flag-no-square) v simplicialcomplexescarryacanonicalCAT(0)(respectively,CAT(-1))metric. An- i X other important local combinatorial condition is local k–largeness, introduced by r Januszkiewicz-S´wia¸tkowski[JS´06] and Haglund [Hag03]. A flag simplicial complex a is locally k–large if its links do not contain ‘essential’ loops of length less than k. In particular, simply connected locally 7–large simplicial complexes, i.e. 7–systolic complexes, are Gromov hyperbolic [JS´06]. The theory of 7–systolic groups, that is groups acting geometrically on 7–systolic complexes, allowed to provide impor- tantexamplesofhighlydimensionalGromovhyperbolicgroups[JS´03,JS´06,Osa13a, OS´15]. However, for groups acting geometrically on CAT(-1) cubical complexes or on 7–systolic complexes, some very restrictive limitations are known. For example, 7–systolicgroupsareinasense‘asymptoticallyhereditarilyaspherical’,i.e.asymp- totically they can not contain essential spheres. This yields in particular that such groups are not fundamental groups of negatively curved manifolds of di- mension above two; see e.g. [JS´07,Osa07,Osa08,OS´15,GO14,Osa15b]. This rises need for other combinatorial conditions, not imposing restrictions as above. In 1 2 [Osa13b,CO15,BCC+13,CCHO14] some conditions of this type are studied – they form a way of unifying CAT(0) cubical and systolic theories. On the other hand, Osajda[Osa15a]introducedalocalcombinatorialconditionof8–location, andused it to provide a new solution to Thurston’s problem about hyperbolicity of some 3–manifolds. In the currentpaperwe undertakea systematic study ofa versionof8–location, suggestedin [Osa15a, Subsection5.1]. This versionis in a sense morenaturalthan the original one (tailored to Thurston’s problem), and none of them is implied by the other. However, in the new 8–location we do allow essential 4–loops. This suggests that it can be used in a much wider context. Roughly (see Section 2 for the precise definition), the new 8–location says that essential loops of length at most 8 admit filling diagrams with at most one internal vertex. We show that this local combinatorial condition is a negative-curvature-type condition, by proving the following main result of this paper. Theorem 1.1. Let X be a simply connected, 8–located simplicial complex. Then the 1–skeleton of X, equipped with the standard path metric, is Gromov hyperbolic. The above theorem was announced without a proof in [Osa15a, Subsection 5.1]. In[Osa15a,Subsection5.2]applicationsconcerningsomeweaklysystoliccomplexes and groups are mentioned. Our proof consists of two steps. In Theorem 3.2 we show that an 8–located ′ simplicial complex satisfying a global condition SD (see Definition 2.2) is Gro- mov hyperbolic. Then, in Theorem 4.1 we show that the universal cover of an ′ 8–located complex satisfies the property SD . The main Theorem 4.3 follows im- mediately from those two results. For proving Theorem 4.1 we use a method of constructing the universal cover introduced in [Osa13b], and then developed in [BCC+13,CCHO14]. Acknowledgements. I am indebted to Damian Osajda for introducing me to the subject, posing the problem, helpful explanations, and patience. I thank Louis Funar for introducing me to systolic geometry a few years ago. I would also like to thank the Mathematical Institute in Wrocl aw,Polandfor hospitality during the winter of 2014. The visit was partially supported by (Polish) Narodowe Centrum Nauki, decision no DEC-2012/06/A/ST1/00259. 