# measure-theoretic equicontinuity and rigidity

Fangzhou Cai  Wu Wen-Tsun Key Laboratory of Mathematics, USTC, Chinese Academy
of Sciences and Department of Mathematics, University of Science and
Technology of China, Hefei, Anhui, 230026, P.R. China<EMAIL_ADDRESS>
###### Abstract.

Let $(X,T)$ be a topological dynamical system and $\mu$ be a invariant
measure, we show that $(X,\mathcal{B},\mu,T)$ is rigid if and only if there
exists some subsequence $A$ of $\mathbb{N}$ such that $(X,T)$ is
$\mu$-$A$-equicontinuous if and only if there exists some IP-set $A$ such that
$(X,T)$ is $\mu$-$A$-equicontinuous. We show that if there exists some
subsequence $A$ of $\mathbb{N}$ with positive upper density such that $(X,T)$
is $\mu$-$A$-mean-equicontinuous, then $(X,\mathcal{B},\mu,T)$ is rigid. We
also give results with respect to a function.

##  1\. introduction

Throughout this paper, a topological dynamical system(t.d.s. for short) is a
pair $(X,T)$, where $X$ is an non-empty compact metric space with a metric $d$
and $T$ is a homeomorphism from $X$ to itself. For a t.d.s. $(X,T)$, denote by
$M(X,T)$ the set of $T$-invariant regular Borel probability measures on $X$.
It is well known that there exists some $\mu\in M(X,T)$, thus
$(X,\mathcal{B},\mu,T)$ can be viewed as a measure preserving system(m.p.s.
for short), where $\mathcal{B}$ is the Borel $\sigma$-algebra.

A t.d.s. $(X,T)$ is called equicontinuous if $\\{T^{n}:n\in\mathbb{N}\\}$ is
uniformly equicontinuous. For a t.d.s., equicontinuity represents
predictability, one may see that the dynamical behaviour of such a system is
very “rigid”. For example, it is well known that a transitive t.d.s. $(X,T)$
is equicontinuous if and only if it is topologically conjugate to a minimal
rotation on a compact abelian metric group, if and only if it has topological
discrete spectrum(see [18]).

The analogous concept of equicontinuity for a m.p.s. was also introduced.
While studying cellular automata(a subclass of t.d.s.), Gilman [7, 8]
introduced a notion of $\mu$-equicontinuity. Later Huang $et\ al$ [11]
introduced a different definition of $\mu$-equicontinuity(which under some
conditions are equivalent [4]) and showed that $\mu$-equicontinuous systems
have discrete spectrum. García-Ramos [5] introduced a weakening of
$\mu$-equicontinuity called $\mu$-mean-equicontinuity and showed that if a
m.p.s. $(X,\mathcal{B},\mu,T)$ is an ergodic system, then it has discrete
spectrum if and only if it is $\mu$-mean-equicontinuous. Recently, Huang $et\
al$ [10] proved that for a general m.p.s. $(X,\mathcal{B},\mu,T)$, it has
discrete spectrum if and only if it is $\mu$-mean-equicontinuous.

The notion of rigidity in ergodic theory was first introduced by Furstenberg
and Weiss in [3]. A m.p.s. $(X,\mathcal{B},\mu,T)$ is called rigid if there
exists a subsequence $\\{t_{k}\\}$ of $\mathbb{N}$ such that
$T^{t_{k}}f\stackrel{{\scriptstyle L^{2}}}{{\longrightarrow}}f$ for all $f\in
L^{2}(\mu)$. So rigidity is a spectral property. A function $f\in L^{2}(\mu)$
is called rigid if there exists a subsequence $\\{t_{k}\\}$ of $\mathbb{N}$
such that $T^{t_{k}}f\stackrel{{\scriptstyle L^{2}}}{{\longrightarrow}}f$.
Note that the sequence $\\{t_{k}\\}$ may depend on $f$. It is clear that if
$(X,\mathcal{B},\mu,T)$ is rigid, then every function in $L^{2}(\mu)$ is
rigid. An interesting result showed that the converse is true (see [2,
Corollary 2.6]).

The notion of uniformly rigidity was introduced by Glasner and Maon in [9] as
a topological analogue of rigidity in ergodic theory. A t.d.s. $(X,T)$ is
uniformly rigid if there exists a subsequence $\\{t_{k}\\}$ of $\mathbb{N}$
such that $T^{t_{k}}\rightarrow id$ uniformly on $X$, where $id$ is the
identity mapping.

The concept of rigidity is close to equicontinuity. In [13], the authors
proved the following theorem:

######  Theorem 1.1.

[13, Lemma 4.1] Let $(X,T)$ be a t.d.s.. If $(X,T)$ is uniformly rigid, then
there exists an IP-set $A$ such that $(X,T)$ is $A$-equicontinuous. If in
addition $(X,T)$ is an E-system, the converse holds.

In fact, in this paper we prove that:

######  Theorem 1.2.

Let $(X,T)$ be a t.d.s.. Then $(X,T)$ is uniformly rigid if and only if there
exists an IP-set $A$ such that $(X,T)$ is $A$-equicontinuous, if and only if
there exists a subsequence $A$ of $\mathbb{N}$ such that $(X,T)$ is
$A$-equicontinuous.

It is natural to consider the corresponding relation between measure-theoretic
rigidity and measure-theoretic equicontinuity. Following this idea, in our
paper we prove that:

######  Theorem 1.3.

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$. Then $(X,\mathcal{B},\mu,T)$ is
rigid if and only if there exists some subsequence $A$ of $\mathbb{N}$ such
that $(X,T)$ is $\mu$-$A$-equicontinuous if and only if there exists some IP-
set $A$ such that $(X,T)$ is $\mu$-$A$-equicontinuous.

Inspired by the work in [10], we also show that:

######  Theorem 1.4.

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$. If there exists a subsequence $A$
of $\mathbb{N}$ with $\bar{D}(A)>0$ such that $(X,T)$ is $\mu$-$A$-mean-
equicontinuous, then $(X,\mathcal{B},\mu,T)$ is rigid.

Since rigidity implies zero entropy, we have following corollary:

######  Corollary 1.5.

Let $(X,T)$ be a t.d.s.. If there exists a subsequence $A$ of $\mathbb{N}$
with $\bar{D}(A)>0$ such that $(X,T)$ is $A$-mean-equicontinuous, then the
topological entropy of $(X,T)$ is zero.

##  2\. preliminaries

In this section we recall some notions and aspects of dynamical systems.

###  2.1. Subsets of non-negative integers

In the article, integers and natural numbers are denoted by $\mathbb{Z}$ and
$\mathbb{N}$ respectively. Let $F$ be a set, denote by $|F|$ the number of
elements of $F$. Let $A$ be a subset of $\mathbb{N}$. Define the lower density
$\underline{D}(A)$ and the upper density $\bar{D}(A)$ of $A$ by:

$\underline{D}(A)=\liminf_{n\rightarrow\infty}\frac{|A\cap[0,n-1]|}{n},$
$\bar{D}(A)=\limsup_{n\rightarrow\infty}\frac{|A\cap[0,n-1]|}{n}.$

If $\underline{D}(A)=\bar{D}(A)$, denote by $D(A)=\underline{D}(A)=\bar{D}(A)$
the density of $A$.

Let $\\{b_{i}:i\in\mathbb{N}\\}\subset\mathbb{N}$. Define

$FS\\{b_{i}\\}=\\{b_{i_{1}}+b_{i_{2}}+\cdots+b_{i_{k}}:i_{1}<i_{2}<\cdots<i_{k},k\in\mathbb{N}\\}.$

We say $F\subset\mathbb{N}$ is an $IP$-set if there exists a subset
$\\{b_{i}:i\in\mathbb{N}\\}$ of $\mathbb{N}$ such that $F=FS\\{b_{i}\\}$.

###  2.2. Discrete spectrum in measurable dynamics

Let $(X,\mathcal{B},\mu,T)$ be a m.p.s., one can define the m.p.s. $(X\times
X,\mathcal{B}\times\mathcal{B},\mu\times\mu,T)$ by $T(x,y)=(Tx,Ty)$ for
$(x,y)\in X\times X$. One can define the Koopman operator $U_{T}$ as an
isometry of $L^{p}(\mu)$ into $L^{p}(\mu)$ by $U_{T}f=f\circ T$, where $1\leq
p<\infty$. We say a function $f\in L^{2}(\mu)$ is almost periodic if
$\\{U_{T}^{n}f:n\in\mathbb{Z}\\}$ is pre-compact in $L^{2}(\mu)$, that is the
closure of $\\{U_{T}^{n}f:n\in\mathbb{Z}\\}$ is compact in $L^{2}(\mu)$. We
say that $(X,\mathcal{B},\mu,T)$ has discrete spectrum if for all $f\in
L^{2}(\mu)$, $f$ is almost periodic.

###  2.3. Measure-theoretic equicontinuity

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$, let $K$ be a subset of $X$ and
$A=\\{a_{i}:i\in\mathbb{N}\\}$ be a subsequence of $\mathbb{N}$.

We say that $K$ is $A$-equicontinuous if for any $\epsilon>0$, there exists
$\delta>0$ such that

$x,y\in K,d(x,y)<\delta\Rightarrow d(T^{a_{n}}x,T^{a_{n}}y)<\epsilon,\forall
n\in\mathbb{N}.$

We say $(X,T)$ is $A$-equicontinuous if $X$ is $A$-equicontinuous.

We say $(X,T)$ is $\mu$-$A$-equicontinuous if for any $\tau>0$, there exists a
compact subset $K$ of $X$ with $\mu(K)>1-\tau$ such that $K$ is
$A$-equicontinuous.

By the regularity of $\mu$, the compactness of $K$ in the definition above can
be ignored. It is clear that the definition of $\mu$-$A$-equicontinuity is
independent on the choice of metric $d$.

###  2.4. Sequence entropy for a measure

Let $(X,\mathcal{B},\mu,T)$ be a m.p.s. and $S=\\{s_{i}:i\in\mathbb{N}\\}$ be
a subsequence of $\mathbb{N}$. Suppose $\xi$ and $\eta$ are two finite
measurable partitions of $X$. Define $\xi\vee\eta=\\{A\cap
B:A\in\xi,B\in\eta\\}$. It is easy to see $\xi\vee\eta$ is also a finite
partition of $X$. The entropy of $\xi$, written $H_{\mu}(\xi)$, is defined by

$H_{\mu}(\xi)=-\sum_{A\in\xi}\mu(A)\log\mu(A),$

and the entropy of $\xi$ given $\eta$, written $H_{\mu}(\xi|\eta)$, is defined
by

$H_{\mu}(\xi|\eta)=H_{\mu}(\xi\vee\eta)-H_{\mu}(\eta)=-\sum_{A\in\xi}\sum_{B\in\eta}\mu(A\cap
B)\log\frac{\mu(A\cap B)}{\mu(B)}.$

The sequence entropy of $\xi$ with respect to $(X,\mathcal{B},\mu,T)$ along
$S$ is defined by

$h_{\mu}^{S}(T,\xi)=\limsup_{n\rightarrow\infty}\frac{1}{n}H_{\mu}(\bigvee_{i=1}^{n}T^{-s_{i}}\xi).$

And the sequence entropy of $(X,\mathcal{B},\mu,T)$ along $S$ is

$h_{\mu}^{S}(T)=\sup_{\alpha}h_{\mu}^{S}(T,\alpha),$

where supremum is taken over all finite measurable partitions. When
$S=\mathbb{N}$, we simply write $h_{\mu}(T)$ and it is called the entropy of
$(X,\mathcal{B},\mu,T)$.

##  3\. measure-theoretic equicontinuity and rigidity

In this section, we study the relation between measure-theoretic
equicontinuity and rigidity. We prove that $(X,\mathcal{B},\mu,T)$ is rigid if
and only if there exists some subsequence $A$ of $\mathbb{N}$ such that
$(X,T)$ is $\mu$-$A$-equicontinuous. In the rest we give an IP-version of our
result.

###  3.1. Measure-theoretic equicontinuity, rigidity and sequence entropy

In this subsection, we prove that $\mu$-$A$-equicontinuity implies rigidity
via sequence entropy.

First we give a characterization of measure-theoretic rigidity. It is worth
pointing out the following theorem:

######  Theorem 3.1.

[12, Theorem 3.10] Let $(X,\mathcal{B},\mu,T)$ be a m.p.s.. Then
$(X,\mathcal{B},\mu,T)$ is rigid if and only if there exists some IP-set $F$
such that $h_{\mu}^{A}(T)=0$ for any subsequence $A$ of $F$.

