# Twisted Rokhlin property for mapping class groups

Pravin Kumar<EMAIL_ADDRESS>Department of Mathematical Sciences,
Indian Institute of Science Education and Research (IISER) Mohali, Sector 81,
S. A. S. Nagar, P. O. Manauli, Punjab 140306, India.  ,  Apeksha Sanghi
<EMAIL_ADDRESS>Department of Mathematical Sciences, Indian
Institute of Science Education and Research (IISER) Mohali, Sector 81, S. A.
S. Nagar, P. O. Manauli, Punjab 140306, India.  and  Mahender Singh
<EMAIL_ADDRESS>Department of Mathematical Sciences, Indian
Institute of Science Education and Research (IISER) Mohali, Sector 81, S. A.
S. Nagar, P. O. Manauli, Punjab 140306, India.

###### Abstract.

In this paper, generalising the idea of the Rokhlin property, we explore the
concept of the twisted Rokhlin property of topological groups. A topological
group is said to exhibit the twisted Rokhlin property if, for each
automorphism $\phi$ of the group, there exists a $\phi$-twisted conjugacy
class that is dense in the group. We provide a complete classification of
connected orientable infinite-type surfaces without boundaries whose mapping
class groups possess the twisted Rokhlin property. Additionally, we prove that
the mapping class groups of the remaining surfaces do not admit any dense
$\phi$-twisted conjugacy class for any automorphism $\phi$. This supplements
the recent work of Lanier and Vlamis on the Rokhlin property of big mapping
class groups. We also prove that the mapping class group of each connected
orientable infinite-type surface without boundary possesses the
$R_{\infty}$-property.

###### Key words and phrases:

Big mapping class group; infinite type surface; $R_{\infty}$-property; Rokhlin
property; twisted conjugacy; twisted Rokhlin property

###### 2020 Mathematics Subject Classification:

Primary 57K20, 54H11; Secondary 20E36

##  1\. Introduction

A topological group is said to exhibit the Rokhlin property if it contains a
dense conjugacy class. From the point of view of dynamics, the Rokhlin
property is equivalent to the existence of a dense orbit under the conjugation
action of the topological group on itself. This perspective allows us to
consider more general actions of the topological group arising from its
structure.

Given an automorphism $\phi$ of a group $G$, the map $(g,h)\mapsto
gh\phi(g)^{-1}$, defines a left action of $G$ on itself. Two elements $x,y\in
G$ are said to be $\phi$-twisted conjugate if they are in the same orbit under
this action. The resulting orbits are referred as $\phi$-twisted conjugacy
classes. The idea of twisted conjugacy arose from the work of Reidemeister
[22]. It has deep connection with Nielsen fixed-point theory [6], and appears
in Arthur–Selberg theory [24], algebraic geometry [12], and representation
theory [10], to name a few. In recent years, there has been a tremendous
amount of work on understanding twisted conjugacy in various classes of
groups. See, for example [4, 8], for recent works on twisted conjugacy classes
in mapping class groups and braid groups.

In this paper, we combine the concept of the Rokhlin property with that of
twisted conjugacy. Let $G$ be a topological group and $\phi$ an automorphism
of the underlying abstract group. We say that $G$ has the $\phi$-twisted
Rokhlin property if $G$ admits a dense $\phi$-twisted conjugacy class.
Further, $G$ is said to have the twisted Rokhlin property if it has the
$\phi$-twisted Rokhlin property for each automorphism $\phi$ of the underlying
abstract group. Clearly, if a topological group has the twisted Rokhlin
property, then it has the Rokhlin property.

The mapping class group of a connected orientable surface is the group of
isotopy classes of orientation-preserving self-homeomorphisms of the surface.
These groups inherit a non-trivial quotient topology from the compact-open
topology on the group of orientation-preserving self-homeomorphisms of the
surface. The aim of this paper is to classify connected orientable surfaces
without boundaries whose mapping class groups have the twisted Rokhlin
property. Our approach is built upon the recent work [18] of Lanier and
Vlamis, where they classify connected orientable surfaces without boundaries
whose mapping class groups admit the Rokhlin property (see also [13] for an
independent such classification).

The paper is organised as follows. In Section 2, we recall basic results on
infinite type surfaces, mapping class groups and curve graphs, that we shall
need in the latter sections. In Section 3, we introduce the twisted Rokhlin
property of topological groups and make some basic observations. We prove that
if $\phi,\psi$ are two automorphisms of a topological group $G$ such that
$\psi\in\operatorname{Inn}(G)\phi$, then $G$ has the $\phi$-twisted Rokhlin
property if and only if it has the $\psi$-twisted Rokhlin property
(Proposition 3.5). In particular, this implies that the twisted Rokhlin
property for an inner automorphism is equivalent to the Rokhlin property of
the group. For Polish topological groups, we prove that the $\phi$-twisted
Rokhlin property is equivalent to the property that, given two non-empty open
subsets $U$ and $V$ of $G$, there exists $g\in G$ such that $U\cap
gV\phi(g)^{-1}\neq\emptyset$ (Proposition 3.10). Section 4 contains our main
results on the twisted Rokhlin property of big mapping class groups. By
employing the approach outlined in the following flowchart, we establish
Theorem 4.2, Proposition 4.4, Theorem 4.7 and Theorem 4.9.

Infinite type surface
---
Every compact subsurface
---
is displaceable
There exists a compact
---
non-displaceable subsurface
Unique maximal end
---
Two maximal ends
---
Space of maximal ends
---
is a Cantor space

As a consequence, we deduce the following classification result.

######  Theorem 1.1.

The mapping class group $\mathrm{MCG}(S)$ of a connected and orientable
surface $S$ without boundary has the $\phi$-twisted Rokhlin property for some
automorphism $\phi$ of $\mathrm{MCG}(S)$ if and only if the surface is either
the $2$-sphere or satisfy the property that every compact subsurface of $S$ is
displaceable and $S$ has a unique maximal end.

Along the way, we also investigate the $R_{\infty}$-property of mapping class
groups of surfaces and prove the following result.

######  Theorem 1.2.

Let $S$ be a connected orientable infinite-type surface without boundary. Then
$\mathrm{MCG}(S)$ possesses the $R_{\infty}$-property.

##  2\. Preliminaries

As the subject resides at the intersection of group theory, topology, and
geometry, we offer relevant background information to enhance its readibility.

###  2.1. Classification of surfaces

A surface is said to be of finite-type if its fundamental group is finitely
generated; otherwise it is said to be of infinite-type. Throughout the paper,
our primary surface $S$ under consideration will be connected, orientable,
without boundary and of infinite-type, unless stated otherwise. At some
occasions, we shall also need finite-type surfaces, and we write $S_{g,n}^{b}$
to denote a finite-type surface of genus $g$ with $n$ punctures and $b$
boundary components.

######  Definition 2.1.

An exiting sequence in a surface $S$ is a sequence $\\{U_{n}\\}_{n\geq 1}$ of
connected open subsets of $S$ admitting the following properties:

  1. (1)

If $m<n$, then $U_{n}\subset U_{m}$.

  2. (2)

For each $n\in\mathbb{N}$, $U_{n}$ is not relatively compact.

  3. (3)

For each $n\in\mathbb{N}$, $U_{n}$ has compact boundaries.

  4. (4)

Each relatively compact subset of $S$ is disjoint from all except finitely
many $U_{n}$’s.

Two exiting sequences $\\{U_{n}\\}_{n\geq 1}$ and $\\{V_{n}\\}_{n\geq 1}$ are
equivalent if for every $n\in\mathbb{N}$, there exists $m\in\mathbb{N}$ such
that $U_{m}\subset V_{n}$ and $V_{m}\subset U_{n}$. An equivalence class of an
exiting sequence is called an end and we denote the set of all ends by
$\mathcal{E}(S)$. The set $\mathcal{E}(S)$ can be topologised as follows.
Given a subset $U\subset S$ with compact boundary, consider the set

$U^{*}=\big{\\{}[\\{U_{n}\\}_{n\geq
1}]\in\mathcal{E}(S)~{}\mid~{}\textrm{there
exists}~{}n\in\mathbb{Z}~{}\textrm{such that}~{}U_{n}\subset U\big{\\}}$

Then the desired topology on $\mathcal{E}(S)$ is the topology generated by the
basis

$\\{U^{*}~{}\mid~{}U\subset S~{}~{}\text{is open with compact boundary}\\}.$

With this topology, the space of ends of a surface is compact, totally
disconnected, second countable and Hausdorff. Consequently, the space of ends
is homeomorphic to a closed subset of the Cantor set. If $U\subset S$ is an
open subset with compact boundary and $e\in U^{*}$, then we refer to $U$ as a
neighborhood of $e$. Further, every homeomorphism $h:S\to S$ induces a
homeomorphism $\mathcal{E}(S)\to\mathcal{E}(S)$ given by

$[\\{U_{n}\\}_{n\geq 1}]\mapsto[\\{h(U_{n})\\}_{n\geq 1}].$

An end is called planar if it has a neighbourhood that can be embedded in the
plane. Otherwise, it is called non-planar. Let $\mathcal{E}_{np}(S)$ be the
subspace of $\mathcal{E}(S)$ consisting of non-planar ends. Clearly,
$\mathcal{E}_{np}(S)$ is a closed set in $\mathcal{E}(S)$. The following
theorem gives a classification of surfaces that we shall need [23, Theorem 1].

