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A Convergence Theorem for Λ-Trees
200, 32]55 Ž1998. JA977228
JOURNAL OF ALGEBRA ARTICLE NO.
A Convergence Theorem for L-Trees Andrew Chermak* Department of Mathematics, Kansas State Uni¨ ersity, Manhattan, Kansas 66506 Communicated by J. Tits Received March 27, 1996
1. INTRODUCTION Our aim in this paper is to refine the arboreal ‘‘Small Cancellation Theory’’ introduced by the author in w2x, with a view toward certain applications. Namely, our results here will be used, in w3x, to obtain results about locally non-spherical Artin groups, in particular, that such groups have a solvable Word Problem. In the Introduction to w2x we explained how ordinary, metric Small Cancellation Theory, as developed Žfor example. in Chapter VI of w4x, can be viewed as the study of certain sets of hyperbolic isometries of trees. We refer the reader to those pages, for the background of the present work. The key concept both here and in w2x is that of a ‘‘convergence property,’’ by which we always mean something roughly analogous to ‘‘solvability of a Word Problem by Dehn’s Algorithm.’’ The convergence property considered here Žsee Ž3.10.. is significantly more flexible than any of those considered in w2x. Another difference between the present work and w2x is that we now set everything in the context of L-trees, L an Žalmost. arbitrary ordered abelian group. ŽThe qualifying adjective ‘‘almost’’ is due to the necessity that L contain an element l which is ‘‘large,’’ in the sense that l lies in no proper convex subgroup of L.. The reason for working in this degree of generality is that, now that the basic properties of L-trees have been carefully established in w1x, it is really no more difficult to work with L-trees than with ordinary simplicial trees, in the present context. Readers who are only interested in the applications in w3x, may ignore any mention of L-trees, and read this paper as concerning only simplicial trees. In any * E-mail address: [email protected]. 32 0021-8693r98 $25.00 Copyright Q 1998 by Academic Press All rights of reproduction in any form reserved.
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33
case, a quick overview of L-trees, sufficient for our purposes, is given in Section 2, below. ŽSee also w5x for an expository treatment of L-trees.. Let X be a L-tree, and let H be a set of hyperbolic isometries of X. Always assume: Ž1. Ž2.
H s H y1 , and hy1 H h s H for all h g H .
Associated with H is the hyperbolic length function, here denoted a: H ª L) 0 , and defined by a Ž h . s min d Ž x, x ? h . 4 xgX , where d is the L-metric on X. We assume, always, that each aŽ h. is ‘‘large’’ in the sense mentioned above, so that for any m g L, N ? aŽ h. ) m for some integer N. Now, to each h g H there is associated the characteristic subtree of h, A Ž h . s x g X d Ž x, x ? h . s a Ž h . 4 and AŽ h. is then L-isometric to the ‘‘linear’’ tree L. In practice, it may happen that AŽ h. has an ² h:-invariant subset which we may wish to distinguish. For example, if L s Z, so that X is a simplicial tree, we can view X as a bipartite graph by separating the vertices of X into two classes D and D9, with no vertex of D adjacent to any vertex in D9. Then AŽ h. l D can be viewed as the image of 2Z under a suitable Z-isometry
a h : Z ª AŽ h . . More generally, we may have a subgroup L h of L, a L-isometry
a h : L ª AŽ h . , and a distinguished subset of AŽ h.: D Ž h. s ahŽ L h . . The function D which assigns to each h g H the triple D Ž h. s Ž ah , L h , D Ž h. . is called here a specialization on H Žsee Definition Ž3.7.., provided that D is suitably compatible with inversion and conjugation. In the application in w3x, H will be a union of conjugacy classes of a certain subgroup G of IsomŽ X ., and it will be necessary to consider
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ANDREW CHERMAK
subsets Hu, ¨ of H Žindexed by pairs of generators of an Artin group. where each Hu, ¨ is itself a union of conjugacy classes of G. Note that the length function a is constant on conjugacy classes. We will associate to H also a modified length function b: H ª L) 0 , defined by bŽ h. s min aŽ h. < h g Hu, ¨ 4 . The notion of a modified length function is formalized below, in Definition Ž3.8., and in Ž3.9. we explain the Artin group example in further detail. Having fixed a specialization D and a modified length function b, we introduce the relevant notion of convergence with Ž3.10.. To put it somewhat roughly: H has the Ž b, D .-con¨ ergence property if whenever x g X and g is an element of the group ² H : generated by H , there is a sequence Ž h 0 , . . . , h n . of elements of H with Ža. h 0 , . . . , h n s g, and Žb. the sequence of points x 0 s x, x 1 s x ? h 0 , . . . , x kq1 s x k ? h k , . . . , x nq1 s x n ? h n s x ? g ‘‘converging’’ to x ? g in the sense that for all k, 0 F k F n, we have d Ž x kq 1 , x ? g . q a Ž h k . F d Ž x k , x ? g . q b Ž h k . ,
Ž Ck .
with strict inequality under certain conditions relating AŽ h k ., DŽ h k ., and the geodesic path from x k to x ? g. The result of this paper is a sufficient condition ŽTheorem Ž3.14.. for H to have the Ž b, D .-convergence property. Namely, we show that it suffices that there be a mapping P Žcalled a polarization. which associates certain collections of closed segments of X with certain sequences of elements of H , in a certain way, satisfying certain rules. The relevant definitions are Ž3.11. through Ž3.13.. Theorem Ž3.14. says that if there is such a polarization of X over H , then H has the Ž b, D .-convergence property. The proof, in Section 4, is very close to the proof of the main result of w2x. If I had known the precise form of the application to Artin groups when writing w2x, this duplication of effort might have been avoided. 2. L-TREES Let L be an ordered abelian group. Denote the set of non-negative elements of L by LG 0 , and the set of strictly positive elements by L) 0 . For any a, b g L with a - b let w a, b x denote the closed segment from a to b:
w a, b x s c g L : a F c F b4 .
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35
A subset of L is bounded if it is contained in a closed segment. A subset K of L is con¨ ex if w a, b x : K for all a, b g K with a F b. An element a of L is large if a G 0 and no proper convex subgroup of L contains a. A L-metric space consists of a set X together with a L-metric d: X = X ª L satisfying the familiar axioms: for all x, y, z g X we have d Ž x, y . s d Ž y, x . , d Ž x, y . s 0
if and only if x s y,
and d Ž x, y . q d Ž y, z . G d Ž x, z . . If X and Y are L-metric spaces then a mapping a : X ª Y is said to be a L-metric morphism if dŽ x, y . s dŽ x a , ya . for all x, y g X. ŽNote: we write mappings to the right of their arguments.. A surjective L-metric morphism is called a L-isometry. Notice that L is itself a L-metric space via d Ž a, b . s < a y b < s max a y b, b y a4 . The translation ta : L ª L, given by xta s x q a, is a L-isometry, as is the inversion map x ¬ yx. The L-metric space X is said to be geodesically linear if, for any points x, y g X, there is a unique L-metric morphism
a : 0, d Ž x, y . ª X such that 0 a s x and dŽ x, y . a s y. We write a s w x, y x, and we say that a is the oriented closed segment from x to y. We write a o p p for the opposite segment w y, x x. Further, we consider x and y to be the initial and terminal boundary points, respectively, of w x, y x, and we write
0 w x, y x s x,
1 w x, y x s y,
and
w x, y x s x, y 4 .
The oriented closed segment a s w x, y x gives rise to the Ž unoriented. closed segment consisting of the image of a in X, which we denote also by w x, y x. This need not be cause for confusion. For w x, y x a closed segment, we define
and
w x, y . s w x, y x y y 4 , Ž x, y x s w x, y x y x 4 , Ž x, y . s w x, y x y x, y 4 .
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ANDREW CHERMAK
A segment of X is a subset of X of the form w x, y x, Ž x, y x, w x, y ., or Ž x, y .. A segment is said to be degenerate if it is either empty or consists of a single point. Let w x, y x and w x, w x be oriented closed segments of the geodesically linear L-metric space X. We write
w x, y x F w z, w x if w x, y x : w z, w x as unoriented closed segments, and with x g w z, y x Žequivalently, y g w x, w x.. Define the length of a closed segment w x, y x by l Ž w x, y x . s d Ž x, y . . DEFINITION 2.1 Žcf. w1x.. A L-tree is a geodesically linear L-metric space X satisfying: Ža. Ž Y-condition. For any x, y, z g X there exists w g X such that w x, z x l w y, z x s w w, z x. Žb. If w x, y x l w y, z x s y4 then w x, y x j w y, z x s w x, z x. Henceforth, X will always denote a L-tree. The point w in Ž2.1.Ža. is uniquely determined, and we denote it by Y Ž x, y, z .. Let us say that a subset J of X is a generalized closed segment if J is isometric to a closed segment of L, or to LG 0 , or to L. The following basic result concerning these objects will be used frequently, without explicit citation. LEMMA 2.2. Let I and J be generalized closed segments of X, with I l J / B. Then I l J is a generalized closed segment. If also I is a closed segment then so is I l J. Proof. Let I0 and J 0 be subsets of L and let a : I0 ª I and b : J 0 ª J be isometries. Put i1 s a g I0 : a a g J 4 and J1 s b g J 0 : bb g I 4 . Then I1 a s J1 b s I l J. As X is geodesically linear, I1 and J1 are convex. If I1 s I0 then I l J is isometric to I0 and we are done. So assume I1 / I0 . Then either there exists u g I0 with u F I1 , or there exists ¨ g I0 with I1 F ¨ . Similarly, we may assume J1 / J 0 , so there exists either a lower bound u9 for J1 in J 0 or an upper bound ¨ 9 for J1 in J 0 . By composing a and or b with suitable isometries of L, we may assume that I0 and J 0 have been chosen so that the lower bounds u and u9 exist. Put x s u a and y s u9b , and let z g I l J. Put w s Y Ž x, y, z .. Then w g I l J by Ž2.1.Ža., and we then have w s a a for some a g I1 , with a F I1. If no upper bound ¨ exists for I1 in I0 we then have I l J isometric to LG 0 , and we are done in this case. On the other hand,
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37
suppose that I1 has an upper bound ¨ in I0 . Then also an upper bound ¨ 1 exists for J1 in J 0 . Applying Ž2.1.Ža. once again, we obtain b g I1 with I1 F b. Thus I l J is isometric to w a, b x, proving Ž2.2.. DEFINITIONS 2.3 ŽŽ2.7. and Ž2.10.Ž3. in w1x.. A subset T of X is con¨ ex if w x, y x : T for all x, y g T. If also T / B then T is a subtree of X. We say that T is closed con¨ ex if the intersection of T with any closed segment is either empty or a closed segment. LEMMA 2.4. Let T be a closed subtree of X, and let x g X. There is then a unique point y of T such that, for e¨ ery z g T, we ha¨ e y g w x, z x. Moreo¨ er, we then ha¨ e w y, z x s w x, z x l T. Proof. The lemma is a special case of the ‘‘Bridge Proposition’’ which is result Ž2.17. of w1x. ŽThe directed closed segment w x, y x is called the bridge from x to T.. DEFINITION 2.5. An isometry h of X is hyperbolic if there exists no closed segment w x, y x with x ? h s y and y ? h s x. ŽIn particular, h fixes no point of X.. For such an isometry h we put a Ž h . s min d Ž x, xh . 4 xgX . It is known w1, Theorem 6.6.x that this minimum exists. Then put A Ž h . s x g X : d Ž x, xh . s a Ž h . 4 . PROPOSITION 2.6 w1, Theorem 6.6x. Let h be a hyperbolic isometry of X. Then the following hold. Ža. aŽ h. exists, and AŽ h. is isometric to the smallest convex subgroup of L containing aŽ h.. Žb. AŽ h. is an ² h:-invariant, closed subtree of X, and AŽ h. is contained in every ² h:-invariant subtree of X. Žc. Let x g X and let w x, y x be the bridge from x to AŽ h.. Then w x, x ? h x l AŽ h. s w y, y ? h x has length aŽ h.. Further, Žsee Fig. 1. we have
w x, x ? h x s w x, y x j w y, y ? h x j w y ? h, x ? h x
and
d Ž x, x ? h . s a Ž h . q 2 d Ž x, y . . COROLLARY 2.7. Let h be a hyperbolic isometry of X and let n g Z, n / 0. Then h n is a hyperbolic isometry, with AŽ h. s AŽ h n ., and aŽ h n . s < n < ? aŽ h.. LEMMA 2.8.
Let h be a hyperbolic isometry of X, and let x, z g X. Put
J s A Ž h . l w x, x ? h x l w x, z x ,
J9 s A Ž h . l w x, x ? h x l w x ? h, z x .
