# On super-rigid and uniformly super-rigid operators

Otmane Benchiheb  ,  Fatimaezzahra Sadek  and  Mohamed Amouch  University
Chouaib Doukkali. Department of Mathematics, Faculty of science Eljadida,
Morocco<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

An operator $T$ acting on a Banach space $X$ is said to be super-recurrent if
for each open subset $U$ of $X$, there exist $\lambda\in\mathbb{K}$ and
$n\in\mathbb{N}$ such that $\lambda T^{n}(U)\cap U\neq\emptyset$. In this
paper, we introduce and study the notions of super-rigidity and uniform super-
rigidity which are related to the notion of super-recurrence. We investigate
some properties of these classes of operators and show that they share some
properties with super-recurrent operators. At the end, we study the case of
finite-dimensional spaces.

###### Key words and phrases:

Hypercyclicity, recurrence, super-recurrence, rigidity, uniform rigidity.

###### 2010 Mathematics Subject Classification:

47A16, 37B20

##  1\. Introduction and preliminaries

Throughout this paper, $X$ will denote a Banach space over the field
$\mathbb{K}$ ($\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$). By an operator, we
mean a linear and continuous map acting on $X$, and we denote by
$\mathcal{B}(X)$ the set of all operators acting on $X.$
A (discrete) linear dynamical system is a pair $(X,T)$ consisting of a Banach
space $X$ and an operator $T$. The most important and studied notions in the
linear dynamical system are those of hypercyclicity and supercyclicity.

We say that $T\in\mathcal{B}(X)$ is hypercyclic if there exists a vector $x$
whose orbit under $T$;

$Orb(T,x):=\\{T^{n}x\mbox{ : }n\in\mathbb{N}\\},$

is dense in $X,$ in which case, the vector $x$ is called a hypercyclic vector
for $T$. The set of all hypercyclic vectors for $T$ is denoted by $HC(T).$

An equivalent notion of the hypercyclicity in context of separable Banach
spaces, called topological transitivity, was introduced by Birkhoff in [5]$:$
an operator $T$ acting on a separable Banach space $X$ is hypercyclic if and
only if it is topologically transitive; that is, for each pair $(U,V)$ of
nonempty open subsets of $X$ there exists $n\in\mathbb{N}$ such that

$T^{n}(U)\cap V\neq\emptyset.$

In $1974$, Hilden and Wallen in [6] introduced the concept of supercyclicity.
An operator $T$ acting on $X$ is said to be supercyclic if there exists $x\in
X$ such that

$\mathbb{K}Orb(T,x):=\\{\lambda T^{n}x\mbox{ : }\lambda\in\mathbb{K}\mbox{,
}n\in\mathbb{N}\\}$

is dense in $X$. Such a vector $x$ is called a supercyclic vector for $T$. The
set of all supercyclic vectors for $T$ is denoted by $SC(T).$
An operator $T$ acting on a separable Banach space $X$ is supercyclic if and
only if for each pair $(U,V)$ of nonempty open subsets of $X$ there exist
$\lambda\in\mathbb{K}$ and $n\in\mathbb{N}$ such that

$\lambda T^{n}(U)\cap V\neq\emptyset,$

see [4, Theorem 1.12].

For more information about hypercyclic and supercyclic operators and their
proprieties, see K.G. Grosse-Erdmann and A. Peris’s book [16] and F. Bayart
and E. Matheron’s book [4], and the survey article [17] by K.G. Grosse-
Erdmann.

Another important notion in the linear dynamical system that has a long story
is that of recurrence which introduced by Poincaré in [20]. Later, it have
been studied by Gottschalk and Hedlund [15] and also by Furstenberg [13].
Recently, recurrent operators have been studied in [10].

We say that $T\in\mathcal{B}(X)$ is recurrent if for each open subset $U$ of
$X$, there exists $n\in\mathbb{N}$ such that

$T^{n}(U)\cap U\neq\emptyset.$

A vector $x\in X$ is called a recurrent vector for $T$ if there exists an
increasing sequence $(n_{k})\subset\mathbb{N}$ such that

$T^{n_{k}}x\longrightarrow x\ \ \mbox{ as }\ \ k\longrightarrow\infty.$

We denoted by $Rec(T)$ the set of all recurrent vectors for $T$. and we have
that $T$ is recurrent if and only if $Rec(T)$ is dense in $X.$ For more
information about this classe of operators, see [1, 7, 8, 9, 11, 12, 14, 18,
19, 21].

Recall that an operator $T\in\mathcal{B}(X)$ is said to be rigid if there
exists a strictly increasing sequence $(n_{k})\subset\mathbb{N}$ such that

$T^{n_{k}}x\longrightarrow x,\ \ \mbox{ as }\ \ k\longrightarrow\infty,\ \
\mbox{ for all }\ \ x\in X.$

This means that each vector of the space $X$ is a recurrent vector for $T$
with respect to the same sequence $(n_{k}).$
An operator $T$ is said to be uniformly rigid if there exists a strictly
increasing sequence $(n_{k})\subset\mathbb{N}$ such that

$\|T^{n_{k}}-I\|=\sup_{\|x\|\leq 1}\|T^{n_{k}}x-x\|\longrightarrow 0,\ \
\mbox{ as }k\longrightarrow\infty.$

Clearly, we have that

$T\mbox{ is uniformly rigid }\Rightarrow T\mbox{ is rigid }\Rightarrow T\mbox{
is recurrent}.$

The converses of those implications does not hold in general, see [10].

Recently in [2], recurrent operators have been generalized to a large class of
operators called super-recurrent operators. We say that $T\in\mathcal{B}(X)$
is super-recurrent if for each open subset $U$ of $X$, there exist
$\lambda\in\mathbb{K}$ and $n\in\mathbb{N}$ such that

$\lambda T^{n}(U)\cap U\neq\emptyset.$

A vector $x\in X$ is called a super-recurrent vector for $T$ if there exist an
increasing sequence $(n_{k})\subset\mathbb{N}$ and a sequence
$(\lambda_{k})\subset\mathbb{K}$ such that

$\lambda_{k}T^{n_{k}}x\longrightarrow x\ \mbox{as}\ k\longrightarrow\infty.$

We denoted by $SRec(T)$ the set of all recurrent vectors for $T$.

Taking into consideration the link between recurrence and their deviation, we
introduce in this paper the notions of super-rigidity and uniform super-
rigidity which are related to the notion of super-recurrence and generalize
the notions of rigidity and uniform rigidity.