######  Theorem 2.2.

Let $S_{1}$ and $S_{2}$ be two surfaces. Let $g_{1},g_{2}$ and $b_{1},b_{2}$
represent the genus and the number of boundary components of $S_{1},S_{2}$,
respectively. Then $S_{1}$ is homeomorphic to $S_{2}$ if and only if
$g_{1}=g_{2}$, $b_{1}=b_{2}$, and there exists a homeomorphism from
$\mathcal{E}(S_{1})$ to $\mathcal{E}(S_{2})$ that restricts to a homeomorphism
from $\mathcal{E}_{np}(S_{1})$ to $\mathcal{E}_{np}(S_{2})$.

Without loss of generality, we shall assume throughout that the surface $S$ is
symmetric about the $xy$-plane when embedded in $\mathbb{R}^{3}$. Since
$\mathcal{E}(S)$ is homeomorphic to a closed subset of the Cantor set, it can
be embedded on the great circle of intersection of the $xy$-plane and the
2-sphere $\mathbb{S}^{2}$. Now, take a sequence of disjoint closed disks
$D_{j}$ on $\mathbb{S}^{2}\setminus\mathcal{E}(S)$ which are symmetric about
the $xy$-plane such that the set of accumulation points of this sequence is
precisely $\mathcal{E}_{np}(S)$. We remove the interiors of the disks $D_{j}$
and suitably identify in pairs the boundary circles by orientation-reversing
homeomorphisms such that the desired resulting surface is again symmetric
about the $xy$-plane. See [23, Theorem 3] for the general construction. For
instance, Figure 1 illustrates the construction of Loch Ness monster surface,
where the infinite sequence of disjoint closed disks
$\\{D_{i}\\}_{i\in\mathbb{N}}$ is chosen such that the radii of the disks
decreases as it approaches a point of $\mathcal{E}(S)$.

$D_{1}$ at 80, 230 $D_{2}$ at 190, 228 $D_{3}$ at 100, 228 $D_{4}$ at 168, 228
$D_{5}$ at 125, 228 $D_{6}$ at 148, 228 $x$ at 305, 118 $y$ at 140, 330

Figure 1. Realization of the Loch Ness monster surface.

For a connected orientable surface $S$ (not necessarility of infinite-type),
let $\operatorname{Homeo}^{+}(S)$ denote the group of orientation-preserving
self-homeomorphisms of $S$. We define a binary relation $\preceq$ on the space
of ends $\mathcal{E}(S)$ of $S$. For $x,y\in\mathcal{E}(S)$, we say that
$y\preceq x$ if for every open neighbourhood $\mathcal{U}$ of $x$, there is an
open neighbourhood $\mathcal{V}$ of $y$ and a homeomorphism
$h\in\operatorname{Homeo}^{+}(S)$ such that
$h(\mathcal{V})\subseteq\mathcal{U}$. An end $\mu$ is called maximal if, for
any end $e$ satisfying $\mu\preceq e$, we must have $e\preceq\mu$. Let
$\mathcal{M}(S)$ denote the set of all maximal ends of $S$. Note that for
every end $e\in\mathcal{E}(S)$, there exists a maximal end
$\mu\in\mathcal{M}(S)$ such that $e\preceq\mu$ [19, Proposition 4.7]. We say
that two ends $e_{1}$ and $e_{2}$ are comparable if either $e_{1}\preceq
e_{2}$ or $e_{2}\preceq e_{1}$. The space $\mathcal{E}(S)$ is said to be self-
similar if given a decomposition
$\mathcal{E}(S)=\mathcal{E}_{1}\sqcup\cdots\sqcup\mathcal{E}_{n}$ into a
finite disjoint union of clopen subsets, there exists an index
$i\in\\{1,\ldots,n\\}$ such that $\mathcal{E}_{i}$ contains an open
homeomorphic copy of $\mathcal{E}(S)$. The idea of maximality and self-
similarity for end spaces is due to [19].

A subsurface $\Sigma$ of $S$ is called displaceable if there exists a
$h\in\operatorname{Homeo}^{+}(S)$ such that $\Sigma\cap h(\Sigma)=\emptyset$.
We shall use the following result, which follows from [1, Theorem 6.1] and
[19].

######  Theorem 2.3.

If $S$ is a surface in which every compact subsurface is displaceable, then
either

  1. (1)

$\mathcal{M}(S)$ has exactly one element,

  2. (2)

$\mathcal{M}(S)$ has exactly two elements, or

  3. (3)

$\mathcal{M}(S)$ is homeomorphic to a Cantor space in which any two maximal
ends are comparable.

Moreover, $\mathcal{M}(S)$ is self-similar in the cases (1) and (3).

###  2.2. Mapping class groups of surfaces

Recall that, the mapping class group $\mathrm{MCG}(S)$ of a connected
orientable surface $S$ (not necessarily of infinite-type) is the group of
isotopy classes of orientation-preserving self-homeomorphisms of $S$, which
preserves the boundary of $S$ point-wise and permutes the punctures, if any.
The mapping class group of an infinite-type surface is also referred as the
big mapping class group.

Let $\operatorname{Homeo}^{+}(S,\partial(S))$ be the group of orientation-
preserving self-homeomorphisms of $S$ fixing the boundary $\partial(S)$ point-
wise, and equipped with the compact-open topology. We equip $\mathrm{MCG}(S)$
with the corresponding quotient topology inherited from
$\operatorname{Homeo}^{+}(S,\partial(S))$. Note that the mapping class group
of a surface with compact boundary is discrete if and only if the surface is
of finite-type. On the other hand, big mapping class groups have uncountably
many elements and inherit a non-discrete topology from
$\operatorname{Homeo}^{+}(S,\partial(S))$. The extended mapping class group of
$S$ is the group $\mathrm{MCG}^{\pm}(S)$ of isotopy classes of, possibly
orientation-reversing, self-homeomorphisms of $S$. We refer the reader to [2]
for a survey on both topological and algebraic aspects of big mapping class
groups.

For infinite-type surfaces without boundary, the outer automorphism group of
the mapping class group is known due to [3, Theorem 1.1].

######  Proposition 2.4.

If $S$ is a connected orientable infinite-type surface without boundary, then

$\operatorname{Out}\left(\mathrm{MCG}^{\pm}(S)\right)=1\quad\textrm{and}\quad\operatorname{Out}(\mathrm{MCG}(S))\cong\mathbb{Z}_{2}.$

As mentioned earlier, we assume that our surface $S$ is symmetric about the
$xy$-plane when embedded in $\mathbb{R}^{3}$. Thus, there is an orientation-
reversing self-homeomorphism

$\mathfrak{t}:S\rightarrow S$

which maps a point on $S$ to its mirror image about the $xy$-plane. Clearly,
$\mathfrak{t}$ is an involution of $S$. For notational convenience, let
$\hat{\mathfrak{t}}$ denote the automorphism of $\mathrm{MCG}(S)$ given by

$\hat{\mathfrak{t}}(h)=\mathfrak{t}h\mathfrak{t}$

for all $h\in\mathrm{MCG}(S)$. Then $\hat{\mathfrak{t}}$ is a representative
of the non-trivial element of $\operatorname{Out}(\mathrm{MCG}(S))$.

For a simple closed curve $c$ on a surface $S$, let $T_{c}$ denote the left-
hand Dehn twist along $c$. By abuse of notation, we denote a curve and its
isotopy class by the same symbol, and use the same convention for
homeomorphisms as well. The following lemma, perhaps well-known, gives a
relation between the Dehn twist and its conjugate by a homeomorphism of $S$.
The proof of the following lemma is a slight modification of the arguments of
[5, Fact 3.7]. See also [16, Lemma 2.1].

######  Lemma 2.5.

If $h$ is a homeomorphism of a surface $S$ and $c$ a simple closed curve on
$S$, then

$hT_{c}h^{-1}=T_{h(c)}^{\epsilon},$

where $\epsilon=1$ when $h$ is orientation-preserving and $\epsilon=-1$ when
$h$ is orientation-reversing.

By an essential curve on a surface $S$, we mean a simple closed curve on $S$
which does not bound a disk, a punctured disk, or an annulus. Let $C(S)$ be
the set of isotopy classes of all essential simple closed curves on $S$. Given
a subset $A$ of $C(S)$, we define the subset $U_{A}$ of $\mathrm{MCG}(S)$ by

$U_{A}=\\{\phi\in\mathrm{MCG}(S)~{}\mid~{}\phi(a)=a\text{ for all }a\in A\\}.$

It is known that if $S$ has empty boundary and $\mathrm{MCG}(S)$ has trivial
center, then the set

$\mathcal{B}=\\{\psi
U_{A}~{}\mid~{}\psi\in\mathrm{MCG}(S)~{}~{}\textrm{and}~{}~{}A~{}~{}\text{is a
finite subset of}~{}~{}C(S)\\}$

is a basis for the topology on $\mathrm{MCG}(S)$. Note that, the center of the
big mapping class group is always trivial [17, Proposition 2]. The following
basic observation shall be used later.

######  Lemma 2.6.

Suppose that $g,h\in\mathrm{MCG}(S)$. Then $h\in gU_{A}$ if and only if
$hU_{A}=gU_{A}$.