2. Preliminaries Let X be a simplicial complex. We denote by X(k) the k-skeletonof X,0≤k < dimX. A subcomplex L in X is called full (in X) if any simplex of X spanned by a set of vertices in L, is a simplex of L. For a set A={v ,...,v } of vertices of X, 1 k by hAi or by hv ,...,v i we denote the span of A, i.e. the smallest full subcomplex 1 k ′ ′ ′ of X that contains A. We write v ∼ v if hv,v i ∈ X (it can happen that v = v ). We write v ≁v′ if hv,v′i∈/ X. We call X flag if anyfinite setof vertices,whichare pairwise connected by edges of X, spans a simplex of X. A cycle (loop) γ in X is a subcomplex of X isomorphic to a triangulation of S1. A k-wheel in X (v ;v ,...,v ) (where v ,i ∈ {0,...,k} are vertices of X) is a 0 1 k i subcomplex of X such that (v ,...,v ) is a full cycle and v ∼v ,...,v . The length 1 k 0 1 k of γ (denoted by |γ|) is the number of edges in γ. 3 Definition 2.1. Asimplicialcomplex ism-located ifitis flagandeveryfullhomo- topically trivial loop of length at most m is contained in a 1-ball. Thelink ofX atσ,denotedX ,isthesubcomplexofX consistingofallsimplices σ ofX whicharedisjointfromσ andwhich,togetherwithσ, spanasimplex ofX. A full cycle in X is a cycle that is full as subcomplex of X. We call a flag simplicial complex k-large if there are no full j-cycles in X, for j < k. We say X is locally k-large if all its links are k-large. We define the metric on the 0-skeleton of X as the number of edges in the shortest 1−skeleton path joining two given vertices and we denote it by d. A ball (sphere) B (v,X) (S (v,X)) of radius i around some vertex v is a full subcomplex i i of X spanned by vertices at distance at most i (at distance i) from v. We introduce further a global combinatorial condition on a flag simplicial com- plex. Definition 2.2. Let X be a flag simplicial complex. For a vertex O of X and ′ a natural number n, we say that X satisfies the property SD (O) if for every n i∈{1,...,n} we have: (1) (T) (triangle condition): for every edge e ∈S (0), the intersection X ∩ i+1 e B (O) is non-empty; i (2) (V) (vertex condition): for every vertex v ∈ S (0), and for every two i+1 vertices u,w∈X ∩B (O), there exists a vertex t∈X ∩B (O) such that v i v i t∼u,w. ′ ′ We say X satisfies the property SD (O) if SD (O) holds for each natural number n ′ ′ n. We say X satisfies the property SD if SD (O) holds for each natural number n n and for each vertex O of X. The following result is given in [Osa15a]. Proposition 2.1. A simplicial complex which satisfies the property SD′(O) for some vertex O, is simply connected. By a covering we meana simplicial covering, i.e. a simplicialmaprestrictingto isomorphisms from 1-balls onto their images. 