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 ϕ 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 ϕ-twisted conjugate if they are in the same orbit under this action. The resulting orbits are referred as ϕ-twisted conjugacy classes. The group G is said to have the 
$R_\infty$-property if the number of ϕ-twisted conjugacy classes is infinite for each automorphism ϕ of G. The idea of twisted conjugacy arose from the work of Reidemeister [Reference Reidemeister21]. It has deep connection with Nielsen fixed-point theory [Reference Fel’shtyn6], and appears in Arthur–Selberg theory [Reference Shokranian23], algebraic geometry [Reference Grothendieck12], and representation theory [Reference Gantmacher10], 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 [Reference Bhunia and Krishna4, Reference Fel’shtyn and Gonçalves8], 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 ϕ an automorphism of the underlying abstract group. We say that G has the ϕ-twisted Rokhlin property if G admits a dense ϕ-twisted conjugacy class. Further, G is said to have the twisted Rokhlin property if it has the ϕ-twisted Rokhlin property for each automorphism ϕ 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 the quotient topology from the compact-open topology on the group of orientation-preserving self-homeomorphisms of the surface. It turns out that this quotient topology is non-discrete if and only if the surface is of infinite-type. The aim of this paper is to classify connected orientable infinite-type surfaces without boundaries whose mapping class groups admit the twisted Rokhlin property. Our approach is built upon the recent work [Reference Lanier and Vlamis17] of Lanier and Vlamis, where they classify connected orientable infinite-type surfaces without boundaries whose mapping class groups admit the Rokhlin property (see also [Reference Hernández, Hrušák, Morales, Randecker, Sedano and Valdez13] 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 ϕ-twisted Rokhlin property if and only if it has the ψ-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 ϕ-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 
$U \cap g V \phi(g)^{-1}\neq \emptyset$ (Proposition 3.10). Section 4 contains our results on the twisted Rokhlin property of big mapping class groups. Our main result is the following classification theorem.
Theorem 1.1. Let 
$S$ be a connected orientable infinite-type surface without boundary and let 
$\mathrm{MCG}(S)$ denote its mapping class group. Then the following assertions are equivalent:
(1)
$\mathrm{MCG}(S)$ has the twisted Rokhlin property.(2)
$\mathrm{MCG}(S)$ has the Rokhlin property.(3)
$\mathrm{MCG}(S)$ has the ϕ-twisted Rokhlin property for some automorphism ϕ of
$\mathrm{MCG}(S)$.(4)
$S$ has a unique maximal end and each of its compact subsurfaces is displaceable.
Along the way, in Section 5, 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 the readability of the paper.
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 
$\{U_n\}_{n \ge 1}$ of connected open subsets of 
$S$ admitting the following properties:
(1) If
$m \lt n$, then
$U_n \subset U_m$.(2) For each
$n \in \mathbb N$,
$\textstyle U_n$ is not relatively compact.(3) For each
$n \in \mathbb N$,
$U_n$ has compact boundaries.(4) Each relatively compact subset of
$S$ is disjoint from all except finitely many
$U_n$’s.
Two exiting sequences 
$\{U_n\}_{n \ge 1}$ and 
$\{V_n\}_{n \ge 1}$ 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
\begin{equation*}U^*=\big\{[\{U_n\}_{n \ge 1}] \in \mathcal{E}(S) ~\mid~\textrm{there exists}~n \in \mathbb{Z}~\textrm{such that}~U_n \subset U \big\}\end{equation*} Then the desired topology on 
$\mathcal{E}(S)$ is the topology generated by the basis
\begin{equation*} \{U^* ~\mid~ U \subset S~~ \text{is open with compact boundary}\}.\end{equation*} 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 neighbourhood of e. Further, every homeomorphism 
$h: S\to S$ induces a homeomorphism 
$\mathcal{E}(S)\to \mathcal{E}(S)$ given by
\begin{equation*}[\{U_n \}_{n \ge 1}] \mapsto [\{h(U_n)\}_{n \ge 1}].\end{equation*} 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 [Reference Richards22, 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 unit 2-sphere 
$\mathbb{S}^2$. Now, take a sequence of disjoint closed disks Dj 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 Dj 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 [Reference Richards22, 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 
$\mathcal{E}(S)$.
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 U of x, there is an open neighbourhood V of y and a homeomorphism 
$h \in \operatorname{Homeo}^{+}(S)$ such that 
$h(V) \subseteq U$. 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 
$e \preceq \mu$ [Reference Mann and Rafi18, 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 [Reference Mann and Rafi18].
A subsurface Σ 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 [Reference Aougab, Patel and Vlamis1, Theorem 6.1] and [Reference Mann and Rafi18].
Theorem 2.3. If 
$S$ is a surface in which every compact subsurface is displaceable, then either
(1)
$\mathcal{M}(S)$ has exactly one element,(2)
$\mathcal{M}(S)$ has exactly two elements, or(3)
$\mathcal{M}(S)$ is homeomorphic to a Cantor space in which any two maximal ends are comparable.
Moreover, 
$\mathcal{E}(S)$ 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 [Reference Aramayona and Vlamis2] 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 [Reference Bavard, Dowdall and Rafi3, Theorem 1.1].
