Morse theory and conjugacy classes of finite subgroups II
Abstract.
We construct a hyperbolic group containing a finitely presented subgroup, which has infinitely many conjugacy classes of finiteorder elements.
We also use a version of Morse theory with high dimensional horizontal cells and use handle cancellation arguments to produce other examples of subgroups of CAT(0) groups with infinitely many conjugacy classes of finiteorder elements.
Introduction
This paper is a continuation of our earlier work in [3]. In that paper we showed how the construction of Leary–Nucinkis [8] fits into a more general framework than that of right angled Artin groups. We used this more general framework to produce a CAT(0) group containing a finitely presented subgroup with infinitely many conjugacy classes of finiteorder elements. Unlike previous examples (which were based on rightangled Artin groups) our ambient CAT(0) group did not contain any rank 3 free abelian subgroups. In the current paper (see Section 4), we produce a hyperbolic group containing a finitely presented subgroup which has infinitely many conjugacy classes of finiteorder elements.
We work in the more general situation of Morse functions with high dimensional horizontal cells in Sections 2 and 3 of this paper. This allows us to see (in Section 2) that the original examples of Feighn–Mess [7] fit into the same general framework as the examples of Leary–Nucinkis [8]. This addresses a remark we made after Example 1.2 of [3].
In Section 3 we use Morse theory with horizontal cells to see that a suitable modification of the Rips’ construction (suggested to the authors by Dani Wise) produces many new examples of hyperbolic groups containing finitely generated subgroups with infinitely many conjugacy classes of finiteorder elements. The Morse arguments used here involve a careful accounting of handle cancellations.
1. Counting conjugacy classes of finiteorder elements
The following proposition describes a general algebraic situation which ensures that a conjugacy class in some group will intersect a subgroup in infinitely many conjugacy classes. In all the examples in this paper, the target group is just , and the result is used to count conjugacy classes of finiteorder elements.
Proposition 1.1.
Suppose is an epimorphism where
for some element . Then the conjugacy class of in intersects in infinitely many conjugacy classes.
Proof.
Let and fix such that . Now is conjugate to in if and only if there is an such that , equivalently . Applying we see this is equivalent to for some . Since , this is equivalent to , in other words .
Therefore, the conjugacy class of in intersects in at least many conjugacy classes. By hypothesis, this index is infinite, completing the proof. ∎
Remark 1.2.
Model Situation.
Let be an epimorphism where is a CAT(0) group. Let be a CAT(0) metric space on which acts properly by isometries, and let be a equivariant (Morse) function,where acts on by integer translations. Let have finite order and the property that is compact. This generalizes our model situation from [3] where we required that had an isolated fixed point. Then implies that acts on and therefore by equivariance of , acts on . Since is compact and acts on by translations, we see that . Therefore, applying Proposition 1.1 to , the conjugacy class of in intersects in infinitely many conjugacy classes, since .
2. The Feighn–Mess examples
In this section we show how the Feighn–Mess examples [7] fit into the more general Morse theory setup in Proposition 1.1. This addresses the remark we made after Example 1.2 in [3]. Recall the Feighn–Mess examples are subgroups of where the factor acts by an involution in each that fixes one generator and sends the other to its inverse.
Let be the wedge of two circles glued together at the point . Label the two circles and . There is an order two isometry on defined by the identity on and by on . The fixed point set of consists of two components: one is the circle , the other is the point . This induces a coordinatewise defined map where is the direct product of copies of . Here we see that the fixed point set of has homeomorphism types of components, a representative of each is of the form , which is isometric to the dimensional torus .
Define by mapping each of the circles labeled homeomorphically around , the circles labeled to and extending linearly over the higher skeleta. Clearly is equivariant.
Let be the universal cover of and lift to . Then is a CAT(0) cubical complex. We can identify with a subgroup of the group of isometries of . There are several different types of lifts of the isometry to an isometry corresponding to the different homeomorphism types of components in the fixed point set of . For each we get a lift whose fixed set is a lift of . This lift is isometric to an dimensional plane . The image of under the map is if and a single point if . Let be such that . For , we have drawn the action of on in Figure 1, for , the maps are induced coordinatewise from this map.
Let , this is the group considered by Feighn and Mess. Then there is a homomorphism induced from and by sending . The map is equivariant. This is our model situation where Proposition 1.1 applies, hence the conjugacy class of in intersects in infinitely many conjugacy classes.
To compute the finiteness properties of we use Morse theory with horizontal cells. The map is a equivariant Morse function in the sense of [4]. Since finiteness properties are virtual notions, it suffices to compute the finiteness properties of . Any horizontal cell for is a face of a cube of the form . The ascending or descending link of an dimensional face is connected as it is an simplex, where we make the convention that connected means “not empty” and a simplex is the empty set. Therefore, is of type F. Furthermore, since the ascending link of a horizontal cell of the form is empty we see that is not of type F. Therefore is of type F but not of type F.
3. Examples arising from Rips’ construction
The idea behind following examples was suggested by Dani Wise. Consider the Rips’ construction of a nonelementary hyperbolic group with quotient and finitely generated kernel. Wise’s suggestion is to take a quotient of by a power of some element of this kernel. One expects to get a new short exact sequence
