# Rigidity results for large displacement quotients of mapping class groups

Giorgio Mangioni  (Giorgio Mangioni) Maxwell Institute and Department of
Mathematics, Heriot-Watt University, Edinburgh, UK<EMAIL_ADDRESS>
###### Abstract.

We consider quotients of mapping class groups of orientable, finite type
surfaces by subgroups whose action on the curve graph has large displacement.
This class includes quotients by the normal closure of a pseudo-Anosov
element, the mapping class group itself, and in view of forthcoming work of
Abbott-Berlyne-Ng-Rasmussen also random quotients. First, we show that every
automorphism of the corresponding quotient of the curve graph is induced by a
mapping class, thus generalising Ivanov’s Theorem about automorphisms of the
curve graph. Then we use this to prove quasi-isometric rigidity under
additional assumptions, satisfied by all aforementioned quotients. In the
process, we clarify a proof of quasi-isometric rigidity of mapping class
groups by Behrstock, Hagen, and Sisto. Finally, we show that the outer
automorphisms groups of our quotients, as well as their abstract
commensurators, are “the smallest possible”.

###### Contents

  1. 1 Large displacement quotients
  2. 2 Combinatorial rigidity
  3. 3 Quasi-isometric rigidity
  4. 4 Algebraic rigidity

## Introduction

In recent years there has been a growing interest in studying quotients and
subgroups of mapping class groups of orientable, finite type surfaces, in
order to establish the analogues of Ivanov’s theorem about automorphisms of
the curve graph and its consequences (see e.g. [BM19, McL19] for a wide class
of normal subgroups, and, among others, [MS23] for quotients by large enough
powers of Dehn Twists).

In this paper, we address the problem for a class of quotients which we call
_large displacement quotients_. The defining property, articulated in
Definition 1.2, is that every non-trivial element in the kernel acts with
large displacement on the curve graph of the surface, that is, the distance
between any vertex and its image is bounded from below by a large enough
constant, which we call the _minimum displacement_. An example of such
behaviour is the quotient by the normal closure of some power of a pseudo-
Anosov element, which was studied in e.g. [DGO17] to answer questions about
the existence of normal free subgroups of $\mathcal{MCG}^{\pm}(S)$.
Furthermore, the definition applies to the mapping class group itself, seen as
the quotient by the trivial subgroup.

We have three main results, which we think of as rigidity properties of our
quotients. The first one is a generalisation of a celebrated theorem of
Ivanov-Korkmaz [Iva97, Kor99], and morally tells us that all properties of a
large displacement quotient are already encoded in the action on the
corresponding quotient of the curve graph. Here $\mathcal{MCG}^{\pm}(S)$
denotes the extended mapping class group, where we allow orientation-reversing
mapping classes.

######  Theorem 1.

Let $S$ be a surface of complexity at least $3$ which is not $S_{2,0}$, and
let $N\unlhd\mathcal{MCG}^{\pm}(S)$ be a normal subgroup. If the minimum
displacement of $N$ is large enough, then the map
$\mathcal{MCG}^{\pm}(S)/N\to\text{Aut}(\mathcal{C}S/N)$, induced by the action
of $\mathcal{MCG}^{\pm}(S)/N$ on $\mathcal{C}S/N$, is an isomorphism.

The bound on the minimum displacement is made precise in Theorem 2.1. A
similar result holds for some surfaces of low complexity too, but in these
cases the map might only be surjective with finite kernel (see Subsection 2.1
for the details).

In the second half of the paper, we restrict our attention to those large
displacement quotients admitting a _hierarchically hyperbolic group_ (HHG)
structure, in the sense of [BHS19]. We also require the structure to be
“inherited” from the standard HHG structure of the mapping class group, in a
sense that is clarified in Convention 3.2. Roughly, we ask that the domain set
of the HHG structure consists almost exclusively of all $N$-orbits of
surfaces, that the top-level coordinate space coincides with $\mathcal{C}S/N$,
and that the coordinate space of the orbit of a subsurface $U\subsetneq S$ is
quasi-isometric to the curve graph of $U$. The class of quotients admitting
such a HHG structure, which we call “surface-inherited”, includes:

  * •

the mapping class group itself (by e.g. [BHS19, Theorem 11.1]);

  * •

the quotient by the normal closure of a power of a pseudo-Anosov element (by
[BHS17a, Corollary F]);

  * •

more generally, quotients by “deep enough” subgroups of _hierarchically
hyperbolically embedded_ subgroups, in the sense of [BHS17a, Definition 6.1]
(see Lemma 3.5 for details);

  * •

in view of forthcoming work of Abbott-Berlyne-Ng-Rasmussen, also random
quotients of $\mathcal{MCG}^{\pm}(S)$.

For such quotients, we can state our second main result, which is quasi-
isometric rigidity. Recall that two groups $G$ and $H$ are _weakly
commensurable_ if there exist two finite normal subgroups $L\unlhd H$ and
$M\unlhd G$ such that the quotients $H/L$ and $G/M$ have two finite index
subgroups that are isomorphic.

######  Theorem 2.

Let $S$ be a surface of complexity at least $2$ which is not a torus with two
punctures, and let $N\unlhd\mathcal{MCG}^{\pm}(S)$ be a normal subgroup. If
the minimum displacement of $N$ is large enough, and if
$\mathcal{MCG}^{\pm}(S)/N$ has a surface-inherited HHG structure, then
$\mathcal{MCG}^{\pm}(S)/N$ is quasi-isometrically rigid, meaning that if a
finitely generated group $G$ and $\mathcal{MCG}^{\pm}(S)/N$ are quasi-
isometric then they are weakly commensurable.

Since $\mathcal{MCG}^{\pm}(S)$ itself is a large displacement quotient, we
also recover quasi-isometric rigidity of mapping class groups, first proven in
[BKMM12].

Finally, combining quasi-isometric rigidity and results on automorphisms of
acylindrically hyperbolic groups from [AMS16], we are able to deduce the
following, which can be regarded as a form of algebraic rigidity and is a
generalisation of a result of Ivanov for mapping class groups [Iva97]:

######  Theorem 3 (Algebraic rigidity).

Let $S$ be either a surface of complexity at least $3$ (excluding the case
$S=S_{2,0}$) or a $S_{0,5}$, and let $\mathcal{MCG}^{\pm}(S)/N$ be a large
displacement quotient with a surface-inherited HHG structure. If the minimum
displacement is large enough, then the following facts hold:

  1. (1)

The map $\mathcal{MCG}^{\pm}(S)/N\to\text{Aut}(\mathcal{MCG}^{\pm}(S)/N)$,
which maps every element to the corresponding inner automorphism, is an
isomorphism. In particular, $\text{Out}(\mathcal{MCG}^{\pm}(S)/N)$ is trivial.

  2. (2)

The abstract commensurator of $\mathcal{MCG}^{\pm}(S)/N$ coincides with
$\mathcal{MCG}^{\pm}(S)/N$ itself. In particular, any isomorphism between
finite index subgroups of $\mathcal{MCG}^{\pm}(S)/N$ is the restriction of an
inner automorphism.

### Sketch of proofs

#### Combinatorial rigidity

To show that any automorphism
$\overline{\phi}:\,\mathcal{C}S/N\to\mathcal{C}S/N$ is induced by some element
$\overline{g}\in\mathcal{MCG}^{\pm}(S)/N$ we proceed as follows. First, we
find an automorphism $\phi:\mathcal{C}S\to\mathcal{C}S$ which “lifts”
$\overline{\phi}$, i.e. that makes the following diagram commute, where
$\pi:\,\mathcal{C}S\to\mathcal{C}S/N$ is the quotient map:

${\mathcal{C}S}$${\mathcal{C}S}$${\mathcal{C}S/N}$${\mathcal{C}S/N}$$\scriptstyle{\phi}$$\scriptstyle{\pi}$$\scriptstyle{\pi}$$\scriptstyle{\overline{\phi}}$

The key observation is that, if we regard both $\mathcal{C}S$ and
$\mathcal{C}S/N$ as simplicial _complexes_ , then $\pi$ is a covering map (see
Lemma 2.2). Moreover, by a result of Harer [Har86], $\mathcal{C}S$ is simply
connected whenever $S$ has large enough complexity, and we can lift
$\overline{\phi}$ by standard arguments of covering theory. Then one can apply
a result of Ivanov-Korkmaz [Iva97, Kor99] to show that the lift $\phi$ is
induced by a mapping class $g\in\mathcal{MCG}^{\pm}(S)$, and therefore its
image $\overline{g}\in\mathcal{MCG}^{\pm}(S)/N$ induces $\overline{\phi}$.

#### Quasi-isometric rigidity

We follow a general strategy, developed by Jason Behrstock, Mark Hagen, and
Alessandro Sisto in [BHS21], to prove quasi-isometric rigidity of certain HHG.
The key idea from that paper is that any self-quasi-isometry $f$ of a HHG,
satisfying some additional assumptions, induces an automorphism of a certain
graph, which should be thought of as encoding the intersection patterns of
maximal _quasiflats_ (that is, quasi-isometric embeddings of $\mathbb{R}^{n}$
of maximum dimension). In the case of $\mathcal{MCG}^{\pm}(S)/N$, such graph
is precisely the quotient of the curve graph, since maximal quasiflats of
$\mathcal{MCG}^{\pm}(S)/N$ morally correspond to subgroups generated by
maximal families of commuting Dehn Twists. But then, by the combinatorial
rigidity results from Section 2, the automorphism of $\mathcal{C}S/N$ is
induced by some element $\overline{g}\in\mathcal{MCG}^{\pm}(S)/N$, and with a
little more effort one can show that $f$ coarsely coincides with the left
multiplication by $\overline{g}$.

In [BHS21], the main result to extract combinatorial data from a self-quasi-
isometry of a HHG is [BHS21, Theorem 5.7], which is then used in [BHS21,
Theorem 5.10] to give a new proof of quasi-isometric rigidity of mapping class
groups. Unfortunately, as pointed out by Jason Behrstock after that paper was
published, [BHS21, Theorem 5.7] does not apply to mapping class groups.
Indeed, if $U$ is a subsurface of complexity $1$, then any maximal collection
of pairwise disjoint subsurfaces that contains $U$ must also contain the
boundary annuli of $U$. Hence, mapping class groups do not satisfy [BHS21,
Assumption 2], which roughly states that such a $U$ should be the only
subsurface shared by two maximal collections of pairwise disjoint subsurfaces.

Behrstock, Hagen, and Sisto, worked out a modification of their Assumptions
which allows them to prove a version of [BHS21, Theorem 5.7] which applies to
the mapping class group. Rather than writing their argument separately, they
instead opted to allow me to incorporate a version of those assumptions into
this work, and I am extremely grateful for this possibility. Then the proof of
[BHS21, Theorem 5.10] runs verbatim once one replaces [BHS21, Theorem 5.7]
with our Lemma 3.18, whose Assumptions (A) - (F) are satisfied by both
$\mathcal{MCG}^{\pm}(S)$ and our quotients (see Lemma 3.20).

We stress that [BHS21, Theorem 5.7] is true as stated, and it is also used in
[DDLS21] to study quasi-isometric rigidity of extensions of Veech groups.

#### Algebraic rigidity

Given an isomorphism $\phi:\,H\to H^{\prime}$ between finite-index subgroups
of $\mathcal{MCG}^{\pm}(S)/N$, one can consider the element
$\overline{g}\in\mathcal{MCG}^{\pm}(S)/N$ whose left-multiplication coarsely
coincides with $\phi$, seen as a self-quasi-isometry of
$\mathcal{MCG}^{\pm}(S)/N$, and then try to prove that the conjugation by
$\overline{g}$ restricts to the given isomorphism on $H$. We do just that in
Theorem 4.1, using tools from [AMS16] for acylindrical group actions on
hyperbolic spaces. We can use such tools since, by [BHS17b, Corollary 14.4],
$\mathcal{MCG}^{\pm}(S)/N$ acts acylindrically on the top-level coordinate
space of the HHG structure, which coincides with $\mathcal{C}S/N$ for those
quotients satisfying Convention 3.2. Notice that finite normal subgroups could
cause the outer automorphism group to be finite rather than trivial, so the
key technical results we need is that $\mathcal{MCG}^{\pm}(S)/N$ does not
contain non-trivial finite normal subgroups, Lemma 4.7.

### Comparison with quotients by large powers of Dehn twists

Very similar results hold for quotients of the form
$\mathcal{MCG}^{\pm}(S)/DT_{K}$, where $DT_{K}$ is the normal subgroup
generated by all $K$-th powers of Dehn Twists. Such quotients can be regarded
as “Dehn filling quotients” of mapping class groups, as pointed out and
explored in [DHS21] and then in [BHMS20], where they are proven to be
hierarchically hyperbolic. Moreover, in view of [MS23], they enjoy the same
combinatorial, quasi-isometric, and algebraic rigidity properties as our
quotients (at least when $S$ is a punctured sphere, and conjecturally also in
the general case). While the proofs there also rely on lifting, the quotient
map $\mathcal{C}S\to\mathcal{C}S/DT_{K}$ is not a covering map, and indeed it
is quite far from being locally injective; therefore lifting properties follow
from different technologies, involving results from [Dah18] about rotating
families and the existence of finite rigid sets as defined in [AL13].
Moreover, in this paper quasi-isometric rigidity relies on the fact that Dehn
twist flats survive in the quotient, while they disappear in the quotient by
$DT_{K}$ for any choice of $K$. In particular, quasi-isometries of
$\mathcal{MCG}^{\pm}(S)/DT_{K}$ induce automorphisms of a graph which is not
$\mathcal{C}S/DT_{K}$, and then one has to relate these two graphs with
further combinatorial considerations.

### Questions

The following question was asked by Jason Behrstock, during an exchange on a
first draft of this paper:

######  Question 1.

Let $S$ be either a surface of complexity at least $3$ (excluding the case
$S=S_{2,0}$) or a $S_{0,5}$, and let $\mathcal{MCG}^{\pm}(S)/N$ be a large
displacement quotient with a surface-inherited HHG structure. Does there exist
$D\geq 0$ such that, if the minimum displacement is at least $D$, _any
injection_ from a finite index subgroup $H\leq\mathcal{MCG}^{\pm}(S)/N$ to
$\mathcal{MCG}^{\pm}(S)/N$ is induced by an inner automorphism?

This would be a strengthening of our Theorem 3, where we also require the
image of the injection to have finite index. Indeed, our proof needs this
further assumption, since we want to apply Theorem 2 in order to find a
candidate element of $\mathcal{MCG}^{\pm}(S)/N$, and an isomorphism between
subgroups is a quasi-isometry of $\mathcal{MCG}^{\pm}(S)/N$ if and only if
both its domain and image are of finite index.

Question 1 would also be a generalisation of the analogue results for mapping
class groups, proved by Irmak in [Irm06a, Irm04, Irm06b] and then extended by
Behrstock-Margalit in [BM06]. The main strategy of all these papers is
articulated in two steps:

  * •

First, one proves that any _superinjective map_ of $\mathcal{C}S$ (that is,
any self-map preserving adjacency and non-adjacency between vertices) is
induced by an element of $\mathcal{MCG}^{\pm}(S)$. This is a generalisation of
Ivanov’s combinatorial rigidity theorem, since a simplicial isomorphism is
clearly superinjective.

  * •

Then one shows that any injection from a finite-index subgroup of
$\mathcal{MCG}^{\pm}(S)$ into $\mathcal{MCG}^{\pm}(S)$ maps powers of Dehn
Twists to powers of Dehn Twists, preserving commutation and non-commutation.
Therefore the injection induces a superinjective map of $\mathcal{C}S$, and
one finds a candidate element of $\mathcal{MCG}^{\pm}(S)$ whose conjugation
restricts to the given injection.