In Section $2$, we study the notion of super-rigidity. We give the
relationship between super-rigid, rigid, super-recurrent, and recurrent
operators. Also we prove that the super-rigidity is preserved under similarity
and we give the relationship between the super-rigidity of an operator $T$ and
its iterates by showing that $T$ is super-rigid if and only for each $p\geq
2$, the operator $T^{p}$ is super-rigid. At the end of the section, several
spectral properties of super-rigid operators will be proven.

In Section $3$, we introduce and study the notion of uniformly super-rigid
operators. As in the super-rigidity’s case, we prove that the property of
being uniform super-rigid is preserved under similarity and that $T$ is
uniformly super-rigid if and only if $T^{p}$ is uniformly super rigid for all
$p.$ Moreover, we prove that the spectrum of a uniform-super-rigid operator
has a specific property. This lead us to show that being invertible is a
necessary condition of being uniformly super-rigid.

In Section $4$, we characterize the super-recurrence in finite-dimensional
spaces, and we show that in this case, the notions of super-recurrent, super-
rigid and uniformly super-rigid are equivalent.

##  2\. Super-rigid operators

In the following, we introduce the notion of super-rigid operators which
generalizes the notion of rigidity and related to the notion of super-
recurrence.

######  Definition 2.1.

An operator $T$ acting on $X$ is called super-rigid if there exist a strictly
increasing sequence of positive integers $(n_{k})_{n\in\mathbb{N}}$ and a
sequence $(\lambda_{k})_{n\in\mathbb{N}}$ of numbers such that

$\lambda_{k}T^{n_{k}}x\longrightarrow x$

for every $x\in X.$

######  Example 2.2.

For $1<p<\infty$, let $X=\ell^{p}(\mathbb{N})$. Let $R$ be a strictly positive
number and $(\lambda_{k})$ a sequence of numbers such that
$\lambda_{k}\in\\{z\in\mathbb{C}\mbox{ : }|z|=R\\}$ for all $k$. Let $T$ be an
operator defined on $\ell^{p}(\mathbb{N})$ by

$T(x_{1},x_{2},\dots)=(\lambda_{1}x_{1},\lambda_{2}x_{2},\dots),\hskip
8.5359pt\mbox{ for all }(x_{k})\in\ell^{p}(\mathbb{N}).$

Then $T$ is super-rigid. Indeed, this is since $R^{-1}T$ is recurrent, see
[10, Theorem 5.4].

######  Remark 2.3.

It is clear that the notion of super-rigidity implies the notion of rigidity,
but the converse does not hold in general. Indeed, let $T$ be the operator
defined in Example 2.2, then $T$ is super-rigid. However, $T$ is recurrent if
and only if $\lambda_{k}=1$, for all $k,$ see [10, Theorem 5.4].

Let $T$ be an operator acting on $X$. It is clear that

$T\mbox{ is super-rigid }\Rightarrow T\mbox{ is super-recurrent. }$

However, the converse does not hold in general even if $T$ is supercyclic as
we show in the next example.

######  Example 2.4.

Let $(e_{n})_{n\in\mathbb{N}}$ be the canonical basis of
$\ell^{2}(\mathbb{N})$ and $\textbf{w}=(\omega_{n})_{n\in\mathbb{N}}$ be a
bounded sequence of positive numbers. Let $B_{\textbf{w}}$ be the weighted
backward shift operator, defined on $\ell^{2}(\mathbb{N})$ by

$B_{\textbf{w}}(e_{0})=0\hskip 5.69046pt\mbox{ and }\hskip
5.69046ptB_{\textbf{w}}(e_{n})=\omega_{n}e_{n-1}\hskip 2.84544pt\mbox{ for all
}n\geq 1.$

$B_{\textbf{w}}$ is super-recurrent since it is supercyclic, see [4, Example
1.15]. However, $B_{\textbf{w}}$ cannot be super-rigid since
$B_{\textbf{w}}(e_{1})=0$.

We have the following diagram showing the relationships among super-rigidity,
rigidity, super-recurrence, and recurrence.

Rigidity  Recurrence  Super-rigiditySuper-recurrence$\not\uparrow_{\mbox{
\tiny{Remark \ref{remp}
}}}$$\nleftarrow_{\mbox{\tiny{\cite[cite]{[\@@bibref{}{CMP}{}{}, Example
5.4]}}}}$$\nleftarrow_{\mbox{\tiny{Example
\ref{exs}}}}$$\not\uparrow_{\mbox{\tiny{\cite[cite]{[\@@bibref{}{ABS}{}{},
Remarks 2.2]}}}}$

In the following proposition, we prove that the super-rigidity is preserved
under similarity.

######  Proposition 2.5.

Let $T\in\mathcal{B}(X)$ and $S\in\mathcal{B}(Y)$. Assume that $T$ and $S$ are
similar. Then $T$ is super-rigid on $X$ if and only if $S$ is super-rigid on
$Y$.

###### Proof.

Since $T$ and $S$ are similar, then there exists a homeomorphisme $\phi$ :
$X\longrightarrow Y$ such that $S\circ\phi=\phi\circ T.$ Let $y\in Y$, then
there exists $x\in X$ such that $y=\phi x.$ Since $T$ is super-rigid in $X$,
there exist a strictly increasing sequence $(n_{k})$ of positive integers and
a sequence $(\lambda_{k})$ of numbers such that
$\lambda_{k}T^{n_{k}}x\longrightarrow x$ as $k\longrightarrow+\infty$. Since
$\phi$ is continuous and $S\circ\phi=\phi\circ T$, it follows that
$\lambda_{k}S^{n_{k}}y\longrightarrow y$ as $k\longrightarrow+\infty$, which
means that $S$ is super-rigid. ∎

Let $p$ be a nonzero fixed positive integer. In $1995$, Ansari proved that $T$
is a hypercyclic (resp, supercyclic) operator on a separable Banach space if
and only if $T^{p}$ is hypercyclic (resp, supercyclic), see [3]. Later, the
same result was proven for recurrent operators, see [10, Proposition 2.3], and
for super-recurrent operators, see [2, Theorem 3.11]. In the following
theorem, we prove that this result remains true for super-rigid operators.

######  Theorem 2.6.

Let $T$ be an operator acting on $X$. Then $T$ is super-rigid if and only if
$T^{p}$ is super-rigid for all $p\geq 2$.