###### Proof.

If $h\in gU_{A}$, then $h=g\phi$ for some $\phi\in U_{A}$. Thus,
$g=h\phi^{-1}\in hU_{A}$, and hence $gU_{A}=hU_{A}$. Conversely, if
$hU_{A}=gU_{A}$, then $h\in gU_{A}$. ∎

###  2.3. Curve graphs of surfaces

We say that a simple closed curve $c$ on a surface $S$ is non-separating if
$S\setminus c$ is connected; otherwise, it is called separating. The
intersection number $i(c,c^{\prime})$ between two curves $c,c^{\prime}$ on $S$
is defined to be the minimum number of points of intersections between the
representatives in their isotopy classes. A multicurve on $S$ is a set of
pairwise (non-isotopic) disjoint curves.

The set $C(S)$ can be viewed as a graph whose vertex set is $C(S)$ and there
is an edge between two vertices if the corresponding curves on $S$ are
disjoint. The graph $C(S)$ is referred as curve graph of $S$. There is a
natural metric $d_{C(S)}$ on $C(S)$ by defining the length of each edge as 1.
We shall need the following result [20, Proposition 4.6].

######  Proposition 2.7.

Let $S_{g,n}$ be a surface of genus $g$ with $n$ punctures such that $3g+n\geq
5$. Then there exists a real number $c>0$ such that, for any pseudo-Anosov
$h\in\mathrm{MCG}(S_{g,n})$, $\gamma\in C(S_{g,n})$ and $k\in\mathbb{Z}$, we
have

$d_{C(S_{g,n})}(h^{k}(\gamma),\gamma)\geq c|k|.$

In particular, for any pseudo-Anosov $h\in\mathrm{MCG}(S_{g,n})$ and
$\gamma\in C(S_{g,n})$, there exists $\ell\in\mathbb{N}$ such that

$d_{C(S_{g,n})}(h^{m}(\gamma),\gamma)>2$

for all $m\geq\ell$.

We say that a subsurface $\Sigma$ and a simple closed curve $c$ on a surface
$S$ have a non-trivial geometric intersection if there exist $a\in C(\Sigma)$
such that $i(a,c)\neq 0$.

######  Definition 2.8.

Let $\Sigma$ be a subsurface of a surface $S$ such that the embedding of
$\Sigma$ into $S$ induces an embedding of $\mathcal{C}(\Sigma)$ into
$\mathcal{C}(S)$. Let $c\in\mathcal{C}(S)$ be such that it has a non-trivial
geometric intersection with $\Sigma$. Assume that the boundary
$\partial(\Sigma)$ and $c$ are in a minimal position. Then a projection of $c$
onto $\mathcal{C}(\Sigma)$ is defined to be any component
$b\in\mathcal{C}(\Sigma)$ of the boundary of a regular neighborhood of
$\alpha\cup\partial(\Sigma)$, where $\alpha$ is a component of $c\cap\Sigma$.

For example, let $S$ be a surface and $\Sigma$ be a genus two subsurface of
$S$ with a single boundary component. Let $c$ be a curve on $S$ as illustrated
in Figure 2. Then the curves $b_{1}$ and $b_{2}$ are two projections of $c$
onto $\Sigma$ (see Figure 2).

$\partial(\Sigma)$ at 205, 140 $c$ at 250, 110 $\Sigma$ at 100, 130 $b_{1}$ at
150, 95 $b_{2}$ at 150, 28

Figure 2. Projections of the curve $c$ on $\Sigma$.

Note that, if each component of $\partial(\Sigma)$ is an essential separating
simple closed curve on $S$, then $\mathcal{C}(\Sigma)$ embeds into
$\mathcal{C}(S)$. Further, for each $c\in\mathcal{C}(S)$ such that it has a
non-trivial geometric intersection with $\Sigma$, such a projection
$b\in\mathcal{C}(\Sigma)$ always exists, and if $c\in\mathcal{C}(\Sigma)$,
then $b=c$.

######  Lemma 2.9.

[21, Lemma 2.2] If $c\in\mathcal{C}(S)$ and
$b,b^{\prime}\in\mathcal{C}(\Sigma)$ are two projections of $c$, then
$d_{\mathcal{C}(\Sigma)}\left(b,b^{\prime}\right)\leq 2$.

##  3\. Twisted Rokhlin property

Recall that, a topological group is said to have Rokhlin property if it
contains a dense conjugacy class. Equivalently, the topological group has a
dense orbit under its conjugation action on itself. See, for example, the
introduction of [15] for a survey on Rokhlin property.

######  Example 3.1.

Some topological groups with the Rokhlin property are:

  1. (1)

The symmetric group on $\mathbb{N}$.

  2. (2)

The homeomorphism group of the Cantor set [11, Theorem 2.6].

  3. (3)

The group of orientation-preserving homeomorphisms of an even-dimensional
sphere [11, Theorem 3.4].

  4. (4)

The (big) mapping class group of a connected orientable surface which is
either the 2-sphere or non-compact with genus either zero or infinite and
whose end space is self-similar with a unique maximal end [18, Theorem 6.1].

######  Example 3.2.

Some topological groups without the Rokhlin property are:

  1. (1)

A non-trivial closed subgroup of $GL_{n}(\mathbb{C})$ [26, Lemma 5.2].

  2. (2)

A non-trivial connected locally compact group [26, Theorem 5.3].

Next we consider the idea of twisted conjugacy.

######  Definition 3.3.

Let $G$ be a group and $\phi$ an automorphism of $G$. Two elements $x,y\in G$
are said to be $\phi$-twisted conjugate, denoted by $x\sim_{\phi}y$, if
$y=gx\phi(g)^{-1}$ for some $g\in G$. The resulting equivalence classes are
called as $\phi$-twisted conjugacy classes, and we denote such a class by
$C_{\phi}(x)$.

Note that, given an automorphism $\phi$ of a group $G$, the map $G\times G\to
G$ given by $(g,h)\mapsto gh\phi(g)^{-1}$ defines a left action of $G$ on
itself. Thus, $\phi$-twisted conjugacy classes are simply orbits under this
action.

######  Definition 3.4.

Let $G$ be a topological group and $\phi$ an automorphism of the underlying
abstract group $G$. We say that $G$ has the $\phi$-twisted Rokhlin property if
$G$ admits a dense $\phi$-twisted conjugacy class. Further, $G$ is said to
have the twisted Rokhlin property if it has $\phi$-twisted Rokhlin property
for each automorphism $\phi$ of the underlying abstract group $G$.

The subsequent proposition significantly simplifies the task of verifying the
twisted Rokhlin property for a topological group.

######  Proposition 3.5.

Let $G$ be a topological group, and $\phi,\psi$ be two automorphisms of $G$
such that $\psi\in\operatorname{Inn}(G)\phi$. Then $G$ has the $\phi$-twisted
Rokhlin property if and only if it has the $\psi$-twisted Rokhlin property. In
particular, if $\psi\in\operatorname{Inn}(G)$, then $G$ has the $\psi$-twisted
Rokhlin property if and only if it has the Rokhlin property.

###### Proof.

Let $\psi=\hat{g}\,\phi$, where $\hat{g}$ is the inner automorphism of $G$
induced by the element $g\in G$. We have

$\displaystyle C_{\phi}(x)$ $\displaystyle=\\{kx\phi(k)^{-1}\mid k\in G\\}$
$\displaystyle=\\{kxg^{-1}(g\phi(k)^{-1}g^{-1})g\mid k\in G\\}$
$\displaystyle=\\{k(xg^{-1})\,\hat{g}(\phi(k)^{-1})g\mid k\in G\\}$
$\displaystyle=\\{k(xg^{-1})\,\psi(k)^{-1}g\mid k\in G\\}$
$\displaystyle=R_{g}(C_{\psi}(xg^{-1})),$

where $R_{g}:G\to G$ is right multiplication by $g$. Since $R_{g}$ is a
homeomorphism of $G$, it follows that $C_{\phi}(x)$ is dense in $G$ if and
only if $C_{\psi}(xg^{-1})$ is dense in $G$. ∎

######  Remark 3.6.

The preceding lemma implies that if
$\operatorname{Aut}(G)=\operatorname{Inn}(G)$, then $G$ has the Rokhlin
property if and only if $G$ has the twisted Rokhlin property. In particular,
if $G$ is the symmetric group on a countably infinite set, then every
automorphism of $G$ is inner, and hence $G$ has the twisted Rokhlin property.

######  Proposition 3.7.

Let $G$ be a topological group admitting a proper open subgroup $H$ such that
$gH\phi(g)^{-1}=H$ for all $g\in G$. Then $G$ does not have the $\phi$-twisted
Rokhlin property. In particular, if there exists a proper open normal subgroup
$H$, then $G$ does not have the Rokhlin property.

###### Proof.

The given condition implies that, for each $x\in G$, either
$C_{\phi}(x)\subseteq H$ or $C_{\phi}(x)\cap H=\emptyset$. Thus,
$C_{\phi}(x)\cap gH=\emptyset$ for some $g\in G$, where $gH$ is open.
Therefore, $C_{\phi}(x)$ is not dense. Hence, $G$ does not have $\phi$-twisted
Rokhlin property. ∎

######  Remark 3.8.

By Proposition 2.4, if $S$ is an infinite-type surface, then
$\operatorname{Aut}(\mathrm{MCG}^{\pm}(S))=\operatorname{Inn}(\mathrm{MCG}^{\pm}(S))$.
Since $\mathrm{MCG}(S)$ is a proper open normal subgroup of
$\mathrm{MCG}^{\pm}(S)$ (being a finite index subgroup), it follows that
$\mathrm{MCG}^{\pm}(S)$ does not have the Rokhlin property. Consequently, it
does not have the twisted Rokhlin property.

