text:
Go Ahead, Wall Street, Make Obama's Day
A high-profile throwdown over the president's Consumer Financial Protection Bureau pick is exactly what the White House needs.
Reid Cramer
Flickr/whitehouse.gov
The Dodd-Frank Wall Street reform bill is now the law of the land. It's hard to know if it can prevent the next financial meltdown. Yet for most families, a better question is whether the law is strong enough to remake the financial services marketplace.
In recent years, too-easy credit indebted much of the middle class, subprime loans were peddled without concern for a borrower's ability to repay, and expensive check cashers and payday lenders became more ubiquitous than the corner saloon. This fringe financial sector not only destabilized family balance sheets, it undermined the larger economy as a whole. Thanks to the new law, for the first time, there will be a single agency—the Consumer Financial Protection Bureau—focused on protecting consumers, shutting down the latest scams and gimmicks, and making sure families can access appropriate and affordable financial services.
But much depends on the strength of this new agency. A weak agency will send the signal that with a few adjustments (maybe redesigned "disclosure statements"), business can proceed as usual. But a strong agency that is ready to flex its regulatory powers can stamp out the worst predatory lending and transform the market in meaningful ways.
No wonder, then, that there's intense interest in whom President Barack Obama will tap to lead this bureau—and whether the president will back them up in a showdown.
Consumer and other progressive groups have proclaimed Harvard law professor Elizabeth Warren, who heads the Congressional Oversight Panel that's been monitoring the TARP bailout, the only person for this post. They have organized petitions and endorsement statements on her behalf. After all, in 2007, she originated the idea of creating a consumer financial protection agency. In the past year and a half, she has led a series of public tutorials on how our financial system skidded off the tracks, screwing average Americans. Check out her appearances on the Daily Show (or her various congressional appearances) to hear her explain in plain language how the proliferation of complex, unfair, and predatory financial products have driven trust and integrity from the marketplace. To her credit, Warren recognizes that job one will be reversing this trend. As a high profile figure during the financial-reform bill debates, she repeatedly emphasized that trust can only be restored if consumers are provided access to appropriate savings and credit products in a fair and transparent manner.
Usually, the confirmation process for financial regulators is a below-the-radar affair—nothing like, say, a Supreme Court nomination. Not this time. The GOP seems eager to head into the campaign season running on a platform of repealing the Wall Street reform bill. Mostly likely, they will be ginned up to oppose any nominee. While some in the White House may calculate that passing over Warren for another qualified candidate may avoid a contentious and distracting spectacle, the reality is that whoever gets the nod is heading into a storm.
Yet this is a fight the president ought to embrace.
A battle against congressional antagonists and an array of financial firms can readily be turned into political advantage. By creating a clear contrast in which Obama and the Democrats line up on the side of families facing off against financiers, the president can embolden his political supporters and rally the base before the fall elections.
Perhaps more consequentially, a high-profile confirmation battle can serve vital policy goals. This is because effective government oversight requires competence and vigilance, but the essential—often overlooked—ingredient is empowerment. Regulatory agencies have to resist forming cozy relationships. At many turns, regulatory law opts for discretion rather than dictates. This means an agency can turn up the heat or turn it down with a large degree of autonomy. History shows that it takes a large dose of political will for an agency and its officials to resist the dangerous phenomenon of regulatory capture. There may be no better way to demonstrate resolve than by commencing the inevitable battle at the get-go, during the confirmation process.
Besides Warren, there are other highly capable contenders for the job: mainly, Michael Barr, the assistant secretary for financial institutions at Treasury, and Justice Department official Gene Kimmelman, a longtime consumer advocate in Washington. Barr, for his part, has long focused on expanding access to responsible financial services, particularly among lower-income families, and he worked hard to create the new consumer agency, drafting much of the legislative language. He worked the inside game when Warren was often the public face.
But the same interest groups will line up against any nominee. And if that's the case, the show ought to begin with a star. Warren's visibility and public credibility certainly can be tremendous assets in such a clash. She's a highly effective communicator and has received high marks in running the Congressional Oversight Panel (which should mitigate some of the whispers that she doesn't have sufficient management experience). She will garner the full throated endorsements of many ready to back her up during the confirmation battle
Winning this showdown is crucial for getting the policy reform process underway. Whoever is nominated, he or she should be expected to explain clearly how the law establishes a new accountability framework and be prepared to fight in the regulatory trenches to implement it. Exerting authority right out of the gate will make it easier for the new agency to play tough. Issuing cease-and-desist orders, filing lawsuits, and seeking damages should become commonplace and signal that there will be no tolerance for the marketing of financial products deemed unfair or deceptive. A high-profile confirmation battle—supported and waged by the White House—will not only be a clear demonstration there's a new cop on the beat, but that the Obama administration is ready and willing to do what it takes to put the financial pirates out of business for good.
A Convincing Case for Elizabeth Warren
So Did Congress Actually Reform Wall Street?
Bank Buster


text:
WEMBLEY: GAFFER'S VERDICT
Smith on Wembley defeat...
SADDLERS' boss, Dean Smith, has reflected upon our 2-0 defeat at the hands of Bristol City in the Johnstone's Paint Trophy Final at Wembley.
"I'm disappointed," he said. "We didn't perform to the level we are capable of and our use of the ball was not as good as theirs.
"We didn't really get started in the opening ten or 15 minutes and when we did, our final ball wasn't quite there.
"We needed that bit of momentum; I felt that had we scored a goal at any time if would have given us that lift that we were searching for.
"It was a special occasion and I wanted us to embrace it. In the end we just came up short against the Champions-elect.
"I have to say that the fans were brilliant. It was great to see so many of them there to support us and I am proud of what this club has achieved, even though we couldn't get the result that we were looking for.
"We're building and we'll move on from this. The players will learn from the experience and it's part of a learning curve.
"We will pick ourselves up, dust ourselves down and be ready to go again for the next game."

redpajama_set_name: RedPajamaCommonCrawl


Dr. Olaf Kraus de Camargo is an Associate Professor in the Department of Paediatrics at McMaster University. He completed his medical education and paediatric training in Brazil, followed by a residency in Germany where he received training in developmental-behavioural paediatrics and child neurology.
Prior to joining the faculty at McMaster, Dr. Kraus de Camargo held positions as a Professor of Social Medicine at the University of Applied Sciences Nordhausen and as CEO and Medical Director of Kinderzentrum Pelzerhaken gGmbH in Germany, an inpatient and outpatient facility for children with developmental behavioural disabilities and chronic neurologic disorders.
Within the Department of Paediatrics he is currently the Education Resource Person for the Division of Developmental Paediatrics.
Since 2001, Olaf Kraus de Camargo has been involved with the implementation of the WHO - International Classification of Functioning, Disability and Health (ICF), initially in neuro-rehabilitation and later in implementing it into Early Intervention Centres in Germany. Together with Judith Hollenweger, he edited the German translation of the ICF-CY (Children and Youth Version) for WHO. With Liane Simon he co-authored the first handbook about clinical applications of the ICF-CY in Germany. He contributes with his expertise as a collaborator to the Functioning and Disability Reference Group of the Family of Classification of WHO.
At McMaster, Dr. Kraus de Camargo is introducing ICF into education models, clinical practice and research. His goal is to develop an electronic version of the ICF (the ICanFunction App) to empower patients to better describe their needs to clinicians. He also is a co-founder and editor of an educational website called ICFEducation.
Dr. Kraus de Camargo's research and publications are available on ResearchGate, a digital network dedicated to science and research.
Dr. Kraus de Camargo's presentations are available to view and download on SlideShare.


text:
\section{Introduction}\label{sec:intro}

Two minimal topological dynamical systems are orbit equivalent if there is a homeomorphism between the phase spaces sending orbits to orbits. They are strongly orbit equivalent if they are orbit equivalent and the {\em cocyle} function is discontinuous at most in one point (see \cref{subsec:oe} for precise definitions). When the strong orbit equivalence property was introduced in topological dynamics, it was not clear its relationship with entropy, in particular, it was asked if entropy was preserved under strong orbit equivalence. This was dramatically disproved by Boyle and Handelman \cite{boylehandelman}, who showed that any Cantor minimal system (a system where the phase space is a Cantor space) is strongly orbit equivalent to a zero entropy Cantor system. Later, in a series of papers, Sugisaki \cite{sugisaki:1998,sugisaki:2003,sugisaki:2007} generalized this result, showing among other things that for any $c\in [0,+\infty]$ any Cantor minimal system is strongly orbit equivalent to a system whose topological entropy equals $c$. All these results make clear that entropy poses no restriction whatsoever to being strongly orbit equivalent. So one could ask, what are the natural restrictions that arise from being strongly orbit equivalent. A fundamental object that appears in this context is the dimension group of a dynamical system, an algebraic concept imported from the study of $C^{\ast}$-algebras \cite{vershik}, \cite{HPS}. In \cite{GPS} Giordano, Putnam and Skau  showed that two Cantor minimal systems are strongly orbit equivalent if and only if they have homeomorphic dimension groups with unit (we refer to \cref{sec:basicdef} for definitions). In particular, entropy does not impose any restriction on the dimension group of a Cantor minimal system.

One could consider a finer notion than entropy and study how it restricts the dimension group of a Cantor minimal system. In the case of a subshift, we focus our attention on the {\em complexity function}, defined as the function $p_X\colon {\mathbb N}\to {\mathbb N}$ that counts  the  number  of  non-empty  cylinders  of  length $n$. The topological entropy of the subshift is nothing but the exponential growth rate of $p_X$.
The complexity function has been widely studied in the past decades, a survey is provided in \cite{fe99}. Note that among zero entropy systems, one can still observe a wide variety of behaviors of $p_X$ for different subshifts, ranging from bounded (when $X$ is a finite system), sublinear (e.g. primitive substitutions, see for example \cite{Que}) till superpolynomial subexponential functions. First, notice that the complexity function does impose some restrictions on the dimension group of a minimal subshift. A trivial example is when $p_X$ is bounded, since that implies that $X$ is finite \cite{HM}, \cite{HM2}, and its dimension group is isomorphic to $({\mathbb Z},{\mathbb Z}^+,1)$\footnote{Here, the minimality assumption implies that the system is a periodic orbit}. A  non-trivial restriction occurs when the minimal subshift has non-superlinear complexity,{\it i.e.}, $\liminf p_X(n)/n<+\infty$. In \cite{DDMP20}, such a system was proved to be of {\em finite rank} (they admit a Bratteli-Vershik representation with a bounded number of vertices at each level) and therefore its dimension group is severely constrained: it is an abelian group whose rational rank is finite.