###### Proof.

Assume that $T$ is super-rigid, then there exist a strictly increasing
sequence $(n_{k})$ of positive integers and a sequence $(\lambda_{k})$ of
numbers such that $\lambda_{k}T^{n_{k}}x\longrightarrow x$ for every $x\in X$.
Let $M:=\sup_{n\in\mathbb{N}}\|\lambda_{k}T^{n_{k}}\|$. By Banach-Steinhaus
theorem we have $M<+\infty.$ Let $x$ be a vector of $X$, then we have

$\displaystyle\|\lambda_{k}^{p}T^{pn_{k}}x-x\|$
$\displaystyle=\|\lambda_{k}^{p}T^{pn_{k}}x-\lambda_{k}^{p-1}T^{(p-1)n_{k}}x+\lambda_{k}^{p-1}T^{(p-1)n_{k}}x-\dots+\lambda_{k}T^{n_{k}}x-x\|$
$\displaystyle\leq\|\lambda_{k}^{p}T^{pn_{k}}x-\lambda_{k}^{p-1}T^{(p-1)n_{k}}x\|+\dots+\|\lambda_{k}T^{n_{k}}x-x\|$
$\displaystyle\leq\|\lambda_{k}^{p-1}T^{p-1}\|\|\lambda_{k}T^{n_{k}}x-x\|+\dots+\|\lambda_{k}T^{n_{k}}x-x\|$
$\displaystyle\leq\left(\sum_{i=0}^{p-1}M^{i}\right)\|\lambda_{k}T^{n_{k}}x-x\|.$

Since $T$ is super-rigid with respect to $(n_{k})$ and $(\lambda_{k})$, it
follows that $\lambda_{k}^{p}T^{pn_{k}}x\longrightarrow x,$ which means that
$T^{p}$ is super-rigid with respect to $(n_{k})$ and $(\lambda_{k}^{p})$.
∎

In the following, we will give a characterization of the spectrum of super-
rigid operator. We begin with the following lemma.

######  Lemma 2.7.

Let $T$ be an operator acting on $X$ and $\lambda$ a nonzero number. Then, $T$
is super-rigid if and only if $\lambda T$ is super-rigid.

###### Proof.

Assume that $T$ is super-rigid. then there exists a strictly increasing
sequence $(n_{k})\subset\mathbb{N}$ and $(\lambda_{k})\subset\mathbb{K}$ such
that $\lambda_{k}T^{n_{k}}x\longrightarrow x$ for every $x\in X$. Let
$\mu_{k}=\lambda^{-n_{k}}\lambda_{k}$. Then

$\mu_{k}(\lambda
T)^{n_{k}}x=\lambda^{-n_{k}}\lambda^{n_{k}}\lambda_{k}T^{n_{k}}x=\lambda_{k}T^{n_{k}}x\longrightarrow
x,$

for every $x\in X$. This implies that $\lambda T$ is super-rigid. ∎

The next proposition shows that the eigenvalues of a super-rigid operators has
the same argument.

######  Proposition 2.8.

Let $T$ be an operator acting on $X$. If $T$ is super-rigid, then

$\sigma_{p}(T)\subset\\{z\in\mathbb{C}\mbox{ : }|z|=R\\},$

for some strictly positive real number $R$.

###### Proof.

Assume that there exist $\lambda_{1}$ and $\lambda_{2}$ in the point spectrum
of $T$ such that $|\lambda_{1}|<|\lambda_{2}|$. Let $m$ be a strictly positive
real number such that $|\lambda_{1}|<m<|\lambda_{2}|$. Since $\lambda_{1}$,
$\lambda_{2}\in\sigma_{p}(T)$, it follows that there exist $x$, $y\in
X\setminus\\{0\\}$ such that $Tx=\lambda_{1}x$ and $Ty=\lambda_{2}y.$ By Lemma
2.7, the operator $\frac{1}{m}T$ is super-rigid. Hence, there exist a sequence
$(\mu_{k})\subset\mathbb{K}$ and a sequence $(n_{k})\subset\mathbb{N}$ such
that

(2.1) $\mu_{k}\left(\frac{\lambda_{1}}{m}\right)^{n_{k}}x\longrightarrow
x\hskip 5.69046pt$

and

(2.2) $\hskip
5.69046pt\mu_{k}\left(\frac{\lambda_{2}}{m}\right)^{n_{k}}y\longrightarrow y.$

By (2.1), we have $|\mu_{k}|\longrightarrow+\infty$, and by (2.2), we have
$|\mu_{k}|\longrightarrow 0$, which is a contradiction. ∎

######  Remark 2.9.

Proposition 2.8 is not true in general for super-recurrent operators which are
not super-rigid. Indeed, let $(e_{n})_{n\in\mathbb{N}}$ be the canonical basis
of $\ell^{2}(\mathbb{N})$ and $\textbf{w}=(\omega_{n})_{n\in\mathbb{N}}$ be a
bounded sequence of positive numbers. Suppose that $B_{\textbf{w}}$ is defined
as in Example 2.4, then $B_{\textbf{w}}$ is super-recurrent. However, if
$\omega_{n}=1$, for all $n\geq 1$, and $B_{\textbf{w}}=B$ is the backward
shift operator, then

$\sigma_{p}(B)=B(0,1).$

Let $T$ be a super-rigid operator acting on a Banach space $X$. Since any
super-rigid operator is super-recurrent, it follows by [2, Theorem 4.2] that
that if $T$ is super-rigid, then the eigenvalues of $T^{*}$; the Banach
adjoint operator of $T$, are of the same argument. This means that there
exists $R_{1}>$ such that $\sigma_{p}(T^{*})\subset\\{z\in\mathbb{C}\mbox{ :
}|z|=R_{1}\\}$. Moreover, by Proposition 2.8, the eigenvalues of $T$ are of
the same argument. This means that there exists $R_{2}>$ such that
$\sigma_{p}(T)\subset\\{z\in\mathbb{C}\mbox{ : }|z|=R_{2}\\}$. The next
proposition shows that we have $R_{1}=R_{2}.$ This means that the eigenvalues
of $T$ and the eigenvalues of his Banach adjoint are of the same argument.

