Genus two Goeritz group of \mathbb{S}^{2}\times\mathbb{S}^{1}

# The genus two Goeritz group of S2×S1

Sangbum Cho Department of Mathematics Education, Hanyang University, Seoul 133-791, Korea  and  Yuya Koda Mathematical Institute
Tohoku University, Sendai, 980-8578, Japan
and
(Temporary) Dipartimento di Matematica
Università di Pisa, Largo Bruno Pontecorvo 5, 56127 Pisa, Italy
July 17, 2019
###### Abstract.

The genus- Goeritz group is the group of isotopy classes of orientation-preserving homeomorphisms of a closed orientable -manifold that preserve a given genus- Heegaard splitting of the manifold. In this work, we show that the genus- Goeritz group of is finitely presented, and give its explicit presentation.

###### 2000 Mathematics Subject Classification:
Primary 57N10, 57M60.
The first-named author is supported by Basic Science Research Program through the National Research Foundation of Korea(NRF) funded by the Ministry of Education, Science and Technology (2012006520).
The second-named author is supported in part by Grant-in-Aid for Young Scientists (B) (No. 20525167), Japan Society for the Promotion of Science, and by JSPS Postdoctoral Fellowships for Research Abroad.

## 1. Introduction

Given a genus- Heegaard splitting of a closed orientable -manifold, the genus- Goeritz group is the group of isotopy classes of orientation-preserving homeomorphisms of the manifold preserving each of the handlebodies of the splitting setwise. When a manifold admits a unique Heegaard splitting of genus up to isotopy, we might define the genus- Goeritz group of the manifold without mentioning a specific splitting. For example, the -sphere, and lens spaces are known to be such manifolds from [17], [2] and [3].

It is natural to study the structures of Goeritz groups and to ask if they are finitely generated or presented, and so finding their generating sets or presentations has been an interesting problem. But the generating sets or the presentations of those groups have been obtained only for few manifolds with their splittings of small genus. A finite presentation of the genus- Goeritz group of the -sphere was obtained from the works of [8], [15], [1] and [4], and recently of each lens space in [5]. In addition, finite presentations of the genus- Goeritz groups of other lens spaces are given in [7]. Also a finite generating set of the genus- Goeritz group of the -torus was obtained in [10]. For the higher genus Georitz groups of the -sphere and lens spaces, it is conjectured that they are all finitely presented but it is still known to be an open problem.

In this work, we show that the genus- Goeritz group of is finitely presented, by giving its explicit presentation as follows.

###### Theorem 1.1.

The genus- Goeritz group has the presentation

 ⟨ ϵ ⟩⊕⟨ α | α2=1 ⟩⊕⟨ β,γ,σ | γ2=σ2=(γβσ)2=1 ⟩.

The generators , , and are illustrated in Figure 2 as orientation-preserving homeomorphisms of a Heegaard surface which can extend to homeomorphisms of the whole . The generator is the Dehn twist about the circle in the figure. In Section 2, we describe those generators in detail.

To find the presentation of , we generalize the method developed in [4]. We construct a tree on which the group acts such that the quotient of the tree by the action is a single edge, and then find the presentations of the stabilizer subgroups of the edge and each of end vertices.

Throughout the paper, we denote by the genus- Heegaard splitting of . That is, and are genus- handlebodies such that and . All disks in a handlebody are always assumed to be properly embedded and their intersections are transverse and minimal up to isotopy. In particular, if two disks intersect each other, then the intersection is a collection of pairwise disjoint arcs that are properly embedded in each disk. Finally, will denote a regular neighborhood of and the closure of for a subspace of a polyhedral space, where the ambient space will always be clear from the context.

