# Isometries of combinatorial Tsirelson spaces

Natalia Maślany  Jagiellonian University, Doctoral School of Exact and Natural
Sciences, Faculty of Mathematics and Computer Science, Institute of
Mathematics, Łojasiewicza 6, 30-348 Kraków, Poland and Institute of
Mathematics, Czech Academy of Sciences, Žitná 25, 115 67 Prague 1, Czech
Republic<EMAIL_ADDRESS>
###### Abstract.

We extend existing results that characterize isometries on the Tsirelson-type
spaces $T\big{[}\frac{1}{n},\mathcal{S}_{1}\big{]}$
($n\in\mathbb{N},n\geqslant 2$) to the class $T[\theta,\mathcal{S}_{\alpha}]$
($\theta\in\big{(}0,\frac{1}{2}\big{]}$, $1\leqslant\alpha<\omega_{1}$), where
$\mathcal{S}_{\alpha}$ denote the Schreier families of order $\alpha$. We
prove that every isometry on $T[\theta,\mathcal{S}_{1}]$
($\theta\in\big{(}0,\frac{1}{2}\big{]}$) is determined by a permutation of the
first $\lceil{\theta}^{-1}\rceil$ elements of the canonical unit basis
followed by a possible sign-change of the corresponding coordinates together
with a sign-change of the remaining coordinates. Moreover, we show that for
the spaces $T[\theta,\mathcal{S}_{\alpha}]$
($\theta\in\big{(}0,\frac{1}{2}\big{]}$, $2\leqslant\alpha<\omega_{1}$) the
isometries exhibit a more rigid character, namely, they are all implemented by
a sign-change operation of the vector coordinates.

###### Key words and phrases:

combinatorial spaces, combinatorial Tsirelson spaces, higher-order Schreier
families, isometry group, regular families, Schreier families

###### 2020 Mathematics Subject Classification:

46B04, 46B25, 46B45

The author was supported by GAČR grant GF20-22230L and received funding in the
scope Research Support Module at the Jagiellonian University in Kraków.

##  1\. Introduction and the main result

The well-known Tsirelson space $T$ (in the setting of Figiel and Johnson [4],
i.e., the dual of the space constructed by Tsirelson [6], the first example of
a space containing no isomorphic copies of $c_{0}$ or $\ell_{p}$ for
$1\leqslant p<\infty$) may be regarded as special instance of a space from a
double-parameter family of Banach spaces $T[\theta,\mathcal{S}_{\alpha}]$
($\theta\in\big{(}0,\frac{1}{2}\big{]}$, $1\leqslant\alpha<\omega_{1}$), where
$\alpha$ is a countable ordinal and $S_{\alpha}$ is the Schreier family of
order $\alpha$. We emphasize that in this paper the Schreier families are
defined to satisfy some natural additional conditions (see the discussion
after Definition 3 in Preliminaries). For brevity, we call members of this
family _combinatorial Tsirelson spaces_ , which appears to be in line with the
terminology used, e.g., in [2]. (These are, of course, special cases of the
so-called mixed Tsirelson spaces whose definition allows the parameter
$\theta$ to vary, but by employing this name we want to emphasize the
underlying family of sets rather than the numeric parameter.)

The aim of this paper is to delineate the structure of isometries on
combinatorial Tsirelson spaces. We refer to the recent excellent survey [1]
for further results and references concerning the problem of characerization
of isometries on Banach (sequence) spaces.

In [1, Theorem 4.1] the authors provide a characterization of (linear)
isometries of the spaces $T\big{[}\frac{1}{n},\mathcal{S}_{1}\big{]}$ for
$n\in\mathbb{N}$, $n\geqslant 2$, which we take as a departure point for our
considerations and extend it to the whole scale of spaces
$T\big{[}\theta,\mathcal{S}_{1}\big{]}$
($\theta\in\big{(}0,\frac{1}{2}\big{]}\big{)}.$ Let us then record the first
main result. (In this paper, all considered Banach spaces are real; by an
_isometry_ we understand a linear isometry.)

######  Theorem A.

Let $\theta\in\big{(}0,\frac{1}{2}\big{]}$. If $U\colon
T\big{[}\theta,\mathcal{S}_{1}\big{]}\to
T\big{[}\theta,\mathcal{S}_{1}\big{]}$ is an isometry, then

$Ue_{i}=\left\\{\begin{array}[]{ll}\varepsilon_{i}e_{\pi(i)},&1\leqslant
i\leqslant\lceil\theta^{-1}\rceil\\\
\varepsilon_{i}e_{i},&i>\lceil\theta^{-1}\rceil\end{array}\right.\quad(i\in\mathbb{N})$

for some $\\{-1,1\\}$-valued sequence $(\varepsilon_{i})_{i=1}^{\infty}$ and a
permutation $\pi$ of $\big{\\{}1,2,\ldots,\lceil\theta^{-1}\rceil\big{\\}}.$

(Here $(e_{i})_{i=1}^{\infty}$ is the standard unit vector basis of
$T\big{[}\theta,\mathcal{S}_{1}\big{]}$ and $\lceil\theta^{-1}\rceil$ is the
ceil of $\theta^{-1}$, i.e., the least integer that $\theta^{-1}$ does not
exceed.)