######  Definition 3.9.

Let $G$ be a topological group and let $\phi$ be an automorphism of $G$. Then
$G$ has the $\phi$-twisted joint embedding property ($\phi$-TJEP for short) if
given any two non-empty open subsets $U$ and $V$ of $G$, there exists $g\in G$
such that $U\cap gV\phi(g)^{-1}\neq\emptyset$.

Note that the set $gV\phi(g)^{-1}$ is open whenever $V$ is open. Recall that,
a topological group is Polish if its underlying topological space is separable
and completely metrizable. The next result extends [18, Theorem 2.2] to the
$\phi$-twisted case.

######  Proposition 3.10.

Let $G$ be a Polish topological group and let $\phi$ be an automorphism of
$G$. Then $G$ has the $\phi$-twisted Rokhlin property if and only if $G$ has
the $\phi$-TJEP.

###### Proof.

Suppose that $G$ has the $\phi$-TJEP. Since $G$ is Polish, it is second
countable. Let $\mathcal{B}$ be a countable basis for $G$. For each
$U\in\mathcal{B}$, define

$D_{U}=\cup_{g\in G}~{}gU\phi(g)^{-1}.$

Note that $D_{U}$ is open. Further, it follows from the $\phi$-TJEP that
$D_{U}$ is dense in $G$. Since $G$ is completely metrizable, it is a Baire
space. Hence, the set $D=\cap_{U\in\mathcal{B}}~{}D_{U}$ is dense in $G$. Let
$x\in D$ be an arbitrary element. Then, for each basis member
$V\in\mathcal{B}$, there exists $g\in G$ (depending on $V$) such that $x\in
gV\phi(g)^{-1}$. In other words, there exists $v\in V$ such that
$v=g^{-1}x\phi(g)$, and hence $v\in V\cap C_{\phi}(x)$. This proves that
$C_{\phi}(x)$ is dense in $G$.

Conversely, suppose that the $\phi$-twisted conjugacy class $C_{\phi}(x)$ is
dense in $G$. Given non-empty open sets $U$ and $V$ in $G$, there exist
$y,z\in G$ such that $yx\phi(y)^{-1}\in U$ and $zx\phi(z)^{-1}\in V$. Setting
$h=yz^{-1}$, it is easy to see that $yx\phi(y)^{-1}\in U\cap hV\phi(h)^{-1}$.
Hence, the group $G$ has the $\phi$-TJEP. ∎

######  Remark 3.11.

It follows from [25, Corollary 5] that if $S$ is an infinite-type surface,
then $\mathrm{MCG}(S)$ is a Polish group. Hence, $\mathrm{MCG}(S)$ has the
$\phi$-twisted Rokhlin property if and only if it has the $\phi$-TJEP.

##  4\. Twisted Rokhlin property of big mapping class groups

It is known that the mapping class group of a surface with compact boundary
(including no boundary) is discrete if and only if the surface is of finite-
type. Thus, the mapping class group of a finite-type surface does not contain
any proper dense subset, and consequently no dense $\phi$-twisted conjugacy
class. Hence, we consider only infinite-type surfaces (without boundary).

Recently, in [18, Theorem 6.1], Lanier and Vlamis have shown that the big
mapping class group of a connected orientable infinite-type surface without
boundary has the Rokhlin property if and only if the surface is non-compact
whose genus is either zero or infinite and whose end space is self-similar
with a unique maximal end.

If a topological group has the twisted Rokhlin property, then it has the
Rokhlin property. Thus, we need to investigate the twisted Rokhlin property
only for a non-compact surface whose genus is either zero or infinite and
whose end space is self-similar with a unique maximal end. Further, in latter
sections, we shall also see whether the big mapping class group of the
remaining surfaces have the $\phi$-twisted Rokhlin property for some
automorphism $\phi$.

###  4.1. When the surface has a unique maximal end

In this subsection, we assume that every compact subsurface of $S$ is
displaceable and $\mathcal{M}(S)$ is a singleton set. Let $\mu$ be the unique
maximal end. Notice that, in this case, $S$ is either planar or has infinite
genus, and the end space $\mathcal{E}(S)$ is self-similar. Consider a
separating simple closed curve $c$ on the surface $S$. Define $\Omega_{c}$ to
be the component of $S\setminus c$ such that $\mu\in{\Omega}_{c}^{*}$ and
$\Sigma_{c}$ to be the complement of $\Omega_{c}$ in $S$. Note that
$\Sigma_{c}$ has $c$ as its unique boundary component. Let $G$ be the subset
of $\mathrm{MCG}(S)$ consisting of elements admitting representatives that
restricts to the identity on $\Omega_{c}$ for some separating simple closed
curve $c$. Note that the set

$\left\\{{\Omega}_{c}^{*}~{}\mid~{}c~{}~{}\textrm{is a separating simple
closed curve on}~{}S\right\\}$

is a neighbourhood basis for $\mu$, and consequently $G$ is a subgroup of
$\mathrm{MCG}(S)$. Since the surface $S$ is assumed to be symmetric about the
$xy$-plane when embedded in $\mathbb{R}^{3}$, we have the orientation-
reversing self-homeomorphism $\mathfrak{t}:S\rightarrow S$. Given a subsurface
$\Sigma$ of $S$, the intersections

$\Sigma^{+}:=\Sigma\cap\\{(x,y,z)\in\mathbb{R}^{3}~{}\mid~{}z\geq
0\\}\quad\textrm{and}\quad\Sigma^{-}:=\Sigma\cap\\{(x,y,z)\in\mathbb{R}^{3}~{}\mid~{}z\leq
0\\}$

are called the positive and the negative parts of $\Sigma$, respectively. We
now present an extension of [18, Proposition 4.2].

######  Proposition 4.1.

Let $S$ be a surface in which every compact subsurface is displaceable and $S$
has a unique maximal end, say $\mu$. Let $G$ be the subgroup of
$\mathrm{MCG}(S)$ defined above. Then the following assertions holds:

  1. (1)

The subgroup $G$ is dense in $\mathrm{MCG}(S)$.

  2. (2)

For each separating simple closed curve $c$ in $S$ such that
$\mathfrak{t}(\Sigma_{c})=\Sigma_{c}$, there exists
$g\in\operatorname{Homeo}^{+}(S)$ such that $g(\Sigma_{c})\subset\Omega_{c}$
and $g\mathfrak{t}g^{-1}\mathfrak{t}$ is identity on $\Sigma_{c}$.

###### Proof.

Assertion (1) follows from [18, Proposition 4.2]. Let $c$ be a separating
simple closed curve in $S$ such that $\mathfrak{t}(\Sigma_{c})=\Sigma_{c}$.
Since $S$ is of infinite-type, we can choose another separating simple closed
curve $c^{\prime}$ such that $\Sigma_{c}\cap\Sigma_{c^{\prime}}=\emptyset$,
$\mathfrak{t}(\Sigma_{c^{\prime}})=\Sigma_{c^{\prime}}$ and
$\Sigma_{c}\cong\Sigma_{c^{\prime}}$. Further, choose a separating simple
closed curve $b$ such that the curves $\\{c,c^{\prime},b\\}$ bounds a pair of
pants (a subsurface homeomorphic to the two-sphere with three disks removed)
and $\\{c,c^{\prime}\\}\subset\Sigma_{b}$. Fix some hyperbolic metric on
$\Sigma_{b}$ and choose an isometry $h:\Sigma_{b}\to\Sigma_{b}$ fixing
$\partial(\Sigma_{b})$ identically such that
$h(\Sigma_{c})=\Sigma_{c^{\prime}}$ and
$h(\Sigma_{c}^{+})=\Sigma_{c^{\prime}}^{+}$. Define $f:S\to S$ by

$f(x)=\begin{cases}h(x)&\text{ if }x\in\Sigma_{c}^{+},\\\
\mathfrak{t}h\mathfrak{t}(x)&\text{ if }x\in\Sigma_{c}^{-},\\\ h(x)&\text{ if
}x\in\Sigma_{b}\setminus\Sigma_{c},\\\ x&\text{ if }x\in
S\setminus\Sigma_{b}.\end{cases}$

Thus $g:=f^{-1}\in\operatorname{Homeo}^{+}(S)$ such that
$g^{-1}(\Sigma_{c})\subset\Omega_{c}$. Consequently, we have
$g(\Sigma_{c})\subset\Omega_{c}$. Now, if $x\in\Sigma_{c}^{+}$, then
$\mathfrak{t}(x)\in\Sigma_{c}^{-}$, and hence

$g\mathfrak{t}g^{-1}\mathfrak{t}(x)=g\mathfrak{t}\mathfrak{t}h\mathfrak{t}(\mathfrak{t}(x))=gh(x)=gf(x)=x.$

Similarly, if $x\in\Sigma_{c}^{-}$, then $\mathfrak{t}(x)\in\Sigma_{c}^{+}$,
and hence $g\mathfrak{t}g^{-1}\mathfrak{t}(x)=x.$ This shows that
$g\mathfrak{t}g^{-1}\mathfrak{t}$ is identity on $\Sigma_{c}$, which
establishes assertion (2). ∎

######  Theorem 4.2.

Let $S$ be a surface in which every compact subsurface is displaceable and $S$
has a unique maximal end. Then $\mathrm{MCG}(S)$ has the twisted Rokhlin
property.

