AAccessibility of partially acylindrical actions
Michael HillFebruary 22, 2021
Abstract
In [16] Weidmann shows that there a bound on the number of orbits of edgesin a tree on which a finitely generated group acts ( k, C )–acylindrically. In thispaper we extend this result to actions which are k –acylindrical except on a family ofgroups with “finite height”. We also give an example which gives a negative resultto a conjecture of Weidmann from the same paper and produce a sharp bound forgroups acting k –acylindrically. Given a group G one question we would like to answer is if there is some bound onthe size of a graph of groups decomposition for it. For example Grushko’s theorem [11]implies that a free decomposition has at most rank( G ) vertices with non-trivial label.Further examples include a result by Dunwoody [9] which gives a bound for the numberof edges of a (reduced) splitting over finite edge groups for a given finitely presentedgroup, and by Bestvina and Feighn [2] who extended this to small edge groups. (Agroup is small if it doesn’t act hyperbolically on any tree.) These results do not extendto finitely generated groups; for instance Dunwoody [8] gives an example of a finitelygenerated group which has splittings over finite edge groups with an arbitrary numberof edges.It’s also possible to obtain bounds by imposing restrictions other than restrictingthe class of edge groups. For example Sela [12] showed that there is a bound for thesize of a (minimal) splitting where the action on the corresponding Bass-Serre tree is k –acylindrical, assuming that G is freely indecomposable and finitely generated. Weidmann[17] later reproved this and gave a nice bound of at most 2 k (rank G −
1) edges for thesplitting. Later Delzant [7] showed that a bound for ( k, C )–acylindrical actions existsprovided that the acting group is finitely presented. Weidmann [16] then extended this tofinitely generated groups and conjectured that there should be a common generalisationbetween this and Bestvina and Feighn’s aforementioned result on actions with small edgestabilisers. More precisely they suggest that the currently known techniques should beenough to show that we can obtain a bound for a finitely presented group acting ona tree where large subgroups fix at most region of bounded diameter [16, pg.213]. (Agroup is large if it’s not small.)In this paper we extend Weidmann’s result on ( k, C )–acylindrical actions to actionswhich are k –acylindrical except on a family of groups with “finite height”. Roughlyspeaking this means that if there is a family of subgroups P with a bound on the length1 a r X i v : . [ m a t h . G R ] F e b f chains of subgroups and the diameter of the fixator of any group not in P is boundedby k , then we obtain a bound on the number of edges. (A more precise statement canbe found in Section 1.) In Section 2 we recall the definition and basic properties ofStallings folds [14]. In Section 3 we construct examples which show that Weidmann’saforementioned conjecture from [16, pg.213] is false. In Section 4 we introduce the notionof a forest of influence and show that it interacts nicely with Stallings folds. We thenapply this machinery in Sections 5 and 6 to prove the main positive results. Finally inSection 7 we improve Weidmann’s bound from [17] for actions which are k –acylindricalto have at most 2 k (cid:0) rank G − (cid:1) edges (for non-cyclic groups) and show that this boundis sharp. We begin by recalling the natural correspondence between graph of groups decomposi-tions of G and trees on which G acts [13]. If we take the “universal cover” of graph ofgroups we obtain a tree on which the fundamental group of the graph of groups acts.We call this tree the Bass-Serre tree . Likewise we can quotient an (orientation preserv-ing) action of G on a tree to obtain a graph of groups whose fundamental group is G .Throughout we will implicitly use this correspondence.We now give a series of basic definitions. Definition 1.1
Let G be a group and k be a non-negative integer. Let Q be a classof subgroups of G which is closed under conjugation. An action of G on a tree T is (partially) k –acylindrical on Q if whenever some H ∈ Q fixes every edge in a reducededge path p then p contains at most k edges.If Q isn’t specified then it’s assumed to contain all the non-trivial subgroups of G andwe say the action is k –acylindrical . An action is ( k, C )–acylindrical if it’s k –acylindricalon the subgroups of G with size strictly greater than C . Definition 1.2
The rank of a finitely generated group G is the minimal size of a gen-erating set for G . Denote this quantity by rank G . There a couple of trivial ways in which we can add arbitrarily many orbits of edgesto a tree. The first is to add additional “hanging” edges which tell us nothing aboutthe structure of the group; for example the splitting G ∼ = G ∗ H H where H (cid:54) G . Thefollowing definition prevents this. Definition 1.3
The action of a group on a tree is said to be minimal if there are noproper subtrees which are invariant under the action.
Another thing that we need to prevent is the possibility of arbitrary subdivision ofedges. The notion of a tree being reduced stops this behaviour; however a tree being k –acylindrical also stops this happening on edges whose stabiliser contains a member of Q . As such we can introduce the following less restrictive notion.2 efinition 1.4 Suppose P is a class of subgroups of G which is closed under conjuga-tion. A minimal action is said to be partially-reduced over P if either • T /G is a circle consisting of a single vertex and edge; or • whenever a vertex v of T has stabiliser equal to that of an edge of T and which iscontained in a subgroup of a member of P then v/G has valence at least in T /G .We say an action is reduced if it’s partially-reduced over the class of all subgroups of G . We also say an action is C –partially-reduced if it’s partially-reduced over the classof all subgroups of size at most C . A critical element of the proof will be the idea of measuring how “large” a givengroup is relative to P . Definition 1.5
Let P be a conjugation invariant set of subgroups of G . For a subgroup K (cid:54) G suppose there is some maximal integer n such that there are H , · · · , H n ∈ P with K (cid:54) H < H < · · · < H n We define the P –weight of K to be n and we denote this quantity by W P ,K . (We saythis is equal to ∞ if chains of arbitrary length exist.) We say that K (cid:54) G is larger than P if it’s not a subgroup of a member of P ; equivalently if W P ,K = 1 .If G acts on T then we define the P –weight of each edge to be the P –weight of itsstabiliser. We say that P has height M if the maximal weight of any K (cid:54) G is M ;equivalently if W P , = 2 M . Remark 1.6
Since we insist that P is conjugation invariant we see that the P -weightof a subgroup is conjugation invariant. As such we define the P -weight of a conjugacyclass of subgroups to be the P -weight of any representative of that class. We now state an easier version of our main results. We will prove this before movingon to the full theorems as it will demonstrate the important ideas of the argumentwithout being obscured by as many technical details.
Theorem 1.7
Let G be a finitely presented group and let P be a set of subgroups for G with height M and which is closed under conjugation and taking subgroups. Suppose G acts on a tree T and that this action is both k –acylindrical on groups larger than P andpartially reduced on groups in P . Then there is some C ( G ) (which depends only on G )such the that number of edges of T /G is bounded above by (2 k + 1)2 M C ( G ) . Our main results are two different generalisations of the above. In the first we extendthe result to certain cases where P isn’t closed under taking subgroups, which is necessaryfor including infinite subgroups in P . In the second we extend the result to groups whichare merely finitely generated instead of just those which are finitely presented. In orderto state the former we first need to make the following definitions.3 efinition 1.8 Let P be a set of subgroups for G . Let K be a subgroup of G . Wesay H ∈ P is a minimal extension of K (to P ) if K (cid:54) H and whenever ˜ H ∈ P with K (cid:54) ˜ H (cid:54) H then ˜ H = H . We say that K is P –closed if every minimal extension ofany subgroup of K is contained in K . We say an action of G on a tree T is P –closed if all its edge stabilisers are P –closed. Remark 1.9 If P has finite height then minimal extensions always exist for any groupwhich isn’t larger than P . Definition 1.10
We say that P satisfies condition ( † ) if the following conditions hold. • P has finite height. • Suppose G acts on a tree T and let e be an edge of T . Then for any subgroup K (cid:54) Stab e which is not larger than P there is a vertex v of T which is fixed bysome minimal extension of K to P . In particular this always holds if each minimalextension is of finite index as a finite index extension of an elliptically acting groupalso fixes a point [13]. • Every member of P is P -closed. Equivalently if H and H are in P then so is H ∩ H . A third equivalent condition is that minimal extensions to P are unique. Example 1.11
Suppose that G is a torsion-free hyperbolic group. Take P to be the setof cyclic subgroups of G which are root-closed. Equivalently P is the set of maximalcyclic subgroups of G . This P satisfies ( † ) since every cyclic subgroup is contained (withfinite index) in a unique maximal cyclic subgroup. Moreover a tree which G acts on is P –closed iff all its edge stabilisers are root-closed. We are now ready to state our main result, which as mentioned before is a pair ofextensions of Theorem 1 . Theorem 1.12
Let G be a finitely generated group and let P be a set of subgroups for G which is closed under conjugation and has height M . Suppose G acts on a tree T andthat this action is both k –acylindrical on groups larger than P and partially reduced onsubgroups of members of P . Then the following statements hold.(a) Suppose G is finitely presented, P satisfies ( † ) and T is P –closed. Then thereis some integer C ( G ) such that the number of edges of T /G is bounded above by (2 k + 1)2 M C ( G ) .(b) If P is closed under taking subgroups then the number of edges of T /G is boundedabove by (cid:0) k +12 (cid:1) M (rank G − . Moreover suppose either of the following condi-tions hold • T is reduced and k > ; or • every edge stabiliser of T is not in P ; hen if G isn’t cyclic the number of edges of T /G is bounded above by M k (rank G − . Remark 1.13
In Theorem . and Theorem . (a) the bound C ( G ) is given by Dun-woody’s resolution lemma (Theorem . ). Dunwoody’s resolution lemma holds for socalled almost finitely presented groups and this extends to both Theorem . and Theo-rem . (a). (A group is almost finitely presented if it’s both finitely generated and actsfreely, simplicially and cocompactly on a simplicial complex X with H ( X, Z ) = 0 .) The following is an immediate consequence of Theorem 1 .
12 (b) and is an extensionof Weidmann’s result on ( k, C )–acylindrical actions [16]. In particular this shows thatthe number of prime factors is the limiting factor, not the absolute size of the group.
Corollary 1.14
Let G be a finitely generated group and M ∈ N . Suppose G acts on atree T and that this action is both k –acylindrical on groups which are infinite or haveat least M prime factors and partially reduced on subgroups with at most M − primefactors. (Where the number of prime factors is counted with multiplicity.) Then thenumber of edges of T /G is bounded above by (cid:0) k +12 (cid:1) M (rank G − . Moreover if either T is reduced and k > or every edge stabiliser of T is either infinite or has at least M prime factors then the number of edges of T /G is bounded above by M k (rank G − . Proof
Let P be the set of finite subgroups of G whose order has at most M − P has height of at most M and is closed under taking subgroups.The result now immediately follows from Theorem 1 .
12 (b). (cid:3)
We also apply Theorem 1 .
12 (a) to a couple of specific cases to get some interestingresults. The first case is a generalisation of Example 1 .
