Cubical geometry in the polygonalisation complex

Cubical geometry in the polygonalisation complex

Mark C. Bell Department of Mathematics, University of Illinois, Urbana, IL mcbell@illinois.edu Valentina Disarlo Mathematical Sciences Research Institute, Berkeley, CA v.disarlo@gmail.com  and  Robert Tang Department of Mathematics, University of Oklahoma, Norman, OK rtang@math.ou.edu
Abstract.

We introduce the polygonalisation complex of a surface, a cube complex whose vertices correspond to polygonalisations. This is a geometric model for the mapping class group and it is motivated by works of Harer, Mosher and Penner. Using properties of the flip graph, we show that the midcubes in the polygonalisation complex can be extended to a family of embedded and separating hyperplanes, parametrised by the arcs in the surface.

We study the crossing graph of these hyperplanes and prove that it is quasi-isometric to the arc complex. We use the crossing graph to prove that, generically, different surfaces have different polygonalisation complexes. The polygonalisation complex is not CAT(0), but we can characterise the vertices where Gromov’s link condition fails. This gives a tool for proving that, generically, the automorphism group of the polygonalisation complex is the (extended) mapping class group of the surface.

Figure 1. The polygonalisation complex of a hexagon .

1. Introduction

In this paper we study the polygonalisation complex of a surface with marked points. This is a cube complex encoding the combinatorics of the polygonalisations of , that is, the multiarcs that decompose into polygons. The polygonalisation complex contains (the barycentric subdivision of) the flip graph as a subcomplex. It can also be regarded as the barycentric subdivision of the contractible CW-complex naturally associated to the Ptolemy groupoid of which appears in quantum Teichmüller theory [8] [19].

The mapping class group acts geometrically on . However, in general, is not a CAT(0) group [13, Theorem 4.2] and so is not a CAT(0) cube complex. We show that, in spite of this, has many of the properties of CAT(0) cube complexes. In particular, it has a rich hyperplane structure that is closely related to the arc complex.

{thrm:pol_sageev}

There is a natural one-to-one correspondence between the hyperplanes of and the arcs on . Moreover, for each arc , the corresponding hyperplane is embedded, two-sided and separates into two connected components: and .

One can encode the combinatorics of the hyperplanes in via its crossing graph : this has a vertex for every hyperplane in , and two hyperplanes are connected via an edge if and only if they cross.

{prop:equivalent_crossing}

Let and be the hyperplanes of corresponding to arcs . Then and cross if and only if and are disjoint and do not bound a (folded) triangle.

Together, Theorem 3.8 and Proposition 3.12 show that there is a natural embedding of into the arc graph . This embedding is in fact a quasi-isometry (Corollary 3.16). In particular, when the arc graph is Gromov hyperbolic [15, Theorem 20.2], hence the crossing graph is also. Proposition 3.12 also enables us to characterise the edges of the arc graph that do not appear in the crossing graph (Lemma 3.17). We show that can be recovered from the combinatorics of . Applying rigidity results for arc graphs [12] [5] [14] we then obtain:

{thrm:pol_isom}

For all but finitely many pairs of surfaces, the complexes and are isomorphic if and only if and are homeomorphic.

We rephrase Gromov’s link condition in terms of curves on (Definition 4.1). We use this to extract the number of components of a given polygonalisation from the local combinatorics of .

{cor:deficiency_combinatorics}

For each , there is a combinatorial criterion that characterises the vertices of corresponding to polygonalisations with exactly arcs. This gives a method for proving the rigidity of the polygonalisation complex via flip graph rigidity. {thrm:pol_aut} For all but finitely many surfaces, the natural homomorphism

is an isomorphism.

We list the exceptions to these theorems in Appendix C. Additionally, we highlight these results do not follow from [1, Theorem 1.1] since does not satisfy the required rigidity axioms.