3. Hyperbolicity In this section we show that the 8-location on a simplicial complex enjoying the ′ SD property, implies Gromov hyperbolicity. Lemma 3.1. Let X be an 8-located simplicial complex which satisfies the SD′(O) n property for some vertex O. Let v ∈ S (O) and let y,z ∈ B (O) be such that n+1 n v ∼ y,z and d(y,z) = 2. Let w ∈ B (O) be a vertex such that w ∼ y,v,z, given n by the vertex condition (V). Consider the vertices u1,u2 ∈ Bn−1(O) such that hy,u ,wi,hw,u ,zi∈ X, given by the triangle condition (T) and such that u ≁ z 1 2 1 and u ≁y. Then u ∼u (possibly with u =u ). 2 1 2 1 2 ′ Proof. Theproofisbycontradiction. Assumethatd(u1,u2)=2. Letu ∈Bn−1(O) ′ be a vertex such that u ∼u ,w,u given by the vertex condition (V). Let t ,t ∈ 1 2 1 2 ′ ′ Bn−2(O) be vertices such that hu1,t1,ui,hu,t2,u2i ∈ X given by the triangle ′ ′ ′ condition (T). Let t ∈Bn−2(O) be a vertex such that t ∼t1,u,t2 (possibly with t′ =t )givenbythevertexcondition(V).Notethatifu ≁t andu ≁t ,thefull 2 1 2 2 1 ′ ′ homotopically trivial loop (v,z,u ,t ,t,t ,u ,y) has length at most 8. If u ∼ t, 2 2 1 1 1 4 ′ considerthefullhomotopicallytrivialloop(v,z,u ,t ,t,u ,y)oflengthatmost7. 2 2 1 If u ∼ t or t = t , consider the full homotopically trivial loop (v,z,u ,t ,u ,y) 1 2 1 2 2 2 1 of length 6. In all three cases it follows, by 8-location, that d(v,t )=2. But since 2 v ∈Sn+1 whereast2 ∈Sn−2,we haved(v,t2)=3. Hence,because wehavereached a contradiction, it follows that u ∼u (possibly with u =u ). 1 2 1 2 v y w z u' u u 1 2 t t 1 t' 2 (cid:3) Theorem 3.2. Let X be an 8-located simplicial complex which satisfies the SD′ property. Then the 0−skeleton of X with a path metric induced from X(1), is δ−hyperbolic, for a universal constant δ. Proof. The proof is similar to the one of the analogous Theorem 3.3 given in [Osa15a]. According to [Pap95], we can prove hyperbolicity of the 0-skeleton by showing that intervals are uniformly thin. ′ Let O,O be two vertices. Denote by I the set of vertices lying on geodesics ′ ′ between O and O and let n = d(O,O ). Let I denote the intersection S (O)∩ k k ′ Sn−k(O ) = Sk(O)∩I. We prove by contradiction that for every k ≤ n and for every two vertices v,w ∈ I ,d(v,w) ≤ 2. This also shows that the hyperbolicity k constant is universal. We build a full path of diameter 3 as in [Osa15a]. Suppose there are vertices v,w ∈ I such that d(v,w) > 2. Let k be the maximal natural number for which k ′ ′ ′ ′ thishappens. Thenthereexistverticesv ,w inI suchthatv ∼v, w ∼w, and k+1 ′ ′ d(v ,w)≤2. ′ ′ By the vertex condition (V), there is a vertex z in I such that z ∼ v ,w , k+1 ′ ′′ ′′ possibly with z = w . By the triangle condition (T) there are vertices v ,w ∈ I k ′ ′′ ′′ ′ ′′ ′′ ′ suchthat hv ,v ,zi,hz,w ,w i∈X (with v =w if z =w ). By the vertexcondi- ′ ′′ ′′ ′′ tion (V) there are vertices s,t and u in I such that s∼v,v ,v ;t∼v ,z,w ;u∼ k ′ ′′ ′′ ′′ ′′ w,w ,w (possiblywiths=v ,t=w ,andu=w). Amongtheverticest,w ,u,w we choose the first one (in the given order), that is at distance 3 from v. Denote ′′′ thisvertexbyv . Inthis wayweobtainafullpath(v ,v ,v ,v )inI ofdiameter 1 2 3 4 k ′′′ 3, with v =v and v =v . We will show that such a path can not exist reaching 1 4 hereby a contradiction and proving the theorem. By the triangle condition (T) there exist vertices wi in Ik−1 such that hvi,wi, v i∈X,1≤i ≤3. We discuss further all possible cases (and the corresponding i+1 subcases) of mutual relations between vertices w ,1 ≤ i ≤ 3 (up to renaming i vertices). Case I is when w =w and w 6=w . Case II is when w 6=w 6=v . 1 2 2 3 1 2 3 Bythetrianglecondition(T)thereareverticesp inI suchthathv ,p ,v i∈ i k+1 i i i+1 X,1 ≤ i ≤ 3 (possibly with p = p ). By the vertex condition (V) there exist 2 3 5 ′ ′′ ′ ′′ vertices p,p in I such that p ∼ p ,v ,p and p ∼ p ,v ,p (possibly with k+1 1 2 2 2 3 3 ′ ′′ p =p and p =p ). 2 3 Note that if p =p or w =w , then d(v ,v )=2 which yields a contradiction 1 3 1 3 1 4 because d(v ,v )=3. 1 4 3.1. Case I. We start treating the case when w = w and w 6= w which 1 2 2 3 has 2 subcases. Case I.1 is when w = w ,w ∼ w . Case I.2 is when w = 1 2 2 3 1 w ,d(w ,w )=2. 2 2 3 According to Lemma 3.1, since d(v ,v )=2, we get d(p ,p )≤1. 1 3 1 2 3.1.1. Case I.1. We treat the case when w =w ,w ∼w . 1 2 2 3 Claim 3.3. Among the vertices p ,p ,p′′,p in this order we choose the last ver- 1 2 3 ∗ ∗ tex p such that v ∼ p . Among the remaining vertices in the same order, 1 ∗ namely among the sequence of vertices starting with p and ending with p , we 3 ∗∗ ∗∗ choose the first vertex p such that p ∼ v . Let γ be the homotopically triv- 4 ∗∗ ∗ ∗ ∗∗ ial loop containing at least the vertices p ,v ,w ,w ,v ,p . If p ∼ p then 4 3 2 1 γ = (p∗∗,v ,w ,w ,v ,p∗). If p∗ 6= p , p∗∗ 6= p and p∗ ≁ p∗∗ then p ∈ γ. If 4 3 2 1 2 2 2 p∗ 6=p′′, p∗∗ 6=p′′ and p∗ ≁p∗∗ then p′′ ∈γ. Then the loop γ is full. ∗ ∗∗ Proof. Note that p 6=p since otherwise we get contradiction with d(v ,v )=3. 1 4 We also get contradiction with d(v ,v )= 3 if v ∼w or v ∼ w . Hence, due to 1 4 1 3 4 2 ∗ ∗∗ the choice of the vertices p and p , the cycle γ is full. (cid:3) Inordertoapply8-location,weconsiderahomotopicallytrivialloopγcontaining ′′ asubsetofthesetofverticesofthecycle(p ,v ,w ,w ,v ,p ,p ,p ),orallvertices 3 4 3 2 1 1 2 of this cycle. Namely, γ contains the vertices v ,w ,w ,v and at least two of the 4 3 2 1 ′′ vertices p ,p ,p ,p chosen as in Claim 3.3. Note that γ is full. Thus, since its 1 2 3 length is at most 8, γ is contained in the link of a vertex. So we get d(v ,v ) = 2 1 4 which yields contradiction with d(v ,v )=3. 1 4 p p" p 2 p 1 3 v1 v2 v3 v4 Ik w w 2 3 Case I.1 3.1.2. Case I.2. We treat further the case when w = w ,d(w ,w ) = 2. By the 1 2 2 3 ′ ′ vertex condition (V) there exists a vertex w ∈ I such that w ∼ w ,v ,w k+1 2 3 3 ′ (possibly with w =w ). 3 ′ ′′ We consider the homotopically trivial loop (p ,v ,w ,w ,w ,v ,p ,p ,p ) of 3 4 3 2 1 1 2 ′ length at most 9. Note that there are no edges between the vertices w ,w ,w 2 3 ′′ and p ,p ,p ,p . But there can be diagonals between the vertices v ,v ,v ,v 1 2 3 1 2 3 4 ′′ ′ and p ,p ,p ,p . Also there can be diagonals joining the vertices w ,w ,w and 1 2 3 2 3 v ,v ,v ,v . Case I.2.a is when such diagonals exist, or some of the vertices 1 2 3 4 ′′ p ,p ,p ,p coincide,ornonconsecutiveverticesinthissequenceareadjacent. Case 1 2 3 I.2.b is when none of these situations occur. 