Since having non-superlinear complexity constrains the dimension group of a minimal subshift we can ask: do other growths of the  complexity function constrain the dimension group of a minimal subshift? For instance, is there anything particular about the dimension group of a minimal subshift with subquadratic complexity? Our main result is that except for non-superlinearity, there is no restriction on the complexity of a minimal subshift with a given dimension group.

\begin{theorem}\label{theo:main1}
Let $(G,G^+,u)$ be a simple dimension group with unit, and let $p_n$ be a sequence of positive real numbers such that $ \lim n/p_n=0$. Then there exists a minimal subshift $(Y,S)$ such that
\begin{enumerate}
    \item $K^0(Y,S)$ and $(G,G^+,u)$ are isomorphic as ordered groups with unit.
    \item The complexity of $(Y,S)$ satisfies $\lim p_{Y}(n)/p_n=0$.
\end{enumerate}
\end{theorem}
In terms of strong orbit equivalence \cref{theo:main1} can be restated as:
\begin{theorem}\label{theo:main2}
Let $(X,T)$ be a Cantor minimal system and let $p_n$ be a sequence of positive real numbers such that $ \lim n/p_n= 0$. Then there exists a minimal subshift $(Y,S)$ such that
\begin{enumerate}
    \item (Y,S) and (X,T) are strong orbit equivalent.
    \item The complexity of $(Y,S)$ satisfies $\lim p_{Y}(n)/p_n=0$.
\end{enumerate}
\end{theorem}

With some extra conditions on the dimension group,  we can build a Toeplitz subshift having low complexity.

\begin{theorem} \label{thm:divisible_toeplitz}
Let $(G,G^+,u)$ be a simple dimension group with unit, and let $p_n$ be a sequence of positive real numbers such that $ \lim n/p_n=0$. Assume that $G$ is a divisible group. Then there exists a minimal Toeplitz subshift $(Y,S)$ such that
\begin{enumerate}
    \item $K^0(Y,S)$ and $(G,G^+,u)$ are isomorphic as ordered groups with unit.
    \item The complexity of $(Y,S)$ satisfies $\lim p_{Y}(n)/p_n=0$.
\end{enumerate}
\end{theorem}

\subsection*{Invariant measures}
The Choquet simplex of invariant measures of a Cantor minimal system is affine homeomorphic to the set of {\it traces} of the dimension group of the system \cite{HPS} (see \cref{subsec:dg} for precise definitions). Thanks to the characterization of strong orbit equivalence through the dimension group \cite{GPS}, this implies that the simplex of invariant measures, modulo affine homeomorphism, is preserved under strong orbit equivalence. Many works have been developed to study the properties of the set of invariant measures of a Cantor minimal system and the connections between those properties and some dynamical properties of the system \cite{Wi, Do91, bezu1, bezu2}. In \cite{Do91} Downarowicz showed that a Toeplitz subshift may have the affine structure of an arbitrary Choquet simplex. This realization problem has been later extended to more general group actions on the Cantor space \cite{CP14}, \cite{CC19}.

Connections between the complexity of a subshift and its set of invariant measures have been explored as well. By a theorem of Boshernitzan \cite{Bosh}, a subshift having a non-superlinear complexity has finitely many ergodic measures. This can also be seen as a consequence of the fact that such systems are of finite rank \cite{DDMP20}. Some recent works have been devoted to getting better upper bounds to the number of ergodic and generic measures \cite{cyrkra1,DF:2017,DOP:2020}. Recently Cyr and Kra \cite{cyrkra2} showed that for any superlinear sequence $(p_n)_{n\in {\mathbb N}}$ there exists a minimal subshift $(X,S)$ such that $\liminf p_X(n)/p_n=0$ having uncountably many ergodic measures (see \cref{theo:ck2}).  As a consequence of \cref{thm:divisible_toeplitz} we generalize this result obtaining that any Choquet simplex can be realized as the set of invariant measures of a Toeplitz subshift with arbitrarily low, superlinear complexity. This can also be seen as a generalization of the result of Downarowicz \cite{Do91} in the sense that Toeplitz subshifts cannot only realize any Choquet simplex as set of invariant measures, but they can do it having an arbitrarily low superlinear complexity.

\begin{theorem}\label{theo:main3}
Let $K$ be a Choquet simplex. Let $(p_n)_{n\in {\mathbb N}}$ be a sequence of positive real numbers such that $ \lim n/p_n=0$. Then, there exists a Toeplitz subshift $(X,S)$ whose complexity $p_X$ satisfies $\lim p_{X}(n)/p_n=0$
and whose set of invariant measures ${\mathcal M}(X,S)$ is affine homeomorphic to $K$.
\end{theorem}

\subsection{Methods and organization} Our main tools are the use of Bratteli-Vershik models for Cantor minimal systems, together with their associated representation as ${\mathcal S}$-adic subshifts, which are systems obtained by performing an infinite composition of substitutions or morphisms (see \cref{sec:brattelidef} for precise definitions). Some estimations on the complexity of minimal ${\mathcal S}$-adic subshifts have been obtained on \cite{DDMP20} and we use them to bound the complexity of the subshift that will realize a given simple dimension group with unit as its dimension group.

The document is organized as follows. In \cref{sec:basicdef} we give the basic background on symbolic dynamics and more specific facts about dimension groups.  In \cref{sec:brattelidef} we introduce the tools from Bratteli-Vershik diagrams and their associated $\mathcal{S}$-adic subshifts together with complexity results in this context. \cref{sec:proofs} is devoted to proving the main results.

Throughout the article, we let ${\mathbb N}, {\mathbb N}^{\ast}$ and ${\mathbb Z}$ denote the set of non-negative integers, the set of positive integers, and the set of integer numbers respectively.

\section{Notation and background}\label{sec:basicdef}

\subsection{Cantor minimal systems.}\label{subsec:cantorms}
Through this paper, a {\it topological dynamical system} or simply a {\it dynamical system} is a pair $(X,T)$ where $X$ is a compact metric space and $T$ is a homeomorphism from $X$ to itself. When $X$ is a {\it Cantor space}, that is, $X$ is a non-empty, metrizable, perfect and totally disconnected compact space, we say that $(X,T)$ is a {\it Cantor dynamical system}. A dynamical system $(X,T)$ is {\it minimal} if it has no non-trivial closed $T$-invariant subsets. Equivalently, $(X,T)$ is minimal if for all $x\in X$, the {\it orbit} of $x$ under $T$, $O_T(x):=\{T^k(x)\colon k\in{\mathbb Z}\}$, is dense in $X$. If $X$ is a Cantor space and $(X,T)$ is minimal, $(X,T)$ is called a {\it Cantor minimal system}. A dynamical system $(X,T)$ is {\it expansive} if there exists $c>0$ such that for every $x\neq y$, $\sup_k d(T^k(x),T^k(y))>c$, where $d$ is a metric on $X$.
Two topological dynamical systems $(X_1,T_1)$ and $(X_2,T_2)$ are {\em conjugate} if there exists a homeomorphism $h\colon X_1\to X_2$ such that $h(T_1(x))=T_2(h(x))$ for all $x\in X_1$.

Given a dynamical system $(X,T)$, let $C(X,{\mathbb R})$ denote the additive group of continuous functions from $X$ to ${\mathbb R}$. The {\it coboundary map} $\beta\colon C(X,{\mathbb R})\to C(X,{\mathbb R})$ is defined by $\beta f\coloneqq f\circ T - f$.

\subsection{Subshifts and complexity}\label{subsec:subshifts}
Let ${\mathcal A}$ be a finite non-empty set with cardinality $|{\mathcal A}|$, which we call an {\it alphabet}.  The free monoid ${\mathcal A}^{\star}$ is the set of all concatenations of symbols of ${\mathcal A}$ (called {\it words}) including the empty word, which we denote $\varepsilon$. For $w \in {\mathcal A}^{\ast}$, $|w|$ stands for the length of $w$, that is, the total number of letters appearing on $w$. We use the convention $|\varepsilon|=0$. The sets ${\mathcal A}^{\mathbb Z}$ and ${\mathcal A}^{\mathbb N}$ endowed with the product topology of the discrete topology on each copy of ${\mathcal A}$ are compact metric spaces. The elements of ${\mathcal A}^{\mathbb Z}$ and ${\mathcal A}^{\mathbb N}$ are sometimes called {\it infinite words} on ${\mathcal A}$. A {\it factor} of a finite word $w\in{\mathcal A}^{\ast}$ is defined as a finite concatenation of some consecutive letters occurring in $w$. A factor of an infinite word is defined in the same way. The set of factors ${\mathcal L}_x$ of $x\in {\mathcal A}^{\mathbb Z}$ or ${\mathcal A}^{\mathbb N}$ is called its {\it language}. If $|{\mathcal A}|\geq 2$, the space ${\mathcal A}^{\mathbb Z}$ is a Cantor space.

Let $S\colon {\mathcal A}^{\mathbb Z}\to {\mathcal A}^{\mathbb Z}$ denote the {\it shift} transformation, which is defined by $S((x_i)_{i \in {\mathbb Z}})=(x_{i+1})_{i \in {\mathbb Z}}$. A {\it subshift} on ${\mathcal A}$ is a dynamical system given by a pair $(X,S|_X)$, where $X$ is a closed shift-invariant subset of ${\mathcal A}^{\mathbb Z}$, endowed with the induced topology. We denote by $(X,S)$ the system $(X,S|_X)$ to avoid an overcharged notation. We use the term {\it subshift} indistinctly to name both the space $X\subseteq {\mathcal A}^{\mathbb Z}$ and the dynamical system $(X,S)$. Given any infinite word $x$ in ${\mathcal A}^{\mathbb Z}$ (or in ${\mathcal A}^{\mathbb N}$), we define $(X_x,S)$ to be the {\it subshift generated by} $x$. That is,
$$X_x=\{y\in {\mathcal A}^{\mathbb Z}: \forall w, w \prec y \implies  w\prec x \},$$
where the symbol $\prec$ means ``is a factor of". Equivalently, $X_x$ is the closure of the orbit of $x$ under the action of the shift $S$.
If $(X,S)$ is a subshift on ${\mathcal A}$, its {\it language} ${\mathcal L}_X$ is defined as the set of factors of elements of $X$.

The {\it factor complexity} or simply the {\it complexity} of a subshift $(X,S)$ is the function $p_X\colon {\mathbb N}\to{\mathbb N}$ given by the number of factors of length $n$ in ${\mathcal L}_X$. A subshift $(X,S)$ is said to have {\it non-superlinear factor complexity} if
\[ \liminf \frac{p_X(n)}{n} < +\infty. \]

\subsection{Toeplitz subshifts}\label{subsec:toeplitz}
An infinite word $(x_i)_{i\in{\mathbb Z}}$ with symbols in the alphabet ${\mathcal A}$ is said to be a {\it Toeplitz sequence} if for all $i\in{\mathbb N}$ there exists $p\in{\mathbb N}$ such that for all $k\in {\mathbb Z}$, $x_i=x_{i+kp}$. A subshift $(X,S)$ is called a {\it Toeplitz subshift} if $X$ is the orbit  closure under the shift of some Toeplitz sequence. Toeplitz subshifts were originally introduced in \cite{JK}. They are known to be minimal (see for example \cite{Do95}).