Therefore, if we want to repeat the same argument for our quotients, we need
to answer the following:

######  Question 2.

In the setting of Question 1, is any superinjective map of $\mathcal{C}S/N$
induced by an element of $\mathcal{MCG}^{\pm}(S)/N$?

This would be a generalisation of Theorem 1, which I believe to be true. One
could try to lift a superinjective map of $\mathcal{C}S/N$ to a superinjective
map of $\mathcal{C}S$, and then conclude by the results of Irmak and
Behrstock-Margalit.

######  Question 3.

In the setting of Question 1, does the injection
$H\to\mathcal{MCG}^{\pm}(S)/N$ map powers of Dehn Twists to powers of Dehn
Twist?

Irmak proof is ultimately a refined version of Ivanov’s characterisation of
Dehn Twists from [Iva88], in terms of the centres of their centralisers in
$\mathcal{MCG}^{\pm}(S)$, which might extend to our quotients. I believe this
is one more reason to try to understand better centralisers in hierarchically
hyperbolic groups.

### Outline of the paper

In Section 1 we gather some examples of the quotients
$\mathcal{MCG}^{\pm}(S)/N$ by subgroups with large displacement, and we
develop the lifting tools that are used throughout the paper.

In Section 2 we prove combinatorial rigidity for the general case, which is
Theorem 2.1, and for some low-complexity surfaces, see Subsection 2.1. This
establishes Theorem 1.

Section 3 is devoted to the proof of Theorem 2. First, we show that a self-
quasi-isometry of a hierarchically hyperbolic space, satisfying certain
properties (which are Assumption (A) - (F)), induce an automorphism of a
certain graph, which encodes the intersections of certain maximal quasiflats
(see Lemma 3.18). Then in Subsection 3.5 we relate such graph to
$\mathcal{C}S/N$, and in the proof of Theorem 3.8 we use the combinatorial
rigidity results from Section 2 to produce an element of
$\mathcal{MCG}^{\pm}(S)/N$ whose left-multiplication is within finite distance
from the quasi-isometry.

We conclude the paper with Section 4, where we use quasi-isometric rigidity
and some tools from [AMS16] about acylindrical actions on hyperbolic spaces to
show that any automorphism between finite index subgroups of
$\mathcal{MCG}^{\pm}(S)/N$ is the restriction of the conjugation by a unique
element $\overline{g}\in\mathcal{MCG}^{\pm}(S)/N$ (see Theorem 4.1 for the
existence of such an element, and Lemma 4.8 for the uniqueness). Therefore,
the outer automorphism group of $\mathcal{MCG}^{\pm}(S)/N$, as well as its
abstract commensurator, are the smallest possible, as clarified in Corollaries
4.9 and 4.10. This proves the two parts of Theorem 3.

### Acknowledgements

I would like to thank Jason Behrstock, Mark Hagen, and Alessandro Sisto for
sharing with me their previous attempts to find a different set of assumptions
under which the conclusion of [BHS21, Theorem 5.7] holds, and for suggesting
many corrections to the first draft of this paper. I am especially grateful to
my supervisor Alessandro Sisto for his constant support. I am also grateful to
Piotr Przytycki for suggesting a way to shorten the proof of Theorem 2.8.
Finally, I thank Carolyn Abbott, Daniel Berlyne, Thomas Ng, and Alexander
Rasmussen for fruitful discussions.

##  1\. Large displacement quotients

Let $S$ be a surface of finite type, that is, a surface obtained from a
closed, connected, oriented surface after removing a finite number of points,
called punctures. When we want to emphasise the genus $g$ and the number of
punctures $b$ we will use the notation $S_{g,b}$, and we will denote by
$\zeta(S)=3g+b-3$ the complexity of $S_{g,b}$. Let $\mathcal{C}S$ be the curve
graph of $S$, whose vertices are (isotopy classes of) curves and adjacency
corresponds to disjointness. Whenever it will not cause ambiguity, we will
always say “curve” to mean its isotopy class. Finally, let
$\mathcal{MCG}^{\pm}(S)$ be the extended mapping class group of $S$, where we
allow orientation-reversing mapping classes. If $x\in\mathcal{C}S^{(0)}$ is a
curve and $f\in\mathcal{MCG}^{\pm}(S)$ is a mapping class, we will denote the
image of $x$ under $f$ simply by $fx$.

Let us introduce which quotients of $\mathcal{MCG}^{\pm}(S)$ we are
considering.

######  Definition 1.1 (Minimum displacement).

The _minimum displacement_ of a mapping class $f\in\mathcal{MCG}^{\pm}(S)$ is
defined as $\min_{x\in\mathcal{C}S^{(0)}}\mathrm{d}_{\mathcal{C}S}(x,fx)$.
Similarly, the minimum displacement of a subgroup
$N\leq\mathcal{MCG}^{\pm}(S)$ is defined as

$\min_{f\in
N-\\{1\\}}\min_{x\in\mathcal{C}S^{(0)}}\mathrm{d}_{\mathcal{C}S}(x,fx),$

where $1\in\mathcal{MCG}^{\pm}(S)$ is the identity.

######  Definition 1.2 (Large displacement quotient).

Let $N\unlhd\mathcal{MCG}^{\pm}(S)$ be a normal subgroup, and let
$D\in\mathbb{N}_{>0}$. We will say that the quotient group
$\mathcal{MCG}^{\pm}(S)/N$ is a _$D$ -large displacement quotient_ if the
minimum displacement of $N$ is at least $D$.

Here are some examples of this behaviour.

######  Example 1.3.

Definition 1.2 applies to the trivial subgroup $N=\\{1\\}$, thus
$\mathcal{MCG}^{\pm}(S)$ itself is a $D$-large displacement quotient for every
$D\in\mathbb{N}_{>0}$.

######  Example 1.4.

Let $f\in\mathcal{MCG}^{\pm}(S)$ be a pseudo-Anosov element. For every
$D\in\mathbb{N}_{>0}$ one can find some large enough $K\in\mathbb{N}_{>0}$
such that $f^{K}$ has minimum displacement at least $D$. Moreover, up to
choosing a larger $K$, the quotient by the normal closure $N=\langle\langle
f^{K}\rangle\rangle$ is a $D$-large displacement quotient. We will give a
proof in Corollary 3.6, after we have developed some more notation.

###  1.1. Isometric projections

Let $\mathcal{MCG}^{\pm}(S)/N$ be a $D$-large displacement quotient. Since $N$
acts on $\mathcal{C}S$ by simplicial automorphisms, we can consider the
quotient $\mathcal{C}S/N$ by this action. Here we gather some properties of
$\mathcal{C}S/N$ and the projection map $\pi:\,\mathcal{C}S\to\mathcal{C}S/N$,
which hold whenever $D$ is large enough:

######  Lemma 1.5.

Let $\mathcal{MCG}^{\pm}(S)/N$ be a $D$-large displacement quotient. If $D\geq
3$ then $\mathcal{C}S/N$ is a simplicial graph.

###### Proof.

By definition of quotient by a simplicial action, there is an edge in
$\mathcal{C}S/N$ between two vertices
$\overline{x},\overline{y}\in\mathcal{C}S/N^{(0)}$ if they admit two adjacent
representatives in $\mathcal{C}S$. First notice that, by the assumption on the
minimum displacement, two vertices in the same $N$-orbit cannot be adjacent,
hence no edge of $\mathcal{C}S/N$ connects a vertex to itself. Moreover, we
claim that no two vertices $\overline{x},\overline{y}$ are connected by more
than one edge. Indeed, if there were two edges between them, we could find two
edges $\\{x,y\\}$ and $\\{x^{\prime},y^{\prime}\\}$ between representatives of
$\overline{x}$ and $\overline{y}$, respectively. We can further assume that
$y=y^{\prime}$, up to replacing $\\{x^{\prime},y^{\prime}\\}$ with one of its
$N$-translates. But then $x$ and $x^{\prime}$ would be $N$-translate within
distance at most $2$, contradicting the fact that the minimum displacement is
at least $3$. ∎

######  Definition 1.6.

Let $\overline{X}$ be a subgraph of $\mathcal{C}S/N$. We say that a subgraph
$X$ of $\mathcal{C}S$ is a _lift_ of $\overline{X}$ if the projection map
$\pi:\,\mathcal{C}S\to\mathcal{C}S/N$ restricts to an isometry between $X$ and
$\overline{X}$.

######  Lemma 1.7 (Isometric lifts and projections).

Let $\mathcal{MCG}^{\pm}(S)/N$ be a $D$-large displacement quotient. If $D\geq
8$ then the following facts hold for the projection
$\pi:\,\mathcal{C}S\to\mathcal{C}S/N$:

  * •

$\pi$ is $1$-Lipschitz;

  * •

For every combinatorial path $\overline{\Gamma}\subset\mathcal{C}S/N$ there
exists a combinatorial path $\Gamma\subset\mathcal{C}S$ such that
$\pi(\Gamma)=\overline{\Gamma}$, and moreover if $\overline{\Gamma}$ is
geodesic then so is $\Gamma$;

  * •

For every $x\in\mathcal{C}S^{(0)}$, $\pi$ restricts to an isometry between the
closed ball of radius $2$ centred at $x$, which we denote by $B(x,2)$, and the
closed ball of radius $2$ centred at its projection $\overline{x}$, which we
denote by $\overline{B}(\overline{x},2)$;

  * •

Every subgraph $\overline{X}$ of $\mathcal{C}S/N$ which is contained in a
closed ball of radius $2$ admits a unique $N$-orbit of lifts.

It is possible that our bound on $D$ can be sharpened. However this is
irrelevant, since in all notable applications one is always allowed to replace
$N$ with a suitable finite-index subgroup, in order to make the minimum
displacement larger.

###### Proof.

The quotient map is $1$-Lipschitz since the action of $N$ on $\mathcal{C}S$ is
simplicial. In order to find a combinatorial path $\Gamma\subset\mathcal{C}S$
which projects to a given path $\overline{\Gamma}\subset\mathcal{C}S/N$ it
suffices to lift one edge at a time, given a lift of its starting point, and
again this can be done since the action is simplicial. Notice that, by
construction, $\Gamma$ has the same length of $\overline{\Gamma}$. Moreover,
suppose that $\overline{\Gamma}$ is a geodesic between its endpoints
$\overline{x},\overline{y}$, which lift to the endpoints $x,y$ of $\Gamma$.
Then, denoting by $\ell(\Gamma)$ the length of $\Gamma$, and similarly for
$\overline{\Gamma}$, we get

$\mathrm{d}_{\mathcal{C}S}(x,y)\leq\ell(\Gamma)=\ell(\overline{\Gamma})=\mathrm{d}_{\mathcal{C}S/N}(\overline{x},\overline{y})\leq\mathrm{d}_{\mathcal{C}S}(x,y),$

where we used that $\overline{\Gamma}$ realises the distance between
$\overline{x}$ and $\overline{y}$, and that the projection map is
$1$-Lipschitz. But then the inequalities above are all equalities, and in
particular $\Gamma$ realises the distance between its endpoints, i.e., it is a
geodesic path.

Now, let $x\in\mathcal{C}S^{(0)}$ and let $\overline{x}$ be its projection.
Combining the previous points we have that
$\pi(B(x,2))=\overline{B}(\overline{x},2)$, so we are left to show that, for
every $y,z\in B(x,2)$, we have that
$\mathrm{d}_{\mathcal{C}S/N}(\overline{y},\overline{z})=\mathrm{d}_{\mathcal{C}S}(y,z)$,
where $\overline{y}$ and $\overline{z}$ are the respective projections. If by
contradiction this was not the case, then we could pick any geodesic
$\overline{\Gamma}$ between $\overline{y}$ and $\overline{z}$. Notice that,
since $y,z\in B(x,2)$ and $\pi$ is $1$-Lipschitz, we have that
$\mathrm{d}_{\mathcal{C}S/N}(\overline{y},\overline{z})\leq 4$, thus
$\overline{\Gamma}$ must have length at most $3$. Now lift $\overline{\Gamma}$
to a geodesic path $\Gamma$ starting at $y$ and ending at some $z^{\prime}$ in
the same orbit of $z$. But then $z$ and $z^{\prime}$ would be within distance
at most $7$, contradicting the assumption on the minimum displacement.

For the last statement, every subgraph $\overline{X}$ which is contained in a
closed $2$-ball admits a lift, by applying the local inverse of $\pi$. Given
any two lifts $X,X^{\prime}$, let $x\in X$ and $x^{\prime}\in X^{\prime}$ be
two points with the same projection, so that there exists an element $n\in N$
such that $x^{\prime}=nx$. Now pick another pair $y\in X$ and $y^{\prime}\in
X^{\prime}$ with the same projection. Since $y^{\prime}$ and $ny$ are within
distance at most $4$, by the assumption on the minimum displacement we must
have that $y^{\prime}=ny$. Hence $nX=X^{\prime}$, that is, all lifts belong to
the same $N$-orbit. ∎

###  1.2. Orbits of subsurfaces

We conclude the Section with a study of how $N$ acts on the subsurfaces of
$S$. We first recall a definition from [BKMM12, Section 2.1.3]:

######  Definition 1.8.

A subsurface $U\subset S$ is _essential_ if it is the disjoint union of some
components of the complement of a collection of disjoint simple closed curves,
so that no component is a pair of pants and no two annuli components are
isotopic.

Let $\mathfrak{S}$ be the collection of (isotopy classes of) essential
subsurfaces. Whenever it will not cause ambiguity, we will always say
“subsurface” to mean its isotopy class. Moreover, if $U\in\mathfrak{S}$ is an
annulus, then it is the regular neighbourhood of some curve
$x\in\mathcal{C}S^{(0)}$, which we call the _core_ of $U$.

Two subsurfaces $U,V\in\mathfrak{S}$ are _nested_ (resp. _orthogonal_), and we
write $U\sqsubseteq V$ (resp. $U\bot V$) if $U$ is contained in $V$ (resp. if
they are disjoint). If $U,V$ are neither $\sqsubseteq$-related nor orthogonal,
then they are _transverse_ and we write $U\pitchfork V$.

Now let $\overline{\mathfrak{S}}=\mathfrak{S}/N$ be the collection of
$N$-orbits of subsurfaces. We will denote the equivalence class of a
subsurface $U\in\mathfrak{S}$ by $\overline{U}$, and similarly if
$\mathcal{U}=\\{U_{i}\\}_{i\in I}$ is a collection of subsurfaces we will set
$\overline{\mathcal{U}}=\\{\overline{U}_{i}\\}_{i\in I}$. Two classes
$\overline{U},\overline{V}\in\overline{\mathfrak{S}}$ are _nested_ (resp.
_orthogonal_), and we write $\overline{U}\sqsubseteq\overline{V}$ (resp.
$\overline{U}\bot\overline{V}$) if there exists _at least one_ pair of
representatives $U,V$ such that $U\sqsubseteq V$ (resp. $U\bot V$). Notice
that we are not requiring that _all_ pairs of representatives are nested
(resp. orthogonal), and indeed this is almost never the case. If
$\overline{U},\overline{V}$ are neither $\sqsubseteq$-related nor orthogonal,
then they are _transverse_ and we write $\overline{U}\pitchfork\overline{V}$.

######  Lemma 1.9.

Let $\mathcal{MCG}^{\pm}(S)/N$ be a $D$-large displacement quotient. If $D\geq
2$ then nesting and orthogonality in $\overline{\mathfrak{S}}$ are mutually
exclusive, meaning that if
$\overline{U}\sqsubseteq\overline{V}\in\overline{\mathfrak{S}}$ then
$\overline{U}\not\bot\overline{V}$.

###### Proof.

Let $U,V$ be two representatives such that $U\sqsubseteq V$, and let $x$ be a
curve which lies in $U$. By the assumption on the minimum displacement, for
every $n\in N-\\{1\\}$ the curve $nx$, which lies in $nU$, must intersect $x$.
This means that any representative for $\overline{U}$ cannot be disjoint from
$x$, and therefore from $V$.