## 2. Primitive disks in a handlebody

Recall that is a genus- handlebody in whose exterior is the handlebody . A non-separating disk in is called a reducing disk if there exists a disk in such that . The disk is also a reducing disk in . An essential disk in is called primitive if there exists an essential disk in such that meets in a single point. Such a disk is called a dual disk of , and is also primitive in with its dual disk . Primitive disks are necessarily non-separating. We remark that and (and and , respectively) are solid tori, which form a genus- Heegaard splitting of . It follows that, for a meridian disk of the solid torus , there exists a meridian disk of the solid torus satisfying . In particular, we can find such disks and so that they are disjoint from the -ball . Therefore, once we have a primitive disk with its dual disk , there exist reducing disks in and in disjoint from such that . See Figure 1.

We call a pair of disjoint, non-isotopic primitive disks in a primitive pair of , and if a disk in is a dual disk of each the two disks of the pair, then call it simply a common dual disk of the pair.

We first introduce six elements , , , , and of the group , which will turn out to form a generating set of . These elements can be described as orientation-preserving homeomorphisms of the surface which extend to homeomorphisms of the whole preserving each of and setwise. Fix a primitive pair of and a primitive pair of common dual disks of and . (The existence of such disks will be shown in Lemmas 2.5 and 2.6.) Figures 2 (a) and (b) illustrate these disks with the unique reducing disks in and in disjoint from them. Notice that . (The existence and uniqueness of such reducing disks will be shown in Lemma 2.4.)

In Figure 2 (a), the order two element of is described as a hyperelliptic involution of . The elements and described in Figure 2 (b) are half-Dehn twists about a separating loop in disjoint from and from . This loop meets each of and transversely at two points, and bounds a disk in each of and . The elements and have infinite order and satisfy . The order two element in Figure 2 (d) exchanges and while fixing each of and . Notice that and are the two unique common dual disks of the primitive pair disjoint from . (The uniqueness will be shown in Lemma 2.6.) In the same manner, the order two element in Figure 2 (c) exchanges and and fixes each of and . Also, and are the two unique common dual disks of the primitive pair disjoint from . Finally, the element is a Dehn twist about . Notice that all of those elements preserve each of and . In addition, we define and for later in the argument. The element has infinite order and preserves each of and , but sends to . The element has order two, and exchanges each of and , and and respectively.

In the remaining of the section, we develop several properties of primitive disks and reducing disks, which will be used in the next sections. The boundary circle of any essential disk in represents an element of , a free group of rank two. In particular, if is a reducing disk in , then represents the trivial element of . For primitive disks, we have the following intepretation, which is a direct consequence of [9].

###### Lemma 2.1.

Let be a non-separating disk in . Then is primitive if and only if represents a primitive element of .

Let and be essential disks in such that cuts into a -ball. Assign symbols and to and respectively after fixing orientations of and . Let be any oriented simple closed curve in such that intersects transversally and minimally. Then determines a word in terms of and which can be read off from the intersections of with and . Thus represents an element of the free group . In this set up, the following is a simple criterion for triviality and primitiveness of the elements represented by , which is found in Lemma 2.2 in [4] with its proof.

###### Lemma 2.2.

If a word determined by contains a sub-word of the form after a suitable choice of orientations, then represents a non-trivial, non-primitive element of .

The idea of the proof is that, once a word determined by contains , any word determined by is cyclically reduced and so nonempty, and any cyclically reduced word containing both and cannot represent a primitive element of .

Let and be essential disks in , and suppose that intersects transversely and minimally. Let be a disk cut off from by an outermost arc of in such that . We call such a an outermost sub-disk of cut off by . The arc cuts into two disks, say and . Then we have two essential disks and in which are isotopic to disks and respectively. We call and the disks from surgery on along the outermost sub-disk of cut off by . Observe that each of and has fewer arcs of intersection with than had, since at least the arc no longer counts.

Since and are assumed to intersect minimally, (and ) is isotopic to neither nor . In particular, if is non-separating, then the resulting disks and are both non-separating and they are not isotopic to each other because, after isotopying and away from , both of them are meridian disks of the solid torus cut off by , and the boundary circles and are not isotopic to each other in the two holed torus cut off by .