Then we answer the question from [1]. Indeed, we characterize the linear
isometries of the spaces $T[\theta,\mathcal{S}_{\alpha}]$
($\theta\in\big{(}0,\frac{1}{2}\big{]}$, $2\leqslant\alpha<\omega_{1}$) by
proving the following second main result:

######  Theorem B.

Let $\theta\in\big{(}0,\frac{1}{2}\big{]}$ and let $\alpha\geqslant 2$ be a
countable ordinal. Then an operator $U\colon T[\theta,\mathcal{S}_{\alpha}]\to
T[\theta,\mathcal{S}_{\alpha}]$ is an isometry if and only if
$Ue_{i}=\varepsilon_{i}e_{i}$ for $i\in\mathbb{N}$ and some
$\\{-1,1\\}$-valued sequence $(\varepsilon_{i})_{i=1}^{\infty}.$

Let us record the following observation that we draw directly from the proofs
of Theorems A–B, which may be of independent interest.

######  Remark 1.

Every isometry on $T[\theta,\mathcal{S}_{\alpha}]$
($\theta\in\big{(}0,\frac{1}{2}\big{]}$, $1\leqslant\alpha<\omega_{1}$) is
surjective.

### Acknowledgement.

We would like to thank Kevin Beanland for helpful conversations and feedback
concerning the present work.

##  2\. Preliminaries

###  2.1. Combinatorial spaces

We will denote by $(e_{i})_{i=1}^{\infty}$ the standard unit vector basis of
$c_{00}$ and by $[\mathbb{N}]^{<\omega}$ the family of finite subsets of
$\mathbb{N}.$ For the sets $F_{1},F_{2}\in[\mathbb{N}]^{<\omega}$ we use the
following notation: $F_{1}<F_{2}$, whenever $\max F_{1}<\min F_{2}$ and in
such case we say that these sets are consecutive. Moreover, for
$n\in\mathbb{N},$ we write $F_{1}<n$ instead of $F_{1}<\\{n\\}.$

######  Definition 2.

A family $\mathcal{F}\subset[\mathbb{N}]^{<\omega}$ is regular, whenever it is
simultaneously

  * $\bullet$

hereditary ($F\in\mathcal{F}$ and $G\subset F\implies G\in\mathcal{F}$);

  * $\bullet$

spreading ($\\{l_{1},l_{2},\dots,l_{n}\\}\in\mathcal{F}$ and $l_{i}\leqslant
k_{i}\implies\\{k_{1},k_{2},\dots,k_{n}\\}\in\mathcal{F}$);

  * $\bullet$

compact as a subset of the Cantor set $\\{0,1\\}^{\mathbb{N}}$ via the natural
identification of $F\in\mathcal{F}$ with

$\chi_{F}=\sum_{i\in F}e_{i}\in\\{0,1\\}^{\mathbb{N}}.$

If $\mathcal{F}$ is a regular family, we say that $F\in\mathcal{F}$ is
maximal, whenever there is no $n\in\mathbb{N}$ with $\max F<n$ so that
$F\cup\\{n\\}\in\mathcal{F}$. The simplest examples of regular families
include

$\mathcal{A}_{n}:=\big{\\{}F\in[\mathbb{N}]^{<\omega}\colon|F|\leqslant
n\big{\\}}\quad(n\in\mathbb{N})$

i.e., for a given $n\in\mathbb{N}$, the family comprising subsets of
$\mathbb{N}$ of cardinality at most $n$. We employ these families to define
the family of Schreier sets in the following way.

######  Definition 3.

Given a countable ordinal $\alpha$, we define inductively the Schreier family
of order $\alpha$ as follows:

  * $\bullet$

$\mathcal{S}_{0}:=\mathcal{A}_{1};$

  * $\bullet$

if $\alpha$ is a successor ordinal, i.e., $\alpha=\beta+1$ for some
$\beta<\omega_{1}$, then

$\mathcal{S}_{\alpha}:=\Bigg{\\{}\bigcup_{i=1}^{d}S^{i}_{\beta}\colon
d\leqslant
S^{1}_{\beta}<S^{2}_{\beta}<\cdots<S^{d}_{\beta},\,\,\big{\\{}S^{i}_{\beta}\big{\\}}_{i=1}^{d}\subset\mathcal{S}_{\beta}\text{
and }d\in\mathbb{N}\Bigg{\\}}\cup\big{\\{}\emptyset\big{\\}};$

  * $\bullet$

if $\alpha$ is a non-zero limit ordinal and
$\left(\alpha_{n}\right)_{n=1}^{\infty}$ is a fixed strictly increasing
sequence of successor ordinals converging to $\alpha$ with
$\mathcal{S}_{\beta_{n}}\subset\mathcal{S}_{\beta_{n+1}}$ for all
$n\in\mathbb{N}$, where $\alpha_{n}=\beta_{n}+1$ for all $n\in\mathbb{N}$, we
set

$\mathcal{S}_{\alpha}:=\big{\\{}S_{\alpha_{n}}\in[\mathbb{N}]^{<\omega}\colon
S_{\alpha_{n}}\in\mathcal{S}_{\alpha_{n}},\,\,n\leqslant\min
S_{\alpha_{n}}\text{ for some n}\big{\\}}\cup\big{\\{}\emptyset\big{\\}}.$

We emphasize that in the case where $\alpha$ is a limit ordinal, we require
the sequence $(\alpha_{n})_{n=1}^{\infty}$ cofinal in $\alpha$ to comprise
successor ordinals as needed in the proof of Theorem B. We may (and do) also
assume that $S_{\alpha_{n}}\subset S_{\alpha_{n+1}}$ for all $n\in\mathbb{N}$,
which we will also utilize in the proof of Theorem B. Indeed, repeating the
proof of [3, Proposition 3.2.] in the case of Schreier families
$\big{\\{}S_{\xi}\big{\\}}_{\xi<\omega_{1}}$ which are multiplicative in the
sense of [3] we obtain the required result also in the case of Schreier
families. Elements of $\mathcal{S}_{\alpha}$ are called
$\mathcal{S}_{\alpha}$_-sets_.

