A note on the curve complex of the 3-holed projective plane

TL;DR

This paper studies the curve complex of a 3-holed projective plane, showing it can be exhausted by finite rigid sets, its automorphism group matches the mapping class group, and it is quasi-isometric to a tree.

Contribution

It establishes a finite rigid set exhaustion of the curve complex and proves the automorphism group equals the mapping class group, also showing the complex's quasi-isometry to a tree.

Findings

Curve complex admits an exhaustion by finite rigid sets

Automorphism group is isomorphic to the mapping class group

Curve complex is quasi-isometric to a simplicial tree

Abstract

Let $S$ be a projective plane with $3$ holes. We prove that there is an exhaustion of the curve complex $\mathcal{C}(S)$ by a sequence of finite rigid sets. As a corollary, we obtain that the group of simplicial automorphisms of $\mathcal{C}(S)$ is isomorphic to the mapping class group $\mathrm{Mod}(S)$. We also prove that $\mathcal{C}(S)$ is quasi-isometric to a simplicial tree.

Equations

$${\mathcal{T}}_n^0=\{T\in {\mathcal{T}}^0\mid d_{\mathcal{T}}(T,T_0)\le n\}.$$

$${\mathcal{Y}}_{n+1}=\bigcup _{T\in {\mathcal{T}}_n^0}{T}^{\ast }.$$

$$d_{\mathcal{C}}(x,y)=2d_{\mathcal{D}}(\varphi (x),\varphi (y))$$

Full text

Supported by grant 2015/17/B/ST1/03235 of National Science Centre, Poland.

1. introduction

The complex of curves $\mathcal{C}(S)$ of a surface $S$, first introduced by Harvey [6], is the simplicial complex with $k$-simplices representing collections of homotopy classes of $k+1$ non-isotopic disjoint simple closed curves in $S$. In this paper we let $S=N_{1,3}$ be the $3$-holed projective plane. Then $\mathcal{C}(S)$ is one-dimensional and its combinatorial structure was described by Scharlemann [16]. The first purpose of this note is to prove some rigidity results about $\mathcal{C}(N_{1,3})$, which are known for most surfaces, but have not been proved in the literature in this particular case. The second purpose is to show that $\mathcal{C}(N_{1,3})$ is quasi-isometric to a simplicial tree.

By the celebrated theorem of Ivanov [10], Korkmaz [11] and Luo [12], the group $\mathrm{Aut}(\mathcal{C}(S))$ of simplicial automorphisms of $\mathcal{C}(S)$ for orientable surface $S$ is, with a few well understood exceptions, isomorphic to the extended mapping class group $\mathrm{Mod}^{\pm}(S)$. A stronger version of this result, due to Shackleton [17], says that every locally injective simplicial map from $\mathcal{C}(S)$ to itself is induced by some element of $\mathrm{Mod}^{\pm}(S)$ (simplicial map is locally injective if its restriction to the star of every vertex is injective). Analogous results for nonorientable surfaces were proved by Atalan-Korkmaz [3] and Irmak [8], omitting the case of $N_{1,3}$.

Aramayona and Laininger introduced in [1] the notion of a rigid set. It is a subcomplex $X\subset\mathcal{C}(S)$, with the property that every locally injective simplicial map $X\to\mathcal{C}(S)$ is induced by some homeomorphism of $S$. They constructed in [1] a finite rigid set in $\mathcal{C}(S)$, for every orientable surface $S$, and in [2] they proved that there is an exhaustion of $\mathcal{C}(S)$ by a sequence of finite rigid sets.

Let $S=N_{g,n}$ be a nonorientable surface of genus $g$ with $n$ holes. Ilbira and Korkmaz [7] constructed a finite rigid set in $\mathcal{C}(S)$ for $g+n\neq 4$. Irmak [9] proved that $\mathcal{C}(S)$ can be exhausted by a sequence of finite rigid sets for $g+n\geq 5$ or $(g,n)=(3,0)$. In this paper we show that the main results of Ilbira-Korkmaz [7] and Irmak [9] are true also for $N_{1,3}$.