11; where a torsion-free hyperbolicgroup acts on a tree with root-closed edge stabilisers. We now allow the group is to havefinite order elements and the root closed condition is replaced by one which says that themaximal virtually Z subgroups of edge stabilisers should be “almost” maximal in G . Inthe second we consider splittings of RAAGs which are k –acylindrical on its non-abeliansubgroups and the maximal abelian subgroups of edge stabilisers should be maximal fortheir rank. Corollary 1.15
Let G be a hyperbolic group. We say a virtually Z subgroup H (cid:54) G is m -almost maximal if whenever we have a virtually cyclic K (cid:54) G with H (cid:54) K then [ H : K ] ≤ m . Let P m be the collection of subgroups of G which are either finite or m -almost maximal.Suppose G acts on a tree T which is partially reduced on P m and k –acylindrical ongroups larger than P m . Suppose also that T is P m -closed. Then the number of edges of T /G is bounded above by (2 k + 1) C (cid:48) ( G ) . (Where C (cid:48) ( G ) = 2 n C ( G ) for some n ∈ N .) Proof
Observe that P m is not closed under intersections and so doesn’t satisfy condition( † ). Instead define P (cid:48) m to be the set of subgroups which are a (finite) intersection ofgroups in P m . It’s clear that if P (cid:48) m has finite height then it satisfies ( † ). As a hyperbolicgroup has only finitely many conjugacy classes of finite subgroups [4] we just need to5how that chains of infinite subgroups in P (cid:48) m have bounded length. It therefore sufficesto show that given any index in N there is a uniform bound on the number of subgroupsof that index for any virtually cyclic subgroup of G .We have that every virtually cyclic H (cid:54) G is either of the form H ∼ = K ∗ K where K is finite or H ∼ = A ∗ C B where A, B, C are finite and C is an index 2 subgroup ofboth A and B [10]. The general description of a subgroup of the fundamental group of agraph of groups [13] now tells us there’s a bound on the number of subgroups of H of agiven index which depends only on the index and the size of either K or C respectively.Again a hyperbolic group has finitely many conjugacy classes of finite subgroups, whichuniformly bounds the order of K and C . This implies the result. (cid:3) Before stating the other application we briefly recall the definition of a RAAG.
Definition 1.16
Let Γ be a finite graph. Let v , · · · , v n be the vertices of Γ . The right-angled Artin group (RAAG) associated to Γ is the group A (Γ) where A (Γ) := (cid:104) v , · · · , v n | [ v i , v j ] wherever there’s an edge between v i and v j in Γ . (cid:105) Corollary 1.17
Let G = A (Γ) be a RAAG. An abelian subgroup H (cid:54) G is said to be rank maximal if whenever we have an abelian subgroup K (cid:54) G which contains H withfinite index we have K = H . Let P be the collection of rank maximal abelian subgroupsof G . Suppose G acts on a tree T which is partially reduced on abelian subgroups and k –acylindrical on non-abelian subgroups. Suppose also that T is P -closed. Then thenumber of edges of T /G is bounded above by (2 k + 1)2 n C ( G ) (where n is the size of thelargest complete subgraph of Γ ). Proof
Recall that a RAAG acts freely and cocompactly on a simply connected CAT(0)-cube complex X Γ whose dimension is equal to the size of the largest complete subgraph ofΓ. The Flat Torus Theorem [6, Theorem II.7.1] says that an abelian subgroup H (cid:54) A (Γ)must act properly and cocompactly by isometries on a Euclidean hyperplane of X Γ . Inparticular H must have rank at most equal to the size of the largest complete subgraphof Γ. Hence P has finite height.Now every rank 2 subgroup (cid:104) u, v (cid:105) (cid:54) A (Γ) is either free abelian or free [1, Theo-rem 1.2]. Suppose H (cid:54) A (Γ) is an abelian subgroup, u is a root of an element of H andpick any v ∈ H . Since a power of u is in H and H is abelian we see that (cid:104) u, v (cid:105) cannotbe non-abelian free and so u and v must commute. Hence every member of P is rootclosed.It remains to check that P satisfies condition ( † ), then we can apply Theorem 1 . H ∈ P and K (cid:54) H and let M ∈ P be a minimal extensionof K . We must have M (cid:54) H as H is root closed and K is a finite index subgroup of M .Hence P is P -closed and hence satisfies ( † ). (cid:3) Remark 1.18
For a general group G it need not be the case that P as defined in Corol-lary . satisfies ( † ) . For example if G ∼ = Z ∗ Z Z then G acts freely and cocompactlyon a CAT(0) space; but contains two rank maximal copies of Z whose intersection isanother copy of Z which is not rank maximal. .
12 (b) also immediately implies Weidmann’s earlier result on k –acylindricalactions [17]; which says that a finitely generated group acting k -acylindrically on a treewithout edges with trivial stabiliser has at most 2(rank G − k orbits of edges. Indeedwith slightly more work we’ll show it’s possible to improve their bound sightly furtherto one which we’ll show is the best possible. Theorem 1.19
Let G be a (non-cyclic) finitely generated group acting k –acylindricallyon a minimal tree T (where k ≥ .) Suppose that each edge of T has non-trivial stabiliser.Then T /G has at most (cid:4)(cid:0) G − (cid:1) k (cid:5) edges. If G is torsion-free then this bound canbe improved to (2 rank G − k . Theorem 1.20
For any k > and r ≥ there is a finitely presented group G with rank G = r which acts k –acylindrically on a minimal tree T where each edge of T hasnon-trivial stabiliser and T /G has exactly (cid:4)(cid:0) G − (cid:1) k (cid:5) edges.Similarly F r admits a k –acylindrical action on a minimal tree T where each edge of T has non-trivial stabiliser and T /F r has exactly (2 r − k edges. Remark 1.21
Unlike in the previous results there is no requirement that T needs tobe reduced. Instead the conditions that T is k –acylindrical and has no edges with triv-ial stabiliser are enough to completely prevent the unrestricted edge subdivision whichmotivated the definition of a reduced action. The idea of a fold will be of vital importance. Recall the following.
Definition 2.1
Let G act on a tree T . Let e , e be distinct edges with a commonendpoint x and let φ : e → e be the linear map which leaves x fixed. Let ∼ be theminimal equivalence relation on T such that x ∼ φ ( x ) for each x ∈ e and such that T / ∼ is a naturally a tree on which G acts. A fold is the map T → T / ∼ . Folds were introduced by Stallings in [14]. In particular they showed that mapsbetween trees with finitely generated edge stabilisers can be decomposed into a finitesequence of folds. We need something similar for trees whose edge groups need not befinitely generated. Fortunately we can “add in” generators of each edge group one at atime, then take a limit to see that our maps are a composition of (potentially infinitelymany) folds. This is formalised into the following theorem, the proof of which is largelythe same as the one given in [2, p.455] with changes to deal with the fact that the edgestabilisers aren’t necessarily finitely generated.
Theorem 2.2
Let G be a countable group. Suppose Ψ : S → T is a surjective simplicalequivariant map between trees which G acts on with S/G finite and where no edge of S gets mapped to a point by Ψ . Then Ψ can be viewed as a (possibly infinite) compositionof folds. i.e. Ψ = · · · α α where each α i is an (orientation preserving) fold. emark 2.3 The codomain of · · · α α is S/ ∼ where ∼ is the equivalence relationgenerated by all the α i . This is a tree which G acts on in the obvious way with vertexand edge stabilisers equal to the natural direct limit of their preimages. Remark 2.4
The condition that no edge of T gets collapsed is not a restrictive onein practice. In particular if Ψ maps an edge of S to a point then let π be the mapwhich collapses each edge of S which is sent to a point by Ψ . Then there is a naturalcomposition Ψ = Ψ (cid:48) ◦ π where no edge in the domain of Ψ (cid:48) is sent to a point in T . Proof
Throughout we’ll let α i fold the tree S i − into the tree S i . Also we let β i := α i ◦ · · · ◦ α and γ i be the map such that Ψ = γ i ◦ β i .First suppose that we have a surjective simplical map m : A → B between finitetrees where no edge gets mapped to a point. Claim that m can be considered to be afinite series of folds; a fact we shall refer to as ( (cid:63) ). If m is injective then this is trivial.Otherwise we have distinct vertices x, y ∈ A with m ( x ) = m ( y ). Let p be the reducededge path from x to y . Since every edge of A is mapped to an edge in B and B is atree we see that there must be a vertex z ∈ p such that m | p is not locally injective at z . Let e and e be the edges in p which contain z as an endpoint and observe that m ( e ) = m ( e ). Thus m factors though the fold with edges e and e . Repeat thisprocess until the map is injective, which must happen as the number of edges is finiteand decreasing at each stage. This completes the decomposition of m into folds.Now suppose we have an equivarient simplical map δ : R → R (cid:48) . Let A be a finitesubtree of R . We can apply ( (cid:63) ) to δ | A to obtain a finite series of folds δ (cid:48) : R → R (cid:48)(cid:48) whichfactors through δ and where the corresponding δ (cid:48)(cid:48) : R (cid:48)(cid:48) → R (cid:48) is injective on δ (cid:48) ( A ). Thisis how we will apply ( (cid:63) ) in practice.Our initial folds of Ψ will be to set S i /G isomorphic as a graph to T /G for allsufficiently large i . Let F be the closure of a fundamental domain of S = S . We nowuse ( (cid:63) ) to find a series of folds α , · · · , α N such that γ | β ( F ) is a homeomorphism onto itsimage. Thus γ N /G is a homeomorphism of graphs.Let K = Ψ( F ). Let v be a vertex in K and g ∈ Stab( v ). Let v i be a preimage of v in F i and let p i be the reduced edge path from v i to gv i . Now we apply ( (cid:63) ) to p i inorder to get folds α i , · · · α j which get g in the relevant vertex group in S j /G . Now repeatthis process for each vertex v ∈ K and g ∈ Stab( v ). This potentially gives an infinitesequence of folds as K has finitely many vertices and each Stab( v ) is countable.Now let ˜ γ be the map such that Ψ = ˜ γ ◦ ( · · · ◦ α ◦ α ). Claim that ˜ γ is a homeom-porphism. Indeed by construction we see that ˜ γ induces a bijection between the orbitsof vertices; moreover the stabiliser of each vertex is the same as that of its image. Hence˜ γ induces a bijection between the vertices and hence is a homeomorphism between trees.Thus · · · ◦ α ◦ α is a decomposition of Ψ into folds. (cid:3) Remark 2.5
It should be straightforward to extend this result to the case where G isuncountable and where S/G is not necessarily finite using the well ordering principle.We do not do this here because it is unnecessary to prove our main results.
8n general a simplicial map is too restrictive of a notion. As such we now introducethe idea of a combinatorial map.