Note that the Schreier families
$\left\\{\mathcal{S}_{\alpha}\right\\}_{\alpha<\omega_{1}}$ do depend on the
choice of the sequences $\left\\{\alpha_{n}\right\\}_{n=1}^{\infty}$
converging to each limit ordinal $\alpha$. It is a well-known fact
([3][Proposition 3.2] or [5]) that they are always regular families.

###  2.2. Combinatorial Tsirelson spaces

For a regular family $\mathcal{F}$ and $\theta\in\big{(}0,\frac{1}{2}\big{]}$,
we define the Banach space $T[\theta,\mathcal{F}]$ that we shall later
specialize to a combinatorial Tsirelson space $T[\theta,\mathcal{S}_{\alpha}]$
for some countable ordinal $\alpha$.

For a vector $x=(a_{1},a_{2},\ldots,a_{n})\in c_{00}$ and a finite set
$E\subset\mathbb{N}$, we use the same symbol $Ex$ to denote the projection of
$x$ onto the space $[e_{i}\colon i\in E],$ given by

(2.1) $E\Bigg{(}\sum_{i=1}^{n}a_{i}e_{i}\Bigg{)}=\sum_{i\in E}a_{i}e_{i}.$

If the cardinality of set $E$ is equal $k,$ then we say that $E$ is a _$k$
-element projection_.

We denote by $\|\cdot\|_{0}$ the supremum norm on $c_{00}.$ Suppose that for
some $n\in\mathbb{N}$ the norm $\|\cdot\|_{n}$ has been defined. Let

$\|x\|_{n+1}=\max\big{\\{}\|x\|_{n},\|x\|_{T_{n}}\big{\\}}\quad(n\in\mathbb{N}),$

where

$\|x\|_{T_{n}}=\sup\bigg{\\{}\theta\sum_{i=1}^{d}\big{\|}E_{i}x\big{\|}_{n}\colon
E_{1}<\cdots<E_{d},\,d\in\mathbb{N},\,\\{E_{i}\\}_{i=1}^{d}\subset[\mathbb{N}]^{<\omega},\,\\{\min
E_{i}\\}_{i=1}^{d}\in\mathcal{F}\bigg{\\}}.$

We define the norm
$\|x\|_{\theta,\mathcal{F}}:=\sup_{n\in\mathbb{N}}\|x\|_{n}$ and denote by
$T[\theta,\mathcal{F}]$ the completion of $c_{00}$ with respect to it.

A proof by induction shows that this norm is majorized by the $\ell_{1}$-norm
and that satisfies the following implicit formula for $x\in
T[\theta,\mathcal{F}]$:

(2.2)
$\|x\|_{\theta,\mathcal{F}}=\max\big{\\{}\|x\|_{\infty},\|x\|_{T}\big{\\}},$

where

$\|x\|_{T}=\sup\bigg{\\{}\theta\sum_{i=1}^{d}\big{\|}E_{i}x\big{\|}_{\theta,\mathcal{F}}:E_{1}<\cdots<E_{d},\,d\in\mathbb{N},\,\\{E_{i}\\}_{i=1}^{d}\subset[\mathbb{N}]^{<\omega},\,\\{\min
E_{i}\\}_{i=1}^{d}\in\mathcal{F}\bigg{\\}}.$

It follows easily from the definition that the unit vectors
$(e_{i})_{i=1}^{\infty}$ form an 1-unconditional basis of the space
$T[\theta,\mathcal{S}_{\alpha}]$ for a countable ordinal $\alpha$.

For $x_{1},x_{2}\in c_{00}$, we write $x_{1}<x_{2}$ whenever
$\operatorname{supp}\,x_{1}<\operatorname{supp}\,x_{2}$ and for
$n\in\mathbb{N}$ we simplify the notation of $\operatorname{supp}\,x_{1}<n$ to
$x_{1}<n.$

In this paper, we will use the following convention: we say that the norm of
an element $x\in T[\theta,\mathcal{F}]$ is given by sets
$E_{1}<E_{2}<\cdots<E_{d}$ for some $d\in\mathbb{N}$ (with $\\{\min
E_{i}\\}_{i=1}^{d}\in\mathcal{F}$) precisely when

$\|x\|_{\theta,\mathcal{F}}=\theta\cdot\sum_{i=1}^{d}\big{\|}E_{i}x\big{\|}_{\theta,\mathcal{F}}.$

It follows easily from the definition that if $(x_{i})_{i=1}^{d}$ is a block
sequence in $T[\theta,\mathcal{F}]$ (i.e., $x_{1}<x_{2}<\ldots<x_{d}$) with
$\\{\min\text{supp}\,x_{i}\\}_{i=1}^{d}\in\mathcal{F}$ we have

(2.3)
$\bigg{\|}\sum_{i=1}^{n}x_{i}\bigg{\|}_{\theta,\mathcal{F}}\geqslant\theta\sum_{i=1}^{n}\|x_{i}\|_{\theta,\mathcal{F}}.$

It follows that in the case of the space $T[\theta,\mathcal{S}_{1}]$, (2.3)
yields that if $d\leqslant x_{1}<\cdots<x_{d}$, then

(2.4)
$\bigg{\|}\sum_{i=1}^{d}x_{i}\bigg{\|}_{\theta,\mathcal{S}_{1}}\geqslant\theta\sum_{i=1}^{d}\|x_{i}\|_{\theta,\mathcal{S}_{1}}$

For brevity, we write $\|\cdot\|$ instead of
$\|\cdot\|_{\theta,\mathcal{S}_{\alpha}},$ where
$\theta\in\big{(}0,\frac{1}{2}\big{]}$, $1\leqslant\alpha<\omega_{1}.$ Let us
record the following lemma that we shall later use extensively.