2. Preliminaries

Let be a connected, compact, orientable surface with a finite non-empty set of marked points. We assume that each boundary component (if any) contains at least one marked point. When , such surfaces are also known in the literature as ciliated [7, Section 2]. Let

where is the genus of , is the number of marked points in its interior, is the number of boundary components and is the number of marked points on .

To avoid pathologies, from now on we will require that . The exceptional surfaces, for which , are listed and discussed in Appendix C.

2.1. Objects

We recall some standard objects that will appear throughout:

2.1.1. Mapping class group

The (extended) mapping class group is the group of homeomorphisms of relative to the marked points up to isotopy. We allow mapping classes to reverse orientation, permute the marked points and permute the boundary components of .

2.1.2. Triangulations

An (essential) arc on is an embedded arc connecting marked points up to isotopy (relative to the set of marked points). Such arcs are not null-homotopic and are not boundary-parallel, that is, they do not cut off a monogon (Figure 1(a)) or cut off a bigon together with part of (Figure 1(b)) respectively. We write for the (geometric) intersection number of and . Arcs and have disjoint interiors, which we refer to simply as being disjoint, if and only . A multiarc is a set of distinct and pairwise disjoint arcs. For example, see Figure 3.

((a)) A monogon arc.

((b)) A bigon arc.
Figure 2. Inessential arcs in .

Figure 3. A multiarc on .

The set of multiarcs is a poset with respect to inclusion. An (ideal) triangulation of is a maximal multiarc. An Euler characteristic argument shows that every triangulation has arcs and faces. Each complementary region of a triangulation is a triangle with vertices on the marked points of . Triangles have embedded interior, but their boundary may be non-embedded. In particular, triangles can be folded as shown in Figure 4.

Figure 4. A folded triangle.

2.1.3. The flip graph

The flip graph has a vertex for each triangulation. Triangulations are connected via an edge (of length one) if and only if they differ by a flip. This move consists of replacing the diagonal of a quadrilateral inside the triangulation with the other diagonal, as shown in Figure 5.

Any arc of a triangulation is either flippable or appears as the arc in Figure 4. In the latter case, the arc is flippable after first flipping the arc . Thus for any arc there is a triangulation in which is flippable.

Flip

Figure 5. Flipping the arc of a triangulation.

The flip graph appears implicitly in works of Harer [10], Mosher [16] and Penner [17]. It is connected [10] and the mapping class group acts on it geometrically, that is, properly, cocompactly and by isometries. Thus by the Švarc–Milnor Lemma [4, Proposition I.8.19] this graph is quasi-isometric to . The geometry of the flip graph was recently studied by the second author and Parlier in [6]. By [14, Theorem 1.2] and [2, Theorem 1.1], we have:

Theorem 2.1.

The natural homomorphism

is an isomorphism. ∎

2.1.4. The arc graph

The arc graph has a vertex for each arc on . Arcs are connected via an edge (of length one) if and only if they are disjoint. The arc graph appeared first in work of Harer [10]. The arc graph of can be naturally extended to a flag simplicial complex called the arc complex of . The one-skeleton of the dual of the arc complex is the flip graph.

Theorem 2.2.

The graphs and are isomorphic if and only if and are homeomorphic.

Proof.

Every isomorphism between and induces an isomorphism between and . If neither nor is the complement of a (possibly empty) multiarc on the four-times marked sphere or the twice-marked torus then the result follows from [2, Theorem 1.4]. Otherwise:

  • If then the result follows from [14, Theorem 1.1].

  • If then and . Since maximal complete subgraphs of have exactly vertices, it follows that and cannot be isomorphic.

  • If then the result follows from [5, Theorem 1.1].

In any case, the result holds. ∎

The mapping class group acts on cocompactly and by isometries but not properly. Generally, the automorphism group of is isomorphic to by [12, Theorem 1.2] and [5, Theorem 1.2].

2.2. The polygonalisation complex

A polygonalisation (or ideal cell decomposition) is a multiarc such that each complementary region of is a polygon, a topological disk with at least three sides and no marked points in its interior. These are dual to fat graphs (also known as ribbon graphs) [18, Section 4.2].