\subsection{Invariant measures.}\label{subsec:im}
Given a topological dynamical system $(X,T)$, an {\it invariant measure} of $(X,T)$ is a Borel probability measure on $X$ such that for every Borel subset $A\subseteq X$, $\mu(T^{-1}(A))=\mu(A)$. The set of all invariant measures of $(X,T)$ is denoted ${\mathcal M}(X,T)$. We say that a measure $\mu\in{\mathcal M}(X,T)$ is {\it ergodic} if whenever $T(A)=A$ for some Borel set $A\subseteq X$, either $\mu(A)=0$ or $\mu(A)=1$. We say that $(X,T)$ is {\it uniquely ergodic} if ${\mathcal M}(X,T)$ consists of one element. The set ${\mathcal M}(X,T)$ is known to be non-empty and a {\it Choquet simplex}, that is, a compact convex metrizable subset $K$ of a locally convex real vector space such that for each $v\in K$ there is a unique probability measure $m$ supported on the extreme points of $K$ with $\int_{ext(K)}xdm(x)=v$. The extreme points of ${\mathcal M}(X,T)$ are exactly the ergodic measures on $X$ (see for example \cite{Glasner}). Here, ${\mathcal M}(X,T)$ is considered as a subspace of the dual $C(X,{\mathbb R})$ endowed with the weak$^\star$ topology.

The following result, due to Downarowicz \cite{Do91} tells us that Toeplitz subshifts can realize any Choquet simplex as set of invariant measures (see also \cite[Theorem 12]{noruegos}).
\begin{theorem}\cite[Theorem 5]{Do91}\label{theo:do91}
Let $K$ be a Choquet simplex. There exists a Toeplitz subshift $(X,S)$ such that ${\mathcal M}(X,S)$ is affine homeomorphic to $K$.
\end{theorem}

\subsubsection{Invariant measures and complexity}\label{subsec:imandcomplexity}
The following condition, called the Boshernitzan condition, states that the factor complexity of a minimal subshift restricts the possible affine structure of its set of invariant measures. More precisely, it states that minimal subshifts with non-superlinear factor complexity have a finite number of ergodic measures.
\begin{theorem}\cite[Corollary 1.3]{Bosh}\label{theo:boshcond}
Let $(X,S)$ be a minimal subshift and let $p_X$ denote the factor complexity function of $X$. If
$$\liminf_{n\to\infty}\frac{p_X(n)}{n}<\alpha<+\infty,$$
then the number of ergodic invariant probability measures of $(X,S)$ is at most $\max(\lfloor\alpha\rfloor,1)$, where $\lfloor\alpha\rfloor$ denotes the greatest integer which is smaller than $\alpha$.
\end{theorem}

Cyr and Kra showed that any subshift (not necessarily minimal) with linear factor complexity has at most finitely many nonatomic ergodic invariant probability measures, and so at most countably many ergodic invariant probability measures \cite{cyrkra1}. In \cite{cyrkra2}, the authors show that it is possible to realize any arbitrarily low superlinear factor complexity with a minimal subshift having uncountably many ergodic invariant probability measures. More precisely, they prove the following theorem.
\begin{theorem}\cite[Theorem 1.1]{cyrkra2}\label{theo:ck2}
If $(p_n)_{n\in{\mathbb N}}$ is a sequence of natural numbers such that
$$\liminf_{n\to\infty}\frac{p_n}{n}=\infty,$$
then there exists a minimal subshift $(X,S)$ which supports uncountably many ergodic measures and is such that
$$\liminf_{n\to\infty}\frac{p_X(n)}{p_n}=0.$$
\end{theorem}
\begin{Rem}
Theorem \ref{theo:main3} is a strengthening of Theorem \ref{theo:ck2} in two senses. The first one is that the conclusion is about the limit of $\frac{p_X(n)}{p_n}$ and not only about the liminf. Here it is worth mentioning that the behaviour of the complexity along different sequences could be very different, for instance, examples in \cite[Section 4]{DDMP16} show that it is possible to have $\liminf p_X(n)/p_n =0$ and $\limsup p_X(n)/p_n=+\infty$ for a superlinear sequence $(p_n)_{n\in {\mathbb N}}$.

The second strengthening is that we are able to realize not only uncountably many invariant ergodic measures, but any Choquet simplex of invariant measures, which at the same time implies that \cref{theo:main3} is a generalization of \cref{theo:do91}.
\end{Rem}
\subsection{Inductive limits.}\label{subsec:inductive}
For a sequence of group homomorphisms $(h_i\colon G_i\to G_{i+1})_{i\in{\mathbb N}}$, the associated {\it inductive limit} is defined by
$$\varinjlim_i(G_i,h_i)=\{(g,i)\in G_i\times {\mathbb N}\}/\sim,$$
where $(g,i)\sim (g',i')$ if and only if there exists $k\geq i,i'$ such that $h_k\circ\cdots\circ  h_i(g)=h_k\circ \cdots \circ h_{i'}(g')$. We denote by $[g,i]$ the equivalence class of the element $(g,i)$. The inductive limit $\varinjlim\limits_{i}(G_i,h_i)$ is also denoted
$$ G_0 \xrightarrow[]{h_0} G_1 \xrightarrow[]{h_1} G_2 \xrightarrow[]{h_2} \cdots, $$
and it has a group structure given by the product
$$[g,i]\cdot [g',i']= [h_k\circ \cdots \circ h_i(g)\cdot h_k\circ \cdots \circ h_{i'}(g'),k+1],$$ where $k\geq i,i'$.

For a $m\times n$ integer matrix $A$, we consider the natural homomorphism it induces, {\it i.e.}, $A\colon {\mathbb Z}^{m}\to {\mathbb Z}^{n}$ (or $A\colon {\mathbb Q}^{m}\to {\mathbb Q}^{n}$ ), $x\mapsto Ax$.

\subsection{Dimension groups.}\label{subsec:dg}
An {\it ordered group} is a pair $(G,G^+)$ where $G$ is a countable abelian group $G$ and $G^+$ is a subset of $G$, called the {\it positive cone}, satisfying
$$G^+ + G^+ \subset G^+, \quad  G^+ \cap (-G^+) = \{ 0 \}, \quad  G^+ - G^+ = G.
$$
We write $a\leq b$ if $b-a \in G^+$, and $ a<b$ if $b-a \in  G^+$ and $b \neq a$. Two ordered groups $(G,G^+)$ and $(H,H^+)$ are {\it isomorphic} if there exists a group isomorphism $\varphi\colon G\to H$ such that $\varphi(G^+)=H^+$. We also say that $G$ and $H$ are {\it order isomorphic} when the positive cones are clear from the context. An ordered group $(G,G^+)$ is a {\it dimension group} if for every $i\geq 0$ there exists an integer $d_i\geq 1$ and a positive homomorphism $A_i\colon {\mathbb Z}^{d_i}\to {\mathbb Z}^{d_{i+1}}$ such that $(G,G^+)$ is isomorphic to the inductive limit
$$ {\mathbb Z}^{d_0}  \xrightarrow[]{A_0} {\mathbb Z}^{d_1}  \xrightarrow[]{A_1} {\mathbb Z}^{d_2} \xrightarrow[]{A_2} \cdots. $$
endowed with the induced order.
The dimension group is called {\it simple} if all matrices $A_i$ can be chosen to be positive. An {\it order unit} for $(G,G^+)$  is an element $u $ in $G^+$  such that, for all $a $ in $G$, there exists  some non-negative integer $k$ with $a  \leq ku$. An {\it ordered group with unit} is a triple $(G,G^+,u)$, where $(G,G^+)$ is an ordered group and $u$ is an order unit for $(G,G^+)$. A {\it dimension group with unit} is a triple $(G,G^+,u)$, where $(G,G^+)$ is a dimension group and $u$ is an order unit for $(G,G^+)$. If $(G,G^+)$ is isomorphic to the inductive limit
$$ {\mathbb Z}^{d_0}  \xrightarrow[]{A_0} {\mathbb Z}^{d_1}  \xrightarrow[]{A_1} {\mathbb Z}^{d_2} \xrightarrow[]{A_2} \cdots $$
endowed with the induced order, any element $[x,i]$, where all the entries of the vector $x\in {\mathbb Z}^{d_i}$ are positive, is an order unit for $(G,G^+)$.
Two ordered groups with unit $(G,G^+,u)$ and $(H,H^+,v)$ are {\em isomorphic} if there exists a group isomorphism $\varphi \colon G \rightarrow H$ such that $\varphi(G^+)=H^+$ and $\varphi (u)=v$.

The following lemmas will be useful in the proofs of the main theorems. Recall that an abelian group $(G,+)$ is called {\it divisible} if for each $g\in G$ and each integer $k\geq 1$, there exists $h\in G$ such that $kh=g$.

\begin{lemma}\label{prop:divisible2}
Let $(A_i)_{i\in{\mathbb N}}$ be a sequence of $d_{i+1}\times d_i$ integer matrices, let $G=\varinjlim\limits_{i}({\mathbb Z}^{d_i},A_i)$. If for all $i\geq 1$, all the entries of $A_i$ are divisible by $i+1$, then $G$ is a divisible group.
\end{lemma}
\begin{proof}
We must show that for all $g\in G$, for all positive integer $k$, there exists $h\in G$ such that $kh=g$. Let $g=[x,i]\in G$ and $k$ a positive integer.

If $k=1$, it suffices to take $h=g$.
Suppose that $k>1$. If $i=0$, let $z=A_{k-1}\cdots A_0x$. Then, $[z,k]=[x,0]$. Let $y=\frac{1}{k}A_{k-1}\cdots A_0x$. Since every entry of $A_{k-1}$ is divisible by $k$, $y\in {\mathbb Z}^{d_k}$. Taking $h=[y,k]$, we have that $kh=k[y,k]=[z,k]=[x,0]=g$. If $i>0$, $ki>i$. In this case, let  $z=A_{ki-1}\cdots A_ix$. Since every entry of $A_{ki-1}$ is divisible by $ki$, they are all divisible by $k$ as well. Let $y=\frac{1}{k}A_{ki-1}\cdots A_ix$. As before, $y\in {\mathbb Z}^{d_{ki}}$. Taking $h=[y,ki]$, we have that $kh=k[y,ki]=[z,ki]=[x,i]=g$.
\end{proof}

\begin{lemma}\label{prop:divisible1}
Let $(A_i)_{i\in{\mathbb N}}$ be a sequence of $d_{i+1}\times d_i$ positive integer matrices, let $G=\varinjlim\limits_{i}({\mathbb Z}^{d_i},A_i)$ and $H=\varinjlim\limits_{i}({\mathbb Q}^{d_i},A_i)$. Let $G^+$ and $H^+$ be the positive cones of $G$ and $H$ respectively, given by the induced order on the inductive limits. Let $u=[{\bf 1}, 0]_G$ and $v=[{\bf 1},0]_H$ be order units for $(G,G^+)$ and $(H,H^+)$ respectively, where $[\cdot,\cdot]_G$ and $[\cdot,\cdot]_H$ denote the equivalence classes modulo $\sim$ in $G$ and $H$ respectively, and ${\bf 1} \in {\mathbb Z}^{d_0}\subseteq {\mathbb Q}^{d_0}$ is the vector $(1,1,\cdots, 1)$. If $G$ is divisible, then $(G,G^+,u)$ and $(H,H^+,v)$ are isomorphic as ordered groups with unit.
\end{lemma}

\begin{proof}
Let $\varphi\colon G\to H$ be the map defined by $[x,i]_G\mapsto [x,i]_H$. This map is a well defined injective group homomorphism. By definition, $\varphi([{\bf 1},0]_G)=[{\bf 1},0]_H$. Let $[y,i]_H \in H$, let $(y,i)\in {\mathbb Q}^{d_i}\times {\mathbb N}$ be a representative. Since $y\in {\mathbb Q}^{d_i}$, we can write it in the form
$$\left(\frac{w_1}{k_1},\cdots, \frac{w_{d_i}}{k_{d_i}}\right),$$
where $w_1,\cdots, w_{d_i}\in {\mathbb Z}$, $k_1,\cdots, k_{d_i}\in {\mathbb N}^{\ast}$. Let $k=lcm\{k_\ell\}$, so that $y=\frac{1}{k}(w_1',\cdots, w_{d_i}')$, some $w_1',\cdots, w_{d_i}'\in{\mathbb Z}$. Let $z=(w_1',\cdots, w_{d_i}')\in {\mathbb Z}^{d_i}$. Since $G$ is divisible, there exists $[x,j]_G\in G$ such that $k[x,j]_G=[z,i]_G$. This means that there exists $s\geq j,i$ such that $kA_s\cdots A_jx=A_s\cdots A_i z$, and since $ky=z$, we obtain that $[x,j]_H=[y,i]_H$. This proves that $\varphi$ is surjective. It is strightforward to show that $\varphi(G^+)=H^+$.
\end{proof}