Theorem 1.1**.**

There exists a sequence $\mathcal{Y}_{1}\subset\mathcal{Y}_{2}\subset\cdots\subset\mathcal{C}(N_{1,3})$ such that:

• (1)

$\mathcal{Y}_{i}$* is a finite rigid set for all $i\geq 1$;*

• (2)

$\mathcal{Y}_{i}$* has trivial pointwise stabilizer in $\mathrm{Mod}(N_{1,3})$ for all $i\geq 1$;*

• (3)

$\bigcup_{i\geq 1}\mathcal{Y}_{i}=\mathcal{C}(N_{1,3})$.

Our proof is independent from [7, 9]. The following corollary is an extension of the main results of Atalan-Korkmaz [3] and Irmak [8]. It follows easily from Theorem 1.1 (see the proof of the analogous corollary in [2]).

Corollary 1.2**.**

If $\phi\colon\mathcal{C}(N_{1,3})\to\mathcal{C}(N_{1,3})$ is a locally injective simplicial map, then there exists a unique $f\in\mathrm{Mod}(N_{1,3})$ such that $\phi=f$.

In particular, the group of simplicial automorphisms of $\mathcal{C}(N_{1,3})$ is isomorphic to $\mathrm{Mod}(N_{1,3})$.

2. Preliminaries.

Let $S$ be a surface of finite type. By a hole in a surface we mean a boundary component. A curve on $S$ is an embedded simple closed curve. A curve is one-sided (resp. two-sided) if its regular neighbourhood is a Möbius band (resp. an annulus). If $\alpha$ is a curve on $S$, then $S\backslash\alpha$ is the subsurface obtained by removing from $S$ an open regular neighbourhood of $\alpha$. A curve $\alpha$ is essential if no boundary component of $S\backslash\alpha$ is a disc or an annulus or a Möbius band.

The curve complex $\mathcal{C}(S)$ is a simplicial complex whose $k$-simplices correspond to sets of $k+1$ isotopy classes of essential curves on $S$ with pairwise disjoint representatives. To simplify the notation, we will confuse a curve with its isotopy class and the corresponding vertex of $\mathcal{C}(S)$. Simplices of dimension $1$, $2$ and $3$ will be called edges, triangles and tetrahedra respectively.

For $\alpha,\beta\in\mathcal{C}^{0}(S)$ we denote by $i(\alpha,\beta)$ their geometric intersection number.

The mapping class group $\mathrm{Mod}(S)$ of a nonorientable surface $S$ (resp. the extendend mapping class group $\mathrm{Mod}^{\pm}(S)$ of an orientable surface $S$) is the group of isotopy classes of all self-homeomorphisms of $S$. If $S$ is orientable, then the mapping class group $\mathrm{Mod}(S)$ is defined to be the group if isotopy classes of orientation preserving homeomorphisms. Note that $\mathrm{Mod}(S)$ and $\mathrm{Mod}^{\pm}(S)$ act on $\mathcal{C}(S)$ by simplicial automorphisms.

If $S$ is a four-holed sphere, then $\mathcal{C}(S)$ is a countable set of vertices. In order to obtain a connected complex, the definition of $\mathcal{C}(S)$ is modified by declaring $\alpha,\beta\in\mathcal{C}^{0}(S)$ to be adjacent in $\mathcal{C}(S)$ whenever $i(\alpha,\beta)=2$. Furthermore, triangles are added to make $\mathcal{C}(S)$ into a flag complex. The complex $\mathcal{C}(S)$ so obtained is isomorphic to the well-known Farey complex. Two adjacent vertices of $\mathcal{C}(S)$ will be called Farey neighbours, and $2$-simplices of $\mathcal{C}(S)$ will be called Farey triangles. If $C$ is a boundary component, then we denote by $\mathrm{Mod}(S,C)$ (resp. $\mathrm{Mod}^{\pm}(S,C)$) the subgroup of $\mathrm{Mod}(S)$ (resp. $\mathrm{Mod}^{\pm}(S)$) consisting of elements fixing $C$.