###### Lemma 2.3.

Let be a reducing disk in disjoint from , where is a primitive disk in and is a dual disk of . Let be any non-separating disk in which is not isotopic to .

1. If is disjoint from , then is a primitive disk, and hence is not a reducing disk.

2. If intersects , then is neither a reducing disk nor a primitive disk.

###### Proof.

Let be a reducing disk in such that . Then is also disjoint from . Let be the -holed sphere obtained by cutting along and . We note that is an arc in connecting two holes coming from .

(1) Suppose that is disjoint from . Consider first the case that is also disjoint from . Then is determined by an arc properly embedded in connecting the holes coming from and . That is, is the frontier of a regular neighborhood of the union of and the two holes connected by in . See Figure 3 (a). If intersects , then take the sub-arc of which connects and and whose interior is disjoint from . Then the band sum of and along this sub-arc is a non-separating disk, denoted by , in . We observe that , and is again a dual disk of and is disjoint from . See Fig 3 (b). We can repeat this process until we find a dual disk of disjoint from so that is disjoint from . Then intersects in a single point, which implies that is primitive.

Next, suppose that intersects . Let be any outermost sub-disk of cut off by . Then is an arc properly embedded in whose end points lie in a single hole coming from . The arc is determined by an arc, say , in connecting the holes coming from and each. Similarly to the case of the arc , the arc is the frontier of a regular neighborhood of the union of and the hole coming from which ends in.

If intersects , then we repeat the band sum constructions along the sub-arc of connecting and to find a dual disk of such that is disjoint from and from . Then the arc is also disjoint from . One of the disks from surgery on along is , and the other one, say , intersects in a single point. See Figure 4 (a). That is, is a primitive disk in with the dual disk . The disk is disjoint from , and further we have . We repeat the process to find a new primitive disk disjoint from , and also from , which has a dual disk disjoint from . Then we go back to the first case.

(2) Suppose that intersects . Let be an outermost sub-disk of cut off by . If intersects , the same argument of (1) for the sub-disk (instead of the disk in (1)) enables us to find a new primitive disk and its dual disk such that and are disjoint from and , and is disjoint from and intersects in a single point.

Giving symbols and to the oriented circles and , respectively, the boundary circle of determines a word in terms of and . In particular, there exists a word determined by containing which is determined by the sub-arc (after changing orientations if necessary). See Figure 4 (b). By Lemma 2.2, is neither a reducing disk nor a primitive disk. ∎

In the proof of Lemma 2.3, we see that if a non-separating disk in is not isotopic to , then represents a non-trivial element of . Thus the reducing disk is the unique non-separating disk in such that represents a trivial element of up to isotopy. The following is also a direct consequence of Lemma 2.3.

###### Lemma 2.4.

There exists a unique non-separating reducing disk in . A non-separating disk in is primitive if and only if is not isotopic to and disjoint from .

Of course, the same result we have for the reducing disk and a primitive disk in .

###### Lemma 2.5.

Any primitive pair has a common dual disk.

###### Proof.

Let be a primitive pair of , and let and be the unique reducing disks in and respectively such that . Any dual disk of is primitive in , and hence is disjoint from by Lemma 2.3. The primitive disk is disjoint from and , thus as in the proof of Lemma 2.3 (a), we can find a common dual disk of and by repeating the band sum constructions. ∎

###### Lemma 2.6.

Let be a common dual disk of a primitive pair of . Then there exist exactly two common dual disks, say and , of disjoint from . Further, intersects in a single arc.