######  Lemma 3.2.

[12, Lemma 3.3] Let $(X,\mathcal{B},\mu,T)$ be a m.p.s. and $B\in\mathcal{B}$.
Let $A$ be a subsequence of $\mathbb{N}$. Then ${\\{U_{T}^{n}1_{B}:n\in A\\}}$
is pre-compact in $L^{2}(\mu)$ if and only if $h_{\mu}^{S}(T,\\{B,B^{c}\\})=0$
for any subsequence $S$ of $A$.

Now we can give a characterization of rigidity:

######  Theorem 3.3.

Let $(X,\mathcal{B},\mu,T)$ be a m.p.s.. Then $(X,\mathcal{B},\mu,T)$ is rigid
if and only if there exists some subsequence $A$ of $\mathbb{N}$ such that
$h_{\mu}^{S}(T)=0$ for any subsequence $S$ of $A$, if and only if
${\\{U_{T}^{n}f:n\in A\\}}$ is pre-compact in $L^{2}(\mu)$ for all $f\in
L^{2}(\mu)$.

###### Proof.

We only need to prove that: If there exists some subsequence $A$ of
$\mathbb{N}$ such that $h_{\mu}^{S}(T)=0$ for any subsequence $S$ of $A$, then
$(X,\mathcal{B},\mu,T)$ is rigid. By Lemma 3.2, we have that for any
$B\in\mathcal{B}$, $\overline{\\{U_{T}^{n}1_{B}:n\in A\\}}$ is compact in
$L^{2}(\mu)$. Let $\\{f_{j}:j\in\mathbb{N}\\}$ be a dense subset of
$L^{2}(\mu)$, then for any $j\in\mathbb{N}$, $\overline{\\{U_{T}^{n}f_{j}:n\in
A\\}}$ is compact in $L^{2}(\mu)$. Hence

$\prod_{j\in\mathbb{N}}\overline{\\{U_{T}^{n}f_{j}:n\in A\\}}\text{ is compact
in }(L^{2}(\mu))^{\mathbb{N}}.$

It follows that

$\\{(U_{T}^{n}f_{1},\cdots,U_{T}^{n}f_{j},\cdots):n\in
A\\}\subset\prod_{j\in\mathbb{N}}\overline{\\{U_{T}^{n}f_{j}:n\in A\\}}$

is totally bounded. For each $k\in\mathbb{N}$, there exist
$n_{1},\ldots,n_{N}\in A$ such that

$\\{(U_{T}^{n}f_{1},\cdots,U_{T}^{n}f_{j},\cdots):n\in
A\\}\subset\bigcup_{i=1}^{N}B((U_{T}^{n_{i}}f_{1},\cdots,U_{T}^{n_{i}}f_{j},\cdots),\frac{1}{k}).$

Without loss of generality, we assume
$B((U_{T}^{n_{1}}f_{1},\cdots,U_{T}^{n_{1}}f_{j},\cdots),\frac{1}{k})$
contains infinite many elements of the left set. Hence there are infinite
$n\in A$ such that

$\begin{split}d((U_{T}^{n}f_{1},\cdots,U_{T}^{n}f_{j},\cdots),(U_{T}^{n_{1}}f_{1},\cdots,U_{T}^{n_{1}}f_{j},\cdots))&=\\\
\sum_{j\in\mathbb{N}}\frac{\|U_{T}^{n}f_{j}-U_{T}^{n_{1}}f_{j}\|_{L^{2}}}{2^{j}}&=\sum_{j\in\mathbb{N}}\frac{\|U_{T}^{n-n_{1}}f_{j}-f_{j}\|_{L^{2}}}{2^{j}}<\frac{1}{k}.\end{split}$

Take $t_{k}=n-n_{1}$, since $n$ can be sufficiently large, we can choose
$\\{t_{k}\\}$ increasing. It follows that
$\|U_{T}^{t_{k}}f_{j}-f_{j}\|_{L^{2}}<\frac{2^{j}}{k}$ for all
$j,k\in\mathbb{N}$. Hence

$U_{T}^{t_{k}}f_{j}\stackrel{{\scriptstyle
L^{2}}}{{\longrightarrow}}f_{j}\text{ as }k\rightarrow\infty\text{ for each
}f_{j}.$

It follows that $U_{T}^{t_{k}}f\stackrel{{\scriptstyle
L^{2}}}{{\longrightarrow}}f$ for each $f\in L^{2}(\mu)$. ∎

Now we study the relation between measure-theoretic equicontinuity and
measure-theoretic sequence entropy.

Let $A=\\{a_{i}:i\in\mathbb{N}\\}$ be a subsequence of $\mathbb{N}$. Let
$\mathcal{U}$ be a finite open cover of $X$ and $K$ be a subset of $X$. Denote
by $N(\mathcal{U}|K)$ the minimum among the cardinalities of the subset of
$\mathcal{U}$ which covers $K$, and set
$C_{A}(\mathcal{U}|K)=\lim\limits_{n\rightarrow\infty}N(\bigvee\limits_{i=1}^{n}T^{-a_{i}}\mathcal{U}|K)$,
where $\mathcal{U}\vee\mathcal{V}=\\{U\cap
V:U\in\mathcal{U},V\in\mathcal{V}\\}$ for two finite open covers
$\mathcal{U},\mathcal{V}$ of $X$. We recall that for two finite covers
$\alpha,\beta$ of $X$, $\alpha\preceq\beta$ means that for each $B\in\beta$,
there exists $A\in\alpha$ such that $B\subset A.$

The following lemma is useful(see [14, Lemma 2.3]):

######  Lemma 3.4.

Let $\alpha=\\{A_{1},A_{2}\ldots,A_{k}\\}$ be a Borel partition of $X$ and
$\epsilon>0$, then there exists a finite open cover $\mathcal{U}$ with $k$
elements such that for any $j\in\mathbb{N}$ and any Borel partition $\beta$
satisfying $T^{-j}\mathcal{U}\preceq\beta$,
$H_{\mu}(T^{-j}\alpha|\beta)\leq\epsilon$.

Now we prove that $\mu$-$A$-equicontinuity implies rigidity. The idea is from
[11, Proposition 5.4].

######  Theorem 3.5.

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$. If there exists a subsequence $A$
of $\mathbb{N}$ such that $(X,T)$ is $\mu$-$A$-equicontinuous, then
$h_{\mu}^{S}(T)=0$ for any subsequence $S$ of $A$, hence
$(X,\mathcal{B},\mu,T)$ is rigid.

###### Proof.

Claim: For any $\tau>0$, there exists a compact subset $K$ of $X$ with
$\mu(K)>1-\tau$ such that for any finite open cover $\mathcal{U}$,
$C_{A}(\mathcal{U}|K)<\infty$.

Proof of Claim: Given $\tau>0$, since $(X,T)$ is $\mu$-$A$-equicontinuous,
there exists a compact $A$-equicontinuous subset $K$ of $X$ with
$\mu(K)>1-\tau$. For any finite open cover $\mathcal{U}$, let $\epsilon>0$ be
a Lebesgue number of $\mathcal{U}$. Since $K$ is $A$-equicontinuous, there
exists $\delta>0$ such that

$x,y\in K,d(x,y)<\delta\Rightarrow d(T^{a}x,T^{a}y)<\epsilon,\ \forall a\in
A.$

As $K$ is compact, there exist $x_{1},\ldots,x_{L}\in K$ such that

$K=\bigcup_{i=1}^{L}(B(x_{i},\frac{\delta}{2})\cap K).$

For any $a\in A$ and $1\leq i\leq L$, we have that

$diam(T^{a}(B(x_{i},\frac{\delta}{2})\cap K))\leq\epsilon.$

Hence there exists $U_{a,i}\in\mathcal{U}$ such that

$T^{a}(B(x_{i},\frac{\delta}{2})\cap K)\subset U_{a,i}.$

Thus

$B(x_{i},\frac{\delta}{2})\cap K\subset\bigcap_{a\in A}T^{-a}U_{a,i},\
\forall\ 1\leq i\leq L.$

This implies $C_{A}(\mathcal{U}|K)\leq L<\infty$. The proof of Claim is
complete.

Now we prove $h_{\mu}^{S}(T)=0$ for any subsequence $S$ of $A$. Assume the
contrary, there exists a subsequence $S$ of $A$ such that $h_{\mu}^{S}(T)>0$,
then there exists a Borel partition $\alpha$ of $X$ and $\epsilon_{0}>0$ such
that $h_{\mu}^{S}(T,\alpha)>\epsilon_{0}$. By Lemma 3.4, there exists a finite
open cover $\mathcal{U}$ of $X$ such that for any $j\in\mathbb{N}$ and any
Borel partition $\beta$ satisfying $T^{-j}\mathcal{U}\preceq\beta$,
$H_{\mu}(T^{-j}\alpha|\beta)\leq\epsilon_{0}$.

For any $\tau>0$, by Claim, there exists a compact set $K$ with
$\mu(K)>1-\tau$ and $C_{A}(\mathcal{U}|K)<\infty$, set
$C_{A}(\mathcal{U}|K)=L$. For any $n\in\mathbb{N}$, we pick out two sub-
collections $\mathcal{E}^{1}_{n},\mathcal{E}^{2}_{n}$ of
$\bigvee\limits_{i=1}^{n}T^{-s_{i}}\mathcal{U}$ such that
$\mathcal{E}^{1}_{n}$ is a cover of $K$ with
$N(\bigvee\limits_{i=1}^{n}T^{-s_{i}}\mathcal{U}|K)$ elements, and
$\mathcal{E}^{2}_{n}$ is a cover of $X$ with
$N(\bigvee\limits_{i=1}^{n}T^{-s_{i}}\mathcal{U})$ elements. Enumerate them by

$\mathcal{E}^{1}_{n}=\\{U_{i}\\}_{i=1}^{m(n,1)},\mathcal{E}^{2}_{n}=\\{V_{i}\\}_{i=1}^{m(n,2)}.$

Set

$\beta_{1}=\\{K\cap(U_{i}-\bigcup_{k=1}^{i-1}U_{k}):i=1,\ldots,m(n,1)\\},$
$\beta_{2}=\\{K^{c}\cap(V_{i}-\bigcup_{k=1}^{i-1}V_{k}):i=1,\ldots,m(n,2)\\}.$

Then $\beta_{1}$ is a partition of $K$ with $|\beta_{1}|\leq
N(\bigvee\limits_{i=1}^{n}T^{-s_{i}}\mathcal{U}|K)$ and $\beta_{2}$ is a
partition of $K^{c}$ with $|\beta_{2}|\leq
N(\bigvee\limits_{i=1}^{n}T^{-s_{i}}\mathcal{U})$. Put
$\beta=\beta_{1}\cup\beta_{2}$, then $\beta$ is a partition of $X$ and
$\beta\succeq\bigvee\limits_{i=1}^{n}T^{-s_{i}}\mathcal{U}$. Hence we have

$\begin{split}H_{\mu}(\beta)&=-\sum_{A\in\beta_{1}}\mu(A)\text{log}\mu(A)-\sum_{B\in\beta_{2}}\mu(B)\text{log}\mu(B)\\\
&\leq(\mu(K)\text{log}|\beta_{1}|-\mu(K)\text{log}\mu(K))+(\mu(K^{c})\text{log}|\beta_{2}|-\mu(K^{c})\text{log}\mu(K^{c}))\\\
&\leq\mu(K)\text{log}N(\bigvee_{i=1}^{n}T^{-s_{i}}\mathcal{U}|K)+\mu(K^{c})\text{log}N(\bigvee_{i=1}^{n}T^{-s_{i}}\mathcal{U})+\text{log}2\\\
&\leq\mu(K)\text{log}L+n\mu(K^{c})\text{log}N(\mathcal{U})+\text{log}2.\end{split}$

On the other hand, since
$\beta\succeq\bigvee\limits_{i=1}^{n}T^{-s_{i}}\mathcal{U}$, by the
construction of $\mathcal{U}$,

$H_{\mu}(T^{-s_{i}}\alpha|\beta)\leq\epsilon_{0},\ \forall\ 1\leq i\leq n.$

It follows that

$\begin{split}&H_{\mu}(\bigvee_{i=1}^{n}T^{-s_{i}}\alpha)\leq
H_{\mu}((\bigvee_{i=1}^{n}T^{-s_{i}}\alpha)\vee\beta)\leq
H_{\mu}(\beta)+H_{\mu}(\bigvee_{i=1}^{n}T^{-s_{i}}\alpha|\beta)\\\
&\leq\mu(K)\text{log}L+n\mu(K^{c})\text{log}N(\mathcal{U})+\text{log}2+\sum_{i=1}^{n}H_{\mu}(T^{-s_{i}}\alpha|\beta)\\\
&\leq\mu(K)\text{log}L+n\mu(K^{c})\text{log}N(\mathcal{U})+\text{log}2+n\epsilon_{0}.\end{split}$

Hence

$h_{\mu}^{S}(T,\alpha)=\limsup_{n\rightarrow\infty}\frac{1}{n}H_{\mu}(\bigvee_{i=1}^{n}T^{-s_{i}}\alpha)\leq\mu(K^{c})\text{log}N(\mathcal{U})+\epsilon_{0}\leq\tau\text{log}N(\mathcal{U})+\epsilon_{0}.$

Let $\tau\rightarrow 0,$ we have $h_{\mu}^{S}(T,\alpha)\leq\epsilon_{0}$, it
is a contradiction. Hence $h_{\mu}^{S}(T)=0$ for any subsequence $S$ of $A$.
By Theorem 3.3, $(X,\mathcal{B},\mu,T)$ is rigid. ∎

###  3.2. Measure-theoretic rigidity and equicontinuity

In this subsection, we prove that if $(X,\mathcal{B},\mu,T)$ is rigid, then we
can find some subsequence $A$ of $\mathbb{N}$ such that $(X,T)$ is
$\mu$-$A$-equicontinuous.