Lemma 2.1**.**

Suppose that $S$ is a $4$-holed sphere, and $C$ is one of the boundary components. For any two Farey triangles $\{\beta_{0},\beta_{1},\beta_{2}\}$ and $\{\beta^{\prime}_{0},\beta^{\prime}_{1},\beta^{\prime}_{2}\}$ in $\mathcal{C}(S)$ there exists a unique $f\in\mathrm{Mod}^{\pm}(S,C)$ such that $f(\beta_{i})=\beta^{\prime}_{i}$ for $i=0,1,2$.

Proof.

By cutting $S$ along Farey neighbours we obtain four annuli, each containing one boundary component of $S$. Therefore there exists an orientation preserving $f^{\prime}\in\mathrm{Mod}(S,C)$ such that $f^{\prime}(\beta_{i})=\beta^{\prime}_{i}$ for $i=1,2$. Furthermore, since $f^{\prime}(C)=C$, such $f^{\prime}$ is easily shown to be unique by the Alaxander method [5, Prop. 2.8]. The pointwise stabilizer of $\{\beta^{\prime}_{1},\beta^{\prime}_{2}\}$ in $\mathrm{Mod}^{\pm}(S,C)$ is a cyclic group of order $2$ generated by an orientation reversing involution $\tau$ fixing every hole and such that $\beta^{\prime}_{0}$ and $\tau(\beta^{\prime}_{0})$ are the unique common Farey neighbours of both $\beta^{\prime}_{1}$ and $\beta^{\prime}_{2}$. By composing $f^{\prime}$ with $\tau$ if necessary we obtain the desired $f$. ∎

We represent the surface $N_{1,n}$ as a sphere with one crosscap and $n$ holes. The following two lemmas are easy to prove, and otherwise, they can be found in [16].

Lemma 2.2**.**

$\mathcal{C}(N_{1,2})$* consists of two one-sided vertices $\alpha,\alpha^{\prime}$ such that $i(\alpha,\alpha^{\prime})=1$ (Figure 1).*

Lemma 2.3**.**

In $\mathcal{C}(N_{1,3})$ every two-sided vertex $\beta$ is connected by an edge with exactly two vertices $\alpha,\alpha^{\prime}$ which are one-sided and $i(\alpha,\alpha^{\prime})=1$. Conversely, for every pair of one-sided vertices $\alpha,\alpha^{\prime}$ such that $i(\alpha,\alpha^{\prime})=1$, there exists exactly one two-sided vertex $\beta$ connected by an edge with $\alpha$ and $\alpha^{\prime}$ (Figure 1).

3. Finite rigid sets

In this section $S$ denotes the three-holed projective plane. The complex $\mathcal{C}(S)$ was studied by Scharlemann [16]. It is a bipartite graph: its vertex set can be partitioned as $\mathcal{C}^{0}(S)=V_{1}\sqcup V_{2}$, where $V_{1}$ and $V_{2}$ denote the sets of one-sided and two-sided vertices respectively, and every edge of $\mathcal{C}(S)$ connects a one-sided vertex with a two-sided one. Furthermore, by Lemma 2.3, every $\beta\in V_{2}$ is connected by an edge with exactly two $\alpha,\alpha^{\prime}\in V_{1}$ such that $i(\alpha,\alpha^{\prime})=1$. We say that $\beta$ is determined by $\alpha$ and $\alpha^{\prime}$.

We define an auxiliary simplicial complex $\mathcal{D}$ whose vertex set is $V_{1}$, and a set of vertices $\{\alpha_{0},\dots,\alpha_{k}\}$ is a simplex if $i(a_{i},a_{j})=1$ for $0\leq i<j\leq k$. It follows from the above discussion that $\mathcal{C}(S)$ is isomorphic to the graph obtained by subdividing every edge of $\mathcal{D}^{1}$ – the $1$-skeleton of $\mathcal{D}$. Indeed, the subdivision of an edge of $\mathcal{D}^{1}$ corresponds to adding the two-sided vertex determined by this edge.

Proposition 3.1**.**

• (a)

The link of each vertex of $\mathcal{D}$ is isomorphic to the Farey complex.