###### Proof.

Let and be the unique reducing disks in and , respectively, such that , and let be the -holed sphere cut off by . Then and are disjoint arcs properly embedded in connecting the two holes coming from . The boundary circle of any common dual disk disjoint from also lies in and intersects each of and in a single point. Thus there exist exactly two such circles and illustrated in Figure 5, which bound two disks intersecting each other in a single arc. ∎

## 3. The complex of primitive disks

Let be an irreducible -manifold with compressible boundary. The disk complex of is a simplicial complex defined as follows. The vertices of are isotopy classes of essential disks in , and a collection of vertices spans a -simplex if and only if it admits a collection of representative disks which are pairwise disjoint. Let be the full subcomplex of spanned by the vertices of non-separating disks, which we will call the non-separating disk complex of . It is well known that and are both contractible. For example, see Theorems 5.3 and 5.4 in [12].

In particular, for a genus- handlebody , the disk complex is a -dimensional simplicial complex which is not locally finite. Further, we have a precise description of the non-separating disk complex of as follows. First, it is easy to verify that is -dimensional, and every edge of is contained in infinitely but countably many -simplices. Next, itself is contractible, and also the link of any vertex of is a tree each of whose vertices has infinite valency. This is a direct consequence of Theorem 4.2 in [4]. Figure 6 illustrates a portion of . We note that the entire disk complex of is constructed by attaching infinitely (but countably) many -simplices to each edge of , where each of the new vertices is represented by an essential separating disk in .

Given a genus- Heegaard splitting of , we define the primitive disk complex to be the full sub-complex of spanned by the vertices of primitive disks in . By Lemma 2.4, is the link in of the vertex represented by the unique reducing disk in (see Figure 6), which implies our key result.

###### Theorem 3.1.

The primitive disk complex is an infinite tree each of whose vertices has infinite valency.

We note here that in [4], [5], [6], [7] and [11], primitive disk complexes are defined in the same way under various settings, and they are used to obtain presentations of Goeritz groups or their generalizations of given manifolds.

Obviously, the primitive disk complex of is isomorphic to . Given a primitive disk of , define the sub-complex of to be the full sub-complex spanned by the vertices of dual disks of . Similarly, is the full sub-complex of spanned by the vertices of common dual disks of a primitive pair of .

###### Theorem 3.2.

The sub-complexes and are both infinite trees. Each vertex of has infinite valency, and each vertex of has valency exactly two.

###### Proof.

It is clear that is an infinite sub-complex of the tree whose vertices have infinite valency. By Lemmas 2.5 and 2.6, the sub-complex is also an infinite sub-complex of the tree , and each vertex of has valency two. Thus it remains to show that both of and are connected.

Let and be any two non-isotopic dual disks of . If is disjoint from , then the two vertices represented by them are joined by a single edge. If intersects , then it is easy to see that one of the disks from surgery on along an outermost sub-disk of cut off by is the unique reducing disk in , and the other one, say is still a dual disk of . Notice that is disjoint from and . By repeating surgery construction, we find a finite sequence of dual disks of from to , which realizes a path in joining the two vertices of and . Thus is connected. The connectivity of also can be shown in a similar way, by considering surgery on a common dual disk. ∎

###### Remark 3.3.

It is interesting to compare the sub-complexes of common dual disks for genus- Heegaard splittings of the -sphere , the lens spaces , , and . The following results are known from [4] and [6]. In the case of , there exist infinitely many common dual disks, and each two of them intersect each other. Thus we see that is a collection of infinitely many vertices. For the lens space , if , then there exist exactly two common dual disks disjoint from each other. Thus is a single edge. If , then there exists a unique common dual disk, and so is a single vertex. For the other lens spaces, is either a vertex or the empty set depending on the choice of , and both exist infinitely many.

## 4. The genus two Goeritz group of S2×S1

In this section, we prove the main theorem, Theorem 1.1. We know that the primitive disk complex is a tree from Theorem 3.1, which is the link in of the vertex of the unique reducing disk . Let be the barycentric subdivision of , which is a bipartite tree. In Figure 7, the black vertices of are the vertices of , while the white ones are the barycenters of the edges of . The black vertices correspond to the primitive disks, and the white ones to the primitive pairs in consisting of the primitive disks representing the two adjoining black vertices. For convenience, we will not distinguish disks (or pairs of disks) and homeomorphisms from their isotopy classes.