First we give a useful characterization of measure-theoretic equicontinuity:

######  Proposition 3.6.

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$. Let
$A=\\{a_{i}:i\in\mathbb{N}\\}$ be a subsequence of $\mathbb{N}$. Then the
following statements are equivalent:

  1. (1)

$(X,T)$ is $\mu$-$A$-equicontinuous.

  2. (2)

For any $\tau>0$ and $\epsilon>0$, there exist a compact subset $K$ of $X$
with $\mu(K)>1-\tau$ and $\delta>0$ such that

$x,y\in K,d(x,y)<\delta\Rightarrow d(T^{a_{n}}x,T^{a_{n}}y)<\epsilon,\ \forall
n\in\mathbb{N}.$

###### Proof.

(1)$\Rightarrow$(2) is obvious.

(2)$\Rightarrow$(1): Given $\tau>0$, for any $l\in\mathbb{N}$, by (2) there
exist a compact set $K_{l}$ with $\mu(K_{l})>1-\frac{\tau}{2^{l}}$ and
$\delta_{l}>0$ such that for any $x,y\in K_{l}$ with $d(x,y)<\delta_{l}$, we
have $d(T^{a_{n}}x,T^{a_{n}}y)<\frac{1}{l}$ for all $n\in\mathbb{N}.$ Let
$K=\bigcap\limits_{l=1}^{\infty}K_{l}$, then $\mu(K)>1-\tau$ and $K$ is
compact. It is easy to see that $K$ is $A$-equicontinuous, hence $(X,T)$ is
$\mu$-$A$-equicontinuous. ∎

Now we prove that rigidity implies $\mu$-$A$-equicontinuity. Note that the
following theorem is stronger.

######  Theorem 3.7.

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$. If there exists a subsequence
$A=\\{a_{i}:i\in\mathbb{N}\\}$ of $\mathbb{N}$ such that
$U_{T}^{a_{n}}d\stackrel{{\scriptstyle a.e.}}{{\longrightarrow}}d$, where $d$
is the metric on $X$, then $(X,T)$ is $\mu$-$A$-equicontinuous.

###### Proof.

It is similar to Theorem 4.10. ∎

Combining Theorem 3.5 we have:

######  Theorem 3.8.

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$. Then $(X,\mathcal{B},\mu,T)$ is
rigid if and only if there exists a subsequence $A$ of $\mathbb{N}$ such that
$(X,T)$ is $\mu$-$A$-equicontinuous.

###  3.3. Results with respect to a function

In this subsection we give corresponding results with respect to a function.

Following the idea in [6, 19], we introduce the notion of
$f$-$\mu$-$A$-equicontinuity. we prove that for a function $f\in L^{2}(\mu)$,
$f$ is rigid if and only if there exists some subsequence $A$ of $\mathbb{N}$
such that $f$ is $\mu$-$A$-equicontinuous.

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$, let $K$ be a subset of $X$, $f\in
L^{2}(\mu)$ and $A=\\{a_{i}:i\in\mathbb{N}\\}$ be a subsequence of
$\mathbb{N}$.

We say that $K$ is $f$-$A$-equicontinuous(or $f$ is $A$-equicontinuous on $K$)
if for any $\epsilon>0$, there exists $\delta>0$ such that

$x,y\in
K,d(x,y)<\delta\Rightarrow|f(T^{a_{n}}x)-f(T^{a_{n}}y)|<\epsilon,\forall
n\in\mathbb{N}.$

We say $f$ is $\mu$-$A$-equicontinuous if for any $\tau>0$, there exists a
compact subset $K$ of $X$ with $\mu(K)>1-\tau$ such that $K$ is
$f$-$A$-equicontinuous.

Similar as Proposition 3.6, we give a characterization of
$f$-$\mu$-$A$-equicontinuity:

######  Proposition 3.9.

Let $(X,T)$ be a t.d.s., $\mu\in M(X,T)$ and $f\in L^{2}(\mu)$. Let
$A=\\{a_{i}:i\in\mathbb{N}\\}$ be a subsequence of $\mathbb{N}$. Then the
following statements are equivalent:

  1. (1)

$f$ is $\mu$-$A$-equicontinuous.

  2. (2)

For any $\tau>0$ and $\epsilon>0$, there exist a compact subset $K$ of $X$
with $\mu(K)>1-\tau$ and $\delta>0$ such that

$x,y\in
K,d(x,y)<\delta\Rightarrow|U_{T}^{a_{n}}f(x)-U_{T}^{a_{n}}f(y)|<\epsilon,\
\forall n\in\mathbb{N}.$

Now we prove that for a function $f\in L^{2}(\mu)$,
$f$-$\mu$-$A$-equicontinuity implies rigidity.

######  Theorem 3.10.

Let $(X,T)$ be a t.d.s., $\mu\in M(X,T)$ and $f\in L^{2}(\mu)$. If there
exists a subsequence $A=\\{a_{i}:i\in\mathbb{N}\\}$ of $\mathbb{N}$ such that
$f$ is $\mu$-$A$-equicontinuous, then $\\{U_{T}^{a_{n}}f:n\in\mathbb{N}\\}$ is
pre-compact in $L^{2}(\mu)$, hence $f$ is rigid.

###### Proof.

Assume that $\\{U_{T}^{a_{n}}f:n\in\mathbb{N}\\}$ is not pre-compact in
$L^{2}(\mu)$, then there exists $\epsilon>0$ and a subsequence $\\{n_{i}\\}$
of $\mathbb{N}$ such that

$||U^{a_{n_{i}}}_{T}f-U^{a_{n_{j}}}_{T}f||^{2}_{L^{2}}>9\epsilon^{2}+4\epsilon,\
\forall i\neq j\in\mathbb{N}.$

Since $f\in L^{2}(\mu)$, there exists $\eta>0$ such that

$\mu(E)<\eta\Rightarrow\int_{E}|f|^{2}d\mu<\epsilon,\ \forall E\subset X.$

Claim: For any subsequence $\\{s_{i}\\}$ of $\mathbb{N}$, we have

$\mu(\\{x:|f(T^{s_{i}}x)|\rightarrow+\infty\\})=0.$

Proof of Claim: Let $M=\\{x:|f(T^{s_{i}}x)|\rightarrow+\infty\\}$. It is easy
to see that $M$ is measurable. By Fatou’s lemma

$\int_{M}\liminf_{i\rightarrow\infty}|f(T^{s_{i}}x)|d\mu(x)\leq\liminf_{i\rightarrow\infty}\int_{M}|f(T^{s_{i}}x)|d\mu(x).$

Note that

$\int_{M}|f(T^{s_{i}}x)|d\mu(x)\leq\int|f(T^{s_{i}}x)|d\mu(x)=\int|f|d\mu<+\infty,$

we have $+\infty\cdot\mu(M)<+\infty$, hence $\mu(M)=0$. The proof of Claim is
complete.

Since $f$ is $\mu$-$A$-equicontinuous, there exists a $f$-$A$-equicontinuous
set $K_{0}$ in $X$ with $\mu(K_{0})>1-\eta$, hence there exists $\delta>0$
such that

$x,y\in K_{0},d(x,y)<\delta\Rightarrow|f(T^{a_{i}}x)-f(T^{a_{i}}y)|<\epsilon,\
\forall i\in\mathbb{N}.$

By the compactness of $X$, there exist

$K_{0}=X_{1}\cup X_{2}\cup\cdots\cup X_{m_{0}},$

where $diamX_{i}<\delta,\ 1\leq i\leq m_{0}$. We assume $\mu(X_{i})>0,\ 1\leq
i\leq m;\ \mu(X_{i})=0,\ m+1\leq i\leq m_{0}$. Let $K=X_{1}\cup
X_{2}\cup\cdots\cup X_{m}$, then $K\subset K_{0}$ and
$\mu(K)=\mu(K_{0})>1-\eta$. Since $\mu(X_{1})>0$ and

$\mu(\\{x:|f(T^{a_{n_{i}}}x)|\rightarrow+\infty\\})=0,$

we can find $y_{1}\in X_{1}$ and $|f(T^{a_{n_{i}}}y_{1})|$ not convergent to
$+\infty$, hence there exists a subsequence
$\\{a_{n_{i}}^{1}\\}\subset\\{a_{n_{i}}\\}$ such that
$\\{f(T^{a_{n_{i}}^{1}}y_{1})\\}$ is a Cauchy sequence. Note that
$\mu(X_{2})>0$ and

$\mu(\\{x:|f(T^{a_{n_{i}}^{1}}x)|\rightarrow+\infty\\})=0,$

we can find $y_{2}\in X_{2}$ and a subsequence
$\\{a_{n_{i}}^{2}\\}\subset\\{a_{n_{i}}^{1}\\}$ such that
$\\{f(T^{a_{n_{i}}^{2}}y_{2})\\}$ is a Cauchy sequence. Working inductively,
we can find $y_{i}\in X_{i},1\leq i\leq m$ and a subsequence
$\\{s_{i}\\}\subset\\{a_{n_{i}}\\}$ such that $\\{f(T^{s_{i}}y)\\}$ is a
Cauchy sequence for all $y\in\\{y_{1},y_{2}\ldots,y_{m}\\}$. Hence there
exists $N\in\mathbb{N}$ such that

$\forall i,j>N,\ |f(T^{s_{i}}y)-f(T^{s_{j}}y)|<\epsilon,\ \forall
y\in\\{y_{1},y_{2}\ldots,y_{m}\\}.$

Choose $i_{0}\neq j_{0}>N$, now we estimate
$||U_{T}^{s_{i_{0}}}f-U_{T}^{s_{j_{0}}}f||_{L^{2}}^{2}$.

$||U_{T}^{s_{i_{0}}}f-U_{T}^{s_{j_{0}}}f||_{L^{2}}^{2}=\int_{K}|f\circ
T^{s_{i_{0}}}-f\circ T^{s_{j_{0}}}|^{2}d\mu+\int_{K^{c}}|f\circ
T^{s_{i_{0}}}-f\circ T^{s_{j_{0}}}|^{2}d\mu.$

We first estimate $\int_{K}|f\circ T^{s_{i_{0}}}-f\circ
T^{s_{j_{0}}}|^{2}d\mu$. If $x\in K$, then there exists
$y\in\\{y_{1},y_{2}\ldots,y_{m}\\}$ such that $d(x,y)<\delta$. Since $x,y\in
K_{0}$, we have

$|f(T^{a_{i}}x)-f(T^{a_{i}}y)|<\epsilon,\ \forall i\in\mathbb{N},$

hence

$|f(T^{s_{i}}x)-f(T^{s_{i}}y)|<\epsilon,\ \forall i\in\mathbb{N}.$

Since $|f(T^{s_{i_{0}}}y)-f(T^{s_{j_{0}}}y)|<\epsilon,$ it follows that

$\begin{split}|f(T^{s_{i_{0}}}x)-f(T^{s_{j_{0}}}x)|&\leq|f(T^{s_{i_{0}}}x)-f(T^{s_{i_{0}}}y)|+|f(T^{s_{j_{0}}}x)-f(T^{s_{j_{0}}}y)|\\\
&+|f(T^{s_{i_{0}}}y)-f(T^{s_{j_{0}}}y)|<3\epsilon.\end{split}$

Hence $\int_{K}|f\circ T^{s_{i_{0}}}-f\circ
T^{s_{j_{0}}}|^{2}d\mu<9\epsilon^{2}$.
Now we estimate $\int_{K^{c}}|f\circ T^{s_{i_{0}}}-f\circ
T^{s_{j_{0}}}|^{2}d\mu$.