Now let $U^{\prime}\in\overline{U}$ and $V^{\prime}\in\overline{V}$ be any two
representatives, and let $n^{\prime}\in N$ such that $n^{\prime}V=V^{\prime}$.
Then the pair $(U^{\prime},V^{\prime})$ is the image under $n^{\prime}$ of
$((n^{\prime})^{-1}U,V)$, and by the previous argument we know that the latter
two subsurfaces intersect. Hence $U^{\prime}$ intersects $V^{\prime}$, that
is, any representative for $\overline{V}$ is not disjoint from any
representative for $\overline{U}$. ∎

######  Lemma 1.10.

Let $\mathcal{MCG}^{\pm}(S)/N$ be a $D$-large displacement quotient for some
$D\geq 4$ and let
$\overline{U}_{1},\ldots,\overline{U}_{k}\in\overline{\mathfrak{S}}$ be a
collection of pairwise orthogonal elements. Then there exist representatives
$U_{1},\ldots,U_{k}\in\mathfrak{S}$ which are pairwise orthogonal.

Notice that this lemma is not at all obvious: we just know that every two
elements of the collection admit disjoint representatives, but this does not
mean, a priori, that this conditions can all be satisfied simultaneously.

###### Proof.

The proof is by induction on $k$, the base case $k=2$ being true by how we
defined orthogonality. Thus let $k>2$ and let $U_{1},U_{2}$ be disjoint
representatives for $\overline{U}_{1},\overline{U}_{2}$. Moreover, let
$U_{3}\in\overline{U}_{3}$ be such that $U_{3}\bot U_{1}$ (such representative
exists since we know that $\overline{U}_{1},\overline{U}_{3}$ admit disjoint
representatives, and up to the $N$-action we can assume that the
representative for $\overline{U}_{1}$ is $U_{1}$). Similarly, there exists
$n\in N$ such that $nU_{3}\bot U_{2}$.

Now fix a curve $x\in U_{1}$. First notice that every curve $y$ lying on
$U_{3}$ is disjoint from $x$. Moreover, $ny$ is disjoint from all curves on
$U_{2}$, which are in turn disjoint from $x$. Hence
$\mathrm{d}_{\mathcal{C}S}(y,ny)\leq 3$, and by the assumption on the minimum
displacement we must have that $n=1$. Thus $U_{3}$ is disjoint from both
$U_{1}$ and $U_{2}$. If we repeat the procedure we can find representatives
$U_{i}\in\overline{U}_{i}$ for every $i=3,\ldots,k$ which are disjoint from
both $U_{1}$ and $U_{2}$. Hence we can replace $U_{1}$ and $U_{2}$ with the
subsurface $U_{1}\sqcup U_{2}$, and now the conclusion follows by induction. ∎

##  2\. Combinatorial rigidity

For the rest of the paper, $\mathcal{MCG}^{\pm}(S)/N$ will be a $D$-large
displacement quotient for some constant $D\geq 8$, so that all results from
Subsections 1.1 and 1.2 hold. The $\mathcal{MCG}^{\pm}(S)$-action on
$\mathcal{C}S$ induces an action of $\mathcal{MCG}^{\pm}(S)/N$ on
$\mathcal{C}S/N$, defined as follows: for every equivalence classes
$\overline{g}\in\mathcal{MCG}^{\pm}(S)/N$ and $\overline{x}\in\mathcal{C}S/N$
we can choose representatives $g$ and $x$ and set
$\overline{g}\cdot\overline{x}=\overline{gx}$. Using that $N$ is a normal
subgroup, one can see that the definition is well-posed. Moreover, the action
is by simplicial automorphism, since it preserves adjacency. Thus we have a
group homomorphism $\mathcal{MCG}^{\pm}(S)/N\to\text{Aut}(\mathcal{C}S/N)$,
where $\text{Aut}(\mathcal{C}S/N)$ is the group of simplicial automorphisms of
$\mathcal{C}S/N$.

The goal of this Section is to prove the following, which is Theorem 1 from
the Introduction. For the sake of simplicity, we will first state the result
for surfaces of complexity at least $3$ which are not $S_{2,0}$, postponing
low-complexity cases to Subsection 2.1.

######  Theorem 2.1 (Combinatorial rigidity).

Let $S$ be a surface of complexity at least $3$ which is not $S_{2,0}$, and
let $\mathcal{MCG}^{\pm}(S)/N$ be a $D$-large displacement quotient. If $D\geq
8$ then the map $\mathcal{MCG}^{\pm}(S)/N\to\text{Aut}(\mathcal{C}S/N)$,
induced by the natural action, is an isomorphism.

For this Section only, it is convenient to see $\mathcal{C}S$ and
$\mathcal{C}S/N$ as simplicial _complexes_ , that is, we consider the flag
simplicial complexes whose $1$-skeleton are the curve graph and its quotient,
respectively. By a result of Harer [Har86], if $S$ has complexity at least $3$
then $\mathcal{C}S$ is simply connected. We can show that $\mathcal{C}S$ is
actually the universal cover of $\mathcal{C}S/N$:

######  Lemma 2.2.

Let $\mathcal{MCG}^{\pm}(S)/N$ be a $D$-large displacement quotient. If $D\geq
8$ then the projection $\pi:\,\mathcal{C}S\to\mathcal{C}S/N$ is a covering
map.

###### Proof.

Just notice that, for every point $p\in\mathcal{C}S$, there exists an open
neighbourhood $U$ whose preimage is the disjoint union of open subsets which
are homeomorphic to $U$ via $\pi$. Indeed, for every $p\in\mathcal{C}S$,
choose a clique $\Delta$ that contains $p$, and let $x$ be any of its
vertices. Then we can choose $U$ to be the interior of the subcomplex spanned
by $B(x,2)$, as a consequence of Lemma 1.7. ∎

###### Proof of Theorem 2.1.

We first argue that the map
$\mathcal{MCG}^{\pm}(S)/N\to\text{Aut}(\mathcal{C}S/N)$ is surjective, i.e.
that, given an automorphism $\overline{\phi}\in\text{Aut}(\mathcal{C}S/N)$, we
can find an element $\overline{g}\in\mathcal{MCG}^{\pm}(S)/N$ which induces
$\overline{\phi}$. Since $\pi:\,\mathcal{C}S\to\mathcal{C}S/N$ is the
universal cover, there exists a continuous map
$\phi:\,\mathcal{C}S\to\mathcal{C}S$ that makes the following diagram commute:

${\mathcal{C}S}$${\mathcal{C}S}$${\mathcal{C}S/N}$${\mathcal{C}S/N}$$\scriptstyle{\phi}$$\scriptstyle{\pi}$$\scriptstyle{\pi}$$\scriptstyle{\overline{\phi}}$

Recall that $\phi$ is defined as follows: fix a vertex $x_{0}\in\mathcal{C}S$
and a lift $x_{1}$ of $\overline{\phi}(\pi(x_{0}))$. Then for every
$x\in\mathcal{C}S$ pick any path $\Gamma$ from $x_{0}$ to $x$, and set
$\phi(x)$ as the endpoint of the lift of $\overline{\phi}(\pi(\Gamma))$
starting at $x_{0}$. In particular, if two vertices $x,y\in\mathcal{C}S^{(0)}$
are adjacent, then their images $\phi(x),\phi(y)$ will either be adjacent or
coincide. Moreover, the second happens precisely when $\pi(x)=\pi(y)$, and by
the assumption on the minimum displacement this cannot happen for adjacent
vertices.

Hence, at the level of the $1$-skeleton, $\phi$ is a simplicial map, and it is
actually an automorphism since we can repeat the same argument replacing
$\overline{\phi}$ with its inverse $\overline{\phi}^{-1}$. Hence, by Ivanov’s
Theorem [Iva97, Kor99] there is an element $g\in\mathcal{MCG}^{\pm}(S)$ which
induces $\phi$, and therefore its image
$\overline{g}\in\mathcal{MCG}^{\pm}(S)/N$ induces $\overline{\phi}$.

Now we turn to injectivity. If an element $g\in\mathcal{MCG}^{\pm}(S)$ induces
the identity on $\mathcal{C}S/N$ when passing to the quotient, then the
$g$-action on $\mathcal{C}S$ is a covering automorphism. Now, the group of
covering automorphisms is the group $N$ itself, since by Definition 1.2 the
$N$-action on $\mathcal{C}S$ is properly discontinuous. Thus there exists an
element $n\in N$ such that the composition $gn$ fixes every curve of
$\mathcal{C}S$, and since $S$ has complexity at least $3$ and is not an
$S_{2,0}$ we have that $gn$ must be the identity (see e.g. the discussion in
[FM12, Section 3.4]). ∎

###  2.1. Some low-complexity cases

In this Subsection we establish an analogue of Theorem 2.1 for some surfaces
of low complexity. Let us first consider the cases when $S$ is either
$S_{0,4}$, $S_{1,0}$ or $S_{1,1}$. Then the curve graph has a slightly
different definition: we connect two (isotopy classes of) curves if and only
if they realise the minimal intersection number among all pairs of curves on
the surface, which is $1$ for $S_{1,0}$ and $S_{1,1}$ and $2$ for $S_{0,4}$.
The resulting simplicial complex is the Farey complex (see e.g. [Min96] for a
proof), which is a triangulation of the compactification of the hyperbolic
plane and therefore is simply connected. Moreover, the definition of a
$D$-large displacement quotient and the relative Lemmas from Section 1 did not
assume any property on the nature of the vertices of $\mathcal{C}S$, thus they
still hold in our context. Then one gets the following:

######  Theorem 2.3.

Let $S$ be either $S_{0,4}$, $S_{1,0}$ or $S_{1,1}$, let $D\geq 8$ and let
$\mathcal{MCG}^{\pm}(S)/N$ be a $D$-large displacement quotient. Then the map
$\mathcal{MCG}^{\pm}(S)/N\to\text{Aut}(\mathcal{C}S/N)$ is surjective and has
finite kernel $\overline{K}$, generated by the images of the _hyperelliptic
involutions_ in Figure 1.

###### Proof.

For the surjectivity part, we can run the exact same proof of Theorem 2.1,
which only uses that the covering space $\mathcal{C}S$ is simply connected.
Moreover, the injectivity part of that proof implies that, if
$\overline{g}\in\mathcal{MCG}^{\pm}(S)/N$ acts trivially on $\mathcal{C}S/N$,
then it is the image of an element $g\in\mathcal{MCG}^{\pm}(S)$ which acts
trivially on $\mathcal{C}S$. Such $g$ must then lie in the finite subgroup
generated by the _hyperelliptic involutions_ in Figure 1 (again, this follows
from e.g. [FM12, Section 3.4]). ∎

With the exact same procedure, one can get the following:

######  Theorem 2.4.

Let $S=S_{2,0}$, and let $\mathcal{MCG}^{\pm}(S)/N$ be a $D$-large
displacement quotient. If $D\geq 8$ then the map
$\mathcal{MCG}^{\pm}(S)/N\to\text{Aut}(\mathcal{C}S/N)$ is surjective, and the
kernel is the $\mathbb{Z}/2\mathbb{Z}$ subgroup generated by the image of the
hyperelliptic involution in Figure 1.

Figure 1. The dots represent the punctures. Rotations by $\pi$ about the
indicated axes fix all curves on the surfaces.

Finally, leaving aside the case $S=S_{1,2}$, which is more involved since the
natural map $\mathcal{MCG}(S_{1,2})\to\text{Aut}(\mathcal{C}S_{1,2})$ is not
surjective (see the main result of [Luo00]), we are left to deal with the case
$S=S_{0,5}$. The curve complex $\mathcal{C}S$ has dimension $1$ and is not
simply connected. However, we shall see that we can repeat the same procedure
as in Theorem 2.1 if we show that the polyhedral complex, obtained from
$\mathcal{C}(S_{0,5})$ by gluing a cell to every $5$-cycle, is simply
connected. The latter is an unpublished result of Piotr Przytycki, which I
thank for explaining its proof to me.

First, by an _arc_ we mean the isotopy class, relative to the punctures, of a
map $\gamma:\,[0,1]\to S_{0,5}$ mapping each endpoint of the interval to a
puncture. For every surface $S$ with at least $2$ punctures, let
$\mathcal{A}^{2}(S)$ be the simplicial complex whose vertices are all arcs
$\gamma:\,[0,1]\to S_{0,5}$ whose endpoints are _distinct_ punctures, and
where a collection of arcs span a simplex if their interiors are pairwise
disjoint.

######  Lemma 2.5.

If $S$ has at least two punctures, $\mathcal{A}^{2}(S)$ is simply-connected.

The same proof will yield that $\mathcal{A}^{2}(S)$ is contractible, though we
will not need it.

###### Proof.

This fact is proven as [Sch20, Claim 3.17], but we spell out the details for
clarity. Let $\gamma:\,\mathbb{S}^{1}\to\mathcal{A}^{2}(S)$ be a simplicial
loop, and fix an arc $x_{0}\in\gamma$. If every other arc $y\in\gamma$ is
disjoint from $x_{0}$, then we can connect $x_{0}$ to $y$, and this defines an
extension of $\gamma$ to some triangulation of the disk $D^{2}$.

Thus, suppose that $\Sigma(\gamma)=\sum_{y\in\gamma}i(x,y)>0$, where $i(x,y)$
is the number of intersection of two representatives of $x$ and $y$ in minimal
position. Fix and endpoint $p_{0}$ of $x_{0}$, and let $y_{0}\in\gamma$ be the
first arc one meets when travelling along $x_{0}$ starting from $p_{0}$. Let
$t$ be the path along $x_{0}$ from $p_{0}$ to the first intersection to
$y_{0}$, and let $z$ and $z^{\prime}$ be the essential arcs obtained as the
boundaries of a regular neighbourhood of $t\cup y_{0}$. Since $y_{0}$ has
distinct endpoints, one of its endpoints is not $p_{0}$, and therefore one of
the new arcs, say $z$, has distinct endpoints. Notice moreover that $z$ is
disjoint from $y_{0}$, and if $y_{1}\in\gamma$ was disjoint from $y_{0}$ then
it is also disjoint from $z$, because we choose $y_{0}$ to have the closest
intersection to $p_{0}$. Thus, we can replace $y_{0}$ with $z$ and obtain a
loop $\gamma^{\prime}$, which is homotopic to $\gamma$ but is such that
$\Sigma(\gamma^{\prime})<\Sigma(\gamma)$. Then we conclude by induction on
$\Sigma(\gamma)$. ∎

######  Definition 2.6.

Let $S=S_{0,5}$. By a _pentagon_ we mean a $5$-cycle $P\subset\mathcal{C}S$.
Let $\mathcal{C}S_{P}$ be the polyhedral complex obtained from $\mathcal{C}S$
by gluing a $2$-cell to every pentagon.

######  Lemma 2.7 (Przytycki).

The complex $\mathcal{C}S_{P}$ is simply connected.

###### Proof.

First, notice that $\mathcal{C}S$ can be seen as a non-complete subgraph of
$\mathcal{A}^{2}(S)$, by mapping every curve $y$ to the unique arc $a(y)$ in
the twice-punctured disk that $y$ cuts out on $S$. The map
$a:\,\mathcal{C}S\to\mathcal{A}^{2}(S)$ is bijective at the level of vertices,
but not at the level of edges since in $\mathcal{A}^{2}(S)$ we allow two arcs
to have the same endpoints. Notice moreover that, if the union of five arcs is
a pentagon on $S$ (meaning that there is a cyclic ordering such that each arc
intersects only the previous and the following, and only at a single
endpoint), then the preimages under $a$ of these arcs form a pentagon, in the
sense of Definition 2.6.