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ANDREW CHERMAK
FIGURE 1
Then J j J9 is the closed segment AŽ h. l w x, x ? h x, of length aŽ h. s l Ž J . q l Ž J9.. Moreo¨ er: Ža. dŽ x, z . y dŽ x ? h, z . - aŽ h. if and only if J9 is non-degenerate, in which case d Ž x, z . y d Ž x ? h, z . F l Ž J . y l Ž J9 . . Žb. dŽ x ? h, z . y dŽ x, z . - aŽ h. if and only if J is non-degenerate, in which case d Ž x ? h, z . y d Ž x, z . F l Ž J9 . y l Ž J . . Proof. In view of Ž2.6., one can be comfortable with a proof by ‘‘pictures.’’ There are three possibilities; see Fig. 2. ŽIn each case one verifies Lemma 2.8.. DEFINITION 2.9. Let h be a hyperbolic isometry of X, let z g X, and let w x, y x be an oriented closed segment of AŽ h.. We say that w x, y x is oriented toward z by h if Ži. w x, z x s w x, y x j w y, z x, and Žii. dŽ x ? hy1 , y . G dŽ x ? h, y .. In particular, w x, y x is oriented towards z by h if x s y. LEMMA 2.10. Let h and h9 be hyperbolic isometries of X. Then the following are equi¨ alent. Ž1. There exists x g X with dŽ x, x ? hh9. - aŽ h. q aŽ h9.. Ž2. AŽ h. s AŽ h9. or AŽ h. l AŽ h9. is a non-degenerate closed segment, and in either case, AŽ h. l AŽ h9. is oriented in opposite directions by h and h9. Proof. The direction Ž1. « Ž2. is given by Propositions Ž8.1.Ža. and Ž8.3.Ža. of w1x. The direction Ž2. « Ž1. is trivial.
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39
FIGURE 2
3. SEQUENCES OF HYPERBOLIC ISOMETRIES This section consists almost entirely of definitions, with a few elementary consequences, and with also a statement of the Main Theorem. Fix an ordered abelian group L, a L-tree X, and a set H of hyperbolic L-isometries of X. Ž3.1. THE BASIC HYPOTHESIS. The following three conditions hold for e¨ ery h g H .
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ANDREW CHERMAK
Ža. aŽ h. is large. That is, no proper con¨ ex subgroup of L contains aŽ h.. Žb. hy1 g H Ž H is symmetric.. Žc. hy1 H h s H Ž H is self-normalizing.. We assume Ž3.1. for the remainder of this paper. An H-sequence is defined to be a finite sequence of elements of H . If h and h9 are H-sequences then h(h9 denotes the H-sequence obtained by concatenating h and h9 in the given order. If h s Ž h 0 , . . . , h k . is an H-sequence then we write y1 hy1 s Ž hy1 k , . . . , h0 . ,
Ž 3.2.
and for any g g ² H : we write h g s Ž gy1 h 0 g , . . . , gy1 h k g . .
Ž 3.3.
Of course, hy1 and h g are H-sequences. DEFINITION 3.4. Let h s Ž h 0 , . . . , h k . and h9 s Ž hX0 , . . . , hXk . be two H-sequences of length k q 1, k G 1. We say that h9 is a simple braiding of h if for some i with 0 F i - k we have h j s hXj
for all j f i , i q 1 4 ,
and either
Ž hXi , hXiq1 . s Ž h iq1 , hy1 iq1 h i h iq1 . , or
Ž hXi , hXiq1 . s Ž h i h iq1 hy1 i , hi . . Notice that the relation ‘‘is a simple braiding of’’ is symmetric. We say that h9 is a braiding of h if there exists a chain h s s 0 , . . . , s l s h9 of H-sequences such that for all i, 1 F i F l, s i is a simple braiding of s iy1. This defines an equivalence relation Žwhose equivalence classes are in fact the orbits of a natural action of the Artin braid group on k q 1 threads. on the H-sequences of length k q 1. The H-sequence h s Ž h 0 , . . . , h k . will be said to be reduced if for all i with 1 F i - k we have h iy1 h i f H j 14 . We put: H * s H-sequences h: every braiding of h is reduced 4 .
Ž 3.5.
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41
LEMMA 3.6. Let h s Ž h 0 , . . . , h k . and h9 s Ž hX0 , . . . , hXk . be H-sequences, with h9 a braiding of h. Then the following hold. Ža. h 0 . . . h k s hX0 . . . hXk . Žb. ² h 0 , . . . , h k : s ² hX0 , . . . , hXk :. Žc. If h is a word of minimal length for the element h 0 , . . . , h k of ² H :, in terms of the generating set H , then h g H *. Žd. If h g H * then h9 g H *, and any prefix Ž h 0 , . . . , h i . of h, 0 F i F k, is in H *. Že. If h g H * then h i h j f H j 14 for any i and j with 0 F i / j F k. Proof. Parts Ža. and Žb. are easily verified in the case that h9 is a simple braiding of h, and the obvious induction argument then yields Ža. and Žb. in general. Part Žc. follows from Ža.. If h g H * then so is h9, by definition. If Ž h 0 , . . . , h i . is a prefix of h and Ž hY0 , . . . , hYi . is a braiding of Ž h 0 , . . . , h i ., then Ž hY0 , . . . , hYi , h iq1 , . . . , h n . is a braiding of h, and this yields Žd.. Suppose that h g H * and let 0 F i - j F k be given. Then the sequence
žh ,..., h 0
iy1 ,
hj
hj
h i , h j , Ž h iq1 . , . . . , Ž h jy1 . , h jq1 , . . . , h k
is a braiding of h, and so h i h j f H j 14 . Since h j h i s Ž h i h j . and Žc. then yields h j h i f H j 14 , proving Že..
/
hy1 j
, Ž3.1.Žb.
DEFINITION 3.7. A specialization on H is a function D which assigns to each h g H a triple D Ž h. s Ž ah , L h , D Ž h. . , where Ža. a h is a L-isometry from L onto AŽ h., Žb. L h s B or L h is a subgroup of L containing aŽ h., Žc. DŽ h. s B or DŽ h. s a hŽ L h ., and Žd. L h s L hy1 s L h g , DŽ h. s DŽ hy1 ., and DŽ h g . s DŽ h. ? g for any g g ² H :. DEFINITION 3.8. Let D be a specialization on H . A modified length function Žon H , relati¨ e to D . is a mapping b: H ª L) 0 such that the following hold for all h g H : Ža. bŽ h. F aŽ h., and bŽ h. g 2 ? L h if L h / B. Žb. bŽ hy1 . s bŽ h.. Žc. bŽ h g . s bŽ h. for all g g ² H :.
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ANDREW CHERMAK
EXAMPLE 3.9. Let G s) i g I Gi be a free product, and let X denote the standard tree of G. Thus, X is a simplicial tree whose vertex set VX is the disjoint union, VX s D@ D9, D s G, D9 s @ i g I Gi R G, where Gi R G denotes the set of right cosets of Gi in G. The edge-set EX consists of pairs of vertices of the form of Ž g, Gi g .. Regard each Gi and each Gi )Gj as a subgroup of G Žfor i / j ., and for Ri, j and all i / j let Ri, j be a subset of Gi )Gj y Ž Gi j Gj .. Put H0 s jR put H s Ž H0 j H0y1 . G . Then H is a symmetric, self-normalizing set of hyperbolic isometries of X. For each i / j, put bi , j s min a Ž h . : h g Ri , j 4 . There is then a well-defined mapping b of H into 2Z given by b Ž h . s bi , j
G
if h, hy1 4 l Ri , j / B.
Put Z h s 2Z for all h g H , and let a h : Z ª AŽ h. be an isometry which sends 2Z to D l AŽ h.. We then have a specialization D given by D Ž h . s Ž a h , 2Z, a h Ž 2Z . . , and b is a modified length function relative to D. This example will be pursued in w3x, where Gr² H : will be an Artin group which is ‘‘locally non-spherical.’’ In what follows, let D be a fixed specialization on H and let b be a fixed, modified length function on H relative to D. DEFINITIONS 3.10. Let h s Ž h 0 , . . . , h n . g H *, and let x g X. Put x 0 s x and, inductively, define x kq 1 s x k ? h k , 0 F k F n. Write z for x nq1. Also, put Ik s AŽ h k . l w x k , x kq1 x l w x k , z x. We say that x con¨ erges to z ¨ ia h Ž relati¨ e to Ž b, D .. if for all k, 0 F k F n, we have d Ž x kq 1 , z . q a Ž h k . F d Ž x k , z . q b Ž h k . , with equality only if Ik : D Ž h k . .
Ž Ck .
We say that H has the Ž b, D .-con¨ ergence property if, whenever h9 g H * and x g X, there exists a braiding h s Ž h 0 , . . . , h n . of h9 such that x converges to x ? h 0 . . . h n via h Žsee Fig. 3.. The aim of this paper is to find suitable, and verifiable, conditions on H which will guarantee that H has the Ž b, D .-convergence property. Definitions Ž3.11. through Ž3.13. will set up such a collection of conditions.
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FIGURE 3
DEFINITION 3.11. Let P be a mapping which associates to each h in H * a collection P Žh. of closed oriented segments of X. We say that P is a polarization Žof X o¨ er H . if the following conditions hold. Ža. P ŽŽB.. is the set of all degenerate closed segments of X. That is, P ŽŽB.. s X. Žb. P Žh. s P Žh9. if h9 is a braiding of h. Žc. If J g P Žh. then J o p p g P Žhy1 ., and J ? g g P Žh g . for any g g ² H :. DEFINITION 3.12. Let P be a fixed polarization of X over H . Let h g H *, J g P Žh.. Then E Žh, J . denotes the set of all h g H satisfying the following three conditions. Ža. Žb. Žc. oriented
h(Ž h. g H *, J l AŽ h. / B, and if J is non-degenerate, then J l AŽ h. is non-degenerate and is toward 1 J by h.
Given h g E Žh, J ., we wish to state conditions which give expression to there being some influence of J on the structure of P Žh(Ž h... As usual, we shall state these conditions in the form of a definition. This will be our most important definition, bringing together all of the notions encountered so far. DEFINITION 3.13. Let P be a polarization of X over H , D a specialization on H , and b a modified length function on H relative to D. We say that P controls con¨ ergence Ž relati¨ e to Ž b, D .. if there exists a large positive element l* of 2L such that bŽ h. G l* for all h g H , and if the following conditions hold whenever we have h g H *, J g P Žh., and h g E Žh, J ..
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ANDREW CHERMAK
FIGURE 4
Ž1. Ž Exclusion. Ža. J AŽ h. if h / ŽB., Žb. l Ž J l AŽ h.. - aŽ h., and Žc. if J l AŽ h. / B then l Ž J l AŽ h.. G 12 ŽbŽ h. y l*.. Ž2. Ž Extension. Ža. We have wŽ 0 J . ? h, 1 J x g P Žh(Ž h.., provided that Žsee Fig. 4.: Ži. J l AŽ h. s B, Žii. l Ž J l AŽ h.. G aŽ h. y 12 bŽ h., with strict inequality if Ž J l AŽ h.. DŽ h.. Žb. For y g AŽ h., we have w y ? h, 1 J x g P Žh(Ž h.., provided that Žsee Fig. 5.: Ži. 0 J g w y, y ? h., and Žii. dŽ 0 J, y ? h. G 12 bŽ h., with strict inequality if 0 J, y4 : DŽ h.. We may now state our main result. THEOREM 3.14. Let P be a polarization of X o¨ er H , D a specialization on H , and b a modified length function on H relati¨ e to D. Suppose that P controls con¨ ergence relati¨ e to Ž b, D .. Then H has the Ž b, D .-con¨ ergence property.
FIGURE 5
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45
Henceforth, fix b, D, and l* as in Ž3.13., and assume that we are given a polarization P such that P controls convergence relative to Ž b, D .. We will take up the proof of Theorem Ž3.14. in Section 4, but first we have two lemmas. LEMMA 3.15. Let h g H and let J be a closed segment of AŽ h., oriented toward 0 J by h. Assume 12 bŽ h. F l Ž J . F aŽ h., and if J : DŽ h. assume l Ž J . ) 12 bŽ h.. Then J g P ŽŽ h... Proof. Put x s 1 J. Then x g P ŽŽB.. by Ž3.11.Ža., and h g E ŽŽB., x .. Put y s Ž 0 J . ? hy1 . As 0 - l Ž J . F aŽ h. we have x g w y, y ? h.. If x, y4 : DŽ h. then Ž3.7. says that L h is a subgroup of L containing aŽ h., and hence x, y ? h4 s J : DŽ h.. Now Ž3.13.Ž2.Žb. applies, with x 4 in the role of J, and this yields the lemma. LEMMA 3.16. Let h, h9 g H with Ž h, h9. g H *, and suppose that aŽ hh9. - aŽ h. q aŽ h9. Ž cf. Ž2.10... Put I s AŽ h. l AŽ h9.. Then: Ža. Žb.
l Ž I . F 12 min bŽ h., bŽ h9.4 y 12 l*, l Ž I . - min aŽ h. y 21 bŽ h., aŽ h9. y 21 bŽ h9.4 .
Proof. By Ž2.1., I contains a non-degenerate segment. Let K be an oriented closed segment of I, oriented toward 1 K by h9, and with l Ž K . F aŽ h9.. Then K g P ŽŽ h9.. by Ž3.15., and we have h g E ŽŽ h9., J .. Then Ž3.13.Ž1.Žb. yields l Ž J . - aŽ h.. In particular, it follows that I is itself a closed segment with l Ž I . - aŽ h.. Take I to be oriented toward 0 I by h. Let J be the closed segment of AŽ h. such that 1 J s 1 I and 0 J s Ž 1 I . ? h. Then I : J or J : I. We have J g P ŽŽ h.. by Ž3.15., and then h9 g E ŽŽ h., J .. Now Ž3.13.Ž1.Žb. gives I : J, I s J l AŽ h9., and J l AŽ h9. / B. Now apply Ž3.13.Ž1.Žc. with h9 in the role of h, and obtain l Ž I . F 12 bŽ h9. y l*. Symmetry then yields Ža., and Žb. follows since aŽ h. G bŽ h., by Ž3.8.. 4. PROOF OF THEOREM 3.14 We proceed by contradiction. Thus, assume that we are given b, D, and l* as in Ž3.13., and a polarization P of H which controls convergence relative to Ž b, D .. But assume that H does not have the Ž b, D .-convergence property. Thus: Ž4.1. There exists h s Ž h 0 , . . . , h n . g H *, and there exists x g X such that, setting z s x ? h 0 . . . h n , x does not converge to z via h9 for any braiding h9 of h. We now chose h and x so that the length, n q 1, of h is as small as possible for Ž4.1.. Fix the notation: z s x ? h 0 . . . h n .