$\begin{split}\int_{K^{c}}|f\circ T^{s_{i_{0}}}-f\circ
T^{s_{j_{0}}}|^{2}d\mu&\leq\int_{K^{c}}2(|f\circ T^{s_{i_{0}}}|^{2}+|f\circ
T^{s_{j_{0}}}|^{2})d\mu\\\
&=2\int_{T^{s_{i_{0}}}K^{c}}|f|^{2}d\mu+2\int_{T^{s_{j_{0}}}K^{c}}|f|^{2}d\mu.\end{split}$

Since

$\mu(T^{s_{i_{0}}}K^{c})=\mu(T^{s_{j_{0}}}K^{c})=\mu(K^{c})<\eta,$

the right side $<4\epsilon$. Hence

$\int_{K^{c}}|f\circ T^{s_{i_{0}}}-f\circ T^{s_{j_{0}}}|^{2}d\mu<4\epsilon.$

It follows that

$||U_{T}^{s_{i_{0}}}f-U_{T}^{s_{j_{0}}}f||_{L^{2}}^{2}<9\epsilon^{2}+4\epsilon.$

Note that $\\{s_{i}\\}\subset\\{a_{n_{i}}\\}$ and

$||U^{a_{n_{i}}}_{T}f-U^{a_{n_{j}}}_{T}f||_{L^{2}}>9\epsilon^{2}+4\epsilon,\
\forall i\neq j\in\mathbb{N},$

it is a contradiction. So $\\{U_{T}^{a_{n}}f:n\in\mathbb{N}\\}$ is pre-compact
in $L^{2}(\mu)$.

Now we proof $f$ is rigid. Since $\\{U_{T}^{a_{n}}f:n\in\mathbb{N}\\}$ is pre-
compact in $L^{2}(\mu)$, for every $k\in\mathbb{N}$, there exists
$N\in\mathbb{N}$ such that

$\\{U_{T}^{a_{n}}f:n\in\mathbb{N}\\}\subset\bigcup_{i=1}^{N}B(U_{T}^{a_{i}}f,\frac{1}{k}).$

For $n\in\mathbb{N}$, there exists $1\leq i\leq N$ such that

$||U_{T}^{a_{n}-a_{i}}f-f||_{L^{2}}=||U_{T}^{a_{n}}f-U_{T}^{a_{i}}f||_{L^{2}}<\frac{1}{k}.$

Let $n_{k}=a_{n}-a_{i}$, since $a_{n}\rightarrow+\infty$ we can choose
$\\{n_{k}:k\in\mathbb{N}\\}$ increasing, hence we have
$||U_{T}^{n_{k}}f-f||_{L^{2}}<\frac{1}{k},\
U_{T}^{n_{k}}f\stackrel{{\scriptstyle L^{2}}}{{\longrightarrow}}f$. ∎

The next theorem shows that for a function $f\in L^{2}(\mu)$, rigidity implies
$f$-$\mu$-$A$-equicontinuity.

######  Theorem 3.11.

Let $(X,T)$ be a t.d.s., $\mu\in M(X,T)$ and $f\in L^{2}(\mu)$. If there
exists a subsequence $A=\\{a_{i}:i\in\mathbb{N}\\}$ of $\mathbb{N}$ such that
$U_{T}^{a_{n}}f\stackrel{{\scriptstyle a.e.}}{{\longrightarrow}}f$, then $f$
is $\mu$-$A$-equicontinuous.

###### Proof.

$\forall\tau>0,\forall\epsilon>0$, let

$A_{N}=\\{x:|U_{T}^{a_{n}}f(x)-f(x)|<\frac{\epsilon}{3},\ \forall n>N\\}.$

It is easy to see that $\\{A_{N}\\}$ is increasing and

$\\{x:U_{T}^{a_{n}}f(x)\rightarrow f(x)\\}\subset\bigcup_{N=1}^{\infty}A_{N}.$

Hence $\mu(\bigcup\limits_{N=1}^{\infty}A_{N})=1$ so there exists
$N\in\mathbb{N}$ such that $\mu(A_{N})>1-\dfrac{\tau}{2}$. Since
$f,U_{T}^{a_{1}}f,\ldots,U_{T}^{a_{N}}f\in L^{2}(\mu)$, by Lusin’s theorem,
there exists a compact subset $K_{0}$ of $X$ with
$\mu(K_{0})>1-\dfrac{\tau}{2}$ such that
$f,U_{T}^{a_{1}}f,\ldots,U_{T}^{a_{N}}f$ are continuous hence uniformly
continuous on $K_{0}$. Then there exists $\delta>0$ such that

$x,y\in
K_{0},d(x,y)<\delta\Rightarrow|f(x)-f(y)|<\dfrac{\epsilon}{3},|U_{T}^{a_{n}}f(x)-U_{T}^{a_{n}}f(y)|<\epsilon,\
1\leq n\leq N.$

Choose a compact set $K\subset A_{N}\cap K_{0}$ with $\mu(K)>1-\tau$. For any
$x,y\in K$ and $d(x,y)<\delta$, if $1\leq n\leq N$, then
$|U_{T}^{a_{n}}f(x)-U_{T}^{a_{n}}f(y)|<\epsilon$; if $n>N$, then

$\begin{split}|U_{T}^{a_{n}}f(x)-U_{T}^{a_{n}}f(y)|\leq|U_{T}^{a_{n}}f(x)-f(x&)|+|U_{T}^{a_{n}}f(y)-f(y)|+|f(x)-f(y)|\\\
<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon.\end{split}$

Hence $|U_{T}^{a_{n}}f(x)-U_{T}^{a_{n}}f(y)|<\epsilon$ for all
$n\in\mathbb{N}$. By Proposition 3.9, $f$ is $\mu$-$A$-equicontinuous. ∎

######  Corollary 3.12.

Let $(X,T)$ be a t.d.s., $\mu\in M(X,T)$ and $f\in L^{2}(\mu)$. Then $f$ is
rigid if and only if there exists a subsequence $A=\\{a_{i}:i\in\mathbb{N}\\}$
of $\mathbb{N}$ such that $f$ is $\mu$-$A$-equicontinuous.

Using the notion of $f$-$\mu$-$A$-equicontinuity, we give a different proof of
Theorem 3.8:

###### alternative proof of Theorem 3.8.

$(\Leftarrow)$: Assume $(X,T)$ is $\mu$-$A$-equicontinuous. For any $f\in
C(X)$, it is clear that $f$ is $\mu$-$A$-equicontinuous. By Theorem 3.10,

$\\{U_{T}^{a_{n}}f:n\in\mathbb{N}\\}\text{ is pre-compact in }L^{2}(\mu).$

Since $C(X)$ is dense in $L^{2}(\mu)$, we have

$\\{U_{T}^{a_{n}}f:n\in\mathbb{N}\\}\text{ is pre-compact in
}L^{2}(\mu),\forall f\in L^{2}(\mu).$

By Theorem 3.3, $(X,\mathcal{B},\mu,T)$ is rigid.

$(\Rightarrow)$: Assume $(X,\mathcal{B},\mu,T)$ is rigid. Then there exists a
subsequence $\\{n_{i}\\}$ of $\mathbb{N}$ such that

$U_{T}^{n_{i}}f\stackrel{{\scriptstyle L^{2}}}{{\longrightarrow}}f,\forall
f\in L^{2}(\mu).$

Let $\\{f_{n}:n\in\mathbb{N}\\}$ be a dense subset of $C(X)$. It is well known
that

$d(x,y)=\sum_{n=1}^{\infty}\dfrac{|f_{n}(x)-f_{n}(y)|}{2^{n+1}||f_{n}||_{\infty}}$

is a compatible metric of $X$. Without loss of generality, assume $d$ is the
metric on $X$. Since $U_{T}^{n_{i}}f_{1}\stackrel{{\scriptstyle
L^{2}}}{{\longrightarrow}}f_{1}$, we can find a subsequence $\\{n_{i}^{1}\\}$
of $\\{n_{i}\\}$ such that $U_{T}^{n_{i}^{1}}f_{1}\stackrel{{\scriptstyle
a.e.}}{{\longrightarrow}}f_{1}$. Note that
$U_{T}^{n_{i}}f_{2}\stackrel{{\scriptstyle L^{2}}}{{\longrightarrow}}f_{2}$,
we have $U_{T}^{n_{i}^{1}}f_{2}\stackrel{{\scriptstyle
L^{2}}}{{\longrightarrow}}f_{2}$, so we can find a subsequence
$\\{n_{i}^{2}\\}$ of $\\{n_{i}^{1}\\}$ such that
$U_{T}^{n_{i}^{2}}f_{2}\stackrel{{\scriptstyle
a.e.}}{{\longrightarrow}}f_{2}$. Working inductively and by diagonal
procedure, we can find a subsequence $A=\\{a_{i}\\}$ of $\\{n_{i}\\}$ such
that $U_{T}^{a_{i}}f_{n}\stackrel{{\scriptstyle
a.e.}}{{\longrightarrow}}f_{n}$ for all $n\in\mathbb{N}$. By Theorem 3.11,
$f_{n}$ is $\mu$-$A$-equicontinuous for all $n\in\mathbb{N}$.

Now we proof $(X,T)$ is $\mu$-$A$-equicontinuous. For any $\tau>0$ and
$\epsilon>0$, there exists $N\in\mathbb{N}$ such that
$\sum\limits_{n=N+1}^{\infty}\dfrac{1}{2^{n}}<\dfrac{\epsilon}{2}$. Since
$f_{1},\ldots,f_{N}$ are $\mu$-$A$-equicontinuous, there exists a compact set
$K$ with $\mu(K)>1-\tau$ such that $K$ is $f_{i}$-$A$-equicontinuous, $1\leq
i\leq N$. Hence there exists $\delta>0$ such that for any $x,y\in K$ and
$d(x,y)<\delta$, we have

$|f_{n}(T^{a_{m}}x)-f_{n}T^{a_{m}}(y)|<\dfrac{\epsilon}{2N}\cdot
2^{n+1}||f_{n}||_{\infty},\ \forall m\in\mathbb{N},\ 1\leq n\leq N.$

For any $m\in\mathbb{N}$, we have

$d(T^{a_{m}}x,T^{a_{m}}y)=\sum_{n=1}^{\infty}\dfrac{|f_{n}(T^{a_{m}}x)-f_{n}(T^{a_{m}}y)|}{2^{n+1}||f_{n}||_{\infty}}=\sum_{n=N+1}^{\infty}+\sum_{n=1}^{N}<\dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}=\epsilon.$

By Proposition 3.6, $(X,T)$ is $\mu$-$A$-equicontinuous. ∎

Now we give the IP-version of our main result:

######  Theorem 3.13.

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$. Then $(X,\mathcal{B},\mu,T)$ is
rigid if and only if there exists an IP-set $F$ such that $(X,T)$ is
$\mu$-$F$-equicontinuous.

###### Proof.

By Theorem 3.8, we only need to prove $(\Rightarrow)$. From the proof above,
there exists a subsequence $A=\\{a_{i}\\}$ of $\mathbb{N}$ such that
$U_{T}^{a_{i}}f_{n}\stackrel{{\scriptstyle a.e.}}{{\longrightarrow}}f_{n}$ for
all $n\in\mathbb{N}$. Hence

$d(T^{a_{i}}x,x)=\sum_{n=1}^{\infty}\dfrac{|f_{n}(T^{a_{i}}x)-f_{n}(x)|}{2^{n+1}||f_{n}||_{\infty}}\rightarrow
0\text{ a.e.}.$

Since $T^{a_{n}}\stackrel{{\scriptstyle a.e.}}{{\longrightarrow}}id$, for each
$i\in\mathbb{N}$, we can find $n_{i}\in\mathbb{N}$ such that

$\mu(A_{i})>1-\dfrac{1}{4^{i}},\text{ where }A_{i}=\\{x\in
X:d(T^{a_{n_{i}}}x,x)<\dfrac{1}{2^{i}}\\}.$

Let $F=FS\\{a_{n_{i}}\\}$, we prove that $(X,T)$ is $\mu$-$F$-equicontinuous.
For any $\tau>0$ and $\epsilon>0$, there exists $N\in\mathbb{N}$ such that
$\dfrac{1}{2^{N-1}}<\min\\{\tau,\dfrac{\epsilon}{3}\\}$. Note that

$\mu(\bigcap_{1\leq i_{1}<\cdots<i_{k}<i}T^{-(a_{n_{i_{1}}}+\cdots
a_{n_{i_{k}}})}A_{i})>1-\dfrac{1}{2^{i}},$

we have

$\mu(\bigcap_{i=N}^{\infty}\bigcap_{1\leq
i_{1}<\cdots<i_{k}<i}T^{-(a_{n_{i_{1}}}+\cdots a_{n_{i_{k}}})}A_{i})>1-\tau.$

Let $K=\bigcap\limits_{i=N}^{\infty}\bigcap\limits_{1\leq
i_{1}<\cdots<i_{k}<i}T^{-(a_{n_{i_{1}}}+\cdots a_{n_{i_{k}}})}A_{i}$, then
$\mu(K)>1-\tau$. Since