######  Proposition 2.10.

Let $T$ be an operator acting on $X$. If $T$ is super-rigid, then the
eigenvalues of $T$ and the eigenvalues of his Banach adjoint are of the same
argument. This means that there exists some $R>0$ such that

$\sigma_{p}(T)\cup\sigma_{p}(T^{*})\subset\\{z\in\mathbb{C}\mbox{ :
}|z|=R\\}.$

###### Proof.

Assume that there exist $\lambda_{1}\in\sigma_{p}(T)$ and
$\lambda_{1}\in\sigma_{p}(T^{*})$ such that $|\lambda_{1}|<m<|\lambda_{2}|,$
where $m$ is a strictly positive real number. Since
$\lambda_{1}\in\sigma_{p}(T)$ and $\lambda_{1}\in\sigma_{p}(T^{*})$, it
follows that there exist $x\in X\setminus\\{0\\}$ and $x^{*}\in
X^{*}\setminus\\{0\\}$ such that $Tx=\lambda_{1}x$ and
$T^{*}x^{*}=\lambda_{2}x^{*}$. By lemma 2.7, the operator $\frac{1}{m}T$ is
super-rigid. Hence, there exist a sequence $(\mu_{k})\subset\mathbb{K}$ and a
sequence $(n_{k})\subset\mathbb{N}$ such that
$\mu_{k}\left(\frac{1}{m}T\right)^{n_{k}}y\longrightarrow y$, for all $y\in
X.$ In particular, for $y=x,$ we have

(2.3)
$\mu_{k}\left(\frac{1}{m}T\right)^{n_{k}}x=\mu_{k}\left(\frac{\lambda_{1}}{m}\right)^{n_{k}}x\longrightarrow
x.$

Since $x^{*}$ is an nonzero linear form on $X$, it follows that there exists
$z\in X$ such that $x^{*}(z)\neq 0.$ Since $\frac{1}{m}T$ is super-rigid and
$x^{*}$ is continuous, we have

(2.4)
$\mu_{k}\left(\frac{\lambda_{2}}{m}\right)^{n_{k}}x^{*}(z)\longrightarrow
x^{*}(z).$

∎

By (2.3), we have $|\mu_{k}|\longrightarrow+\infty$ and by (2.4), we have
$|\mu_{k}|\longrightarrow 0$, which is a contradiction.

######  Example 2.11.

Let $T$ be the operator defined in Example 2.2. If $p=2,$ then it is easy to
prove that $\sigma_{p}(T)=\\{\lambda_{1},\lambda_{2},\dots\\}.$ Moreover, a
simple verification shows that the adjont of $T$ is defined by

$T^{*}(x_{1},x_{2},\dots)=(\overline{\lambda_{1}}x_{1},\overline{\lambda_{2}}x_{2},\dots).$

Hence,
$\sigma_{p}(T^{*})=\\{\overline{\lambda_{1}},\overline{\lambda_{2}},\dots\\}.$
This shows that

$\sigma_{p}(T)\cup\sigma_{p}(T^{*})\subset\\{z\in\mathbb{C}\mbox{ :
}|z|=R\\}.$

##  3\. Uniformly super-rigid operators

In the following, we introduce the notion of uniform super-rigidity which
generalizes the notion of uniform rigidity.

######  Definition 3.1.

An operator $T$ acting on $X$ is called uniformly super-rigid if there exist a
strictly increasing sequence of positive integers $(n_{k})_{n\in\mathbb{N}}$
and a sequence $(\lambda_{k})_{n\in\mathbb{N}}$ of numbers such that

$\|\lambda_{k}T^{n_{k}}-I\|=\sup_{x\neq
0}\|\lambda_{k}T^{n_{k}}x-x\|\longrightarrow 0.$

######  Example 3.2.

Let $T$ be an operator defined on $\mathbb{C}^{n}$ by$:$
$T(x_{1},\dots,x_{n})=(\lambda_{1}x_{1},\dots,\lambda_{n}x_{n})$, where
$\lambda_{i}\in\mathbb{K}$ and $\lambda_{i}=\lambda_{j}=R>0$, for $1\leq
i,j\leq n.$ Let $(x_{1},\dots x_{n})\neq(0,\dots 0)$ and $m\geq 0$, we have

$R^{-m}T^{m}(x_{1},\dots x_{n})-(x_{1},\dots
x_{n})=(R^{-m}\lambda_{1}^{m}x_{1},\dots R^{-m}\lambda_{n}^{m}x_{n}).$

Since $\lambda_{i}=\lambda_{j}=R>0$, for $1\leq i,j\leq n,$ it follows that
there exists a strictly increasing sequence $(m_{k})$ of positive integers
such that

$\|R^{m_{k}}T^{m_{k}}(x_{1},\dots x_{n})-(x_{1},\dots x_{n})\|\longrightarrow
0.$

This means that $T$ is a uniformly super-rigid operator.

######  Remark 3.3.

It is clear that the uniform super-rigidity implies the uniform rigidity.
However, the converse does not hold in general. Indeed, let $T$ be the
operator defined as in the Example 3.2, then $T$ is a uniformly super-rigid
whenever $\lambda_{i}=\lambda_{j}=R>0$, for $1\leq i,j\leq n.$ But the
operator $T$ is uniform rigid if and only if $\lambda_{i}=\lambda_{j}=1$, for
$1\leq i,j\leq n,$ see [10, Section 4].

######  Lemma 3.4.

Let $T$ be an operator acting on $X$. If $T$ is uniformly super-rigid, then
$\lambda T$ is uniformly super-rigid for all nonzero number $\lambda.$

###### Proof.

If $T$ is uniformly super-rigid with respect to a sequence $(\mu_{k})$ and a
sequence $(n_{k})$, then it is easy to show that $\lambda T$ is uniformly
super-rigid with respect to $(\mu_{k}\lambda^{-n_{k}})$ and $(n_{k}).$ ∎

######  Remark 3.5.

It is clear that each uniform super-rigid operator is super-rigid. However,
the converse does not hold in general. Indeed, let
$(\theta_{k})_{k\in\mathbb{N}}$ be a sequence of real numbers such that

$\liminf_{n\longrightarrow+\infty}\left(\sup_{k\in\mathbb{N}}\lvert e^{2\pi
in\theta_{k}}-1\rvert\right)\nrightarrow 0.$

Let $R>0$, and $\lambda_{k}=Re^{2\pi in\theta_{k}}$, for all $k\in\mathbb{N}$.
Let $T$ be the operator defined as in Example 2.2. Then $T$ is super-rigid. On
the other hand, assume that if $T$ is uniformly super-rigid, then $R^{-1}T$ is
uniformly rigid, which is not possible by [10, Theorem 5.4]. Hence, $T$ is
super-rigid but not uniformly super-rigid.