Remark 2.3.

If is a polygonalisation and is a multiarc such that then is a polygonalisation. Additionally, suppose that and are polygonalisations:

  • If then is a polygonalisation.

  • If then is a polygonalisation.

Definition 2.4.

The polygonalisation complex is a cube complex in which vertices correspond to polygonalisations. Two polygonalisations are connected by an edge (of length one) if and only if they differ by a single arc, that is, their symmetric difference is a single arc. Inductively, a –cube is added whenever its –skeleton appears. See Appendix A for examples.

Since any polygonalisation can be extended to a triangulation and is connected, is also connected. Again, acts geometrically on and so they are quasi-isometric.

We now establish some definitions and notation that will be used throughout.

Definition 2.5.

For an arc , its stratum is the subcomplex of induced by the subset . Similarly, define to be the subcomplex induced by .

Definition 2.6.

An arc is removable from if and is also a polygonalisation. An arc addable to if and is a polygonalisation.

Observe that is removable if and only if its interior meets two distinct polygons of . Let (resp. ) be the subcomplex of (resp. ) induced by the polygonalisations in which is removable (resp. addable).

Notation 2.7.

For , write if and .

For an edge in let denote the arc that appears in but not (or vice versa).

Remark 2.8.

For any edge in , observe that if and only if and (or vice versa).

3. The cube complex structure of

We describe some of the key properties of the cube complex structure of . We begin by recalling some of the standard terms for cube complexes. For a complete reference see [21].

An –cube is the Euclidean cube . For every a face is a subspace obtained restricting coordinates to . A –face is also a –cube. A cube complex is a cell complex obtained by gluing cubes along their faces by isometries.

The link of a vertex of is the complex induced on a small sphere about by . A flag complex is a simplicial complex where vertices span an –simplex if and only if they are pairwise adjacent. A cube complex is non-positively curved if it satisfies Gromov’s link condition: The link of each vertex is a flag complex [9, Section 4.2.C]. If is non-positively curved and simply connected then is CAT(0).

A midcube is a subspace of a cube obtained by restricting exactly one coordinate to . A hyperplane is the union of midcubes that meet parallel edges. The carrier of a hyperplane is the union of all the cubes such that intersect in a midcube. Sageev showed that CAT(0) cube complexes come equipped with a family of “nice” hyperplanes.

Theorem 3.1 ([20, Theorem 1.1]).

If is a CAT(0) cube complex then:

  1. each midcube lies in an embedded hyperplane;

  2. every hyperplane is two-sided, that is, ;

  3. every hyperplane separates , that is, consists of two connected components and called halfspaces;

  4. every hyperplane is a CAT(0) cube complex; and

  5. every hyperplane and its carrier are convex in .

Despite the fact that generically is not CAT(0), we show that Properties 1, 2 and 3 still hold for (Theorem 3.8). However, generically, Properties 4 and 5 do not hold for . For example, see Table 2 and in Table 1 respectively.

3.1. The cubes of

The following lemma will also be useful in Section 4.

Lemma 3.2 (Square lemma).

Suppose that is an embedded –cycle. If then . Moreover, .

Proof.

Suppose that instead . By considering the cardinalities of the ’s, we see that . Since , by Remark 2.3 we have . Since , we deduce . Similarly, contradicting the fact that this cycle is embedded. Furthermore, if then counting the number of components shows that . Again, this contradicts this cycle being embedded. ∎

The cube graph is the 1–skeleton of the standard –cube. In particular, an embedded –cycle is a . By induction on , the square lemma shows that any embedded cube graph in contains a unique source and a unique sink . That is, for any polygonalisation in we have that .

For polygonalisations , let

In fact these are all the cubes that appear in :

Lemma 3.3 (Characterisation of cubes).