Proposition 2.4. If 
$S$ is a connected orientable infinite-type surface without boundary, then
\begin{equation*}\operatorname{Out}\left(\mathrm{MCG}^{\pm}(S)\right)=1 \quad \textrm{and} \quad \operatorname{Out}(\mathrm{MCG}(S)) \cong \mathbb{Z}_2.\end{equation*} As mentioned earlier, we assume that our surface S is symmetric about the xy-plane when embedded in 
$\mathbb{R}^3$. Thus, the map 
$\mathfrak{t}: \mathbb{R}^3\to \mathbb{R}^3$ which sends a point to its mirror image about the xy-plane restricts to an orientation-reversing self-homeomorphism
\begin{equation*}\mathfrak{t}: S \rightarrow S.\end{equation*} Clearly, 
$\mathfrak{t}$ is an involution of S. For notational convenience, let 
$\hat{\mathfrak{t}}$ denote the automorphism of 
$\mathrm{MCG}(S)$ given by
\begin{equation*}\hat{\mathfrak{t}}(h )= \mathfrak{t} h \mathfrak{t},\end{equation*} 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 Tc 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 [Reference Farb and Margalit5, Fact 3.7]. See also [Reference Korkmaz15, Lemma 2.1].
Lemma 2.5. If 
$h$ is a homeomorphism of a surface 
$S$ and 
$c$ a simple closed curve on 
$S$, then
\begin{equation*}
h T_c h^{-1} = T_{h(c)}^{\epsilon},
\end{equation*} where ϵ = 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 UA of 
$\mathrm{MCG}(S)$ by
\begin{equation*}U_A=\{\phi\in\mathrm{MCG}(S)\mid\phi(a)=a\,\,\text{for all }a\in A\}.\end{equation*} It is known that if S has empty boundary and 
$\mathrm{MCG}(S)$ has trivial center, then the set
\begin{equation*}\mathcal B=\{\psi U_A\mid\psi\in\mathrm{MCG} (S)\;\text{and}\;A\;\text{is a finite subset of}\;C(S)\},\end{equation*} is a basis for the topology on 
$\mathrm{MCG}(S)$. Note that, the center of the big mapping class group is always trivial [Reference Lanier and Loving16, Proposition 2]. The following basic observation shall be used later.
Lemma 2.6. Suppose that 
$g, h \in \mathrm{MCG}(S)$. Then 
$h \in g U_A$ if and only if 
$h U_A = g U_A$.
Proof. If 
$h \in g U_A$, then 
$h = g \phi$ for some 
$\phi \in U_A$. Thus, 
$g = h \phi^{-1} \in h U_A$, and hence 
$g U_A = h U_A$. Conversely, if 
$h U_A = g U_A$, then 
$h \in g U_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')$ between two curves 
$c, c'$ 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 [Reference Masur and Minsky19, Proposition 4.6].
Proposition 2.7. Let 
$S_{g,n}$ be a surface of genus 
$g$ with 
$n$ punctures such that 
$3g + n \ge 5$. Then there exists a real number 
$c \gt 0$ 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
\begin{equation*}
d_{C(S_{g,n})}(h^k(\gamma), \gamma) \geq c |k|.
\end{equation*} 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
\begin{equation*}
d_{C(S_{g,n})}(h^m(\gamma), \gamma) \gt 2,
\end{equation*} for all 
$m\geq \ell$.
We say that a subsurface Σ 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 Σ be a subsurface of a surface 
$S$ such that the embedding of Σ into 
$S$ induces an embedding of 
$C(\Sigma)$ into 
$C(S)$. Let 
$c \in C(S)$ be such that it has a non-trivial geometric intersection with Σ. Assume that the boundary 
$\partial (\Sigma)$ and 
$c$ are in a minimal position. Then a projection of 
$c$ onto 
$C(\Sigma)$ is defined to be any component 
$b \in C(\Sigma)$ of the boundary of a regular neighbourhood of 
$\alpha \cup \partial(\Sigma)$, where α is a component of 
$c \cap \Sigma$.
For example, let S be a surface and Σ 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 Σ (see Figure 2).
Projections of the curve c on Σ.
Note that, if each component of 
$\partial (\Sigma)$ is an essential separating simple closed curve on S, then 
$C(\Sigma)$ embeds into 
$C(S)$. Further, for each 
$c \in C(S)$ such that it has a non-trivial geometric intersection with Σ, such a projection 
$b \in C(\Sigma)$ always exists, and if 
$c \in C(\Sigma)$, then b = c.
Lemma 2.9. [Reference Masur and Minsky20, Lemma 2.2] Let Σ be a subsurface of a surface 
$S$ such that the embedding of Σ into 
$S$ induces an embedding of 
$C(\Sigma)$ into 
$C(S)$. If 
$c \in C(S)$ and 
$b,b' \in C(\Sigma)$ are two projections of 
$c$, then 
$d_{C(\Sigma)}\left(b, b'\right) \leq 2$.
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 [Reference Kechris and Rosendal14] for a survey on Rokhlin property.
Example 3.1. Some topological groups with the Rokhlin property are:
(1) The symmetric group on
$\mathbb{N}$.(2) The homeomorphism group of the Cantor set [Reference Glasner and Weiss11, Theorem 2.6].
(3) The group of orientation-preserving homeomorphisms of an even-dimensional sphere [Reference Glasner and Weiss11, Theorem 3.4].
(4) The mapping class group of a connected orientable surface which is either the 2-sphere, 2-sphere with one puncture or infinite-type with genus either zero or infinite, and whose end space is self-similar with a unique maximal end (see [Reference Hernández, Hrušák, Morales, Randecker, Sedano and Valdez13, Theorem 0.2] or [Reference Lanier and Vlamis17, Theorem 6.1]).
Example 3.2. Some topological groups without the Rokhlin property are:
(1) A non-trivial closed subgroup of
${\rm GL}_n(\mathbb C)$ [Reference Wesolek25, Lemma 5.2].(2) A non-trivial connected locally compact group [Reference Wesolek25, Theorem 5.3].
Next we consider the idea of twisted conjugacy.