• (b)

$\dim\mathcal{D}=3$**

• (c)

every triangle of $\mathcal{D}$ is contained in exactly two different tetrahedra.

Proof.

Fix a vertex $\alpha\in\mathcal{D}$ and consider the four-holed sphere $S\backslash\alpha$. Recall that $\mathcal{C}(S\backslash\alpha)$ is the Farey complex. We define a map $\theta_{\alpha}\colon\mathrm{Lk}(\alpha)\to\mathcal{C}(S\backslash\alpha)$, where $\mathrm{Lk}(\alpha)$ is the link of $\alpha$ in $\mathcal{D}$. For a vertex $\alpha^{\prime}\in\mathrm{Lk}(\alpha)$, $\theta_{\alpha}(\alpha^{\prime})$ is the two-sided curve determined by $\alpha$ and $\alpha^{\prime}$. It follows from Lemma 2.3 that $\theta_{\alpha}$ is a bijection on vertices, and we claim that it is a simplicial isomorphism. Indeed, note that for $\alpha^{\prime},\alpha^{\prime\prime}\in\mathrm{Lk}(\alpha)$ we have $i(\alpha^{\prime},\alpha^{\prime\prime})=1\iff i(\theta_{\alpha}(\alpha^{\prime}),\theta_{\alpha}(\alpha^{\prime\prime}))=2$. This proves (a). The other assertions are consequences of (a) and well-known properties of the Farey complex; namely $\dim\mathcal{C}(S\backslash\alpha)=2$ and every edge of $\mathcal{C}(S\backslash\alpha)$ is contained in exactly two different triangles. ∎

Given a one-sided curve $\alpha_{0}$ we can construct infinitely many tetrahedra of $\mathcal{D}$ containing $\alpha_{0}$ as a vertex. Let $\{\beta_{i}\}_{1}^{3}$ be any Farey triangle of $\mathcal{C}(S\backslash\alpha_{0})$ and, for $1\leq i\leq 3$, let $\alpha_{i}$ be the one-sided curve such that $\beta_{i}$ is determined by $\alpha_{i}$ and $\alpha_{0}$. Then $\{\alpha_{i}\}_{0}^{3}$ is a tetrahedron of $\mathcal{D}$.

We define a “dual” graph $\mathcal{T}$ whose vertices are tetrahedra of $\mathcal{D}$. Two tetrahedra are connected by an edge in $\mathcal{T}$ if their intersection is a triangle. The following theorem follows from [16, Theorem 3.1].

Theorem 3.2** (Sharlemann).**

$\mathcal{T}$* is a $4$-regular tree.*

Lemma 3.3**.**

For any two tetrahedra $\{\alpha_{i}\}_{i=0}^{3}$ and $\{\alpha^{\prime}_{i}\}_{i=0}^{3}$ of $\mathcal{D}$ there exists a unique $f\in\mathrm{Mod}(S)$ such that $f(\alpha_{i})=\alpha^{\prime}_{i}$ for $0\leq i\leq 3$.

Proof.

For $1\leq i\leq 3$ let $\beta_{i}$ (resp. $\beta^{\prime}_{i}$) be the two-sided curve determined by $\alpha_{0}$ and $\alpha_{i}$ (resp. $\alpha^{\prime}_{0}$ and $\alpha^{\prime}_{i}$). Note that $\{\beta_{1},\beta_{2},\beta_{3}\}$ and $\{\beta^{\prime}_{1},\beta^{\prime}_{2},\beta^{\prime}_{3}\}$ are Farey triangles in $\mathcal{C}(S\backslash\alpha_{0})$ and $\mathcal{C}(S\backslash\alpha^{\prime}_{0})$ respectively. By Lemma 2.1 there exists a unique $f\in\mathrm{Mod}(S)$ such that $f(\alpha_{0})=\alpha^{\prime}_{0}$ and $f(\beta_{i})=\beta^{\prime}_{i}$ for $1\leq i\leq 3$. Since $\alpha^{\prime}_{i}$ is the unique vertex of $\mathcal{C}(S)$ different from $\alpha^{\prime}_{0}$ and adjacent to $\beta^{\prime}_{i}$, we have $f(\alpha_{i})=\alpha^{\prime}_{i}$ for $1\leq i\leq 3$. ∎

Let $T$ be a tetrahedron of $\mathcal{D}$. We denote by $T^{\ast}$ the full subcomplex of $\mathcal{C}(S)$ spanned by the four vertices of $T$ and the six two-sided vertices determined by the edges of $T$ (Figure 2). The following proposition says that $T^{\ast}$ is rigid. It is thus an extension of the main result of [7].