###### Proof.

It follows from [18, Theorem 2.2 and Lemma 4.3] that $\mathrm{MCG}(S)$ has the
Rokhlin property. Consequently, by Proposition 3.5, $\mathrm{MCG}(S)$ has the
$\phi$-twisted Rokhlin property for each inner automorphism $\phi$ of
$\mathrm{MCG}(S)$. By Proposition 2.4, we have
$\operatorname{Out}(\mathrm{MCG}(S))\cong\mathbb{Z}_{2}$. Thus, in view of
Proposition 3.5 and Proposition 3.10, it is enough to prove that
$\mathrm{MCG}(S)$ has the $\hat{\mathfrak{t}}$-TJEP.

Let $U_{1}$ and $U_{2}$ be non-empty open subsets of $\mathrm{MCG}(S)$.
Without loss of generality, we assume that $U_{1}$ and $U_{2}$ are the basis
elements of $\mathrm{MCG}(S)$. Thus, there exist finite subsets
$A_{1},A_{2}\subset C(S)$ and $h_{1},h_{2}\in\mathrm{MCG}(S)$ such that
$U_{1}=h_{1}U_{A_{1}}$ and $U_{2}=h_{2}U_{A_{2}}$. By Proposition 4.1(1), the
subgroup $G$ is dense in $\mathrm{MCG}(S)$. Thus, by Lemma 2.6, we can assume
that $h_{1},h_{2}\in G$. Since $h_{1},h_{2}\in G$, there exist separating
simple closed curves $c_{1},c_{2}$ on $S$ such that $h_{1}$ and $h_{2}$ admit
representatives that restrict to the identity on $\Omega_{c_{1}}$ and
$\Omega_{c_{2}}$, respectively. Next, we choose a separating simple closed
curve $b$ such that $c_{1},c_{2}\subset\Sigma_{b}$ and
$\mathfrak{t}(\Sigma_{b})=\Sigma_{b}$. Moreover, by enlarging $\Sigma_{b}$, if
necessary, we can assume that each curve in $A_{1}\cup A_{2}$ has a
representative in $\Sigma_{b}$ and $\mathfrak{t}(\Sigma_{b})=\Sigma_{b}$. By
Proposition 4.1(2), there exists $g\in\operatorname{Homeo}^{+}(S)$ such that
$g(\Sigma_{b})\subset\Omega_{b}$ and $g\mathfrak{t}g^{-1}\mathfrak{t}$ is
identity on $\Sigma_{b}$. Note that $g(\Sigma_{b})\subset\Omega_{b}$ implies
that $g^{-1}(\Sigma_{b})\subset\Omega_{b}$. We have

(4.1.1)
$gU_{2}\hat{\mathfrak{t}}(g)^{-1}=gh_{2}U_{A_{2}}\hat{\mathfrak{t}}(g)^{-1}=gh_{2}\hat{\mathfrak{t}}(g)^{-1}U_{\hat{\mathfrak{t}}(g)(A_{2})}.$

The element $h_{1}$ has a representative that restricts to the identity on
$\Omega_{b}$. Similarly, since $g\mathfrak{t}=\mathfrak{t}g$ on $\Sigma_{b}$
and $g^{-1}(\Sigma_{b})\subset\Omega_{b}$, it follows that the element
$gh_{2}\hat{\mathfrak{t}}(g)^{-1}$ also has a representative that restricts to
the identity on $\Sigma_{b}$. Thus, we have the following:

  * •

$h_{1}$ and $gh_{2}\hat{\mathfrak{t}}(g)^{-1}$ commute on $S$,

  * •

$h_{1}\in U_{\hat{\mathfrak{t}}(g)(A_{2})}$ (since
$\hat{\mathfrak{t}}(g)(A_{2})\subset\hat{\mathfrak{t}}(g)(\Sigma_{b})\subset\Omega_{b}$),

  * •

$gh_{2}\hat{\mathfrak{t}}(g)^{-1}\in U_{A_{1}}$ (since
$A_{1}\subset\Sigma_{b}$).

Taking $h=h_{1}(gh_{2}\hat{\mathfrak{t}}(g)^{-1})$, we see that

$h\in h_{1}U_{A_{1}}=U_{1}$

and

$h\in
gh_{2}\hat{\mathfrak{t}}(g)^{-1}U_{\hat{\mathfrak{t}}(g)(A_{2})}=gU_{2}\hat{\mathfrak{t}}(g)^{-1}\quad(\textrm{by}~{}\eqref{gu2phig}).$

Thus, $h\in U_{1}\cap gU_{2}\hat{\mathfrak{t}}(g)^{-1}$, which completes the
proof. ∎

###  4.2. When the surface has exactly two maximal ends

In this subsection, we assume that every compact subsurface of our surface $S$
is displaceable and $\mathcal{M}(S)$ has exactly two elements. We claim that
$\mathrm{MCG}(S)$ does not have the $\phi$-twisted Rokhlin property for any
$\phi\in\operatorname{Aut}(\mathrm{MCG}(S))$. It is enough to consider the
automorphism $\hat{\mathfrak{t}}$ of $\mathrm{MCG}(S)$. Recall that, there is
an action of $\mathrm{MCG}(S)$ on $\mathcal{E}(S)$ that keeps $\mathcal{M}(S)$
invariant.

######  Lemma 4.3.

The stabiliser $\operatorname{Stab}(\mu)$ of a maximal end is a proper open
normal subgroup of $\mathrm{MCG}(S)$ such that
$h\operatorname{Stab}(\mu)\hat{\mathfrak{t}}(h)^{-1}=\operatorname{Stab}(\mu)$
for all $h\in\mathrm{MCG}(S)$.

###### Proof.

It follows from [18, Lemma 5.4] that $\operatorname{Stab}(\mu)$ is a proper
open normal subgroup of $\mathrm{MCG}(S)$. If $f\in
h\operatorname{Stab}(\mu)\hat{\mathfrak{t}}(h)^{-1}$, then
$f=hf^{\prime}\mathfrak{t}h^{-1}\mathfrak{t}$ for some
$f^{\prime}\in\operatorname{Stab}(\mu)$. Since $\operatorname{Stab}(\mu)$ is
normal, we see that

$f(\mu)=hf^{\prime}\mathfrak{t}h^{-1}\mathfrak{t}(\mu)=hf^{\prime}\mathfrak{t}h^{-1}(\mu)=hf^{\prime}h^{-1}(\mu)=\mu.$

Hence,
$h\operatorname{Stab}(\mu)\hat{\mathfrak{t}}(h)^{-1}\subseteq\operatorname{Stab}(\mu)$
for all $h\in\mathrm{MCG}(S)$. The converse inclusion can be proved in a
similar way. ∎

######  Proposition 4.4.

Let $S$ be a surface in which every compact subsurface is displaceable and
$\mathcal{M}(S)$ has exactly two elements. Then $\mathrm{MCG}(S)$ does not
have the $\phi$-twisted Rokhlin property for any
$\phi\in\operatorname{Aut}(\mathrm{MCG}(S))$.

###### Proof.

The proof follows directly from Proposition 3.5, Proposition 3.7 and Lemma
4.3. ∎

###  4.3. When the space of maximal ends is a Cantor space

This subsection deals with the case when every compact subsurface of $S$ is
displaceable and $\mathcal{M}(S)$ is a Cantor space. The following result is a
special case of [5, Theorem 14.1].

######  Proposition 4.5.

Let $S_{g,n}$ be a surface of genus $g$ with $n$ punctures and let $\\{a,b\\}$
be a filling system of curves for $S_{g,n}$. Consider the subgroup $\langle
T_{a},T_{b}\rangle$ of $\mathrm{MCG}(S)$ and the representation $\rho:\langle
T_{a},T_{b}\rangle\to PSL(2,\mathbb{R})$ given by

$\rho(T_{a})=\left(\begin{matrix}1&i(a,b)\\\
0&1\end{matrix}\right)\quad\textrm{and}\quad\rho(T_{b})=\left(\begin{matrix}1&0\\\
-i(a,b)&1\end{matrix}\right).$

Let $f\in\langle T_{a},T_{b}\rangle$ such that $\rho(f)$ is hyperbolic in
$PSL(2,\mathbb{R})$. Then $f$ is pseudo-Anosov.

The following lemma will be used in the main result of this subsection.

######  Lemma 4.6.

Consider the surface $S_{0,6}$ such that it is symmetric about the $xy$-plane
and all its punctures lie on the $xy$-plane. Then there exists a mapping class
$f\in\mathrm{MCG}(S_{0,6})$ such that $f$ permutes the punctures cyclically
and $f\mathfrak{t}f\mathfrak{t}$ is pseudo-Anosov.

###### Proof.

Let $b_{1},\ldots,b_{6}$ be the punctures on $S_{0,6}$. Let $r$ be the order
six homeomorphism of $S_{0,6}$ such that $r(b_{i})=b_{i+1}$ for $1\leq i\leq
5$ and $r(b_{6})=b_{1}$. Let $\mathfrak{t}$ denote a map that sends a point of
$S_{0,6}$ to its mirror image about the $xy$-plane, $c$ be the simple closed
curve on $S_{0,6}$ as shown in Figure 3 and $d=\mathfrak{t}(c)$. Then
$f=rT_{c}T_{d}^{-2}T_{c}^{3}T_{d}^{-4}$ is an orientation-preserving
homeomorphism of $S_{0,6}$. Note that $r(c)=\mathfrak{t}(c)=d$ and
$r\mathfrak{t}=\mathfrak{t}r$ on $S_{0,6}$. Furthermore, $f(b_{i})=b_{i+1}$
for $1\leq i\leq 5$ and $f(b_{6})=b_{1}$.