In the following proposition, we prove that the uniform super-rigidity is
preserved under similarity.

######  Proposition 3.6.

Let $T\in\mathcal{B}(X)$ and $S\in\mathcal{B}(Y)$. Assume that $T$ and $S$ are
similar. Then $T$ is uniformly super-rigid on $X$ if and only if $S$ is
uniformly super-rigid on $Y$.

###### Proof.

Since $T$ and $S$ are similar, then there exists a homeomorphisme $\phi$ :
$X\longrightarrow Y$ such that $S\circ\phi=\phi\circ T.$
Since $T$ is uniformly super-rigid, it follows that a strictly increasing
sequence of positive integers $(n_{k})_{n\in\mathbb{N}}$ and a sequence
$(\lambda_{k})_{n\in\mathbb{N}}$ of numbers such that

$\|\lambda_{k}T^{n_{k}}-I\|=\sup_{x\neq
0}\|\lambda_{k}T^{n_{k}}x-x\|\longrightarrow 0.$

Let $y$ be a nonzero vector of $Y$ and pick $x$ a nonzero vector of $X$ such
that $y=\phi(x)$. Then

$\displaystyle\|\lambda_{k}S^{n_{k}}-I\|$ $\displaystyle=\sup_{y\neq
0}\|\lambda_{k}S^{n_{k}}y-y\|$ $\displaystyle=\sup_{x\neq
0}\|\lambda_{k}S^{n_{k}}\circ\phi(x)-\phi(x)\|$ $\displaystyle=\sup_{x\neq
0}\|\phi(\lambda_{k}T^{n_{k}}x-x)\|$ $\displaystyle\leq\|\phi\|\sup_{x\neq
0}\|\lambda_{k}T^{n_{k}}x-x\|\longrightarrow 0.$

Hence $S$ is uniformly super-rigid on $Y$. ∎

In the following theorem, we give the relationship between the uniform super-
rigidity of an operator and its iterations.

######  Theorem 3.7.

Let $T$ be an operator acting on $X$. Then $T$ is uniformly super-rigid if and
only if $T^{p}$ is uniformly super-rigid, for all $p\geq 2$.

###### Proof.

Let $p$ be a strictly positive integer. Assume that $T$ is uniformly super-
rigid, then there exist a strictly increasing sequence of positive integers
$(n_{k})_{k\in\mathbb{N}}$ and a sequence $(\lambda_{k})_{k\in\mathbb{N}}$ of
numbers such that $\|\lambda_{k}T^{n_{k}}-I\|\longrightarrow 0.$ By Banach-
Steinhaus theorem we have
$\sup_{n\in\mathbb{N}}\|\lambda_{k}T^{n_{k}}\|=M:=<+\infty.$ It follows that

$\displaystyle\|\lambda_{k}^{p}T^{pn_{k}}-I\|$
$\displaystyle\leq\|\lambda_{k}^{p-1}T^{(p-1)n_{k}}+\lambda_{k}^{p-2}T^{(p-2)n_{k}}+\dots+I\|\|\lambda_{k}T^{n_{k}}-I\|$
$\displaystyle\leq\left(\sum_{i=0}^{p-1}M^{i}\right)\|\lambda_{k}T^{n_{k}}-I\|.$

This shows that $T^{p}$ is uniformly super-rigid whenever $T$ is uniformly
super-rigid. ∎

Let $T$ be an operator acting on $X$. If $T$ is super-rigid, then by
Proposition 2.10, there exists some $R>0$ such that
$\sigma_{p}(T)\cup\sigma_{p}(T^{*})\subset\\{z\in\mathbb{C}\mbox{ :
}|z|=R\\}.$ For uniformly super-rigid operators we have a significant
strengthening of this result.

######  Theorem 3.8.

Let $T$ be an operator acting on $X$. If $T$ is uniformly super-rigid, then
there exists $R>$ such that

$\sigma(T)\subset\\{z\in\mathbb{C}\mbox{ : }|z|=R\\}.$

###### Proof.

Assume that $T$ is uniformly super-rigid. Then there exist a sequence
$(\mu_{k})\subset\mathbb{K}$ and a sequence $(n_{k})\subset\mathbb{N}$ such
that $\|\mu_{k}T^{n_{k}}-I\|\longrightarrow 0$ as $k\longrightarrow\infty.$

Without loss of generality, by Lemma 3.4, we may suppose that $\|T\|=1.$ By
Proposition 2.10, we have
$\sigma_{p}(T)\cup\sigma_{p}(T^{*})\subset\\{z\in\mathbb{C}\mbox{ : }|z|=R\\}$
for some $0<R\leq 1$. Since $\sigma_{p}(T)\cap\mathbb{D}=\emptyset$, it
follows that $R=1.$

If $\lambda\in\sigma(T)\cap\mathbb{D}$, then by the previous discussion,
$\lambda$ is necessarily in $\sigma_{a}(T)$, the approximate point spectrum of
$T.$

By contradiction, assume that there exists $\lambda\in\mathbb{K}$ such that
$\lambda\in\sigma(T)\cap\mathbb{D}$. Since $\lambda\in\sigma(T)$, it follows
by the spectral theorem that $\lambda^{p}\in\sigma(T^{p}),$ for all $p\geq 2.$
On the other hand, by Theorem 3.7, the operator $T^{p}$ is uniformly super
rigid for all $p\geq 2.$ Hence $\lambda^{p}$ is in the approximate point
spectrum of $T^{p}$. Thus, for all $p\geq 2,$ there exists a sequence
$(x_{k}^{(p)})_{k\in\mathbb{N}}\subset\\{z\in\mathbb{C}\mbox{ : }|z|=1\\}$
such that $\|Tx_{k}^{(p)})-\lambda x_{k}^{(p)})\|\longrightarrow 0$ as
$k\longrightarrow\infty.$ Using this, one can find sequence
$(y_{k})_{k\in\mathbb{N}}$ of $X$ such that
$(y_{k})_{k\in\mathbb{N}}\subset\\{|z|=1\\}$ and $\|T^{k}y_{k}-\lambda
y_{k}\|<a_{k}$, where $(a_{k})$ is such that
$(|\mu_{k}|(a_{k}+|\lambda|^{k})\longrightarrow 0$ as
$k\longrightarrow\infty$. This implies that

$\|\mu_{k}T^{k}y_{k}\|\leq|\mu_{k}|(a_{k}+|\lambda|^{k})\longrightarrow
0\mbox{ as }k\longrightarrow\infty.$