where is hyperbolic, is finitely generated but not finitely presentable, and has infinitely many conjugacy classes of elements of order . We show that this is the case in Theorem 3.3 below.
We work with Wise’s CAT() version of the Rips’ construction, and use handle cancellation techniques to see that the finiteness properties of the kernel subgroups follow from Morse theory.
A key component of Wise’s version of Rips’ construction is the long word with no twoletter repetitions. We work with a slight variant of Wise’s construction in our definition of the nonelementary hyperbolic group .
Definition 3.1 (Wise’s long word and the group ).
Given the set of letters , define to be the following word
It is easy to see that is a positive word of length , with no two letter subword repetitions. For we can, starting at the left hand side, partition into at least disjoint subwords; of length , and two of length . Construct the positive word (respectively ) by adding as a prefix to one of the length subwords (respectively as a suffix to the other length subword). Call the remaining length subwords and for .
Now define by the presentation
(1) 
where are given above.
Remark 3.2.
Note that the group surjects to , taking and () to the identity, and to a generator of . Thus, the group maps onto taking to a generator, and all the other generators to the identity. This implies that the element has order in .
Theorem 3.3.
The group defined by (2) is CAT(1). Let denote the kernel of the map which takes each to the identity, and to a generator of . Then is finitely generated but not finitely presentable. Furthermore, the conjugacy class of the element in intersects in infinitely many conjugacy classes.
Proof.
The proof is presented in several steps. First, we establish that the groups and are CAT(1). Then we define the equivariant Morse function, and show how the conjugacy result follows from Section 1. Finally, we use the Morse theory and an analysis of handle cancellations to establish the finiteness properties of .
Step 1. The structure for . First one sees that the group is CAT(), by subdividing each relator cell in its presentation complex into rightangled hyperbolic pentagons as shown in Figure 2. The presentation complex of satisfies the large link condition. This is a consequence of the fact that the ’s and the ’s are positive words with no twoletter repetitions. The argument is identical to that of [9]. We highlight the key points for the reader’s convenience.
The link is obtained from a subgraph on the by adding the vertices and new edges of length connecting the to the . This is large if and only if the subgraph on the is large. The latter graph is large because it is bipartite (since the and the are all positive words), has no bigons (since the collection of all and have no two letter repetitions by design) and the edge lengths are all at least .
Step 2. The structure for . Next, we show that is a CAT() group for . First attach a cell to the loop in the locally CAT() presentation complex for . The resulting complex is a presentation complex for .
By Remark 3.2, the preimage of the loop in the universal cover of consists of a disjoint collection of embedded circles of length (of the form ). The preimage of in consists of distinct families of cells, one for each component of the preimage of . Each of the cells in a family is attached to the same embedded loop .
For each loop labeled by in collapse the attached cells in the preimage of down to a single cell. The resulting cell complex can be given a locally CAT() structure, by giving each cell in the preimage of the geometry of a regular hyperbolic gon whose side lengths are equal to the side length of a rightangled hyperbolic pentagon. The proof that is locally CAT() involves a slight modification of the argument given above to show that is CAT(). In particular, we add a single edge from to of length at least , and note that the subgraph on the is still bipartite with no bigons.
Since is connected and locally CAT(), it is CAT(), and since acts properly discontinuously and cocompactly on , we conclude that is a CAT() group. Hence is hyperbolic.
Step 3. The Morse function and counting conjugacy classes. We define a circle valued Morse function on the original (unsubdivided) presentation complex for as follows. Send the edge homeomorphically around the circle, and send the remaining edges in the skeleton (namely the ) to the basepoint of the target circle. Extend linearly over the cells. Thus the cell is mapped to the base point of the target circle, and the remaining cells are mapped onto the edge (horizontal projection in Figure 3) and then to the circle as described above.
This lifts to a equivariant Morse function , which factors through . We denote the equivariant factor map by . Note that the cells labeled by are all horizontal, and that the cells of in the preimage of are also horizontal.
The element fixes a unique point of : namely, the center of the gon with vertices . We are now in the model situation of Section 1 above, and so the conjugacy class of in intersects in infinitely many conjugacy classes.
Step 4. Finiteness properties of the kernel . Finally, we use Morse theory on the complex to show that is finitely generated but not finitely presentable.
To this end, we first subdivide the cells and into triangular cells. This subdivision is indicated in Figure 3. Writing , define new edges for inductively by
For each and each , let denote the triangular cell in the subdivision of which has boundary .
Similarly, writing , define new edges for inductively by
For each , let denote the triangular cell in the subdivision of with boundary .
We are in the setting of [4], where has horizontal cells, cells and cells. The Morse theory argument will be essentially that of [4], but we will have to take care of handle cancellations. Note that for any cell of , its image under will either be an integer or an interval of length one (between successive integers) in .
Ascending links. A schematic of the ascending link of a vertex is given in Figure 4. It consists of a graph with components. One component is a graph which is a subdivision of the cone on the points and for . The cone vertex is , and each of the edges from to () is subdivided into segments by for . The remaining points are labeled by ( and ).
The handle additions. For integers we obtain from by a succession of three types of coning operations (equivalently, handle addition operations).