Definition 3.3. Let 
$G$ be a group and ϕ an automorphism of 
$G$. Two elements 
$x, y \in G$ are said to be ϕ-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 ϕ-twisted conjugacy classes, and we denote such a class by 
$C_\phi(x)$.
Note that, given an automorphism ϕ 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, ϕ-twisted conjugacy classes are simply orbits under this action.
Definition 3.4. Let 
$G$ be a topological group and ϕ an automorphism of the underlying abstract group 
$G$. We say that 
$G$ has the ϕ-twisted Rokhlin property if 
$G$ admits a dense ϕ-twisted conjugacy class. Further, 
$G$ is said to have the twisted Rokhlin property if it has ϕ-twisted Rokhlin property for each automorphism ϕ 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 ϕ-twisted Rokhlin property if and only if it has the ψ-twisted Rokhlin property. In particular, if 
$\psi \in \operatorname{Inn}(G)$, then 
$G$ has the ψ-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
\begin{align*}
C_{\phi}(x) &= \{k x \phi(k)^{-1} \mid k \in G\}\\
&=\{k x g^{-1} (g \phi(k)^{-1} g^{-1}) g \mid k\in G \}\\
&=\{k (x g^{-1}) \,\hat{g} (\phi(k)^{-1}) g \mid k \in G \}\\
&=\{k (x g^{-1}) \,\psi(k)^{-1} g \mid k \in G\}\\
&=R_g(C_{\psi}(xg^{-1} )),
\end{align*} where 
$R_g:G \to G$ is right multiplication by g. Since Rg 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 and 
$\phi \in \operatorname{Aut}(G)$. Suppose that 
$G$ admits a proper open subgroup 
$H$ such that 
$gH\phi(g)^{-1} = H$ for all 
$g \in G$. Then 
$G$ does not have the ϕ-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 ϕ-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 ϕ be an automorphism of 
$G$. Then 
$G$ has the ϕ-twisted joint embedding property (ϕ-TJEP for short) if given any two non-empty open subsets 
$U$ and 
$V$ of 
$G$, there exists 
$g \in G$ such that 
$U \cap g V \phi(g)^{-1}\neq \emptyset$.
Note that the set 
$g V \phi(g)^{-1}$ 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 [Reference Lanier and Vlamis17, Theorem 2.2] to the ϕ-twisted case.
Proposition 3.10. Let 
$G$ be a Polish topological group and let ϕ be an automorphism of 
$G$. Then 
$G$ has the ϕ-twisted Rokhlin property if and only if 
$G$ has the ϕ-TJEP.
Proof. Suppose that G has the ϕ-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
\begin{equation*}
D_U = \cup_{g\in G} ~g U \phi(g)^{-1}.
\end{equation*} Note that DU is open. Further, it follows from the ϕ-TJEP that DU 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 
$x \in g V \phi(g)^{-1}$. 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 ϕ-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 
$y x \phi(y)^{-1}\in U$ and 
$z x \phi(z)^{-1}\in V$. Setting 
$h= y z^{-1}$, it is easy to see that 
$y x \phi(y)^{-1} \in U \cap h V \phi(h)^{-1}$. Hence, the group G has the ϕ-TJEP.
Remark 3.11. It follows from [Reference Vlamis24, Corollary 5] that if S is an infinite-type surface, then 
$\mathrm{MCG}(S)$ is a Polish group. Hence, 
$\mathrm{MCG}(S)$ has the ϕ-twisted Rokhlin property if and only if it has the ϕ-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 ϕ-twisted conjugacy class. Hence, we consider only infinite-type surfaces (without boundary).
Recently, in [Reference Lanier and Vlamis17, Theorem 6.1], Lanier and Vlamis have shown that the mapping class group of a connected orientable infinite-type surface without boundary has the Rokhlin property if and only if the surface has genus 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 an infinite-type surface whose genus is either zero or infinite, and whose end space is self-similar with a unique maximal end. Further, we shall also see whether the big mapping class group of the remaining surfaces have the ϕ-twisted Rokhlin property for some automorphism ϕ. Employing the approach outlined in the following flowchart, we establish Theorem 4.2, Proposition 4.4, Theorem 4.7 and Theorem 4.9, which collectively lead to our main theorem.
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, say, 
$\mathcal{M}(S)= \{\mu\}$. It follows from Theorem 2.3 that the end space 
$\mathcal{E}(S)$ is self-similar. Notice that, in this case, S is either planar or has infinite genus. 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
\begin{equation*}
\left\{{\Omega}_c^* ~\mid~ c~~ \textrm{is a separating simple closed curve on}~ S\right\},
\end{equation*} is a neighbourhood basis for 
$\mu$, and consequently G is a subgroup of 
$\mathrm{MCG}(S)$.