######  Lemma 4.

Let $\theta\in\big{(}0,\frac{1}{2}\big{]}$ and $1\leqslant\alpha<\omega_{1}$.
Suppose that $x\in T[\theta,\mathcal{S}_{\alpha}]$ is a vector whose
coordinates are either $0$ or $1$. Fix $k\geqslant 2.$ Let $E$ be a
$k$-element set. Then the norm given by the set $E$ is not greater than the
norm given by the $k-1$ many singleton projections. (For simplicity we assume
we only project to non-zero coordinates and all of these projections are
admissible.)

###### Proof.

Fix $k\geqslant 2.$ The norm given by $k-1$ many $1$-element projections is
$\theta(k-1)$. Let $E$ be a $k$-element set and let $E_{1}<E_{2}<\dots<E_{d}$.
Then

$\begin{split}\theta\cdot\|Ex\|&=\theta\cdot\max\bigg{\\{}1,\,\theta\cdot\sum_{i=1}^{d}\|E_{i}x\|\bigg{\\}}\\\
&\leqslant\theta\cdot\max\bigg{\\{}1,\,\theta\cdot\sum_{i=1}^{d}\|E_{i}x\|_{\ell_{1}}\bigg{\\}}\\\
&=\theta\cdot\max\bigg{\\{}1,\,\theta\cdot\sum_{i=1}^{d}|E_{i}|\bigg{\\}}\\\
&\leqslant\theta\cdot\max\\{1,\,\theta k\\}.\end{split}$

Since $\theta\in\big{(}0,\frac{1}{2}\big{]}$ and $k\geqslant 2$, so
$\theta(k-1)\geqslant\theta^{2}k.$ ∎

######  Lemma 5.

Let $\theta\in\big{(}0,\frac{1}{2}\big{]}$ and $1\leqslant\alpha<\omega_{1}$.
Suppose that $x\in T[\theta,\mathcal{S}_{\alpha}]$ is given by the formula

$x=e_{i}+\sum_{j\in A}e_{j},$

where $A=\big{\\{}j_{1},j_{2},\ldots,j_{|A|}\big{\\}}$ is an
$\mathcal{S}_{\alpha}$-set with $A>i$ and
$|A|\geqslant\lceil\theta^{-1}\rceil$. If $A\cup\\{i\\}$ is not an
$\mathcal{S}_{\alpha}$-set, then $\|x\|=\theta\cdot|A|.$

###### Proof.

Since $A$ is an $\mathcal{S}_{\alpha}$-set, we have

$\|x\|\geqslant\theta\cdot|A|\geqslant 1=\|x\|_{\infty}.$

Assume that there are sets $E_{1}<E_{2}<\dots<E_{d}$ which give norm of $x$
greater than $\theta\cdot|A|$. If $\min E_{1}>i,$ then by Lemma 4 we arrive at
a contradiction. If $\min E_{1}\leqslant i,$ then we have at least one
$k$-element set ($k\geqslant 2$), because by the hypothesis,
$A\cup\\{i\\}\notin S_{\alpha}.$ This contradicts Lemma 4 likewise.

∎

##  3\. Isometries on $T[\theta,\mathcal{S}_{1}]$ spaces for
$\theta\in\big{(}0,\frac{1}{2}\big{]}$

For the space $T[\theta,\mathcal{S}_{1}]$ $\big{(}\theta\in(0,1)\big{)}$ the
norm given by the formula (2.2) takes the following form:

(3.1) $\|x\|=\max\big{\\{}\|x\|_{\infty},\|x\|_{T}\big{\\}},$

where

(3.2)
$\|x\|_{T}=\sup\bigg{\\{}\theta\sum_{i=1}^{d}\big{\|}E_{i}x\big{\|}:d\leqslant
E_{1}<E_{2}<\cdots<E_{d},\,d\in\mathbb{N},\,\\{E_{i}\\}_{i=1}^{d}\subset[\mathbb{N}]^{<\omega}\bigg{\\}}.$

We are now ready to prove Theorem A; the proof emulates the one of [1, Theorem
4.1].

###### Proof.

Let $Ue_{n}:=\sum_{i=1}^{\infty}a^{n}_{i}e_{i}$ ($n\in\mathbb{N}$).

  * _Claim 1_. For any $n\geqslant\lceil\theta^{-1}\rceil$ we have $U([e_{1},e_{2},\ldots,e_{n}])\subset[e_{1},e_{2},\ldots,e_{n}].$

Let $n\geqslant\lceil\theta^{-1}\rceil$ and
$j\in\big{\\{}1,2,\dots,n\big{\\}}.$ Define
$x:=\sum_{i={n}+1}^{\infty}a^{j}_{i}e_{i}$ and fix $\varepsilon>0.$

As $(Ue_{i})_{i=1}^{\infty}$ is weakly null we may find indices
$n<j_{1}<j_{2}<\cdots<j_{n}$ and vectors

$n+1\leqslant x^{\prime}<y_{1}<y_{2}<\cdots<y_{n},$

so that $\|x^{\prime}-x\|<\varepsilon$ and
$\big{\|}Ue_{j_{i}}-y_{i}\big{\|}<\varepsilon$ for $1\leqslant i\leqslant{n}$.