Now let $\gamma:\,\mathbb{S}^{1}\to\mathcal{C}S_{P}$ be a simplicial loop,
whose image lies in the $1$-skeleton $\mathcal{C}S_{P}^{(1)}=\mathcal{C}S$.
The image $a(\gamma)\subset\mathcal{A}^{2}(S)$ can be filled with triangles,
since $\mathcal{A}^{2}(S)$ is simply connected. Let
$x_{0},x_{1},x_{2}\in\mathcal{A}^{2}(S)$ be the vertices of such a triangle
$T$. If $x_{i}$ and $x_{j}$ have disjoint endpoints, then the corresponding
curves $a^{-1}(x_{i})$ and $a^{-1}(x_{j})$ are disjoint, so that they are
connected by an edge in $\mathcal{C}S$. Otherwise, there exists a unique arc
$\varepsilon_{ij}\in\mathcal{A}^{2}(S)$ such that $a^{-1}(\varepsilon_{ij})$
is disjoint from both $a^{-1}(x_{i})$ and $a^{-1}(x_{j})$, as in Figure 2 (it
is the only arc with distinct endpoints in the complement of $x_{i}\cup
x_{j}$).

Figure 2. The possible configurations for two arcs $x_{0}$ and $x_{1}$, and
the corresponding arcs $\varepsilon_{01}$, if needed.

Now by construction, the curves

$a^{-1}\left(\\{x_{i}\\}_{i=0,1,2}\cup\\{\varepsilon_{ij}\\}_{0\leq i<j\leq
2}\right)$

form a loop $\delta$ in $\mathcal{C}S$. Therefore we are left to prove that
$\delta$ can be filled with finitely many pentagons.

If there is a pair of arcs, say, $x_{0}$, $x_{1}$, which have distinct
endpoints, then $x_{2}$ must share an endpoint with each of them. This shows
that $\delta$ has precisely $5$ vertices, and therefore is filled by a
pentagon in $\mathcal{C}S_{P}$. Otherwise, suppose that all pairs $x_{i}$,
$x_{j}$ share an endpoint, that is, $\delta$ has six vertices. There are five
possible cases, as in Figure 3.

Figure 3. The possible configurations for three arcs $x_{0},x_{1},x_{2}$, and
the corresponding arcs $\varepsilon_{ij}$.

Let us analyse all of them separately.

  1. (1)

The $\varepsilon$-arcs coincide, thus
$\delta=a^{-1}\left(\\{x_{i}\\}_{i=0,1,2}\cup\\{\varepsilon\\}\right)$ is a
tripod.

  2. (2)

Add the arc $z$ in the Figure. Now
$\\{x_{0},x_{1},z,\varepsilon_{01},\varepsilon_{12}\\}$ form a pentagon, as
well as $\\{x_{0},x_{2},z,\varepsilon_{02},\varepsilon_{12}\\}$. Therefore we
can fill $\delta$ with the preimage under $a$ of these arcs, which is a union
of two pentagons joined along two edges.

  3. (3)

The $\varepsilon$-arcs coincide, so we are in the same situation as case 1.

  4. (4)

Similarly to case 2, adding $z$ creates two pentagons.

  5. (5)

This time we need four auxiliary curves, in order to form four pentagons
$P_{1}=\\{x_{0},a,\varepsilon_{01},\varepsilon_{12},b\\}$,
$P_{2}=\\{x_{2},a,\varepsilon_{12},\varepsilon_{02},c\\}$,
$P_{3}=\\{x_{1},a,\varepsilon_{12},d,c\\}$ and
$P_{4}=\\{x_{1},a,\varepsilon_{01},d,b\\}$.

∎

Now we are ready to prove quasi-isometric rigidity for
$\mathcal{C}(S_{0,5})/N$:

######  Theorem 2.8.

Let $S=S_{0,5}$, and let $\mathcal{MCG}^{\pm}(S)/N$ be a $D$-large
displacement quotient. If $D\geq 8$ then the map
$\mathcal{MCG}^{\pm}(S)/N\to\text{Aut}(\mathcal{C}S/N)$ is an isomorphism.

###### Proof of Theorem 2.8.

Notice first that pentagons in $\mathcal{C}S/N$ can be lifted and pentagons in
$\mathcal{C}S$ can be projected isometrically, as a consequence of Lemma 1.7.
Thus we can consider the quotient $\mathcal{C}S_{P}/N$, which is again a
polygonal complex. Moreover, any automorphism
$\overline{\phi}:\,\mathcal{C}S/N\to\mathcal{C}S/N$ maps pentagons to
pentagons, and therefore extends to an automorphism of $\mathcal{C}S_{P}/N$.
Now we can argue as in Theorem 2.1, this time using the covering
$\mathcal{C}S_{P}\to\mathcal{C}S_{P}/N$, to show that the map
$\mathcal{MCG}^{\pm}(S)/N\to\text{Aut}(\mathcal{C}S/N)$ is an isomorphism
(again, injectivity follows from [FM12, Section 3.4]). ∎

##  3\. Quasi-isometric rigidity

This section is devoted to the proof of Theorem 2 from the Introduction, that
is quasi-isometric rigidity of certain large displacement quotients, which
admit a particular _hierarchically hyperbolic group_ (HHG) structure (see
Theorem 3.8 for the exact statement). The idea is to show that, from any self-
quasi-isometry $f$ of a HHG satisfying some additional assumptions, one can
extract an automorphism of a certain graph, which in our case will coincide
with $\mathcal{C}S/N$. Then we will be able to apply the combinatorial
rigidity results from Section 2 to produce a candidate element
$\overline{g}\in\mathcal{MCG}^{\pm}(S)/N$, and with a little more effort we
will show that $f$ coarsely coincides with the left multiplication by
$\overline{g}$. Once again, I am grateful to Jason Behrstock, Mark Hagen, and
Alessandro Sisto for their contribution, in particular to the proof of Lemma
3.15.

###  3.1. Background on hierarchical hyperbolicity

First, we recall the intuition behind some notions from the world of
_hierarchically hyperbolic spaces and groups_ , first introduced by Behrstock,
Hagen, and Sisto in [BHS17b]. We refer to [BHS19] for the details of the
definitions.

####  3.1.1. HHS and HHG

A _hierarchically hyperbolic space_ (HHS for short) is a metric space $X$ that
comes with certain additional data, most importantly a family of uniformly
hyperbolic spaces $\\{\mathcal{C}Y\\}_{Y\in\mathfrak{S}}$, called _coordinate
spaces_ , and uniformly coarsely Lipschitz maps $\pi_{Y}:X\to\mathcal{C}Y$,
which shall be thought of as coordinates for points in $X$. For mapping class
groups, these are curve graphs of subsurfaces and maps coming from subsurface
projections. Moreover, the _domain set_ $\mathfrak{S}$, that is, the set which
indexes the family of coordinate spaces, has a partial ordering $\sqsubseteq$,
called _nesting_ , with a unique maximal element, and an equivalence relation
$\bot$, called _orthogonality_. For mapping class groups, these are
containment and disjointness of subsurfaces (up to isotopy). When two domains
$U,V$ of $\mathfrak{S}$ are not $\sqsubseteq$\- nor $\bot$-related, one says
that they are _transverse_ , and one writes $U\pitchfork V$. Finally, whenever
$U\not\bot V$ there is a _projection_
$\rho^{U}_{V}:\,\mathcal{C}U\to\mathcal{C}V$, which for mapping class group is
again defined using subsurface projection.

An _action_ of a finitely generated group $G$ on a hierarchically hyperbolic
space $(X,\mathfrak{S})$ is the data of:

  * •

an action $G\circlearrowleft X$ by isometries;

  * •

an action $G\circlearrowleft\mathfrak{S}$, preserving nesting and
orthogonality;

  * •

for every $g\in G$ and every $Y\in\mathfrak{S}$, an isometry
$g_{Y}:\,\mathcal{C}Y\to\mathcal{C}g(Y)$.

Moreover, one requires that the two actions are compatible, meaning that for
every $g\in G$ and every non-orthogonal domains $U,V\in\mathfrak{S}$ the
following diagrams commute:

${X}$${X}$${\mathcal{C}U}$${\mathcal{C}g(U)}$$\scriptstyle{g}$$\scriptstyle{\pi_{U}}$$\scriptstyle{\pi_{U}}$$\scriptstyle{g_{U}}$${\mathcal{C}U}$${\mathcal{C}g(U)}$${\mathcal{C}V}$${\mathcal{C}g(V)}$$\scriptstyle{g_{U}}$$\scriptstyle{\rho^{U}_{V}}$$\scriptstyle{\rho^{g(U)}_{g(V)}}$$\scriptstyle{g_{V}}$

We will often slightly abuse notation and drop the subscript for the
isometries $g_{U}$. If the action on $X$ is metrically proper and cobounded,
and the action on $\mathfrak{S}$ is cofinite, then $G$ is a _hierarchically
hyperbolic group_ (HHG for short), and any quasi-isometry between $G$ and $X$
given by the Milnor-Švarc lemma can be used to endow $G$ with the HHS
structure of $X$.

For later purposes, we point out that hierarchically hyperbolic groups are
examples of _asymphoric_ HHS, in the sense of [BHS21, Definition 1.14], though
we will not need the actual definition of this property.

####  3.1.2. Product, flats, orthants

The idea of orthogonality is that it corresponds to products, in the following
sense. Given any $U\in\mathfrak{S}$, there is a corresponding space $F_{U}$
associated to it, which is quasi-isometrically embedded in $X$ ($F_{U}$ is a
HHS itself with domain set $\mathfrak{S}_{U}=\\{Y\in\mathfrak{S}:Y\sqsubseteq
U\\}$, and in mapping class groups these essentially correspond to mapping
class groups of subsurfaces). Given a maximal set $\\{U_{i}\\}$ of pairwise
orthogonal elements of $\mathfrak{S}$, there is a corresponding _standard
product region_ $P_{\\{U_{i}\\}}$ which is quasi-isometric to the product of
the $F_{U_{i}}$ (think of a Dehn twist flat as, coarsely, a product of annular
curve graphs).

Moreover, inside $X$ and the $F_{U}$s there are special (uniform) quasi-
geodesics, called _hierarchy paths_ , which are those that project
monotonically (with uniform constants) to all $\mathcal{C}Y$. Similarly, there
are hierarchy rays and hierarchy lines (which are quasigeodesic rays and
lines, respectively). Given a product region $P_{\\{U_{i}\\}}$ and hierarchy
lines (resp., rays) in some of the $F_{U_{i}}$, the product region contains a
product of those (where for the indices $i$ for which a line/ray has not been
assigned one chooses a point in $F_{U_{i}}$ instead). This is what we will
refer to as _standard $k$-flats (resp., orthants)_, where $k$ is the number of
lines/rays. The _support_ of a standard $k$-flat (resp., orthant) is the set
of $U_{i}$ for which a line/ray has been assigned. Related to this, a
_complete support set_ is a subset
$\\{U_{i}\\}_{i=1}^{\nu}\subseteq\mathfrak{S}$ of pairwise orthogonal domains
with all $\mathcal{C}U_{i}$ unbounded, and with maximal possible cardinality
$\nu$ among sets with these properties. This definition is relevant here as
$\nu$, which is called the _rank_ of $X$, is then the maximal dimension of
standard orthants (if the $\mathcal{C}U_{i}$ are not only unbounded, but also
have non-empty Gromov boundary, as will be the case for us). Moreover, in this
case points in the Gromov boundary $\partial C(U_{i})$ each determine a
hierarchy ray, and similarly pairs of points determine hierarchy lines. In
particular, for every complete support set $\\{U_{i}\\}_{i=1}^{\nu}$ and for
every choice of distinct points $p_{i}^{\pm}\in\partial\mathcal{C}U_{i}$ there
exists an associated standard $\nu$-flat
$\mathfrak{F}_{\\{(U_{i},p_{i}^{\pm})\\}}$, which is the product of the
hierarchy lines in $F_{U_{i}}$ whose projections to $\mathcal{C}U_{i}$ have
endpoints $p_{i}^{\pm}$. We will refer to standard $\nu$-flats (resp. standard
$\nu$-orthants) simply as _standard flats_ (resp. _standard orthants_).

The following Lemma, which follows from carefully inspecting the proofs of
[BHS21, Lemmas 4.11 and 4.12], describes the (coarse) intersection of standard
flats and orthants. Recall that, given subsets $A,B$ of a metric space $X$,
the coarse intersection $A\tilde{\cap}B$, if well-defined, is a subspace of
$X$ within bounded Hausdorff distance of all intersections $N_{R}(A)\cap
N_{R}(B)$ of their $R$-neighbourhoods, for $R$ sufficiently large.

######  Lemma 3.1 (Coarse intersection of standard orthants and flats).

Let $(X,\mathfrak{S})$ be an asymphoric HHS of rank $\nu$.

  * •

Let $\mathcal{O},\mathcal{O}^{\prime}$ be standard orthants in $X$ with
supports $\\{U_{i}\\}_{i=1}^{\nu}$ and $\\{V_{i}\\}_{i=1}^{\nu}$. Then
$\mathcal{O}\tilde{\cap}\mathcal{O}^{\prime}$ is well-defined, and coarsely
coincides with a standard $k$–orthant whose support is contained in
$\\{U_{i}\\}_{i=1}^{\nu}\cap\\{V_{i}\\}_{i=1}^{\nu}$.

  * •

Let $\mathcal{F},\mathcal{F}^{\prime}$ be standard orthants in $X$ with
supports $\\{U_{i}\\}_{i=1}^{\nu}$ and $\\{V_{i}\\}_{i=1}^{\nu}$. Then
$\mathcal{F}\tilde{\cap}\mathcal{F}^{\prime}$ is well-defined, and coarsely
coincides with a product $\prod_{T}\ell_{T}$, where $T$ varies in
$\\{U_{i}\\}_{i=1}^{\nu}\cap\\{V_{i}\\}_{i=1}^{\nu}$ and every $\ell_{T}$ is
either a point, a standard $1$-orthant or a standard $1$-flat supported on
$T$.

###  3.2. HHG structures inherited from the surface

From now on, we will further restrict our attention to those large
displacement quotients admitting a HHG structure whose “interesting” domains
are orbits of subsurfaces:

######  Convention 3.2 (Surface-inherited HHG structure).

$\mathcal{MCG}^{\pm}(S)/N$ is a $D$-large displacement quotient for some
$D\geq 8$, and it admits a HHG structure with the following properties:

  * •

The domain set is
$\mathfrak{T}=\overline{\mathfrak{S}}\sqcup\mathfrak{T}^{\prime}$, where
$\overline{\mathfrak{S}}=\mathfrak{S}/N$ is the collection of $N$-orbits of
essential subsurfaces.

  * •

Two domains $\overline{U},\overline{V}\in\overline{\mathfrak{S}}$ are nested
(resp. orthogonal) if and only if they admit representatives
$U,V\in\mathfrak{S}$ which are nested (resp. orthogonal).

  * •

The $\sqsubseteq$-maximal element of $\mathfrak{T}$ is (the $N$-orbit of) the
whole surface $S$, and its coordinate space, which we refer to as the _top-
level coordinate space_ , is $\mathcal{MCG}^{\pm}(S)/N$-equivariantly quasi-
isometric to $\mathcal{C}S/N$. In particular, $\mathcal{C}S/N$ is hyperbolic.

  * •

Whenever $\overline{U}\in\overline{\mathfrak{S}}$ is not the
$\sqsubseteq$-maximal element, the coordinate space associated to
$\overline{U}$ is $\left(\bigcup_{U\in\overline{U}}\mathcal{C}U\right)/N$.
Furthermore, the quotient projection
$\bigcup_{U\in\overline{U}}\mathcal{C}U\to\mathcal{C}\overline{U}$ restricts
to a quasi-isometry $\mathcal{C}U\to\mathcal{C}\overline{U}$, for any
representative $U\in\overline{U}$. Here $\mathcal{C}U$ is the usual curve
graph of $U$, unless $U$ has complexity $1$ and $\mathcal{C}U$ is isomorphic
to the Farey complex, or $U$ is an annulus and $\mathcal{C}U$ is its annular
curve graph.