Proposition 3.4**.**

Suppose that $T$ is a tetrahedron $\mathcal{D}$ and $\phi\colon T^{\ast}\to\mathcal{C}(S)$ is a locally injective simplicial map. Then there exists a unique $f\in\mathrm{Mod}(S)$ such that $\phi=f$ on $T^{\ast}$.

Proof.

First note that $\phi$ is injective because it is locally injective and $T^{\ast}$ has diameter $2$. Let $T=\{\alpha_{i}\}_{i=0}^{3}$. We claim that $\{\phi(\alpha_{i})\}_{i=0}^{3}$ is a tetrahedron of $\mathcal{D}$. Indeed, for $1\leq i\leq 3$, $\phi(\alpha_{i})$ is adjacent in $\mathcal{C}(S)$ to three different vertices, and hence it is one-sided, as two-sided vertices of $\mathcal{C}(S)$ have degree $2$. For $i\neq j$, the distance in $\mathcal{C}(S)$ between $\phi(\alpha_{i})$ and $\phi(\alpha_{j})$ is $2$, and hence $\phi(\alpha_{i})$ and $\phi(\alpha_{j})$ are adjacent in $\mathcal{D}$.

By Lemma 3.3 there exists a unique $f\in\mathrm{Mod}(S)$ such that $f(\alpha_{i})=\phi(\alpha_{i})$ for $0\leq i\leq 3$. Let $\beta$ be a two-sided vertex of $T^{\ast}$ determined by $\alpha_{i}$ and $\alpha_{j}$. Then $\phi(\beta)$ is adjacent to $\phi(\alpha_{i})$ and $\phi(\alpha_{j})$, and since such a curve is unique, $\phi(\beta)=f(\beta)$. ∎

We denote by $\mathcal{T}^{0}$ the vertex set of $\mathcal{T}$, that is the set of tetrahedra of $\mathcal{D}$. Let $d_{\mathcal{T}}$ denote the path metric on $\mathcal{T}$. We fix a reference tetrahedron $T_{0}$ and define

[TABLE]

In other words, $\mathcal{T}^{0}_{n}$ is the set of tetrahedra within distance at most $n$ from $T_{0}$ in the path metric on $\mathcal{T}$.

Proof of Theorem 1.1..

Let $\mathcal{Y}_{1}=T_{0}^{\ast}$ and for $n\geq 1$:

[TABLE]

We prove by induction that $\mathcal{Y}_{n}$ is rigid for all $n\geq 1$. By Proposition 3.4 $\mathcal{Y}_{1}$ is rigid. Assume that $\mathcal{Y}_{n}$ is rigid and let $\phi\colon\mathcal{Y}_{n+1}\to\mathcal{C}(S)$ be a locally injective simplicial map. Since $\mathcal{Y}_{n}$ is rigid, there exists a unique $f\in\mathrm{Mod}(S)$ such that $f=\phi$ on $\mathcal{Y}_{n}$. Let $\phi^{\prime}=f^{-1}\circ\phi$.