Let S be a surface where every compact subsurface is displaceable, and let S have a unique maximal end, denoted 
$\mu$. We choose a separating simple closed curve b 1 on S such that 
$\mathfrak{t} (\Sigma_{b_1})= \Sigma_{b_1}$. In particular, 
$\mathfrak{t} (b_1)= b_1$, and hence b 1 meets the xy-plane in precisely two points. By the proof of [Reference Lanier and Vlamis17, Proposition 4.2], 
$\Omega_{b_1}$ has a homeomorphic copy of 
$\Sigma_{b_1}$. Let 
$\Sigma_{\tilde{b}_1}$ be this homeomorphic copy, where 
$\tilde{b}_1$ is a separating simple closed curve in 
$\Omega_{b_1}$. If 
$\mathfrak{t} (\Sigma_{\tilde{b}_1}) \neq \Sigma_{\tilde{b}_1}$, then we cut S along the curves b 1 and 
$\tilde{b}_1$. Thus, we can write 
$S=\Sigma_{b_1}\cup \Sigma_{\tilde{b}_1} \cup \Omega_{b_1, \tilde{b}_1}$, where 
$\Omega_{b_1, \tilde{b}_1}=\Omega_{b_1}\setminus \Sigma_{\tilde{b}_1}$. We choose realisations Σ and Ω of 
$\Sigma_{\tilde{b}_1}$ and 
$\Omega_{b_1, \tilde{b}_1}$ respectively, on the unit 2-sphere in 
$\mathbb{R}^3$ such that the following hold:
•
$\mathfrak{t}(\Sigma)=\Sigma$ and
$\mathfrak{t} (\Omega) = \Omega$.•
$\Sigma_{b_1}, \Sigma, \Omega$ have disjoint interiors.• If α and β are the two boundary components of Ω, then
$\mathfrak{t}(\alpha)=\alpha$ and
$\mathfrak{t}(\beta)=\beta$.• If γ is the boundary component of Σ, then we have
$\mathfrak{t}(\gamma)=\gamma$.
This implies that the curves 
$\alpha,\beta,$ and 
$\gamma$ meet the xy-plane in precisely two points. Thus, we can assume that 
$b_1=\alpha$ and 
$\beta= \gamma$ as curves in 
$\mathbb{R}^3$. Let 
$S_{b_1}$ be the surface obtained from 
$\Sigma_{b_1}$, Σ and Ω by identifying b 1 with α (denote the resulting curve again by b 1) and β with γ (denote the resulting curve by 
$b_1'$). Then we see that 
$S_{b_1}$ is homeomorphic to S with 
$\mathfrak{t}(S_{b_1})=S_{b_1}$, 
$\mathfrak{t}(\Sigma_{b_1})=\Sigma_{b_1}$ and 
$\mathfrak{t}(\Sigma_{b_1'})=\Sigma_{b_1'}$. See below Figure 3.
Different realisation of infinite type surface with unique maximal end.
Now, choose a separating simple closed curve b 2 on 
$S_{b_1}$ such that 
$\Sigma_{b_1} \cup \Sigma_{b_1'} \subset \Sigma_{b_2}$ and 
$\mathfrak{t}(\Sigma_{b_2})=\Sigma_{b_2}$. By the proof of [Reference Lanier and Vlamis17, Proposition 4.2], 
$\Omega_{b_2}$ has a homeomorphic copy, say 
$\Sigma_{\tilde{b}_2}$, of 
$\Sigma_{b_2}$. By repeating the preceding process, we get another surface 
$S_{b_2}$ from 
$S_{b_1}$ without perturbing 
$\Sigma_{b_2}$ such that 
$\mathfrak{t}(S_{b_2})=S_{b_2}$ and 
$\mathfrak{t}(\Sigma_{b_2'})=\Sigma_{b_2'}$. By repeating this process, for each i, we get a surface 
$S_{b_i}$ from 
$S_{b_{i-1}}$ without perturbing 
$\Sigma_{b_i}$ such that 
$\mathfrak{t}(S_{b_i})=S_{b_i}$ and 
$\mathfrak{t}(\Sigma_{b_i'})=\Sigma_{b_i'}$. Define 
$\widetilde{S}=\cup_{i=1}^{\infty} \Sigma_{b_i}$. Then, we see that 
$\widetilde{S}$ is homeomorphic to S such that 
$\mathfrak{t}(\Sigma_{b_i})=\Sigma_{b_i}$, 
$\mathfrak{t}(\Sigma_{b_i'})=\Sigma_{b_i'}$, 
$\Sigma_{b_i} \cap \Sigma_{b_i'} = \emptyset$, 
$\Sigma_{b_i} \cong \Sigma_{b_i'}$ and 
$\Sigma_{b_i} \cup \Sigma_{b_i'}\subset \Sigma_{b_{i+1}}$ for each i. We now present an extension of [Reference Lanier and Vlamis17, Proposition 4.2].
Proposition 4.1. Let 
$S$ be a surface where every compact subsurface is displaceable, and let 
$S$ have a unique maximal end, denoted 
$\mu$. Assume that 
$S$ has a realisation such that 
$S=\cup_{i=1}^{\infty} \Sigma_{b_i}$ as above. Let 
$G$ be the subgroup of 
$\mathrm{MCG}(S)$ defined above. Then the following assertions holds:
(1) The subgroup
$G$ is dense in
$\mathrm{MCG}(S)$.(2) For each separating simple closed curve
$b_i$, there exists
$g \in \operatorname{Homeo}^+(S)$ such that
$g(\Sigma_{b_i}) \subset \Omega_{b_i}$ and
$g \mathfrak{t} g^{-1}\mathfrak{t}$ is identity on
$\Sigma_{b_i}$.