46
ANDREW CHERMAK
The positive element l* lies in no proper convex subgroup of L, so there is a positive integer N with N ? l* ) dŽ x, z .. It follows that h can be chosen so that, for any braiding Ž hX0 , . . . , hXn . of h we have dŽ x ? hX0 , z . ) dŽ x ? h 0 , z . y l*. Indeed, the alternative is that after composing N successive braidings we obtain dŽ x ? hX0 , z . - 0, which is absurd. ŽThus, no matter what braiding h9 of h we look at, x ? hX0 will not be much closer to z than x ? h 0 is.. Having chosen h as above, we next observe that, by the minimality of n q 1, x ? h 0 converges to z via some braiding Ž hX1 , . . . , hXn . of Ž h1 , . . . , h n .. But Ž h 0 , hX1 , . . . , hXn . is then a braiding of h, so we may assume to begin with, without disturbing the choice in the preceding paragraph, that x ? h 0 converges to z via Ž h1 , . . . , h n .. For easy reference, we record these results as follows: Ž4.2. x ? h 0 converges to z s x ? h 0 . . . h n via Ž h1 , . . . , h n .. Ž4.3. For any braiding Ž hX0 , . . . , hXn . of h, we have dŽ x ? h 0 , z . - dŽ x ? X h 0 , z . q l*. The next result tells us that h 0 doesn’t move x very much closer to z than x already is, and this remains true even if h is replaced by a braiding h9. Ž4.4.
For any braiding Ž hX0 , . . . , hXn . of h, we ha¨ e d Ž x, z . q b Ž hX0 . F d Ž x ? hX0 , z . q a Ž hX0 .
with strict inequality if I0X : DŽ hX0 ., and where I0X is defined by I0X s A Ž hX0 . l w x, x ? hX0 x l w x, z x . Proof. Let h9 s Ž hX0 , . . . , hXn . be a braiding of h, and put hX0 s hXn .. The minimality of n q 1 in our choice of h implies that x ? hX0 converges to z via some braiding hY0 of hX0 . Put h0 s Ž hX0 .(hY0 . Then h0 is a braiding of h9, hence also of h, and so x does not converge to z via h0. But this means that condition Ž C0 . in Ž3.10. fails to hold, with hX0 in the role of h 0 , and with I0X in place of I0 . This yields Ž4.4.. Ž hX1 , . . . ,
We now establish notation, to be followed from now on. Set x 0 s x and, as in Ž3.10., set x iq1 s x i ? h i , 0 F i F n. Thus, z s x nq1. Now, for any i in the above range, put K i s A Ž h i . l w x i , x iq1 x , Ii s K i l w x i , z x , and yi s Y Ž x i , x iq1 , z .
Ž cf. Ž 2.1. Ž a . . .
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47
Notice that, by Ž2.6.Žc., K i is a closed segment of length aŽ h i ., and we may write K i s w pi , ri x , with ri s pi ? h i . Define a point qi in K i by K i s w pi , qi x j w qi , ri x ,
d Ž qi , ri . s 12 b Ž h i . .
We also put J0 s K 0 l w x 1 , z x . Applying Ž4.4. to Ž h 0 , . . . , h n ., we obtain dŽ x, z . y dŽ x 1 , z . - aŽ h 0 ., and then J 0 is non-degenerate by Ž2.8.Ža.. Write J0 s w u 0 , ¨ 0 x ,
with ¨ 0 g Ž u 0 , z x
Ž 4.5.
so that J 0 is an oriented closed segment, oriented away from z by h 0 . Ž4.6. AŽ h 0 . l w x, z x is non-degenerate. In particular, we ha¨ e ¨ 0 g w x, z x, and x / z. Proof. Suppose false, so that AŽ h 0 . l w x, z x is either empty or is the point p 0 s 0 K 0 . It may be helpful to have the following picture in mind Žpossibly with x s z, and possibly with p 0 s y 0 . ŽFig. 6.. Suppose first that p 0 f w x 1 , q1 x. Then q1 f w p 0 , z x, and hence q1 g w x 1 , p 0 . by Ž4.2.. If now p1 g w r 0 , p 0 x then w p1 , q1 x : AŽ h 0 . l AŽ h1 .. But dŽ p1 , q1 . s aŽ h1 . y 12 bŽ h1 ., so we contradict Ž3.16.Žb.. Thus p1 f w r 0 , p 0 x, so p1 g w x 1 , r 0 x. Put d s 12 min bŽ h 0 ., bŽ h1 .4 y 12 l*. Since aŽ h 0 . G bŽ h 0 ., by Ž3.8., we certainly have d - aŽ h 0 .. Also, recall from Ž3.13. that bŽ h. G l* for any h g H , so we have d G 0. Let w be the point in J 0 at distance d from r 0 . By Ž3.16.Ža. we then have y 1 g w x 1 , w x. Set xX1 s x 2 ? hy1 0 . Then d Ž xX1 , p 0 . s d Ž x 2 , r 0 . F d Ž x 2 , w . q d Ž w, r 0 . s d Ž x 2 , w . q d.
FIGURE 6
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ANDREW CHERMAK
Now Ž4.2. and condition Ž C1 . in Ž3.10. yield d Ž x 2 , w . F d Ž x 1 , w . y a Ž h1 . q b Ž h1 . s d Ž x 1 , r 0 . y a Ž h1 . q b Ž h1 . q d, and so d Ž xX1 , p 0 . F d Ž x 1 , r 0 . y a Ž h1 . q b Ž h1 . q 2 d. On the other hand, we have d Ž x 1 , p0 . s d Ž x 1 , r 0 . q aŽ h 0 . , so we get d Ž x 1 , p 0 . y d Ž xX1 , p 0 . G a Ž h 0 . q a Ž h1 . y b Ž h1 . y 2 d G Ž a Ž h 0 . y b Ž h 0 . . q Ž a Ž h1 . y b Ž h1 . . q l* G l*. As p 0 g w x 1 , z x we then have dŽ x 1 , z . y dŽ xX1 , z . G l*. Setting hX0 s h 0 h1 hy1 we see that xX1 s x ? hX0 , and Ž hX0 , h 0 , h 2 , . . . , h n . is a braiding of h. 0 Thus, we contradict Ž4.3.. Hence p 0 g w x 1 , q1 x. Now switch notation and set xX1 s x ? h1 , and take w to be the point in J 0 at distance d from p 0 , where d is defined as above. Since p 0 g w x 1 , q1 x, Ž4.2. implies that q1 g w p 0 , z x. Then Ž3.16.Ža. shows that p1 g w x, z x, and Ž3.16.Žb. yields q1 g Ž p 0 , z x. Now d Ž x 1 , q1 . s d Ž x 1 , r 0 . q a Ž h 0 . q d Ž p 0 , q1 . s d Ž x, p 0 . q a Ž h 0 . y d q d Ž w, q1 . s d Ž x, p 0 . q a Ž h 0 . y d q d Ž w, p1 . q d Ž p1 , q1 . s d Ž x, p 0 . q a Ž h 0 . y d q d Ž x, p1 . q a Ž h1 . y 12 b Ž h1 . . On the other hand, d Ž xX1 , q1 . s d Ž x, q1 ? hy1 1 . F d Ž x, p1 . q d Ž p1 , q1 ? hy1 1 . s d Ž x, p1 . q d Ž q1 , r 1 . s d Ž x, p1 . q 12 b Ž h1 . F d Ž x, p 0 . q d Ž p 0 , p1 . q 12 b Ž h1 . F d Ž x, p 0 . q d q d Ž x, p1 . q 12 b Ž h1 . .
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We then have d Ž x 1 , q1 . y d Ž xX1 , q1 . G a Ž h 0 . q a Ž h1 . y b Ž h1 . y 2 d G Ž a Ž h 0 . y b Ž h 0 . . q Ž a Ž h1 . y b Ž h1 . . q l* G l*. . Ž . Since Ž h1 , hy1 1 h 0 h1 , h 2 , . . . , h n is a braiding of h, we again contradict 4.3 . Ž . This proves 4.6 . Ž4.7. We ha¨ e J 0 s w r 0 , y 0 x g P ŽŽ h 0 .., and J 0 is oriented away from z by h 0 . Moreo¨ er, l Ž J 0 . G 12 bŽ h 0 ., and this inequality is strict if J 0 : DŽ h 0 .. Proof. Since x 1 converges to z via Ž h1 , . . . , h n ., by Ž4.2., condition Ž C0 . in Ž3.10. must fail to hold, as otherwise x converges to z via h. Thus, l Ž J 0 . G 12 bŽ h 0 ., and this inequality is strict if I0 : DŽ h 0 .. Now, I0 j J 0 is a segment of AŽ h 0 . of length aŽ h 0 ., and it then follows from Ž3.7. that I0 : DŽ h 0 . if J 0 : DŽ h 0 .. By Ž3.15. we get J 0 g P ŽŽ h 0 ... Evidently J 0 s w r 0 , y 0 x, oriented toward r 0 by h 0 . This yields Ž4.7.. Ž4.8. Assume now that we are given k with 1 F k F n, such that for each i with 0 F i - k, we have a non-degenerate segment Ji in P ŽŽ h 0 , . . . , h i .. satisfying the following conditions: Ža. Ji s w u i , ¨ i x : w x iq1 , z x, with ¨ i g Ž u i , z x. Žb. One of the following holds if i ) 0: Ži. u iy1 g Ii and Ji s w ri , ¨ iy1 x. Žii. ¨ iy1 g Ii and Ji s w u iy1 ? h i , yi x. Žiii. Ii : Ž u iy1 , ¨ iy1 . and Ji s w u iy1 ? h i , ¨ iy1 x. Our goal will be to show that we can also define Jk g P ŽŽ h 0 , . . . , h k .. so as to satisfy Ž4.8.Ža. and Žb., with k in place of i. That is, we aim to show that Ž4.8. amounts to a well-defined inductive procedure for defining J1 , . . . , Jn , starting with the segment J 0 which we have already constructed. Ž4.9. We ha¨ e ¨ i g w x, z x for all i, 0 F i - k. Proof. First, ¨ 0 g w x, z x by Ž4.6.. By Ž4.8.Žb. we have ¨ i s ¨ iy1 , or else ¨ i s yi g w ¨ iy1 , z x. So, if ¨ iy1 g w x, z x then ¨ i g w x, z x, and we are done by induction. Ž4.10. We ha¨ e y k g Ž u ky1 , z x and qk g w u ky1 , z x. Moreo¨ er, if pk g DŽ h k . then qk / u ky1. Proof. Define a point sk by sk s
½
yk
if y k g w x k , u ky1 x
u ky 1
if y k g w u ky1 , z x .
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ANDREW CHERMAK
Of course, u ky 1 g w x k , z x by Ž4.8.Ža., so sk is well-defined. Now define s0 , . . . , sky1 by the formula sky i s skyiq1 ? hy1 kyi
Ž1 F i F k. . pX0 , . . . ,
In a similar manner we define points pXk s pk , pXky i s pkyiq1 ? hy1 kyi
pXk :
Ž1 F i F k. .
Put g s h 0 . . . h ky1 , hX0 s gh k gy1 , xX1 s x ? hX0 , and finally, set uy1 s p 0 . Ž . Thus uy1 s u 0 ? hy1 0 , since u 0 s r 0 by 4.6 . Assume that Ž4.10. is false. We will then show that the following condition holds. Ž4.10.1. w pXky i , skyi x is a subinterval of w x kyi , u kyiy1 x for all i, 0 F i F k. We will also show: Ž4.10.2. There exists a braiding of h which begins with hX0 . Suppose, for the moment, that both Ž4.10.1. and Ž4.10.2. hold. Notice that qk g w x k , y k x by Ž4.2.. Then, since Ž4.10. is false we get qk g w x k , sk x. Then the definition of qk yields dŽ x k , sk . q bŽ h k . G dŽ x kq1 , sk . q aŽ h k ., with strict inequality unless qk s sk . Applying gy1 , we then have d Ž x 0 , s0 . q b Ž hX0 . G d Ž xX1 , s0 . q a Ž hX0 . ,
Ž ).
again with strict inequality unless qk s sk . Since sk g w x k , x kq1 x, we have s0 g w x 0 , xX1 x. Also, taking i s k in Ž4.10.1. we see that s0 g w x 0 , p 0 x : w x 0 , z x, and it follows from Ž). that d Ž x 0 , z . q b Ž hX0 . G d Ž xX1 , z . q a Ž hX0 . .
Ž )) .