Let $T\in\mathcal{T}_{n+1}\backslash\mathcal{T}_{n}$. We need to show that $\phi^{\prime}$ fixes every vertex of $T^{\ast}$. It suffices to show that $\phi^{\prime}$ fixes every vertex of $T$ because then it also has to fix the two-sided vertices of $T^{\ast}$ determined by edges of $T$. The tetrahedron $T$ has a common face with some (unique) tetrahedron $T^{\prime}\in\mathcal{T}_{n}$. Let $T=\{\alpha_{0},\alpha_{1},\alpha_{2},\alpha_{3}\}$ and $T^{\prime}=\{\alpha^{\prime}_{0},\alpha_{1},\alpha_{2},\alpha_{3}\}$. Let $\beta$ (resp. $\beta^{\prime}$) be the two-sided vertex of $T^{\ast}$ (resp. $(T^{\prime})^{\ast}$) determined by $\alpha_{0}$ and $\alpha_{1}$ (resp. $\alpha^{\prime}_{0}$ and $\alpha_{1}$). By local injectivity of $\phi^{\prime}$, $\phi^{\prime}(\beta)\neq\phi^{\prime}(\beta^{\prime})=\beta^{\prime}$, and hence also $\phi^{\prime}(\alpha_{0})\neq\phi^{\prime}(\alpha^{\prime}_{0})=\alpha^{\prime}_{0}$. By Proposition 3.4, $\phi^{\prime}(T)$ is a tetrahedron different from $T^{\prime}$ and having a common face with $T^{\prime}$. Since such a tetrahedron is unique by (c) of Proposition 3.1, $\phi^{\prime}(T)=T$ and $\phi^{\prime}(\alpha_{0})=\alpha_{0}$. We have shown that $\phi^{\prime}$ pointwise fixes $T^{\ast}$, and it follows that it pointwise fixes $\mathcal{Y}_{n+1}$. Hence $\phi=f$ on $\mathcal{Y}_{n+1}$.

Since $\mathcal{Y}_{n}$ contains $T_{0}^{\ast}$ for all $n\geq 1$, it has trivial pointwise stabilizer in $\mathrm{Mod}(S)$. Finally, it follows from the connectedness of $\mathcal{T}$ that $\bigcup_{n\geq 1}\mathcal{Y}_{n}=\mathcal{C}(S)$. ∎

4. Coarse geometry

In this section we consider $\mathcal{C}(S)$ and $\mathcal{D}^{1}$ as metric graphs with all edges of length $1$. We denote the metrics on these graphs by $d_{\mathcal{C}}$ and $d_{\mathcal{D}}$ respectively.

There is a natural piecewise-linear homeomorphism $\phi\colon\mathcal{C}(S)\to\mathcal{D}^{1}$ equal to the identity on one-sided vertices and which forgets the two-sided vertices. That is, if $\beta$ is the two-sided vertex of $\mathcal{C}(S)$ determined by $\alpha$ and $\alpha^{\prime}$, then $\phi(\beta)$ is defined to be the midpoint of the edge of $\mathcal{D}$ connecting $\alpha$ and $\alpha^{\prime}$. We have

[TABLE]

for all $x,y\in\mathcal{C}(S)$. In particular, $\phi$ is a quasi-isometry.

Since $\mathcal{T}$ is a tree, every triangle of $\mathcal{D}$ is separating, i.e. the space obtained by removing a triangle from $\mathcal{D}$ has two connected components. If $\Delta$ is a triangle of $\mathcal{D}$, and $x$ and $y$ are points lying in different connected components of $\mathcal{D}\backslash\Delta$, then we say that $\Delta$ separates $x$ from $y$.

Lemma 4.1**.**

Let $p$ be a vertex on a geodesic in $\mathcal{D}^{1}$ from $x$ to $y$, such that $d_{\mathcal{D}}(p,x)\geq 1$ and $d_{\mathcal{D}}(p,y)\geq 1$. There exists a triangle $\Delta$ of $\mathcal{D}$ such that $p\in\Delta$ and $\Delta$ separates $x$ from $y$.

Proof.