On the other hand

$\left\lvert\mbox{
}\|\mu_{k}T^{n_{k}}y_{n_{k}}\|-1\right\rvert\leq\|\mu_{k}T^{n_{k}}y_{n_{k}}-y_{n_{k}}\|\leq\|\mu_{k}T^{n_{k}}-I\|\longrightarrow
0\mbox{ as }k\longrightarrow\infty,$

which is a contradiction. ∎

######  Remark 3.9.

The inclusion in Theorem 3.8 could be a strict inclusion. Indeed, let
$\lambda$ be an nonzero number, then $\lambda I$ is a uniformly super-rigid
operator. On the other hand, let $|\lambda|=R$. Then

$\sigma(T)=\\{\lambda\\}\varsubsetneq\\{z\in\mathbb{C}\mbox{ : }|z|=R\\}.$

######  Proposition 3.10.

Let $T$ be an operator acting on $X$. If $T$ is uniformly super-rigid, then it
is invertible.

###### Proof.

Assume that $T$ is uniformly super-rigid. By Theorem 3.8, there exists $R>$
such that $\sigma(T)\subset\\{z\in\mathbb{C}\mbox{ : }|z|=R\\}$. Hence,
$0\notin\sigma(T)$, which means that $T$ is invertible. ∎

##  4\. Finite Dimensional Spaces

In this section, we will characterize the super-recurrence, super-rigidity,
and uniform super-rigidity in finite-dimensional space.

In the following, if $T\in\mathcal{B}(\mathbb{K}^{d})$, then we denote by $A$
a matrix of $T$. Moreover, if $T$ is super-recurrent, let $R>0$ be such that
each component of the spectrum of $T$ intersects the circle
$\\{z\in\mathbb{C}\mbox{ : }|z|=R\\}$, see [2, Theorem 4.1].

In the complex case, we have then the following theorem.

######  Theorem 4.1.

Let $T\in\mathcal{B}(\mathbb{C}^{d})$. Then the following statements are
equivalent$:$

  1. (1)

$T$ is super-recurrent;

  2. (2)

$T$ is super-rigid;

  3. (3)

$T$ is uniformly super-rigid;

  4. (4)

$A$ is similar to a diagonal matrix with entries of the same argument.

###### Proof.

We need only to prove that $(4)\Rightarrow(3)$ and $(1)\Rightarrow(4).$

$(4)\Rightarrow(3)$ : Assume that $A$ is similar to a diagonal matrix with
entries $\lambda_{1},\dots,\lambda_{d}\in\\{z\in\mathbb{C}\mbox{ : }|z|=R\\}$
for some $R>0$. Then there exists an increasing sequence $(n_{k})$ such that
$(R^{-1}\lambda_{i})^{n_{k}}\longrightarrow 1$, for $1\leq i\leq d$. This
implies that that $T$ is uniformly super-rigid.

$(1)\Rightarrow(4)$ : Assume that $T$ is super-recurrent. Using the fact that
$\sigma_{p}(T)=\sigma(T)$ and [2, Theorem 4.1], we conclude that
$\sigma(T)=\\{\lambda_{1},\dots,\lambda_{M}\\}\subset\\{z\in\mathbb{C}\mbox{ :
}|z|=R\\},$ where each $\lambda_{i}$ with multiplicities $m_{i}.$ By Jordan
decomposition theorem, the matrix $A$ is similar to a matrix of the form

$\begin{pmatrix}A_{1}&0&\dots&0\\\ 0&A_{2}&\dots&0\\\
\vdots&\vdots&\ddots&\vdots\\\ 0&0&\dots&A_{M}\end{pmatrix},$

where each $A_{j}$ has the form $A_{j}=\lambda_{j}I_{m_{j}}$ or the form

(4.1) $A_{j}=\begin{pmatrix}\lambda_{j}&1&0&\dots&0\\\
0&\lambda_{j}&1&\dots&0\\\ \vdots&\vdots&\ddots&\ddots&\vdots\\\
0&0&\dots&\lambda_{j}&1\\\ 0&0&\dots&0&\lambda_{j}\end{pmatrix}.$

Assume that there exists a block $A_{j_{0}}$ has the form $(\ref{eq13})$, then
$m_{j}\geq 2$. Using [2, Proposition 3.7], we conclude that the operator
presented by the matrix

$B=\begin{pmatrix}\lambda_{j_{0}}&1\\\ 0&\lambda_{j_{0}}\end{pmatrix}$

is super-recurrent on $\mathbb{C}^{2}$. By straightforward induction, we have
for all $n\in\mathbb{N}$

$B^{n}=\begin{pmatrix}\lambda_{j_{0}}^{n}&n\lambda_{j_{0}}^{n-1}\\\
0&\lambda_{j_{0}}^{n}\end{pmatrix}.$

Let $(z_{1},z_{2})\in\mathbb{C}^{2}$ be a super-recurrent vector for $B$ with
$z_{2}\neq 0$. Then there exist $(\mu_{k})\subset\mathbb{C}$ and a strictly
increasing sequence of positive integers $(n_{k})$ such that

$\mu_{k}\begin{pmatrix}\lambda_{j_{0}}^{n_{k}}&n\lambda_{j_{0}}^{n_{k}-1}\\\
0&\lambda_{j_{0}}^{n_{k}}\end{pmatrix}\begin{pmatrix}z_{1}\\\
z_{2}\end{pmatrix}\longrightarrow(z_{1},z_{2}).$

which implies that

$\mu_{k}\lambda_{j_{0}}^{n_{k}}z_{2}\longrightarrow z_{2}\hskip
5.69046pt\mbox{ and }\hskip
5.69046pt\mu_{k}\lambda_{j_{0}}^{n_{k}}z_{1}+\mu_{k}n_{k}\lambda_{j_{0}-1}^{n_{k}}z_{2}\longrightarrow
z_{1}.$

This is impossible since by hypothesis $z_{2}\neq 0.$ ∎

In the real case, we have the following theorem.