Definition 2.6 A combinatorial map Ψ : S → T is a G –equivariant map where eachvertex gets sent to a vertex and each edge e = [ u, v ] gets sent to the reduced edge pathfrom Ψ( u ) to Ψ( v ) . Observe that a combinatorial map can be viewed as a simplical map after subdividingedges in the domain. As such Theorem 2 . Definition 2.7
Suppose that
Ψ : T → T (cid:48) is a combinatorial map. We say that anothercombinatorial map α : R → R (cid:48) factors through Ψ if there is some β : T → R and γ : R (cid:48) → T (cid:48) where Ψ = γ ◦ α ◦ β . Depending which of the vertices and edges are in common G –orbits there are a fewdifferent cases that can arise from a fold. The following classification of folds is the sameas the one found in [2].First we make the distinction between whether x is in the same G orbit as one ofthe y i . If x is not in the same G –orbit as either y i we say that the fold is of type A.Otherwise WLOG we have gx = y for some g ∈ G and we say that the fold is of type B.Note that such a g must act hyperbolicily on T , (with translation length 1,) as it movesa vertex an odd distance.Additionally we split each of these cases into three additional categories. We say thefold is of type I if y and y are in distinct orbits of G . We say the fold is of type II if e and e are in a common orbit of G . Finally we say the fold is of type III if y and y are in a common G –orbit, but e and e are not. We will now go into the specifics ofeach type of fold. Throughout we let e i be the vertex between the vertices x and y i anduse capital letters to denote the group associated to the corresponding vertex or edge. Remark 2.8
The following diagrams represent what happens to the relevant subgraphof a particular graph of groups decomposition. Crucially the pictures for type I and IIIfolds only give the correct groups if both e and e are in the fundamental domain forthis decomposition. In general we need to conjugate certain groups in the decompositionbefore these pictures become accurate. Type I
We have y and y in distinct orbits of G . In this case the number of verticesand the number of edges of the graph of groups decomposition both decrease by one sothe Euler characteristic of the underlying graph stays the same. Type II
We have e and e in a common orbit of G , suppose that he = e . Observethat if h acts hyperbolicily on T then the action of G after the fold is not orientationpreserving and so we will ignore this case. Thus we can assume that h ∈ X . In this casethe underlying graph of the graph of groups decomposition doesn’t change. Instead theelement h gets “pulled” along the edge in the graph of groups decomposition.9ype IA Y X Y E E (cid:104) E , E (cid:105) X (cid:104) Y , Y (cid:105) Type IB Y E Xg E (cid:104) E , E (cid:105)(cid:104) X, g − Y g (cid:105) g Figure 1: A typical example of the effects of a type I fold on a graph of groups. Thevertices y and y are inequivalent so the fold reduces the number of vertices by 1.Likewise for the edges e and e . Type IIA (cid:104) E , h (cid:105) X (cid:104) Y , h (cid:105) E X Y h ∈ X Type IIB E Xg (cid:104) E , h (cid:105)(cid:104) X, g − hg (cid:105) gh ∈ X Figure 2: A typical example of the effects of a type II fold on a graph of groups. Thevertices y and y are equivalent so the fold keeps the number of vertices the same.Likewise for the edges e and e . Type III
We have y and y in a common orbit, but e and e are not. Suppose that hy = y . Observe that h has to act hyperbolicily on T with translation length 2. Afterthe fold this h now fixes the image of y and y thus no longer acts hyperbolicily. Thistype of fold reduces the number of edges of the graph of groups by one while keeping thenumber of vertices fixed. Thus the Euler characteristic of the underlying graph increasesby one. In [2] Bestvina and Feighn showed that a reduced tree with small edge stabilisers hasa bound on the number of edges as long as the underlying group is (almost) finitelypresented. Recall the following.
Definition 3.1
A group G is large if it acts on a tree T hyperbolicly. That is to say X Y E E (cid:104) E , E (cid:105) X (cid:104) Y , h (cid:105) h Type IIIB E Xg E (cid:104) E , E (cid:105)(cid:104) X, h (cid:105) ghg − Figure 3: A typical example of the effects of a type III fold on a graph of groups. Thevertices y and y are equivalent so the fold keeps the number of vertices the same.However the edges e and e are inequivalent so the fold reduces the number of edgesby 1. that there are two elements g , g ∈ G which don’t fix a point of T and their axes (linesof minimal displacement) have bounded intersection [13]. A group is small if it’s notlarge. Note that the ping-pong lemma implies that any large group must contain F asa subgroup. The converse is not true; for example SL ( Z ) contains many subgroupsisomorphic to F and has Serre’s property (FA) [13], so any tree it acts on has fixedpoint.Later Weidmann [16] showed that the action of a finitely generated group acting( k, C )–acylindrically on a C –partially reduced tree also has a bound on the number ofedges depending only on the rank of the group and k . In the same paper Weidmannthen goes on to conjecture that some sort of common generalisation between their resultand the aforementioned result of Bestvina and Feighn might exist. More precisely theysuggest it should be possible to give a positive answer to the following using knowntechniques. Question [16, pg.213]
Given a finitely presented group G and k > C ( G, k ) such that any reduced action of G which is k –acylindrical on large subgroupshas at most C ( G, k ) orbits of edges?The purpose of this section is to construct an example which shows that the answer tothe above question is no. In fact we will construct a counterexample with even strongerproperties.
Theorem 3.2
There is a finitely presented group G which for any N > acts on areduced tree which is –acylindrical on infinite subgroups and has N orbits of edges. Proof
Let D := (cid:10) a , a , · · · | a = 1 , a i +1 = a i ∀ i ≥ (cid:11) ∼ = Z (cid:2) (cid:3) / Z ; the additive groupof dyadic rationals modulo Z . Let A be any finitely presented group into which D embeds; for example we can take A to be Thompson’s group T [3]. Let B := (cid:104) b (cid:105) ∼ = Z .11ake G := A ∗ B and pick any N >
0. Start by taking the one edge splitting correspondingto G ∼ = A ∗ B and subdividing this edge into N subedges. (In the diagrams we take N = 4.) A B N folds of type II. The first “pulls” a N across the first edge, the second“pulls” a N − across the second edge and so on, so that the i th fold “pulls” a N +1 − i acrossthe i th edge. A (cid:104) a (cid:105) (cid:104) a (cid:105) (cid:104) a (cid:105) (cid:104) a , b (cid:105)(cid:104) a (cid:105)(cid:104) a (cid:105)(cid:104) a (cid:105)(cid:104) a (cid:105) Let b := b and for 1 ≤ i ≤ N we define b i := b i − a i b i − . We now apply N − b across the first edge, the second “pulls” b across the second edge and so on, so that the i th fold “pulls” b i across the i th edge. A (cid:104) a , b (cid:105) (cid:104) a , b (cid:105) (cid:104) a , b (cid:105) (cid:104) a , b (cid:105)(cid:104) a , b (cid:105)(cid:104) a , b (cid:105)(cid:104) a , b (cid:105)(cid:104) a (cid:105) It’s clear that this is a reduced decomposition. It remains to show that the action onthe corresponding Bass-Serre tree is 1–acylindrical on infinite subgroups. In other wordsit suffices to show that the stabilisers of any two distinct edges with a common end vertexare finite. Observe that a generic vertex of this decomposition has label (cid:10) a (cid:48) , b (cid:48) | a (cid:48) r (cid:11) ∼ = Z ∗ ( Z / r Z ) with two edges with labels (cid:10) a (cid:48) , b (cid:48) (cid:11) and (cid:10) a (cid:48) , b (cid:48) a (cid:48) b (cid:48) (cid:11) respectively. (cid:104) a (cid:48) , b (cid:48) | a (cid:48) r (cid:105) (cid:104) a (cid:48) , b (cid:48) (cid:105)(cid:104) a (cid:48) , b (cid:48) a (cid:48) b (cid:48) (cid:105) So there are three different pairs of edges we need to consider.
Case 1
The intersection of (cid:10) a (cid:48) , b (cid:48) (cid:11) and (cid:10) a (cid:48) , b (cid:48) (cid:11) g for g ∈ (cid:104) a (cid:48) , b (cid:48) (cid:105) \ (cid:10) a (cid:48) , b (cid:48) (cid:11) .Let ˜ w = α α · · · α n and ˜ w h be cyclicly reduced words in (cid:8) a (cid:48)± , b (cid:48)± (cid:9) for some h ∈ (cid:104) a (cid:48) , b (cid:48) (cid:105) . Observe by cycling letters in (cid:8) a (cid:48)± , b (cid:48)± (cid:9) that either h ∈ (cid:10) a (cid:48) , b (cid:48) (cid:11) or α i = a (cid:48)± for all i and h is an (odd) power of a (cid:48) . It follows that every element in the intersectionis conjugate to a power of a (cid:48) . Moreover we see that w ∈ (cid:10) a (cid:48) , b (cid:48) (cid:11) ∩ (cid:10) a (cid:48) , b (cid:48) (cid:11) g if and onlyif there are g , g ∈ (cid:10) a (cid:48) , b (cid:48) (cid:11) and n ∈ Z such that g = g a (cid:48) g and w = ( a (cid:48) n ) g .If we can show that g can only be expressed in the form g a (cid:48) g in an “essentiallyunique” way then it follows that the intersection is cyclic and hence finite as a (cid:48) hasfinite order. More precisely it suffices to show that whenever g a (cid:48) g = g (cid:48) a (cid:48) g (cid:48) (where g , g (cid:48) , g , g (cid:48) ∈ (cid:10) a (cid:48) , b (cid:48) (cid:11) ) then a (cid:48) g = a (cid:48) g (cid:48) . By the rigidity of reduced words in (cid:10) a (cid:48) , b (cid:48) | a (cid:48) r (cid:11) observe that this equality only happens if g (cid:48) = g a (cid:48)− r and g (cid:48) = a (cid:48) r g for some r ∈ Z .Thus a (cid:48) g (cid:48) = a (cid:48) ( a (cid:48) r g ) = (cid:16) a (cid:48) a (cid:48) r (cid:17) g = a (cid:48) g As required. 12 ase 2
The intersection of (cid:10) a (cid:48) , b (cid:48) a (cid:48) b (cid:48) (cid:11) and (cid:10) a (cid:48) , b (cid:48) a (cid:48) b (cid:48) (cid:11) g for g ∈ (cid:104) a (cid:48) , b (cid:48) (cid:105) \ (cid:10) a (cid:48) , b (cid:48) a (cid:48) b (cid:48) (cid:11) .Let ˜ w = α α · · · α n and ˜ w h be cyclicly reduced words in (cid:8) a (cid:48)± , ( b (cid:48) a (cid:48) b (cid:48) ) ± (cid:9) for some h ∈ (cid:104) a (cid:48) , b (cid:48) (cid:105) . Observe by cycling letters in (cid:8) a (cid:48)± , b (cid:48)± (cid:9) that we must have h ∈ (cid:10) a (cid:48) , b (cid:48) a (cid:48) b (cid:48) (cid:11) .It follows that every element in the intersection trivial unless g ∈ (cid:10) a (cid:48) , b (cid:48) a (cid:48) b (cid:48) (cid:11) . Case 3
The intersection of (cid:10) a (cid:48) , b (cid:48) (cid:11) and (cid:10) a (cid:48) , b (cid:48) a (cid:48) b (cid:48) (cid:11) g for g ∈ (cid:104) a (cid:48) , b (cid:48) (cid:105) .Let ˜ w = α α · · · α n be a cyclicly reduced word in (cid:8) a (cid:48)± , b (cid:48)± (cid:9) and let ˜ w = β β · · · β m be a cyclicly reduced word in (cid:8) a (cid:48)± , ( b (cid:48) a (cid:48) b (cid:48) ) ± (cid:9) . Suppose that ˜ w and ˜ w conjugate toeach other. Observe that ( b (cid:48) a (cid:48) b (cid:48) ) ± can’t be a subword of any cyclic permutation (in (cid:8) a (cid:48)± , b (cid:48)± (cid:9) ) of ˜ w and so β j (cid:54) = ( b (cid:48) a (cid:48) b (cid:48) ) ± for any j . Hence ˜ w and ˜ w are (even) powersof a (cid:48) . Essentially the same argument also shows that (cid:10) a (cid:48) , b (cid:48) (cid:11) ∩ (cid:10) a (cid:48) , b (cid:48) a (cid:48) b (cid:48) (cid:11) = (cid:10) a (cid:48) (cid:11) .Now the above says that if w ∈ (cid:10) a (cid:48) , b (cid:48) (cid:11) ∩ (cid:10) a (cid:48) , b (cid:48) a (cid:48) b (cid:48) (cid:11) g then we must have w = ( a (cid:48) n ) h for some h ∈ (cid:104) a (cid:48) , b (cid:48) (cid:105) . Now arguments from case 1 imply that either h or a (cid:48) h must bein (cid:10) a (cid:48) , b (cid:48) (cid:11) as w ∈ (cid:10) a (cid:48) , b (cid:48) (cid:11) . Likewise arguments from case 2 implies that k − := hg − ∈ (cid:10) a (cid:48) , b (cid:48) a (cid:48) b (cid:48) (cid:11) as w g − ∈ (cid:10) a (cid:48) , b (cid:48) (cid:11) . So g = kh and WLOG we have h ∈ (cid:10) a (cid:48) , b (cid:48) (cid:11) ; as if a (cid:48) h ∈ (cid:10) a (cid:48) , b (cid:48) (cid:11) we can just replace h with a (cid:48)− h and k with ka (cid:48) .If we can show that g can only be expressed in the form kh in an “essentially unique”way then it follows that the intersection is cyclic and hence finite as a (cid:48) has finite order.More precisely we wish to show that if kh = k (cid:48) h (cid:48) (where k, k (cid:48) ∈ (cid:10) a (cid:48) , b (cid:48) a (cid:48) b (cid:48) (cid:11) and h, h (cid:48) ∈ (cid:10) a (cid:48) , b (cid:48) (cid:11) ) then a (cid:48) h = a (cid:48) h (cid:48) . Since (cid:10) a (cid:48) , b (cid:48) (cid:11) ∩ (cid:10) a (cid:48) , b (cid:48) a (cid:48) b (cid:48) (cid:11) = (cid:10) a (cid:48) (cid:11) and by the rigidity ofreduced words in (cid:10) a (cid:48) , b (cid:48) | a (cid:48) r (cid:11) we see that this only happens if k (cid:48) = ka (cid:48)− r and h (cid:48) = a (cid:48) r h for some r ∈ Z . Once again we get a (cid:48) h = a (cid:48) h (cid:48) in the same way as in case 1. (cid:3) Note that a hyperbolic group G cannot satisfy Theorem 3 .