6 p p" p 2 p 1 3 v v v v I 1 2 3 4 k w w' w 2 3 Case I.2.b ′ ′ Note that if v ∼ w and v ∼ w , or if v ∼ w , or if v ∼ w , we have 1 4 1 3 4 2 contradictionwith d(v ,v )=3. If v ∼w′ but v ≁w′, or if v ∼w′ but v ≁w′, 1 4 1 4 4 1 we have reached case I.1 treated above. Note that if v ∼ w Lemma 3.1 implies d(p ,p ) ≤ 1. We choose a loop β 2 3 2 3 ′ as in Claim 3.3 containing at most all vertices of γ = (p ,v ,w ,w ,w ,v ,p ,p ). 3 4 3 2 1 1 2 Namely, β contains the vertices v ,v , at least two of the vertices p ,p ,p , and at 1 4 1 2 3 ′ least two of the vertices w ,w ,w . Because β is full and has length at most 8, by 2 3 8-location we get contradiction with d(v ,v )=3. 1 4 IncaseI.2.aweconsiderahomotopicallytrivialloopγ containingasubsetofthe ′ ′′ setofverticesofthecycle(p ,v ,w ,w,w ,v ,p ,p ,p )chosenasinClaim3.3,or 3 4 3 2 1 1 2 allofits vertices. Namely, γ containsthe verticesv ,v , atleasttwoofthe vertices 4 1 ′′ ′ p ,p ,p ,p ,andatleasttwooftheverticesw ,w ,w . Notethatγ isfull. Inorder 1 2 3 2 3 to apply8-location,the lengthofγ shouldbe atmost8. Ifso,since γ is contained, by 8-location, in the link of a vertex, we get contradiction with d(v ,v )=3. 1 4 For the rest of case I.2.a, we consider the situation when there is no full ho- motopically trivial loop γ as in the paragraph above of length at most 8. This happens if d(p ,p )+d(p ,p )=3 (if d(p ,p )+d(p ,p )≤2,the lengthof γ is at 1 2 2 3 1 2 2 3 most 8) and if there are no diagonals between the vertices v , v and the vertices 1 4 ′′ p ,p ,p ,p (if such diagonals exist, we can find a full loop γ as in the paragraph 1 2 3 aboveoflengthatmost8). ButbecausewearestillincaseI.2.a,eitherthereexists ′′ at least one diagonal between the vertices v , v and the vertices p ,p ,p ,p , or 2 3 1 2 3 ′′ nonconsective vertices in the sequence p ,p ,p ,p are adjacent, or both. 1 2 3 •Considerfirstthe cased(p ,p )+d(p ,p )=3, p ≁v ,p ≁p′′. Bythetrian- 1 2 2 3 1 3 1 ′′ glecondition(T)thereareverticesq ,q inI suchthathp ,q ,p i,hp ,q ,p i∈ 1 2 k+2 1 1 2 2 2 ′ ′ X. By the vertex condition(V) there is a vertex q in I such that q ∼q ,p ,q k+2 1 2 2 ′ ′′ (possiblywithq =q ). Note thatifp ∼v ,Lemma3.1impliesthatd(q ,q )≤1. 2 2 1 2 Consider a homotopically trivial loop γ containing a subset of the set of vertices ′′ ′ of the cycle (q ,p ,v ,w ,v ,p ,q ,q ) chosenas in Claim 3.3, or allof its vertices. 2 3 2 1 1 1 ′′ Namely, γ contains the vertices p ,v ,w ,v ,p , and at least one of the vertices 3 2 1 1 ′ ∗ ∗∗ q ,q ,q chosen as in Claim 3.3 (note that here the vertices p and p as denoted 1 2 inClaim3.3maycoincide). Becauseγ isfullandhaslengthatmost8,by8-location we get contradiction with d(q ,w )=3. 