If then forms the vertex set of an –cube in where . Conversely, if is an embedded –cube in then the vertex set of is . ∎

When , this characterisation of cubes enables us to apply Penner’s argument for the contractibility of the fatgraph complex [17, Theorem 2.5] to . Hence, in this case the polygonalisation complex is contractible.

3.2. The hyperplanes of

Let denote the equivalence relation on the edges of generated by if and only if and are opposite edges of some square in . We say that and are parallel if .

Definition 3.4 ([20, Definition 1.4]).

A hyperplane is the set of midcubes of that meet edges in , for some equivalence class of edges. We write for the set of all hyperplanes of .

By the square lemma, if edges are parallel then . Thus, the map descends to a well-defined map . This map is surjective: for any there is a triangulation from which is removable and so . One of the aims of this section is to prove that this map is also injective. To achieve this we shall require some technical results.

Lemma 3.5 (Pentagon detour lemma).

Let and be adjacent triangulations in and . Suppose there is an arc that is removable from and but not from . Then there is a path such that is removable from each and from each .

Proof.

Let . Since is not removable from but is removable from and , it must appear as opposite sides of the unique square of . Since has at least three triangles, this square has another side that is removable from . The arcs and are therefore a pair of chords of a pentagon in , as shown in Figure 6. The path is the one obtained going around the natural five-cycle in defined by and in the other direction. ∎

Figure 6. A detour around a pentagon to stay in .
Proposition 3.6.

For each arc , the subcomplex is connected.

Proof.

Suppose are polygonalisations and let and be triangulations containing and respectively. Note that is removable from and . Furthermore, and are in the same path-component of (and likewise for and ). Thus to show is connected it suffices to show that and are in the same path-component.

There is a path in such that [6, Corollary 2.15]. Call good if can be removed from it and bad otherwise. We describe how to modify this path to avoid any bad triangulations.

Suppose that is the first bad triangulation in this path. Note that since both and are good. Let and be the arcs of that can be flipped to obtain and respectively. There are three possibilities to consider:

  • If is supported on two triangles then, since cuts off a once-marked monogon, . Thus and so we simplify our path by removing and .

  • If is supported on three triangles then they fill a pentagon in . Thus we may replace with the good triangulations as shown in Figure 6(a).

  • If is supported on four triangles then the flips commute. Thus we may replace with the good triangulation as shown in Figure 6(b).

Performing any of these modifications will reduce the number of bad triangulations. Hence, by induction, we may assume is removable from each along this path. Taking the barycentric subdivision gives a path in , where . Replacing subpaths with pentagon detours if necessary, we may assume that is also removable from each . This yields a path in connecting to as required. ∎

((a)) When flips overlap.

((b)) When flips are disjoint.
Figure 7. Diverting a path around a bad triangulation.

Observe that are adjacent if and only if they form a square with . Hence, by the above proposition, is also connected.

Lemma 3.7.

Let be an arc. Then the subcomplexes and in are both connected.

Proof.

Suppose that and . For any edge-path in from to , the arc must be removed from a polygonalisation at some point along this path. Hence such a path in must cross . Therefore there is a path within from to and a path within from to . Since and are both connected, and are too. ∎

Theorem 3.8.

The map is a bijection. Moreover, for each arc , the corresponding hyperplane is embedded, two-sided and separates into two connected components: and .

Proof.

To show that the map is a bijection we must prove that edges in are parallel if and only if . As described above, the forwards direction holds trivially by the square lemma.

For the backwards direction, suppose that are edges of such that . By Remark 2.8, we deduce that and for some and . By Proposition 3.6 there is a path from to in . Now can be removed from each polygonalisation in to obtain a parallel path from to in . This gives a sequence of squares from which we can deduce that and are parallel.

It is easy to verify that hyperplanes are embedded. If a hyperplane self-intersects then a self-intersection must occur in some square. Then all four edges in this square must correspond to adding/removing the same arc, which is impossible.

As in the proof of Lemma 3.7, any path from a vertex in to a vertex in must cross , hence separates and . By Lemma 3.7, these are both connected and so separates into and .