2. This is because thereare only finitely many conjugacy classes of finite subgroups of a hyperbolic group [5];thus there is some bound on the order of finite subgroups. We can then apply the boundfor ( k, C )–acylindrical actions to get a bound here. One may then wonder if Weidmann’sconjecture holds for hyperbolic groups; however a slight tweak to our example shows thatthis isn’t true either, even for free groups.
Theorem 3.3
For any
N > there is an action of F on a reduced tree which is –acylindrical on non-cyclic subgroups and has N orbits of edges. Proof
The construction is mostly the same as Theorem 3 . A := (cid:104) a (cid:105) ∼ = Z so that G = (cid:104) a, b (cid:105) ∼ = F . Pick any N > a i = a i . We now define the tree and see that it satisfies the necessaryconditions in the same way as before. (cid:3) In both of these constructions we exploit chains of subgroups with arbitrary length.More precisely we have the chain of subgroups (cid:104) a (cid:105) > (cid:104) a (cid:105) > · · · > (cid:104) a N (cid:105) and build thetree in such a way that each group in this chain fixes a vertex which isn’t fixed by any ofthe larger ones. Forcing P to have finite height and insisting the tree is P -closed ensuresthat we cannot use these long chains to make arbitrarily complicated decompositions inthe same way. 13 Forests of Influence
We begin by stating Dunwoody’s resolution lemma.
Theorem 4.1 ([9])
Suppose that G is an (almost) finitely presented group. Then thereis some C ( G ) ∈ N with the following properties. Whenever G acts on a minimal tree T there is some minimal tree T (cid:48) with at most C ( G ) orbits of edges and a combinatorialmap Ψ : T (cid:48) → T where no edge gets mapped to a point. We’ll now give an extremely rough outline of the core ideas of the argument. Supposethat G acts on a minimal tree T which is k –acylindrical on groups larger than P . UseDunwoody’s resolution lemma to obtain a tree T (cid:48) which has a bound on the number ofedges and a map Ψ : T (cid:48) → T . If some edge of T (cid:48) (before subdividing) has a stabiliserlarger than P then its image in T cannot have more than k edges because of the acylin-drical condition. Thus we can collapse this edge in T (cid:48) and only collapse at most k edgesof T .Now subdivide T (cid:48) to make Ψ simplicial, but note that the initial vertices are ‘moreimportant’ in the sense that every vertex stabiliser is contained in one of these. So wecan build a collection of disjoint subtrees for T (cid:48) by starting with this set of initial verticesand then iteratively expanding to include vertices whose stabiliser is contained in thestabiliser of the corresponding initial vertex.Now we subdivide Ψ into folds using Stallings’ folding theorem (Theorem 2 .
2) andapply the first fold. If every vertex stabilizer is still contained in a stabilizer for oneof the initial vertices then we have still have a collection of subtrees with the sameproperties as before. Otherwise some vertex stabiliser isn’t contained in one of theinitial ones. This only happens if two of our subtrees gets folded together in some waywhich is unavoidable. We then add this vertex to our set of “initial” ones and thenrebuild our collection of subtrees with the same properties as before. However we willsee that the intersections of the stabilisers between one of the original initial verticesand this “new initial vertex” is strictly larger than the intersection of the original initialvertices. (See Figure 4 for an example or Lemma 4 .
12 for a more precise statement.) If P has finite height this means that this can only happen boundedly often before one ofthese intersections is larger than P and so can collapse down a path of length at most k . So either we can keep doing this until we are left with a single point or we get a setof “initial” vertices for T . In the latter case if T is P –partially reduced we can find abound for the number of edges using our set “initial” vertices. (See Lemma 5 . Definition 4.2
Suppose G acts on a tree T . We call a subset of vertices S a set of seedvertices if it’s G –invariant and for every vertex v (with non-trivial stabiliser) there issome u ∈ S with Stab v (cid:54) Stab u . In particular if the action on T is free we also allowthe empty set to be a set of seed vertices, otherwise S is necessarily non-empty. Definition 4.3
Suppose G acts on a tree T . A G –invariant subgraph Γ ⊆ T is a forestof influence if the following conditions hold. X X EX X E E (cid:104) E , E (cid:105) Figure 4: After applying a series of type II folds to an edge (with subdivisions) there maya vertex whose stabiliser isn’t contained in the stabiliser of either of the initial vertices.If this happens we see that the intersection of stabilisers between this vertex and eitherof the initial vertices must contain the original edge group as a proper subgroup. • Γ deformation retracts to a non-empty set of seed vertices S , equivalently everycomponent of Γ contains exactly one member of S . We say that Γ is grown from S . • If vertices u and v are in the same connected component of Γ with u ∈ S then Stab v (cid:54) Stab u . We call such a component the tree of influence of u , say that v is influenced by u and call the reduced edge path from v to u the branch of v . • Every vertex of T is contained in Γ . Remark 4.4
The branch of any vertex v is stabilised by Stab v . As such the first edgeon the branch of v must have the same stabiliser as v as any edge cannot be fixed bymore than either of its endpoints. Definition 4.5
Suppose G acts on a tree T and that Γ ⊆ T is a forest of influence.We call the edges of T \ Γ the connecting edges of Γ . The connecting groups are theconjugacy classes of (a set of representatives for) the connecting edges, counted withmultiplicity. In general there is not a distinguished choice for a forest of influence. However thefollowing proposition says there is something canonical lurking underneath. This willallow us to move between different choices with minimal difficulties.
Proposition 4.6
Suppose that T has finitely many orbits of vertices. Suppose also that Γ and Γ are forests of influence which are both grown from the same set of seed vertices S . Then Γ and Γ have the same connecting groups. In other words the connectinggroups are determined by S . Before proving this we’ll first we’ll define an elementary transformation of a forestof influence. Take a forest of influence Γ and pick a vertex v ∈ Γ \ S . Suppose that v is contained in the tree of influence of u and let e be the first edge on the branchof v . Observe that Stab e = Stab v and pick some connecting edge e with endpoint v and with Stab e = Stab v . We now define Γ (cid:48) := (Γ \ G { e } ) ∪ G { e } . In otherwords we replace the orbit of e in Γ with the orbit of e in Γ (cid:48) . (See Figure 5.) SinceStab e = Stab e = Stab v we see that Γ (cid:48) is also a forest of influence grown from S andthat both Γ and Γ (cid:48) have the same connecting groups.Proposition 4 . ve e u Γ u ve e u Γ (cid:48) Figure 5: An example of an elementary transformation. The edge e is removed andreplaced with e . For Γ (cid:48) to be a forest of influence we must have Stab e = Stab e =Stab v . Lemma 4.7
Suppose that T has finitely many orbits of vertices and that Γ and Γ areforests of influence which are both grown from the same set of seed vertices S . Then wecan apply a finite series of elementary transformations to Γ to obtain Γ . Proof
Let d (Γ , Γ ) be the number of (orbits of) vertices which are in trees of influenceof different seed vertices in Γ and Γ . If d (Γ , Γ ) = 0 then Γ = Γ and there is nothingto show.If d (Γ , Γ ) > v which is in the tree of influence of u in Γ and of u (cid:54) = u in Γ . Let e (cid:48) be the final edge in the branch of v (in Γ ) which is not containedin Γ and so is a connecting edge of Γ . Let v (cid:48) be the endpoint of e (cid:48) which is not in thetree of influence of u in Γ . Observe that v (cid:48) is in the tree of influence of u in Γ as v is. Suppose that v (cid:48) is in the tree of influence of u (cid:48) in Γ and let e (cid:48) be the first edge onthe branch of v (cid:48) (in Γ ). Since Stab v (cid:48) = Stab e (cid:48) = Stab e (cid:48) we can apply an elementarytransformation to Γ by removing the orbit of e (cid:48) and adding the orbit of e (cid:48) to get Γ (cid:48) .(See Figure 6.) v u u v (cid:48) u (cid:48) e e Γ e (cid:48) e (cid:48) Γ Figure 6: An example of a situation where d (Γ , Γ ) >
0. By replacing e (cid:48) with e (cid:48) in Γ we can make it “more similar” to Γ . This idea of how similar two forests of influenceare is formalised by the metric d . 16f we can show that d (Γ (cid:48) , Γ ) < d (Γ , Γ ) then we are done by induction. Observethat v (cid:48) is in the tree of influence of u in Γ (cid:48) and Γ but not in Γ . Thus we just need toshow that any vertex which is influenced by the same seed vertex in Γ and Γ is alsoinfluenced by the same one in Γ (cid:48) . This holds because the only vertices whose influencingvertex changed under the elementary transformation were those in (the orbit of) the treeof influence of u (cid:48) in Γ at and beyond v (cid:48) . These can’t be influenced by u (cid:48) in Γ as v (cid:48) isinfluenced by u in Γ and so the tree of influence of u (cid:48) in Γ cannot contain them. (cid:3) Recall the definition of the P –weight of a subgroup K (cid:54) G from Definition 1 . W P ,K ≤ M where M is the length of the longest chain of groups in P which contain K .From this definition we note that the following properties are all obvious. Proposition 4.8
Let P be a conjugation invariant set of subgroups of G .(a) If P has height M then W P ,K ≤ M for any K (cid:54) G .(b) K (cid:54) G has P –weight if and only if it’s larger than P .(c) If H ∈ P and H < K ≤ G then W P ,H ≤ W P ,K . We will now extend our definition of P –weight to sets of seed vertices. Proposition 4 . Definition 4.9 If G acts on a tree T and S is a non-empty set of seed vertices for T then we define its P –weight W P ,S to be the sum of the P –weights of the correspondingconnecting groups (and ∞ if any of the connecting groups have infinite P –weight). If S is empty then we instead define W P ,S := ( β ( T /G ) − W P , . Remark 4.10