Let $[x,y]$ be a geodesic in $\mathcal{D}^{1}$ from $x$ to $y$ containing $p$, and let $q$ be the vertex preceding $p$ on $[x,y]$. Let $(T_{i})_{0}^{n}$ be any sequence of tetrahedra forming a geodesic in $\mathcal{T}$ such that $q\in T_{0}$ and $y\in T_{n}$. Note that $q\notin T_{n}$ since $d_{\mathcal{D}}(q,y)=1+d_{\mathcal{D}}(p,y)\geq 2$. Let $T_{i}$ be the first tetrahedron in this sequence such that $q\notin T_{i}$. Then $\Delta=T_{i}\cap T_{i-1}$ is a triangle separating $q$ from $y$. The segment $[q,y]$ must pass through a vertex of $\Delta$, which gives $p\in\Delta$. Finally notice that $\Delta$ separates $x$ from $y$, for otherwise $[x,y]$ could not contain $q$ (there would be shorter path from $x$ to $y$ avoiding $q$). ∎

Theorem 4.2**.**

The curve graph $\mathcal{C}(S)$ is quasi-isometric to a simplicial tree.

Proof.

Since $\mathcal{C}(S)$ is quasi-isometric to $\mathcal{D}^{1}$, it suffices to show that $\mathcal{D}^{1}$ is quasi-isometric to a simplicial tree. By [13, Throem 4.6], this is equivalent to $\mathcal{D}^{1}$ satisfying the following Bottleneck Property: There is some $L>0$ so that for all $x,y\in\mathcal{D}^{1}$ there is a midpoint $m=m(x,y)$ with $d(x,m)=d(y,m)=\frac{1}{2}d(x,y)$ and the property that any path from $x$ to $y$ must pass within less than $L$ of the point $m$.

Let $L>\frac{3}{2}$ and define $m=m(x,y)$ to be the midpoint of any geodesic from $x$ to $y$. Clearly we can assume $d_{\mathcal{D}}(x,m)\geq L$. Let $p$ be a vertex on a geodesic from $x$ to $y$ such that $d_{\mathcal{D}}(m,p)\leq\frac{1}{2}$. By Lemma 4.1, there exists a triangle $\Delta$ separating $x$ from $y$ such that $p\in\Delta$. Any path from $x$ to $y$ must pass through $\Delta$, and hence within at most $\frac{3}{2}$ of the point $m$. ∎

By using the Bottleneck Property from the proof above we can determine the hyperbolicity constant of $\mathcal{C}(S)$. Recall that a geodesic metric space $(X,d)$ is $\delta$-hyperbolic if, for any geodesic triangle $[x,y]\cup[x,z]\cup[y,z]$ and any $p\in[x,y]$ there exists some $q\in[x,z]\cup[y,z]$ with $d(p,q)\leq\delta$. A triangle satisfying the condition above is called $\delta$-thin. Masur and Minsky [14] proved that $\mathcal{C}(S)$ is hyperbolic for orientable $S$. Their result was extended to nonorientable surfaces by Bestvina-Fujiwara [4] and Masur-Schleimer [15].

Proposition 4.3**.**

The graph $\mathcal{C}(S)$ is $3$-hyperbolic.

Proof.

First we prove that $\mathcal{D}^{1}$ is $\frac{3}{2}$-hyperbolic. Let $[x,y]\cup[x,z]\cup[y,z]$ be geodesic triangle in $\mathcal{D}^{1}$ and $p\in[x,y]$. Clearly we can assume $d_{\mathcal{D}}(x,p)\geq\frac{3}{2}$. Let $p^{\prime}$ be a vertex on $[x,y]$ such that $d_{\mathcal{D}}(p,p^{\prime})\leq\frac{1}{2}$. By Lemma 4.1, there exists a triangle $\Delta$ separating $x$ from $y$ such that $p^{\prime}\in\Delta$. It follows that $[x,z]\cup[y,z]$ has non-empty intersection with $\Delta$, and for any point $q$ in this intersection $d_{\mathcal{D}}(q,p)\leq\frac{3}{2}$.

To finish the proof we use the homeomorphism $\phi\colon\mathcal{C}(S)\to\mathcal{D}^{1}$. Observe that $\phi$ maps geodesics triangles to geodesic triangles and $d_{\mathcal{C}}(x,y)=2d_{\mathcal{D}}(\phi(x),\phi(y))$ for all $x,y\in\mathcal{C}(S)$. Since geodesic triangles in $\mathcal{D}^{1}$ are $\frac{3}{2}$-thin, geodesic triangles in $\mathcal{C}(S)$ are $3$-thin. ∎