The next lemma concerning dimension groups will be useful in the sequel. We first introduce a definition
\begin{definition} let $(A_i)_{i\in {\mathbb N}}$ be a sequence of $d_{i+1}\times d_i$ positive integer matrices. We say that the sequence of $d_i\times d_i$ matrices $(J_{i})_{i\in {\mathbb N}}$, is {\it adapted} to $(A_i)_{i\in {\mathbb N}}$ if
\begin{enumerate}
\item \label{item:adaptedcond1} $J_i, J_i^{-1}$ have positive rational entries and for all $i\in {\mathbb N}$, there exists $M\geq i$ such that $A_{M} A_{M-1}\cdots A_{i} J_{i}^{-1}$ has positive integer entries.
\item \label{item:adaptedcond2} $B_i:=J_{i+1}A_iJ_i^{-1}$ has positive integer entries for all $i\in {\mathbb N}$.
\end{enumerate}
\end{definition}

\begin{lemma} \label{lema:AdaptedMatrices}
 	Let $(A_i)_{i\in {\mathbb N}}$ be a sequence of positive integer matrices and let $(J_{i})_{i\in {\mathbb N}}$ be  a sequence adapted to $(A_{i})_{i\in {\mathbb N}}$. Let $G=\varinjlim\limits_{i}({\mathbb Z}^{d_i},A_i)$ and $H=\varinjlim\limits_{i}({\mathbb Z}^{d_i},B_i)$. Consider the positive cones $G^+$ and $H^+$ given by the induced order, and the order units $u=[A_{M_0} A_{M_0-1}\cdots A_{0} J_{0}^{-1}{\bf 1},M_0+1]_G$ and $v=[{\bf 1},0]_H$, where $[\cdot,\cdot]_G$ and $[\cdot,\cdot]_H$ denote the equivalence classes modulo $\sim$ in $G$ and $H$ respectively, ${\bf 1} \in {\mathbb Z}^{d_0}$ is the vector $(1,1,\cdots, 1)$, and $M_0\geq 0$ is such that $A_{M_0} A_{M_0-1}\cdots A_{0} J_{0}^{-1}$ has positive integer entries. Then, $(G,G^+,u)$ and $(H,H^+,v)$ are isomorphic as ordered groups with unit.
\end{lemma}

\begin{proof}
Note that condition \ref{item:adaptedcond2} allows us to define morphisms $A_i\colon J_{i}^{-1}{\mathbb Z}^{d_i}\to J_{i+1}^{-1}{\mathbb Z}^{d_{i+1}}$. Note also that $H$ is order isomorphic to the inductive limit $K=\varinjlim\limits_{i}(J_i^{-1}{\mathbb Z}^{d_i},A_i)$, via the map $\varphi\colon H\to K$, $[x,i]_H\mapsto [J_i^{-1}x,i]_K$. This order isomorphism maps the unit $v$ into $[J_0^{-1}{\bf 1},0]_K$. On the other hand, condition \ref{item:adaptedcond1} ensures that $K$ is order isomorphic to $G$ via the map $\theta\colon K\to G$, $[y,i]_K\mapsto [A_{M} A_{M-1}\cdots A_{i} J_{i}^{-1}y, M+1]_G$, where $M\geq i$ is such that $A_{M} A_{M-1}\cdots A_{i} J_{i}^{-1}$ has positive integer entries. We conclude by noticing that $\theta\circ\varphi(H^+)=G^+$ and $\theta\circ\varphi(v)=u$.
\end{proof}

Given a dimension group with unit $(G,G^+,u)$, a {\it trace} of $(G,G^+,u)$ is a group homomorphism $p\colon G\to \mathbb{R}$ such that $p$ is non-negative ($p(G^+)\geq 0$) and $p(u)=1$. The collection of all traces of $(G,G^+,1)$ is denoted by $S(G,G^+,u)$. An element $g \in G$ is said to be {\it infinitesimal} if $p(a) = 0$ for every trace $p \in S(G, G^+,u)$. The collection of all infinitesimals of $(G,G^+,u)$ form a subgroup of $G$, called the infinitesimal subgroup of $(G,G, u)$ and denoted $\Inf(G,G^+,u)$.

The following result connects simple dimension groups and Choquet simplices, its proof can be found in \cite{effros}. We use the formulation of \cite{malen-hoynes}.
\begin{theorem}\cite[Theorem 3.22]{malen-hoynes}\label{theo:simplexanddg}
Let $K$ be a Choquet simplex and $\Aff(K)$ the additive group of real, affine and continuous functions on $K$. Let $H$ be a countable dense subgroup of $\Aff(K)$, and suppose there exists a torsion-free abelian group $G$ and a homomorphism $\theta:G\to H$. Then, letting
$$G^+=\{g\in G \mid \theta(g)(p)>0 \forall p\in K\}\cup \{0\},$$
we get that $(G,G^+)$ is a simple dimension group such that $\Inf(G)=\ker(\theta)$. In particular, if $G=H$ (with $\theta$ the identity map) and $G$ contains the constant function $1$, then $\Inf(G)=\{0\}$ and $S(G,G^+,1)$ is affine homeomorphic to $K$ by the map sending $k\in K$ to $\hat{k}:G\to {\mathbb R}$, where $\hat{k}(g)=g(k)$ for all $g\in G$.
\end{theorem}

Let $(X,T)$ be a Cantor minimal system. The {\it dynamical dimension group} of $(X,T)$ or simply the {\it dimension group} of $(X,T)$ is the following triple,
$$K^0(X,T)=(H(X,T), H^+(X,T), [1]),$$
where $H(X,T)=C(X,{\mathbb Z})/\beta C(X,{\mathbb Z})$, $\beta$ is the coboundary map, $[\cdot]$ denote the class modulo $\beta C(X,{\mathbb Z})$ of an element in $H(X,T)$, $H^+(X,T)$ is the set of classes of non-negative functions and $1$ is the constant function equal to $1$.
\begin{theorem}\cite[Corollary 6.3]{HPS}\label{theo:isdg}
If $(X,T)$ is a Cantor minimal system, the triple $K^0(X,T)$ is a simple dimension group with unit. Furthermore, if $(G,G^+,u)$ is a simple dimension group with unit, then there exists a Cantor minimal system $(X,T)$ such that $K^0(X,T)$ is isomorphic to $(G,G^+,u)$ as ordered groups with unit.
\end{theorem}
Given an invariant measure $\mu \in {\mathcal M}(X,T)$, we define the trace $\tau_{\mu}$ on $K^0 (X,T)$  by  $\tau_{\mu} ([f]):= \int f d\mu$. The correspondence $\mu\mapsto \tau_\mu$ is an affine isomorphism from ${\mathcal M}(X,T)$ to $S(K^0(X,T))$, so that traces of the dynamical dimension group $K^0(X,T)$ correspond to the Choquet simplex of invariant measures of the system $(X,T)$ (see for instance \cite[Theorem 5.5]{HPS}).

\subsection{Orbit equivalence.}\label{subsec:oe}
Two minimal dynamical systems $(X_1,T_1)$ and $(X_2,T_2)$ are {\it orbit equivalent} if there exists a homeomorphism $\phi\colon X_1 \to X_2$ sending orbits of the $T_1$-action onto orbits of the $T_2$-action, {\it i.e.},
$$\phi( \{ T_1^k (x)  : k \in {\mathbb Z} \})= \{ T_2^k \phi (x) : k \in {\mathbb Z}\} \text{ for all } x \in X_1.$$
Orbit equivalence implies the existence of two maps $n_1\colon X_1  \to {\mathbb Z}$ and $n_2\colon X_2 \to {\mathbb Z}$ (uniquely defined by minimality) such that, for all $x \in X_1$,
$$\phi \circ T_1 (x)= T_2^{n_1(x)} \circ \phi(x) \mbox { and }  \phi \circ T_1^{n_2(x)} (x)= T_2 \circ \phi (x).$$
The dynamical systems $(X_1,T_1)$ and $(X_2,T_2)$ {\it strong orbit equivalent}  if $n_1$ and $n_2$ both have at most one point  of discontinuity. Such notion is natural since it was shown  in \cite{Boyle} that if $n_1$ (or $n_2$) is continuous, then the two systems are {\it flip conjugate}, that is, $(X_1,T_1)$ is either conjugate to $(X_2,T_2)$ or to its inverse $(X_2,T_2^{-1})$.

The following result is one of the most important theorems connecting dimension groups and orbit equivalence.
\begin{theorem}\cite[Theorem 2.1]{GPS}\label{theo:caracterizationoe}
Let $(X_1,T_1)$ and $(X_2,T_2)$ two Cantor minimal systems. Then $(X_1,T_1)$ and $(X_2,T_2)$ are strong orbit equivalent if and only if $K^0(X_1,T_1)$ and $K^0(X_2,T_2)$ are isomorphic as ordered groups with unit.
\end{theorem}

\section{Bratteli diagrams and ${\mathcal S}$-adic subshifts}\label{sec:brattelidef}

\subsection{Bratteli diagrams and Bratteli-Vershik systems.}\label{subsec:bratteli}
A {\it Bratteli diagram} is an infinite directed graph $(V,E)$ where the set of vertices $V$ and the set of edges $E$ can be written as a countable disjoint union of non-empty finite sets,
$$V=V_0\cup V_1\cup V_2\cup\cdots \quad \mbox{ and } \quad E=E_1\cup E_2\cup\cdots,$$
with the property that $V_0$ is a single point and there exist a {\it range map} $r\colon E\to V$ and a {\it source map} $s\colon E\to V$ so that $r(E_i)\subseteq V_i$ and $s(E_i)\subseteq V_{i-1}$. Also, we assume that $s^{-1}(v)\neq \emptyset$ for all $v\in V$ and $r^{-1}(v)\neq \emptyset$ for all $v\in V\setminus V_0$.