The case of a free action is special because the stabiliser of each vertexis trivial. As there are no “interesting” stabilisers we aren’t really missing anything byjust forgoing seed vertices entirely. If the action is free and S is non empty then we seethat W P ,S := ( β ( T /G ) + | S/G | − W P , . This justifies the definition of W P ,S for empty S by setting | S/G | = 0 in the aboveequation. On a more practical level we allow the empty set to be a set of seed verticesfor a free action to prevent an otherwise guaranteed drop in P –weight if a fold causes afree action to become non-free. (See Lemma . .) With this in hand we are ready to state the key lemma. From this Theorem 1 . Lemma 4.11
Suppose G is a non-cyclic countable group. Let P be a conjugation in-variant set of subgroups of G which is closed under taking subgroups. Let G act on atree T where this action is both P –partially-reduced and k –acylindrical on a subgroupslarger than P . Let G act on another tree T (cid:48) and suppose that there is a G -equivarientcombinatorial map Ψ : T (cid:48) → T . Suppose also that T (cid:48) has a set of seed vertices S withfinite P –weight W P ,S . Then T /G has at most (cid:0) k +12 (cid:1) W P ,S edges. Lemma 4.12
Suppose that α : R → ˜ R . Suppose that S is a non-empty set of seedvertices for R where all of the connecting groups are in P . Then there is a set of seedvertices ˜ S for ˜ R with α ( S ) ⊆ ˜ S and W P , ˜ S ≤ W P ,S . Moreover if W P , ˜ S = W P ,S then α | S is injective. Proof
Suppose that α folds together the edges e = [ x, y ] and e = [ x, y ]. Suppose α ( e ) = α ( e ) = e (cid:48) and α ( y ) = α ( y ) = y (cid:48) . Let y i be in the tree of influence of u i and if y i (cid:54) = u i we also let f i be the first edge in the branch of y i . Throughout we will assumethat y i (cid:54) = u i and so f i exists. The cases where y i = u i turn out to be essentially thesame except the lack of f i sometimes causes W P , ˜ S to be smaller. We will split into casesdepending on if there is a forest of influence containing e and/or e . Case 1
There is a forest of influence Γ containing both e and e . y y e e u x u x y (cid:48) e (cid:48) α f f x y (cid:48) u e (cid:48) αy u y xe e f f Figure 7: The two pictures that can arise when Γ contains both e and e . In eithercase Stab u contains Stab x , Stab y and Stab y and hence also contains Stab y (cid:48) .The fold cannot be of type III as otherwise y and y = hy would be in the sametree of influence. So α (Γ) is a forest of influence for ˜ R which is grown from α ( S ). Theconnecting edges of Γ are untouched by α and so S and α ( S ) have the same connectinggroups. 18 ase 2 There is no forest of influence containing either e or e . αy u y xe e u f f u u y (cid:48) x e (cid:48) f f Figure 8: The picture that arises when Γ cannot possibly contain either e or e . Observethat Stab e i < Stab u i . After applying the fold we remove the edges f i (if they exist)from the forest of influence. Separate arguments depending on the type of fold are nowrequired to show that this doesn’t increase the P -weight.Pick any forest of influence Γ. Observe that ˜ S = α ( S ) ∪ G { e (cid:48) } is a set of seed verticeswhich grows into α (Γ) \ G { f , f } . Since e i is not contained in a forest of influence wemust have e i < f i as otherwise we could apply an elementary transformation to get itinto one. So W f i , P ≤ W e i , P by Proposition 4 . α is a fold of type I then e , e , f and f are pairwise inequivalent. Moreover e (cid:48) con-tains the image of both e and e , so W e (cid:48) , P ≤ max( W e , P , W e , P ) by Proposition 4 . W S, P − W ˜ S, P = W e , P + W e , P − W f , P − W f , P − W e (cid:48) , P ≥ W e , P + 12 W e , P − max( W e , P , W e , P ) ≥ α is a fold of type II then e is equivalent to e and f is equivalent to f . Addi-tionally Stab e (cid:48) > Stab e . So by Proposition 4 . W S, P − W ˜ S, P = W e , P − W f , P − W e (cid:48) , P ≥ W e , P − W e , P − W e , P = 0Now assume α is a fold of type III. We see that f and f are equivalent, while e and e are inequivalent. Thus W S, P − W ˜ S, P = W e , P + W e , P − W f , P − W e (cid:48) , P ≥ W e , P − W f , P ≥ W e , P − W e , P > ase 3 There is a forest of influence Γ containing e but not e ; also there isn’t onewhich contains both of them.Note that α cannot be a fold of type II (as then the e i are equivalent) or type IIIB (soboth e i are always connecting edges). We will split into four subcases; corresponding tocombinations whether or not e is equal to f and whether or not Stab y is a subgroupof Stab y . Case 3ai
We have e = f and Stab y (cid:54) Stab y . y y e e u x u x y (cid:48) e (cid:48) αu f f f f u Figure 9: The picture that arises when Γ contains e but not e , the branch of y contains x and Stab y contains Stab y . Here we take the image of each connecting edge to stillbe a connecting edge.If the fold is of type I then observe that ˜ S := α ( S ) is a set of seed vertices for ˜ R .Observe that the image of the the connecting edges of Γ are the connecting edges of aforest of influence grown from ˜ S . If instead the fold is of type IIIA then ˜ S := α ( S ) ∪ G { y (cid:48) } is a set of seed vertices. In this case we have a forest of influence ˜Γ := α (Γ) \ G { f } .The connecting edges of ˜Γ are the image of the connecting edges of Γ with the orbit of e removed and the orbit of α ( f ) added. Observe that Stab e (cid:54) Stab f . Hence the P –weight can’t increase in either case. Case 3aii
We have e = f and Stab y is not contained in Stab y .If the fold if type IIIA then proceed as in case 3ai. Otherwise observe ˜ S = α ( S ) ∪ G { y (cid:48) } is a set of seed vertices for ˜ R and that ˜Γ = α (Γ) \ G { e (cid:48) , f } is a forest of influencegrown from ˜ S . Since Stab y is not contained in Stab y and Stab y = Stab e we haveStab e < Stab e (cid:48) . We also have Stab e < Stab f because otherwise we could apply anelementary transformation to Γ to get a new forest of influence which is in case 1. Henceby Proposition 4 . W S, P − W ˜ S, P = W e , P − W f , P − W e (cid:48) , P ≥ W e , P − W e , P − W e , P = 0 20 y e e u x u x y (cid:48) e (cid:48) αu f f f f u Figure 10: The picture that arises when Γ contains e but not e , the branch of y contains x and Stab y doesn’t contain Stab y . Here we take the new connecting edgesto be the image of the old connecting edges together with f (if it exists). Case 3bi
We have e (cid:54) = f and Stab y (cid:54) Stab y . αy u y xe e u f f u u y (cid:48) x e (cid:48) f f Figure 11: The picture that arises when Γ contains e but not e , the branch of y doesn’t contain x and Stab y contains Stab y . Here we take the new forest of influenceto be the image of the old one with f removed (if it exists).If the fold is of type I then ˜ S := α ( S ) is a set of seed vertices for ˜ R . If instead thefold is of type IIIA then ˜ S := α ( S ) ∪ G { y (cid:48) } is a set of seed vertices instead. In either caseobserve that α (Γ) \ G { f } is a forest of influence grown from ˜ S . (Note that if the foldis of type IIIA then f and f are in a common orbit, so f also becomes a connectingedge.) Since Stab e (cid:54) Stab x (cid:54) Stab f we have W ˜ S,C ≤ W S,C . Case 3bii
We have e (cid:54) = f and Stab y is not contained in Stab y .If the fold is of type IIIA then proceed as in case 3bi. Otherwise observe that˜ S = α ( S ) ∪ G { y (cid:48) } is a set of seed vertices which grows into a forest of influence ˜Γ = α (Γ) \ G { f , f } . If Stab e = Stab f then we could apply an elementary transformationto get both e and e in the same forest of influence and so we are in case 1; henceStab f < Stab e . Also since Stab y is not contained in Stab y and Stab x (cid:54) Stab y y u y xe e u f f u u y (cid:48) x e (cid:48) f f Figure 12: The picture that arises when Γ contains e but not e , the branch of y doesn’t contain x and Stab y doesn’t contain Stab y . Here we take the new forest ofinfluence to be the image of the old one with both f and f removed (if they exist).we have Stab e = Stab y ∩ Stab y < Stab y = Stab f . Hence by Proposition 4 . W S, P − W ˜ S, P = W e , P − W f , P − W f , P ≥ W e , P − W e , P − W e , P = 0 (cid:3) Recall that a free action is a special case as we allow the set of seed vertices to beempty. Thus we must deal with this case separately.
Lemma 4.13
Suppose that α : R → ˜ R is a fold and S is a set of seed vertices for R .Suppose also that P has finite height. If G acts freely on R but not on ˜ R then there is aset of seed vertices ˜ S for ˜ R such that W P , ˜ S ≤ W P ,S . Proof
Recall from Definition 1 . W P ,S ≥ ( β ( R/G ) − W P , . Asthe action on ˜ R is non-free and α is a fold it follows that α is a fold of type III. Moreoverall the edge stabilisers of ˜ R are trivial and there is a single vertex u (up to equivalence)with a non-trivial stabiliser. We define ˜ S := G { u } . Since all the connecting groups of˜ S are trivial we get that W P , ˜ S = β ( ˜ R/G ) W P , . Since α is a fold of type III we have β ( ˜ R/G ) = β ( R/G ) − (cid:3) Now suppose that we have a map Ψ : T → T (cid:48) where T has a set of seed vertices S and the action on T (cid:48) is k -acylindrical on groups larger than P . If a connecting groupin S is larger than P then there are seed vertices, say u and u , whose images in T (cid:48) are separated by distance at most k . Since we have control of the length of the22ath between these images we wish to collapse it to avoid unnecessary extra counting.However in general the image of the path between u and u need not lie in the pathbetween Ψ( u ) and Ψ( u ). For our core argument to work we require this containmentand the following says we can do this with some extra folds. Lemma 4.14
Let
Ψ : T → T (cid:48) be a simplical map where the action on T (cid:48) is k -acylindricalon groups larger than P . Let S be a set of seed vertices for T where at least one of theconnecting groups are larger than P . Then there are T , T (cid:48) and a simplical Ψ : T → T (cid:48) such that the action on T (cid:48) is k -acylindrical on groups larger than P , there’s a set of seedvertices S for T with W P ,S ≤ W P ,S (cid:48) − and T (cid:48) has at most k more edges than T (cid:48) . Proof
Now suppose that S has a connecting edge e of P –weight 1 (in some forest ofinfluence Γ). Suppose this edge connects the trees of influence of u and u . Let γ be the reduced edge path between u and u . Recall statement ( (cid:63) ) from the proof ofStallings folding theorem (Theorem 2 .