On the other hand, with Ž4.10.2., Ž4.4. yields dŽ x 0 , z . q bŽ hX0 . F dŽ xX1 , z . q aŽ hX0 ., with strict inequality if I0X : DŽ hX0 ., and where I0X s AŽ hX0 . l w x 0 , xX1 x l w x 0 , z x. Thus, equality holds in Ž))., and I0X DŽ hX0 ., and also
qk s s k . Next, put J 0X s AŽ hX0 . l w xX1 , z x, and note that dŽ x 0 , z . y dŽ xX1 , z . F l Ž I0X . y l Ž J 0X ., with equality if I0X is non-empty. Note also that l Ž J 0X . G 1 Ž X. Ž . Ž . 2 b h 0 , by 4.4 , and then )) yields l Ž I0X . G a Ž hX0 . y 12 b Ž hX0 . . Again, Ž4.4. applies and yields equality here. Since also equality holds in Ž)., we conclude that s0 s Y Ž x 0 , xX1 , z . g I0X . Also, d Ž pX0 , s0 . s d Ž pXk , sk . s d Ž pk , qk . s a Ž hX0 . y 12 b Ž hX0 . ,
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51
and so Ž4.10.1. yields I0X s w pX0 , s0 x. Then pX0 , s0 4 DŽ hX0 ., and so pk , qk 4 DŽ h k .. If now pk g DŽ h k . then L h k / B, and Ž3.8.Ža. yields qk g DŽ h k . for a contradiction. Thus, we conclude that pk f DŽ h k .. By Ž4.2. we then have qk / y k , and since sk s qk the definition of sk yields qk s u ky1 , so that Ž4.10. holds. It remains to prove Ž4.10.1. and Ž4.10.2.. Since Ž4.10. is assumed false, we have pk g w x k , u ky1 x, and then w pk , sk x is a subinterval of w x k , u ky1 x, by definition of sk . Thus, Ž4.10.1. holds for i s 0. Proceed by induction on i. Thus, suppose that w pXky iq1 , skyiq1 x is a subinterval of w x kyiq1 , u kyi x for y1 x w X x w some i, 0 F i - k. Apply hy1 ky i and obtain p kyi , s kyi F x kyi , u kyi ? h kyi . y1 Ž .Ž . If u ky iy1 s u kyi ? h kyi , we are done. So, referring to 4.8 b , we see that Jky i s w r kyi , ¨ kyiy1 x and u kyiy1 g Ikyi s w pkyi , y kyi x. Here u kyi s r kyi , so u ky i ? hy1 kyi s p kyi . That is, we have
w pXky i , skyi x F w x kyi , pkyi x F w x kyi , u kyiy1 x , which proves Ž4.10.1.. In order to prove Ž4.10.2., it will suffice to show that the H-sequence h9 given by
žŽ h
k
.
Ž h 0 , . . . , h ky 1 .y1
, Ž h ky1 .
Ž h 0 , . . . , h ky 2 .y1
, . . . , h 0 h1 hy1 0 , h 0 , h kq1 , . . . , h n
/
is a braiding of h. Indeed, one may ‘‘perform’’ this braiding as follows. First, move h 0 from the 0-position of h into the k-position, via k simple braidings. Then h 0 h1 hy1 is the new occupant of the 0-position, which we 0 now move into the Ž k y 1.-position via k-1 simple braidings. Continuing on in this way we obtain h9, proving Ž4.10.. Ž4.11. Suppose qk f Jky1. Then pk g Ž u ky1 , ¨ ky1 ., y k g w qk , r k ., and dŽ qk , y k . F dŽ pk , ¨ ky1 ., with strict inequality if y k g DŽ h k .. Proof. Set u s u ky 1 , ¨ s ¨ ky1 , J s Jky1 , xX1 s x ? h k , and put y1 h9 s Ž h k , hy1 k h 0 h k , . . . , h k h ky1 h k , h kq1 , . . . , h n . .
Notice that h9 is a braiding of h. It then follows from Ž4.4. that d Ž xX1 , z . q a Ž h k . G d Ž x, z . q b Ž h k . with strict inequality if I0X : DŽ h k ., where I0X s A Ž h k . l w x, xX1 x l w x, z x . We have qk g w u, z x by Ž4.10., and we are assuming that qk f J, so qk g Ž ¨ , z x. Suppose first that pk g w ¨ , z x. By Ž4.9. we then have d Ž x, z . s d Ž x, pk . q d Ž pk , z . s d Ž xX1 , r k . q d Ž pk , z . .
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ANDREW CHERMAK
Also, by the triangle inequality, d Ž xX1 , z . F d Ž xX1 , r k . q d Ž r k , z . , and this is strict if r k f w xX1 , z x. Then d Ž x, z . y d Ž xX1 , z . G d Ž pk , z . y d Ž r k , z . , and so a Ž h k . y b Ž h k . G d Ž pk , z . y d Ž r k , z . . s d Ž x k , z . y d Ž x kq1 , z . . By Ž4.2. we then have equality throughout, and so Ž4.4. yields I0X DŽ h k ., while r k g w xX1 , z x. Since also pk g w x, z x it follows that I0X s w pk , y k x s Ik , and thus Ik DŽ h k .. This contradicts condition Ž Ck . in Ž3.10., so we now conclude that pk f w ¨ , z x. But pk g w x k , z x, and J AŽ h k . by Ž3.13.Ž1.Ža., so we get pk g Ž u, ¨ .. Notice that then also h k g E ŽŽ h 0 , . . . , h ky1 ., J .. Suppose next that r k g w x k , z x. Then r k g w qk , z x F w ¨ , z x, so d Ž x, z . s d Ž x, ¨ . q d Ž ¨ , r k . q d Ž r k , z . . Also, we have dŽ pk , ¨ . s l Ž J l AŽ h k .. F 12 bŽ h k . y 12 l* by Ž3.13.Ž1.Žc.. Then d Ž ¨ , r k . G a Ž h k . y 12 b Ž h k . q 12 l*. On the other hand, we have d Ž xX1 , z . F d Ž xX1 , r k . q d Ž r k , z . F d Ž x, pk . q d Ž r k , z . F d Ž x, ¨ . q d Ž pk , ¨ . q d Ž r k , z . F d Ž x, ¨ . q d Ž r k , z . q 12 b Ž h k . y 12 l*. This yields d Ž x, z . y d Ž xX1 , z . G a Ž h k . y b Ž h k . q l* ) aŽ h k . y b Ž h k . which contradicts Ž4.4.. Hence r k f w x k , z x, and so y k g w qk , r k .. Put d s dŽ pk , ¨ ., d9 s dŽ qk , y k .. Then Ž4.9. gives d Ž x, z . G d Ž x, pk . q d Ž pk , z . y 2 d, s d Ž xX1 , r k . q d Ž pk , z . y 2 d, with a strict inequality unless J l w x, z x s ¨ 4 . That is, we have strict inequality unless I0X s w ¨ , y k x.
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53
Next, we have dŽ xX1 , z . F dŽ xX1 , r k . q dŽ r k , z ., so that d Ž x, z . y d Ž xX1 , z . G d Ž pk , z . y d Ž r k , z . y 2 d. We also have d Ž pk , z . s d Ž pk , qk . q d Ž y k , z . q d9 d Ž r k , z . s d Ž r k , qk . q d Ž y k , z . y d9 and so d Ž pk , z . y d Ž r k , z . s a Ž h k . y b Ž h k . q 2 d9. Thus, we now have d Ž x, z . y d Ž xX1 , z . G a Ž h k . y b Ž h k . q 2 Ž d9 y d . , with strict inequality unless I0X s w ¨ , y k x. On the other hand, Ž4.4. yields d Ž x, z . y d Ž xX1 , z . F a Ž h k . y b Ž h k . , with strict inequality if I0X : DŽ h k .. Thus we obtain d9 F d, and if d9 s d then ¨ , y k 4 DŽ h k .. Suppose finally that d9 s d and that y k g DŽ h k .. Then ¨ f DŽ h k .. But now dŽ ¨ , y k . s dŽ pk , qk . s aŽ h k . y 12 bŽ h k . g L h k, by Ž3.7.Žb. and Ž3.8.Ža., whence ¨ g DŽ h k .. This contradiction completes the proof of Ž4.11.. Ž4.12. There exist a non-degenerate segment J k s w u k , ¨ k x in P ŽŽ h 0 , . . . , h k .., with J : w x kq1 , z x, oriented so that ¨ k g Ž u k , z x, and defined as follows: Ži. Žii. Žiii.
If u ky 1 g Ik then Jk s w r k , ¨ ky1 x. If ¨ ky 1 g Ik then Jk s w u ky1 ? h k , y k x. If Ik : Ž u ky1 , ¨ ky1 . then Jk s w u ky1 ? h k , ¨ ky1 x.
Proof. Put J s Jky 1 , u s u ky1 , ¨ s ¨ ky1 , u s Ž h 0 , . . . , h ky1 ., and h s h k . We have u(Ž h. g H * as h g H *. By Ž4.10. and Ž4.11., J l aŽ h. is non-degenerate, so we have h g E Žu, J .. Now J AŽ h. by Ž3.13.Ž1.Ža., so precisely one of the definitions Ži., Žii., or Žiii. for Jk applies. That is to say, Jk is well-defined in Ž4.12.. We proceed case by case. Suppose u g Ik , so that Jk s w r k , ¨ x. By Ž4.10. we then have dŽ u, r k . G dŽ qk , r k ., with strict inequality if pk g DŽ h k .. Thus, the conditions Ži. through Žiii. in Ž3.13.Ž2.Žb. all hold, with pk in the role of y, and this yields w r k , ¨ x g P Žu(Ž h.., as desired. Evidently w r k , ¨ x F w x kq 1 , z x, and w r k , ¨ x is non-degenerate since J AŽ h.. Thus, Ž4.12. holds this case.
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ANDREW CHERMAK
Suppose next that Ik : Ž u, ¨ . so that Jk s w u ? h, ¨ x. Here we have l Ž Ik . - aŽ h. by Ž3.13.Ž1.Žb., so Ik s AŽ h. l w x k , z x. As J : w x k , z x by Ž4.8., we then have Ik s J l AŽ h., and so J l AŽ h. s B. By Ž4.2. we then have l Ž J l AŽ h.. G aŽ h. y 12 bŽ h., with restrict inequality if Ž J l AŽ h.. DŽ h.. Now Ž3.13.Ž2.Ža. yields w u ? h, ¨ x g P Žu(Ž h.., as desired. Further, w u ? h, ¨ x F w x k ? h, z x by Ž4.9., so Ž4.12. holds in this case. Finally, suppose that ¨ g Ik , so that Jk s w u ? h, y k x. Since dŽ pk , qk . s aŽ h. y 12 bŽ h. G 12 bŽ h., Ž3.13.Ž1.Žc. implies that qk f Jky1. Put y s y k . By Ž4.11. we have pk g Ž u, ¨ . and y g w qk , r k ., so we get ¨ g w y, y ? hy1 .. We have J o p p g P Žuy1 ., by Ž3.11.Žc.. Also, notice that h
uy1 ( Ž hy1 . s Ž uy1 . ( Ž hy1 .
ž
hy1
/
y1
and this is a braiding of ŽŽ hy1 .(uy1 . h , which is an H-conjugate of Žu(Ž h..y1 . Since u(Ž h. g H * it then follows that uy1 (Ž hy1 . g H *. Since J o p p l AŽ hy1 . s J l AŽ h. s w pk , ¨ x, which is non-degenerate, we then have hy1 g E Žuy1 , J o p p .. Further, Ž4.11. yields dŽ qk , y . F dŽ pk , ¨ ., so d Ž ¨ , y . F d Ž pk , ¨ . q d Ž ¨ , qk . s a Ž h . y 12 b Ž h . , and Žagain referring to Ž4.11.. the inequality is strict if y g DŽ h.. On the other hand, we have dŽ ¨ , y . s aŽ h. y dŽ ¨ , y ? hy1 . since ¨ g w y, y ? hy1 .. This yields dŽ ¨ , y ? hy1 . G 12 bŽ h., with strict inequality if y g DŽ h.. Now Ž3.13.Ž2.Žb. applies, with J o p p and uy1 in the roles of J and h, and the result is that w y ? hy1 , u x g P Žuy1 (Ž hy1 ... Then w u, y ? hy1 x g P ŽŽ h.(u. and w u ? h, y x g P ŽŽ h.(u h ., by Ž3.11.Žc.. But
Ž h . (u h s Ž h, h 0h , . . . , h hky1 . is a braiding of Ž h 0 , . . . , h ky1 , h., so we arrive at w u ? h, y x g P Žu(Ž h.., as desired. Since u g w x k , z x and y g w u, z x l w r kq1 , z x, we have w u ? h, y x F w x kq 1 , z x. Also, w u ? h, y x is non-degenerate, since w r k , y x F w u ? h, y x, where r k / y by Ž4.11.. This completes the proof of Ž4.12.. With Ž4.12. we have completed the induction step that began with Ž4.8.. We therefore conclude that there is a non-degenerate segment Jn g P Žh. with Jn : w x nq1 , z x. But this contradicts the plain fact that x nq1 is by definition equal to z. The proof of Theorem Ž3.14. is thereby complete.
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55
REFERENCES 1. R. Alperin and H. Bass, Length functions of group actions on L-trees, in ‘‘Combinatorial Group Theory and Topology, Alta, Utah, 1984,’’ Ann. of Math. Stud., Vol. 111, pp. 265]378, Princeton Univ. Press, Princeton, NJ, 1987. 2. A. Chermak, R-trees, small cancellation, and convergence, Trans. Amer. Math. Soc. 347, No. 11 Ž1995.. 3. A. Chermak, Locally non-spherical Artin groups, to appear in J. Alg.. 4. R. C. Lyndon and P. E. Schupp, ‘‘Combinatorial Group Theory,’’ Springer-Verlag, New YorkrBerlin, 1977. 5. J. W. Morgan, L-trees and their applications, Bull. Amer. Math. Soc. (N.S. ) 26, No. 1 Ž1992..
JOURNAL OF ALGEBRA ARTICLE NO.