For $i\in{\mathbb N}$, let $A_i$ denote the $i$th {\it incidence matrix} of $(V,E)$, that is, the $|V_{i+1}|\times|V_i|$ matrix whose coefficient $A_i(k,j)$ is the number of edges connecting $u_j\in V_i$ with $v_k\in V_{i+1}$.

Let $(i_k)_{k\in{\mathbb N}}$ be a subsequence of ${\mathbb N}$. Define a new Bratteli diagram $(V',E')$ by letting $V_k'=V_{i_k}$ and $A_k'=A_{i_{k+1}-1}A_{i_{k+1}-2}\cdots A_{i_k}$. The sets of edges $E_k'$ and the range and source maps are obtained from the new incidence matrices. The diagram $(V',E')$ is called a {\it telescoping} of $(V,E)$. If a Bratteli diagram $(V',E')$ can be telescoped to obtain $(V,E)$, then we say that $(V',E')$ is a {\it microscoping} of $(V,E)$.

A Bratteli diagram is {\it simple} if it can be telescoped so that all its incidence matrices have only positive entries. An {\it ordered Bratteli diagram} is a Bratteli diagram together with a linear ordering on $r^{-1}(v)$ for each $v\in V\setminus V_0$. This defines a partial order $\geq$ on $E$. An ordered Bratteli diagram $(V,E)$ together with a partial order $\geq$ on $E$ is denoted $(V,E,\geq)$. Let $E_{min}$ and $E_{max}$ denote the sets of minimal and maximal edges, respectively. An {\it infinite path} in $(V,E)$, that is, a sequence of the form $(e_1,e_2,\cdots)$ where $e_i\in E_i$ and  $r(e_i)=s(e_{i+1})$ for all $i\in{\mathbb N}$, is said to be {\it minimal} (resp. {\it maximal}) if all its edges belong to $E_{min}$ (resp. $E_{max}$). An ordered Bratteli diagram is {\it properly ordered} if it is simple and it has a unique minimal path and a unique maximal path, which we denote by $x_{min}$ and $x_{max}$, respectively. An ordered Bratteli diagram is {\it left/right ordered} if whenever two edges $e_1$, $e_2$ with the same range verify $e_1\geq e_2$ and $s(e_1)=v_i$, $s(e_2)=v_j$, we have $i\geq j$.

Given a properly ordered Bratteli diagram $(V,E,\geq)$, it is possible to put a dynamic on it in the following way: let $X_B$ denote the space of infinite paths on $E$ (where we assume that $X_B$ is infinite). We endow $X_B$ with a topology by giving a basis of open sets, namely the family of {\it cylinder sets},
$$[e_1 , e_2 , \cdots, e_k ]_B = \{(f_1 , f_2 , \cdots) \in X_B: f_i=e_i \mbox{ for all } 1\leq i\leq k\}.$$
The cylinder sets are also closed. The space $X_B$ endowed with this topology is called the {\it Bratteli compactum} associated with $(V,E)$ and it is a Cantor space. We define the {\it Vershik map} $V_B$ on $X_B$ as follows: $V_B(x_{max})=x_{min}$; if $x=(e_1,e_2,\cdots)$ is not the maximal path, let $k$ be the smallest integer such that $e_k$ is not a maximal edge, let $f_k$ be the sucesor of $e_k$ on $E_k$, and define $V_B(x)=(f_1,f_2,\cdots, f_{k-1},f_k,e_{k+1},e_{k+2},\cdots)$, where $(f_1,\cdots, f_{k-1})$ is the minimal finite path on $E_1\circ E_2\circ \cdots \circ E_{k-1}$ with range equal to $s(f_k)$. The system $(X_B,V_B)$ is called the {\it Bratteli-Vershik dynamical system} associated with $(V,E,\geq)$.

It was shown by Herman, Putnam and Skau \cite{HPS} that for any Cantor minimal system $(X,T)$, there exists a properly ordered Bratteli diagram $(V,E,\geq)$ such that $(X,T)$ and $(X_B,V_B)$ are conjugate. In this case, $(V,E,\geq)$ is called a Bratteli-Vershik {\it model} for $(X,T)$.

To a Bratteli diagram $(V,E)$ is associated a dimension group with unit denoted $K_0(V,E)$. By definition, $K_0(V,E)$ is the inductive limit of the system
$$ {\mathbb Z}^{|V_0|}  \xrightarrow[]{A_0} {\mathbb Z}^{|V_1|}   \xrightarrow[]{A_1} {\mathbb Z}^{|V_2|}\xrightarrow[]{A_2} \cdots, $$
endowed with the induced order, where $(A_i)_{i\in{\mathbb N}}$ is the sequence of incidence matrices of $(V,E)$. The order unit in $K_0(V,E)$ is the element $[1,0]\in \varinjlim\limits_{i}({\mathbb Z}^{|V_i|},A_i)$. Two Bratteli diagrams $(V,E)$ and $(V',E')$ have isomorphic dimension groups with unit if and only if $(V',E')$ can be obtained from $(V,E)$ by a finite number of telescopings and microscopings. If $(V,E,\geq)$ is a properly ordered Bratteli diagram and $(X_B,V_B)$ is its associated Bratteli-Vershik system, then one has $K_0(V,E)\cong K^0(X_B,V_B)$ as ordered groups with unit.

A simple Bratteli diagram $(V,E)$ has the {\it equal path number property} if for all $i\geq 1$, for all $u,v\in V_i$, $|r^{-1}(u)|=|r^{-1}(v)|$. Equivalently, $(V,E)$ has the equal path number property if for all $i\geq 1$, for all $u,v\in V_i$, the number of finite paths from $V_0$ to $u$ and from $V_0$ to $v$ are equal. In terms of the matrices, this means that for all $i\in{\mathbb N}$, the sum of the entries on a row of $A_i$ is constant.  This property is called the {\it equal row sum (ERS)} property of the matrix $A_i$. The equal path number property was introduced in \cite{noruegos}, where the authors prove the following result.
\begin{theorem}\cite[Theorem 8]{noruegos}\label{teo:ers}
The family of expansive Bratteli-Vershik systems associated with Bratteli diagrams with the equal path number property coincides with the family of Toeplitz flows up to conjugacy.
\end{theorem}

\subsection{Morphisms and ${\mathcal S}$-adic systems}\label{subec:Sadic}
\noindent Let ${\mathcal A}$, ${\mathcal B}$ be two finite alphabets. Let $\tau \colon {\mathcal A}^{\ast} \to {\mathcal B}^{\ast}$ be a morphism. We say that $\tau$ is {\it non-erasing} if the image of any letter is a non-empty word. The {\it incidence matrix} of $\tau$ is the $|{\mathcal B}|\times|{\mathcal A}|$ matrix whose entry at a position $(b,a)$ is the number of times that $b$ appears in $\tau(a)$. For $a\in{\mathcal A}$, the length of the word $\tau(a)\in {\mathcal B}^{\ast}$ is denoted $|\tau(a)|$. The morphism $\tau$ is {\em positive} if the entries of $|{\mathcal B}|\times|{\mathcal A}|$ are all positive and is {\it left proper} (resp. {\it right proper}) if there exists a letter $b\in{\mathcal B}$ such that for every $a\in{\mathcal A}$, $\tau(a)$ starts with $b$ (resp. ends with $b$); it is {\it proper} if is both left and right proper. We say that $\tau$ is a {\it hat morphism} if for all $a,b\in {\mathcal A}$, the letters appearing in $\tau(a)$ and $\tau(b)$ are all distinct. By concatenation, a morphism $\tau\colon {\mathcal A}^{\ast}\to {\mathcal B}^{\ast}$ can be extended to ${\mathcal A}^{\mathbb N}$ and ${\mathcal A}^{\mathbb Z}$.\\

For a non-erasing morphism $\tau \colon {\mathcal A}^{\ast} \to {\mathcal B}^{\ast}$, define $\|\tau\|=\max_{a\in {\mathcal A}}\{|\tau(a)|\}$ and $\langle \tau \rangle= \min_{a\in {\mathcal A}}\{|\tau(a)|\}$.
The following notion was introduced in \cite[Section 6]{DDMP20} in the context of finite topological rank minimal subshifts. Assume that ${\mathcal B}=\{b_1,\ldots,b_{|{\mathcal B}|}\}$ and for $a\in {\mathcal A}$, write $\tau(a)=b_{i_1}^{\ell_1}b_{i_2}^{\ell_2}\cdots b_{i_{k(a)}}^{\ell_{k(a)}}$, for some $i_1,\ldots,i_{k(a)}\in \{1,\ldots,|{\mathcal B}|\}$ and where $b_{i_{j+1}}\neq b_{i_j}$ for all $j=1,\ldots,k(a)-1$. The integer $k(a)$ represents the number of times one needs to switch letters while writing $\tau(a)$, plus one.
The {\it repetition complexity} of $\tau$ is defined as
\[\textrm{r-comp}(\tau)= \sum_{a\in {\mathcal A}} k(a). \]

Note that if $\tau$ is positive then $k(a)\geq |{\mathcal B}|$ for all $a\in {\mathcal A}$ and $\textrm{r-comp}(\tau)\geq |{\mathcal A}||{\mathcal B}|$. For a morphism $\tau\colon {\mathcal A}\to {\mathcal B}$, we let $\tau({\mathcal A}^{{\mathbb Z}})$ denote the smallest subshift containing $\tau((x_i)_{i\in {\mathbb Z}})$ for all $(x_i)_{i\in {\mathbb Z}}\in {\mathcal A}^{{\mathbb Z}}$.

The following result is a consequence of Lemma 6.10 and Theorem 6.11 in \cite{DDMP20}.

\begin{theorem}\label{theo:boundedrepetition}
Let $\sigma_1 \colon{\mathcal C}^{\ast} \to {\mathcal D}^{\ast}$, $\sigma_2 \colon  {\mathcal B}^{\ast} \to {\mathcal C}^{\ast}$, and $\sigma_3 \colon  {\mathcal A}^{\ast} \to {\mathcal B}^{\ast}$ be three morphisms and assume that $\sigma_2$ and $\sigma_3$ are positive.
Then we have that
\[{ p_{\sigma_1 \circ \sigma_2\circ \sigma_3 ({\mathcal A}^{{\mathbb Z}}) } (n) \leq \begin{cases} (|{\mathcal C}| + (|{\mathcal B}|+1)\textrm{r-comp}(\sigma_2))n  &  \text{ if } n \in I \\ (|{\mathcal B}|+|{\mathcal C}|+\textrm{r-comp}(\sigma_2) + (|{\mathcal A}|+1)\textrm{r-comp}(\sigma_3))  n & \text{ if } n \in J \end{cases} }\]
where $I=[\| \sigma_1 \| , \langle \sigma_1 \circ \sigma_2\rangle)$ and $J=[\langle  \sigma_1 \circ \sigma_2  \rangle , \|\sigma_1 \circ \sigma_2\|)$.
\end{theorem}

We briefly recall the definition of ${\mathcal S}$-adic subshifts (see \cite{BSTY} for more details). A {\it directive sequence} ${\boldsymbol\tau}=(\tau_i)_{i\geq 0}$ is a sequence of non-erasing morphisms $\tau_i\colon {\mathcal A}_{i+1}^{\ast}\to {\mathcal A}_i^{\ast}$, $i\in {\mathbb N}$. We let $\tau_{[i,k)}$ denote the composition $\tau_i\circ\tau_{i+1}\circ\cdots\circ \tau_{k-1}$. We say that $\boldsymbol\tau$ is {\it everywhere growing} if $\langle \tau_{[0,i)} \rangle=\min_{a\in {\mathcal A}_i}\{|\tau_{[0,i)}(a)|\}$ tends to $\infty$ as $i\to\infty$. We say that $\boldsymbol\tau$ is {\it primitive} if for every $i\geq 0$, there exists $k\geq i$ such that $\tau_{[i,k)}$ has a positive incidence matrix.