By Lemma 5 and since $U$ is an isometry we have

$\theta
n=\Bigg{\|}e_{j}+\sum_{i=1}^{n}e_{j_{i}}\Bigg{\|}=\Bigg{\|}\sum_{i=1}^{n}a^{j}_{i}e_{i}+x+\sum_{i=1}^{n}Ue_{j_{i}}\Bigg{\|}$

Hence by triangle inequality we obtain

(3.3)
$\Bigg{\|}\sum_{i=1}^{n}a^{j}_{i}e_{i}+x^{\prime}+\sum_{i=1}^{n}y_{i}\Bigg{\|}\leqslant\theta
n+({n}+1)\varepsilon$

On the other hand

(3.4)
$\Bigg{\|}\sum_{i=1}^{n}a^{j}_{i}e_{i}+x^{\prime}+\sum_{i=1}^{n}y_{i}\Bigg{\|}\geqslant\theta\Bigg{(}\|x^{\prime}\|+\sum_{i=1}^{n}\|y_{i}\|\Bigg{)}>\theta\Big{(}\|x^{\prime}\|+n(1-\varepsilon)\Big{)}.$

The first inequality follows from (2.4) and second by the fact that
$\|Ue_{i}\|=1$ for any $i\in\mathbb{N}.$ Therefore by (3.3) and (3.4) we
obtain

$\|x\|\leqslant\|x^{\prime}-x\|+\|x^{\prime}\|<\varepsilon+\Big{(}\big{(}{n}+1\big{)}\theta^{-1}+{n}\Big{)}\varepsilon.$

Since $\varepsilon$ was arbitrary, we get $\|x\|=0.$ Consequently,

$U\big{[}e_{1},e_{2},\ldots,e_{n}\big{]}\subset\big{[}e_{1},e_{2},\ldots,e_{n}\big{]}.$

  * _Claim 2_. There exists a permutation $\pi$ of $\big{\\{}1,2,\ldots,{\lceil\theta^{-1}\rceil}\big{\\}}$ such that $Ue_{n}=\pm e_{\pi(n)}$ for $n\in\big{\\{}1,2,\ldots,{\lceil\theta^{-1}\rceil}\big{\\}}.$

First we will show that the norm on
$\big{[}e_{1},e_{2},\ldots,e_{\lceil\theta^{-1}\rceil}\big{]}$ is the supremum
norm. Indeed, suppose that the norm of some

$x=\sum_{i=1}^{\lceil\theta^{-1}\rceil}a_{i}e_{i}$

is given by certain sets $d\leqslant E_{1}<E_{2}<\cdots<E_{d}$ for some
$d\in\mathbb{N}$ in the sense that

(3.5) $\|x\|=\theta\cdot\sum_{i=1}^{d}\big{\|}E_{i}x\big{\|}.$

Suppose that $\min E_{1}<\lceil{\theta^{-1}}\rceil.$ Then

$d\leqslant\min E_{1}\leqslant\lceil{\theta^{-1}}\rceil-1<{\theta^{-1}},$

so

$\theta\cdot\sum_{i=1}^{d}\big{\|}E_{i}x\big{\|}\leqslant\theta\cdot
d\cdot\|x\|<\|x\|.$

Hence (3.5) cannot hold; a contradiction. Suppose that $\min
E_{1}\geqslant\lceil\theta^{-1}\rceil.$ Then the only non-zero coordinate of
$(E_{1}\cup E_{2}\cup\ldots\cup E_{d})x$ is $a_{\lceil\theta^{-1}\rceil}$, so

$\|x\|\leqslant\theta\cdot|a_{\lceil\theta^{-1}\rceil}|\leqslant\theta\cdot\|x\|_{\infty}<\|x\|_{\infty}.$

This contradiction ends the proof that $\|x\|=\|x\|_{\infty}.$ This means that
for each $n\in\big{\\{}1,2,\ldots,{\lceil\theta^{-1}\rceil}\big{\\}}$ there is
at least one index $\pi(n)$ so that $\big{|}a^{n}_{\pi(n)}\big{|}=1.$ By the
very definition of the norm and since $U$ is an isometry we have

$\qquad\quad\,1=\max\\{\,1,\,2\cdot\theta\,\\}=\|e_{n}\pm
e_{i}\|=\big{\|}U(e_{n}\pm e_{i})\big{\|}\geqslant\big{\|}U(e_{n}\pm
e_{i})\big{\|}_{\infty}\geqslant\big{|}a_{\pi(n)}^{n}\pm
a_{\pi(n)}^{i}\big{|}$

for any $i\neq n$ in $\big{\\{}1,\ldots,{\lceil\theta^{-1}\rceil}\big{\\}}.$
Therefore $\big{|}a^{i}_{\pi(n)}\big{|}=0$ for any $i\neq n,$ so $\pi$ is the
desired permutation.

  * _Claim 3_. $Ue_{n}=\pm e_{n}$ for $n>\lceil\theta^{-1}\rceil.$

Let $n=\lceil\theta^{-1}\rceil+1$. Then, by Claim 1,
$Ue_{n}=a^{n}_{1}e_{1}+a^{n}_{2}e_{2}+\dots+a^{n}_{{n}}e_{n}$ and, by Claim 2,
for $1\leqslant j<n$ we have

$1=\max\\{\,1,\,2\cdot\theta\,\\}=\|e_{j}\pm e_{n}\|=\big{\|}e_{\pi(j)}\pm
Ue_{n}\big{\|}\geqslant\big{|}1\pm a^{n}_{\pi(j)}\big{|}.$

Consequently, $a^{n}_{\pi(j)}=0$ for all such $j$, so $Ue_{n}=\pm e_{n}.$

We now proceed to the inductive step. Fix $n\in\mathbb{N},$
$n>\lceil\theta^{-1}\rceil$ and assume that for $k\in\mathbb{N}$ with
$\lceil\theta^{-1}\rceil<k<n$ one has $Ue_{k}=\pm e_{k}$. Then by Claim 1 we
have $Ue_{n}=a^{n}_{1}e_{1}+a^{n}_{2}e_{2}+\dots+a^{n}_{{n}}e_{n}.$ By Claim
2, for $1\leqslant j\leqslant\lceil\theta^{-1}\rceil$ we have