  * •

There exists a constant $E$ such that, for every $T\in\mathfrak{T}^{\prime}$,
either $\mathrm{diam}(\mathcal{C}T)\leq E$ or $T$ is not orthogonal to any
other element of $\mathfrak{T}$.

Here are some examples of quotients with a surface-inherited structure.

######  Example 3.3.

$\mathcal{MCG}^{\pm}(S)$ itself has a HHG structure whose domain set coincides
with $\mathfrak{S}$ and whose coordinate spaces are curve graphs of
subsurfaces (see e.g. [BHS19, Theorem 11.1]).

######  Example 3.4 (Hierarchically hyperbolically embedded subgroups).

Let $H$ be a finitely generated subgroup of $\mathcal{MCG}^{\pm}(S)$ which is
_hierarchically hyperbolically embedded_ , as defined in [BHS17a, Definition
6.1] building on the notion of _hyperbolically embedded_ subgroups from
[DGO17, Definition 2.1]. Lemma 3.5 describes conditions on a subgroup $M\leq
H$ ensuring that the quotient by the normal closure of $M$ satisfies
Convention 3.2.

######  Lemma 3.5.

Let $H\leq\mathcal{MCG}^{\pm}(S)$ be a finitely generated subgroup which is
hierarchically hyperbolically embedded. There exists a finite set $F\subset
H-\\{1\\}$ such that, if a normal subgroup $M\unlhd H$ avoids $F$ and $H/M$ is
hyperbolic, then the quotient $\mathcal{MCG}^{\pm}(S)/N$ by the normal closure
of $M$ satisfies Convention 3.2.

###### Proof.

Consider any Cayley graph for $H$, with respect to any finite set of
generators. As noted in [BHS17a, Remark 6.3], the orbit maps of $H$ to
$\mathcal{C}S$ are quasi-isometric embeddings (actually, the image is also
quasi-convex). Hence there exists $R\geq 0$ such that, denoted by $B(1,R)$ the
ball of radius $R$ centred at the identity of $H$, if $M\cap B(1,R)=\\{1\\}$
then $M$ acts on $\mathcal{C}S$ with minimum displacement at least $8$. This
in turn means that, for $M$\- to act with large minimum displacement, it
suffices that it avoids finitely many elements, since balls in a Cayley graph
of a finitely generated group are finite.

Now we claim that, if $M$ avoids a possibly bigger finite set, then its normal
closure $N$ acts on $\mathcal{C}S$ with minimum displacement at least $8$ as
well. This follows from results of [DGO17] on hyperbolically embedded
subgroups, but we can give a proof based on the machinery from [CM22]. There
the authors take as input our setting and output a _projection complex_ ,
which we shall think of as the data of:

  * •

a graph $\mathcal{P}$, whose vertices correspond in our case to orbits of the
form $gH(x_{0})\subset\mathcal{C}S$ for all $g\in\mathcal{MCG}^{\pm}(S)$ and
for a fixed basepoint $x_{0}\in\mathcal{C}S^{(0)}$;

  * •

a notion of projection between vertices of $\mathcal{P}$, which for us is
given by projections onto the quasi-convex subsets
$gH(x_{0})\subset\mathcal{C}S$;

  * •

an action $N\circlearrowleft\mathcal{P}$, such that the stabiliser of
$gH(x_{0})$ is precisely $gMg^{-1}$ (this follows from [DGO17, Theorem 2.14],
where $N$ is described as an infinite free product of conjugates of $M$).

An example of the construction of the projection complex for the case when $M$
is generated by a pseudo-Anosov mapping class can be found in [CM22, Section
8.1].

Now, for every $x\in\mathcal{C}S^{(0)}$ there exists
$g\in\mathcal{MCG}^{\pm}(S)$ such that the orbit $gH(x_{0})$ is uniformly
close to $x$, since
$\bigcup_{g\in\mathcal{MCG}^{\pm}(S)}gH=\mathcal{MCG}^{\pm}(S)$ and the action
of $\mathcal{MCG}^{\pm}(S)$ on the curve graph is cofinite. Now let $n\in
N-\\{1\\}$, and we want to say that $\mathrm{d}_{\mathcal{C}S}(x,nx)\geq 8$.
Notice that we can think of $gH(x_{0})$ and $ngH(x_{0})$ as vertices of
$\mathcal{P}$. Therefore, by [CM22, Proposition 3.2] one of the following must
hold:

  * •

$n$ fixes $gH(x_{0})$, and therefore it belongs to $gMg^{-1}$. In particular,
$n$ has minimum displacement at least $8$, since we already know that every
non-trivial element in $M$ does and minimum displacement is preserved under
conjugation.

  * •

$n$ fixes $ngH(x_{0})$, and for the same reason it has minimum displacement at
least $8$.

  * •

There exists a third vertex of $\mathcal{P}$, that is, a third orbit
$g^{\prime}H(x_{0})$, on which $gH(x_{0})$ and $ngH(x_{0})$ have distant
projection. Now, the projection of $x$ is close to the projection of
$gH(x_{0})$, and similarly $nx$ projects close to $ngH(x_{0})$. Therefore $x$
and $nx$ have distant projections to $g^{\prime}H(x_{0})$, and in turn this
means that $x$ and $nx$ are far from each other in $\mathcal{C}S$.

Now, [BHS17a, Theorem 6.2] states that, if $M$ avoids a (possibly larger)
finite set and $H/M$ is hyperbolic, then $\mathcal{MCG}^{\pm}(S)/N$ admits a
hierarchically hyperbolic structure with the following properties:

  * •

The index set is
$\mathfrak{T}=\overline{\mathfrak{S}}\cup\\{NgH\\}_{g\in\mathcal{MCG}^{\pm}(S)}$;

  * •

Orthogonality and nesting in $\overline{\mathfrak{S}}$ correspond to
orthogonality and nesting between representatives;

  * •

If $\overline{U}\in\overline{\mathfrak{S}}$ is not the maximal element, then
$\mathcal{C}\overline{U}$ is quasi-isometric to $\mathcal{C}U$ via the
projection map, for any representative $U$ for $\overline{U}$;

  * •

If $T\in\mathfrak{T}$ is of the form $NgH$ then $T$ has no orthogonal, and
$\mathcal{C}T$ is isometric to $H/M$;

  * •

The top-level coordinate space is obtained from $\mathcal{C}S/N$ after coning
off every $H/M$-orbit.

Thus all requirements of Convention 3.2 are satisfied, except for the quasi-
isometry type of the top-level coordinate space. Now one can apply the
procedure from [ABD21, Theorem 3.7] to remove the domains without orthogonals,
and get a new structure $(\mathcal{MCG}^{\pm}(S)/N,\mathfrak{R})$ such that:

  * •

The new index set coincides with the maximal element, together with all
elements $\mathfrak{T}$ admitting an orthogonal domain with unbounded
coordinate space. Hence $\mathfrak{R}=\overline{\mathfrak{S}}$;

  * •

Orthogonality and nesting are inherited from the original structure;

  * •

If $\overline{U}\in\overline{\mathfrak{S}}$ is not the maximal element, then
$\mathcal{C}\overline{U}$ is unchanged;

  * •

The top-level coordinate space is the space $\widehat{X}$ obtained from
$\mathcal{MCG}^{\pm}(S)/N$ after coning off the factors $F_{\overline{U}}$ for
every non-maximal $\overline{U}\in\overline{\mathfrak{S}}$. These factors
correspond to the stabilisers of the action of $\mathcal{MCG}^{\pm}(S)/N$ on
$\mathcal{C}S/N$. Therefore, by a version of the Milnor-Švarc Lemma described
in e.g. [CC07, Theorem 5.1], $\mathcal{C}S/N$ is quasi-isometric to
$\widehat{X}$.

Therefore the new structure satisfies Convention 3.2. ∎

Now we can finally prove that the quotient by a high enough power of a pseudo-
Anosov element is a large displacement quotient:

######  Corollary 3.6.

Let $f\in\mathcal{MCG}^{\pm}(S)$ be pseudo-Anosov element. There exists
$K_{0}\in\mathbb{N}$ such that, if $K\in\mathbb{Z}-\\{0\\}$ is a multiple of
$K_{0}$, then the quotient $\mathcal{MCG}^{\pm}(S)/N$ by the normal closure of
$f^{K}$ satisfies Convention 3.2, and in particular is a $D$-large
displacement for some $D\geq 8$.

###### Proof.

The _elementary closure_ $E(f)$ of a pseudo-Anosov element
$f\in\mathcal{MCG}^{\pm}(S)$, that is, the unique maximal virtually-cyclic
subgroup containing $f$, is hierarchically hyperbolically embedded (this
ultimately follows from [DGO17, Theorem 6.50]). Moreover, by inspection of the
proof of [DGO17, Lemma 6.5], the cyclic group $\langle f\rangle$ generated by
$f$ has finite index in $E(f)$, therefore there exists $K_{0}$ such that
$\langle f^{K_{0}}\rangle$ is normal in $E(f)$. We can further assume that
$\langle f^{K_{0}}\rangle$ avoids the finite set $F\subset E(f)$ given by
Lemma 3.5. Hence, the hypothesis of Lemma 3.5 are satisfied by $\langle
f^{K}\rangle$ whenever $K$ is a non-trivial multiple of $K_{0}$. ∎

For later purposes, we also notice that the existence of a surface-inherited
HHG structure implies that $\mathcal{C}/N$ is unbounded, though in all known
examples this is already true for less artificial reasons:

######  Lemma 3.7.

Let $S$ be either a surface of complexity at least $2$, and let
$\mathcal{MCG}^{\pm}(S)/N$ satisfy Convention 3.2. Then $\mathcal{C}S/N$ is
unbounded.

###### Proof.

By [PS23, Theorem 3.2], with holds for any HHG, there exists a maximal family
$\\{T_{1},\ldots,T_{k}\\}\subset\mathfrak{T}$ of pairwise orthogonal domains
with unbounded coordinate spaces, such that any other domain
$R\in\mathfrak{T}$ with unbounded coordinate space is nested in one of the
$T_{i}$. If one of the $T_{i}$ is the $\sqsubseteq$-maximal element then we
are done, because then its coordinate space, which by Convention 3.2 is quasi-
isometric to $\mathcal{C}S/N$, must be unbounded.

Otherwise, pick an annulus $U$, and let $V$ be the subsurface obtained from
$S-U$ after removing all connected components which are pants. Since the
complexity is at least $2$, $V$ is actually an element of $\mathfrak{S}$,
which is nested only inside $S$ and itself. Moreover, if we denote its
$N$-orbit by $\overline{V}$, then $\mathcal{C}\overline{V}$ is quasi-isometric
to $\mathcal{C}V$ and is therefore unbounded. Thus, without loss of generality
$\overline{V}\sqsubseteq T_{1}$, and since $T_{1}$ is not the maximal element,
we must have that $T_{1}=\overline{V}$. In particular, since any other $T_{i}$
is orthogonal to $T_{1}$, the collection $\\{T_{1},\ldots,T_{k}\\}$ can
consist only of $\overline{V}$ and $\overline{U}$.

Now it is enough to produce an annulus $W$ whose $N$-orbit $\overline{W}$ is
not nested in either $\overline{V}$ or $\overline{U}$, and this will lead to a
contradiction because $\mathcal{C}\overline{W}$ is unbounded by our
Convention. Let $x$ be the core of $U$, which is also the boundary of $V$
inside $S$, pick a curve $y\in\mathcal{C}S^{(0)}$ at distance $2$ from $x$,
and let $W$ be the annulus with core $y$. Then all $N$-translates of $x$ must
be at distance at least $2$ from $y$, by the assumption on the minimum
displacement, and this means that all $N$-translates of $W$ must cross $U$. In
particular $\overline{W}$ cannot be nested in either $\overline{U}$ or
$\overline{V}$. ∎

###  3.3. Statement of the results

The goal of this Section is to prove that, if $\mathcal{MCG}^{\pm}(S)/N$
satisfies Convention 3.2, then it is quasi-isometrically rigid in the
following sense. Recall that two groups $G$ and $H$ are said to be _weakly
commensurable_ if there exist two finite normal subgroups $L\unlhd H$ and
$M\unlhd G$ such that the quotients $H/L$ and $G/M$ have two finite index
subgroups that are isomorphic.

######  Theorem 3.8 (Quasi-isometric rigidity).

Let $S$ be a surface of complexity at least $2$ which is not a torus with two
punctures, and let $\mathcal{MCG}^{\pm}(S)/N$ satisfy Convention 3.2. Then
$MCG/N$ is quasi-isometrically rigid, meaning that if a finitely generated
group $G$ and $\mathcal{MCG}^{\pm}(S)/N$ are quasi-isometric then they are
weakly commensurable.

As we will see, the assumption on the topological type of $S$ is needed to
ensure that every automorphism of $\mathcal{C}S/N$ is induced by a mapping
class, as a consequence of Theorems 2.1, 2.4 and 2.8. Moreover, we require the
complexity to be at least $2$, so that in $\mathcal{MCG}^{\pm}(S)/N$ we can
find “Dehn Twist flats”, that is, standard flats supported on the orbits of
pairwise disjoint annuli, of dimension at least $2$. See Lemma 3.20 for more
details.

Since $\mathcal{MCG}^{\pm}(S)$ itself is a $D$-large displacement quotient for
every choice of $D$, as pointed out in Example 1.3, we also recover quasi-
isometric rigidity of mapping class groups:

######  Corollary 3.9 (QI-rigidity of $\mathcal{MCG}^{\pm}(S)$).

Let $S$ be a surface of complexity at least $2$ which is not a torus with two
punctures. Then $\mathcal{MCG}^{\pm}(S)$ is quasi-isometrically rigid, meaning
that if a finitely generated group $G$ and $\mathcal{MCG}^{\pm}(S)$ are quasi-
isometric then they are weakly commensurable.

Moreover, in view of Lemma 3.5 we get the following:

######  Corollary 3.10.

Let $S$ be a surface of complexity at least $2$ which is not a torus with two
punctures. Let $H\leq\mathcal{MCG}^{\pm}(S)$ be a hierarchically
hyperbolically embedded subgroup, let $M\leq H$ and let $N$ be the normal
closure of $M$. There exists a finite set $F\subset H-\\{1\\}$ such that, if
$M$ avoids $F$ and $H/M$ is hyperbolic, then $\mathcal{MCG}^{\pm}(S)/N$ is
quasi-isometrically rigid, meaning that if a finitely generated group $G$ and
$\mathcal{MCG}^{\pm}(S)/N$ are quasi-isometric then they are weakly
commensurable.

###  3.4. Quasi-isometries induce maps on hinges

For this Subsection only, we will work in the general framework of an
asymphoric HHS $(X,\mathfrak{S})$ of rank $\nu$. We will show that, if $X$
satisfies some additional properties (see Assumptions (A) - (F)), then any
self-quasi-isometry of $X$ induces an automorphism of a certain graph, which
we shall think of as encoding certain standard flats and their intersections.

Recall that a complete support set is a collection
$\\{U_{1}\\}_{i=1}^{\nu}\subset\mathfrak{S}$ of pairwise orthogonal domains
with unbounded coordinate spaces, with maximal cardinality $\nu$ among all
such collections.

######  Definition 3.11 ([BHS21, Definition 5.2]).

A _hinge_ is a pair $(U,p)$ where $U$ belongs to a complete support set and
$p\in\partial CU$. We say that $U$ is the _support_ of $(U,p)$. Let
$\bf{Hinge}(\mathfrak{S})$ be the set of hinges.