For $i\geq 0$, the {\it language of order $i$} $L_{{\boldsymbol\tau}}^{(i)}$ associated with ${\boldsymbol\tau}$ is defined as
$$L_{{\boldsymbol\tau}}^{(i)}=\{w\in{\mathcal A}_i^{\ast}: \exists k>i, \exists a\in {\mathcal A}_k, w\prec \tau_{[i,k)}(a)\}.$$
For each $i\geq 0$, the set $X_{{\boldsymbol\tau}}^{(i)}$ is the set of infinite words $x\in{\mathcal A}_i^{\mathbb Z}$  whose factors belong to $L_{{\boldsymbol\tau}}^{(i)}$.
We set $X_{{\boldsymbol\tau}}=X_{{\boldsymbol\tau}}^{(0)}$, $L_{{\boldsymbol\tau}}=L_{{\boldsymbol\tau}}^{(0)}$ and call $(X_{{\boldsymbol\tau}}, S)$ the {\it ${\mathcal S}$-adic system generated by the directive sequence} ${\boldsymbol\tau}$, where $S$ is the shift transformation. For all $\ell \geq 1$, we denote by $L_{{\boldsymbol\tau},\ell}^{(i)}$ the subset of length $\ell$ factors of $L_{{\boldsymbol\tau}}^{(i)}$. When the directive sequence $\boldsymbol\tau$ is primitive, $(X_{\boldsymbol\tau},S)$ is a minimal subshift.

As a consequence of Theorem \ref{theo:boundedrepetition}, we get the following result.

\begin{proposition} \label{prop:Complexitybound}
Let $(X_{\boldsymbol\tau},S)$ be the ${\mathcal S}$-adic system generated by the directive sequence $\boldsymbol\tau = (\tau_i \colon  \mathcal{A}_{i+1}^{\ast}\to \mathcal{A}_i^{\ast})_{i\geq 0}$. For $n\in {\mathbb N}, n\geq \|\tau_0\|$, let $i=i(n)$ such that
$n\in [\|\tau_{[0,i)}\|, \|\tau_{[0,i+1)} \| )$.
Then
\[ p_{X_{\boldsymbol\tau}} (n) \leq \begin{cases} (|{\mathcal A}_{i}| + (|{\mathcal A}_{i+1}|+1)\textrm{r-comp}(\tau_i))n  & \text{ if }  n \in  I_i \\ (|{\mathcal A}_{i+1}|+|{\mathcal A}_{i}|+\textrm{r-comp}(\tau_i) + (|{\mathcal A}_{i+2}|+1)\textrm{r-comp}(\tau_{i+1}))  n & \text{ if } n \in J_i \end{cases} \]
where $I_i=[\| \tau_{[0,i)} \| , \langle \tau_{[0,i+1)}\rangle)  $ and $J_i= [\langle  \tau_{[0,i+1)} \rangle , \|\tau_{[0,i+1)}\|) $.

\end{proposition}

\begin{proof}
It follows from applying \cref{theo:boundedrepetition} for the morphisms $\sigma_3=\tau_{i+1}\colon{\mathcal A}_{i+2}^{\ast}\to {\mathcal A}_{i+1}^{\ast}$, $\sigma_2=\tau_{i}\colon{\mathcal A}_{i+1}^{\ast}\to {\mathcal A}_{i}^{\ast}$ and $\sigma_1=\tau_{[0,i)}\colon{\mathcal A}_{i}^{\ast}\to {\mathcal A}_{0}^{\ast}$, and recalling that $X_{\boldsymbol\tau}\subseteq \sigma_1\circ \sigma_2 \circ \sigma_3({\mathcal A}_{i+2}^{{\mathbb Z}})$.
\end{proof}

The following notion connecting ordered Bratteli diagrams and ${\mathcal S}$-adic systems was originally  introduced in \cite{fdurand}. Given an ordered Bratteli diagram $(V,E,\geq)$, let $i\geq 1$ and consider $V_i$, $V_{i+1}$ as finite alphabets. For every $u\in V_{i+1}$, consider the ordered list $(e_1,e_2,\cdots, e_k)$ of edges on $E_{i+1}$ arriving to $u$, and let $(v_1,v_2,\cdots, v_k)$ the list of labels of the sources of these edges in $V_i$. This defines a morphism $\tau_i\colon V_{i+1}^{\ast} \to V_{i}^{\ast}$,  $u\mapsto v_1v_2\cdots v_k$. For $i=0$, we let $\tau_0\colon V_1^{\ast} \to E_1^{\ast}$ denote the morphism such that
$\tau(v)=e_1(v)\ldots e_\ell(v)$, where  $e_1(v),\ldots,e_\ell(v)$ are the edges connecting $V_0$ with $v$ according to the order of the diagram. For $i\geq 0$ we say that $\tau_i$ is {\it the morphism read on $(V,E,\geq)$ at level $i$} and that the directive sequence ${\boldsymbol \tau}=(\tau_i)_{i\geq 0} $ is the {\it sequence of morphisms read  on} $(V,E,\geq)$. Note that the morphism read on $(V,E,\geq)$ at level $0$ is a hat morphism.\\

The previous notion gives an ${\mathcal S}$-adic subshift naturally associated with each Bratteli-Vershik system. The following result, which is a slight modification of \cite[Proposition 4.5]{DDMP20}, states when these two representations are conjugate as dynamical systems (see also \cite[Proposition 2.2]{DurandLeroy2012}).
\begin{proposition}\cite[Proposition 4.5]{DDMP20}\label{prop:isom}
Let $(X_{\boldsymbol\tau},S)$ be the minimal ${\mathcal S}$-adic subshift defined by the directive sequence $\boldsymbol\tau  = (\tau_i)_{i\geq 0}$, where $\tau_0$ is a hat morphism and $\tau_i$ is proper for all $i\geq 1$. Suppose that all morphisms $\tau_i$ extend by concatenation to a one-to-one map from $X^{(i+1)}_{\boldsymbol\tau}$ to $X^{(i)}_{\boldsymbol\tau}$. Then, $(X_{\boldsymbol\tau},S)$ is conjugate to the Bratteli-Vershik system $(X_{B} , V_{B})$ associated with the ordered Bratteli diagram $(V,E,\geq)$, where $\boldsymbol\tau$ is the sequence of morphisms read  on $(V,E,\geq)$.
\end{proposition}
In the following lemma we give sufficient conditions for a simple Bratteli diagram to have an order such that the sequence of morphisms read on the ordered diagram satisfies the hypothesis of \cref{prop:isom}.

\begin{lemma}\label{lema:injective}
Let $(V,E)$ be a simple Bratteli diagram with sequence of incidence matrices $(A_i)_{i\in{\mathbb N}}$. For all $i\in{\mathbb N}$, let $V_i$ be the set of vertices at level $i$ and $m_i=|V_i|$. Suppose that for all $i\geq 1$, $A_i(j,1)> m_{i+1}$ for all $1\leq j\leq m_{i+1}$. Then it is possible to give $(V,E)$ an order $\geq$ such that the sequence $\boldsymbol{\tau}=(\tau_i)_{i\in{\mathbb N}}$ of morphisms read on $(V,E,\geq)$ verifies that for all $i\geq 1$, $\tau_i:V_{i+1}^{\ast}\to V_i^{\ast}$ is proper, injective and extends by concatenation to a one-to-one map from $X^{(i+1)}_{\boldsymbol{\tau}}$ to $X^{(i)}_{\boldsymbol{\tau}}$.
\end{lemma}

\begin{proof}
For the edges of $E_1$, choose any order. Let $i\geq 1$, let $V_i=\{u_1,\cdots, u_{m_i}\}$ and $V_{i+1}=\{v_1,\cdots, v_{m_{i+1}}\}$. For every $v_j\in V_{i+1}$, we order the edges $r^{-1}(v_j)$ in the following manner. The first $j$ edges have source $u_1$ (this is possible since we assume that $A_i(j,1)> m_{i+1}$ for all $1\leq j\leq m_{i+1}$). The next $A_i(j,2)$ edges have source $u_2$. The remaining edges are ordered left/right.

The previous order gives the following morphism $\tau_i\colon V_{i+1}^{\ast}\to V_i^{\ast}$,
$$\tau_i(v_j)=u_1^{\ell_{1,j}}u_2^{\ell_{2,j}}u_1^{\ell'_{1,j}}u_3^{\ell_{3,j}}u_4^{\ell_{4,j}}\cdots u_{m_i}^{\ell_{m_i,j}} \quad \forall 1\leq j\leq m_{i+1},$$
where $\ell_{1,j}=j,\ell'_{1,j}=A_i(j,1)-j$ and $\ell_{k,j}=A_i(j,k)$ for all $2\leq k\leq m_i$. It is clear that $\tau_i$ is proper. Since $\ell_{1,j}$ is different for each $v_j\in V_{i+1}$, $\tau_i$ is injective on $V_{i+1}$.

Let us show that $\tau_i$ extends by concatenation to a one-to-one map from $X^{(i+1)}_{\boldsymbol{\tau}}$ to $X^{(i)}_{\boldsymbol{\tau}}$. For $x\in X^{(i+1)}_{\boldsymbol{\tau}}$, we define the set $C_{\tau_i}(x)$ of {\it cutting points} in $\tau_i(x)$ as the following set,
$$C_{\tau_i}(x)=\{|\tau_i(x_{[0,\ell)})|:\ell >0\}\cup\{0\}\cup \{-|\tau_i(x_{[\ell,0)})|:\ell <0\}.$$
Note that for the order we have considered, for any word $x\in X^{(i+1)}_{\boldsymbol{\tau}}$, the cutting points of $\tau_i(x)$ are located exactly in those places where a letter $u_{m_i}$ is followed by a letter $u_1$.