Finally, each hyperplane must be two-sided since it is separating. ∎

Corollary 3.9.

Let be an arc. Then the hyperplane separates polygonalisations if and only if . In particular, there are at most hyperplanes separating any given pair of polygonalisations. ∎

Remark 3.10.

The hyperplanes of cannot self-osculate, but they can interosculate. For example, see in Table 1.

3.3. The crossing graph of

Definition 3.11.

Two hyperplanes and cross, denoted , if

are all non-empty.

There are several ways of characterising when hyperplanes cross.

Proposition 3.12.

The following are equivalent:

  1. The hyperplanes and cross.

  2. The arcs and are distinct and disjoint but do not form a folded triangle.

  3. There is a triangulation containing such that is a polygonalisation.

Proof.

We follow a cycle of implications:

: There is a polygonalisation containing and and so these arcs must be disjoint. If and form a folded triangle then every polygonalisation that contains must contain or vice versa. Therefore either or is empty and so and do not cross.

: Since and do not form a folded triangle, there is a triangulation containing and in which both are removable. Let . Suppose that is not removable from . Then, as in the proof of Lemma 3.5, a polygon of is a square, two sides of which are . Since has at least three triangles, this square has another side that is removable from . We flip in to obtain a new triangulation in which both and are simultaneously removable.

: The polygonalisations , , and show that the four sets required by Definition 3.11 are all non-empty. ∎

Definition 3.13.

The crossing graph is the graph with a vertex for each hyperplane in . Two hyperplanes are connected via an edge (of length one) if and only if .

By Proposition 3.12, we immediately deduce that the crossing graph embeds into the arc graph . To get control over this embedding we will need to consider the paths in without folds, that is, the ones in which no consecutive pair of arcs form a folded triangle.

Lemma 3.14.

If then there is a geodesic

in that is without folds.

Proof.

Let

be a geodesic in from to .

Suppose that form a folded triangle for some . Then either or , which contradicts this path being a geodesic. Hence we need only consider folded triangles formed by and . We show that if the former occurs then there are arcs such that

is a geodesic without folds. We will take care to ensure that and are both arcs, that is, that they are not null-homotopic nor boundary-parallel. Similarly we can replace the end of the geodesic if forms a folded triangle.

Now note that if forms a folded triangle then must cut off a once-marked monogon. Furthermore, must have at least one endpoint on , the inner marked point of this monogon, since it must intersect .

If has exactly one endpoint on then let and be as shown in Figure 7(a). If or is null-homotopic then , which contradicts this path being a geodesic. If and are both boundary-parallel then , which is again a contradiction. Hence without loss of generality is an arc. Since this arc is disjoint from and and does not form a folded triangle with either, forms the required geodesic without folds.

On the other hand, if has two endpoints on , we construct by surgery as in Figure 7(b). We note that cannot be null-homotopic. If it were then cuts off a once-marked monogon and so must also be disjoint from as it is disjoint from . Again, this contradicts this path being a geodesic. Hence we need only consider the case in which is boundary-parallel, since otherwise is an arc and is a geodesic without folds.

If is boundary-parallel then must cut off an annulus with one marked point on each boundary component. Since it must meet , the arc must be contained in the annulus cut off by . However, as is essential, there is an arc in the other connected component with one endpoint on . By construction does not cut off an annulus and it is disjoint from . Hence is a geodesic and has a single endpoint on . Thus we can repeat the above argument to find a new arc such that

is a geodesic without folds. ∎

((a)) When has one end on .

((b)) When has both ends on .
Figure 8. A detour to avoid a once-marked monogon.

By Proposition 3.12, we have . On the other hand, Lemma 3.14 shows that if then . Thus, on the large scale, the natural map preserves distances. Furthermore, from Proposition 3.12, if then . The final possibility is the following:

Lemma 3.15.

If then .

Proof.