As in [BHS21, Definition 5.3], one can associate to a hinge $\sigma=(U,p)$ a
standard $1$-orthant, denoted $\mathfrak{h}_{\sigma}$. This is a hierarchy ray
whose projection to $\mathcal{C}U$ is a quasigeodesic ray asymptotic to $p$,
and whose projections to all other coordinate spaces are uniformly bounded.
Another property of $\mathfrak{h}_{\sigma}$, stated in [BHS21, Remark 5.4], is
that if $\sigma\neq\sigma^{\prime}$ are two different hinges then
$\mathrm{d}_{Haus}(\mathfrak{h}_{\sigma},\mathfrak{h}_{\sigma^{\prime}})=\infty$.

Let us introduce the first set of hypothesis on $(X,\mathfrak{S})$.

######  Assumption A.

For every $U$ which belongs to a complete support set, the Gromov boundary
$\partial\mathcal{C}U$ of its coordinate space consists of at least $2$
points.

######  Definition 3.12 (Good domain).

A domain $U\in\mathfrak{S}$ is _good_ if there exist two complete support sets
whose intersection is $\\{U\\}$. Let $\mathfrak{S}_{G}$ be the set of good
domains, and let $\mathbf{Hinge}_{G}(\mathfrak{S})$ be the set of hinges
supported on good domains.

######  Assumption B.

For every $U,V\in\mathfrak{S}_{G}$ which are orthogonal, there exist two
support sets whose intersection is $\\{U,V\\}$.

######  Assumption C.

For every $U\in\mathfrak{S}$ which belongs to a complete support set, there
exists a (possibly empty) collection of good domains
$\\{V_{1},\ldots,V_{n}\\}$ and two complete support sets whose intersection is
$\\{U,V_{1},\ldots,V_{n}\\}$.

######  Remark 3.13.

Let $f:\,X\to X$ be a quasi-isometry. With the same arguments as in the first
half of the proof of [BHS21, Theorem 5.7], one can define a map
$f_{Hin}:\,\mathbf{Hinge}_{G}(\mathfrak{S})\to\mathbf{Hinge}(\mathfrak{S})$ by
mapping $\sigma\in\mathbf{Hinge}_{G}(\mathfrak{S})$ to the unique hinge
$\sigma^{\prime}\in\mathbf{Hinge}(\mathfrak{S})$ such that
$\mathrm{d}_{\text{Haus}}(f(\mathfrak{h}_{\sigma}),\mathfrak{h}_{\sigma^{\prime}})<+\infty$.
Moreover, by inspection of the construction one sees that $\sigma^{\prime}$ is
itself good, since $\mathfrak{h}_{\sigma^{\prime}}$ arises as the coarse
intersection of two standard orthants (whose supports must therefore coincide
only on the support of $\sigma^{\prime}$ by Lemma 3.1). Hence $f_{Hin}$ maps
$\mathbf{Hinge}_{G}(\mathfrak{S})$ to itself, and therefore is a bijection
whose inverse is the map $g_{Hin}$, induced by any quasi-inverse $g$ for $f$.

######  Definition 3.14.

We say that two hinges are _co-supported_ if they are supported on the same
domain, i.e. if the union of their hinge rays is (coarsely) a 1-dimensional
standard flat. We say that two hinges are _orthogonal_ if they are supported
on orthogonal domains.

######  Lemma 3.15 (Behrstock, Hagen, Sisto).

Under Assumptions (A) - (C), if two good hinges
$\sigma,\sigma^{\prime}\in\mathbf{Hinge}_{G}(\mathfrak{S})$ are orthogonal
then $f_{Hin}(\sigma)$ and $f_{Hin}(\sigma^{\prime})$ are either orthogonal or
co-supported.

###### Proof.

Let $\sigma=(U,p)$ and $\sigma^{\prime}=(U^{\prime},p^{\prime})$. By
Assumption (B) there exist two standard flats
$\mathcal{F}_{1},\mathcal{F}_{2}$ whose coarse intersection is a standard
2-flat containing $\mathfrak{h}_{\sigma}$ and
$\mathfrak{h}_{\sigma^{\prime}}$. Let $\mathcal{O}$ be the orthant spanned by
$\mathfrak{h}_{\sigma}$ and $\mathfrak{h}_{\sigma^{\prime}}$ in this standard
2-flat.

As a consequence of the Quasiflats Theorem [BHS21, Theorem A] and of Lemma
3.1, $f(\mathcal{F}_{1})\tilde{\cap}f(\mathcal{F}_{2})$ is a union of standard
2-orthants, whose “boundary” 1-orthants, which we call _coordinate rays_ , can
be ordered cyclically. Moreover, $f(\mathfrak{h}_{\sigma})$ must arise as some
coordinate ray, since it coarsely coincides with some hinge
$\mathfrak{h}_{f_{Hin}(\sigma)}$, and if a hinge is contained in a standard
2-orthant then it must be one of its boundary 1-orthants by [BHS21, Lemma
4.11]. The same is true for $f(\mathfrak{h}_{\sigma^{\prime}})$.

Now, if $f(\mathfrak{h}_{\sigma})$ and $f(\mathfrak{h}_{\sigma^{\prime}})$ are
adjacent in the cyclic ordering, then they belong to a common 2-orthant, and
therefore they are orthogonal. Thus suppose that there is a coordinate ray
$r\in f(\mathcal{O})$ between $f(\mathfrak{h}_{\sigma})$ and
$f(\mathfrak{h}_{\sigma^{\prime}})$, and let $V\in\mathfrak{S}$ be its
support.

If $V$ is good then we can proceed as in the proof of [BHS21, Theorem 5.7] to
get a contradiction. Indeed, in this case $f^{-1}(r)$ must coincide with some
hinge ray $\mathfrak{h}_{\sigma^{\prime\prime}}$, which belongs to
$\mathcal{O}$ but lies at infinite Hausdorff distance from both
$\mathfrak{h}_{\sigma}$ and $\mathfrak{h}_{\sigma^{\prime}}$, and this
contradicts [BHS21, Lemma 4.11].

Then we can assume that no ray between $f(\mathfrak{h}_{\sigma})$ and
$f(\mathfrak{h}_{\sigma^{\prime}})$ is supported on a good domain, and in
particular that $V$ is not good. By Assumption (C) there exists a collection
of good domains $\\{V_{1},\ldots,V_{n}\\}$ and two complete support sets whose
intersection is $\\{V,V_{1},\ldots,V_{n}\\}$.

Hence, there exist two standard flats $\mathcal{G}_{1},\mathcal{G}_{2}$ whose
intersection is a flat supported on $\\{V,V_{1},\ldots,V_{n}\\}$ and
containing $r$. Now consider
$Y=\mathcal{G}_{1}\tilde{\cap}\mathcal{G}_{2}\tilde{\cap}f(\mathcal{F}_{1})\tilde{\cap}f(\mathcal{F}_{2})$.
This coarse intersection is well-defined, since both $f(\mathcal{F}_{1})$ and
$f(\mathcal{F}_{2})$ are finite unions of standard orthants by the Quasiflat
Theorem, and the coarse intersection of standard orthants and flats is well-
defined by Lemma 3.1. Moreover, let $Z=f^{-1}(Y)$, which is a union of
standard 1- and 2-orthants supported on $\\{U,U^{\prime}\\}$ (again, this
follows from combining the Quasiflat Theorem, applied to
$f^{-1}(\mathcal{G}_{1})$ and $f^{-1}(\mathcal{G}_{2})$, and Lemma 3.1). Now,
$Z$ contains the ray $f^{-1}(r)$, which lies inside $\mathcal{O}$ but cannot
coarsely coincide with $\mathfrak{h}_{\sigma}$ nor with
$\mathfrak{h}_{\sigma^{\prime}}$, since their image lie at infinite Hausdorff
distance. Therefore, $Z$ must contain the whole 2-orthant $\mathcal{O}$. In
turn, this means that $Y$ must contain all 2-orthants of
$f(\mathcal{F}_{1})\tilde{\cap}f(\mathcal{F}_{2})$ which lie between
$f(\mathfrak{h}_{\sigma})$ and $f(\mathfrak{h}_{\sigma^{\prime}})$ in the
cyclic ordering, and moreover the support of every such 2-orthant must be
contained in $\\{V,V_{1},\ldots,V_{n}\\}$.

Now, since $f(\mathfrak{h}_{\sigma})$ and $f(\mathfrak{h}_{\sigma^{\prime}})$
are boundary 1-orthants of some 2-orthants of $Y$, their supports must belong
to $\\{V,V_{1},\ldots,V_{n}\\}$, and therefore they are either orthogonal or
co-supported. ∎

For the rest of the Section, we will work under the following additional
assumptions:

######  Assumption D.

If $U\in\mathfrak{S}_{G}$ is a good domain, then $|\partial\mathcal{C}U|=2$.

######  Assumption E.

For every two good domains $U\neq V\in\mathfrak{S}_{G}$ there exists a good
domain $W\in\mathfrak{S}_{G}$ which is orthogonal to $U$ but not to $V$. In
other words, every good domain is uniquely determined by the set of good
domains which are orthogonal to it.

######  Lemma 3.16.

Under Assumptions (A) - (E), if
$\sigma,\sigma^{\prime}\in\mathbf{Hinge}_{G}(\mathfrak{S})$ are co-supported
then $f_{Hin}(\sigma)$ and $f_{Hin}(\sigma^{\prime})$ are either orthogonal or
co-supported.

###### Proof.

Let $\sigma,\sigma^{\prime}\in\mathbf{Hinge}_{G}(\mathfrak{S})$ be two co-
supported good hinges. For every good hinge $\theta$ which is orthogonal to
$f_{Hin}(\sigma)$ we can consider its preimage $(f_{Hin})^{-1}(\theta)$. By
Lemma 3.15, which also applies to $(f_{Hin})^{-1}$ (again, because it
coincides with the map $g_{Hin}$ for any quasi-inverse $g$ for $f$), we have
that $(f_{Hin})^{-1}(\theta)$ and $\sigma$ are either orthogonal or co-
supported. Moreover, by Assumption (D) the unique hinge which has the same
support of $\sigma$ is $\sigma^{\prime}$, thus $(f_{Hin})^{-1}(\theta)$ is
orthogonal to $\sigma$ and therefore also to $\sigma^{\prime}$. Then again
Lemma 3.15 tells us that $\theta$ and $f_{Hin}(\sigma^{\prime})$ are either
co-supported or orthogonal.

We can repeat the argument for every good hinge $\theta$ which is orthogonal
to $f_{Hin}(\sigma)$, and we get that either $f_{Hin}(\sigma^{\prime})$ has
the same support of some $\theta$, and therefore is orthogonal to
$f_{Hin}(\sigma)$, or $f_{Hin}(\sigma)$ and $f_{Hin}(\sigma^{\prime})$ are
orthogonal to the same hinges, and therefore they are co-supported by
Assumption (E). ∎

To conclude the proof, we also need good domains to satisfy the following
strengthening of Definition 3.12, which roughly says that good hinges arise as
intersection of “good” standard orthants:

######  Definition 3.17 (Very good domain).

A domain $U\in\mathfrak{S}$ is _very good_ if there exist two complete support
sets, _made of good domains_ , whose intersection is $\\{U\\}$.

######  Assumption F.

Good domains are very good.

######  Lemma 3.18.

Under Assumptions (A) - (F), $f_{Hin}$ induces an automorphism $f_{G}$ of the
graph $(\mathfrak{S}_{G},\bot)$, whose vertex set is $\mathfrak{S}_{G}$ and
where adjacency corresponds to orthogonality.

###### Proof.

We can give $\mathbf{Hinge}_{G}(\mathfrak{S})$ a simplicial graph structure,
by saying that two hinges are adjacent if and only if they are orthogonal or
co-supported. Combining Remark 3.13, Lemma 3.15, and Lemma 3.16, we obtain
that $f$ induces a simplicial automorphism of
$\mathbf{Hinge}_{G}(\mathfrak{S})$. With a slight abuse of notation, we will
keep denoting such automorphism as $f_{Hin}$, since at the level of vertices
it maps every good hinge $\sigma$ to $f_{Hin}(\sigma)$.

Now notice that any maximal clique of $\mathbf{Hinge}_{G}(\mathfrak{S})$ is
the (infinite) subgraph spanned by all hinges supported on some complete
support set, made of good domains. Moreover, we claim that two good hinges
$\sigma$ and $\sigma^{\prime}$ are co-supported if and only if they belong to
the same maximal cliques. Indeed, if two hinges are co-supported then they
clearly belong to the same maximal cliques; conversely, if $\sigma$ and
$\sigma^{\prime}$ are supported on $U\neq V$, respectively, then one of the
following holds:

  * •

$U\pitchfork V$, and therefore $\sigma$ and $\sigma^{\prime}$ cannot belong to
the same maximal cliques since they are not adjacent in
$\mathbf{Hinge}_{G}(\mathfrak{S})$;

  * •

$U\bot V$, and in view of Assumption (F) we can find a complete support set,
made of good domains, which contains $U$ but not $V$.

Since we just showed that being co-supported is a purely combinatorial
property, it must be preserved by the automorphism $f_{Hin}$. Moreover, the
same argument works for the inverse $(f_{Hin})^{-1}$, hence two good hinges
$\sigma,\sigma^{\prime}\in\mathbf{Hinge}_{G}(\mathfrak{S})$ are co-supported
if and only if $f_{Hin}(\sigma)$ and $f_{Hin}(\sigma^{\prime})$ are co-
supported. This also implies that two good hinges
$\sigma,\sigma^{\prime}\in\mathbf{Hinge}_{G}(\mathfrak{S})$ are orthogonal if
and only if $f_{Hin}(\sigma)$ and $f_{Hin}(\sigma^{\prime})$ are orthogonal.

Now, $f_{Hin}$ induces an automorphism $f_{G}$ of $(\mathfrak{S}_{G},\bot)$ as
follows. For every $U\in\mathfrak{S}$ choose a hinge $\sigma$ supported on
$U$, and set $f_{G}(U)$ as the support of $f_{Hin}(\sigma)$. This map is well-
defined since $f_{Hin}$ maps co-supported hinges to co-supported hinges, and
it preserves orthogonality since $f_{Hin}$ does. ∎

The last Lemma gives a uniform bound between the image of a maximal flat,
supported on good domains, and the maximal flat supported on the image of the
supports. In the case of $\mathcal{MCG}^{\pm}(S)/N$, this will tell us that
the quasi-isometry permutes “Dehn twist flats”, up to uniformly bounded
distance.

######  Lemma 3.19 (Flats go to flats).

Let $(X,\mathfrak{S})$ be an asymphoric HHS of rank $\nu$, satisfying
Assumptions (A) - (F). For every $L\geq 0$ there exists $C\geq 0$ such that
the following holds. Let $f:\,X\to X$ be a $(L,L)$-quasi-isometry. Let
$\\{U_{i}\\}_{i=1}^{\nu}$ be a complete support set, made of good domains, and
let $\mathcal{F}_{\\{U_{i}\\}}$ be the standard flat it supports. Similarly,
let $\mathcal{F}_{\\{f_{G}(U_{i})\\}}$ the standard flat supported on
$\\{f_{G}(U_{i})\\}_{i=1}^{\nu}$. Then
$\mathrm{d}_{Haus}\left(f\left(\mathcal{F}_{\\{U_{i}\\}}\right),\mathcal{F}_{\\{f_{G}(U_{i})\\}}\right)\leq
C$.