1 2 •Considerfurtherthecased(p ,p )+d(p ,p )=3,p ≁v ,p ∼p′′. Bythetri- 1 2 2 3 1 3 1 ′′ ′′ anglecondition(T)thereareverticesq ,q inI suchthathp ,q ,p i,hp ,q ,p i∈ 1 2 k+2 1 1 2 3 ′ ′ ′′ X. By the vertexcondition (V) there is a vertex q inI suchthat q ∼q ,p ,q k+2 1 2 ′ (possiblywithq =q ). Considerahomotopicallytrivialloopγ containingasubset 2 ′ ofthe set ofvertices ofthe cycle (q ,p ,v ,w ,v ,p ,q ,q )chosenas in Claim3.3, 2 3 3 2 1 1 1 or all of its vertices. Namely, γ contains the vertices p ,v ,w ,v ,p , and at least 3 3 2 1 1 7 q q' q 1 2 p p" p 2 p 1 3 v1 v2 v3 v4 Ik w w' w 2 3 Case I.2.a: d(p ,p )+d(p ,p )=3,p ≁v ,p ≁p′′ 1 2 2 3 1 3 1 ′ oneofthe verticesq ,q ,q chosenasinClaim3.3. Becauseγ is fullandhaslength 1 2 at most 8, by 8-location we get contradiction with d(q ,w )=3. 1 2 q' q q 1 2 1 3 v1 v2 v3 v4Ik w2 w' w3 Case I.2.a: d(p ,p )+d(p ,p )=3,p ≁v ,p ∼p′′ 1 2 2 3 1 3 1 • Consider further the situation when d(p ,p )+d(p ,p )=3 and p ∼v . By 1 2 2 3 1 3 the triangle condition (T) there are vertices q ,q ,q ∈ I such that hp ,q ,p i, 1 2 3 k+2 1 1 2 ′′ ′′ hp ,q ,p i,hp ,q ,p i ∈ X. Note that according to Lemma 3.1, d(q ,q ) ≤ 1. By 2 2 3 3 2 3 ′ ′ thevertexcondition(V)thereexistsavertexq ∈I suchthatq ∼q ,p ,q (pos- k+2 1 2 2 ′ siblywithq =q ). Inordertoapply8-location,weconsiderahomotopicallytrivial 2 ′ loop γ containing a subset of the set of vertices of the cycle (q ,p ,v ,p ,q ,q ,q ) 3 3 3 1 1 2 chosenas in Claim3.3, or all verticesof this cycle. Namely, γ containsthe vertices ′ p ,v ,p , and at least one of the vertices q ,q ,q ,q . Since γ is full, by 8-location, 1 3 3 1 2 3 there exists a vertex p adjacent to all vertices of γ. Note that p ∈ I . We 02 02 k+1 chooseahomotopicallytrivialloopasinClaim3.3containingtheverticesv ,v ,at 1 4 ′ least two of the vertices p ,p ,p , and at least two of the vertices w ,w ,w . By 1 02 3 2 3 8-location, we obtain contradiction with d(v ,v )=3. 1 4 q q'q q 1 2 3 p1 p2 p3 p v1 v2 v0(cid:0)3 v4 Ik w w 2 w' 3 Case I.2.a: d(p ,p )+d(p ,p )=3,p ∼v 1 2 2 3 1 3 8 Note that case I.2.b is similar to case d(p ,p )+d(p ,p )=3, p ≁v ,p ≁p′′ 1 2 2 3 1 3 1 treated above. 3.2. Case II. We treat next the case when w 6= w 6= w which has 3 subcases. 1 2 3 Case II.1 is when w ∼ w ∼ w . Case II.2 is when w ∼ w and d(w ,w ) = 2. 1 2 3 1 2 2 3 Case II.3 is when d(w ,w )=d(w ,w )=2. 1 2 2 3 Note that if w ∼ v , Lemma 3.1 implies that d(p ,p ) ≤ 1. Also note that if 1 3 1 2 w ∼v ,Lemma3.1impliesthatd(p ,p )≤1. Soincasew ∼v andw ∼v ,we 3 2 2 3 1 3 3 2 choose a loop γ containing at most all vertices of the homotopically trivial (p ,v , 3 4 w ,w ,w ,v ,p ,p ). Namely, γ contains the vertices v ,v , at least two of the 3 2 1 1 1 2 1 4 vertices p ,p ,p , and at least two of the vertices w ,w ,w . Because γ is full and 1 2 3 1 2 3 has length at most 8, we get, by 8−location, contradiction with d(v ,v )=3. 