$\\{T^{a_{n_{i_{1}}}+\cdots a_{n_{i_{k}}}}:1\leq i_{1}<\cdots<i_{k}<N\\}$

are uniformly equicontinuous, there exists $\delta>0$ such that

$d(x,y)<\delta\Rightarrow d(T^{a_{n_{i_{1}}}+\cdots
a_{n_{i_{k}}}}x,T^{a_{n_{i_{1}}}+\cdots
a_{n_{i_{k}}}}y)<\dfrac{\epsilon}{3},1\leq i_{1}<\cdots<i_{k}<N.$

For $x,y\in K,d(x,y)<\delta$, we prove that $d(T^{a}x,T^{a}y)<\epsilon$ for
all $a\in F$. Let

$a=a_{n_{i_{1}}}+\cdots a_{n_{i_{k}}}+a_{n_{i_{k+1}}}+\cdots a_{n_{i_{t}}},$

where $1\leq i_{1}<\cdots<i_{k}<N\leq i_{k+1}<\cdots<i_{t}$. Since $i_{t}\geq
N$, we have

$x\in\bigcap_{1\leq i_{1}<\cdots<i_{k}<i_{t}}T^{-(a_{n_{i_{1}}}+\cdots
a_{n_{i_{k}}})}A_{i_{t}},$

hence $T^{a_{n_{i_{1}}}+\cdots a_{n_{i_{t-1}}}}x\in A_{i_{t}}$. It follows
that

$d(T^{a_{n_{i_{1}}}+\cdots a_{n_{i_{t}}}}x,T^{a_{n_{i_{1}}}+\cdots
a_{n_{i_{t-1}}}}x)<\dfrac{1}{2^{i_{t}}}.$

Similarly

$d(T^{a_{n_{i_{1}}}+\cdots a_{n_{i_{t-1}}}}x,T^{a_{n_{i_{1}}}+\cdots
a_{n_{i_{t-2}}}}x)<\dfrac{1}{2^{i_{t-1}}},$ $\vdots$
$d(T^{a_{n_{i_{1}}}+\cdots a_{n_{i_{k+1}}}}x,T^{a_{n_{i_{1}}}+\cdots
a_{n_{i_{k}}}}x)<\dfrac{1}{2^{i_{k+1}}}.$

Hence

$d(T^{a_{n_{i_{1}}}+\cdots a_{n_{i_{t}}}}x,T^{a_{n_{i_{1}}}+\cdots
a_{n_{i_{k}}}}x)<\dfrac{1}{2^{i_{t}}}+\dfrac{1}{2^{i_{t-1}}}+\cdots+\dfrac{1}{2^{i_{k+1}}}\leq\dfrac{1}{2^{N-1}}<\dfrac{\epsilon}{3}.$

Similarly

$d(T^{a_{n_{i_{1}}}+\cdots a_{n_{i_{t}}}}y,T^{a_{n_{i_{1}}}+\cdots
a_{n_{i_{k}}}}y)<\dfrac{\epsilon}{3}.$

Then

$\begin{split}d(T^{a}x,T^{a}y)&\leq d(T^{a_{n_{i_{1}}}+\cdots
a_{n_{i_{t}}}}x,T^{a_{n_{i_{1}}}+\cdots
a_{n_{i_{k}}}}x)+d(T^{a_{n_{i_{1}}}+\cdots
a_{n_{i_{t}}}}y,T^{a_{n_{i_{1}}}+\cdots a_{n_{i_{k}}}}y)\\\
&+d(T^{a_{n_{i_{1}}}+\cdots a_{n_{i_{k}}}}x,T^{a_{n_{i_{1}}}+\cdots
a_{n_{i_{k}}}}y)<\dfrac{\epsilon}{3}+\dfrac{\epsilon}{3}+\dfrac{\epsilon}{3}=\epsilon.\end{split}$

By Proposition 3.6, $(X,T)$ is $\mu$-$F$-equicontinuous. ∎

Similarly we have:

######  Theorem 3.14.

Let $(X,T)$ be a t.d.s., $\mu\in M(X,T)$ and $f\in L^{2}(\mu)$. Then $f$ is
rigid if and only if there exists an IP-set $F$ such that $f$ is
$\mu$-$F$-equicontinuous.

###  3.4. The topological case

In [13], the authors proved the following theorem:

######  Theorem 3.15.

[13, Lemma 4.1] Let $(X,T)$ be a t.d.s.. If $(X,T)$ is uniformly rigid, then
there exists an IP-set $A$ such that $(X,T)$ is $A$-equicontinuous. If in
addition $(X,T)$ is an E-system, the converse holds.

In fact we have:

######  Theorem 3.16.

Let $(X,T)$ be a t.d.s.. Then $(X,T)$ is uniformly rigid if and only if there
exists an IP-set $A$ such that $(X,T)$ is $A$-equicontinuous, if and only if
there exists a subsequence $A$ of $\mathbb{N}$ such that $(X,T)$ is
$A$-equicontinuous.

###### Proof.

We only need to prove that: If there exists a subsequence $A$ of $\mathbb{N}$
such that $(X,T)$ is $A$-equicontinuous, then $(X,T)$ is uniformly rigid.

Assume there exists a subsequence $A=\\{a_{i}:i\in\mathbb{N}\\}$ of
$\mathbb{N}$ such that $(X,T)$ is $\mu$-$A$-equicontinuous. By Ascoli’s
Theorem, $\\{T^{a_{n}}:n\in\mathbb{N}\\}$ is pre-compact in $C(X,X)$ with
uniform topology. For every $k\in\mathbb{N}$, there exists $N\in\mathbb{N}$
such that

$\\{T^{a_{n}}:n\in\mathbb{N}\\}\subset\bigcup_{i=1}^{N}B(T^{a_{i}},\frac{1}{k}).$

For $n\in\mathbb{N}$, there exists $1\leq i\leq N$ such that

$||T^{a_{n}-a_{i}}-id||_{\infty}=||T^{a_{n}}-T^{a_{i}}||_{\infty}<\frac{1}{k}.$

Let $n_{k}=a_{n}-a_{i}$, since $a_{n}\rightarrow+\infty$ we can choose
$\\{n_{k}:k\in\mathbb{N}\\}$ increasing, hence we have
$||T^{n_{k}}-id||_{\infty}<\frac{1}{k},\ T^{n_{k}}\rightarrow id$ uniformly. ∎

##  4\. measure-theoretic mean equicontinuity and rigidity

In [5], the author introduced the concept of measure-theoretic mean
equicontinuity and defined a notion called $\mu$-mean-equicontinuity. In [10],
the authors introduced a similar notion called $\mu$-equicontinuity in the
mean and showed that they are equivalent. In fact, it is shown in [10] that
they are equivalent to discrete spectrum for a general $\mu\in M(X,T)$.

Following the idea, in this section we consider the measure-theoretic mean
equicontinuity along some subsequence $A$ of $\mathbb{N}$. We study the
relation between measure-theoretic mean equicontinuity and rigidity.

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$, let $K$ be a subset of $X$ and
$A=\\{a_{i}:i\in\mathbb{N}\\}$ be a subsequence of $\mathbb{N}$.

For $x,y\in X$, define

$\bar{d}_{n}^{A}(x,y)=\frac{1}{n}\sum_{i=0}^{n-1}d(T^{a_{i}}x,T^{a_{i}}y).$

We say that $K$ is $A$-equicontinuous in the mean if for any $\epsilon>0$,
there exists $\delta>0$ such that

$x,y\in K,d(x,y)<\delta\Rightarrow\bar{d}_{n}^{A}(x,y)<\epsilon,\forall
n\in\mathbb{N}.$

We say $(X,T)$ is $A$-equicontinuous in the mean if $X$ is $A$-equicontinuous
in the mean.

We say $(X,T)$ is $\mu$-$A$-equicontinuous in the mean if for any $\tau>0$,
there exists a compact subset $K$ of $X$ with $\mu(K)>1-\tau$ such that $K$ is
$A$-equicontinuous in the mean.

Similar as Propositin 3.6, we have:

######  Proposition 4.1.

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$. Let
$A=\\{a_{i}:i\in\mathbb{N}\\}$ be a subsequence of $\mathbb{N}$. Then the
following statements are equivalent:

  1. (1)

$(X,T)$ is $\mu$-$A$-equicontinuous in the mean.

  2. (2)

For any $\tau>0$ and $\epsilon>0$, there exist a compact subset $K$ of $X$
with $\mu(K)>1-\tau$ and $\delta>0$ such that

$x,y\in K,d(x,y)<\delta\Rightarrow\bar{d}_{n}^{A}(x,y)<\epsilon,\ \forall
n\in\mathbb{N}.$

We say that $K$ is $A$-mean-equicontinuous if for any $\epsilon>0$, there
exists $\delta>0$ such that

$x,y\in
K,d(x,y)<\delta\Rightarrow\limsup_{n\rightarrow\infty}\bar{d}_{n}^{A}(x,y)<\epsilon.$

We say $(X,T)$ is $A$-mean-equicontinuous if $X$ is $A$-mean-equicontinuous.

We say $(X,T)$ is $\mu$-$A$-mean-equicontinuous  if for any $\tau>0$, there
exists a compact subset $K$ of $X$ with $\mu(K)>1-\tau$ such that $K$ is
$A$-mean-equicontinuous.

The notions of $\mu$-$A$-equicontinuous in the mean and $\mu$-$A$-mean-
equicontinuous are equivalent:

######  Theorem 4.2.

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$. Let
$A=\\{a_{i}:i\in\mathbb{N}\\}$ be a subsequence of $\mathbb{N}$, then the
following statements are equivalent:

  1. (1)

$(X,T)$ is $\mu$-$A$-equicontinuous in the mean.

  2. (2)

$(X,T)$ is $\mu$-$A$-mean-equicontinuous.

###### Proof.

The proof is similar to Theorem 4.8. ∎

######  Remark 4.3.

As the same in [15], it is easy to see that the definition of $\mu$-$A$-mean-
equicontinuity is independent on the choice of metric $d$.

######  Remark 4.4.

The property of mean-equicontinuity in a topological dynamical system has also
been studied, we refer readers to [1, 15, 16].

Now we prove the main theorem of this section. The idea is from [17, Theorem
2.21].

######  Theorem 4.5.

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$. If there exists a subsequence
$A=\\{a_{i}:i\in\mathbb{N}\\}$ of $\mathbb{N}$ with $\bar{D}(A)>0$ such that
$(X,T)$ is $\mu$-$A$-mean-equicontinuous, then $(X,\mathcal{B},\mu,T)$ is
rigid.

###### Proof.

Without loss of generality, we assume the metric $d\leq 1$ and
$\bar{D}(A)>\frac{1}{M},\ M\in\mathbb{N}$. By Theorem 3.7 and Theorem 3.8, we
only need to prove ${\\{U_{T}^{a_{n}}d:n\in\mathbb{N}\\}}$ is pre-compact in
$L^{1}(\mu\times\mu)$. Assume the contrary, then there exists $\epsilon>0$ and
a subsequence $\\{n_{i}\\}$ of $\mathbb{N}$ such that

$\|U_{T}^{a_{n_{i}}}d-U_{T}^{a_{n_{j}}}d\|_{L^{1}}>17M\epsilon\text{ for any
}i\neq j\in\mathbb{N}.$

Since $(X,T)$ is $\mu$-$A$-equicontinuous in the mean, there exist a compact
set $K$ with $\mu(K)>1-\epsilon$ and $\delta>0$ such that

$x,y\in K,d(x,y)<\delta\Rightarrow\bar{d}_{n}^{A}(x,y)<\frac{\epsilon}{2},\
\forall n\in\mathbb{N}.$

By the compactness of $X$, there exist $K=\bigcup\limits_{i=1}^{N}X_{i}$,
where $diamX_{i}<\delta,1\leq i\leq N$, $X_{i}\cap X_{j}=\emptyset,1\leq i\neq
j\leq N$. Without loss of generality, we assume $X_{i}\neq\emptyset$ for all
$1\leq i\leq N$ and choose $x_{i}\in X_{i},1\leq i\leq N$.