$1=\max\\{\,1,\,2\cdot\theta\,\\}=\|e_{j}\pm e_{n}\|=\big{\|}e_{\pi(j)}\pm
Ue_{n}\big{\|}\geqslant\big{|}1\pm a^{n}_{\pi(j)}\big{|}.$

Therefore $a^{n}_{\pi(j)}=0$ for all such $j$. Similarly, by the inductive
hypothesis, we have

$1=\max\\{\,1,\,2\cdot\theta\,\\}=\|e_{j}\pm e_{n}\|=\|e_{j}\pm
Ue_{n}\|\geqslant\big{|}1\pm a^{n}_{j}\big{|}$

for $\lceil\theta^{-1}\rceil<j<n,$ so $a^{n}_{j}=0$ for all such $j.$ Hence
$Ue_{n}=\pm e_{n}.$ This finishes the proof that the isometry has the desired
form.

∎

######  Remark 6.

The reverse implication in Theorem A need not hold. Indeed, let
${\theta^{-1}}=2.1.$ Then, of course, $\lceil{\theta^{-1}}\rceil=3.$ Let us
consider the vectors:

  1. (1)

$x=\left(1,0,0,1,1,0,\ldots\right)$

  2. (2)

$y=\left(0,0,1,1,1,0,\ldots\right)$

The only $\mathcal{S}_{1}$-set with a minimum equal to $1$ is $\\{1\\}.$ Since
$\\{4,5\\}\in\mathcal{S}_{1},$ so from (3.2) we have

$\|x\|_{T}\geqslant\frac{10}{21}\cdot\left(1+1\right)=\frac{20}{21}.$

Now that vector $x$ has only $3$ ones, so by Lemma 4 this inequality is in
fact equality. Since $\\{3,4,5\\}\in\mathcal{S}_{1},$ so again by (3.2) and
Lemma 4 we obtain

$\|y\|_{T}=\frac{10}{21}\cdot\left(1+1+1\right)=\frac{30}{21}.$

Consequently, by (3.1), we have

  1. (1)

$\|x\|=\max\\{\|x\|_{\infty},\|x\|_{T}\\}=\max\\{1,\frac{20}{21}\\}=1,$

  2. (2)

$\|y\|=\max\\{\|y\|_{\infty},\|y\|_{T}\\}=\max\left\\{1,\tfrac{30}{21}\right\\}=\frac{30}{21}.$

Let us define an operator $U$ such that $Ux=y$. More formally,

$Ue_{1}=e_{3},\,\,Ue_{2}=e_{2},\,\,Ue_{3}=e_{1},\,\,Ue_{i}=e_{i}\text{ for
}i\geqslant 4.$

It is clear that $U$ is not an isometry.

##  4\. Isometries on $T[\theta,\mathcal{S}_{\alpha}]$ for
$\theta\in\big{(}0,\frac{1}{2}\big{]}$ and $2\leqslant\alpha<\omega_{1}$

We are now ready to prove Theorem B.

###### Proof.

Let $Ue_{n}:=\sum_{i=1}^{\infty}a^{n}_{i}e_{i}$ ($n\in\mathbb{N}$).

_Claim 1_. For any ordinal $2\leqslant\alpha<\omega_{1}$ and $n\in\mathbb{N}$
we have $U([e_{1},e_{2},\ldots,e_{n}])\subset[e_{1},e_{2},\ldots,e_{n}].$

Fix an ordinal $2\leqslant\alpha<\omega_{1}$,
$j\in\big{\\{}1,2,\dots,n\big{\\}},$ where $n\in\mathbb{N}$ and
$\varepsilon>0.$ Define $x:=\sum_{i=n+1}^{\infty}a^{j}_{i}e_{i}$ and
$t:=\max\\{n,\lceil\theta^{-1}\rceil\\}$. Take vector $x^{\prime}\in c_{00}$
such that $x^{\prime}\geqslant n+1$ and $\|x^{\prime}-x\|<\varepsilon$.

  * _Step 1_. There exits a sequence of indieces $\\{j_{i}\\}_{i=1}^{\infty}$ and a block sequence $\\{y_{i}\\}_{i=1}^{\infty}$ such that $j_{1}>t$, $y_{1}>x^{\prime}$ and for any $k\in\mathbb{N}$ holds $j_{k+1}>j_{k}$, $y_{k+1}>\max\\{j_{k},\max\operatorname{supp}\,y_{k}\\}$ and $\|Ue_{j_{k}}-y_{k}\|<\varepsilon$.

We will show this inductively.

Fix $k=1$. As $(Ue_{i})_{i=1}^{\infty}$ is weakly null we find index $j_{1}>t$
and vector $y_{1}>x^{\prime}$ so that
$\big{\|}Ue_{j_{1}}-y_{1}\big{\|}<\varepsilon$.

Assume that for any $s\leqslant k$, $s\in\mathbb{N}$ the inductive hypothesis
is satisfied. First, applying the inductive hypothesis, determine indieces
$t<j_{1}<\cdots<j_{k}$ and vectors

$\begin{split}x^{\prime}&<y_{1}\leqslant\max\\{\,j_{1},\max\operatorname{supp}\,y_{1}\,\\}<y_{2}\leqslant\cdots\\\
&\cdots\leqslant\max\\{\,j_{k-1},\max\operatorname{supp}\,y_{k-1}\,\\}<y_{k}.\end{split}$

As $(Ue_{i})_{i=1}^{\infty}$ is weakly null, we can choose $j_{k+1}>j_{k}$ and
a vector $y_{k+1}>\max\\{j_{k},\max\operatorname{supp}\,y_{k}\\}$ with
$\big{\|}Ue_{j_{k+1}}-y_{k+1}\big{\|}<\varepsilon$.