For each cell , let denote the union of all the cells of , each of which contains and which maps to the interval under . Then is a geometric realization of .
Attaching the to for each cell is equivalent to coning off each copy of to the corresponding vertex . Denote the resulting complex by .
Now has components, each of which are contractible. Thus coning the copy of in off to is equivalent to attaching a wedge of onehandles to . Thus is obtained from by attaching an infinite family of distinct wedges of onehandles, indexed by the cells of .

For each cell in , let denote the union of cells, each of which contains and which maps to under . Then is a geometric realization of . This is a discrete set of points. We see that its cardinality is at least one, by writing for some , and noting that is a subset of . Recall that is the simplex containing the bottom edge in the subdivision of in Figure 3.
Note also that is isomorphic to . Thus, attaching to is equivalent to coning off to the barycenter of . Since has two points, this is equivalent to attaching a wedge of twohandles to . This even makes sense in the case that has only one point. Then the coning operation is a homotopy equivalence, which is equivalent to attaching 0 twohandles to .
Let denote the result of attaching to for each cell in . As in the previous section, is obtained from by attaching an infinite family of wedges of twohandles to , indexed by cells in .

Finally, attach the cells to to obtain . Note that for each such cell, and so attaching is equivalent to coning off to the barycenter of .
The handle cancellations. We show that a subset (described in Definition 3.4 below) of the twohandles from the coning operation (2) above cancel with the collection of all onehandles from the coning operation (1). Note that is the result of attaching all the onehandles to . By cancel we simply mean that the space obtained from by attaching this subset of twohandles is homotopy equivalent to .
Definition 3.4 (Canceling set).
For each and for each horizontal edge , consider the union of cells of , each of which contains and which maps to under . Let denote the subcomplex of such that is isomorphic to
That is, we are not considering contributions to which correspond to occurrences of as the first letter in any . The canceling set is defined as the following subcomplex of
The shaded portion of Figure 5 shows the intersection of a cell in with .
Lemma 3.5 (Handle cancelation for ascending links).
Let be as defined above. Given integers , let be the space obtained from by attaching handles as described in coning operation (1) above, and let be the canceling set of Definition 3.4.
Then is homotopy equivalent to .
Proof.
The homotopy equivalence is fairly easy to see. First, consider relator cells of the form . Figure 5 shows the intersection of with one of these relators. The only cell of this relator which does not belong to is the unshaded (open) cell labeled . There is an obvious deformation retraction of this intersection onto the boundary skeleton . Perform all these deformation retractions for each and each , to get the space .
Now, we turn our focus to horizontal cells at height in the set . Each is contained in just one cell of . This cell is labeled . Push across the free edge to deformation retract the relator onto the subword of its boundary word. Do this equivariantly for all and all . The resulting space can now be deformed to by collapsing the edges labeled in onto their vertices at level . ∎
Continue with the usual Morse argument. We obtain from by first attaching all the new cells of . Lemma 3.5 ensures that is homotopy equivalent to . Now we obtain from by attaching all cells of which are labeled by for , and by attaching all cells of . The boundary of each of these cells is contained in , and so each such attachment can be viewed as a handle attachment (coning boundary off to barycenter).
In a similar fashion (working with descending links) one can argue that is obtained from up to homotopy by only attaching handles.
Now, the usual Morse theory arguments of [1] apply to conclude that the level set is connected, and hence that is finitely generated. Since we are attaching twohandles, the inclusioninduced map for any integers is always a surjection. If this inclusioninduced homomorphism were ever the zero homomorphism, then we would have . The twohandles attached to obtain would produce nontrivial cycles in which contradicts the fact that is a contractible complex. Now Theorem 2.2 of [6] (taking and taking to be ) implies that is not FP. In particular, is not finitely presented. ∎
4. Conjugacy classes in finitely presented subgroups of hyperbolic groups
In this section we construct a hyperbolic group with a finitely presented subgroup which has infinitely many conjugacy classes of finiteorder elements. Our construction is a modification of the construction in [2] of a hyperbolic group with a finitely presented subgroup which is not hyperbolic. The hyperbolic group in [2] is torsionfree, and does not admit any obvious finiteorder automorphisms.
Theorem 4.1.
There exist hyperbolic groups containing finitely presented subgroups which have infinitely many conjugacy classes of finite order elements.
As indicated above, the proof consists in constructing a variation of the branched cover complex in [2]. The variation will have an added symmetry that the construction in [2] lacked. This symmetry will enable one to extend the groups under consideration by a finiteorder automorphism, and to apply Proposition 1.1. An overview of the branched cover which parallels the construction in [2] is provided in subsection 4.1 below, and the proof that this branched cover does indeed admit a symmetry is given in subsection 4.2.
4.1. The branched cover
Let be the graph in Figure 6, where each edge is isometric to the unit interval. Then , with the product metric, is a piecewise Euclidean cubical complex of nonpositive curvature. The hyperbolic group in [2] is defined to be the fundamental group of a particular branched cover of with branching locus
We refer to Section 5 of [2] for background on branched covers. Recall from [2] that the branched cover is obtained by the following procedure.