Let $C=(\frac{1}{[\epsilon]}+1)^{N^{2}}$. Choose $n_{0}\in\mathbb{N}$
sufficiently large such that

$n_{0}>a_{n_{10MC}}\text{ and }|A\cap[1,n_{0}]|>\frac{n_{0}}{M}.$

Note that $diam_{\bar{d}_{n_{0}}^{A}}X_{i}\leq\epsilon,1\leq i\leq N$. Let
$X_{0}=K^{c},$ then $\\{X_{0},X_{1},\ldots,X_{N}\\}$ is a partition of $X$
with

$\mu(X_{0})<\epsilon,\ diam_{\bar{d}_{n_{0}}^{A}}X_{i}\leq\epsilon,1\leq i\leq
N.$

For any $1\leq s\leq n_{0}$, define

$d_{s}(x,y)=\left\\{\begin{array}[]{ll}0,&(x,y)\in X_{0}\times X\cup X\times
X_{0}\hbox{;}\\\ U^{a_{s}}_{T}d(x_{i},x_{j}),&(x,y)\in X_{i}\times X_{j}\
1\leq i,j\leq N\hbox{.}\end{array}\right.$

For any $(x,y)\in X_{i}\times X_{j}$, we have

$|d_{s}(x,y)-U^{a_{s}}_{T}d(x,y)|=|U^{a_{s}}_{T}d(x,y)-U^{a_{s}}_{T}d(x_{i},x_{j})|\leq
U^{a_{s}}_{T}d(x,x_{i})+U^{a_{s}}_{T}d(x_{j},y).$

Sum $s$ from $1$ to $n_{0}$, we have

$\begin{split}\sum_{s=1}^{n_{0}}|d_{s}(x,y)-U^{a_{s}}_{T}d(x,y)|&\leq\sum_{s=1}^{n_{0}}U^{a_{s}}_{T}d(x,x_{i})+\sum_{s=1}^{n_{0}}U^{a_{s}}_{T}d(x_{j},y)\\\
&=n_{0}\bar{d}_{n_{0}}^{A}(x,x_{i})+n_{0}\bar{d}_{n_{0}}^{A}(x_{j},y).\end{split}$

Hence

$\sum_{s=1}^{n_{0}}|d_{s}(x,y)-U^{a_{s}}_{T}d(x,y)|\leq 2n_{0}\epsilon$

since $diam_{\bar{d}_{n_{0}}^{A}}X_{i}\leq\epsilon,1\leq i\leq N.$ Integrate
on $X_{i}\times X_{j}$, we have

$\int_{X_{i}\times X_{j}}\sum_{s=1}^{n_{0}}|d_{s}-U^{a_{s}}_{T}d|\leq
2n_{0}\epsilon(\mu\times\mu)(X_{i}\times X_{j}).$

Note that

$\begin{split}\int_{X_{0}\times X\cup X\times
X_{0}}\sum_{s=1}^{n_{0}}|d_{s}-U^{a_{s}}_{T}d|&=\int_{X_{0}\times X\cup
X\times X_{0}}\sum_{s=1}^{n_{0}}|U^{a_{s}}_{T}d|\\\ &\leq
n_{0}(\mu\times\mu)(X_{0}\times X\cup X\times X_{0})\leq
2n_{0}\epsilon,\end{split}$

we have

$\sum_{s=1}^{n_{0}}||d_{s}-U^{a_{s}}_{T}d||_{L^{1}}\leq 4n_{0}\epsilon.$

It follows that

$|\\{1\leq s\leq n_{0}:||d_{s}-U^{a_{s}}_{T}d||_{L^{1}}\leq 8M\epsilon\\}|\geq
n_{0}-[\frac{n_{0}}{2M}].$

Note that

$\begin{split}|A\cap[1,n_{0}]|=|\\{s\in\mathbb{N}:a_{s}\leq n_{0}\\}|&=\\\
|\\{1\leq s\leq n_{0}:a_{s}\leq n_{0}\\}|&>\frac{n_{0}}{M}\geq
2[\frac{n_{0}}{2M}],\end{split}$

we have

$|\\{1\leq s\leq n_{0}:||d_{s}-U^{a_{s}}_{T}d||_{L^{1}}\leq 8M\epsilon\text{
and }a_{s}\leq n_{0}\\}|\geq[\frac{n_{0}}{2M}].$

Denote the left set by $S$. Divide $[0,1]$ into $\frac{1}{[\epsilon]}+1$
intervals averagely, then $[0,1]^{N^{2}}$ is divided into
$(\frac{1}{[\epsilon]}+1)^{N^{2}}=C$ cubes. Since

$d_{s}(x,y)=\sum_{1\leq i,j\leq N}b_{ij}1_{X_{i}\times X_{j}},0\leq b_{ij}\leq
1,$

$\\{b_{ij}\\}_{1\leq i,j\leq N}$ must be in one of the $C$ cubes. By
pigeonhole principle, there exists $S_{0}\subset S$ with
$|S_{0}|\geq[\frac{[\frac{n_{0}}{2M}]}{C}]\geq\frac{n_{0}}{4MC}$ such that

$||d_{s}-d_{s^{\prime}}||_{L^{1}}<\epsilon,\ \forall s\neq s^{\prime}\in
S_{0}.$

It follows that for any $s\neq s^{\prime}\in S_{0}$,

$\begin{split}||U^{a_{s}}_{T}d-U^{a_{s^{\prime}}}_{T}d||_{L^{1}}&\leq||d_{s}-U^{a_{s}}_{T}d||_{L^{1}}+||d_{s^{\prime}}-U^{a_{s^{\prime}}}_{T}d||_{L^{1}}+||d_{s}-d_{s^{\prime}}||_{L^{1}}\\\
&\leq 8M\epsilon+8M\epsilon+\epsilon\leq 17M\epsilon.\end{split}$

Enumerate $S_{0}$ as $1\leq s_{1}<s_{2}<\cdots<s_{k}\leq n_{0},\
k\geq\frac{n_{0}}{4MC}.$ Consider the set

$\\{a_{s_{i}}+a_{n_{j}}:1\leq i\leq k,1\leq j\leq 10MC\\},$

it contains $10kMC>2n_{0}$ elements. On the other hand, it has an upper bound
of $2n_{0}$, hence there must be some $i\neq j,\ p\neq q$ with
$a_{s_{p}}+a_{n_{i}}=a_{s_{q}}+a_{n_{j}}$. It follows that

$||U^{a_{n_{i}}}_{T}d-U^{a_{n_{j}}}_{T}d||_{L^{1}}=||U^{a_{s_{p}}}_{T}d-U^{a_{s_{q}}}_{T}d||_{L^{1}}\leq
17M\epsilon.$

A contradiction. ∎

Since rigidity implies zero entropy, we have following corollary:

######  Corollary 4.6.

Let $(X,T)$ be a t.d.s.. If there exists a subsequence $A$ of $\mathbb{N}$
with $\bar{D}(A)>0$ such that $(X,T)$ is $A$-mean-equicontinuous, then the
topological entropy of $(X,T)$ is zero.

###  4.1. Results with respect to a function

Let $(X,T)$ be a t.d.s., let $\mu\in M(X,T)$ and $f\in L^{2}(\mu)$, let $K$ be
a subset of $X$ and $A=\\{a_{i}:i\in\mathbb{N}\\}$ be a subsequence of
$\mathbb{N}$.

For $x,y\in X$, define

$d_{f}(x,y)=|f(x)-f(y)|$

and

$\bar{d}_{n,f}^{A}(x,y)=\frac{1}{n}\sum_{i=0}^{n-1}d_{f}(T^{a_{i}}x,T^{a_{i}}y).$

Then $d_{f}$ and $\bar{d}_{n,f}^{A}$ are pseudo-metrics.

We say that $K$ is $f$-$A$-equicontinuous in the mean if for any $\epsilon>0$,
there exists $\delta>0$ such that

$x,y\in K,d(x,y)<\delta\Rightarrow\bar{d}_{n,f}^{A}(x,y)<\epsilon,\forall
n\in\mathbb{N}.$

We say $f$ is $A$-equicontinuous in the mean if $X$ is $f$-$A$-equicontinuous
in the mean.

We say $f$ is $\mu$-$A$-equicontinuous in the mean if for any $\tau>0$, there
exists a compact subset $K$ of $X$ with $\mu(K)>1-\tau$ such that $K$ is
$f$-$A$-equicontinuous in the mean.

######  Proposition 4.7.

Let $(X,T)$ be a t.d.s., $\mu\in M(X,T)$ and $f\in L^{2}(\mu)$. Let
$A=\\{a_{i}:i\in\mathbb{N}\\}$ be a subsequence of $\mathbb{N}$. Then the
following statements are equivalent:

  1. (1)

$f$ is $\mu$-$A$-equicontinuous in the mean.

  2. (2)

For any $\tau>0$ and $\epsilon>0$, there exist a compact subset $K$ of $X$
with $\mu(K)>1-\tau$ and $\delta>0$ such that

$x,y\in K,d(x,y)<\delta\Rightarrow\bar{d}_{n,f}^{A}(x,y)<\epsilon,\ \forall
n\in\mathbb{N}.$

We say that $K$ is $f$-$A$-mean-equicontinuous if for any $\epsilon>0$, there
exists $\delta>0$ such that

$x,y\in
K,d(x,y)<\delta\Rightarrow\limsup_{n\rightarrow\infty}\bar{d}_{n,f}^{A}(x,y)<\epsilon.$

We say that $f$ is $A$-mean-equicontinuous if $X$ is $f$-$A$-mean-
equicontinuous.

We say that $f$ is $\mu$-$A$-mean-equicontinuous  if for any $\tau>0$, there
exists a compact subset $K$ of $X$ with $\mu(K)>1-\tau$ such that $K$ is
$f$-$A$-mean-equicontinuous.

The notions of $f$-$\mu$-$A$-equicontinuous in the mean and
$f$-$\mu$-$A$-mean-equicontinuous are equivalent. Note that the proof of
Theorem 4.8 is similar to [10, Theorem 4.3].

######  Theorem 4.8.

Let $(X,T)$ be a t.d.s., $\mu\in M(X,T)$ and $f\in L^{2}(\mu)$. Let
$A=\\{a_{i}:i\in\mathbb{N}\\}$ be a subsequence of $\mathbb{N}$, then the
following statements are equivalent:

  1. (1)

$f$ is $\mu$-$A$-equicontinuous in the mean.

  2. (2)

$f$ is $\mu$-$A$-mean-equicontinuous.

###### Proof.

(1)$\Rightarrow$(2) is obvious.

(2)$\Rightarrow$(1): Fix $\tau>0$ and $\epsilon>0$. There exists a compact set
$K_{0}$ with $\mu(K_{0})>1-\dfrac{\tau}{2}$ such that $K_{0}$ is $f$-$A$-mean-
equicontinuous. Then there exists $\delta_{0}>0$ such that for all $x,y\in
K_{0}$ with $d(x,y)<\delta_{0}$, we have

$\limsup_{n\rightarrow\infty}\frac{1}{n}\sum_{i=0}^{n-1}|f(T^{a_{i}}x)-f(T^{a_{i}}y)|<\frac{\epsilon}{2}.$

By the compactness of $X$, there exist $K_{0}=\bigcup\limits_{i=1}^{m}X_{i}$,
where $diamX_{i}<\delta_{0},1\leq i\leq m,\ m\in\mathbb{N}.$ Without loss of
generality, assume $X_{i}\neq\emptyset$ and choose $x_{i}\in X_{i},1\leq i\leq
m.$ For $1\leq j\leq m$ and $N\in\mathbb{N}$, let

$A_{N}(x_{j})=\\{y\in
X_{j}:\frac{1}{n}\sum_{i=0}^{n-1}|f(T^{a_{i}}x_{j})-f(T^{a_{i}}y)|<\frac{\epsilon}{2},\forall
n>N\\}.$

It is easy to see that for each $1\leq j\leq m$,
$\\{A_{N}(x_{j})\\}_{N=1}^{\infty}$ is increasing and

$X_{j}=\bigcup\limits_{N=1}^{\infty}A_{N}(x_{j}).$

Choose $N\in\mathbb{N}$ such that

$\mu(\bigcup\limits_{j=1}^{m}A_{N}(x_{j}))>1-\dfrac{\tau}{2}.$

Since $f,U_{T}^{a_{1}}f,\ldots,U_{T}^{a_{N}}f\in L^{2}(\mu)$, there exists a
compact set $K_{1}$ with $\mu(K_{1})>1-\dfrac{\tau}{2}$ such that
$f,U_{T}^{a_{1}}f,\ldots,U_{T}^{a_{N}}f$ are uniformly continuous on $K_{1}$.
Then there exists $\delta_{1}>0$ such that for any $x,y\in K_{1}$ and
$d(x,y)<\delta_{1}$, we have

$|f(x)-f(y)|<\epsilon,|f(T^{a_{n}}x)-f(T^{a_{n}}y)|<\epsilon,1\leq n\leq N.$

Note that

$\mu(K_{1}\cap\bigcup_{j=1}^{m}A_{N}(x_{j}))>1-\tau,$

by the regularity of $\mu$, we can find pairwise disjoint compact sets

$E_{j}\subset A_{N}(x_{j}),1\leq j\leq m$

such that

$\bigcup_{i=j}^{m}E_{j}\subset K_{1}\cap\bigcup_{j=1}^{m}A_{N}(x_{j}),\
\mu(\bigcup_{j=1}^{m}E_{j})>1-\tau.$