  * _Step 2_.

_Case 1_. Suppose that $\alpha=\beta+1$ for some $\beta<\omega_{1}$.

Note that every $\mathcal{S}_{\alpha}$-set whose minimum is $n$ is the union
of at most $n$ many $\mathcal{S}_{\beta}$-sets, so the idea of the proof of
this case is to choose the indices
$\lceil\theta^{-1}\rceil<j_{1}<j_{2}<\cdots<j_{m},$ for some $m\in\mathbb{N},$
so that they creates $n$ many maximal and consecutive
$\mathcal{S}_{\beta}$-sets. At the same time, we must ensure that the set

$\\{\min\operatorname{supp}\,x^{\prime},\,\min\operatorname{supp}\,y_{1},\,\min\operatorname{supp}\,y_{2},\dots,\min\operatorname{supp}\,y_{m}\\}$

associated with these indices was also $\mathcal{S}_{\alpha}$-set, i.e., it
was the union of at most (not necessarily maximal) $n+1$ many
$\mathcal{S}_{\beta}$-sets. We proceed as follows. By the Step $1$ we may
choose a maximal $\mathcal{S}_{\beta}$-set created from the indices
$j_{1}<j_{2}<\dots<j_{m_{1}},$ for some $m_{1}\in\mathbb{N}$ where $j_{1}>t$.
At the same time, we get vectors $x^{\prime}<y_{1}<y_{2}<\dots<y_{m_{1}}$ so
that $\|x^{\prime}-x\|<\varepsilon$ and
$\big{\|}Ue_{k}-y_{k}\big{\|}<\varepsilon$ for $k={1,\dots,m_{1}}.$ Next, we
apply again the Step $1$ to find the second maximal $\mathcal{S}_{\beta}$-set
with minimum $j_{m_{1}+1}$ greater than $j_{m_{1}}$ and block vectors

$\begin{split}x^{\prime}&<y_{1}\leqslant\max\\{\,j_{1},\max\operatorname{supp}\,y_{1}\,\\}<y_{2}\leqslant\cdots\\\
&\cdots\leqslant\max\\{\,j_{m_{2}-1},\max\operatorname{supp}\,y_{m_{2}-1}\,\\}<y_{m_{2}},\end{split}$

for some $m_{2}\in\mathbb{N}$, so that $\|x^{\prime}-x\|<\varepsilon$ and
$\big{\|}Ue_{k}-y_{k}\big{\|}<\varepsilon$ for $k={1,\dots,m_{2}}$. Then, the
set

$\\{\min\operatorname{supp}\,y_{2},\,\min\operatorname{supp}\,y_{3},\dots,\min\operatorname{supp}\,y_{m_{1}+1}\\}$

is $\mathcal{S}_{\beta}$-set because the Schreier family (of order $\beta$) is
spreading (see Definition 2).

Proceeding analogously, we finally arrive at indices

$j_{1}<j_{2}<\cdots<j_{m},$

for some $m\in\mathbb{N}$, that form a union of $n$ maximal
$\mathcal{S}_{\beta}$-sets with minima

$\big{\\{}j_{1},\,j_{m_{1}+1},\dots,j_{m_{n-1}+1}\big{\\}},$

so we got the conclusion because we may choose maximal
$\mathcal{S}_{\beta}$-sets

(4.1) $n+1\leqslant S^{1}_{\beta}<S^{2}_{\beta}<\dots<S^{n+1}_{\beta},$

where

    * –

$S^{1}_{\beta}=\\{\min\operatorname{supp}\,x^{\prime},\,\min\operatorname{supp}\,y_{1}\\},$

    * –

$S^{2}_{\beta}=\\{\min\operatorname{supp}\,y_{2},\,\min\operatorname{supp}\,y_{3},\ldots,\min\operatorname{supp}\,y_{m_{1}+1}\\},$

    * –

$\vdots$

    * –

$S^{n+1}_{\beta}=\\{\min\operatorname{supp}\,y_{m_{n-1}+2},\,\min\operatorname{supp}\,y_{m_{n-1}+3},\ldots,\min\operatorname{supp}\,y_{m}\\}.$

Note that we profit from the assumption $\alpha\geqslant 2$ in the fact that
$S_{\beta}^{1}\in\mathcal{S}_{\beta}$.

Since the indices $j_{1}<j_{2}<\cdots<j_{m}$, chosen as in Step 2 above, form
a set that is the union of $n$ maximal and consecutive
$\mathcal{S}_{\beta}$-sets with minimum greater than $n$, by the spreading
property of $\mathcal{S}_{\beta}$ and Lemma 5 (as $\\{j,j_{1},j_{2},\ldots
j_{m}\\}$ is not an $\mathcal{S}_{\alpha}$-set), we have

(4.2) $\Bigg{\|}e_{j}+\sum_{i=1}^{m}e_{j_{i}}\Bigg{\|}=\theta\cdot m.$

Since $U$ is an isometry, we obtain

(4.3)
$\Bigg{\|}e_{j}+\sum_{i=1}^{m}e_{j_{i}}\Bigg{\|}=\Bigg{\|}\sum_{i=1}^{n}a^{j}_{i}e_{i}+x+\sum_{i=1}^{m}Ue_{j_{i}}\Bigg{\|}.$