Proof. Assertion (1) follows from [Reference Lanier and Vlamis17, Proposition 4.2]. Recall, for each bi, there exists a separating curve 
$b_i'$ such that 
$\Sigma_{b_i}\cap \Sigma_{b_i'} =\emptyset$, 
$\Sigma_{b_i}\cong \Sigma_{b_i'}$, 
$\mathfrak{t} (\Sigma_{b_i}) = \Sigma_{b_i}$ and 
$\mathfrak{t} (\Sigma_{b_i'}) = \Sigma_{b_i'}$. Further, choose a separating simple closed curve ci such that the curves 
$\{b_i, b_i', c_i\}$ bounds a pair of pants (a subsurface homeomorphic to the two-sphere with three disks removed) and 
$\{b_i,b_i'\} \subset \Sigma_{c_i}$. For each subsurface Σ of S with 
$\mathfrak{t}(\Sigma)=\Sigma$, define
\begin{equation*}\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\}.\end{equation*} We choose a homeomorphism 
$h:\Sigma_{c_i} \to \Sigma_{c_i}$ fixing 
$\partial(\Sigma_{c_i})$ identically such that 
$h(\Sigma_{b_i})=\Sigma_{b_i'}$ and 
$h(\Sigma_{b_i}^+)=\Sigma_{b_i'}^+$. In fact, such a homeomorphism can be defined by extending a homeomorphism on the pair of pants bounded by the curves 
$\{b_i, b_i', c_i\}$ which takes the curve bi to 
$b_i'$ and fix ci point-wise. Define 
$f:S \to S$ by
\begin{equation*}f(x) = \begin{cases}
h(x) & \text{if } x \in \Sigma_{b_i}^+,\\
\mathfrak{t} h \mathfrak{t}(x) & \text{if } x \in \Sigma_{b_i}^-,\\
h(x) & \text{if } x \in \Sigma_{c_i} \setminus \Sigma_{b_i},\\
x & \text{if } x \in S \setminus \Sigma_{c_i}.
\end{cases}\end{equation*} Thus 
$g:=f^{-1} \in \operatorname{Homeo}^+(S)$ such that 
$g^{-1}(\Sigma_{b_i}) \subset \Omega_{b_i}$. Consequently, we have 
$g(\Sigma_{b_i}) \subset \Omega_{b_i}$. Now, if 
$x\in \Sigma_{b_i}^{+}$, then 
$\mathfrak{t}(x) \in \Sigma_{b_i}^{-}$, and hence
\begin{equation*}g\mathfrak{t} g^{-1} \mathfrak{t} (x) = g \mathfrak{t} \mathfrak{t} h \mathfrak{t} (\mathfrak{t}(x)) =g h(x)=g f(x)=x.\end{equation*} Similarly, if 
$x\in \Sigma_{b_i}^{-}$, then 
$\mathfrak{t}(x) \in \Sigma_{b_i}^{+}$, and hence 
$ g\mathfrak{t} g^{-1} \mathfrak{t} (x) = x.$ This shows that 
$g \mathfrak{t} g^{-1}\mathfrak{t}$ is identity on 
$\Sigma_{b_i}$, which establishes assertion (2).
Theorem 4.2. Let 
$S$ be a surface where every compact subsurface is displaceable and 
$S$ has a unique maximal end. Then 
$\mathrm{MCG}(S)$ possesses the twisted Rokhlin property.
Proof. We write 
$S=\cup_{i=1}^{\infty} \Sigma_{b_i}$ as above. It follows from [Reference Lanier and Vlamis17, 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 ϕ-twisted Rokhlin property for each inner automorphism ϕ of 
$\mathrm{MCG}(S)$. By Proposition 2.4, we have 
$\operatorname{Out}(\mathrm{MCG}(S)) \cong \mathbb{Z}_2$. Thus, in view of Propositions 3.5 and 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 
$U_1 = h_1U_{A_1}$ and 
$U_2 = h_2U_{A_2}$. 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. Choose a separating simple closed curve bi such that 
$c_1, c_2 \subset \Sigma_{b_i}$ and each curve in 
$A_1 \cup A_2$ has a representative in 
$\Sigma_{b_i}$. By Proposition 4.1(2), there exists 
$g \in \operatorname{Homeo}^+(S)$ such that 
$g(\Sigma_{b_i}) \subset \Omega_{b_i}$ and 
$g\mathfrak{t} g^{-1}\mathfrak{t}$ is identity on 
$\Sigma_{b_i}$. Note that 
$g(\Sigma_{b_i}) \subset \Omega_{b_i}$ implies that 
$g^{-1}(\Sigma_{b_i}) \subset \Omega_{b_i}$. We have
\begin{equation}
gU_2\hat{\mathfrak{t}}(g)^{-1} = gh_2U_{A_2}\hat{\mathfrak{t}}(g)^{-1} =gh_2\hat{\mathfrak{t}}(g)^{-1} \big(\hat{\mathfrak{t}}(g)U_{A_2}\hat{\mathfrak{t}}(g)^{-1} \big)= gh_2\hat{\mathfrak{t}}(g)^{-1}U_{\hat{\mathfrak{t}}(g)(A_2)}.
\end{equation} Recall that, the element h 1 admits a representative that restricts to the identity on 
$\Omega_{c_1}$. Since 
$\Omega_{{b_i}} \subset \Omega_{c_1}$, this representative of h 1 restricts to the identity on 
$\Omega_{{b_i}}$. Similarly, since 
$g\mathfrak{t} =\mathfrak{t} g$ on 
$\Sigma_{b_i}$ and 
$g^{-1}(\Sigma_{b_i}) \subset \Omega_{b_i}$, it follows that the element 
$gh_2\hat{\mathfrak{t}} (g)^{-1}$ also has a representative that restricts to the identity on 
$\Sigma_{b_i}$. Thus, we have the following:
(i) h 1 and
$gh_2\hat{\mathfrak{t}} (g)^{-1}$ commute (since they admit representatives with complementary supports),(ii)
$h_1 \in U_{\hat{\mathfrak{t}}(g)(A_2)}$ (since
$\hat{\mathfrak{t}} (g)(A_2) \subset \hat{\mathfrak{t}} (g)(\Sigma_{b_i}) \subset \Omega_{b_i}$),(iii)
$gh_2\hat{\mathfrak{t}} (g)^{-1} \in U_{A_1}$ (since
$A_1 \subset \Sigma_{b_i}$).