Let $x,x'\in X^{(i+1)}_{\boldsymbol{\tau}}$ be two infinite words such that $\tau_i(x)=\tau_i(x')$. We want to prove that $x=x'$. Since $\tau_i(x)=\tau_i(x')$, each time we see the word $u_{m_i}u_1$ in $\tau_i(x)$, we see it in $\tau_i(x')$ as well, and {\it vice versa}, so the cutting points of $x$ and $x'$ are located exactly in the same places. Therefore $\tau_i(x_\ell)=\tau_i(x'_\ell)$ for all $\ell\in{\mathbb Z}$. Since $\tau_i$ is injective in $V_{i+1}$, we conclude that $x_\ell=x'_\ell$ for all $\ell\in{\mathbb Z}$, thus $x=x'$.
\end{proof}

\section{Proof of the main results.}\label{sec:proofs}

\subsection{Building a suitable representation}
Given a dimension group with unit, the main idea is to realize it as the dimension group of a  Bratteli-Vershik system, on which we have a precise control of growth of the number of vertices at level $k$, versus the number of paths between the level $k$ and the root of the Bratteli diagram. To this end we introduce the following notion.
\subsubsection*{Splitting the levels}
Let $A$ be a $n\times m$ integer matrix with positive entries and  $d$ a positive integer. For each entry $a_{k,j}$ of $A$, write $a_{k,j}=dq_{k,j}+r_{k,j}$ where $q_{k,j}\in {\mathbb N}$ and $0\leq r_{k,j}< d$. Let $Q=(q_{k,j})_{k,j}$ and $R=(r_{k,j})_{k,j}$. Then
\[ A=dQ+R.\]
Let $B$ be the $n\times 2m$ matrix such that $B_{\bullet,2i-1}=dQ_{\bullet,i}$ and $B_{\bullet,2i}=R_{\bullet,i}$ for $1\leq i \leq m$. That is, in $B$ we write the columns of $dQ$ and $R$ interleaved.
Notice that \[  A= B C \]
where $C$ is the $2m\times m$ matrix given by $C_{\bullet,i}=e_{2i-1}+e_{2i}$, for $1\leq i \leq m$, $e_i$ being the $i$th canonical vector of ${\mathbb R}^{2m}$.

For convenience we write $B=B(A,d)$, $C=C(A,d)$ to stress the dependence on $A$ and $d$.

\begin{proof}[Proof of \cref{theo:main1}]

Let $(G,G^+,u)$ be a simple dimension group with unit, let $(V,E)$ be a Bratteli diagram such that $(G,G^+,u)\cong K_0(V,E)$. Let $(A_i)_{i\in {\mathbb N}}$ denote the sequence of adjacency matrices of $(V,E)$, where $A_i$ is of $n_i\times m_i$ (so $n_i=m_{i+1}$). We may assume that the entries of each $A_i$ are positive. Let $h_i$ be such that $h_i^2+h_i<\min_{k,j}\{(A_i)_{k,j}\}$ and denote $B_i=B(A_i,h_i)$, $C_i=C(A_i,h_i)$ for $i\in {\mathbb N}$.

We may telescope the diagram to guarantee that for all $i\geq 1$, $h_i$ is large enough to satisfy
\begin{equation} \label{equation:growthh's} \frac{m_i^3 t}{p_{t}} <\frac{1}{i}\end{equation}
for all $t\geq h_i$. This is because for a fixed $i$, there exists $t_i$ such that $\forall t\geq t_i$,
$$\frac{t}{p_t}<\frac{1}{i m_i^3},$$
so we can telescope up to a deeper level $i'$ such that $h_i>\max\{t_i,2m_{i-1}\}$ and at the same time $h_i+h_i^2<\min_{k,j}\{(A_{i'}A_{i'-1}\cdots A_i)_{k,j}\}$, and we define $i+1=i'$. The condition $h_i>2m_{i-1}$ will be used later in the proof.
We define the diagram $(V',E')$ as the one given by the matrices $(A_i)'_{i\in {\mathbb N}}$ with $A_0'= C_0$ and $A_i'=C_iB_{i-1}$ for $i\geq 1$.
Note that the diagrams $(V,E)$ and $(V',E')$ satisfy $K_0(V,E)\cong K_0(V',E')$ since  we have the equations $A_i=B_iC_i$ and $A_i'=C_iB_{i-1}$. Note also that $|V_0'|=|V_0|=1$ and $|V_i'|=2|V_{i-1}|=2m_{i-1}$ for all $i\geq 1$.

Let $J_0=1$, and for all $i\geq 1$, let $J_i$ be the $2m_{i-1}\times 2m_{i-1}$ integer matrix given by
\[  J_i=S_iD_i\]

where:
\begin{itemize}
\item $D_i$ is the $2m_{i-1}\times 2m_{i-1}$ diagonal matrix with positive integer coefficients such that $(D_i)_{k,k}=h_{i-1}$ if $k$ is odd and  $(D_i)_{k,k}=1$ if $k$ is even.

\item $S_i$ is a row addition matrix of $2m_{i-1}\times 2m_{i-1} $, such that for a matrix $X$ of $2m_{i-1}\times k$, $S_iX$ is the $2m_{i-1}\times k$ matrix such that
\[ (S_iX)_{2j-1,\bullet}=X_{2j-1,\bullet} \text{ and } (S_iX)_{2j,\bullet}=X_{2j-1,\bullet}+X_{2j,\bullet} \ \text{ for all } 1\leq i\leq m_{i-1}.  \]

\end{itemize}
For all $i\geq 0$, let $A''_i=J_{i+1}A_i'J_{i}^{-1}$.\\
{\bf Claim:} We claim that $A_i'J_{i}^{-1}$ is a matrix with entries in ${\mathbb N}^{\ast}$. To see this, note  that $A_i'J_{i}^{-1}=C_iB_{i-1}D_{i}^{-1}S_{i}^{-1}$ and that when we multiply $B_{i-1}$ from the right by the matrix $D_{i}^{-1}S_{i}^{-1}$, we first divide the odd columns of $B_{i-1}$ by $h_{i-1}$ and then substract the  $2j$-th column to the divided ($2j-1$)th column of $B_{i-1}$. Therefore, the entries of the odd columns of $B_{i-1}D_{i}^{-1}S_{i}^{-1}$ have the form $q_{i-1}(k,j)-r_{i-1}(k,j)$, and those of the even columns have the form $r_{i-1}(k,j)$, where $q_{i-1}(k,j)$ and $r_{i-1}(k,j)$ are the coefficients of the matrices $Q$ and $R$ used in the splitting of $A_{i-1}$ with respect to $h_{i-1}$. This implies that $B_{i-1}D_{i}^{-1}S_{i}^{-1}$ has integer entries. Let us show that they are positive. This is clear for the even columns. For the odd columns, note that since $h_{i-1}q_{i-1}(k,j)>h_{i-1}^2+h_{i-1}-r_{i-1}$ and $r_{i-1}(k,j)< h_{i-1}$, $q_{i-1}(k,j)>h_{i-1}$, we have tnat $q_{i-1}(k,j)-r_{i-1}(k,j)>0$. This shows the Claim.

Let $(V'',E'')$ be the Bratteli diagram associated with the matrices $(A_i'')_{i\in {\mathbb N}}$. By the Claim above, the sequence $(J_i)_{i\in{\mathbb N}}$ is adapted to $(A_i')_{i\in{\mathbb N}}$, and we have that $(V'',E'')\sim (V', E')$ by \cref{lema:AdaptedMatrices}. This implies that $(G,G^+,u)\cong K_0(V'',E'')$. Note that $|V_i''|=|V_i'|$ for all $i\in {\mathbb N}$, so $|V_0''|=1$ and $|V_i''|=2m_{i-1}$ for all $i\geq 1$.

For  the diagram $(V'',E'')$ we  may consider an order as in \cref{lema:injective}. That is, we put any order in $E_1''$, and for all $i\geq 1$, we consider the order given by the morphisms $\tau_i:V_{i+1}''\to V_i''$, defined by
$$\tau_i(v_j)=u_1^{\ell_{1,j}}u_2^{\ell_{2,j}}u_1^{\ell'_{1,j}}u_3^{\ell_{3,j}}u_4^{\ell_{4,j}}\cdots u_{n_i}^{\ell_{n_i,j}} \quad \text{ for all } 1\leq j\leq n_{i+1},$$
where $n_i=|V_i''|$, $n_{i+1}=|V_{i+1}''|$, $\ell_{1,j}=j,\ell'_{1,j}=A_{i}''(j,1)-j$ and $\ell_{k,j}=A_{i}''(j,k)$ for all $2\leq k\leq n_i$. Indeed, the condition required to define this order is that the minimum number of edges connecting the first vertex in $V_i''$ with any vertex in $V_{i+1}''$, is strictly larger than $n_{i+1}=2m_{i}$, that is, $A_{i}''(j,1)> 2m_{i}$. In our case, all entries of $A_{i}''$ are greater than $h_{i+1}$ by construction, in particular those of its first column. Since we have chosen $h_{i+1}$ greater than $2m_{i}$, the condition is fulfilled.

Let $\geq$ denote the partial order considered on $E''$. Let $(Y,S)$ be the ${\mathcal S}$-adic subshift generated by the directive sequence $\boldsymbol\tau=(\tau_i)_{i\geq 0}$ of morphisms read on $(V'',E'',\geq)$. Combining \cref{prop:isom} and \cref{lema:injective} we obtain that the Bratteli-Vershik system associated with $(V'',E'', \geq)$ is conjugate to $(Y,S)$. We obtain that $(G,G^+,u)\cong K^0(Y,S)$, which proves the first part of the theorem.

Remark that by construction we have that $\langle \tau_{[0,i)}\rangle \geq h_{i+1}$ for all $i\geq 1$. Note also that $\textrm{r-comp}(\tau_{i})=|V_{i+2}''|(|V_{i+1}''|+1)$ so we can assume that $\textrm{r-comp}(\tau_{i})\leq |V''_{i+2}|^2=4m_{i+1}^2$.

For $n\in {\mathbb N}$, $n\geq \|\tau_0\|$, let $i=i(n)$ such that $n\in [\|\tau_{[0,i)}\|, \| \tau_{[0,i+1)}\|)$. Then
\cref{prop:Complexitybound} gives us that

\[p_Y(n) \leq \begin{cases}   3|V_{i+2}''|^3   \cdot n  & \text{ if }  n\in [\|\tau_{[0,i)}\|, \langle  \tau_{[0,i+1)}\rangle)    \\
5|V_{i+3}''|^3\cdot n & \text{ if }  n\in [\langle \tau_{[0,i+1)}\rangle , \| \tau_{[0,i+1)}\|)
\end{cases}  \]
For $n\in [\|\tau_{[0,i)}\|, \langle  \tau_{[0,i+1)}\rangle)$, since $n\geq h_{i+1}$, by \eqref{equation:growthh's} we have that

\[\frac{p_{Y}(n)}{p_n} \leq 24m_{i+1}^3\frac{n}{p_n} \leq 24\frac{1}{i+1}  \]
while for $ n\in [\langle \tau_{[0,i+1)}\rangle , \| \tau_{[0,i+1)}\|)$, since $n\geq h_{i+2}$, by \eqref{equation:growthh's}

\[\frac{p_{Y}(n)}{p_n} \leq 40m_{i+2}^3\frac{n}{p_n}\leq 40\frac{1}{i+2} \]
From this, we deduce that $p_Y(n)/p_n \to 0$ since $n\to \infty$ implies that $i(n)\to \infty$. This proves the second part of the theorem.