By (4.2), (4.3), and the triangle inequality

(4.4)
$\Bigg{\|}\sum_{i=1}^{n}a^{j}_{i}e_{i}+x^{\prime}+\sum_{i=1}^{m}y_{i}\Bigg{\|}\leqslant\theta\cdot
m+(m+1)\varepsilon.$

On the other hand,

(4.5)
$\Bigg{\|}\sum_{i=1}^{n}a^{j}_{i}e_{i}+x^{\prime}+\sum_{i=1}^{m}y_{i}\Bigg{\|}\geqslant\theta\Bigg{(}\|x^{\prime}\|+\sum_{i=1}^{m}\|y_{i}\|\Bigg{)}>\theta\big{(}\|x^{\prime}\|+m-m\varepsilon\big{)};$

the former inequality follows from (2.3) as we may choose sets as in (4.1),
whereas the latter one holds because $\big{\|}Ue_{i}\big{\|}=1$
($i\in\mathbb{N}$). Thus, by (4.4) and (4.5), we have

$\|x\|\leqslant\|x^{\prime}-x\|+\|x^{\prime}\|<\varepsilon+\big{(}\theta^{-1}(m+1)+m\big{)}\varepsilon,$

so $\|x\|=0$. Consequently,
$Ue_{j}=a^{j}_{1}e_{1}+a^{j}_{2}e_{2}+\dots+a^{j}_{n}e_{n}.$

  * _Case 2_ : Suppose that $\alpha$ is a limit ordinal.

We proceed as in Case $1$ for $\alpha=\alpha_{n},$ where
$(\alpha_{i})_{i=1}^{\infty}$ is a fixed strictly increasing sequence of
successor ordinals converging to $\alpha$ with
$\mathcal{S}_{\beta_{i}}\subset\mathcal{S}_{\beta_{k}}$ for $i\leqslant k$,
where $\alpha_{n}:=\beta_{n}+1$ for each $n\in\mathbb{N}$, choosing suitable
sequences $(j_{k})_{k=1}^{m}$ and $(y_{k})_{k=1}^{m}$. Indeed, an
$\mathcal{S}_{\alpha}$-set whose minimum is $n$ must be an
$\mathcal{S}_{\alpha_{n}}$-set and $\mathcal{S}_{\beta_{n}}$-sets
$n+1\leqslant S^{1}_{\beta_{n}}<S^{2}_{\beta_{n}}<\dots<S^{n+1}_{\beta_{n}}$,
where

    * –

$S^{1}_{\beta_{n}}=\\{\min\operatorname{supp}\,x^{\prime},\,\min\operatorname{supp}\,y_{1}\\},$

    * –

$S^{2}_{\beta_{n}}=\\{\min\operatorname{supp}\,y_{2},\,\min\operatorname{supp}\,y_{3},\ldots,\min\operatorname{supp}\,y_{m_{1}+1}\\},$

    * –

$\vdots$

    * –

$S^{n+1}_{\beta_{n}}=\\{\min\operatorname{supp}\,y_{m_{n-1}+2},\,\min\operatorname{supp}\,y_{m_{n-1}+3},\ldots,\min\operatorname{supp}\,y_{m}\\},$

give rise to an $\mathcal{S}_{\alpha}$-set (even an
$\mathcal{S}_{\alpha_{n}}$-set). Moreover, the set
$A:=\\{j,j_{1},j_{2},\ldots,j_{m}\\}$ is not an $\mathcal{S}_{\alpha}$-set, as
we ensured that $\mathcal{S}_{\beta_{i}}\subset\mathcal{S}_{\beta_{k}}$ for
$i\leqslant k$. Indeed, suppose $A\in\mathcal{S}_{\alpha}$, then
$A\in\mathcal{S}_{\alpha_{j}}$ i.e. $A$ is the union of at most $j$-many
successive $\mathcal{S}_{\beta_{j}}$-sets, i.e. $\mathcal{S}_{\beta_{n}}$-sets
by the assumption on $(\beta_{n})_{n}$. This contradicts the choice of
$\\{j_{1},j_{2},\ldots,j_{m}\\}$ as the union of $n$-many maximal
$S_{\beta_{n}}$-sets by the spreading property of $S_{\beta_{n}}$.

_Claim 2_. $Ue_{n}=\pm e_{n}$ for $n\in\mathbb{N}.$

  * Set $n=1$. Then by Claim 1 we have $Ue_{1}=a^{1}_{1}e_{1}.$ Since $\big{\|}Ue_{1}\big{\|}=1$, so $Ue_{1}=\pm e_{1}.$

We now proceed to the inductive step. Fix $n\in\mathbb{N}$ and assume that for
$k\in\mathbb{N}$ with $k<n$ one has $Ue_{k}=\pm e_{k}$. Then by Claim 1 we
have $Ue_{n}=a^{n}_{1}e_{1}+a^{n}_{2}e_{2}+\dots+a^{n}_{{n}}e_{n}.$ By the
very definition of the norm and the inductive hypothesis we have

$1=\max\\{1,\,2\cdot\theta\\}=\|e_{k}\pm e_{n}\|=\big{\|}e_{k}\pm
Ue_{n}\big{\|}\geqslant\big{\|}e_{k}\pm
Ue_{n}\big{\|}_{\infty}\geqslant\big{|}1\pm a^{n}_{k}\big{|}$

for $k\in\\{1,2,\dots,n-1\\}.$ Hence $a^{n}_{k}=0$ for all such $k.$ Since
$\big{\|}Ue_{i}\big{\|}=1$ ($i\in\mathbb{N}$), so $Ue_{n}=\pm e_{n}.$

∎

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