2) and apply it to γ . We get a composition offolds ρ : T → ˜ T so that the induced map ˜Ψ : ˜ T → T (cid:48) is locally injective on ρ ( γ ). Theintersection of the stabilisers for u and u is larger than P since e has P –weight 1. Sosince the action on T is k –acylindrical on groups larger than P the distance between the ρ ( u i ) is at most k . Hence we can collapse at most k edges of T (cid:48) to get a new tree T (cid:48) andan induced simplicial map Ψ : T → T (cid:48) . Observe that T has a set of seed vertices S , theimage of S , with W P ,S ≤ W P ,S − e is a set of connecting edges for S . It remains to bound the number of edges of a k –acylindrical action on a tree given a setof seed vertices. Lemma 5.1
Let P be a class of subgroups for a group G which is closed under conju-gation. Suppose G acts on a tree T and that this action is both partially-reduced on P and k –acylindrical on groups larger than P . Suppose that S is a non-empty set of seedvertices for T with n orbits of connecting edges. Then T /G has at most (2 k + 1) n edges.Furthermore if T is reduced and k > then T /G has at most kn edges. Proof
First observe that we can assume that each connecting group is a subgroup ofa group in P . Indeed suppose that there are r > P . For each of the corresponding connecting edges we see that path consisting of ittogether with the branches of both its endpoints must be fixed by the connecting group,which is larger than P . Since the action is k –acylindrical on groups larger than P eachof these paths have length at most k . Thus we can collapse these paths to get a new treewith at most kr fewer edges and a set of seed vertices with r fewer connecting groups.Let F ⊂ T be the forest consisting of S together with every edge and vertex whosestabiliser is larger than P . Since all of the connecting groups are contained in a memberof P we see that F must deformation retract to S . Let R ⊆ T be a maximal subtree23here every edge stabiliser is contained in a member of P . Let A = R ∩ F , the verticeswith stabiliser larger than P and seed vertices which are in R . We define ˜ R as the unionof R and the branches of each v ∈ A . F R A S
Figure 13: An example of what this construction may look like in the graph of groups.Here we see that p := | A | = 3, χ ( R/G ) = 0 and that R contains three orbits of connectingedges; the bottom right one and any two of the other three in R/G . Note the leftmostmember of A has valence 1 in R/G and isn’t in S , so if the tree is reduced F must extendbeyond it. Because of this the length of the branch of this vertex is actually boundedabove by k − k .Let | A/G | = p and suppose R contains q connecting edges (up to equivalence) of someforest of influence Γ grown from S . Now the branch of each vertex of R must containa vertex in A . Moreover this is unique as two distinct members of A are influenced bydifferent seed vertices and so must lie in different components of Γ. Therefore R ∩ Γdeformation retracts to A and so χ ( R/G ) = p − q .Recall from Remark 4 . R/G with valence 1 or 2 (in
R/G ) must be in A as T is partially reduced on P . Suppose that R/G has n i vertices of valence i and ob-serve that p ≥ n + n . Observe that χ ( R/G ) = ( n − n − n − · · · ) and hence (cid:80) i ≥ n i ≤ n − χ ( R/G ). So
R/G ) =
R/G ) − χ ( R/G ) ≤ n + n − χ ( R/G )Now the length of the branch of each v ∈ A is at most k as the action is k –acylindricalon P . (If v ∈ A \ S then Stab v is lager than P and fixes its branch.) Hence we see that R/G ) ≤ R/G ) + kp ≤ n + n + kp − χ ( R/G ) ≤ ( k + 2) p − χ ( R/G )We now split into cases depending on the value of χ ( R/G ). First suppose that χ ( R/G ) ≤
24. Then q ≥ p and so R/G ) ≤ ( k + 2) p − χ ( R/G )= ( k − p + 3( p − χ ( R/G )) ≤ ( k − q + 3 q = ( k + 2) q Otherwise χ ( R/G ) = 1 and we have q = p − R/G ) ≤ ( k + 2) p − χ ( R/G ) ≤ ( k + 2)( q + 1) −
3= ( k + 2) q + ( k − ≤ ( k + 2) q + ( k − q = (2 k + 1) q Let ˜ T be the tree obtained by collapsing each edge of G ˜ R and let π : T → ˜ T . Observethat ˜ S := π ( S ) is a set of seed vertices for ˜ T and that the number of connecting edgesof ˜ S is n − q . Hence by induction ˜ T /G has at most (2 k + 1)( n − q ) edges. Combiningthis with the above we see that T /G has at most (2 k + 1) n edges as required.It remains to show the improved bound if T is reduced and k >
1. In this case any v ∈ A \ S where v/G has valence 1 in R/G must be the endpoint of at least 2 edges notcontained in
R/G . This means that the path from v to the corresponding u ∈ S actuallyhas length at most k −
1. Hence in this case
R/G ) ≤ R/G ) + k ( p − n ) + ( k − n ≤ ( k + 1) n + n + k ( p − n ) − χ ( R/G ) ≤ ( k + 1) p − χ ( R/G )The rest of the calculations are essentially the same as before and so are omitted forthe sake of brevity. We will note however that we only actually obtain the improvedbound if k >
1. (In the case where χ ( ˜ R/G ) = 1 we need ( k − ≤ ( k − q . Since q ≥ k ≥ (cid:3) We now have all the pieces we need to prove Lemma 4 .
11 and hence Theorem 1 . Proof of Lemma .
11 We will proceed by induction on W P ,S (cid:48) . If W P ,S (cid:48) = 0 then since G isn’t isomorphic to Z there is a single seed vertex in T (cid:48) , which must be fixed by G . Sothe image of this vertex in T is fixed by G and so as T is minimal it must just consistof a single vertex.So WLOG W P ,S (cid:48) >
0. Start by using Stallings folding theorem (Theorem 2 .
2) todecompose α into folds α i : T i − → T i and let S := S (cid:48) . Recursively for each i > S i − is defined and has connecting edges contained in P we obtain a set of seed vertices S i for T i at each step using either Lemma 4 .
12 or Lemma 4 .
13 (depending on if the action25n T i − is free). If the P –weight at any step decreases then we are done by inductionon W P ,S (cid:48) .If instead S i − has a connecting edge which is larger than P we apply Lemma 4 . T i − → T where the action on T is k -acylindrical on groups larger than P . There’s a set of seed vertices S i − for T i − with W P ,S i − ≤ W P ,S (cid:48) − T has atmost k more edges than T . Hence by induction on W P ,S (cid:48) we see that T /G has at most( W P ,S (cid:48) − k edges, hence T /G has at most W P ,S (cid:48) k edges as desired.So WLOG we can assume that S i is always defined and that both the P –weight is con-stant and that we never have a connecting edge of P –weight 1. Recall that Lemma 4 . α i ( S i − ) ⊆ S i and since the number of connecting edges is bounded above by W P ,S (cid:48) we must have S i = α i ( S i − ) for all sufficiently large i . Thus by taking limitswe see that there is a set of seed vertices S for T with W P ,S = W P ,S (cid:48) . Now Lemma 5 . T /G is bounded above by (cid:0) k +12 (cid:1) W P ,S (cid:48) . (Since eachconnecting edge has weight of at least 2.) (cid:3) Proof of Theorem . .
1) toget G acting on a tree T (cid:48) which has at most α ( G ) orbits of edges together with acombinatorial map Ψ : T (cid:48) → T . Let S be the set of vertices of T (cid:48) before subdividing.Observe that S is a set of seed vertices for T (cid:48) and that it has P –weight of at most2 M C ( G ) since P has height M . Hence by Lemma 4 .
11 we see that T has as most(2 k + 1)2 M − C ( G ) edges. (cid:3) Now that we have finished proving our simplified result it’s time to extend it to get ourmain theorems. The first way we’re going to do this is to show that we don’t require P to be closed under taking subgroups; although it still must satisfy condition ( † ). (SeeSection 1 for the statement of ( † ).) The following is the analogue to Lemma 4 .
11 in thiscontext.
Lemma 6.1
Let G be a non-cyclic group and let P be a conjugation invariant set ofsubgroups of G which satisfies ( † ) . Let G act on a tree T and suppose this action is P –partially-reduced and k –acylindrical on a subgroups larger than P . Let G act on anothertree T (cid:48) and there is a G –equivarient combinatorial map Ψ : T (cid:48) → T . Suppose that T (cid:48) has a set of seed vertices S with finite P –weight W P ,S and T is P –closed. Then T /G has at most (cid:0) k +12 (cid:1) W P ,S edges. The added difficulty is that Lemma 4 .
12 requires every connecting group to be in P . Previously this was not an issue as every subgroup of G was either in P or largerthan it. We will solve this problem by adding extra folds at each step which forces theconnecting groups to be in P . Lemma 6.2
Suppose
Ψ : T (cid:48) → T and β : T (cid:48) → R are G -equivarient combinatorial mapswhere β factors through Ψ . Let S be a set of seed vertices for R with finite P –weight W P ,S nd where none of the connecting groups are larger than P . Suppose that P satisfies ( † ) and T is P –closed. Then there is a combinatorial map ρ : R → R (cid:48) which factors through Ψ such that R (cid:48) has a set of seed vertices S (cid:48) := ρ ( S ) such that W P ,S (cid:48) ≤ W P ,S and all itsconnecting groups are in P . Proof
Let Γ be any forest of influence which is grown from S . If each connecting groupof S is in P then we are done; so WLOG there is some connecting edge e of Γ which isnot in P . Since P satisfies ( † ) we have H ∈ P which is a minimal extension of Stab e to P and acts elliptically on R . Suppose H fixes the vertex v in R . Let p be the reducededge path which starts at v and has final edge e . Let ˜ p be the union of p together withthe branch of each vertex on p . (See Figure 14.) Since T is P –closed and the stabiliserof each edge f in ˜ p contains Stab e we see that image of f in T must be stabilised by H . Let ρ be the (possibly infinite) composition of type II folds which “pulls” H ontoeach edge of ˜ p and observe that this factors through Ψ since T is P –closed. Henceif e (cid:48) is a connecting edge of Γ with stabiliser in P then either Stab e (cid:48) = Stab ρ ( e (cid:48) ) or W P ,e (cid:48) < W P ,e . Moreover ρ (Γ) is a forest of influence grown from the seed vertices ρ ( S )with W P ,S (cid:48) ≤ W P ,S . Hence we can apply this process finitely many times until we getthe result. (cid:3) e S v Figure 14: An example of the domain we need to apply extra folds to. The reduced pathfrom v to e is p while the entire diagram is ˜ p . We need to apply folds to get H fixingthis whole region and not just p as otherwise we would need additional seed vertices. Proof of Lemma . W P ,S . If W P ,S = 0then there is some vertex of T (cid:48) which is fixed by G . So the image of this vertex in T isfixed by G and so as T is minimal it must just consist of just this single vertex.Use Stallings folding theorem (Theorem 2 .
2) to decompose α into folds α (0) i : T (0) i − → T (0) i .We will iteratively define trees T ( j ) i (for i ≥ j −
1) and sets of seed vertices S ( j ) i for T ( j ) i (for i = j and i = j + 1) together with maps α ( j ) i : T ( j ) i − → T ( j ) i and ρ ( j ) i : T ( j ) i → T ( j +1) i
27s follows for each j > T (0)0 T (0)1 T (0)2 · · · T (1)0 T (1)1 T (1)2 · · · T (2)1 T (2)2 · · · ... . . . α (0)1 ρ (0)0 α (0)2 ρ (0)1 α (0)3 ρ (0)2 α (1)1 α (1)2 ρ (1)1 α (1)3 ρ (1)2 α (2)2 α (2)3 ρ (2)2 First use define β j := α ( j − j − ◦ ρ ( j − j − ◦ · · · ◦ ρ (1)1 ◦ α (1)1 ◦ ρ (0)0 . (Part of the red path along thebottom of the diagram which ends with a right facing arrow.) Applying Lemma 6 . β j we obtain a map ρ ( j ) j : T ( j − j − → T ( j − j − where S ( j ) j − := ρ ( j ) ( S ( j − j − ) is a set of seed verticeswhere the connecting groups are in P and W P ,S ( j ) j − ≤ W P ,S ( j − j − . If α ( j ) i folds togetherthe edges e and e we define α ( j +1) i (for i > j ) to be the fold (or identity) obtained byidentifying the ρ ( j ) i +1 images of the e i . Finally define ρ ( j +1) i (for i > j + 1) to make theabove diagram commute. Finally Lemma 4 .