Let $K=\bigcup\limits_{j=1}^{m}E_{j}$, then $K$ is compact and
$\mu(K)>1-\tau$. Choose $\delta$ with

$0<\delta<\min\\{\delta_{1},d(E_{i},E_{j}),1\leq i\neq j\leq m\\}.$

For any $x,y\in K$ and $d(x,y)<\delta$, if $1\leq n\leq N$, since $x,y\in
K_{1}$, we have
$\dfrac{1}{n}\sum\limits_{i=0}^{n-1}|f(T^{a_{i}}x)-f(T^{a_{i}}y)|<\epsilon$;
if $n>N$, since $d(x,y)<\delta$, there exists $1\leq j_{0}\leq m$ such that
$x,y\in E_{j_{0}}$. Then

$\frac{1}{n}\sum_{i=0}^{n-1}|f(T^{a_{i}}x_{j_{0}})-f(T^{a_{i}}x)|<\frac{\epsilon}{2},\
\frac{1}{n}\sum_{i=0}^{n-1}|f(T^{a_{i}}x_{j_{0}})-f(T^{a_{i}}y)|<\frac{\epsilon}{2}.$

Hence
$\dfrac{1}{n}\sum\limits_{i=0}^{n-1}|f(T^{a_{i}}x)-f(T^{a_{i}}y)|<\epsilon.$
By Proposition 4.7, $f$ is $\mu$-$A$-equicontinuous in the mean. ∎

The main theorem of this subsection is:

######  Theorem 4.9.

Let $(X,T)$ be a t.d.s. $\mu\in M(X,T)$ and $f\in L^{2}(\mu)$. If there exists
a subsequence $A=\\{a_{i}:i\in\mathbb{N}\\}$ of $\mathbb{N}$ with
$\bar{D}(A)>0$ such that $f$ is $\mu$-$A$-mean-equicontinuous, then $f$ is
rigid.

In order to prove Theorem 4.9, we need the following theorem. The idea is from
[10, Theorem 4.4].

######  Theorem 4.10.

Let $(X,T)$ be a t.d.s. $\mu\in M(X,T)$ and $f\in L^{2}(\mu)$. If there exists
a subsequence $A=\\{a_{i}:i\in\mathbb{N}\\}$ of $\mathbb{N}$ such that
$U_{T}^{a_{n}}d_{f}\stackrel{{\scriptstyle a.e.}}{{\longrightarrow}}d_{f}$,
then $f$ is $\mu$-$A$-equicontinuous.

###### Proof.

Given $\tau>0$ and $\epsilon>0$. Since $f\in L^{2}(\mu)$, there exists a
compact set $K_{0}$ with $\mu(K_{0})>1-\dfrac{\tau}{4}$ such that $f$ is
uniformly continuous on $K_{0}$. Then there exists $\delta_{0}>0$ such that
for any $x,y\in K_{0}$ and $d(x,y)<\delta_{0}$, we have
$d_{f}(x,y)<\dfrac{\epsilon}{8}$. By the compactness of $X$, there exist
$K_{0}=\bigcup\limits_{i=1}^{C}X_{i}$, where $diamX_{i}<\delta_{0},1\leq i\leq
C,\ C\in\mathbb{N}$. Without loss of generality, assume $X_{i}\neq\emptyset$
and choose $x_{i}\in X_{i},\ 1\leq i\leq C.$ It is easy to see that
$K_{0}\subset\bigcup\limits_{i=1}^{C}B_{d_{f}}(x_{i},\dfrac{\epsilon}{8}).$ It
follows that

$\mu(\bigcup_{i=1}^{C}B_{d_{f}}(x_{i},\dfrac{\epsilon}{8}))>1-\dfrac{\tau}{4}.$

For any $N\in\mathbb{N}$, let

$A_{N}=\\{(x,y):|U_{T}^{a_{n}}d_{f}(x,y)-d_{f}(x,y)|<\dfrac{\epsilon}{4},\
\forall n>N\\}.$

It is easy to see that

$\\{(x,y):U_{T}^{a_{n}}d_{f}(x,y)\rightarrow
d_{f}(x,y)\\}\subset\bigcup_{N=1}^{\infty}A_{N}.$

Note that $\\{A_{N}\\}$ is increasing, so there exists $N\in\mathbb{N}$ such
that $\mu\times\mu(A_{N})>1-(\dfrac{\tau}{4C})^{2}$. By Fubini’s theorem there
exists a subset $D$ of $X$ with $\mu(D)>1-\dfrac{\tau}{4C}$ such that $\forall
x\in D,\mu((A_{N})_{x})>1-\dfrac{\tau}{4C},$ where $(A_{N})_{x}=\\{y\in
X:(x,y)\in A_{N}\\}$.

Note that

$\mu(\bigcup_{i=1}^{C}(B_{d_{f}}(x_{i},\dfrac{\epsilon}{8})\cap
D))>1-\dfrac{\tau}{2},$

without loss of generality, assume $B_{d_{f}}(x_{i},\dfrac{\epsilon}{8})\cap
D\neq\emptyset$ and choose

$y_{i}\in B_{d_{f}}(x_{i},\dfrac{\epsilon}{8})\cap D,1\leq i\leq C.$

Then

$\mu((A_{N})_{y_{i}})>1-\dfrac{\tau}{4C},1\leq i\leq C.$

Since $B_{d_{f}}(x_{i},\dfrac{\epsilon}{8})\subset
B_{d_{f}}(y_{i},\dfrac{\epsilon}{4})$, we have

$\mu(\bigcup_{i=1}^{C}B_{d_{f}}(y_{i},\dfrac{\epsilon}{4}))\geq\mu(D\cap\bigcup_{i=1}^{C}B_{d_{f}}(y_{i},\dfrac{\epsilon}{4}))>1-\dfrac{\tau}{2}.$

Since $U_{T}^{a_{1}}f,\ldots,U_{T}^{a_{N}}f\in L^{2}(\mu)$, there exists a
compact set $K_{1}$ with $\mu(K_{1})>1-\dfrac{\tau}{4}$ such that
$U_{T}^{a_{1}}f,\ldots,U_{T}^{a_{N}}f$ are uniformly continuous on $K_{1}$.
Then there exists $\delta_{1}>0$ such that for any $x,y\in K_{1}$ and
$d(x,y)<\delta_{1}$, we have

$U_{T}^{a_{n}}d_{f}(x,y)<\epsilon,1\leq n\leq N.$

Note that

$\mu(K_{1}\cap\bigcup_{i=1}^{C}B_{d_{f}}(y_{i},\dfrac{\epsilon}{4})\cap\bigcap_{i=1}^{C}(A_{N})_{y_{i}})>1-\tau,$

by the regularity of $\mu$, we can find pairwise disjoint compact sets

$E_{i}\subset B_{d_{f}}(y_{i},\dfrac{\epsilon}{4}),1\leq i\leq C$

such that

$\bigcup_{i=1}^{C}E_{i}\subset
K_{1}\cap\bigcup_{i=1}^{C}B_{d_{f}}(y_{i},\dfrac{\epsilon}{4})\cap\bigcap_{i=1}^{C}(A_{N})_{y_{i}}$

and $\mu(\bigcup\limits_{i=1}^{C}E_{i})>1-\tau.$ Let
$K=\bigcup\limits_{i=1}^{C}E_{i}$, then $K$ is compact and $\mu(K)>1-\tau$.
Choose $\delta$ with

$0<\delta<\min\\{\delta_{1},d(E_{i},E_{j}),1\leq i\neq j\leq C\\}.$

For any $x,y\in K$ and $d(x,y)<\delta$, if $1\leq n\leq N$, since $x,y\in
K_{1}$, $U_{T}^{a_{n}}d_{f}(x,y)<\epsilon$; if $n>N$, since $d(x,y)<\delta$,
there exists $1\leq i_{0}\leq C$ such that $x,y\in E_{i_{0}}$. Then

$U_{T}^{a_{n}}d_{f}(x,y)\leq
U_{T}^{a_{n}}d_{f}(y_{i_{0}},x)+U_{T}^{a_{n}}d_{f}(y_{i_{0}},y).$

Note that $x,y\in(A_{N})_{y_{i_{0}}}$, we have $(y_{i_{0}},x),(y_{i_{0}},y)\in
A_{N}$, hence

$U_{T}^{a_{n}}d_{f}(y_{i_{0}},x)<d_{f}(y_{i_{0}},x)+\dfrac{\epsilon}{4}.$

Since $E_{i_{0}}\subset B_{d_{f}}(y_{i_{0}},\dfrac{\epsilon}{4}),$ we have
$d_{f}(y_{i_{0}},x)<\dfrac{\epsilon}{4},$ it follows that
$U_{T}^{a_{n}}d_{f}(y_{i_{0}},x)<\dfrac{\epsilon}{2}$. Similarly
$U_{T}^{a_{n}}d_{f}(y_{i_{0}},y)<\dfrac{\epsilon}{2}$, so
$U_{T}^{a_{n}}d_{f}(x,y)<\epsilon$. By Proposition 3.9, $f$ is
$\mu$-$A$-equicontinuous. ∎

Now we prove Theorem 4.9:

###### proof of Theorem 4.9.

Replacing $U_{T}^{a_{n}}d$ by $U_{T}^{a_{n}}d_{f}$, the proof of Theorem 4.5
still works, so we have ${\\{U_{T}^{a_{n}}d_{f}:n\in\mathbb{N}\\}}$ is pre-
compact in $L^{1}(\mu\times\mu)$. Then there exists a subsequence
$S=\\{s_{i}:i\in\mathbb{N}\\}$ of $\mathbb{N}$ such that
$U_{T}^{s_{n}}d_{f}\stackrel{{\scriptstyle a.e.}}{{\longrightarrow}}s_{f}$. By
Theorem 4.10, $f$ is $\mu$-$S$-equicontinuous. By Theorem 3.10, $f$ is rigid.
∎

######  Remark 4.11.

When considering the measure-theoretic mean equicontinuity along some
subsequence $A$ of $\mathbb{N}$, things may get relatively complicated, since
some useful tools and methods in ergodic theory do not work. But there are
still some interesting questions to consider.

######  Question 4.12.

Is the condition $\bar{D}(A)>0$ in Theorem 4.5 necessary?

######  Question 4.13.

If we take $A$ as $\mathbb{P}$, the set of all prime numbers, can we obtain
some interesting results?

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# A Information-Theoretic View on Spacetime

Frank Saueressig Institute for Mathematics, Astrophysics and Particle Physics
(IMAPP)
Radboud University, Heyendaalseweg 135
Nijmegen, 6525 AJ,The<EMAIL_ADDRESS>
author@emailaddress  Amir Khosravi  Institute for Mathematics, Astrophysics
and Particle Physics (IMAPP)
Radboud University, Heyendaalseweg 135
Nijmegen, 6525 AJ,The Netherlands

###### Abstract

We argue in a quantitative way that the unitarity principle of quantum field
theory together with the quantum information bound on correlation functions
are in tension with a space which is made out of disconnected patches at
microscopic scales.

###### keywords:

Gravity; Entanglement; Unitarity Principle.

PACS Nos.: 04.60.-m, 11.10.Hi, 03.65.Ud

##  1 Introduction

The quest to unravel the nature of space and time is one of the oldest and
most challenging ones in the history of intellectual endeavors. Quite
intriguingly, combining fundamental principles of quantum mechanics with an
information theoretic approach to spacetime allows making statements about the
structure of space and time way beyond experimentally accessible scales [1, 2,
3]. In this work we use arguments based on entanglement entropy and unitarity
to conclude that spatial slices of spacetime must be path-connected
topological spaces in order to avoid running into conflicts with these
concepts.

In Sect. 2.1 we review the notion of entanglement entropy and quantum mutual
information. In Sect. 2.2 we derive a constraint that is set by the unitarity
principle of quantum mechanics on the mutual information shared by two regions
of space. In Sect. 2.3, we argue that space cannot consist of (microscopic)
disconnected structures without violating this principle. Finally, a summary
is provided.