$c$ at 110, 210 $x$ at 350, 155 $y$ at 170, 310 $b_{1}$ at 90, -10 $b_{2}$ at
230, -10 $b_{3}$ at 330, 125 $b_{4}$ at 220, 285 $b_{5}$ at 90, 285 $b_{6}$ at
-13, 140

Figure 3. The curve $c$ on $S_{0,6}$ such that $\\{c,\mathfrak{t}(c)\\}$ fills
$S_{0,6}$.

We see that

$\displaystyle(f\mathfrak{t}f\mathfrak{t})^{3}$
$\displaystyle=(rT_{c}T_{d}^{-2}T_{c}^{3}T_{d}^{-4}rT_{d}^{-1}T_{c}^{2}T_{d}^{-3}T_{c}^{4})^{3},~{}~{}\text{since}~{}~{}r\mathfrak{t}=\mathfrak{t}r\text{
and }\mathfrak{t}T_{c}\mathfrak{t}=T_{\mathfrak{t}(c)}^{-1}$
$\displaystyle=(T_{d}T_{c}^{-2}T_{d}^{3}T_{c}^{-4}T_{d}^{-1}T_{c}^{2}T_{d}^{-3}T_{c}^{4}r^{2})^{3},~{}~{}\text{since}~{}~{}rT_{c}{r}^{-1}=T_{r(c)}=T_{d}$
$\displaystyle=(T_{d}T_{c}^{-2}T_{d}^{3}T_{c}^{-4}T_{d}^{-1}T_{c}^{2}T_{d}^{-3}T_{c}^{4})^{3}.$

Since $i(c,d)=6$, by Proposition 4.5, we have

$\rho(T_{c})=\left(\begin{matrix}1&6\\\
0&1\end{matrix}\right)\quad\textrm{and}\quad\rho(T_{d})=\left(\begin{matrix}1&0\\\
-6&1\end{matrix}\right).$

This gives

$\rho((f\mathfrak{t}f\mathfrak{t})^{3})=\left(\begin{matrix}-6768504263453271167830031&-162818421281863319251229760\\\
41172474347137850845138560&990416347916847555740563729\end{matrix}\right),$

which is hyperbolic since the trace of
$\rho\big{(}(f\mathfrak{t}f\mathfrak{t})^{3}\big{)}$ is greater than 2. Hence,
$f\mathfrak{t}f\mathfrak{t}$ is a pseudo-Anosov map. ∎

######  Theorem 4.7.

Let $S$ be a surface in which every compact subsurface is displaceable and
$\mathcal{M}(S)$ is homeomorphic to a Cantor space. Then $\mathrm{MCG}(S)$
does not have the $\phi$-twisted Rokhlin property for any
$\phi\in\operatorname{Aut}(\mathrm{MCG}(S))$.

###### Proof.

By Proposition 3.5 and [18, Lemma 4.8], it is enough to prove that
$\mathrm{MCG}(S)$ does not have the $\hat{\mathfrak{t}}$-twisted Rokhlin
property. By results in [18, Proposition 4.6] (or [19, Lemma 4.18]), we choose
a compact subsurface $\Sigma$ of $S$ such that

  1. (1)

$\Sigma$ is planar,

  2. (2)

$\Sigma$ has six boundary components, say
$b_{1},b_{2},b_{3},b_{4},b_{5},b_{6},$

  3. (3)

each $b_{i}$ is a separating curve of $S$,

  4. (4)

$\Omega_{1},\Omega_{2},\Omega_{3},\Omega_{4},\Omega_{5},\Omega_{6}$ are the
six components of $S\setminus\Sigma$ such that $\Omega_{i}^{*}$ is
homeomorphic to $\mathcal{E}(S)$,

  5. (5)

$\mathfrak{t}(\Sigma)=\Sigma,~{}\mathfrak{t}(b_{i})=b_{i}$ and
$\mathfrak{t}(\Omega_{i})=\Omega_{i}$ for each $1\leq i\leq 6$.

By Lemma 4.6, let $f^{\prime}\in\operatorname{Homeo}^{+}(\Sigma)$ such that
$f^{\prime}(b_{i})=b_{i+1}$ for $1\leq i\leq 5$ and $f^{\prime}(b_{6})=b_{1}$
and $f^{\prime}\mathfrak{t}f^{\prime}\mathfrak{t}$ is pseudo-Anosov. Using
condition (4), there is an extension $f\in\mathrm{MCG}(S)$ of $f^{\prime}$.

Recall that a subset $A$ of $C(S)$ is called a stable Alexander system if
$U_{A}$ is the center of $\mathrm{MCG}(S)$. Choose a stable Alexander system
$A$ for $\Sigma$. Let us take $U=fU_{A}$ and $V=U_{\\{b_{1}\\}}$. We claim
that $U\cap gV\hat{\mathfrak{t}}(g)^{-1}=\emptyset$ for all
$g\in\mathrm{MCG}(S)$. For an element $g\in\mathrm{MCG}(S)$, we consider the
following cases:

Case 1: Suppose that $g(b_{1})$ has a representative which is disjoint from
$\Sigma$. In this case, $g(b_{1})\in\Omega_{i}$ for some $i$. If $h\in
gV\hat{\mathfrak{t}}(g)^{-1}$, then using condition (5), we obtain
$\hat{\mathfrak{t}}(g)(b_{1})\in\Omega_{i}$ and
$h(\hat{\mathfrak{t}}(g)(b_{1}))=g(b_{1})\in\Omega_{i}$. Hence, we have
$h(\Omega_{i})\cap\Omega_{i}\neq\emptyset$. And, if $h\in U$, then
$h(\Omega_{i})\cap\Omega_{i}=\emptyset$ since $h(b_{i})=b_{i+1}$. This proves
that $U\cap gV\hat{\mathfrak{t}}(g)^{-1}=\emptyset$.

Case 2: Suppose that $g(b_{1})$ has a non-trivial geometric intersection with
$\Sigma$. Let $c\in C(\Sigma)$ be a projection of $g(b_{1})$ onto $\Sigma$. If
$h\in U\cap gV\hat{\mathfrak{t}}(g)^{-1}$, then
$h\mathfrak{t}g(b_{1})=h(\hat{\mathfrak{t}}(g)(b_{1}))=g(b_{1})$. This gives

$h\mathfrak{t}h\mathfrak{t}g(b_{1})=h\mathfrak{t}g(b_{1})=g(b_{1}),$

and consequently $(h\mathfrak{t}h\mathfrak{t})^{n}(g(b_{1}))=g(b_{1})$ for all
$n\in\mathbb{N}$. Since $h\in U$, it follows that
$(h\mathfrak{t}h\mathfrak{t})^{n}(c)\in C(\Sigma)$ is also a projection of
$g(b_{1})$ onto $\Sigma$ for all $n\in\mathbb{N}$. Hence, by Lemma 2.9, we
have

(4.3.1) $d_{C(\Sigma)}((h\mathfrak{t}h\mathfrak{t})^{n}(c),c)\leq 2$

for all $n\in\mathbb{N}$. Since $h\in U$, we have
$h\mathfrak{t}h\mathfrak{t}=f\mathfrak{t}f\mathfrak{t}$ on $\Sigma$, and hence
the map $h\mathfrak{t}h\mathfrak{t}$ is pseudo-Anosov on $\Sigma$. By
Proposition 2.7, there exists $m\in\mathbb{N}$ such that

$d_{C(\Sigma)}\big{(}(f\mathfrak{t}f\mathfrak{t})^{m}(c),c\big{)}>2$

for all $c\in C(\Sigma)$. This contradicts (4.3.1). Hence, we must have $U\cap
gV\hat{\mathfrak{t}}(g)^{-1}=\emptyset$ for all $g\in\mathrm{MCG}(S)$. Hence,
by Proposition 3.10, the proof of the theorem is complete. ∎

###  4.4. When the surface admits a compact non-displaceable subsurface

Finally, we consider the case when $S$ admits a compact non-displaceable
subsurface.

######  Lemma 4.8.

Consider the surface $S_{g,n}$ of genus $g\geq 0$ with $n$ punctures, where
$n>5$ is even when $g=0$. Suppose that $S_{g,n}$ is symmetric about the
$xy$-plane with all the punctures lying on the $xy$-plane. Then there exists
an essential simple closed curve $c$ on $S_{g,n}$ such that the collection
$\\{c,\mathfrak{t}(c)\\}$ fills the surface. Moreover, there exists an
orientation-preserving homeomorphism $f$ of $S_{g,n}$ such that
$f\mathfrak{t}f\mathfrak{t}$ is pseudo-Anosov.

###### Proof.

If $g>0$, then the curve $c$ as shown in Figure 4 together with
$\mathfrak{t}(c)$ fills the surface $S_{g,n}$.

Figure 4. The curve $c$ on $S_{g,n}$ such that $\\{c,\mathfrak{t}(c)\\}$ fills
$S_{g,n}$.