###### Proof.

One can argue as in [BHS21, Lemma 5.9], whose proof only relies on
asymphoricity of $X$ and the fact that $f_{Hin}$ maps hinges on the same
support to hinges on the same support. ∎

###  3.5. Application to large displacement quotients

Now let $S$ be a surface of complexity at least $2$. It is easy to see that,
if $U\in\mathfrak{S}$ belongs to a complete support set for
$\mathcal{MCG}^{\pm}(S)$, then $U$ is either an annulus, a $S_{0,4}$ or a
$S_{1,1}$. Moreover, if $U$ has complexity one then every complete support set
containing $U$ must also contain all its boundary annuli, thus $U$ is not
good; on the other hand, annuli are very good, since every annulus can be
obtained as intersection of two _pants decompositions_ , that is, two
collections of pairwise disjoint annuli. Finally, the rank of
$\mathcal{MCG}^{\pm}(S)$ is precisely the complexity of $S$.

The next lemma shows that, if $\mathcal{MCG}^{\pm}(S)/N$ satisfies Convention
3.2, then it fits the framework of Subsection 3.4:

######  Lemma 3.20.

Let $S$ be a surface of complexity at least $2$, and let
$\mathcal{MCG}^{\pm}(S)/N$ satisfy Convention 3.2. Then good domains
correspond to $N$-orbits of annuli, and Assumptions (A) - (F) hold.

###### Proof.

First, let $\\{U_{i}\\}_{i=1}^{\zeta(S)}$ be a complete support set for
$\mathcal{MCG}^{\pm}(S)$. The set of all curves lying on
$\bigcup_{i=1}^{\zeta(S)}U_{i}$ has diameter at most $2$ inside
$\mathcal{C}S$, hence the subsurfaces must belong to pairwise distinct orbits
$\\{\overline{U}_{i}\\}_{i=1}^{\zeta(S)}$ by the assumption on the minimum
displacement. Moreover, $\\{\overline{U}_{i}\\}_{i=1}^{\zeta(S)}$ are again
pairwise orthogonal, by how orthogonality in $\mathfrak{T}$ is defined, and
have unbounded coordinate spaces, since by assumption
$\mathcal{C}\overline{U}_{i}$ is quasi-isometric to $\mathcal{C}U_{i}$. Hence
the projection of a complete support set for $\mathcal{MCG}^{\pm}(S)$ is a
support set for $\mathcal{MCG}^{\pm}(S)/N$ with the same number of domains,
and in particular the rank $\nu$ of $\mathcal{MCG}^{\pm}(S)/N$ is at least
${\zeta(S)}$.

Conversely, let $\mathcal{T}=\\{T_{1},\ldots,T_{\nu}\\}\subset\mathfrak{T}$ be
a complete support set. If $T_{i}\in\mathcal{T}$ belonged to
$\mathfrak{T}-\overline{\mathfrak{S}}$ then $\mathcal{T}$ would only consist
of $T_{i}$, because elements of $\mathfrak{T}-\overline{\mathfrak{S}}$ with
unbounded coordinate spaces have no orthogonals by Convention 3.2. Then
$\mathcal{T}$ could not be complete, since $\nu\geq{\zeta(S)}\geq 2$. Hence,
for every $i=1,\ldots,\nu$, we have that $T_{i}=\overline{U}_{i}$, and by
Lemma 1.10 we can find pairwise disjoint representatives
$U_{i}\in\overline{U}_{i}$. Thus $\nu$ is at most $\zeta(S)$, and if we
combine this with the previous inequality we get $\nu=\zeta(S)$. We just
proved that every complete support set for $\mathcal{MCG}^{\pm}(S)/N$ is of
the form $\\{\overline{U}_{i}\\}_{i=1}^{\zeta(S)}$, for some complete support
set $\\{U_{i}\\}_{i=1}^{\zeta(S)}$ for $\mathcal{MCG}^{\pm}(S)$.

Now, we claim that $\overline{U}\in\overline{\mathfrak{S}}$ is a good domain
if and only if it is the orbit of an annulus. Indeed, if $\overline{U}$ is the
orbit of an annulus $U$ we can find two pants decompositions $\mathcal{U}$ and
$\mathcal{V}$ whose intersection is $\\{U\\}$. Since all curves in
$\mathcal{U}\cup\mathcal{V}$ lie in the ball of radius $1$ centred at $U$
inside $\mathcal{C}S$, the assumption on the large displacement implies that
any two domains in $\mathcal{U}\cup\mathcal{V}$ lie in different $N$-orbits.
Hence $\overline{\mathcal{U}}\cap\overline{\mathcal{V}}=\\{\overline{U}\\}$.
On the other hand, if $\overline{U}$ is the class of a subsurface $U$ of
complexity one, then every complete support set $\overline{\mathcal{U}}$
containing $\overline{U}$ lifts to a complete support set containing $U$,
which must also contain the boundary annuli of $U$. Hence
$\overline{\mathcal{U}}$ must contain the $N$-orbit of the boundary annuli of
$U$.

We conclude the proof by checking the Assumptions:

(A) Whenever $\overline{U}$ belongs to a complete support set,
$\mathcal{C}\overline{U}$ is isometric to $\mathcal{C}U$ for any of its
representatives $U$, and in particular has at least two points at infinity.

(B) For every two $\overline{U}$, $\overline{V}$ which lift to disjoint annuli
$U,V$, we can find two pants decompositions $\mathcal{U}$ and $\mathcal{V}$
whose intersection is $\\{U,V\\}$. Moreover, any two annuli in
$\mathcal{U}\cup\mathcal{V}$ belong to different $N$-orbits, because their
core curves are disjoint, so $\\{\overline{U},\overline{V}\\}$ is the
intersection of $\overline{\mathcal{U}}$ and $\overline{\mathcal{V}}$.

(C) Let $\overline{U}$ belong to a complete support set. If $\overline{U}$ is
not good, we can pick one of its lift $U$ and find two complete support sets
$\mathcal{U}$ and $\mathcal{V}$ whose intersection is
$\\{U,V_{1},\ldots,V_{n}\\}$, where $\\{V_{1},\ldots,V_{n}\\}$ are the
boundary annuli of $U$. Now, all curves in $\mathcal{U}\cup\mathcal{V}$ lie in
the ball of radius $1$ centred at any curve $x$ which lies in $U$. Therefore,
by the assumption on the large displacement, any two domains in
$\mathcal{U}\cup\mathcal{V}$ must belong to different $N$-orbits, and in turn
this means that $\\{\overline{U},\overline{V}_{1},\ldots,\overline{V}_{n}\\}$
is the intersection of $\overline{\mathcal{U}}$ and $\overline{\mathcal{V}}$.

(D) If $\overline{U}$ is good then $\mathcal{C}\overline{U}$ is quasi-
isometric to an annular curve graph, hence
$|\partial\mathcal{C}\overline{U}|=2$.

(E) It is enough to prove that, if
$\overline{x},\overline{y}\in\mathcal{C}S/N^{(0)}$ have the same _link_ inside
$\mathcal{C}S/N$, i.e., if they are adjacent to the same vertices, then they
coincide. Indeed, the subgraph spanned by
$\\{\overline{x},\overline{y}\\}\cup\text{Lk}_{\mathcal{C}S/N}(\overline{x})$
has diameter at most $2$, and therefore can be lifted by Lemma 3.16. Thus we
get two lifts $x,y$ such that $\text{Lk}_{\mathcal{C}S}(x)$ is the lift of
$\text{Lk}_{\mathcal{C}S/N}(\overline{x})$, and therefore coincides with
$\text{Lk}_{\mathcal{C}S}(y)$. Now we are left to prove that, if two curves
$x,y$ have the same link in $\mathcal{C}S$, then they must coincide. Now, if
the curves are disjoint then $x\in\text{Lk}(y)-\text{Lk}(x)$, and we are done.
Otherwise, let $S^{\prime}$ be a component of $S-{x}$ which is not a pair of
pants (here we are using that $S$ has complexity at least $2$), and let $z$ be
the subsurface projection of $y$ inside $S^{\prime}$. By applying a partial
pseudo-Anosov of $S^{\prime}$ to $z$ we can find a curve $z^{\prime}$ inside
$S$ which crosses $z$, and therefore also $y$.

(F) As we saw before, every $\overline{U}$ which is the orbit of an annulus
arises as the intersection of two complete support sets which come from two
pants decompositions. Therefore good domains are very good. ∎

Now, good domains correspond to orbits of annuli, and orthogonality
corresponds to having disjoint representatives. Hence, as a consequence of
Lemma 3.18 we get:

######  Corollary 3.21.

Let $S$ be a surface of complexity at least $2$, and let
$\mathcal{MCG}^{\pm}(S)/N$ satisfy Convention 3.2. Every quasi-isometry
$f:\,\mathcal{MCG}/N\mapsto\mathcal{MCG}/N$ induces an automorphism
$f_{\mathcal{C}S/N}$ of $\mathcal{C}S/N$.

Now we are finally ready to prove quasi-isometric rigidity:

###### Proof of Theorem 3.8.

Let $S$ be a surface of complexity at least $2$ which is not a $S_{1,2}$, and
let $\mathcal{MCG}^{\pm}(S)/N$ satisfy Convention 3.2. By [MS23, Lemma 9.15],
which follows from standard arguments (see e.g. [Sch95, Section 10.4]), to
prove quasi-isometric rigidity of $\mathcal{MCG}^{\pm}(S)/N$ it is enough to
verify that, for every $L\geq 0$, there exists $R\geq 0$ such that:

  1. (a)

Every $(L,L)$-self-quasi-isometry of $\mathcal{MCG}^{\pm}(S)/N$ lies within
distance $R$ from the left multiplication by some element of
$\mathcal{MCG}^{\pm}(S)/N$;

  2. (b)

If a $(L,L)$-self-quasi-isometry of $\mathcal{MCG}^{\pm}(S)/N$ lies within
finite distance from the identity, then it lies within distance $R$ from the
identity.

a Let $f:\,\mathcal{MCG}^{\pm}(S)/N\to\mathcal{MCG}^{\pm}(S)/N$ be a
$(L,L)$-quasi-isometry, and let $f_{\mathcal{C}S/N}$ be the induced
automorphism of $\mathcal{C}S/N$. By either Theorem 2.1, Theorem 2.4 or
Theorem 2.8, according to the homeomorphism type of $S$, we get that
$f_{\mathcal{C}S/N}$ is induced by some element
$\overline{g}\in\mathcal{MCG}^{\pm}(S)/N$. Moreover, by Lemma 3.19 there exist
a constant $C$ such that, whenever $\\{\overline{U}_{i}\\}_{i=1}^{\nu}$ is a
maximal collection of pairwise orthogonal orbits of annuli, $f$ maps the
standard flat $\mathcal{F}_{\\{\overline{U}_{i}\\}}$ within Hausdorff distance
at most $C$ from
$\mathcal{F}_{\\{f_{\mathcal{C}S/N}(\overline{U}_{i})\\}}=\mathcal{F}_{\\{\overline{g}(\overline{U}_{i})\\}}$,
which in turn is within Hausdorff distance at most $C$ from
$\overline{g}\left(\mathcal{F}_{\\{\overline{U}_{i}\\}}\right)$. Therefore
$\mathrm{d}_{Haus}\left(f(\mathcal{F}_{\\{\overline{U}_{i}\\}}),\overline{g}(\mathcal{F}_{\\{\overline{U}_{i}\\}})\right)\leq
2C$.

Now, to show that $f$ and $\overline{g}$ uniformly coarsely coincide it is
enough to say that, for every point $x\in\mathcal{MCG}^{\pm}(S)/N$, there
exist two standard flats $\mathcal{F},\mathcal{F}^{\prime}$, supported on
orbits of annuli, whose coarse intersection is within finite Hausdorff
distance from $x$, and all constants do not depend on $x$. Indeed, if this is
the case then $f$ and $\overline{g}$ should uniformly coarsely agree on the
coarse intersection $\mathcal{F}\tilde{\cap}\mathcal{F}^{\prime}$, and
therefore on $x$. In turn, since $\mathcal{MCG}^{\pm}(S)/N$ acts transitively
on itself and maps orbits of annuli to orbits of annuli, it is enough to
exhibit a pair $\mathcal{F},\mathcal{F}^{\prime}$ of standard flats, supported
on orbits of annuli, whose coarse intersection is bounded.

As a consequence of Lemma 3.1, if $\mathcal{F}$ is supported on
$\\{\overline{U}_{i}\\}_{i=1}^{\nu}$, $\mathcal{F}^{\prime}$ is supported on
$\\{\overline{V}_{i}\\}_{i=1}^{\nu}$, and
$\\{\overline{U}_{i}\\}_{i=1}^{\nu}\cap\\{\overline{V}_{i}\\}_{i=1}^{\nu}=\emptyset$,
then the coarse intersection is bounded. Thus, we are left to find two
collections $\\{\overline{U}_{i}\\}_{i=1}^{\nu}$ and
$\\{\overline{V}_{i}\\}_{i=1}^{\nu}$ of pairwise orthogonal orbits of annuli
whose intersection is empty. By Lemma 1.7, it is enough to find two pants
decompositions $\\{U_{i}\\}_{i=1}^{\nu}$ and $\\{V_{i}\\}_{i=1}^{\nu}$ such
that $\\{U_{i}\\}_{i=1}^{\nu}\cap\\{V_{i}\\}_{i=1}^{\nu}=\emptyset$ and the
subgraph of $\mathcal{C}S$ spanned by the core curves of
$\\{U_{i}\\}_{i=1}^{\nu}\cup\\{V_{i}\\}_{i=1}^{\nu}$ has diameter at most $2$.
This is easy: just pick any pants decomposition $\\{U_{i}\\}_{i=1}^{\nu}$, and
then replace every $U_{i}$ with an annulus $V_{i}$ which only intersects
$U_{i}$. This proves Item a.

b Let $f:\,\mathcal{MCG}^{\pm}(S)/N\to\mathcal{MCG}^{\pm}(S)/N$ be a
$(L,L)$-quasi-isometry which lies within finite distance from the identity. By
Item a we know that $f$ lies within distance $R$ from the left multiplication
by some $\overline{g}$, which depends only on the induced map
$f_{\mathcal{C}S/N}$. In turn $f_{\mathcal{C}S/N}$ is induced by $f_{Hin}$,
thus if we show that this map is the identity then $\overline{g}$ can be
chosen to be the identity, and the corollary follows. Now, for every good
hinge $\sigma$, $f_{Hin}(\sigma)$ was defined in Remark 3.13 as the unique
hinge such that $d_{Haus}(h_{f_{Hin}(\sigma)},f(h_{\sigma}))<\infty$. But
then, since $d_{Haus}(f(h_{\sigma}),h_{\sigma})<\infty$ we must have that
$f_{Hin}(\sigma)=\sigma$, that is, $f_{Hin}$ is the identity. ∎

##  4\. Algebraic rigidity

We end the paper by showing that, whenever $\mathcal{MCG}^{\pm}(S)/N$
satisfies Convention 3.2, then the automorphism group of
$\mathcal{MCG}^{\pm}(S)/N$ and its abstract commensurator are both isomorphic
to $\mathcal{MCG}^{\pm}(S)/N$ itself, via the action of
$\mathcal{MCG}^{\pm}(S)/N$ on itself by conjugation. This is Theorem 3 from
the introduction, which is covered by Corollaries 4.9 and 4.10 below. The main
result of this Section is the following:

######  Theorem 4.1.

Let $S$ be either a surface of complexity at least $3$ (excluding the case
$S=S_{2,0}$) or a $S_{0,5}$, and let $\mathcal{MCG}^{\pm}(S)/N$ satisfy
Convention 3.2. Then any isomorphism $\phi:\,H\to H^{\prime}$ between finite
index subgroups of $\mathcal{MCG}^{\pm}(S)/N$ is the restriction of an inner
automorphism.

We first recall some definitions and a theorem from [AMS16].

######  Definition 4.2.

Two elements $h$ and $g$ of a group $G$ are _commensurable_ , and we write
$h\stackrel{{\scriptstyle G}}{{\approx}}g$, if there exist
$m,n\in\mathbb{Z}\setminus\\{0\\}$, $k\in G$ such that $kg^{m}k^{-1}=h^{n}$
(that is, if they have non-trivial conjugate powers).