Taking 
$h = h_1(gh_2\hat{\mathfrak{t}} (g)^{-1})$, we see that
\begin{equation*}h\in h_1U_{A_1}=U_1\;\;(\text{using }(iii))\end{equation*}and
\begin{equation*}h\in gh_2\widehat{\mathfrak t}(g)^{-1}U_{\widehat{\mathfrak t}(g)(A_2)}\;\;(\text{using }(i)\,\,\text{and }(ii)).\end{equation*} In view of (1), we have 
$h\in gU_2\hat{\mathfrak{t}}(g)^{-1}$. Hence, 
$h \in U_1 \cap gU_2 \hat{\mathfrak{t}}(g)^{-1}$, 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 ϕ-twisted Rokhlin property for any 
$\phi \in \operatorname{Aut}(\mathrm{MCG}(S))$. In view of Proposition 3.5, 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 [Reference Lanier and Vlamis17, 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 
$f= h f' \mathfrak{t} h^{-1} \mathfrak{t}$ for some 
$f' \in \operatorname{Stab}(\mu)$. Since 
$\operatorname{Stab}(\mu)$ is normal, we see that
\begin{equation*}f(\mu)= h f' \mathfrak{t} h^{-1} \mathfrak{t}(\mu)=h f' \mathfrak{t} h^{-1} (\mu)=h f' h^{-1} (\mu)=\mu.\end{equation*} 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 where every compact subsurface is displaceable, and 
$\mathcal{M}(S)$ contains exactly two elements. Then 
$\mathrm{MCG}(S)$ does not have the ϕ-twisted Rokhlin property for any 
$\phi \in \operatorname{Aut}(\mathrm{MCG}(S))$.
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 [Reference Farb and Margalit5, 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
\begin{equation*}
\rho(T_a)= \left(\begin{matrix}
1 & i(a,b) \\
0 & 1
\end{matrix}\right) \quad {and} \quad \rho(T_b )= \left(\begin{matrix}
1 & 0 \\
-i(a,b) & 1
\end{matrix}\right).
\end{equation*} 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 4 and 
$d= \mathfrak{t}(c)$. Then 
$f = r T_c T_{d}^{-2} T_{c}^3 T_{d}^{-4}$ 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 
$f(b_6) = b_1$.
The curve c on 
$S_{0,6}$ such that 
$\{c, \mathfrak{t}(c)\}$ fills 
$S_{0,6}$.
We see that
\begin{align*}
(f\mathfrak{t} f\mathfrak{t})^3 &= (r T_c T_{d}^{-2} T_{c}^3 T_{d}^{-4} r T_{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}\\
& = (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\\
& = (T_{d} T_{c}^{-2} T_{d}^{3} T_{c}^{-4}T_{d}^{-1} T_{c}^{2} T_{d}^{-3} T_{c}^{4})^3.
\end{align*} Since 
$i(c,d)=6$, by Proposition 4.5, we have
\begin{equation*}
\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).
\end{equation*}This gives
\begin{equation*}
\rho((f\mathfrak{t} f\mathfrak{t})^3)= \left(\begin{matrix}
-6768504263453271167830031 & -162818421281863319251229760 \\
41172474347137850845138560 & 990416347916847555740563729
\end{matrix}\right),\end{equation*} 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 where every compact subsurface is displaceable, and 
$\mathcal{M}(S)$ is homeomorphic to a Cantor space. Then, 
$\mathrm{MCG}(S)$ does not possess the ϕ-twisted Rokhlin property for any 
$\phi\in \operatorname{Aut}(\mathrm{MCG}(S))$.
Proof. By Proposition 3.5 and [Reference Lanier and Vlamis17, 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 [Reference Lanier and Vlamis17, Proposition 4.6] (or [Reference Mann and Rafi18, Lemma 4.18]), we choose a compact subsurface Σ of S such that
(i) Σ is planar,
(ii) Σ has six boundary components, say
$b_1, b_2, b_3, b_4, b_5,b_6,$(iii) each bi is a separating curve of S,
(iv)
$\Omega_1, \Omega_2, \Omega_3, \Omega_4, \Omega_5, \Omega_6$ are the six components of
$S \setminus \Sigma$ such that
$\Omega_i^*$ is homeomorphic to
$\mathcal{E}(S)$,(v)
$\mathfrak{t} (\Sigma) = \Sigma, ~ \mathfrak{t} (b_i) = b_i$ and
$\mathfrak{t} (\Omega_i) = \Omega_i$ for each
$1 \leq i \leq 6$.
By Lemma 4.6, let 
$f' \in \operatorname{Homeo}^{+}(\Sigma)$ such that 
$f'(b_i) = b_{i+1}$ for 
$1 \leq i \leq 5$ and 
$f'(b_6) = b_1$ and 
$f'\mathfrak{t} f'\mathfrak{t}$ is pseudo-Anosov. Using condition (iv), there is an extension 
$f \in \mathrm{MCG}(S)$ of fʹ. Recall that a subset A of C(S) is called a stable Alexander system if UA is the center of 
$\mathrm{MCG}(S)$. Choose a stable Alexander system A for Σ. Let us take 
$U = fU_A$ and 
$V = U_{\{b_1\}}$. We claim that 
$U \cap g V \hat{\mathfrak{t}}(g)^{-1} = \emptyset$ 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 Σ. In this case, 
$g(b_1) \in \Omega_i$ for some i. If 
$h \in g V\hat{\mathfrak{t}}(g)^{-1}$, then using condition (v), 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 
$U \cap g V\hat{\mathfrak{t}}(g)^{-1} = \emptyset$.