A Convergence Theorem for L-Trees Andrew Chermak* Department of Mathematics, Kansas State Uni¨ ersity, Manhattan, Kansas 66506 Communicated by J. Tits Received March 27, 1996
1. INTRODUCTION Our aim in this paper is to refine the arboreal ‘‘Small Cancellation Theory’’ introduced by the author in w2x, with a view toward certain applications. Namely, our results here will be used, in w3x, to obtain results about locally non-spherical Artin groups, in particular, that such groups have a solvable Word Problem. In the Introduction to w2x we explained how ordinary, metric Small Cancellation Theory, as developed Žfor example. in Chapter VI of w4x, can be viewed as the study of certain sets of hyperbolic isometries of trees. We refer the reader to those pages, for the background of the present work. The key concept both here and in w2x is that of a ‘‘convergence property,’’ by which we always mean something roughly analogous to ‘‘solvability of a Word Problem by Dehn’s Algorithm.’’ The convergence property considered here Žsee Ž3.10.. is significantly more flexible than any of those considered in w2x. Another difference between the present work and w2x is that we now set everything in the context of L-trees, L an Žalmost. arbitrary ordered abelian group. ŽThe qualifying adjective ‘‘almost’’ is due to the necessity that L contain an element l which is ‘‘large,’’ in the sense that l lies in no proper convex subgroup of L.. The reason for working in this degree of generality is that, now that the basic properties of L-trees have been carefully established in w1x, it is really no more difficult to work with L-trees than with ordinary simplicial trees, in the present context. Readers who are only interested in the applications in w3x, may ignore any mention of L-trees, and read this paper as concerning only simplicial trees. In any * E-mail address: [email protected]. 32 0021-8693r98 $25.00 Copyright Q 1998 by Academic Press All rights of reproduction in any form reserved.
A CONVERGENCE THEOREM FOR L-TREES
33
case, a quick overview of L-trees, sufficient for our purposes, is given in Section 2, below. ŽSee also w5x for an expository treatment of L-trees.. Let X be a L-tree, and let H be a set of hyperbolic isometries of X. Always assume: Ž1. Ž2.
H s H y1 , and hy1 H h s H for all h g H .
Associated with H is the hyperbolic length function, here denoted a: H ª L) 0 , and defined by a Ž h . s min d Ž x, x ? h . 4 xgX , where d is the L-metric on X. We assume, always, that each aŽ h. is ‘‘large’’ in the sense mentioned above, so that for any m g L, N ? aŽ h. ) m for some integer N. Now, to each h g H there is associated the characteristic subtree of h, A Ž h . s x g X d Ž x, x ? h . s a Ž h . 4 and AŽ h. is then L-isometric to the ‘‘linear’’ tree L. In practice, it may happen that AŽ h. has an ² h:-invariant subset which we may wish to distinguish. For example, if L s Z, so that X is a simplicial tree, we can view X as a bipartite graph by separating the vertices of X into two classes D and D9, with no vertex of D adjacent to any vertex in D9. Then AŽ h. l D can be viewed as the image of 2Z under a suitable Z-isometry
a h : Z ª AŽ h . . More generally, we may have a subgroup L h of L, a L-isometry
a h : L ª AŽ h . , and a distinguished subset of AŽ h.: D Ž h. s ahŽ L h . . The function D which assigns to each h g H the triple D Ž h. s Ž ah , L h , D Ž h. . is called here a specialization on H Žsee Definition Ž3.7.., provided that D is suitably compatible with inversion and conjugation. In the application in w3x, H will be a union of conjugacy classes of a certain subgroup G of IsomŽ X ., and it will be necessary to consider
34
ANDREW CHERMAK
subsets Hu, ¨ of H Žindexed by pairs of generators of an Artin group. where each Hu, ¨ is itself a union of conjugacy classes of G. Note that the length function a is constant on conjugacy classes. We will associate to H also a modified length function b: H ª L) 0 , defined by bŽ h. s min aŽ h. < h g Hu, ¨ 4 . The notion of a modified length function is formalized below, in Definition Ž3.8., and in Ž3.9. we explain the Artin group example in further detail. Having fixed a specialization D and a modified length function b, we introduce the relevant notion of convergence with Ž3.10.. To put it somewhat roughly: H has the Ž b, D .-con¨ ergence property if whenever x g X and g is an element of the group ² H : generated by H , there is a sequence Ž h 0 , . . . , h n . of elements of H with Ža. h 0 , . . . , h n s g, and Žb. the sequence of points x 0 s x, x 1 s x ? h 0 , . . . , x kq1 s x k ? h k , . . . , x nq1 s x n ? h n s x ? g ‘‘converging’’ to x ? g in the sense that for all k, 0 F k F n, we have d Ž x kq 1 , x ? g . q a Ž h k . F d Ž x k , x ? g . q b Ž h k . ,
Ž Ck .
with strict inequality under certain conditions relating AŽ h k ., DŽ h k ., and the geodesic path from x k to x ? g. The result of this paper is a sufficient condition ŽTheorem Ž3.14.. for H to have the Ž b, D .-convergence property. Namely, we show that it suffices that there be a mapping P Žcalled a polarization. which associates certain collections of closed segments of X with certain sequences of elements of H , in a certain way, satisfying certain rules. The relevant definitions are Ž3.11. through Ž3.13.. Theorem Ž3.14. says that if there is such a polarization of X over H , then H has the Ž b, D .-convergence property. The proof, in Section 4, is very close to the proof of the main result of w2x. If I had known the precise form of the application to Artin groups when writing w2x, this duplication of effort might have been avoided. 2. L-TREES Let L be an ordered abelian group. Denote the set of non-negative elements of L by LG 0 , and the set of strictly positive elements by L) 0 . For any a, b g L with a - b let w a, b x denote the closed segment from a to b:
w a, b x s c g L : a F c F b4 .
A CONVERGENCE THEOREM FOR L-TREES
35
A subset of L is bounded if it is contained in a closed segment. A subset K of L is con¨ ex if w a, b x : K for all a, b g K with a F b. An element a of L is large if a G 0 and no proper convex subgroup of L contains a. A L-metric space consists of a set X together with a L-metric d: X = X ª L satisfying the familiar axioms: for all x, y, z g X we have d Ž x, y . s d Ž y, x . , d Ž x, y . s 0
if and only if x s y,
and d Ž x, y . q d Ž y, z . G d Ž x, z . . If X and Y are L-metric spaces then a mapping a : X ª Y is said to be a L-metric morphism if dŽ x, y . s dŽ x a , ya . for all x, y g X. ŽNote: we write mappings to the right of their arguments.. A surjective L-metric morphism is called a L-isometry. Notice that L is itself a L-metric space via d Ž a, b . s < a y b < s max a y b, b y a4 . The translation ta : L ª L, given by xta s x q a, is a L-isometry, as is the inversion map x ¬ yx. The L-metric space X is said to be geodesically linear if, for any points x, y g X, there is a unique L-metric morphism
a : 0, d Ž x, y . ª X such that 0 a s x and dŽ x, y . a s y. We write a s w x, y x, and we say that a is the oriented closed segment from x to y. We write a o p p for the opposite segment w y, x x. Further, we consider x and y to be the initial and terminal boundary points, respectively, of w x, y x, and we write
0 w x, y x s x,
1 w x, y x s y,
and
w x, y x s x, y 4 .
The oriented closed segment a s w x, y x gives rise to the Ž unoriented. closed segment consisting of the image of a in X, which we denote also by w x, y x. This need not be cause for confusion. For w x, y x a closed segment, we define
and
w x, y . s w x, y x y y 4 , Ž x, y x s w x, y x y x 4 , Ž x, y . s w x, y x y x, y 4 .
36
ANDREW CHERMAK
A segment of X is a subset of X of the form w x, y x, Ž x, y x, w x, y ., or Ž x, y .. A segment is said to be degenerate if it is either empty or consists of a single point. Let w x, y x and w x, w x be oriented closed segments of the geodesically linear L-metric space X. We write
w x, y x F w z, w x if w x, y x : w z, w x as unoriented closed segments, and with x g w z, y x Žequivalently, y g w x, w x.. Define the length of a closed segment w x, y x by l Ž w x, y x . s d Ž x, y . . DEFINITION 2.1 Žcf. w1x.. A L-tree is a geodesically linear L-metric space X satisfying: Ža. Ž Y-condition. For any x, y, z g X there exists w g X such that w x, z x l w y, z x s w w, z x. Žb. If w x, y x l w y, z x s y4 then w x, y x j w y, z x s w x, z x. Henceforth, X will always denote a L-tree. The point w in Ž2.1.Ža. is uniquely determined, and we denote it by Y Ž x, y, z .. Let us say that a subset J of X is a generalized closed segment if J is isometric to a closed segment of L, or to LG 0 , or to L. The following basic result concerning these objects will be used frequently, without explicit citation. LEMMA 2.2. Let I and J be generalized closed segments of X, with I l J / B. Then I l J is a generalized closed segment. If also I is a closed segment then so is I l J. Proof. Let I0 and J 0 be subsets of L and let a : I0 ª I and b : J 0 ª J be isometries. Put i1 s a g I0 : a a g J 4 and J1 s b g J 0 : bb g I 4 . Then I1 a s J1 b s I l J. As X is geodesically linear, I1 and J1 are convex. If I1 s I0 then I l J is isometric to I0 and we are done. So assume I1 / I0 . Then either there exists u g I0 with u F I1 , or there exists ¨ g I0 with I1 F ¨ . Similarly, we may assume J1 / J 0 , so there exists either a lower bound u9 for J1 in J 0 or an upper bound ¨ 9 for J1 in J 0 . By composing a and or b with suitable isometries of L, we may assume that I0 and J 0 have been chosen so that the lower bounds u and u9 exist. Put x s u a and y s u9b , and let z g I l J. Put w s Y Ž x, y, z .. Then w g I l J by Ž2.1.Ža., and we then have w s a a for some a g I1 , with a F I1. If no upper bound ¨ exists for I1 in I0 we then have I l J isometric to LG 0 , and we are done in this case. On the other hand,
A CONVERGENCE THEOREM FOR L-TREES
37
suppose that I1 has an upper bound ¨ in I0 . Then also an upper bound ¨ 1 exists for J1 in J 0 . Applying Ž2.1.Ža. once again, we obtain b g I1 with I1 F b. Thus I l J is isometric to w a, b x, proving Ž2.2.. DEFINITIONS 2.3 ŽŽ2.7. and Ž2.10.Ž3. in w1x.. A subset T of X is con¨ ex if w x, y x : T for all x, y g T. If also T / B then T is a subtree of X. We say that T is closed con¨ ex if the intersection of T with any closed segment is either empty or a closed segment. LEMMA 2.4. Let T be a closed subtree of X, and let x g X. There is then a unique point y of T such that, for e¨ ery z g T, we ha¨ e y g w x, z x. Moreo¨ er, we then ha¨ e w y, z x s w x, z x l T. Proof. The lemma is a special case of the ‘‘Bridge Proposition’’ which is result Ž2.17. of w1x. ŽThe directed closed segment w x, y x is called the bridge from x to T.. DEFINITION 2.5. An isometry h of X is hyperbolic if there exists no closed segment w x, y x with x ? h s y and y ? h s x. ŽIn particular, h fixes no point of X.. For such an isometry h we put a Ž h . s min d Ž x, xh . 4 xgX . It is known w1, Theorem 6.6.x that this minimum exists. Then put A Ž h . s x g X : d Ž x, xh . s a Ž h . 4 . PROPOSITION 2.6 w1, Theorem 6.6x. Let h be a hyperbolic isometry of X. Then the following hold. Ža. aŽ h. exists, and AŽ h. is isometric to the smallest convex subgroup of L containing aŽ h.. Žb. AŽ h. is an ² h:-invariant, closed subtree of X, and AŽ h. is contained in every ² h:-invariant subtree of X. Žc. Let x g X and let w x, y x be the bridge from x to AŽ h.. Then w x, x ? h x l AŽ h. s w y, y ? h x has length aŽ h.. Further, Žsee Fig. 1. we have
w x, x ? h x s w x, y x j w y, y ? h x j w y ? h, x ? h x
and
d Ž x, x ? h . s a Ž h . q 2 d Ž x, y . . COROLLARY 2.7. Let h be a hyperbolic isometry of X and let n g Z, n / 0. Then h n is a hyperbolic isometry, with AŽ h. s AŽ h n ., and aŽ h n . s < n < ? aŽ h.. LEMMA 2.8.
Let h be a hyperbolic isometry of X, and let x, z g X. Put
J s A Ž h . l w x, x ? h x l w x, z x ,
J9 s A Ž h . l w x, x ? h x l w x ? h, z x .