1 4 We consider the situation when w ∼ w . We consider a homotopically trivial 1 3 ′ loop γ containing at most all vertices of the cycle (p ,v ,w ,w ,v ,p ,p) chosen 2 3 3 1 1 1 as in Claim 3.3. Namely, γ contains the vertices v ,w ,w ,v , and at least one of 3 3 1 1 ′ the vertices p ,p ,p. Because γ is full and has length at most 7, it is contained, 1 2 by 8-location, in the link of a vertex v . Note that v ∈ I . Hence, because the 02 02 k path (v ,v ,v ,v ) is full, and v ∼ v ,v , the path (v ,v ,v ,v ) is also full. 1 2 3 4 02 1 3 1 02 3 4 Moreover, since hv ,w ,v i, hv ,w ,v i, hv ,w ,v i ∈ X and w ∼ w , we have 1 1 02 02 3 3 3 3 4 1 3 reached case I.1 treated above. p' p p p 2 3 I 1 v v v v 1 2 3 4 I k w w 1 w 3 2 Case II: w ∼w 1 3 3.2.1. Case II.1. We treat the case when w ∼w ∼w . 1 2 3 ′ ′′ We consider the homotopically trivialloop (p ,v ,w ,w ,w ,v ,p ,p,p ,p ) of 3 4 3 2 1 1 1 2 lengthatmost10. Notethattherearenodiagonalsbetweentheverticesw ,w ,w 1 2 3 ′ ′′ andp ,p ,p ,p,p . Buttherecanbediagonalsjoiningtheverticesw ,w ,w tothe 1 2 3 1 2 3 vertices v ,v ,v ,v . Also there can be diagonals between the vertices v ,v ,v ,v 1 2 3 4 1 2 3 4 ′ ′′ andp ,p ,p ,p,p . CaseII.1.aiswhensuchdiagonalsexist,orsomeofthevertices 1 2 3 ′ ′′ p ,p,p ,p ,p coincide, or nonconsecutive vertices in this sequence are adjacent. 1 2 3 Case II.1.b is when none of these situations occur. Note that if v ∼ w and v ∼ w , or if v ∼ w , or if v ∼ w , we have 1 2 4 2 1 3 4 1 contradiction with d(v ,v ) = 3. If v ∼ w but v ≁ w , or if v ∼ w but 1 4 1 2 4 2 4 2 v ≁w , we have reached case I.1. 1 2 In case II.1.a, we consider a homotopically trivial loop γ containing a subset of ′ ′′ thesetofverticesofthecycle(p ,v ,w ,w ,w ,v ,p ,p,p ,p )chosenasinClaim 3 4 3 2 1 1 1 2 9 p' p p 2 p 1 3 v v v v I 1 2 3 4 k w w w 1 2 3 Case II.1.b 3.3, or all of its vertices. Namely, γ contains the vertices v ,v , at least two of the 4 1 ′ ′′ vertices p ,p ,p ,p,p chosen as in Claim 3.3, and at least two of the vertices 1 2 3 w ,w ,w chosenas in Claim 3.3. Note that γ is full. In orderto apply 8-location, 1 2 3 the length of γ should be at most 8. If so, since γ is contained, by 8-location, in the link of a vertex, we get contradiction with d(v ,v )=3. 1 4 For the rest of case II.1.a, we consider the situation when there is no full ho- motopically trivial loop γ as in the paragraph above of length at most 8. This happens if d(p ,p )+d(p ,p )≥3 (if d(p ,p )+d(p ,p )≤2,the lengthof γ is at 1 2 2 3 1 2 2 3 most 8) and if there are no diagonals between the vertices v , v and the vertices 1 4 ′ ′′ p ,p ,p ,p,p such that we can find a full loop γ as in the paragraph above of 1 2 3 length at most 8. Because we are still in case II.1.a, either one, or some of the fol- lowing situations occur. Namely, either there exists at least one diagonal between ′ ′′ the vertices v ,v and the vertices p ,p ,p ,p,p such that we can find a full loop 1 4 1 2 3 γ as in the paragraph above of length greater than 8, or else nonconsecutive ver- ′ ′′ tices in the sequence p ,p,p ,p ,p are adjacent, or else there exists at least one 1 2 3 ′ ′′ diagonal between the vertices v , v and the vertices p ,p ,p ,p,p . 