Case 2: Suppose that 
$g(b_1)$ has a non-trivial geometric intersection with Σ. Let 
$c \in C(\Sigma)$ be a projection of 
$g(b_1)$ onto Σ. If 
$h \in U \cap g V \hat{\mathfrak{t}}(g)^{-1}$, then 
$h\mathfrak{t} g(b_1)= h(\hat{\mathfrak{t}}(g)(b_1)) = g(b_1)$. This gives
\begin{equation*}h\mathfrak{t} h\mathfrak{t} g(b_1)= h \mathfrak{t} g(b_1) = g(b_1),\end{equation*} 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 Σ for all 
$n\in \mathbb N$. Hence, by Lemma 2.9, we have
\begin{equation}
d_{C(\Sigma)}((h\mathfrak{t} h\mathfrak{t})^n (c),c) \leq 2
\end{equation} 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 Σ, and hence the map 
$h\mathfrak{t} h\mathfrak{t}$ is pseudo-Anosov on Σ. By Proposition 2.7, there exists 
$m\in \mathbb N$ such that
\begin{equation*}
d_{C(\Sigma)}\big((f\mathfrak{t} f\mathfrak{t})^m(c), c \big) \gt 2
\end{equation*} for all 
$c \in C(\Sigma)$. This contradicts (2). Hence, we must have 
$U \cap g V \hat{\mathfrak{t}}(g)^{-1} = \emptyset$ 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 
$g \ge 0$ with 
$n$ punctures, where 
$n \gt 5$ 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 5 together with 
$\mathfrak{t}(c)$ fills the surface 
$S_{g,n}$.
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 4 for 
$S_{0,6}$. Take 
$f = T_c$, where c is the curve as above. By using Penner’s construction [Reference Farb and Margalit5, 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 possess the ϕ-twisted Rokhlin property for any 
$\phi\in \operatorname{Aut}(\mathrm{MCG}(S))$.
Proof. In view of Proposition 3.5 and [Reference Lanier and Vlamis17, Theorem 3.1], it is enough to prove that 
$\mathrm{MCG}(S)$ does not have the 
$\hat{\mathfrak{t}}$-twisted Rokhlin property. Let Σ be a compact non-displaceable subsurface of S. By enlarging Σ, 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, Σ remains a finite-type subsurface of S.
We claim that there exist open subsets U and V of 
$\mathrm{MCG}(S)$ such that 
$U \cap g V\hat{\mathfrak{t}}(g)^{-1}=\emptyset$ for all 
$g \in \mathrm{MCG}(S)$. Let 
$A \subset C(S)$ be a stable Alexander system for Σ. It follows from the proof of Lemma 4.8 that there is an orientation-preserving homeomorphism 
$f': \Sigma \to \Sigma$ which fixes 
$\partial(\Sigma)$ point-wise such that 
$f'\mathfrak{t} f'\mathfrak{t}$ is pseudo-Anosov. Thus, by Proposition 2.7, there exists 
$m\in \mathbb N$ such that
\begin{equation}
d_{C(\Sigma)}\big((f'\mathfrak{t} f'\mathfrak{t})^m(a), a \big) \gt 2
\end{equation} for all 
$a \in C(\Sigma)$. Let 
$f \in \mathrm{MCG}(S)$ denote the extension of fʹ. Take 
$U=fU_A$ and 
$V=U_A$. Suppose that there exists 
$g \in \mathrm{MCG}(S)$ such that 
$h \in f U_A \cap gU_A\hat{\mathfrak{t}}(g)^{-1}$. Since 
$h \in gU_A\hat{\mathfrak{t}} (g)^{-1}=g\hat{\mathfrak{t}}(g)^{-1}U_{\hat{\mathfrak{t}}(g)(A)}$, we have
\begin{equation*}
h \mathfrak{t} g \mathfrak{t}(c)= g(c)
\end{equation*} for all 
$c \in A$. Thus, we obtain
\begin{equation}
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)
\end{equation} for all 
$c \in A$. Since Σ is non-displaceable in S, there exists a curve 
$c \in C(\Sigma)$ such that g(c) and Σ have a non-trivial geometric intersection. Let b denote a projection of g(c) onto Σ. By (4), 
$(h\mathfrak{t} h \mathfrak{t})^n (b)$ is also a projection of g(c) onto 
$(h \mathfrak{t} h \mathfrak{t})^n (\Sigma)=\Sigma$. By Lemma 2.9, we have
\begin{equation*}
d_{C(\Sigma)}((h\mathfrak{t} h \mathfrak{t})^n (b), b)\leq 2
\end{equation*} for all 
$n\in \mathbb N$. Since 
$h \in f U_A$, by (3), we have
\begin{equation*}
d_{C(\Sigma)}((h\mathfrak{t} h \mathfrak{t})^m (b), b) = d_{C(\Sigma)}((f\mathfrak{t} f \mathfrak{t})^m (b), b) \gt 2
\end{equation*} 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. Let 
$S$ be a connected orientable infinite-type surface without boundary. Then the following assertions are equivalent:
(1)
$\mathrm{MCG}(S)$ has the twisted Rokhlin property.(2)
$\mathrm{MCG}(S)$ has the Rokhlin property.(3)
$\mathrm{MCG}(S)$ has the ϕ-twisted Rokhlin property for some automorphism ϕ of
$\mathrm{MCG}(S)$.(4)
$S$ has a unique maximal end and each of its compact subsurfaces is displaceable.
Proof. 
$(1) \implies (2)$ follows from the definition of the twisted Rokhlin property. 
$(2) \implies (3)$ follows by taking ϕ to be the identity map. For 
$(3) \implies (4)$, suppose that 
$\mathrm{MCG}(S)$ has the ϕ-twisted Rokhlin property for some automorphism ϕ of 
$\mathrm{MCG}(S)$. Since S is an infinite-type surface, by Theorem 4.9, each compact subsurface of S must be displaceable. Further, by Theorem 2.3, Theorem 4.2, Proposition 4.4, and Theorem 4.7, it follows that S has a unique maximal end. Finally, 
$(4) \implies (1)$ follows directly from Theorem 4.2.