Suppose that is a geodesic. There are now three cases to consider. First, if this geodesic is without folds then is a geodesic in and so . Second, if and are both folded triangles then necessarily and both cut off once-marked monogons. Since , there are disjoint arcs and such that

and the path in is without folds. Hence by Proposition 3.12 it pulls back to a path in and so . Third, if the geodesic contains a unique folded triangle then without loss of generality it is formed by . In this case must cut off a once-marked monogon (otherwise and would be disjoint). Hence again there is an arc , disjoint from both and , such that is a path without folds in . Again this pulls back to a path in and so . ∎

Corollary 3.16.

The map is a –quasi-isometry. ∎

Hence has the same large scale geometry as . For example, when , Masur–Schleimer showed that is hyperbolic [15, Theorem 20.2]. Hensel–Przytycki–Webb later showed that in this case in fact every geodesic triangle in has a –centre [11, Theorem 1.2] and so the same is true for .

For ease of notation, for a hyperplane let denote the subgraph of induced by

We will now use the crossing graph to prove that, generically, different surfaces have different polygonalisation complexes.

Lemma 3.17.

Arcs and form a folded triangle, as shown in Figure 4, if and only if .

Proof.

We use Proposition 3.12 repeatedly to determine whether one hyperplane lies in the link of another.

Suppose and form a folded triangle. If then and are distinct, disjoint and do not form a folded triangle. Thus and must also be disjoint and distict. Additionally, since only forms a folded triangle with we have that and do not form a folded triangle and so . On the other hand, let be a triangulation without folded triangles. Then there is an arc such that but . Thus but and so .

Suppose does not form a folded triangle with . Let be a triangulation with one or zero folded triangles depending on whether does or does not cut off a once-marked monogon respectively. There are now three cases to consider:

  1. If then . Therefore, since does not form a folded triangle with , we have that but .

  2. If and there is an arc such that and are distinct and do not form a folded triangle and then but .

  3. If and any arc of that intersects is either or forms a folded triangle with then and are as shown in Figure 9. Hence there is an arc which appears as exactly one side of the unique square of . Flipping in gives an arc . Since does not cut off a once-marked monogon, is disjoint from and we have that but .

In any case, the inclusion does not hold. ∎

((a)) When meets a folded triangle.

((b)) When only meets .
Figure 9. When only meets and arcs that form a folded triangle with .
Theorem 3.18.

The cube complexes and are isomorphic if and only if and are homeomorphic.

Proof.

If and are isomorphic as cube complexes then and are isomorphic graphs. By Proposition 3.12 and Lemma 3.17, adding the edges

to produces a graph isomorphic to . Similarly, by adding the edges

to we obtain a graph isomorphic to . This rule for adding edges is purely combinatorial, that is, it depends only on the graph structure. Since and are isomorphic, we deduce . By rigidity of the arc graph (Theorem 2.2), this in turn implies that is homeomorphic to .

The reverse direction is straightforward. ∎

In fact by examining the possible cases shown in Appendix C, we see that this theorem also holds when .

4. Failure of Gromov’s link condition

In general, contains many vertices that fail Gromov’s link condition [9, Section 4.2.C], which we rephrase as follows. When this is the only reason why fails to be CAT(0), since in this case is contractible.

Definition 4.1.

A positive curvature system (based at ) is a set of edges in incident to such that:

  • every subset of of size is contained in an embedded –cube, and

  • the set is not contained in an embedded –cube.

See Figure 10.

Figure 10. A positive curvature system based at .

Positive curvature systems are useful as they allow us to determine the direction of inclusion of the edges involved.

Proposition 4.2 (Positive curvature criterion).

If is a positive curvature system where then .

Proof.

Let . For a contradiction, suppose that . Then is disjoint from all arcs of . By assumption, lie in an embedded –cube that contains . Recall that has a unique source and a unique sink with respect to inclusion and that . Observe that for all , the edges and are contained in a common –cube since is a positive curvature system. This means that