If $g=0$, then choose the simple closed curve $c$ to be as depicted in Figure
3 for $S_{0,6}$. Take $f=T_{c}$, where $c$ is the curve as above. By using
Penner’s construction [5, Theorem 14.4], the homeomorphism
$f\mathfrak{t}f\mathfrak{t}=T_{c}T_{\mathfrak{t}(c)}^{-1}$ is pseudo-Anosov
since the collection $\\{c,\mathfrak{t}(c)\\}$ fills $S_{g,n}$. ∎

######  Theorem 4.9.

Let $S$ be a surface containing a compact non-displaceable subsurface. Then
$\mathrm{MCG}(S)$ does not have the $\phi$-twisted Rokhlin property for any
$\phi\in\operatorname{Aut}(\mathrm{MCG}(S))$.

###### Proof.

In view of Proposition 3.5 and [18, Theorem 3.1], it is enough to prove that
$\mathrm{MCG}(S)$ does not have the $\hat{\mathfrak{t}}$-twisted Rokhlin
property. Let $\Sigma$ be a compact non-displaceable subsurface of $S$. By
enlarging $\Sigma$, we can ensure that it is connected, has a stable Alexander
system, satisfies $\mathfrak{t}(\Sigma)=\Sigma$ and its boundary
$\partial(\Sigma)$ consists of at least six components with an even number of
boundary components, each being essential and separating. Furthermore,
$\Sigma$ remains a finite-type subsurface of $S$.

We claim that there exist open subsets $U$ and $V$ of $\mathrm{MCG}(S)$ such
that $U\cap gV\hat{\mathfrak{t}}(g)^{-1}=\emptyset$ for all
$g\in\mathrm{MCG}(S)$. Let $A\subset C(S)$ be a stable Alexander system for
$\Sigma$. It follows from the proof of Lemma 4.8 that there is an orientation-
preserving homeomorphism $f^{\prime}:\Sigma\to\Sigma$ which fixes
$\partial(\Sigma)$ point-wise such that
$f^{\prime}\mathfrak{t}f^{\prime}\mathfrak{t}$ is pseudo-Anosov. Thus, by
Proposition 2.7, there exists $m\in\mathbb{N}$ such that

(4.4.1)
$d_{C(\Sigma)}\big{(}(f^{\prime}\mathfrak{t}f^{\prime}\mathfrak{t})^{m}(a),a\big{)}>2$

for all $a\in C(\Sigma)$. Let $f\in\mathrm{MCG}(S)$ denote the extension of
$f^{\prime}$. Take $U=fU_{A}$ and $V=U_{A}$. Suppose that there exists
$g\in\mathrm{MCG}(S)$ such that $h\in fU_{A}\cap
gU_{A}\hat{\mathfrak{t}}(g)^{-1}$. Since $h\in
gU_{A}\hat{\mathfrak{t}}(g)^{-1}=g\hat{\mathfrak{t}}(g)^{-1}U_{\hat{\mathfrak{t}}(g)(A)}$,
we have

$h\mathfrak{t}g\mathfrak{t}(c)=g(c)$

for all $c\in A$. Thus, we obtain

(4.4.2)
$h\mathfrak{t}h\mathfrak{t}(g(c))=h\mathfrak{t}h\mathfrak{t}g\mathfrak{t}\mathfrak{t}(c)=h\mathfrak{t}g(\mathfrak{t}(c))=g(c)$

for all $c\in A$. Since $\Sigma$ is non-displaceable in $S$, there exists a
curve $c\in C(\Sigma)$ such that $g(c)$ and $\Sigma$ have a non-trivial
geometric intersection. Let $b$ denote a projection of $g(c)$ onto $\Sigma$.
By (4.4.2), $(h\mathfrak{t}h\mathfrak{t})^{n}(b)$ is also a projection of
$g(c)$ onto $(h\mathfrak{t}h\mathfrak{t})^{n}(\Sigma)=\Sigma$. By Lemma 2.9,
we have

$d_{C(\Sigma)}((h\mathfrak{t}h\mathfrak{t})^{n}(b),b)\leq 2$

for all $n\in\mathbb{N}$. Since $h\in fU_{A}$, by (4.4.1), we have

$d_{C(\Sigma)}((h\mathfrak{t}h\mathfrak{t})^{m}(b),b)=d_{C(\Sigma)}((f\mathfrak{t}f\mathfrak{t})^{m}(b),b)>2$

for some $m$, which is the contradiction. Hence, we conclude that $fU_{A}\cap
gU_{A}\hat{\mathfrak{t}}(g)^{-1}=\emptyset$ for all $g\in\mathrm{MCG}(S)$.
Hence, by Proposition 3.10, the proof of the theorem is complete. ∎

The outcomes from this section lead to the following theorem.

######  Theorem 4.10.

The mapping class group $\mathrm{MCG}(S)$ of a connected and orientable
surface $S$ without boundary has the $\phi$-twisted Rokhlin property for some
automorphism $\phi$ of $\mathrm{MCG}(S)$ if and only if the surface is either
the $2$-sphere or satisfy the property that every compact subsurface of $S$ is
displaceable and $S$ has a unique maximal end.

###### Proof.

The mapping class group of the $2$-sphere obviously satisfies the desired
property being a trivial group. If the surface $S$ is such that every compact
subsurface of $S$ is displaceable and $S$ has a unique maximal end, then by
Theorem 4.2, it has the twisted Rokhlin property, and hence $\phi$-twisted
Rokhlin property for each $\phi$. Conversely, suppose that $\mathrm{MCG}(S)$
has the $\phi$-twisted Rokhlin property for some automorphism $\phi$ of
$\mathrm{MCG}(S)$. If $S$ is compact, then it must be the 2-sphere. If $S$ is
non-compact, then by Theorem 4.9, every compact subsurface of $S$ must be
displaceable. Finally, by Theorem 2.3, Theorem 4.2, Proposition 4.4, and
Theorem 4.7, $S$ must satisfy the property that every compact subsurface of
$S$ is displaceable and $S$ has a unique maximal end. ∎

##  5\. $R_{\infty}$-property of big mapping class groups

In this section, we investigate the number of twisted conjugacy classes in
mapping class groups.

######  Definition 5.1.

Let $G$ be a group and $\phi\in\operatorname{Aut}(G)$. The Reidemeister number
of $\phi$, denoted by $R(\phi)$, is the number of $\phi$-twisted conjugacy
classes in $G$. The group $G$ is said to have the $R_{\infty}$-property if
$R(\phi)$ is infinite for each automorphism $\phi$ of $G$.

The $R_{\infty}$-property for the mapping class groups of finite-type surfaces
has been investigated in [8, Theorem 4.3]. It has been proven that the mapping
class group of a closed orientable surface has the $R_{\infty}$-property if
and only if the surface is not the 2-sphere. A partial result for infinite-
type surfaces has been established in [4, Theorem 3.4], for instance, it is
proven that if a connected orientable infinite-type surface without boundary
admits a non-displaceable compact subsurface, then its mapping class group
possesses the $R_{\infty}$-property. The aim of this section is to provide a
complete answer for infinite-type surfaces through a general approach.

Let $S$ be a connected orientable infinite-type surface without boundary.
Recall that, $\operatorname{Out}(\mathrm{MCG}(S))\cong\mathbb{Z}_{2}$. In
fact, the group is generated by the coset of the restriction of the inner
automorphism $\hat{f}$ of $\mathrm{MCG}^{\pm}(S)$ induced by an orientation
reversing homeomorphism $f$ of $S$ of order two (see [3, Theorem 1.1] and
[14]). We use the same notation $\hat{f}$ to denote its restriction on
$\mathrm{MCG}(S)$.

The subsequent lemma provides a connection between $\hat{f}$-twisted conjugacy
classes in $\mathrm{MCG}(S)$ and conjugacy classes of $\mathrm{MCG}^{\pm}(S)$
lying in the coset $\mathrm{MCG}(S)f$. See [7, Lemma 2.1] for a general
result.

######  Lemma 5.2.

Let $S$ be a connected orientable infinite-type surface without boundary and
$f$ an orientation reversing homeomorphism of $S$. Then, two elements
$g,h\in\mathrm{MCG}(S)$ are $\hat{f}$-twisted conjugate in $\mathrm{MCG}(S)$
if and only if $gf$ and $hf$ are conjugate in $\mathrm{MCG}^{\pm}(S)$.

###### Proof.

Suppose that there exists $r\in\mathrm{MCG}(S)$ such that
$h=rg\hat{f}(r)^{-1}=rgfr^{-1}f^{-1}$. But, this gives $hf=r(gf)r^{-1}$, that
is, $gf$ and $hf$ are conjugate in $\mathrm{MCG}^{\pm}(S)$.

Conversely, suppose that there exists $s\in\mathrm{MCG}^{\pm}(S)$ such that
$hf=s(gf)s^{-1}$. If $s\in\mathrm{MCG}(S)$, then
$h=sgfs^{-1}f^{-1}=sg\hat{f}(s)^{-1}$, and we are done. If
$s\not\in\mathrm{MCG}(S)$, then $s=t\,f$ for some $t\in\mathrm{MCG}(S)$. In
this case, we have

$h=t(fgf)f^{-1}t^{-1}f^{-1}=t(fgf^{-1})\hat{f}(t)^{-1}.$

Hence, $h$ and $(fgf^{-1})$ are $\hat{f}$-twisted conjugate in
$\mathrm{MCG}(S)$. Since
$(fgf^{-1})=g^{-1}g(fgf^{-1})=g^{-1}g\hat{f}(g^{-1})^{-1}$, the elements
$(fgf^{-1})$ and $g$ are $\hat{f}$-twisted conjugate in $\mathrm{MCG}(S)$.
Consequently, $h$ and $g$ are $\hat{f}$-twisted conjugate in
$\mathrm{MCG}(S)$, and the proof is complete. ∎

######  Lemma 5.3.

Let $S$ be a connected orientable surface.