12 says that the fold α ( j ) j induces a set ofseed vertices S ( j ) j on T ( j ) j with W P ,S ( j ) j ≤ W P ,S ( j ) j − . (If a separating edge is larger than P then we can reduce the P –weight by collapsing at most k edges using Lemma 4 .
14 andthen proceeding by induction on W P ,S .)As α = · · · α (0)2 ◦ α (0)1 we see that α = · · · α (2)2 ◦ ρ (1) ◦ α (1)1 ◦ ρ (0) by the definitions of α ( j ) i and ρ ( j ) . Moreover at each step we have a set of seed vertices with non-increasing P –weight, the number of orbits of connecting edges are non-decreasing and hence thatall but finitely many of the sets of seed vertices are the image of the seed vertices at theprevious level. At this point the proof is exactly the same as the proof of Lemma 4 . (cid:3) Proof of Theorem . (a) First we use Dunwoody’s resolution lemma (Theorem 4 . G acting on a tree T (cid:48) which has at most α ( G ) orbits of edges together with acombinatorial map Ψ : T (cid:48) → T . Let S be the set of vertices of T (cid:48) before subdividing.Observe that S is a set of seed vertices for T (cid:48) and that it has P –weight of at most2 M C ( G ) since P has height M . Hence by Lemma 4 .
11 we see that T has as most(2 k − M − C ( G ) edges. (cid:3) It remains to extend Theorem 1 . . X for G and28onsider the free group F ( X ) acting freely on a tree T (cid:48) . Whenever G acts on a tree T wesee that there is an F ( X )–equivarient combinatorial map T (cid:48) → T . It’s this map whichwe intend to decompose into folds and apply our prior methods to.Before stating the analogue to Lemma 4 .
11 we first need to extend the definition of P –weights. Definition 6.3
Let φ : H → G be a surjective homomorphism of groups. Let P be aset of subgroups for G which is closed under conjugation. We define the P –weight of a K ≤ H , (denoted W P ,φ,K or W P ,K if φ is understood,) to be equal to W P ,φ ( K ) . If H actson a tree with a set of seed vertices S then we define its P –weight W P ,φ,K (or W P ,K if φ is understood) to be equal to the sum of the P –weights of the connecting groups. Lemma 6.4
Let P be a conjugation invariant set of subgroups of G which is closedunder taking subgroups. Let G act on a tree T and suppose this action is P –partially-reduced and k –acylindrical on a subgroups larger than P . Suppose also that G (cid:48) is acountable group acting on a tree T (cid:48) and that the following conditions hold. • There is a surjective homomorphism φ : G (cid:48) → G . • The kernel of φ has trivial intersection with every edge stabiliser of T (cid:48) . • There is a G (cid:48) -equivarient combinatorial map Ψ : T (cid:48) → T . (Where the action of G (cid:48) on T is the natural one given by φ .) • T (cid:48) has a set of seed vertices S (cid:48) with P –weight W P ,S (cid:48) .Then T /G has at most (cid:0) k +12 (cid:1) W P ,S (cid:48) edges. Furthermore if either T is reduced and k > or all of the edges of T have stabiliser of size greater than P then T /G has atmost kW P ,S (cid:48) edges. First observe that the following variation of Lemma 4 .
12 and Lemma 4 .
13 holds withthe exact same proof as before.
Lemma 6.5
Let φ : H → G be a surjective homomorphism and P is a conjugationinvariant set of subgroups for G . Suppose that there is a H –equivariant map α : R → ˜ R which is a fold. Suppose that the kernel of φ has trivial intersection with each vertexstabiliser of R . Suppose that S is a set of seed vertices for R where the image of each con-necting group is in P . Then there is a set of seed vertices ˜ S for ˜ R with W P ,φ, ˜ S ≤ W P ,φ,S .Moreover if W P ,φ, ˜ S = W P ,φ,S then α | S is injective. After each step elements in the kernel of φ may end up acting elliptically on theintermediate tree. As such we need a way of modifying a group G i and tree T i whichessentially keeps the action and map Ψ i : T i → T but removes problematic groupelements found in the kernel of φ : G i → G .29 emma 6.6 Let φ : G (cid:48) → G be a surjective group homomorphism and suppose that G (cid:48) acts on a tree T (cid:48) . Then there’s a group G (cid:48)(cid:48) acting on a tree T (cid:48)(cid:48) together with surjectivehomomorphisms φ (cid:48) : G (cid:48)(cid:48) → G and σ : G (cid:48) → G (cid:48)(cid:48) and a G (cid:48) –equivarient simplical map ρ : T (cid:48) → T (cid:48)(cid:48) . (The action of G (cid:48) on T (cid:48)(cid:48) is given by σ .) Additionally ρ/G : T (cid:48) /G → T (cid:48)(cid:48) /G is a homeomorphism of graphs with φ having trivial intersection with each edge stabiliserof T (cid:48) and σ (Stab e ) = σ (Stab ρ ( e )) . Moreover the kernel of φ (cid:48) has trivial intersectionwith every vertex stabiliser of T (cid:48)(cid:48) .Hence if T (cid:48) has a set of seed vertices S (cid:48) then S (cid:48)(cid:48) := ρ ( S (cid:48) ) is a set of seed vertices for T (cid:48)(cid:48) with W P ,S (cid:48)(cid:48) = W P ,S (cid:48) . Proof
We define G (cid:48)(cid:48) as the fundamental group of a graph of groups decompositioncorresponding to T (cid:48) but with each vertex label replaced with its image under φ and let T (cid:48)(cid:48) be the corresponding Bass-Serre tree. This naturally induces maps φ (cid:48) : G (cid:48)(cid:48) → G and˜ φ : G (cid:48) → G (cid:48)(cid:48) and ˜Ψ : T (cid:48)(cid:48) → T . Moreover we naturally get a set of seed vertices S (cid:48)(cid:48) for T (cid:48)(cid:48) with W P φ (cid:48) ,S (cid:48)(cid:48) = W P φ ,S (cid:48) . (cid:3) Proof of Lemma . W P ,S . If W P ,S = 0then there is some vertex of T (cid:48) which is fixed by G . So the image of this vertex in T isfixed by G and so as T is minimal it must just consist of this single vertex.Use Stallings folding theorem (Theorem 2 .
2) to decompose α into folds α (0) i : T (0) i − → T (0) i and let G (0) := G (cid:48) . We will iteratively define groups G ( j ) which act on trees T ( j ) i withsets of seed vertices S ( j ) i for T ( j ) i . Also we will define maps σ ( j ) : G ( j ) → G ( j +1) and σ ( j ) : G ( j ) → G together with α ( j ) i : T ( j ) i − → T ( j ) i and ρ ( j ) i : T ( j ) i → T ( j +1) i (for i ≥ j − T (0)0 T (0)1 T (0)2 · · · T (1)0 T (1)1 T (1)2 · · · T (2)1 T (2)2 · · · ... . . . α (0)1 ρ (0)0 α (0)2 ρ (0)1 α (0)3 ρ (0)2 α (1)1 α (1)2 ρ (1)1 α (1)3 ρ (1)2 α (2)2 α (2)3 ρ (2)2 First we define β j := α ( j ) j ◦ ρ ( j − j − ◦ · · · ◦ ρ (1)1 ◦ α (1)1 ◦ ρ (0)0 . (Part of the red path alongthe bottom of the diagram which ends with a right facing arrow.) First use Lemma 6 . β j we obtain a map ρ ( j ) : T ( j ) j → T ( j +1) j together with group homomorphisms σ ( j ) : G ( j ) → G ( j +1) and φ ( j +1) : G ( j +1) → G . Also S ( j +1) j := ρ ( j ) ( S ( j ) j ) is a set of seed verticeswhere the images of the connecting groups are in P and W P ,S ( j ) j − ≤ W P ,S ( j − j − . If α ( j ) i folds together the edges e and e we define α ( j +1) i (for i > j ) to be the fold (or identity)30btained by identifying the ρ ( j ) i +1 images of the e i . Finally define ρ ( j +1) i (for i > j + 1) tomake the above diagram commute.Finally Lemma 6 . α ( j ) j induces a set of seed vertices S ( j ) j on T ( j ) j with W P ,S ( j ) j ≤ W P ,S ( j ) j − . (If a separating edge is larger than P then we can reducethe P –weight by collapsing at most k edges using Lemma 4 .
14 and then proceeding byinduction.)As α = · · · α (0)2 ◦ α (0)1 we see that α = · · · α (2)2 ◦ ρ (1)1 ◦ α (1)1 ◦ ρ (0)0 by the definitions of α ( j ) i and ρ ( j ) i . Moreover at each step we have a set of seed vertices with non-increasing P –weight, the number of orbits of connecting edges are non-decreasing and hence thatall but finitely many of the sets of seed vertices are the image of the seed vertices at theprevious level. At this point the proof is exactly the same as the proof of Lemma 4 . (cid:3) Proof of Theorem . (b) Pick a minimal generating set X for G and let G (cid:48) := F ( X ). Let φ : G (cid:48) → G be the natural projection and let T (cid:48) be the tree correspondingto the rose with rank G petals labelled by the elements of X . Let Ψ : T (cid:48) → T beany G (cid:48) –equivarient combinatorial map. If G acts freely on T then (make new lemma)implies that T has at most 3(rank G −
1) edges. Let S be the set of vertices of T (cid:48) before subdividing. Observe that S is a set of seed vertices for T (cid:48) and that it has P –weight of at most 2 M rank( G ) since P has height M . Hence by Lemma 6 . T has asmost (2 k + 1)2 M − (rank( G ) −
1) edges. If all of the edges of T have stabiliser largerthan P or T is reduced with k > k M (rank( G ) − (cid:3) We will now restrict our attention to the case where P = , the collection whichonly contains the trivial subgroup. In other words we are to consider actions whichare k –acylindrical. In [17] Weidmann showed that a finitely generated group acting k –acylindrically on a tree where all the edges have non-trivial stabiliser has at most2 k (rank G −
1) orbits of edges. The purpose of this section is to prove Theorem 1 . k (cid:0) rank G − (cid:1) edges, and to construct an example whichshows that this is the best possible bound. (Theorem 1 .
20) Additionally we’ll refine thisfurther to 2 k (cid:0) rank G − (cid:1) the the case where the group is torsion-free. Proof of Theorem .