######  Theorem 4.2.

Let $T\in\mathbb{R}^{d}$. The following assertions are equivalent$:$

  1. (1)

$T$ is super-recurrent;

  2. (2)

$T$ is super-rigid;

  3. (3)

$T$ is uniformly super-rigid;

  4. (4)

$A$ is similar to a matrix of the form

(4.2) $\begin{pmatrix}A_{1}&0&\dots&0\\\ 0&A_{2}&\dots&0\\\
\vdots&\vdots&\ddots&\vdots\\\ 0&0&\dots&A_{M}\end{pmatrix},$

where each $A_{k}$, $1\leq k\leq M$, is $1\times 1$ matrix with entry $R$ or
$-R$, or $2\times 2$ matrix of the form

(4.3) $\begin{pmatrix}a&b\\\ -b&a\end{pmatrix},\hskip 8.5359pta\mbox{,
}b\in\mathbb{R}.$

###### Proof.

We need only to prove that $(4)\Rightarrow(3)$ and $(1)\Rightarrow(4).$

$(1)\Rightarrow(4)$ : Assume that $T$ is super-recurrent. By the Jordan
decomposition, the matrix $A$ is similar to a matrix which has the form of
$(\ref{eq24}).$ We have then two cases.

(1) : Each $A_{k}$ has the form $\lambda_{k}I_{m_{k}}$. In this case, it
suffices to do the same method which was used in the complex case to conclude.

(2) : Each $A_{k}$ has the form

(4.4) $\begin{pmatrix}B&I_{2}&0&\dots&0\\\ 0&B&I_{2}&\dots&0\\\
\vdots&\vdots&\ddots&\ddots&\vdots\\\ 0&0&\dots&B&I_{2}\\\
0&0&\dots&0&B\end{pmatrix},$

where $B$ has the form of $(\ref{eq34})$. We claim that $A_{k}=B$ for all $k$,
$1\leq k\leq M.$ Indeed, if there exist $k_{0}$ which different to $B$, then
by [2, Proposition 3.7], the operator presented by the matrix

$C=\begin{pmatrix}B&I_{2}\\\ 0&B\end{pmatrix}$

is super-recurrent in $\mathbb{R}^{4}$. The same argument used in the complex
case lead us to a contradiction.

Hence, if $T$ is super-recurrent, then $A$ is similar to a matrix which has
the form of $(\ref{eq24})$.
$(4)\Rightarrow(3)$ : It is not difficult that each matrix of that form is
uniformly super-rigid. ∎

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considered above, we have

$\mathbf{K}\cap(\mathrm{U}^{-}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}=\begin{cases}(\mathrm{U}^{-}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}&\text{if
$\mathbf{K}\cap(\mathrm{U}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}=(\mathrm{U}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}$;}\\\
(\mathrm{U}^{-}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}_{\mathrm{red}}&\text{if
$\mathbf{K}\cap(\mathrm{U}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}=(\mathrm{U}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}_{\mathrm{red}}$.}\end{cases}$

As a conclusion, we have either

$\mathbf{K}\cap(\mathrm{C}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}=(\mathrm{C}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}\quad\text{or}\quad\mathbf{K}\cap(\mathrm{C}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}=(\mathrm{C}_{3,\mathbb{F}_{2}})_{\mathrm{red}}^{\mathbb{Z}/2\mathbb{Z}}.$

Using once again the fact that $\mathbf{K}(\mathbb{F}_{2})$ contains the image
of $n$, and the description of the open cover of
$(\mathrm{SL}_{3,\mathbb{Z}})^{\mathbb{Z}/2\mathbb{Z}}$ considered above, in
the first case we deduce that
$\mathbf{K}=(\mathrm{SL}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}$, and in
the second case that
$\mathbf{K}=(\mathrm{SL}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}_{\mathrm{red}}$.
∎

The second lemma uses the notion of dilatation (or affine blow-up) from
[MRR20]. We will apply this construction to the scheme
$(\mathrm{SL}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}$, the principal
subscheme $(\mathrm{SL}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}$, and
either the closed subscheme
$(\mathrm{SL}_{3,\mathbb{F}_{2}})_{\mathrm{red}}^{\mathbb{Z}/2\mathbb{Z}}\subset(\mathrm{SL}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}$
or the closed subscheme
$(\mathrm{T}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}\subset(\mathrm{SL}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}$.
The construction of [MRR20] provides two affine schemes

$\mathrm{Bl}^{(\mathrm{SL}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}}_{(\mathrm{SL}_{3,\mathbb{F}_{2}})_{\mathrm{red}}^{\mathbb{Z}/2\mathbb{Z}}}\Bigl{(}(\mathrm{SL}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}\Bigr{)}\quad\text{and}\quad\mathrm{Bl}^{(\mathrm{SL}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}}_{(\mathrm{T}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}}\Bigl{(}(\mathrm{SL}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}\Bigr{)}$

endowed with canonical morphisms to
$(\mathrm{SL}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}$. Moreover, by the
universal property of dilatations (see [MRR20, Proposition 2.6]) there exists
a canonical morphism

(6.5)
$\mathrm{Bl}^{(\mathrm{SL}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}}_{(\mathrm{T}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}}\Bigl{(}(\mathrm{SL}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}\Bigr{)}\to\mathrm{Bl}^{(\mathrm{SL}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}}_{(\mathrm{SL}_{3,\mathbb{F}_{2}})_{\mathrm{red}}^{\mathbb{Z}/2\mathbb{Z}}}\Bigl{(}(\mathrm{SL}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}\Bigr{)}$

over $(\mathrm{SL}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}$.

######  Lemma 6.11.

The morphism (6.5) restricts to an isomorphism over
$(\mathrm{C}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}$.

###### Proof.

By compatibility of dilatations with flat base change (see [MRR20, Lemma
2.7]), we have

$\mathrm{Bl}^{(\mathrm{SL}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}}_{(\mathrm{SL}_{3,\mathbb{F}_{2}})_{\mathrm{red}}^{\mathbb{Z}/2\mathbb{Z}}}\Bigl{(}(\mathrm{SL}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}\Bigr{)}\times_{(\mathrm{SL}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}}(\mathrm{C}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}\cong\mathrm{Bl}^{(\mathrm{C}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}}_{(\mathrm{C}_{3,\mathbb{F}_{2}})_{\mathrm{red}}^{\mathbb{Z}/2\mathbb{Z}}}\Bigl{(}(\mathrm{C}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}\Bigr{)},$

with

$\mathrm{Bl}^{(\mathrm{C}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}}_{(\mathrm{C}_{3,\mathbb{F}_{2}})_{\mathrm{red}}^{\mathbb{Z}/2\mathbb{Z}}}\Bigl{(}(\mathrm{C}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}\Bigr{)}=\\\
\mathrm{Bl}^{\mathrm{Spec}(\mathbb{F}_{2}[x,y,z^{\pm
1},x^{\prime},y^{\prime}]/(x^{2},(x^{\prime})^{2}))}_{\mathrm{Spec}(\mathbb{F}_{2}[y,z^{\pm
1},y^{\prime}])}\Bigl{(}\mathrm{Spec}(\mathbb{Z}_{2}[x,y,z^{\pm
1},x^{\prime},y^{\prime}]/(x^{2}-2y,(x^{\prime})^{2}-2y^{\prime}))\Bigr{)}.$