##  2 Quantum Information, Entanglement and Spacetime

###  2.1 A Brief Introduction

Consider a 3-dimensional spatial region $C$ together with its complement $D$
in a 4-dimensional spacetime. Thus $C$ and $D$ are disjoint except for the
2-dimensional boundary shared by them. Suppose the total information about the
geometry of region $E=C\cup D$ is stored in a density matrix $\rho_{\rm CD}$.
The entangelement entropy between the two regions can then be defined via the
von Neumann formula as

$S_{C}=-{\rm Tr}[\rho_{C}\log{\rho_{C}}]\,\,,\qquad S_{D}=-{\rm
Tr}[\rho_{D}\log{\rho_{D}}]\,.$ (1)

Here $\rho_{C}$ and $\rho_{D}$ are the reduced density matrices constructed
out of $\rho_{CD}$ by tracing out $D$ and $C$:

$\rho_{C}={\rm Tr}_{D}[\rho_{CD}]\,\,,\qquad\rho_{D}={\rm
Tr}_{C}[\rho_{CD}]\,.$ (2)

The mutual information $I_{\rm CD}$ is given by [4]

$I_{\rm CD}\equiv S_{C}+S_{D}-S_{\rm CD}\geq 0$ (3)

where $S_{\rm CD}\equiv-{\rm Tr}[\rho_{CD}\log{\rho_{CD}}]$ is the total
entanglement of the $(C,\,D)$-system. In a general state, $I_{CD}$ is a
measure of correlation between $C$ and $D$. It essentially indicates how much
region $C$ knows about $D$ and vice versa. If all the correlation is quantum,
i.e. $S_{\rm CD}=0$, then we have a pure state $S_{C}=S_{D}$.

###  2.2 Unitarity and Mutual Information

The question we seek to answer is: ”How does unitarity constrain the mutual
information shared by two regions of space?”. To address this question we
start from the Callan-Symanzik equation 222For example one could consider a
massive scalar field living on a predescribed background spacetime.. This
equation implies the following differential equation for $n$-point correlation
functions $\Gamma^{(n)}(p_{j},g(\mu),\mu)$ [5]

$\left[\mu\,\partial_{\mu}+\beta(g(\mu))\frac{\partial}{\partial{g(\mu)}}-\frac{n}{2}\eta\left(g(\mu)\right)\right]\Gamma^{(n)}(p_{j},g(\mu),\mu)=0.$
(4)

Here $\mu$ is the renormalization scale, $\eta\left(g(\mu)\right)$ is the
anomalous dimension of the fundamental fields appearing in the action,
$g(\mu)$ is a set of coupling constants and $g(\mu)$ satisfies the
renormalization group equation $\beta(g(\mu))=\mu\frac{d}{d\mu}g(\mu)$. The
background field formalism allows to extend these renormalization group
techniques also to the deep quantum gravity regime [6] by treating quantum
fluctuations of the spacetime metric akin to a quantum field theory in curved
spacetime [7, 8]. The general solution to “(4)” can be written as

$\Gamma^{(n)}(p_{j},g(\mu),\mu)=exp\left[\frac{n}{2}\int_{\mu_{0}}^{\mu}\frac{d\mu^{{}^{\prime}}}{\mu^{{}^{\prime}}}\eta\left(g(\mu^{{}^{\prime}})\right)\right]\,\Gamma^{(n)}(p_{j},g(\mu_{0}),\mu_{0})\,.$
(5)

The crucial observation is that for a generic unitary quantum field theory
(QFT) it is safe to assume $\eta(g(\mu))\geq 0$ [9, 10]. If we then define
operators $O_{C}$ and $O_{D}$ with support on $C$ and $D$ respectively, the
connected two-point function $\langle O_{C}O_{D}\rangle_{c}$ is positive semi-
definite $\braket{O_{C}\,O_{D}}_{c}\geq 0$. Consequently, using “(5)”, we get
$\partial_{\mu}\braket{O_{C}\,O_{D}}_{c}\geq 0$. A rigorously proven example
of this feature is the celebrated c-theorem in conformal field theory where
unitarity implies the positivity of the central charge and consequently
$\partial_{\mu}\braket{O_{C}\,O_{D}}_{c}\geq 0$ [11, 12].

On the other hand, for any two operators $O_{C}$ and $O_{D}$, one can use the
well-established properties of the conditional probabilities and certain
results of quantum information science to show that the mutual information
between the two regions $C$ and $D$ obeys [13]:

$I_{\rm
CD}\geqslant\frac{\braket{O_{C}\,O_{D}}_{c}^{2}}{2|O_{C}|^{2}\,|O_{D}|^{2}}.$
(6)

In more physical terms, “(6)” indicates that the quantum mutual information
shared by two regions of space is the maximum amount of correlation they can
have with each other.

Regarding “(6)” it is crucial to distinguish two cases: firstly, “(6)” can be
applied to ”discrete” spaces coming with a hard UV-cutoff. Such spaces would
be path-disconnected by definition and the operator norms $|O|$ are finite by
construction. This is the case considered in van Raamsdonk’s work [15], where
it was concluded that if the mutual information between $C$ and $D$ goes to
zero, then all correlations must decrease to zero also. Secondly, one can
consider continuum space which is path connected and leads to infinite values
of $|O|$. As a consequence vanishing mutual information $I_{CD}$ does not
automatically imply the vanishing of the correlation function. In the sequel,
we will work in the first setting and construct a contradiction between the
unitarity principle and a discrete space.

Any physically interesting theory should have at least one correlation
function which has support on both $C$ and $D$ and is non-zero at some
(finite) scale. Focusing on such a correlation function, we can make the
following argument: assume the value of shared information between regions $C$
and $D$ is zero in the ultraviolet (UV) limit, i.e. we have $I_{\rm
CD}^{UV}=0$. From unitarity we have $\braket{O_{C}\,O_{D}}_{c}\geq 0$ and thus
eq. “(6)” entails that $\langle O_{C}O_{D}\rangle_{c}^{UV}=0$. Furthermore, if
the unitarity principle is satisfied, we should have
$\partial_{\mu}\braket{O_{C}\,O_{D}}_{c}\geq 0$. If we take
$\partial_{\mu}\braket{O_{C}\,O_{D}}_{c}=0$ then the correlation function
vanishes on all scales contradicting our premise above. So unitarity and the
existence of non-vanishing correlations imply
$\partial_{\mu}\braket{O_{C}\,O_{D}}_{c}>0$. This results in $\langle
O_{C}O_{D}\rangle_{c}^{\rm IR}<0$ which conflicts with unitarity. The bottom
line of this argument is that the requirement of unitarity and eq. “(6)”
implies $I_{\rm CD}^{UV}>0$. In Sect. 2.3 we argue that $I_{\rm CD}^{UV}>0$ is
in conflict with a disconnected space.333 Our statement is actually compatible
with the situation where both regions $C$ and $D$ support non-trivial, unitary
QFTs which do not interact with each other. In this case the two-point
correlator
$\langle\mathcal{O}_{C}\,\mathcal{O}_{D}\rangle=\langle\mathcal{O}_{C}\rangle\langle\mathcal{O}_{D}\rangle$
factorizes, implying the vanishing of the connected correlation function
$\langle\mathcal{O}_{C}\mathcal{O}_{D}\rangle_{c}=\langle\mathcal{O}_{C}\,\mathcal{O}_{D}\rangle-\langle\mathcal{O}_{C}\rangle\langle\mathcal{O}_{D}\rangle\stackrel{{\scriptstyle!}}{{=}}0$.

An alternative derivation of this conclusion builds on the $I_{theorem}$ [16].
In this case one can use $\partial_{\mu}I_{CD}>0$ to argue that
$I_{CD}^{UV}=0$ implies $I_{CD}^{IR}<0$ and hence leads to the violation of
entanglement subadditivity 444Subadditivity of the entanglement of two regions
$C$ and $D$ means $\rm{Entanglement}_{C\cup
D}\leq\rm{Entanglement}_{C}+\rm{Entanglement}_{D}$. The entanglement entropy
always satisfies this property [4]. in the IR. At this stage, there is no
consensus on the existence of the $I_{theorem}$ though.

###  2.3 Space Must be Path-Connected

According to Van Raamsdonk [15] the relation between the entanglement entropy
of two regions of space, their shared area and the geodesic distance between
them has a very remarkable interpretation. The original argument is presented
in an AdS/CFT [14, 15] setup; here we present a simpler version using
properties of the entanglement entropy in QFT. References [17, 18, 19] suggest
that to leading order in the UV cutoff the entanglement entropy is
proportional to the shared area (area of the entangling surface) between a
region and its complement in space. Thus, reducing the entanglement entropy
and mutual information is equivalent to reducing the shared area.

Furthermore, for a typical massive field theory with mass $m$ we have
$\braket{O_{C}(\vec{x})\,O_{D}(\vec{y})}_{c}\propto e^{-m\,L}$ where $\vec{x}$
and $\vec{y}$ are spatial positions in $C$ and $D$ and $L$ indicates the
geodesic distance between these points. In the light of “(6)”, this means
$I\ll 1$ if $L\gg\frac{1}{m}$.

Combining last two paragraphs, one can summarize Van Raamsdonk’s result as
[15]555Notably, the holographic entanglement entropy upon which the original
Van Raamsdonk argument is based produces the ordinary QFT area law to leading
order in the UV cutoff [20].:

Zero entanglement entropy is equivalent to $C$ and $D$ being disconnected
regions of space. Hence, the connectivity of the classical geometry has its
roots in the properties of quantum entanglement entropy. (S1)

The core of our argument, essentially Sect. 2.2, survives the transition to a
fundamental theory of nature as long as this theory admits an effective QFT
description that is unitary and respects the basic rules of quantum
information physics. If QFT itself survives the generalization, we have a
scenario like asymptotic safety [21, 8, 22]. If QFT is an effective
description, we have something like a string theory effective description of
the bulk [14, 2]. Regarding a disconnected space, according to “(S1)” we
should have $I_{CD}^{UV}=0$ if regions $C$ and $D$ are disconnected atoms of
space sharing no boundary. On the other hand, we know from the previous
section that according to “(6)” the only way to reconcile $I_{CD}^{UV}=0$ with
unitarity is by all correlation functions involving products of operators with
support on $C$ and $D$ vanishing at all scales. If the disconnected regions
are microscopic, this is of no interest in any physics scenario. So if one
insists on a disconnected space in the UV and a QFT describing the universe,
one has only two options: 1) Accept unitarity and consequently violate “(6)”
which means certain basic rules of quantum information physics are violated.
2) Drop the unitarity principle and accept the possibility of information
loss. Both options are in tension with principles of quantum mechanics. This
suggests that space cannot be disconnected. A prototypical example for a
physical process ruled out by these considerations is the creation and
pinching-off of baby universes from our spacetime: if a quantum field has
support on the region of spacetime that is pinched-off the information is lost
entailing a violation of unitarity.

##  3 Summary

In this manuscript we assumed that the universe admits a unitary QFT
description, at least at the effective level. On this basis we argued that
such a description together with rules of conditional probability theory and
quantum information science implies that space cannot emerge from the coarse
graining of a patchwork of disconnected structures. This is a consequence of
“(S1)”, the unitarity principle, “(6)” and the requirement of a non-trivial
correlation function. To put it in other words, in the light of “(S1)” it is
obvious that $I_{CD}^{UV}=0$ is a necessary condition for having a
disconnected space. Thus unitarity and “(6)” put conditions on the topological
properties of space, demanding $I_{CD}^{UV}>0$. Notably, our analysis does not
imply that on a space made of path-disconnected regions, any unitary QFT is
necessary trivial: every path-connected region may support its own unitary QFT
coming with non-trivial correlation functions. If the regions are Planck-sized
this is not a physically interesting scenario though.

An interesting situation motivating a non-trivial generalization of the ideas
advocated in this work is the BKL scenario [23]. In this model spatial points
decouple close to the cosmic singularity. This should lead to vanishing
correlation functions in the UV. At the same time, spacetime remains
connected, thereby violating the assumption of fundamental discreteness made
in our derivation. It would be interesting to understand the interplay between
the information theoretic concepts underlying our work and unitarity in such a
situation as well. We also note that our argument may not be applicable to
(apparently discrete) models of quantum gravity as, e.g., causal dynamical
triangulations[24, 25] or group field theories [26, 27], since in this case
the “discreteness scale” plays the role of a UV regulator which is to be
removed by taking a suitable continuum limit.

As a final remark, we note that the incompatibility between the unitarity of
QFT and “(6)” with a disconnected space only depends on a set of basic
properties of conditional probabilities and the unitarity of QFT. This result
may restrict the class of (quantum) theories of gravity that are compatible
with unitarity and basic rules of quantum information theory.

## Acknowledgement

We would like to thank Carlo Rovelli, Slava Rychkov, Tatsuma Nishioka and
Horacio Casini for fruitful discussions and Francesco Caravelli, Andrea
Stergiou and Andrea Trombettoni for critical comments. Special thanks to Alfio
Bonanno for providing insightful comments on the draft. A.K. profited from
illuminating discussion during the Functional and Renormalization-Group
methods (FRGIM 2019) conference in Trento, Italy where the work was presented.
Attending the FRGIM 2019 was made possible by the generous financial support
from the “ACRI-Young Investigators Training Program”.

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