The group acts on the tree as a simplicial automorphism, and further we have the following.

###### Lemma 4.1.

The group acts transitively on each of the collections of black vertices and of white vertices of .

###### Proof.

The baricentric subdivision of is a tree and so connected. Thus, given any two primitive disks in , there exists a sequence of primitive disks from one to another in which any two consecutive disks form a primitive pair in . Therefore, to see the transitivity of the action of on the black vertices, it suffices to find an element of sending a disk to a disk for any primitive pair in . Such an element does exist, since we have the element in Section 2. (The element sends to for the primitive pair in . In fact, exchanges and .)

Next, since is connected again, given any two primitive pairs in , there exists a sequence of primitive pairs from one to another in which any two consecutive pairs share a common primitive disk. Thus, to see the transitivity of the action of on the white vertices, it suffices to find an element of sending a disk to a disk and fixing a disk for any two primitive pairs and in .

Choose any common dual disks and of the pairs and respectively. Since is also a tree by Theorem 3.2, there exists a sequence of dual disks of in which any two consecutive disks form a primitive pair in . Let be an element of sending to and fixing . Such an element does exist, since we have the element in Section 2. (In the description, the element fixes and exchanges two disjoint dual disks and of .) Then the composition sends to and fixes . We observe that and are primitive pairs and is a common dual disk of them. Then there exists an element of sending to and fixing . (The element will be a power of the element in Section 2, when we take , in and in instead of , in and in respectively.) Then the composition sends to and fixes as desired. ∎

From the lemma, we see that the quotient of by the action of is a single edge whose one vertex is black and another one white. By the theory of groups acting on trees due to Bass and Serre [16], the group can be expressed as the free product of the stabilizer subgroups of two end vertices with amalgamated stabilizer subgroup of the edge.

Throughout the section, will denote the subgroup of of elements preserving each of setwise, where will be (the isotopy classes of) disks or union of disks in or in . Then, given a primitive pair of , the subgroups , and are the stabilizer subgroups of a black vertex, of a white vertex and of the edge of joining them respectively. Thus the Goeritz group is the free product of and amalgamated by .

We first find presentations of the three stabilizer subgroups , and . As mentioned above, the basic idea is to describe the generators of the subgroups as homeomorphisms of the surface preserving the boundary circles of some primitive disks and reducing disks in and .

###### Lemma 4.2.

The stabilizer subgroup has the presentation

 ⟨ ϵ ⟩⊕⟨ α | α2=1 ⟩⊕⟨ β,γ | γ2=1 ⟩.
###### Proof.

By Theorem 3.2, is the sub-tree of the tree spanned by the vertices of the dual disks of . The barycentric subdivision of , which we denote by , is a bipartite tree, each of whose vertices corresponds to either a dual disk of or a pair of disjoint dual disks of . The subgroup acts on , and its quotient is again a single edge. Thus, fixing a primitive pair of dual disks of as in Figure 2, the subgroup can be expressed as the free product of stabilizer subgroups and amalgamated by .

First, consider the subgroup . This group also preserves each of and , and so . This group is generated by the elements , and . Thus we have a presentation of as .

Next, the subgroup is generated by , and , and so it has the presentation . In a similar way, the index- subgroup of has the presentation . Observing , we have the desired presentation of . ∎

###### Lemma 4.3.

The stabilizer subgroup has the presentation

 ⟨ ϵ ⟩⊕⟨ α | α2=1 ⟩⊕⟨ σ,τ | σ2=1, (τσ)2=1 ⟩.
###### Proof.

By Theorem 3.2 again, the full subcomplex of spanned by the vertices of common dual disks of the pair is a tree. Denote by the barycentric subdivision of . Then the quotient of by the action of is a single edge. Thus, fixing a primitive pair of common dual disks of and , the subgroup can be expressed as the free product of and amalgamated by .