38
ANDREW CHERMAK
FIGURE 1
Then J j J9 is the closed segment AŽ h. l w x, x ? h x, of length aŽ h. s l Ž J . q l Ž J9.. Moreo¨ er: Ža. dŽ x, z . y dŽ x ? h, z . - aŽ h. if and only if J9 is non-degenerate, in which case d Ž x, z . y d Ž x ? h, z . F l Ž J . y l Ž J9 . . Žb. dŽ x ? h, z . y dŽ x, z . - aŽ h. if and only if J is non-degenerate, in which case d Ž x ? h, z . y d Ž x, z . F l Ž J9 . y l Ž J . . Proof. In view of Ž2.6., one can be comfortable with a proof by ‘‘pictures.’’ There are three possibilities; see Fig. 2. ŽIn each case one verifies Lemma 2.8.. DEFINITION 2.9. Let h be a hyperbolic isometry of X, let z g X, and let w x, y x be an oriented closed segment of AŽ h.. We say that w x, y x is oriented toward z by h if Ži. w x, z x s w x, y x j w y, z x, and Žii. dŽ x ? hy1 , y . G dŽ x ? h, y .. In particular, w x, y x is oriented towards z by h if x s y. LEMMA 2.10. Let h and h9 be hyperbolic isometries of X. Then the following are equi¨ alent. Ž1. There exists x g X with dŽ x, x ? hh9. - aŽ h. q aŽ h9.. Ž2. AŽ h. s AŽ h9. or AŽ h. l AŽ h9. is a non-degenerate closed segment, and in either case, AŽ h. l AŽ h9. is oriented in opposite directions by h and h9. Proof. The direction Ž1. « Ž2. is given by Propositions Ž8.1.Ža. and Ž8.3.Ža. of w1x. The direction Ž2. « Ž1. is trivial.
A CONVERGENCE THEOREM FOR L-TREES
39
FIGURE 2
3. SEQUENCES OF HYPERBOLIC ISOMETRIES This section consists almost entirely of definitions, with a few elementary consequences, and with also a statement of the Main Theorem. Fix an ordered abelian group L, a L-tree X, and a set H of hyperbolic L-isometries of X. Ž3.1. THE BASIC HYPOTHESIS. The following three conditions hold for e¨ ery h g H .
40
ANDREW CHERMAK
Ža. aŽ h. is large. That is, no proper con¨ ex subgroup of L contains aŽ h.. Žb. hy1 g H Ž H is symmetric.. Žc. hy1 H h s H Ž H is self-normalizing.. We assume Ž3.1. for the remainder of this paper. An H-sequence is defined to be a finite sequence of elements of H . If h and h9 are H-sequences then h(h9 denotes the H-sequence obtained by concatenating h and h9 in the given order. If h s Ž h 0 , . . . , h k . is an H-sequence then we write y1 hy1 s Ž hy1 k , . . . , h0 . ,
Ž 3.2.
and for any g g ² H : we write h g s Ž gy1 h 0 g , . . . , gy1 h k g . .
Ž 3.3.
Of course, hy1 and h g are H-sequences. DEFINITION 3.4. Let h s Ž h 0 , . . . , h k . and h9 s Ž hX0 , . . . , hXk . be two H-sequences of length k q 1, k G 1. We say that h9 is a simple braiding of h if for some i with 0 F i - k we have h j s hXj
for all j f i , i q 1 4 ,
and either
Ž hXi , hXiq1 . s Ž h iq1 , hy1 iq1 h i h iq1 . , or
Ž hXi , hXiq1 . s Ž h i h iq1 hy1 i , hi . . Notice that the relation ‘‘is a simple braiding of’’ is symmetric. We say that h9 is a braiding of h if there exists a chain h s s 0 , . . . , s l s h9 of H-sequences such that for all i, 1 F i F l, s i is a simple braiding of s iy1. This defines an equivalence relation Žwhose equivalence classes are in fact the orbits of a natural action of the Artin braid group on k q 1 threads. on the H-sequences of length k q 1. The H-sequence h s Ž h 0 , . . . , h k . will be said to be reduced if for all i with 1 F i - k we have h iy1 h i f H j 14 . We put: H * s H-sequences h: every braiding of h is reduced 4 .
Ž 3.5.
A CONVERGENCE THEOREM FOR L-TREES
41
LEMMA 3.6. Let h s Ž h 0 , . . . , h k . and h9 s Ž hX0 , . . . , hXk . be H-sequences, with h9 a braiding of h. Then the following hold. Ža. h 0 . . . h k s hX0 . . . hXk . Žb. ² h 0 , . . . , h k : s ² hX0 , . . . , hXk :. Žc. If h is a word of minimal length for the element h 0 , . . . , h k of ² H :, in terms of the generating set H , then h g H *. Žd. If h g H * then h9 g H *, and any prefix Ž h 0 , . . . , h i . of h, 0 F i F k, is in H *. Že. If h g H * then h i h j f H j 14 for any i and j with 0 F i / j F k. Proof. Parts Ža. and Žb. are easily verified in the case that h9 is a simple braiding of h, and the obvious induction argument then yields Ža. and Žb. in general. Part Žc. follows from Ža.. If h g H * then so is h9, by definition. If Ž h 0 , . . . , h i . is a prefix of h and Ž hY0 , . . . , hYi . is a braiding of Ž h 0 , . . . , h i ., then Ž hY0 , . . . , hYi , h iq1 , . . . , h n . is a braiding of h, and this yields Žd.. Suppose that h g H * and let 0 F i - j F k be given. Then the sequence
žh ,..., h 0
iy1 ,
hj
hj
h i , h j , Ž h iq1 . , . . . , Ž h jy1 . , h jq1 , . . . , h k
is a braiding of h, and so h i h j f H j 14 . Since h j h i s Ž h i h j . and Žc. then yields h j h i f H j 14 , proving Že..
/
hy1 j
, Ž3.1.Žb.
DEFINITION 3.7. A specialization on H is a function D which assigns to each h g H a triple D Ž h. s Ž ah , L h , D Ž h. . , where Ža. a h is a L-isometry from L onto AŽ h., Žb. L h s B or L h is a subgroup of L containing aŽ h., Žc. DŽ h. s B or DŽ h. s a hŽ L h ., and Žd. L h s L hy1 s L h g , DŽ h. s DŽ hy1 ., and DŽ h g . s DŽ h. ? g for any g g ² H :. DEFINITION 3.8. Let D be a specialization on H . A modified length function Žon H , relati¨ e to D . is a mapping b: H ª L) 0 such that the following hold for all h g H : Ža. bŽ h. F aŽ h., and bŽ h. g 2 ? L h if L h / B. Žb. bŽ hy1 . s bŽ h.. Žc. bŽ h g . s bŽ h. for all g g ² H :.
42
ANDREW CHERMAK
EXAMPLE 3.9. Let G s) i g I Gi be a free product, and let X denote the standard tree of G. Thus, X is a simplicial tree whose vertex set VX is the disjoint union, VX s D@ D9, D s G, D9 s @ i g I Gi R G, where Gi R G denotes the set of right cosets of Gi in G. The edge-set EX consists of pairs of vertices of the form of Ž g, Gi g .. Regard each Gi and each Gi )Gj as a subgroup of G Žfor i / j ., and for Ri, j and all i / j let Ri, j be a subset of Gi )Gj y Ž Gi j Gj .. Put H0 s jR put H s Ž H0 j H0y1 . G . Then H is a symmetric, self-normalizing set of hyperbolic isometries of X. For each i / j, put bi , j s min a Ž h . : h g Ri , j 4 . There is then a well-defined mapping b of H into 2Z given by b Ž h . s bi , j
G
if h, hy1 4 l Ri , j / B.
Put Z h s 2Z for all h g H , and let a h : Z ª AŽ h. be an isometry which sends 2Z to D l AŽ h.. We then have a specialization D given by D Ž h . s Ž a h , 2Z, a h Ž 2Z . . , and b is a modified length function relative to D. This example will be pursued in w3x, where Gr² H : will be an Artin group which is ‘‘locally non-spherical.’’ In what follows, let D be a fixed specialization on H and let b be a fixed, modified length function on H relative to D. DEFINITIONS 3.10. Let h s Ž h 0 , . . . , h n . g H *, and let x g X. Put x 0 s x and, inductively, define x kq 1 s x k ? h k , 0 F k F n. Write z for x nq1. Also, put Ik s AŽ h k . l w x k , x kq1 x l w x k , z x. We say that x con¨ erges to z ¨ ia h Ž relati¨ e to Ž b, D .. if for all k, 0 F k F n, we have d Ž x kq 1 , z . q a Ž h k . F d Ž x k , z . q b Ž h k . , with equality only if Ik : D Ž h k . .
Ž Ck .
We say that H has the Ž b, D .-con¨ ergence property if, whenever h9 g H * and x g X, there exists a braiding h s Ž h 0 , . . . , h n . of h9 such that x converges to x ? h 0 . . . h n via h Žsee Fig. 3.. The aim of this paper is to find suitable, and verifiable, conditions on H which will guarantee that H has the Ž b, D .-convergence property. Definitions Ž3.11. through Ž3.13. will set up such a collection of conditions.
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43
FIGURE 3
DEFINITION 3.11. Let P be a mapping which associates to each h in H * a collection P Žh. of closed oriented segments of X. We say that P is a polarization Žof X o¨ er H . if the following conditions hold. Ža. P ŽŽB.. is the set of all degenerate closed segments of X. That is, P ŽŽB.. s X. Žb. P Žh. s P Žh9. if h9 is a braiding of h. Žc. If J g P Žh. then J o p p g P Žhy1 ., and J ? g g P Žh g . for any g g ² H :. DEFINITION 3.12. Let P be a fixed polarization of X over H . Let h g H *, J g P Žh.. Then E Žh, J . denotes the set of all h g H satisfying the following three conditions. Ža. Žb. Žc. oriented
h(Ž h. g H *, J l AŽ h. / B, and if J is non-degenerate, then J l AŽ h. is non-degenerate and is toward 1 J by h.
Given h g E Žh, J ., we wish to state conditions which give expression to there being some influence of J on the structure of P Žh(Ž h... As usual, we shall state these conditions in the form of a definition. This will be our most important definition, bringing together all of the notions encountered so far. DEFINITION 3.13. Let P be a polarization of X over H , D a specialization on H , and b a modified length function on H relative to D. We say that P controls con¨ ergence Ž relati¨ e to Ž b, D .. if there exists a large positive element l* of 2L such that bŽ h. G l* for all h g H , and if the following conditions hold whenever we have h g H *, J g P Žh., and h g E Žh, J ..
44
ANDREW CHERMAK
FIGURE 4
Ž1. Ž Exclusion. Ža. J AŽ h. if h / ŽB., Žb. l Ž J l AŽ h.. - aŽ h., and Žc. if J l AŽ h. / B then l Ž J l AŽ h.. G 12 ŽbŽ h. y l*.. Ž2. Ž Extension. Ža. We have wŽ 0 J . ? h, 1 J x g P Žh(Ž h.., provided that Žsee Fig. 4.: Ži. J l AŽ h. s B, Žii. l Ž J l AŽ h.. G aŽ h. y 12 bŽ h., with strict inequality if Ž J l AŽ h.. DŽ h.. Žb. For y g AŽ h., we have w y ? h, 1 J x g P Žh(Ž h.., provided that Žsee Fig. 5.: Ži. 0 J g w y, y ? h., and Žii. dŽ 0 J, y ? h. G 12 bŽ h., with strict inequality if 0 J, y4 : DŽ h.. We may now state our main result. THEOREM 3.14. Let P be a polarization of X o¨ er H , D a specialization on H , and b a modified length function on H relati¨ e to D. Suppose that P controls con¨ ergence relati¨ e to Ž b, D .. Then H has the Ž b, D .-con¨ ergence property.
FIGURE 5
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45
Henceforth, fix b, D, and l* as in Ž3.13., and assume that we are given a polarization P such that P controls convergence relative to Ž b, D .. We will take up the proof of Theorem Ž3.14. in Section 4, but first we have two lemmas. LEMMA 3.15. Let h g H and let J be a closed segment of AŽ h., oriented toward 0 J by h. Assume 12 bŽ h. F l Ž J . F aŽ h., and if J : DŽ h. assume l Ž J . ) 12 bŽ h.. Then J g P ŽŽ h... Proof. Put x s 1 J. Then x g P ŽŽB.. by Ž3.11.Ža., and h g E ŽŽB., x .. Put y s Ž 0 J . ? hy1 . As 0 - l Ž J . F aŽ h. we have x g w y, y ? h.. If x, y4 : DŽ h. then Ž3.7. says that L h is a subgroup of L containing aŽ h., and hence x, y ? h4 s J : DŽ h.. Now Ž3.13.Ž2.Žb. applies, with x 4 in the role of J, and this yields the lemma. LEMMA 3.16. Let h, h9 g H with Ž h, h9. g H *, and suppose that aŽ hh9. - aŽ h. q aŽ h9. Ž cf. Ž2.10... Put I s AŽ h. l AŽ h9.. Then: Ža. Žb.
l Ž I . F 12 min bŽ h., bŽ h9.4 y 12 l*, l Ž I . - min aŽ h. y 21 bŽ h., aŽ h9. y 21 bŽ h9.4 .
Proof. By Ž2.1., I contains a non-degenerate segment. Let K be an oriented closed segment of I, oriented toward 1 K by h9, and with l Ž K . F aŽ h9.. Then K g P ŽŽ h9.. by Ž3.15., and we have h g E ŽŽ h9., J .. Then Ž3.13.Ž1.Žb. yields l Ž J . - aŽ h.. In particular, it follows that I is itself a closed segment with l Ž I . - aŽ h.. Take I to be oriented toward 0 I by h. Let J be the closed segment of AŽ h. such that 1 J s 1 I and 0 J s Ž 1 I . ? h. Then I : J or J : I. We have J g P ŽŽ h.. by Ž3.15., and then h9 g E ŽŽ h., J .. Now Ž3.13.Ž1.Žb. gives I : J, I s J l AŽ h9., and J l AŽ h9. / B. Now apply Ž3.13.Ž1.Žc. with h9 in the role of h, and obtain l Ž I . F 12 bŽ h9. y l*. Symmetry then yields Ža., and Žb. follows since aŽ h. G bŽ h., by Ž3.8.. 4. PROOF OF THEOREM 3.14 We proceed by contradiction. Thus, assume that we are given b, D, and l* as in Ž3.13., and a polarization P of H which controls convergence relative to Ž b, D .. But assume that H does not have the Ž b, D .-convergence property. Thus: Ž4.1. There exists h s Ž h 0 , . . . , h n . g H *, and there exists x g X such that, setting z s x ? h 0 . . . h n , x does not converge to z via h9 for any braiding h9 of h. We now chose h and x so that the length, n q 1, of h is as small as possible for Ž4.1.. Fix the notation: z s x ? h 0 . . . h n .