2 3 1 2 3 • We treat first the case d(p ,p ) = d(p ,p ) = 2, p ≁ v . By the triangle 1 2 2 3 1 3 ′ ′ condition (T) there are vertices q ,q in I such that hp ,q ,pi,hp,q ,p i∈X. 1 2 k+2 1 1 2 2 Lemma 3.1 implies that d(q ,q ) ≤ 1. We consider a homotopically trivial loop 1 2 γ containing at most all vertices of the cycle (q ,p ,v ,w ,w ,v ,p ,q ) chosen 2 2 3 2 1 1 1 1 as in Claim 3.3. Namely, γ contains the vertices p ,v ,v ,p , at least one of the 2 3 1 1 verticesq ,q chosenas inClaim3.3,and atleastone ofthe verticesw ,w chosen 1 2 1 2 as in Claim 3.3. Because γ is full and has length at most 8, by 8-location we get contradiction with d(q ,w )=3. 1 1 q q 1 2 p p 2 p 1 p' 3 v1 v2 v3 v4 Ik w w w 1 2 3 Case II.1.a: d(p ,p )=d(p ,p )=2,p ≁v 1 2 2 3 1 3 Note that if p ∼ p , the loop γ chosen in the previous paragraph is no longer 1 2 full. Thissituationwillbetreatedseparatelyinthenexttwoparagraphs. Notethat the case when there exists a diagonal between the vertices v ,v and the vertices 1 4 10 ′ ′′ p ,p ,p ,p,p suchthatwecanfindafullloopγ oflength9containingthevertices 1 2 3 ′ ′′ v ,w ,w ,w ,v and 4 of the vertices p ,p ,p ,p,p , can be treated similarly. 4 3 2 1 1 1 2 3 • We treat further the case d(p ,p )=1,d(p ,p )=2, p ≁v . By the triangle 1 2 2 3 3 2 ′′ ′′ condition (T) there are vertices q ,q ∈I such that hp ,q ,p i,hp ,q ,p i∈X. 2 3 k+2 2 2 3 3 By Lemma 3.1 we get d(q ,q ) ≤ 1. We consider a homotopically trivial loop γ 2 3 containing a subset of the set of vertices of the cycle (q ,p ,v ,w ,w ,v ,p ,q ) 3 3 4 3 2 2 2 2 chosen as in Claim 3.3, or all of its vertices. Namely, γ contains the vertices p ,v ,v ,p ,atleastoneoftheverticesq ,q ,andatleastoneoftheverticesw ,w . 3 4 2 2 2 3 2 3 Because γ is full and has length at most 8, by 8-locationwe get contradictionwith d(q ,w )=3. 2 2 q q 2 3 p1 p2 p" p3 v1 v2 v3 v4 Ik w w w 1 2 3 Case II.1.a: d(p ,p )=1,d(p ,p )=2,p ≁v 1 2 2 3 3 2 • We treat further the case d(p ,p )=1,d(p ,p )=2, p ∼v . By the triangle 1 2 2 3 3 2 ′′ condition (T) there are vertices q ,q ,q ∈ I such that hp ,q ,p i,hp ,q ,p i, 1 2 3 k+2 1 1 2 2 2 ′′ hp ,q ,p i ∈ X. Note that according to Lemma 3.1, d(q ,q ) ≤ 1. By the vertex 3 3 2 3 ′ ′ condition(V) there existsa vertexq ∈I suchthatq ∼q ,p ,q (possibly with k+2 1 2 2 ′ ′′ q = q ). Note that if v ∼ p , Lemma 3.1 implies that d(q ,q )≤ 1. We consider 2 2 1 2 a homotopically trivialloopγ containinga subset ofthe setof vertices of the cycle ′ (q ,p ,v ,p ,q ,q ,q ) chosen as in Claim 3.3, or all of its vertices. Namely, γ 3 3 2 1 1 2 ′ contains the vertices p ,v ,p , and at least one of the vertices q ,q ,q ,q chosen 3 2 1 1 2 3 asin Claim3.3. Becauseγ is full andhas lengthat most7,by 8-locationthere is a vertex p adjacent to all vertices of γ. So p ∈I . We choose a homotopically 02 02 k+1 trivialloopasinClaim3.3containingtheverticesv ,v ,atleasttwoofthevertices 1 4 p ,p ,p , and at least two of the vertices w ,w ,w . By 8-location, we obtain 1 02 3 1 2 3 contradiction with d(v ,v )=3. 1 4 q q' q q 1 2 3 p1 p2 p" p3 p (cid:1)(cid:2) v1 v2 v3 v4 Ik w w w 1 2 3 Case II.1.a: d(p ,p )=1,d(p ,p )=2,p ∼v 1 2 2 3 3 2 • Consider further the case d(p ,p ) = d(p ,p ) = 2,p ∼ v . By the triangle 1 2 2 3 1 3 ′ ′ condition (T) there are vertices q ,q ,q ∈ I such that hp ,q ,pi,hp,q ,p i, 1 2 3 k+2 1 1 2 2