Remove the branching locus from the nonpositively curved cubed complex . The resulting space has an incomplete piecewise Euclidean metric.

Take a finite cover of . The piecewise Euclidean metric on lifts to an incomplete piecewise Euclidean metric on this cover.

Complete this metric to form the branched cover .
In Lemma 5.5 of [2] it is shown that if is a nonpositively curved piecewise Euclidean cubical complex and if is any reasonable branching locus, then every finite branched cover of over is itself a nonpositively curved piecewise Euclidean cubical complex. In particular, every finite branched cover of over above is a nonpositively curved piecewise Euclidean cubical complex.
Now we define a Morse function on , and hence on finite branched covers of . There is a map which maps the vertices and to a base vertex of , and maps the edges isometrically around the circle, with orientations specified by the arrows in Figure 6. This defines a map . Composition with the standard linear map (the one covered by ) gives a map . Finally, the composition is a circlevalued Morse function on the branched cover .
The majority of the work in [2] comes from constructing a specific finite branched cover having the following properties.
 Hyperbolic:

is a hyperbolic group, and
 F–not–F:

the kernel of the map is finitely presented, but not of type F.
These properties are guaranteed by items (3) and (4) of Theorem 6.1 of [2]. The construction of the branched cover is described in detail in the proof of Theorem 6.1. We sketch the main points below, and indicate our variation on that construction. The key point is that our variation is equivariant with respect to a particular isometry (see Section 4.2) and so the branched cover admits an isometry induced by .

For each there are projection maps
where is taken modulo , and we use in place of .
These restrict to projection maps
for .

Let be the graph (depicted in Figure 7)
By Lemma 6.2 of [2], the inclusion of in is a homotopy equivalence. We choose the following basis for , which is a free group of rank :

Define a homomorphism (when convenient, we shall think of as a homomorphism ) by
where and are the permutations
The reader can verify that is a cycle for . This implies that the images under of the 36 commutators obtained by taking a pair of loops in (one composed of two cells with endpoints and , and the second composed of two cells with endpoints and ) are all cycles.

For define homomorphisms by
where is a change of basepoint isomorphism. Specifically, is the isomorphism induced by conjugating an edge path in based at by the edge . For each , this gives an action of on the set . Define an action of on the set via
The fold cover of associated to is connected. Lift the local metric from to , and complete it to obtain the branched covering .
Remark 4.2.
Our version of satisfies the key Property 1 of Theorem 6.1 of [2]. Thus items (3) and (4) of Theorem 6.1 hold, and so the branched covering will satisfy properties Hyperbolic and F–not–F above.
4.2. The isometry
Let denote the isometry which fixes vertices and and transposes the edges and , and the edges and . This induces coordinatewise defined isometries
and
These restrict to isometries of the incomplete spaces and .
Then induces a map . The action of on is given by , , . Moreover, for , we have
(3) 
Let be the inner automorphism .
Claim. The following two maps are
equal:
(4) 
Indeed, using the fact that are cycles, we can check equality on a free basis as follows:
and so the claim is established.
The symmetry in equation (4) is key to showing that lifts to an automorphism of . First, equation (4) is used to show that lifts to a symmetry of the fold cover, and then the isometry is obtained by continuous extension to the completion of this cover.
Recall from item (iv) of subsection 4.1 that the fold cover of corresponds to the subgroup
of . By construction, , where
The isometry lifts to this cover provided that . This will follow if for . If , (i.e. ) then
definition of  