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ANDREW CHERMAK
The positive element l* lies in no proper convex subgroup of L, so there is a positive integer N with N ? l* ) dŽ x, z .. It follows that h can be chosen so that, for any braiding Ž hX0 , . . . , hXn . of h we have dŽ x ? hX0 , z . ) dŽ x ? h 0 , z . y l*. Indeed, the alternative is that after composing N successive braidings we obtain dŽ x ? hX0 , z . - 0, which is absurd. ŽThus, no matter what braiding h9 of h we look at, x ? hX0 will not be much closer to z than x ? h 0 is.. Having chosen h as above, we next observe that, by the minimality of n q 1, x ? h 0 converges to z via some braiding Ž hX1 , . . . , hXn . of Ž h1 , . . . , h n .. But Ž h 0 , hX1 , . . . , hXn . is then a braiding of h, so we may assume to begin with, without disturbing the choice in the preceding paragraph, that x ? h 0 converges to z via Ž h1 , . . . , h n .. For easy reference, we record these results as follows: Ž4.2. x ? h 0 converges to z s x ? h 0 . . . h n via Ž h1 , . . . , h n .. Ž4.3. For any braiding Ž hX0 , . . . , hXn . of h, we have dŽ x ? h 0 , z . - dŽ x ? X h 0 , z . q l*. The next result tells us that h 0 doesn’t move x very much closer to z than x already is, and this remains true even if h is replaced by a braiding h9. Ž4.4.
For any braiding Ž hX0 , . . . , hXn . of h, we ha¨ e d Ž x, z . q b Ž hX0 . F d Ž x ? hX0 , z . q a Ž hX0 .
with strict inequality if I0X : DŽ hX0 ., and where I0X is defined by I0X s A Ž hX0 . l w x, x ? hX0 x l w x, z x . Proof. Let h9 s Ž hX0 , . . . , hXn . be a braiding of h, and put hX0 s hXn .. The minimality of n q 1 in our choice of h implies that x ? hX0 converges to z via some braiding hY0 of hX0 . Put h0 s Ž hX0 .(hY0 . Then h0 is a braiding of h9, hence also of h, and so x does not converge to z via h0. But this means that condition Ž C0 . in Ž3.10. fails to hold, with hX0 in the role of h 0 , and with I0X in place of I0 . This yields Ž4.4.. Ž hX1 , . . . ,
We now establish notation, to be followed from now on. Set x 0 s x and, as in Ž3.10., set x iq1 s x i ? h i , 0 F i F n. Thus, z s x nq1. Now, for any i in the above range, put K i s A Ž h i . l w x i , x iq1 x , Ii s K i l w x i , z x , and yi s Y Ž x i , x iq1 , z .
Ž cf. Ž 2.1. Ž a . . .
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47
Notice that, by Ž2.6.Žc., K i is a closed segment of length aŽ h i ., and we may write K i s w pi , ri x , with ri s pi ? h i . Define a point qi in K i by K i s w pi , qi x j w qi , ri x ,
d Ž qi , ri . s 12 b Ž h i . .
We also put J0 s K 0 l w x 1 , z x . Applying Ž4.4. to Ž h 0 , . . . , h n ., we obtain dŽ x, z . y dŽ x 1 , z . - aŽ h 0 ., and then J 0 is non-degenerate by Ž2.8.Ža.. Write J0 s w u 0 , ¨ 0 x ,
with ¨ 0 g Ž u 0 , z x
Ž 4.5.
so that J 0 is an oriented closed segment, oriented away from z by h 0 . Ž4.6. AŽ h 0 . l w x, z x is non-degenerate. In particular, we ha¨ e ¨ 0 g w x, z x, and x / z. Proof. Suppose false, so that AŽ h 0 . l w x, z x is either empty or is the point p 0 s 0 K 0 . It may be helpful to have the following picture in mind Žpossibly with x s z, and possibly with p 0 s y 0 . ŽFig. 6.. Suppose first that p 0 f w x 1 , q1 x. Then q1 f w p 0 , z x, and hence q1 g w x 1 , p 0 . by Ž4.2.. If now p1 g w r 0 , p 0 x then w p1 , q1 x : AŽ h 0 . l AŽ h1 .. But dŽ p1 , q1 . s aŽ h1 . y 12 bŽ h1 ., so we contradict Ž3.16.Žb.. Thus p1 f w r 0 , p 0 x, so p1 g w x 1 , r 0 x. Put d s 12 min bŽ h 0 ., bŽ h1 .4 y 12 l*. Since aŽ h 0 . G bŽ h 0 ., by Ž3.8., we certainly have d - aŽ h 0 .. Also, recall from Ž3.13. that bŽ h. G l* for any h g H , so we have d G 0. Let w be the point in J 0 at distance d from r 0 . By Ž3.16.Ža. we then have y 1 g w x 1 , w x. Set xX1 s x 2 ? hy1 0 . Then d Ž xX1 , p 0 . s d Ž x 2 , r 0 . F d Ž x 2 , w . q d Ž w, r 0 . s d Ž x 2 , w . q d.
FIGURE 6
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ANDREW CHERMAK
Now Ž4.2. and condition Ž C1 . in Ž3.10. yield d Ž x 2 , w . F d Ž x 1 , w . y a Ž h1 . q b Ž h1 . s d Ž x 1 , r 0 . y a Ž h1 . q b Ž h1 . q d, and so d Ž xX1 , p 0 . F d Ž x 1 , r 0 . y a Ž h1 . q b Ž h1 . q 2 d. On the other hand, we have d Ž x 1 , p0 . s d Ž x 1 , r 0 . q aŽ h 0 . , so we get d Ž x 1 , p 0 . y d Ž xX1 , p 0 . G a Ž h 0 . q a Ž h1 . y b Ž h1 . y 2 d G Ž a Ž h 0 . y b Ž h 0 . . q Ž a Ž h1 . y b Ž h1 . . q l* G l*. As p 0 g w x 1 , z x we then have dŽ x 1 , z . y dŽ xX1 , z . G l*. Setting hX0 s h 0 h1 hy1 we see that xX1 s x ? hX0 , and Ž hX0 , h 0 , h 2 , . . . , h n . is a braiding of h. 0 Thus, we contradict Ž4.3.. Hence p 0 g w x 1 , q1 x. Now switch notation and set xX1 s x ? h1 , and take w to be the point in J 0 at distance d from p 0 , where d is defined as above. Since p 0 g w x 1 , q1 x, Ž4.2. implies that q1 g w p 0 , z x. Then Ž3.16.Ža. shows that p1 g w x, z x, and Ž3.16.Žb. yields q1 g Ž p 0 , z x. Now d Ž x 1 , q1 . s d Ž x 1 , r 0 . q a Ž h 0 . q d Ž p 0 , q1 . s d Ž x, p 0 . q a Ž h 0 . y d q d Ž w, q1 . s d Ž x, p 0 . q a Ž h 0 . y d q d Ž w, p1 . q d Ž p1 , q1 . s d Ž x, p 0 . q a Ž h 0 . y d q d Ž x, p1 . q a Ž h1 . y 12 b Ž h1 . . On the other hand, d Ž xX1 , q1 . s d Ž x, q1 ? hy1 1 . F d Ž x, p1 . q d Ž p1 , q1 ? hy1 1 . s d Ž x, p1 . q d Ž q1 , r 1 . s d Ž x, p1 . q 12 b Ž h1 . F d Ž x, p 0 . q d Ž p 0 , p1 . q 12 b Ž h1 . F d Ž x, p 0 . q d q d Ž x, p1 . q 12 b Ž h1 . .
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We then have d Ž x 1 , q1 . y d Ž xX1 , q1 . G a Ž h 0 . q a Ž h1 . y b Ž h1 . y 2 d G Ž a Ž h 0 . y b Ž h 0 . . q Ž a Ž h1 . y b Ž h1 . . q l* G l*. . Ž . Since Ž h1 , hy1 1 h 0 h1 , h 2 , . . . , h n is a braiding of h, we again contradict 4.3 . Ž . This proves 4.6 . Ž4.7. We ha¨ e J 0 s w r 0 , y 0 x g P ŽŽ h 0 .., and J 0 is oriented away from z by h 0 . Moreo¨ er, l Ž J 0 . G 12 bŽ h 0 ., and this inequality is strict if J 0 : DŽ h 0 .. Proof. Since x 1 converges to z via Ž h1 , . . . , h n ., by Ž4.2., condition Ž C0 . in Ž3.10. must fail to hold, as otherwise x converges to z via h. Thus, l Ž J 0 . G 12 bŽ h 0 ., and this inequality is strict if I0 : DŽ h 0 .. Now, I0 j J 0 is a segment of AŽ h 0 . of length aŽ h 0 ., and it then follows from Ž3.7. that I0 : DŽ h 0 . if J 0 : DŽ h 0 .. By Ž3.15. we get J 0 g P ŽŽ h 0 ... Evidently J 0 s w r 0 , y 0 x, oriented toward r 0 by h 0 . This yields Ž4.7.. Ž4.8. Assume now that we are given k with 1 F k F n, such that for each i with 0 F i - k, we have a non-degenerate segment Ji in P ŽŽ h 0 , . . . , h i .. satisfying the following conditions: Ža. Ji s w u i , ¨ i x : w x iq1 , z x, with ¨ i g Ž u i , z x. Žb. One of the following holds if i ) 0: Ži. u iy1 g Ii and Ji s w ri , ¨ iy1 x. Žii. ¨ iy1 g Ii and Ji s w u iy1 ? h i , yi x. Žiii. Ii : Ž u iy1 , ¨ iy1 . and Ji s w u iy1 ? h i , ¨ iy1 x. Our goal will be to show that we can also define Jk g P ŽŽ h 0 , . . . , h k .. so as to satisfy Ž4.8.Ža. and Žb., with k in place of i. That is, we aim to show that Ž4.8. amounts to a well-defined inductive procedure for defining J1 , . . . , Jn , starting with the segment J 0 which we have already constructed. Ž4.9. We ha¨ e ¨ i g w x, z x for all i, 0 F i - k. Proof. First, ¨ 0 g w x, z x by Ž4.6.. By Ž4.8.Žb. we have ¨ i s ¨ iy1 , or else ¨ i s yi g w ¨ iy1 , z x. So, if ¨ iy1 g w x, z x then ¨ i g w x, z x, and we are done by induction. Ž4.10. We ha¨ e y k g Ž u ky1 , z x and qk g w u ky1 , z x. Moreo¨ er, if pk g DŽ h k . then qk / u ky1. Proof. Define a point sk by sk s
½
yk
if y k g w x k , u ky1 x
u ky 1
if y k g w u ky1 , z x .
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ANDREW CHERMAK
Of course, u ky 1 g w x k , z x by Ž4.8.Ža., so sk is well-defined. Now define s0 , . . . , sky1 by the formula sky i s skyiq1 ? hy1 kyi
Ž1 F i F k. . pX0 , . . . ,
In a similar manner we define points pXk s pk , pXky i s pkyiq1 ? hy1 kyi
pXk :
Ž1 F i F k. .
Put g s h 0 . . . h ky1 , hX0 s gh k gy1 , xX1 s x ? hX0 , and finally, set uy1 s p 0 . Ž . Thus uy1 s u 0 ? hy1 0 , since u 0 s r 0 by 4.6 . Assume that Ž4.10. is false. We will then show that the following condition holds. Ž4.10.1. w pXky i , skyi x is a subinterval of w x kyi , u kyiy1 x for all i, 0 F i F k. We will also show: Ž4.10.2. There exists a braiding of h which begins with hX0 . Suppose, for the moment, that both Ž4.10.1. and Ž4.10.2. hold. Notice that qk g w x k , y k x by Ž4.2.. Then, since Ž4.10. is false we get qk g w x k , sk x. Then the definition of qk yields dŽ x k , sk . q bŽ h k . G dŽ x kq1 , sk . q aŽ h k ., with strict inequality unless qk s sk . Applying gy1 , we then have d Ž x 0 , s0 . q b Ž hX0 . G d Ž xX1 , s0 . q a Ž hX0 . ,
Ž ).
again with strict inequality unless qk s sk . Since sk g w x k , x kq1 x, we have s0 g w x 0 , xX1 x. Also, taking i s k in Ž4.10.1. we see that s0 g w x 0 , p 0 x : w x 0 , z x, and it follows from Ž). that d Ž x 0 , z . q b Ž hX0 . G d Ž xX1 , z . q a Ž hX0 . .
Ž )) .