Similarly to the case of in the proof of Lemma 4.2, is generated by , and , and hence has the presentation . Next, is a subgroup of , which does not contain since does not preserve . Thus has the presentation . Finally, recall the order two element of exchanging and , and and respectively. The subgroup is the extension of by , so it has the presentation , and so the desired presentation of is obtained. ∎

###### Lemma 4.4.

The stabilizer subgroup has the presentation

 ⟨ ϵ ⟩⊕⟨ τ ⟩⊕⟨ α | α2=1 ⟩.
###### Proof.

The presentation of is obtained directly from the fact that is the extension of by the order-2 element exchanging and . ∎

###### Remark 4.5.

The subgroup acts on the tree . Although this action is transitive on the collections of vertices of common dual disks and of pairs of common dual disks of and respectively, the quotient is not a single edge. In fact, one can verify that the quotient is a single loop and can obtain the same presentation of to the above using this loop.

Combining Lemmas 4.2, 4.3 and 4.4 and using , we obtain Theorem 1.1.

Acknowledgments. The second-named author wishes to thank Università di Pisa for its kind hospitality. The authors are grateful to the referee for helping them to improve the presentation.

## References

• [1] E. Akbas, A presentation for the automorphisms of the -sphere that preserve a genus two Heegaard splitting, Pacific J. Math. 236 (2008), no. 2, 201–222.
• [2] F. Bonahon, Difféotopies des espaces lenticulaires, Topology 22 (1983), no. 3, 305–314.
• [3] F. Bonahon, J.-P. Otal, Scindements de Heegaard des espaces lenticulaires, Ann. Sci. École Norm. Sup. (4) 16 (1983), 451–466.
• [4] S. Cho, Homeomorphisms of the -sphere that preserve a Heegaard splitting of genus two, Proc. Amer. Math. Soc. 136 (2008), no. 3, 1113–1123.
• [5] S. Cho, Genus two Goeritz groups of lens spaces, Pacific J. Math. 265 (2013), 1–16.
• [6] S. Cho, Y. Koda, Primitive disk complexes for lens spaces, arXiv:1206.6243.
• [7] S. Cho, Y. Koda, Connected primitive disk complexes and Goeritz groups of lens spaces, in preparation.
• [8] L. Goeritz, Die Abbildungen der Berzelfläche und der Volbrezel vom Gesschlect , Abh. Math. Sem. Univ. Hamburg 9 (1933), 244–259.
• [9] C. McA. Gordon, On primitive sets of loops in the boundary of a handlebody, Topology Appl. 27 (1987), no. 3, 285–299.
• [10] J. Johnson, Automorphisms of the three-torus preserving a genus-three Heegaard splitting, Pacific J. Math. 253 (2011), no. 1, 75–94.
• [11] Y. Koda, Automorphisms of the -sphere that preserve spatial graphs and handlebody-knots, arXiv:1106.4777.
• [12] D. McCullough, Virtually geometrically finite mapping class groups of -manifolds, J. Differential Geom. 33 (1991), no. 1, 1–65.
• [13] R.P. Osborne, H. Zieschang, Primitives in the free group on two generators, Invent. Math. 63 (1981), no. 1, 17–24.
• [14] D. Rolfsen, Knots and links, Mathematics Lecture Series, No. 7. Publish or Perish, Inc., 1976. ix+439 pp.
• [15] M. Scharlemann, Automorphisms of the -sphere that preserve a genus two Heegaard splitting, Bol. Soc. Mat. Mexicana (3) 10 (2004), Special Issue, 503–514.
• [16] J. Serre, Trees, Springer-Verlag, 1980. ix+142 pp.
• [17] F. Waldhausen, Heegaard-Zerlegungen der -Sphäre, Topology 7 (1968), 195–203.
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