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 ϕ, denoted by 
$R(\phi)$, is the number of ϕ-twisted conjugacy classes in 
$G$. The group 
$G$ is said to have the 
$R_\infty$-property if 
$R(\phi)$ is infinite for each automorphism ϕ of 
$G$.
The 
$R_{\infty}$-property for the mapping class groups of finite-type surfaces has been investigated in [Reference Fel’shtyn and Gonçalves8, 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 [Reference Bhunia and Krishna4, 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 [Reference Bavard, Dowdall and Rafi3, Theorem 1.1]. 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 [Reference Fel’shtyn7, 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 
$h = r g \hat{f}(r)^{-1} = r g f r^{-1} f^{-1}$. 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 
$h = s g f s^{-1} f^{-1} = s g \hat{f}(s)^{-1}$, 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
\begin{equation*}h = t (f g f) f^{-1} t^{-1} f^{-1} = t (fgf^{-1}) \hat{f}(t)^{-1}.\end{equation*} Hence, h and 
$(f g f^{-1})$ are 
$\hat{f}$-twisted conjugate in 
$\mathrm{MCG}(S)$. Since 
$(f g f^{-1})= g^{-1} g(f g f^{-1})= g^{-1} g \hat{f}(g^{-1})^{-1}$, the elements 
$(f g f^{-1})$ 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) 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
\begin{equation*}
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}
\end{equation*}in
$\mathrm{MCG}(S)$, then
$l=k$ and
$\left\{T_{a_1}^{p_1}, \ldots, T_{a_l}^{p_l}\right\}=\left\{T_{b_1}^{q_1}, \ldots, T_{b_k}^{q_k}\right\}$ [Reference Farb and Margalit5, Lemma 3.17].(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
$k=l$ [Reference Farb and Margalit5, Section 3.3].
Lemma 5.4. [Reference Fel’shtyn, Leonov and Troitsky9, 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 6 have the property that 
$\{c, \mathfrak{t}(c)\}$ forms a multicurve.
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 [Reference Lanier and Vlamis17, Proposition 4.2], there exist curves 
$\{c, c', b\}$ which bound a pair of pants P such that 
$\{c,c'\} \subset \Sigma_b$ and 
$\Sigma_c\cong \Sigma_{c'}$. We realise the subsurface P to be symmetric about the xz-plane such that c and cʹ do not intersect the xz-plane. We then glue the subsurfaces 
$\Sigma_c$, 
$\Sigma_{c'}$, and 
$\Omega_b$ to P along 
$c,c',$ and 
$b,$ respectively, such that the resulting surface S is symmetric about the xz plane (see Figure 7). 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 
$\{c, f(c)=c' \}$ forms a multicurve.
The curve c such that 
$\{c, f(c)=c'\}$ forms a multicurve.
If S has two maximal ends, then by [Reference Lanier and Vlamis17, Lemma 5.4], there is a homeomorphism of S permuting the two maximal ends. In particular, such a homeomorphism maps an open neighbourhood of one end onto an open neighbourhood of the other. Without loss of generality, we can choose an open neighbourhood U of one of the ends to be such that its boundary is a separating simple closed curve, and U and its image V have disjoint closures. We can realise our surface 
$S\setminus(U\cup V)$ in such a way that it is symmetric about the xz-plane and 
$\partial(S\setminus(U\cup V))$ does not intersect the xz-plane. Now, we glue the neighbourhoods U and V back onto 
$S\setminus(U\cup V)$ along the boundary curves such that the resulting surface S is symmetric about the xz-plane. Let f be the orientation reversing homeomorphism of S, which maps a point on S to its mirror image about the xz-plane. Then f is of order two and 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 by the proof of Theorem 4.7, there exist curves 
$\{b_1, \ldots, b_6\}$ bounding 
$S_{0}^6$, which is symmetric about the xz-plane (see Figure 4). By using a similar argument as above, we can realise our surface S to be symmetric about the xz-plane. 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. 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 class 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 
$m,n \ge 1$. 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
\begin{equation*}T_c^{n}T_d^{-n} = gT_c^{m} T_d^{-m}g^{-1} = T_{g(c)}^{\epsilon m} T_{g(d)}^{- \epsilon m},\end{equation*}where
\begin{equation*}\epsilon = \begin{cases}
1 & \text{if } g \,\,\text{is orientation preserving,}\\
-1 & \text{if } g \,\,\text{is orientation reversing.}
\end{cases}\end{equation*}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 Lemmas 2.5 and 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 [Reference Bhunia and Krishna4, 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) Our findings indicate that when S is an infinite-type surface without boundary, the mapping class group
$\mathrm{MCG}(S)$ either exhibits the ϕ-twisted Rokhlin property for all automorphisms ϕ, or it lacks the ϕ-twisted Rokhlin property for each automorphism ϕ. We do not know an example of a non-abelian topological group G which admits a pair of automorphisms
$\phi,\psi$ such that G has the ϕ-twisted Rokhlin property, but not the ψ-twisted Rokhlin property? In general, it would be interesting to find such examples in topological groups that admit ‘sufficiently large’ outer automorphism groups.(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) Which other interesting topological groups admit the twisted Rokhlin property?
Acknowledgements
The authors are grateful to the anonymous referee(s) for many valuable suggestions which have improved the exposition of the paper. 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.
Competing interests
The authors declare that they have no competing interests and no data are associated with this paper.