On the other hand, with Ž4.10.2., Ž4.4. yields dŽ x 0 , z . q bŽ hX0 . F dŽ xX1 , z . q aŽ hX0 ., with strict inequality if I0X : DŽ hX0 ., and where I0X s AŽ hX0 . l w x 0 , xX1 x l w x 0 , z x. Thus, equality holds in Ž))., and I0X DŽ hX0 ., and also
qk s s k . Next, put J 0X s AŽ hX0 . l w xX1 , z x, and note that dŽ x 0 , z . y dŽ xX1 , z . F l Ž I0X . y l Ž J 0X ., with equality if I0X is non-empty. Note also that l Ž J 0X . G 1 Ž X. Ž . Ž . 2 b h 0 , by 4.4 , and then )) yields l Ž I0X . G a Ž hX0 . y 12 b Ž hX0 . . Again, Ž4.4. applies and yields equality here. Since also equality holds in Ž)., we conclude that s0 s Y Ž x 0 , xX1 , z . g I0X . Also, d Ž pX0 , s0 . s d Ž pXk , sk . s d Ž pk , qk . s a Ž hX0 . y 12 b Ž hX0 . ,
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51
and so Ž4.10.1. yields I0X s w pX0 , s0 x. Then pX0 , s0 4 DŽ hX0 ., and so pk , qk 4 DŽ h k .. If now pk g DŽ h k . then L h k / B, and Ž3.8.Ža. yields qk g DŽ h k . for a contradiction. Thus, we conclude that pk f DŽ h k .. By Ž4.2. we then have qk / y k , and since sk s qk the definition of sk yields qk s u ky1 , so that Ž4.10. holds. It remains to prove Ž4.10.1. and Ž4.10.2.. Since Ž4.10. is assumed false, we have pk g w x k , u ky1 x, and then w pk , sk x is a subinterval of w x k , u ky1 x, by definition of sk . Thus, Ž4.10.1. holds for i s 0. Proceed by induction on i. Thus, suppose that w pXky iq1 , skyiq1 x is a subinterval of w x kyiq1 , u kyi x for y1 x w X x w some i, 0 F i - k. Apply hy1 ky i and obtain p kyi , s kyi F x kyi , u kyi ? h kyi . y1 Ž .Ž . If u ky iy1 s u kyi ? h kyi , we are done. So, referring to 4.8 b , we see that Jky i s w r kyi , ¨ kyiy1 x and u kyiy1 g Ikyi s w pkyi , y kyi x. Here u kyi s r kyi , so u ky i ? hy1 kyi s p kyi . That is, we have
w pXky i , skyi x F w x kyi , pkyi x F w x kyi , u kyiy1 x , which proves Ž4.10.1.. In order to prove Ž4.10.2., it will suffice to show that the H-sequence h9 given by
žŽ h
k
.
Ž h 0 , . . . , h ky 1 .y1
, Ž h ky1 .
Ž h 0 , . . . , h ky 2 .y1
, . . . , h 0 h1 hy1 0 , h 0 , h kq1 , . . . , h n
/
is a braiding of h. Indeed, one may ‘‘perform’’ this braiding as follows. First, move h 0 from the 0-position of h into the k-position, via k simple braidings. Then h 0 h1 hy1 is the new occupant of the 0-position, which we 0 now move into the Ž k y 1.-position via k-1 simple braidings. Continuing on in this way we obtain h9, proving Ž4.10.. Ž4.11. Suppose qk f Jky1. Then pk g Ž u ky1 , ¨ ky1 ., y k g w qk , r k ., and dŽ qk , y k . F dŽ pk , ¨ ky1 ., with strict inequality if y k g DŽ h k .. Proof. Set u s u ky 1 , ¨ s ¨ ky1 , J s Jky1 , xX1 s x ? h k , and put y1 h9 s Ž h k , hy1 k h 0 h k , . . . , h k h ky1 h k , h kq1 , . . . , h n . .
Notice that h9 is a braiding of h. It then follows from Ž4.4. that d Ž xX1 , z . q a Ž h k . G d Ž x, z . q b Ž h k . with strict inequality if I0X : DŽ h k ., where I0X s A Ž h k . l w x, xX1 x l w x, z x . We have qk g w u, z x by Ž4.10., and we are assuming that qk f J, so qk g Ž ¨ , z x. Suppose first that pk g w ¨ , z x. By Ž4.9. we then have d Ž x, z . s d Ž x, pk . q d Ž pk , z . s d Ž xX1 , r k . q d Ž pk , z . .
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Also, by the triangle inequality, d Ž xX1 , z . F d Ž xX1 , r k . q d Ž r k , z . , and this is strict if r k f w xX1 , z x. Then d Ž x, z . y d Ž xX1 , z . G d Ž pk , z . y d Ž r k , z . , and so a Ž h k . y b Ž h k . G d Ž pk , z . y d Ž r k , z . . s d Ž x k , z . y d Ž x kq1 , z . . By Ž4.2. we then have equality throughout, and so Ž4.4. yields I0X DŽ h k ., while r k g w xX1 , z x. Since also pk g w x, z x it follows that I0X s w pk , y k x s Ik , and thus Ik DŽ h k .. This contradicts condition Ž Ck . in Ž3.10., so we now conclude that pk f w ¨ , z x. But pk g w x k , z x, and J AŽ h k . by Ž3.13.Ž1.Ža., so we get pk g Ž u, ¨ .. Notice that then also h k g E ŽŽ h 0 , . . . , h ky1 ., J .. Suppose next that r k g w x k , z x. Then r k g w qk , z x F w ¨ , z x, so d Ž x, z . s d Ž x, ¨ . q d Ž ¨ , r k . q d Ž r k , z . . Also, we have dŽ pk , ¨ . s l Ž J l AŽ h k .. F 12 bŽ h k . y 12 l* by Ž3.13.Ž1.Žc.. Then d Ž ¨ , r k . G a Ž h k . y 12 b Ž h k . q 12 l*. On the other hand, we have d Ž xX1 , z . F d Ž xX1 , r k . q d Ž r k , z . F d Ž x, pk . q d Ž r k , z . F d Ž x, ¨ . q d Ž pk , ¨ . q d Ž r k , z . F d Ž x, ¨ . q d Ž r k , z . q 12 b Ž h k . y 12 l*. This yields d Ž x, z . y d Ž xX1 , z . G a Ž h k . y b Ž h k . q l* ) aŽ h k . y b Ž h k . which contradicts Ž4.4.. Hence r k f w x k , z x, and so y k g w qk , r k .. Put d s dŽ pk , ¨ ., d9 s dŽ qk , y k .. Then Ž4.9. gives d Ž x, z . G d Ž x, pk . q d Ž pk , z . y 2 d, s d Ž xX1 , r k . q d Ž pk , z . y 2 d, with a strict inequality unless J l w x, z x s ¨ 4 . That is, we have strict inequality unless I0X s w ¨ , y k x.
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Next, we have dŽ xX1 , z . F dŽ xX1 , r k . q dŽ r k , z ., so that d Ž x, z . y d Ž xX1 , z . G d Ž pk , z . y d Ž r k , z . y 2 d. We also have d Ž pk , z . s d Ž pk , qk . q d Ž y k , z . q d9 d Ž r k , z . s d Ž r k , qk . q d Ž y k , z . y d9 and so d Ž pk , z . y d Ž r k , z . s a Ž h k . y b Ž h k . q 2 d9. Thus, we now have d Ž x, z . y d Ž xX1 , z . G a Ž h k . y b Ž h k . q 2 Ž d9 y d . , with strict inequality unless I0X s w ¨ , y k x. On the other hand, Ž4.4. yields d Ž x, z . y d Ž xX1 , z . F a Ž h k . y b Ž h k . , with strict inequality if I0X : DŽ h k .. Thus we obtain d9 F d, and if d9 s d then ¨ , y k 4 DŽ h k .. Suppose finally that d9 s d and that y k g DŽ h k .. Then ¨ f DŽ h k .. But now dŽ ¨ , y k . s dŽ pk , qk . s aŽ h k . y 12 bŽ h k . g L h k, by Ž3.7.Žb. and Ž3.8.Ža., whence ¨ g DŽ h k .. This contradiction completes the proof of Ž4.11.. Ž4.12. There exist a non-degenerate segment J k s w u k , ¨ k x in P ŽŽ h 0 , . . . , h k .., with J : w x kq1 , z x, oriented so that ¨ k g Ž u k , z x, and defined as follows: Ži. Žii. Žiii.
If u ky 1 g Ik then Jk s w r k , ¨ ky1 x. If ¨ ky 1 g Ik then Jk s w u ky1 ? h k , y k x. If Ik : Ž u ky1 , ¨ ky1 . then Jk s w u ky1 ? h k , ¨ ky1 x.
Proof. Put J s Jky 1 , u s u ky1 , ¨ s ¨ ky1 , u s Ž h 0 , . . . , h ky1 ., and h s h k . We have u(Ž h. g H * as h g H *. By Ž4.10. and Ž4.11., J l aŽ h. is non-degenerate, so we have h g E Žu, J .. Now J AŽ h. by Ž3.13.Ž1.Ža., so precisely one of the definitions Ži., Žii., or Žiii. for Jk applies. That is to say, Jk is well-defined in Ž4.12.. We proceed case by case. Suppose u g Ik , so that Jk s w r k , ¨ x. By Ž4.10. we then have dŽ u, r k . G dŽ qk , r k ., with strict inequality if pk g DŽ h k .. Thus, the conditions Ži. through Žiii. in Ž3.13.Ž2.Žb. all hold, with pk in the role of y, and this yields w r k , ¨ x g P Žu(Ž h.., as desired. Evidently w r k , ¨ x F w x kq 1 , z x, and w r k , ¨ x is non-degenerate since J AŽ h.. Thus, Ž4.12. holds this case.
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Suppose next that Ik : Ž u, ¨ . so that Jk s w u ? h, ¨ x. Here we have l Ž Ik . - aŽ h. by Ž3.13.Ž1.Žb., so Ik s AŽ h. l w x k , z x. As J : w x k , z x by Ž4.8., we then have Ik s J l AŽ h., and so J l AŽ h. s B. By Ž4.2. we then have l Ž J l AŽ h.. G aŽ h. y 12 bŽ h., with restrict inequality if Ž J l AŽ h.. DŽ h.. Now Ž3.13.Ž2.Ža. yields w u ? h, ¨ x g P Žu(Ž h.., as desired. Further, w u ? h, ¨ x F w x k ? h, z x by Ž4.9., so Ž4.12. holds in this case. Finally, suppose that ¨ g Ik , so that Jk s w u ? h, y k x. Since dŽ pk , qk . s aŽ h. y 12 bŽ h. G 12 bŽ h., Ž3.13.Ž1.Žc. implies that qk f Jky1. Put y s y k . By Ž4.11. we have pk g Ž u, ¨ . and y g w qk , r k ., so we get ¨ g w y, y ? hy1 .. We have J o p p g P Žuy1 ., by Ž3.11.Žc.. Also, notice that h
uy1 ( Ž hy1 . s Ž uy1 . ( Ž hy1 .
ž
hy1
/
y1
and this is a braiding of ŽŽ hy1 .(uy1 . h , which is an H-conjugate of Žu(Ž h..y1 . Since u(Ž h. g H * it then follows that uy1 (Ž hy1 . g H *. Since J o p p l AŽ hy1 . s J l AŽ h. s w pk , ¨ x, which is non-degenerate, we then have hy1 g E Žuy1 , J o p p .. Further, Ž4.11. yields dŽ qk , y . F dŽ pk , ¨ ., so d Ž ¨ , y . F d Ž pk , ¨ . q d Ž ¨ , qk . s a Ž h . y 12 b Ž h . , and Žagain referring to Ž4.11.. the inequality is strict if y g DŽ h.. On the other hand, we have dŽ ¨ , y . s aŽ h. y dŽ ¨ , y ? hy1 . since ¨ g w y, y ? hy1 .. This yields dŽ ¨ , y ? hy1 . G 12 bŽ h., with strict inequality if y g DŽ h.. Now Ž3.13.Ž2.Žb. applies, with J o p p and uy1 in the roles of J and h, and the result is that w y ? hy1 , u x g P Žuy1 (Ž hy1 ... Then w u, y ? hy1 x g P ŽŽ h.(u. and w u ? h, y x g P ŽŽ h.(u h ., by Ž3.11.Žc.. But
Ž h . (u h s Ž h, h 0h , . . . , h hky1 . is a braiding of Ž h 0 , . . . , h ky1 , h., so we arrive at w u ? h, y x g P Žu(Ž h.., as desired. Since u g w x k , z x and y g w u, z x l w r kq1 , z x, we have w u ? h, y x F w x kq 1 , z x. Also, w u ? h, y x is non-degenerate, since w r k , y x F w u ? h, y x, where r k / y by Ž4.11.. This completes the proof of Ž4.12.. With Ž4.12. we have completed the induction step that began with Ž4.8.. We therefore conclude that there is a non-degenerate segment Jn g P Žh. with Jn : w x nq1 , z x. But this contradicts the plain fact that x nq1 is by definition equal to z. The proof of Theorem Ž3.14. is thereby complete.
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REFERENCES 1. R. Alperin and H. Bass, Length functions of group actions on L-trees, in ‘‘Combinatorial Group Theory and Topology, Alta, Utah, 1984,’’ Ann. of Math. Stud., Vol. 111, pp. 265]378, Princeton Univ. Press, Princeton, NJ, 1987. 2. A. Chermak, R-trees, small cancellation, and convergence, Trans. Amer. Math. Soc. 347, No. 11 Ž1995.. 3. A. Chermak, Locally non-spherical Artin groups, to appear in J. Alg.. 4. R. C. Lyndon and P. E. Schupp, ‘‘Combinatorial Group Theory,’’ Springer-Verlag, New YorkrBerlin, 1977. 5. J. W. Morgan, L-trees and their applications, Bull. Amer. Math. Soc. (N.S. ) 26, No. 1 Ž1992..