Now one checks that the right-hand side identifies with

$\mathrm{Spec}(\mathbb{Z}_{2}[u,y,z^{\pm
1},u^{\prime},y^{\prime}]/(2u^{2}-y,2(u^{\prime})^{2}-y^{\prime})),$

with the morphism to $\mathrm{Spec}(\mathbb{Z}_{2}[x,y,z^{\pm
1},x^{\prime},y^{\prime}]/(x^{2}-2y,(x^{\prime})^{2}-2y^{\prime}))$
corresponding to the ring homomorphism

$\mathbb{Z}_{2}[x,y,z^{\pm
1},x^{\prime},y^{\prime}]/(x^{2}-2y,(x^{\prime})^{2}-2y^{\prime})\to\mathbb{Z}_{2}[u,y,z^{\pm
1},u^{\prime},y^{\prime}]/(2u^{2}-y,2(u^{\prime})^{2}-y^{\prime})$

defined by $x\mapsto 2u$ and $x^{\prime}\mapsto 2u^{\prime}$. (In fact the
given scheme satisfies the universal property of [MRR20, Proposition 2.6].)
This description shows that the restriction of the canonical morphism

$\mathrm{Bl}^{(\mathrm{C}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}}_{(\mathrm{C}_{3,\mathbb{F}_{2}})_{\mathrm{red}}^{\mathbb{Z}/2\mathbb{Z}}}\Bigl{(}(\mathrm{C}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}\Bigr{)}\to(\mathrm{C}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}$

to $\mathrm{Spec}(\mathbb{F}_{2})$ factors through
$\mathrm{Spec}(\mathbb{F}_{2}[z^{\pm 1}])$ since $x,x^{\prime},y,y^{\prime}$
get sent to $0$; by the universal property of dilatations (and again
compatibility of dilatations with flat base change) we deduce a canonical
morphism

$\mathrm{Bl}^{(\mathrm{SL}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}}_{(\mathrm{SL}_{3,\mathbb{F}_{2}})_{\mathrm{red}}^{\mathbb{Z}/2\mathbb{Z}}}\Bigl{(}(\mathrm{SL}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}\Bigr{)}\times_{(\mathrm{SL}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}}(\mathrm{C}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}\to\\\
\mathrm{Bl}^{(\mathrm{SL}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}}_{(\mathrm{T}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}}\Bigl{(}(\mathrm{SL}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}\Bigr{)}\times_{(\mathrm{SL}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}}(\mathrm{C}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}.$

We claim that this morphism is an inverse to the restriction of (6.5). Indeed,
both are flat affine schemes over $\mathrm{Spec}(\mathbb{Z}_{2})$ with generic
fiber equal to $(\mathrm{SL}_{3,\mathbb{Q}_{2}})^{\mathbb{Z}/2\mathbb{Z}}$, so
to prove the claim it is enough to see that their global sections inside those
of $(\mathrm{SL}_{3,\mathbb{Q}_{2}})^{\mathbb{Z}/2\mathbb{Z}}$ coincide. But
each containment is implied by the existence of a map of spectra in the
opposite direction. ∎

###### Proof of Proposition 6.9.

Let $\pi$ be as in the statement, and assume that the schematic image of
$\pi_{|\mathrm{Spec}(\mathbb{F}_{2})}$ is not
$(\mathrm{SL}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}$. By Lemma 6.10, we
conclude that the schematic image of this morphism is
$(\mathrm{SL}_{3,\mathbb{F}_{2}})_{\mathrm{red}}^{\mathbb{Z}/2\mathbb{Z}}$. By
the universal property of dilatations (see [MRR20, Proposition 2.6]), we then
have a unique factorization

$\pi:\mathbf{H}\to\mathrm{Bl}^{(\mathrm{SL}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}}_{(\mathrm{SL}_{3,\mathbb{F}_{2}})_{\mathrm{red}}^{\mathbb{Z}/2\mathbb{Z}}}\Bigl{(}(\mathrm{SL}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}\Bigr{)}\to(\mathrm{SL}_{3,\mathbb{Z}_{2}})^{\mathbb{Z}/2\mathbb{Z}}.$

Using Lemma 6.11, we deduce that the restriction of
$\pi_{|\mathrm{Spec}(\mathbb{F}_{2})}$ to
$(\mathrm{C}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}$ factors through
$(\mathrm{T}_{3,\mathbb{F}_{2}})^{\mathbb{Z}/2\mathbb{Z}}$, yielding a
contradiction. ∎

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  *[IoT]: Internet of Things
  *[SATCOMs]: Satellite-based Communication
  *[SATCOM]: Satellite-based Communication
  *[GNSS]: Global Navigation Satellite System
  *[ESA]: European Space Agency
  *[AES]: Advanced Encryption Standard
  *[SNR]: Signal-to-Noise Ratio
  *[SOP]: Secrecy Outage Probability
  *[CSI]: Channel State Information
  *[GPS]: Global Positioning System
  *[NOAA]: National Oceanic and Atmospheric Administration
  *[SDR]: Software Defined Radio
  *[ML]: Machine Learning
  *[CNR]: Carrier-to-Noise Ratio
  *[AGC]: Automatic Gain Control
  *[AWGN]: Additive White Gaussian Noise
  *[CDMA]: Code Division Multiple Access
  *[IRS]: Intelligent Reflecting Surfaces
  *[IRSs]: Intelligent Reflecting Surfaces
  *[AI]: Artificial Intelligence
  *[VSAT]: Very Small Aperture Terminal
  *[ECC]: Elliptic Curve Cryptography
  *[RSA]: Rivest-Shamir-Adleman
  *[QKD]: Quantum Key Distribution
  *[OTP]: One-Time-Pad
  *[P2P]: Point to Point
  *[MITM]: Man-In-The-Middle
  *[DoS]: Denial of Service
  *[UAV]: Unmanned Aerial Vehicles
  *[SDN]: Software Defined Networking
  *[IoST]: Internet of Space Things
  *[NIST]: National Institute of Standards and Technology


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