20 We need to show that for any integers k ≥ r ≥ r acting k -acylindrically on a tree with (cid:4) k (cid:0) rank G − (cid:1)(cid:5) orbitsof edges, none of which have trivial stabilisers. Pick distinct primes p and q such that( p − q − ≥ r − G := (cid:104) a, h , · · · , h r − | a pq = 1 (cid:105) ∼ = (cid:16) Z pq Z (cid:17) ∗ F r − and notethat rank G = r . We will now construct a tree T for G to act on. Start with the graphof groups decomposition consisting of the rose with r − h i andwith a single vertex on the loop representing h with label (cid:104) a (cid:105) . (In the diagrams we take k = 4 and r = 3.) 31 a (cid:105) h h h so that it consists of (cid:6) k (cid:7) + 2 edges. Apply foldsof type II to “pull” a onto each vertex on the loop except the central one. (cid:104) a (cid:105) h (cid:104) a (cid:105)(cid:104) a (cid:105) h h which are adjacent to thecentral vertex into (cid:4) k (cid:5) sub-edges. Apply folds of type II which “pull” a p along one ofthese series of edges and “pull” a q along the other. We see that the central vertex hasstabiliser H := (cid:104) b, c | b q = c p = 1 (cid:105) ∼ = (cid:16) Z p Z (cid:17) ∗ (cid:16) Z q Z (cid:17) where b = a p and c = ( a q ) h . (cid:104) a (cid:105) (cid:104) b, c (cid:105) h (cid:104) a (cid:105)(cid:104) a (cid:105) (cid:104) a p (cid:105)(cid:104) a q (cid:105) h i ≤ r −
2) we define g i = b x +1 c y +1 where i = ( p − x + y for x, y ∈ Z with0 ≤ y ≤ p −
2. Since ( p − q − ≥ r −
2) we see that these g i represent pairwisenon-conjugate elements of H . Subdivide the loop representing each h j (for 2 ≤ j ≤ r − k sub-edges, then apply folds of type II which “pulls” g j − along k edges startingat one end and “pulls” g j − along k edges starting at the other. (cid:104) a (cid:105) (cid:104) b, c (cid:105) h (cid:104) a (cid:105)(cid:104) a (cid:105) (cid:104) a p (cid:105)(cid:104) a q (cid:105) h (cid:104) g , g (cid:105)(cid:104) g (cid:105)(cid:104) g (cid:105) (cid:104) g (cid:105)(cid:104) g (cid:105) (cid:104) g (cid:105) (cid:104) g (cid:105) Observe that this decomposition has (cid:4) (2 r − ) k (cid:5) edges. (It’s not reduced in general,but recall that this isn’t a condition for this result.) It remains to check that thecorresponding Bass-Serre tree is k –acylindrical. The elements of G which act elliptically(upto conjugacy) are powers of a and elements of H ; so these are the ones we need tocheck fix a region of bounded diameter.First consider elements of H . The elements which fix an edge of our tree are (powersof) the g i , b and c (upto conjugacy). As b and c are conjugate to powers of a we’ll leavethese for now. Now observe that each g i has a different image in the ablieanisation of H ; hence distinct g i are in different conjugacy classes. Moreover each cyclic root-closed32ubgroup of H is malnormal in it. Hence each (power of) g i only fixes k edges. (cid:104) a (cid:105) (cid:104) a q (cid:105)(cid:104) a q (cid:105)(cid:104) a q (cid:105)(cid:104) a p (cid:105)(cid:104) a p (cid:105) (cid:104) a p (cid:105) Figure 15: The region of the Bass-Serre tree which is fixed by some power of a , togetherwith their stabilisers. The central arc with stabiliser (cid:104) a (cid:105) has length (cid:6) k (cid:7) while each“offshoot” with stabiliser either (cid:104) a p (cid:105) or (cid:104) a q (cid:105) has length (cid:4) k (cid:5) .We now need to consider powers of a . Let 1 ≤ m < pq and look at the fixator of a m .If p | m then the fixator of a m consists of a central vertex with q + 1 “offshoots”, one oflength (cid:6) k (cid:7) and the rest of length (cid:4) k (cid:5) . In other words the fixator consists of the left andthe centre parts of Figure 15. This region has diameter k and so we are fine. Likewisefor the case where q | m . We cannot have pq | m as m < pq . Finally if m is coprime to pq then p m just fixes a path of length (cid:6) k (cid:7) ; the middle section of Figure 15.Building an example which is maximal for torsion-free groups is similar. First weneed a to have infinite order and so G ∼ = F r . The initial splitting is defined in thesame way as before. Next we subdivide the loop representing h into k subedges andapply folds of type II so that each edge in this loop has label (cid:104) a (cid:105) . (If k = 1 then wecollapse either of the initial edges of the loop instead.) The central vertex now has labelisomorphic to the free group of rank 2 which is generated by b := a and c := a h . Wenow subdivide and fold onto the loops representing the rest of the h j as before. (cid:3) (cid:104) a (cid:105) (cid:104) b, c (cid:105) h (cid:104) a (cid:105)(cid:104) a (cid:105) h (cid:104) g , g (cid:105)(cid:104) g (cid:105)(cid:104) g (cid:105) (cid:104) g (cid:105)(cid:104) g (cid:105) (cid:104) g (cid:105) (cid:104) g (cid:105) Before proving Theorem 1 .
19, which will show the above examples are the bestpossible, it’s useful to compare their constructions to the proof of Lemma 6 .
4. We startwith a single orbit of seed vertices with representative stabiliser (cid:104) a (cid:105) . Our initial foldsinduce a new orbit of seed vertices on the central vertex. Moreover the two connectingedges on the loop representing h are now non-trivial and so we collapse it. In doing thiswe’ll reduce the –weight by two but only collapse either (cid:4) k (cid:5) or k edges, dependingon which construction we’re talking about. This is less than the 2 k edges theoreticallyallowed by the lemma. Continuing we then successively collapse each loop; getting ridof the maximally possible 2 k edges each time.With this comparison in mind we will now show that such an inefficiency must occurat a particular point in Lemma 6 .
4. Specifically whenever a vertex first obtains a non-cyclic stabiliser. 33 emma 7.1
Let α : R → ˜ R be a fold which factors through Ψ : T (cid:48) → T and let S be aset of seed vertices for R . Suppose that the action on T is k –acylindrical. Suppose alsothat both every vertex stabiliser of R is cyclic and every connecting group of S is trivial.Then one of the following holds • Every vertex stabiliser of R is cyclic and we can find a set of seed vertices ˜ S for ˜ R such that every connecting group of ˜ S is trivial and W , ˜ S ≤ W ,S . • There is a simplicial map ρ : ˜ R → ˜ R (cid:48) which factors through Ψ and we can collapseat most (cid:4) k (cid:5) orbits of edges of T to get a new tree T such that the following holds.Let R be the tree obtained by collapsing the edges of ˜ R (cid:48) corresponding to T → T .There is a set of seed vertices S for R with W ,S ≤ W ,S − . Moreover if G istorsion-free then we can obtain T by collapsing at most k edges of T . Proof
Suppose that α folds together the edges e = [ x, y ] and e = [ x, y ] to an edge e (cid:48) = [ x (cid:48) , y (cid:48) ]. If there’s a forest of influence containing both e and e then we can justtake ˜ S := α ( S ) and we end up in the first outcome listed in the statement. The sameapplies the fold is of type I or II and either of the y i have trivial stabiliser. Similarly wecan take ˜ S := α ( S ) ∪ G { y (cid:48) } if α is a fold of type III and the y i have trivial stabilisers.Consider the case where α is a fold of type III, the stabiliser of the y i are non-trivialand hy = y . Suppose that y is influenced by u and u (cid:48) := α ( u ). Both u (cid:48) and y (cid:48) arefixed by the stabiliser of y ; hence as in the proof of Theorem 1 . ρ : ˜ R → ˜ R (cid:48) which factors through Ψ such that the reduced edge path f (cid:48) from u (cid:48) to y (cid:48) consists of at most k edges and is injective under δ : ˜ R (cid:48) → T . Now we define T bycollapsing the image of f (cid:48) in T and R as in the statement. Note that the image of S in T is a set of seed vertices with W ,S ≤ W ,S − χ ( R/G ) = χ ( R/G ) + 2. ρ ◦ α type III xy u f x (cid:48) y (cid:48) u (cid:48) f (cid:48) Figure 16: The case where the fold α is of type III. The edge path labelled f is fixedby some non-trivial member of G , hence its image in ˜ R has diameter at most k . Inparticular if we replace f with its image in ˜ R we see that the path f (cid:48) has length at most k . So now we can assume that the fold is of type I or II where the stabiliser of both y and y are non-trivial. If α is a fold of type I we say that y i is influenced by u i and f i be the branch of y i . If instead α is a fold of type II we say that x is influenced by u and y is influenced by u . We also let f be the union of the branch of x together with e and let f be the branch of y . 34irst consider the case where u is inequivalent to u . Let ρ : ˜ R → ˜ R (cid:48) be thecomposition of folds on f and f (separately) which causes γ : R (cid:48) → T to be locallyinjective on f and f . Let u (cid:48) i be the vertex closest to y (cid:48)(cid:48) := ρ ( y (cid:48) ) with stabiliser equal toStab y i and let f (cid:48) i be the reduced edge path from u (cid:48) i to y (cid:48)(cid:48) . We now define T by collapsingthe (orbits of the) images of f (cid:48) and f (cid:48) in T and R as in the statement. Observe that wehave a set of seed vertices S for R defined to be the union of the image of S \ G { u , u } together with the image of G { y (cid:48)(cid:48) } and observe that W ,S ≤ W ,S −
2. It remains tobound the number of edges we’ve collapsed. If f (cid:48) i consists of more than a single vertexlet g i be a group element which fixes u (cid:48) i but no edge of f (cid:48) i . Then since Stab u (cid:48) i = Stab u i is cyclic we see that Stab y i fixes the reduced edge path f (cid:48) i ∪ g i f (cid:48) i . So since the action on T is k –acylindrical we see that f (cid:48) and f (cid:48) each consist of at most k edges each and so wehave collapsed at most k edges total. u x y y u f f f (cid:48) x (cid:48) y (cid:48) u (cid:48) u (cid:48) f (cid:48) ρ ◦ α type I ρ ◦ α type II u u x y f f u (cid:48) u (cid:48) x (cid:48) y (cid:48) f (cid:48) f (cid:48) Figure 17: The graph of groups in the cases where the u i aren’t equivalent.Now consider the case u = hu for some h ∈ G . Define ρ : ˜ R → ˜ R (cid:48) , u (cid:48) i , y (cid:48)(cid:48) and f (cid:48) i asbefore. Let ˜ f be the path from u (cid:48) to h − u (cid:48) . Define T by collapsing the images of f (cid:48) , f (cid:48) and ˜ f in T . We have a set of seed vertices S for R defined to be the union of the imageof S \ G { u } together with the image of G { y (cid:48)(cid:48) } and again we have W ,S ≤ W ,S − f (cid:48) i haveat most k edges. If ˜ f consists of just a single vertex u (cid:48) then we are done as before. Ifnot then observe that f (cid:48) ∪ ˜ f ∪ h − f (cid:48) is a reduced edge path from y (cid:48)(cid:48) to h − y (cid:48)(cid:48) . If G istorsion-free then as Stab u is cyclic then there is some non-trivial subgroup which fixes f (cid:48) ∪ ˜ f ∪ h − f (cid:48) and so we’ve collapsed at most k edges. If G isn’t torsion-free then we areonly guaranteed to have a non-trivial subgroup which fixes f (cid:48) ∪ ˜ f . Thus f (cid:48) ∪ ˜ f ∪ h − f (cid:48) has at most k + k = k edges. (cid:3) Proof of Theorem .
19 Proceed as in the proof of Theorem 1 .
12 (b). We have ahomomorphism φ : F ( X ) → G where X is a minimal generating set for G and combina-torial map Ψ : T (cid:48) → T . As in the proof of Lemma 6 . ◦ α type I x y y u f f x (cid:48) y (cid:48) ρ ( u ) f (cid:48) f (cid:48) ˜ fu (cid:48) u (cid:48) ρ ◦ α type II x y u f f x (cid:48) y (cid:48) ρ ( u ) f (cid:48) f (cid:48) ˜ fu (cid:48) u (cid:48) Figure 18: The graph of groups in the cases where u and u are equivalent. α i : T i − → T i . We then apply Lemma 7 . α i in turn until one of them causes usto collapse edges. (This must happen eventually as all the edges of T have non-trivialstabiliser.) Then apply Theorem 1 .
12 (b) to the collapsed tree to get the desired bound. (cid:3)
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