######  Definition 4.3.

If a group $G$ acts by isometries on a hyperbolic space $\mathcal{S}$, an
element $g\in G$ is _loxodromic_ if for some $x\in\mathcal{S}$ the map
$\mathbb{Z}\to\mathcal{S}$, $n\mapsto g^{n}(x)$ is a quasi-isometric
embedding. In the same setting, an element $g\in G$ is _weakly properly
discontinuous_ , or _WPD_ , if for every $\varepsilon>0$ and any
$x\in\mathcal{S}$ there exists $N_{0}=N_{0}(\varepsilon,x)$ such that whenever
$N\geq N_{0}$ we have

$\left|\left\\{h\in
G|\max\left\\{d_{S}\left(x,h(x)\right),d_{S}\left(g^{N}(x),hg^{N}(x)\right)\right\\}\leq\varepsilon\right\\}\right|<\infty$

We denote by $\mathcal{L}_{WPD}$ the set of loxodromic WPD elements.

The following result is a special case of [AMS16, Theorem 7.1]. Roughly
speaking, the theorem says that if $H$ is a subgroup of $G$ and both act
“interestingly enough” on some hyperbolic space, then any homomorphism
$\phi:H\to G$ is either (the restriction of) an inner automorphism or it maps
some loxodromic WPD to an element which is not commensurable to it.

######  Theorem 4.4.

Let $G$ be a group acting coboundedly and by isometries on a hyperbolic space
$\mathcal{S}$, with loxodromic WPD elements. Let $H\leq G$ be a non-virtually-
cyclic subgroup such that $H\cap\mathcal{L}_{WPD}\neq\emptyset$, and let
$E_{G}(H)$ be the unique maximal finite subgroup of $G$ normalised by $H$,
whose existence is proven in [AMS16, Lemma 5.6]. Let $\phi:\,H\to G$ be a
homomorphism such that whenever $h\in H\cap\mathcal{L}_{WPD}$ then
$\phi(h)\stackrel{{\scriptstyle G}}{{\approx}}h$. If $E_{G}(H)=\\{1\\}$ then
$\phi$ is the restriction of an inner automorphism.

Our proof of Theorem 4.1 will be very similar to that of [MS23, Theorem 10.1],
from which we now abstract a general statement, both for clarity and for later
purpose. Recall that a finitely generated group is _acylindrically hyperbolic_
if it is not virtually-cyclic and it acts coboundedly and by isometries on a
hyperbolic space $\mathcal{S}$, with loxodromic WPD elements.

######  Theorem 4.5 (Algebraic rigidity from quasi-isometric rigidity).

Let $G$ be an acylindrically hyperbolic group. Suppose that $G$ has no non-
trivial finite normal subgroups, and that every self-quasi-isometry of $G$ is
within bounded distance from the left multiplication by some element of $G$.
Then any isomorphism between finite index subgroups of $G$ is the restriction
of an inner automorphism.

###### Proof.

We just need to verify that the hypotheses of Theorem 4.4 are satisfied for
any isomorphism $\phi:\,H\to H^{\prime}$ between subgroups of finite index of
$G$. By definition of acylindrical hyperbolicity, the hypotheses on the action
are satisfied. Moreover, $G$ is not virtually-cyclic, and so also $H$ is not
because its index is finite.

Next, we show that $\phi$ has the required commensurating property, that is,
for every $h\in H\cap\mathcal{L}_{WPD}$ we have that
$\phi(h)\stackrel{{\scriptstyle G}}{{\approx}}h$. Indeed, $\phi$ can be
extended to a quasi-isometry $\Phi:\,G\to G$ (for example, by precomposing
$\phi$ with any closest-point projection $G\to H$). Hence we can find an
element $g\in G$ whose left-multiplication is uniformly close to $\Phi$. From
here, one can argue as in [MS23, Lemma 10.5], which only uses some basic
properties of Cayley graphs of finitely generated groups, to prove that
$\phi(h)\stackrel{{\scriptstyle G}}{{\approx}}h$ whenever $h\in H$ has
infinite order, and in particular whenever $h$ is loxodromic WPD.

We are left to prove that $E_{G}(H)=\\{1\\}$. In view of Lemma [MS23, Lemma
10.6], which holds for any acylindrically hyperbolic group, this follows from
the fact that $E_{G}(G)=\\{1\\}$, i.e. that $G$ has no non-trivial finite
normal subgroups. ∎

Thus to prove Theorem 4.1 we reduced ourselves to verify the hypotheses of
Theorem 4.5 when $G$ is $\mathcal{MCG}^{\pm}(S)/N$, and $S$ is as in Theorem
4.1. We know that $\mathcal{MCG}^{\pm}(S)/N$ is quasi-isometrically rigid, by
Theorem 3.8, so just need to show that it is acylindrically hyperbolic (Lemma
4.6) and that it has no non-trivial finite normal subgroups (Lemma 4.7). For
the first result, we can actually make milder requirements on the topological
type of $S$:

######  Lemma 4.6.

Let $S$ be either a surface of complexity at least $2$, and let
$\mathcal{MCG}^{\pm}(S)/N$ satisfy Convention 3.2. Then
$\mathcal{MCG}^{\pm}(S)/N$ is acylindrically hyperbolic.

###### Proof.

First of all, $\mathcal{MCG}^{\pm}(S)/N$ is non-virtually-cyclic. Indeed, if
$x,y\in\mathcal{C}S^{(0)}$ are disjoint curves, then the $\mathbb{Z}^{2}$
subgroup generated by the Dehn twists $T_{x}$ and $T_{y}$ around $x$ and $y$,
respectively, intersects $N$ trivially, since it fixes $x$ and therefore its
minimum displacement is zero. Therefore $\langle T_{x},T_{y}\rangle$ injects
in $\mathcal{MCG}^{\pm}(S)/N$, which is therefore non-virtually-cyclic.

Now, $\mathcal{MCG}^{\pm}(S)/N$ acts coboundedly on $\mathcal{C}S/N$, because
the action is induced by the cobounded action of $\mathcal{MCG}^{\pm}(S)$ over
$\mathcal{C}S$. Moreover, our Convention 3.2 says that $\mathcal{C}S/N$ is
hyperbolic and $\mathcal{MCG}^{\pm}(S)/N$-equivariantly quasi-isometric to the
main coordinate space of the HHG structure. Therefore, by [BHS17b, Corollary
14.4] the action admits loxodromic WPD elements provided that $\mathcal{C}S/N$
is unbounded, as we noticed in Lemma 3.7. ∎

######  Lemma 4.7.

Let $S$ be either a surface of complexity at least $3$ (excluding the case
$S=S_{2,0}$) or a $S_{0,5}$, and let $\mathcal{MCG}^{\pm}(S)/N$ satisfy
Convention 3.2. Then $\mathcal{MCG}^{\pm}(S)/N$ has no non-trivial finite
normal subgroup.

###### Proof.

Let $K\leq\mathcal{MCG}^{\pm}(S)/N$ be a finite normal subgroup, and by
contradiction let $\overline{f}\in K$ be a non-trivial element. If
$\overline{f}$ acts trivially on $\mathcal{C}S/N$ then $\overline{f}$ is the
identity, by the injectivity part of either Theorem 2.1 or Theorem 2.8. Thus
let $\overline{x}\in\mathcal{C}S/N^{(0)}$ be a vertex which is not fixed by
$\overline{f}$. Let $x\in\mathcal{C}S^{(0)}$ be a representative for
$\overline{x}$, let $f\in\mathcal{MCG}^{\pm}(S)$ be a representative for
$\overline{f}$, let $T_{x}\in\mathcal{MCG}^{\pm}(S)$ be the Dehn Twist around
$x$, and let $\overline{T_{x}}\in\mathcal{MCG}^{\pm}(S)/N$ be its image in the
quotient. Since $K$ is finite, we can find $m\in\mathbb{N}_{>0}$ such that
$\overline{f}=(\overline{T_{x}})^{-m}\overline{f}(\overline{T_{x}})^{m}$.
Hence, denoting by $T_{f(x)}$ the Dehn twist around $f(x)$, and by
$\overline{1}\in\mathcal{MCG}^{\pm}(S)/N$ the identity element of the
quotient, we have

$\overline{1}=(\overline{T_{x}})^{-m}\left(\overline{f}(\overline{T_{x}})^{m}\overline{f}^{-1}\right)=(\overline{T_{x}})^{-m}(\overline{T_{f(x)}})^{m},$

where we used how Dehn twists behave under conjugation. In other words, we
have that $T_{x}^{-m}T_{f(x)}^{m}\in N$.

Now, if $\overline{x}$ and $\overline{f}(\overline{x})$ are adjacent in
$\mathcal{C}S/N$, then we could have chosen representatives $x$ and $f$ such
that $x$ and $f(x)$ are disjoint curves. Then $T_{x}^{-m}T_{f(x)}^{m}$ cannot
be an element of $N$, since it does not have positive displacement.

Hence suppose that $\overline{x}$ and $\overline{f}(\overline{x})$ are not
adjacent in $\mathcal{C}S/N$. Let $U\in\mathfrak{S}$ be the annulus with core
curve $x$, let $\overline{U}\in\overline{\mathfrak{S}}$ be its $N$-orbit and
let $\rho_{\overline{U}}:\,\mathcal{C}S/N\to\mathcal{C}\overline{U}$ be the
projection given by the HHG structure. Notice that $\overline{T_{x}}$ acts on
$\mathcal{C}\overline{U}$, since $T_{x}$ fixes $x$.

Now, $(\overline{T_{x}})^{-m}(\overline{T_{f(x)}})^{m}=\overline{1}$, so it
must act as the identity on $\mathcal{C}S/N$, and in particular it must fix
$\overline{f}(\overline{x})$. Hence, when we project to
$\mathcal{C}\overline{U}$ we get

$\rho_{\mathcal{C}\overline{U}}\left(\overline{f}(\overline{x})\right)=\rho_{\mathcal{C}\overline{U}}\left((\overline{T_{x}})^{-m}(\overline{T_{f(x)}})^{m}\overline{f}(\overline{x})\right)=\rho_{\mathcal{C}\overline{U}}\left((\overline{T_{x}})^{-m}(\overline{f}(\overline{x})\right),$

where we used that $T_{f(x)}$ fixes $f(x)$. Moreover

$\rho_{\mathcal{C}\overline{U}}\left((\overline{T_{x}})^{-m}(\overline{f}(\overline{x})\right)=(\overline{T_{x}})^{-m}\rho_{\mathcal{C}\left((\overline{T_{x}})^{-m}(\overline{U})\right)}\left(\overline{f}(\overline{x})\right)=(\overline{T_{x}})^{-m}\rho_{\mathcal{C}\overline{U}}\left(\overline{f}(\overline{x})\right),$

where we used that projections in a HHG are equivariant with respect to the
action of the group. Hence, the action of $(\overline{T_{x}})^{-m}$ on
$\mathcal{C}\overline{U}$ must fix the element
$\rho_{\mathcal{C}\overline{U}}\left(\overline{f}(\overline{x})\right)$.

However, the action of $\overline{T_{x}}$ on $\mathcal{C}\overline{U}$ was
defined via the action of $T_{x}$ on $\mathcal{C}U$, so that the following
diagram commutes:

${\mathcal{C}U}$${\mathcal{C}U}$${\mathcal{C}\overline{U}}$${\mathcal{C}\overline{U}}$$\scriptstyle{T_{x}}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$$\scriptstyle{\overline{T_{x}}}$

Here the vertical arrows are the restriction to $\mathcal{C}U$ of the quotient
projection

$\bigcup_{U^{\prime}\in\overline{U}}\mathcal{C}U^{\prime}\to\mathcal{C}\overline{U}=\left(\bigcup_{U^{\prime}\in\overline{U}}\mathcal{C}U^{\prime}\right)/N.$

Such restriction is a quasi-isometry by Convention 3.2. Hence, no power of
$\overline{T_{x}}$ can fix a point in $\mathcal{C}\overline{U}$, since the
action of the Dehn twist $T_{x}$ on its annular curve graph $\mathcal{C}U$ is
loxodromic. ∎

As a by-product of Theorem 4.1, we get that any automorphism of
$\mathcal{MCG}^{\pm}(S)/N$ is the conjugation by some element
$\overline{g}\in\mathcal{MCG}^{\pm}(S)/N$. The following lemma states that
such $\overline{g}$ is also unique:

######  Lemma 4.8.

Let $S$ be either a surface of complexity at least $3$ (excluding the case
$S=S_{2,0}$) or $S=S_{0,5}$, and let $\mathcal{MCG}^{\pm}(S)/N$ satisfy
Convention 3.2. Then $\mathcal{MCG}^{\pm}(S)/N$ is centerless.

###### Proof.

By either Theorem 2.1 or Theorem 2.8, it is enough to show that $\overline{g}$
acts as the identity on $\mathcal{C}S/N$. Notice that $\overline{g}$ commutes
with the image of the Dehn twist $\overline{T_{x}}$ for any curve
$x\in\mathcal{C}S^{(0)}$. In particular, by how Dehn twists behave under
conjugation, we have that $\overline{T_{x}}=\overline{T_{g(x)}}$, where
$g\in\mathcal{MCG}^{\pm}(S)$ is any representative for $\overline{g}$. Now one
can argue as in the proof of Theorem 4.1 to get that
$\overline{g}(\overline{x})=\overline{x}$, and the conclusion follows. ∎

Combining Theorem 4.1 and Lemma 4.8 we get:

######  Corollary 4.9.

Let $S$ be either a surface of complexity at least $3$ (excluding the case
$S=S_{2,0}$) or $S=S_{0,5}$, and let $\mathcal{MCG}^{\pm}(S)/N$ satisfy
Convention 3.2. Then the map
$\mathcal{MCG}^{\pm}(S)/N\to\text{Aut}(\mathcal{MCG}^{\pm}(S)/N)$, which maps
an element $\overline{g}\in\mathcal{MCG}^{\pm}(S)/N$ to the conjugation by
$\overline{g}$, is an isomorphism. Therefore
$\text{Out}(\mathcal{MCG}^{\pm}(S)/N)$ is trivial.

Finally, we recall the definition of the _abstract commensurator_ of a group
$G$. Consider the set of all isomorphisms $H\to H^{\prime}$ between finite-
index subgroups of $G$. Let $\text{Comm}(G)$ be the quotient of this set by
the following equivalence relation: two isomorphisms are identified if they
coincide on a finite index subgroup of $G$. Then $\text{Comm}(G)$ can be
endowed with a group structure, induced by the composition. Our final result
shows that $\text{Comm}(\mathcal{MCG}^{\pm}(S)/N)$ is “the smallest possible”:

######  Corollary 4.10.

Let $S$ be either a surface of complexity at least $3$ (excluding the case
$S=S_{2,0}$, or $S=S_{0,5}$ and let $\mathcal{MCG}^{\pm}(S)/N$ satisfy
Convention 3.2. Then the map
$\mathcal{MCG}^{\pm}(S)/N\to\text{Comm}(\mathcal{MCG}^{\pm}(S)/N)$, which maps
an element $\overline{g}$ to the conjugation by $\overline{g}$, is an
isomorphism.

###### Proof.

The map is surjective by Theorem 4.1. Moreover, if the conjugation by
$\overline{g}$ coincides with the identity on a finite-index subgroup $H$,
then $\overline{g}$ commutes with the $m$-th power of the image of any Dehn
twist, where $m$ only depends on the index of $H$ in
$\mathcal{MCG}^{\pm}(S)/N$. Then one can argue as in the proof of Lemma 4.8 to
get that $\overline{g}$ acts as the identity on $\mathcal{C}S/N$, and
therefore $\overline{g}=\overline{1}$ by either Theorem 2.1 or Theorem 2.8. ∎

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