  1. (1)

Let $\left\\{a_{1},\ldots,a_{l}\right\\}$ and
$\left\\{b_{1},\ldots,b_{k}\right\\}$ be multicurves on $S$. Let
$p_{1},\ldots,p_{l}$ and $q_{1},\ldots,q_{k}$ be non-zero integers. If

$T_{a_{1}}^{p_{1}}T_{a_{2}}^{p_{2}}\cdots
T_{a_{l}}^{p_{l}}=T_{b_{1}}^{q_{1}}T_{b_{2}}^{q_{2}}\cdots T_{b_{k}}^{q_{k}}$

in $\mathrm{MCG}(S)$, then $l=k$ and
$\left\\{T_{a_{1}}^{p_{1}},\ldots,T_{a_{l}}^{p_{l}}\right\\}=\left\\{T_{b_{1}}^{q_{1}},\ldots,T_{b_{k}}^{q_{k}}\right\\}$
[5, Lemma 3.17].

  2. (2)

Let $a$ and $b$ be essential simple closed curves on $S$. Then
$T_{a}^{k}=T_{b}^{l}$ if and only if $a=b$ and $k=l$ [5, Section 3.3].

######  Lemma 5.4.

[9, Corollary 3.2] Let $G$ be a group, $\phi\in\operatorname{Aut}(G)$ and
$g\in G$. Then $R(\phi\,\hat{g})=R(\phi)$. In particular, $R(\hat{g})$ is the
number of conjugacy classes in $G$.

######  Proposition 5.5.

Let $S$ be a connected orientable infinite-type surface without boundary such
that either $g\geq 2$ (can be infinite) or every compact subsurface of $S$ is
displaceable. Then there exists an essential simple closed curve $c$ and an
orientation reversing homeomorphism $f$ of $S$ of order two such that
$\\{c,f(c)\\}$ forms a multicurve.

###### Proof.

We have the following two cases:

Case 1: Suppose that $g\geq 2$ (including infinite genus). We assume that the
surface $S$ is symmetric about the $xy$-plane when embedded in
$\mathbb{R}^{3}$. Then the orientation reversing homeomorphism $\mathfrak{t}$
and the curve $c$ shown in Figure 5 have the property that
$\\{c,\mathfrak{t}(c)\\}$ forms a multicurve.

$c$ at 300, 80 $x$ at 380, 65 $y$ at 195, 140

Figure 5. The curve $c$ such that $\\{c,\mathfrak{t}(c)\\}$ forms a
multicurve.

Case 2: Suppose that every compact subsurface of $S$ is displaceable. In this
case, the genus of $S$ is either zero or infinite. The latter possibility is
already considered in Case 1. So, we assume that the genus is zero, that is,
$S$ is planar. In view of Theorem 2.3, we have three possibilities. If $S$ has
a unique maximal end, then by the proof of Proposition 4.1, there exist curves
$\\{c,c^{\prime},b\\}$ which bound a pair of pants and
$\\{c,c^{\prime}\\}\subset\Sigma_{b}$ as shown in Figure 6. Now, we take $f$
to be the orientation reversing homeomorphism of $S$ which maps a point on $S$
to its mirror image about the $xz$-plane. Then the collection
$\\{c,f(c)=c^{\prime}\\}$ forms a multicurve.

$x$ at 365, 120 $c^{\prime}$ at 67, 170 $c$ at 70, 65 $b$ at 185, 140 $y$ at
125, 290 $\Omega_{b}$ at 235, 180 $\Sigma_{c}$ at 50, 0 $\Sigma_{c^{\prime}}$
at 50, 230

Figure 6. The curve $c$ such that $\\{c,c^{\prime}\\}$ forms a multicurve.

If $S$ has two maximal ends, then by [18, Lemma 5.4], we can realise our
surface $S$ such that there exists an orientation reversing homeomorphism $f$
of $S$ of order two, which permutes the two maximal ends. If $c$ is a curve in
a neighbourhood of one of the maximal ends, then $\\{c,f(c)\\}$ forms a
multicurve.

If $S$ is such that the space $\mathcal{M}(S)$ of maximal ends is homeomorphic
to the cantor space, then we take $f$ to be the orientation reversing
homeomorphism of $S$ which maps a point of $S$ to its mirror image about the
$xz$-plane (see Figure 3). In this case, $\\{b_{1},f(b_{1})=b_{5}\\}$ forms a
multicurve. ∎

######  Proposition 5.6.

Let $S$ be a connected orientable infinite-type surface without boundary. Let
$f$ be an orientation reversing homeomorphism of $S$ of order two such that
there exists an essential simple closed curve $c$ for which $\\{c,f(c)\\}$ is
a multicurve. Then the right coset $\mathrm{MCG}(S)f$ contains infinitely many
conjugacy classes of the group $\mathrm{MCG}^{\pm}(S)$.

###### Proof.

We claim that, for each $n\in\mathbb{N}$, the element $T_{c}^{n}f$ represents
a distinct conjugacy classes in $\mathrm{MCG}^{\pm}(S)$. We set $d:=f(c)$.
Suppose that $T_{c}^{n}f$ is conjugate to $T_{c}^{m}f$ in
$\mathrm{MCG}^{\pm}(S)$ for some $m,n\geq 1$. This implies that
$(T_{c}^{n}f)^{2}=T_{c}^{n}T_{d}^{-n}$ is conjugate to
$(T_{c}^{m}f)^{2}=T_{c}^{m}T_{d}^{-m}$ in $\mathrm{MCG}^{\pm}(S)$. Thus, there
exists $g\in\mathrm{MCG}^{\pm}(S)$ such that

$T_{c}^{n}T_{d}^{-n}=gT_{c}^{m}T_{d}^{-m}g^{-1}=T_{g(c)}^{\epsilon
m}T_{g(d)}^{-\epsilon m},$

where

$\epsilon=\begin{cases}1&\text{ if }g\text{ is orientation preserving,}\\\
-1&\text{ if }g\text{ is orientation reversing.}\end{cases}$

By Lemma 5.3, this is possible only when $m=n$. ∎

######  Theorem 5.7.

Let $S$ be a connected orientable infinite-type surface without boundary. Then
$\mathrm{MCG}(S)$ has the $R_{\infty}$-property.

###### Proof.

It follows from Lemma 2.5 and Lemma 5.3 that if $c$ is an essential simple
closed curve on $S$, then $T_{c}^{m}$ is conjugate to $T_{c}^{n}$ in
$\mathrm{MCG}(S)$ if and only if $m=n$. Thus, $\mathrm{MCG}(S)$ has infinitely
many conjugacy classes. If $S$ admits a non-displaceable compact subsurface of
genus $0$ or $1$, then the theorem follows from [4, Theorem 3.4]. So, we
assume that either every compact subsurface of $S$ is displaceable or $S$ has
genus at least two. Since
$\operatorname{Out}(\mathrm{MCG}(S))\cong\mathbb{Z}_{2}$, in view of Lemma
5.4, it is enough to show that $R(\hat{f})$ is infinite for an orientation
reversing homeomorphism $f$ of $S$ of order two. By Proposition 5.5, we can
take $f$ to be such that there exists an essential simple closed curve $c$ for
which $\\{c,f(c)\\}$ forms a multicurve. By Proposition 5.6,
$\mathrm{MCG}(S)f$ contains infinitely many conjugacy classes of the group
$\mathrm{MCG}^{\pm}(S)$. Finally, Lemma 5.2 implies that $R(\hat{f})$ is
infinite, which completes the proof. ∎

##  6\. Further directions

We conclude with the following questions that arose naturally during our
investigation.

  1. (1)

Our findings indicate that when $S$ is an infinite-type surface without
boundary, the mapping class group $\mathrm{MCG}(S)$ either exhibits the
$\phi$-twisted Rokhlin property for all automorphisms $\phi$, or it lacks the
$\phi$-twisted Rokhlin property for each automorphism $\phi$. We do not know
an example of a topological group $G$ which admits a pair of automorphisms
$\phi,\psi$ such that $G$ has the $\phi$-twisted Rokhlin property, but not the
$\psi$-twisted Rokhlin property? In general, it would be interesting to find
such examples in topological groups that admit ‘sufficiently large’ outer
automorphim groups.

  2. (2)

Suppose that $H$ is a (closed) subgroup of a topological group $G$ such that
each automorphism of $H$ is the restriction of an inner automorphism of $G$.
Is it true that $H$ has the Rokhlin property if and only if it has the twisted
Rokhlin property?

  3. (3)

Which other interesting topological groups admit the twisted Rokhlin property?

######  Acknowledgement.

Pravin Kumar is supported by the PMRF fellowship at IISER Mohali. Mahender
Singh is supported by the Swarna Jayanti Fellowship grants
DST/SJF/MSA-02/2018-19 and SB/SJF/2019-20/04. Apeksha Sanghi acknowledges
post-doctoral fellowship from the grant SB/SJF/2019-20/04.

##  7\. Declaration

The authors declare that they have no conflicts of interest and that there is
no data associated with this paper.

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