\end{proof}

\begin{proof}[Proof of \cref{theo:main2}]
It suffices to take $(G,G^+,u)$ equal to $K^0(X,T)$ and apply \cref{theo:main1}. The resulting subshift $(Y,S)$ has the desired property on the complexity and is strong orbit equivalent to $(X,T)$ by \cref{theo:caracterizationoe}.
\end{proof}

For the proof of \cref{thm:divisible_toeplitz} we proceed in a slightly different way, since, as we will see in the proof, having a divisible dimension group gives more flexibility when modifying the Bratteli diagram. Our strategy follows the ideas in the proof of \cite[Theorem 12]{noruegos}, tailored to our situation in order to obtain a Toeplitz subshift $(Y,S)$ having low enough complexity. For the sake of completeness we provide all the details, repeating some arguments of \cite[Theorem 12]{noruegos}.

\begin{proof}[Proof of \cref{thm:divisible_toeplitz}] Suppose that $G$ is a divisible group and let $(V,E)$ be a Bratteli diagram such that $(G,G^+,u)\cong K_0(V,E)$. Let $(A_i)_{i\in {\mathbb N}}$ be the sequence of adjacency matrices of $(V,E)$, where $A_i$ is of $m_{i+1}\times m_i$. We may assume that all the entries of each $A_i$ are positive, that $m_i\geq 2$ for all $i\geq 1$ and that $m_1=2$ with a single edge connecting $v_0$ with each of the two vertices of $V_1$. This is because we can always split the levels where there is only one node in a similar way as in the non-divisible case, to get a new Bratteli diagram with the same dimension group with unit: write the first adjacency matrix as $A_0=B_0C_0$, where $C_0=(1,1)^t$ and $B_0$ is any $|m_1|\times 2$ integer matrix verifying $B_0(j,1)+B_0(j,2)=A_0(j,1)$ for all $1\leq j\leq m_1$; for $i\geq 1$, if $m_i\geq 2$, define $C_i=\Id_{m_i\times m_i}$ and $B_i=A_i$, if $m_i=1$, define $C_i=(1,1)^t$ and $B_i$ as any $|m_{i+1}|\times 2$ integer matrix verifying $B_i(j,1)+B_i(j,2)=A_i(j,1)$ for all $1\leq j\leq m_{i+1}$. Then, define $\Tilde{A_0}=C_0$ and $\Tilde{A_i}=C_iB_{i-1}$ for all $i\geq 1$. The Bratteli diagram associated with the sequence $(\Tilde{A_i})_{i\in{\mathbb N}}$ has the desired property.

Note that since we assumed that $A_0=(1,1)^t$, the first adjacency matrix has the ERS property. The group $G$ corresponds to the inductive limit $\varinjlim_i ({\mathbb Z}^{m_i},A_i)$ and being $G$ divisible, we also have that $\varinjlim_i ({\mathbb Z}^{m_i},A_i)\cong \varinjlim_i ({\mathbb Q}^{m_i},A_i)$ by \cref{prop:divisible1}.

For each $i\geq 1$, let $t_i$ be a positive integer such that $\forall t\geq t_i$,
\begin{equation} \label{equation:growtht} \frac{m_{i+1}^3 t}{p_{t}} <\frac{1}{i+1} \end{equation}
We will define a new sequence of matrices $(B_i)_{i\in{\mathbb N}}$ in the following way. Let $B_0=A_0$, $J_0=1$, $J_1=\Id_{m_1\times m_1}$ and $k_1=1$. For $1\leq j\leq m_2$, let $\alpha_j$ be the sum of the $j$th row of $A_1$. Let $J_2'$ be the diagonal $m_2\times m_2$ matrix
$$\diag(\alpha_1^{-1},\alpha_2^{-1},\cdots, \alpha_{m_2}^{-1}).$$
Let $s_2\in {\mathbb N}^{\ast}$ be such that $s_2J_2'A_1$ has integer entries (for instance, take $s_2$ to be the least common multiple of the denominators of the entries of $J_2'A_1$), and $\ell_2\in {\mathbb N}^{\ast}$ chosen so that the two following conditions hold,
\begin{itemize}
\item $2s_2\ell_2>t_2$,
\item $2s_2\ell_2\cdot \min_{p,q}\{(J_2'A_1)(p,q)\}>m_2$.
\end{itemize}
Let $k_2=2s_2\ell_2$ and define $J_2=k_2J_2'$, $B_1=J_2A_1$. Note that $J_2$ is invertible over ${\mathbb Q}$ and that $B_1$ is a positive integer matrix whose entries are all divisible by $2$ and strictly greater that $m_2$. Note also that $B_1$ has the ERS property, with $B_1(1,1,\cdots, 1)^t=(k_2,k_2,\cdots, k_2)^t$.

Suppose we have defined $J_2,J_3,\cdots, J_{i-1}$. Let $J_i'$ the diagonal invertible matrix over ${\mathbb Q}$ such that $J_i'A_{i-1}J_{i-1}^{-1}(1,1,\cdots, 1)^t=(1,1,\cdots, 1)^t$. Let $s_i\in {\mathbb N}^{\ast}$ such that  $s_iJ_i'A_{i-1}J_{i-1}^{-1}$ has integer entries, and $\ell_i\in {\mathbb N}^{\ast}$  so that the two following conditions hold,
\begin{itemize}
\item $is_i\ell_i>t_i$,
\item $is_i\ell_i\cdot \min_{p,q}\{(J_i'A_{i-1}J_{i-1}^{-1})(p,q)\}>m_i$.
\end{itemize}
Let $k_i=is_i\ell_i$. Define $J_i=k_iJ_i'$, $B_{i-1}=J_iA_{i-1}J_{i-1}^{-1}$.
Note that for all $i\geq 0$, the matrix $B_i$ has the ERS property with $B_i(1,1,\cdots, 1)^t=(k_{i+1},k_{i+1},\cdots, k_{i+1})^t$. Note also that for all $i\geq 1$, $B_i$ satisfies the following properties,
\begin{itemize}
    \item $B_i$ is a positive integer matrix whose entries are all divisible by $i+1$.
    \item All the entries of $B_i$ are strictly greater that $m_{i+1}$.
\end{itemize}
Note also that the sequence $(k_i)_{i\geq 1}$ satisfies that $k_i>t_i$ for all $i\geq 2$.
Since all matrices $J_i$ are positive and invertible over ${\mathbb Q}$, the inductive limits $\varinjlim_i ({\mathbb Q}^{m_i},A_i)$ and $\varinjlim_i ({\mathbb Q}^{m_i},B_i)$ are order isomorphic, and the latter is at the same time order isomorphic to $\varinjlim_i ({\mathbb Z}^{m_i},B_i)$, since all the entries of each $B_i$ are divisible by $i+1$ (see \cref{prop:divisible2}).

Let $(V',E')$ be the Bratteli diagram associated with $\varinjlim_i ({\mathbb Z}^{m_i},B_i)$. By construction, $(G,G^+,u)\cong K_0(V',E')$. Note that for all $i\in{\mathbb N}$, $|V_i'|=|V_i|=m_i$. Since for all $i\geq 1$, each entry of $B_i$ is strictly greater than $m_{i+1}$, we  may endow $(V',E')$ with the order given in \cref{lema:injective}. As before, let $\geq$ denote the partial order considered on $E'$, and let $(Y,S)$ be the ${\mathcal S}$-adic subshift generated by the directive sequence $\boldsymbol{\tau}=(\tau_i)_{i\geq 0}$ of morphisms read on $(V',E')$. Combining \cref{prop:isom} and \cref{lema:injective} we obtain that the Bratteli-Vershik system associated with $(V',E', \geq)$ is conjugate to $(Y,S)$. Since the adjacency matrices of $(V',E')$ have the ERS property, \cref{teo:ers} gives us that $(Y,S)$ is a Toeplitz subshift.
We now estimate the complexity $p_Y(n)$. Remark that since the $B_i$'s have the ERS property,
$$\|\tau_{[0,i)}\|=\langle \tau_{[0,i)}\rangle=k_1k_2\cdots k_{i+1}\geq k_{i+1}.$$
Note also that $\textrm{r-comp}(\tau_{i})=|V_{i+2}'|(|V_{i+1}|'+1)=m_{i+2}(m_{i+1}+1)$, so we can assume that $\textrm{r-comp}(\tau_{i})\leq |V_{i+2}'|^2=m_{i+2}^2$.

For $n\in {\mathbb N}$, $n\geq \|\tau_0\|$, let $i=i(n)$ such that $n\in [\|\tau_{[0,i)}\|, \| \tau_{[0,i+1)}\|)$. \cref{prop:Complexitybound} gives us that
\[p_Y(n) \leq    3m_{i+2}^3   \cdot n. \]
Since $n\geq \|\tau_{[0,i)}\| \geq  k_{i+1}$, and $k_{i+1}\geq t_{i+1}$, by \eqref{equation:growtht} we have that
\[\frac{p_{Y}(n)}{p_n} \leq 3m_{i+2}^3\frac{n}{p_n} \leq \frac{3}{i+2}.  \]
From this, we deduce that $p_Y(n)/p_n \to 0$ since $n\to \infty$ implies that $i(n)\to \infty$. The proof is completed.
\end{proof}

\begin{proof}[Proof of \cref{theo:main3}]
Given any Choquet simplex $K$, we can choose a countable, dense subgroup $G$ of $\Aff(K)$, such that $G$ is a ${\mathbb Q}$-vector space and contains the constant function $1$. Then, $G$ is divisible and by \cref{theo:simplexanddg} $K$ is affine homeomorphic to $S(G,G^+,1)$, where the positive cone is defined as in \cref{theo:simplexanddg}. Applying \cref{thm:divisible_toeplitz} to $(G,G^+,1)$, we get a Toeplitz subshift $(X,S)$ such that $\lim p_{X}(n)/p_n=0$ and $K^0(X,S)\cong (G,G^+,1)$. By \cite[Theorem 5.5]{HPS}, ${\mathcal M}(X,S)\cong S(G,G^+,1)$, and then $K\cong {\mathcal M}(X,S)$.
\end{proof}

We suspect that in \cref{thm:divisible_toeplitz} the condition of $G$ being divisible is superfluous, and that it could be replaced with ``$G$ is the dimension group associated with a Toeplitz subshift", but currently we do not know how to prove it.

It would be interesting to additionally consider ergodic theoretical properties that we can obtain in superlinear complexity subshifts. This has been recently explored in \cite{CKJS}, where the authors constructed examples of subshifts with superlinear complexity admitting loosely Bernoulli and non-loosely Bernoulli ergodic measures.

redpajama_set_name: RedPajamaArXiv


It wasn't originally planned that Hayden Christensen would wear the Darth Vader costume. He asked the producers […] and they had to change the costume, put in shoulder pads and high heels… I think he sees through Darth Vader's mouth.
It's very cute and all (newsflash: I fall for tiny twinks), but IMDb measures him at 6'1. Has there been, like on gay chatrooms, an inflation over the last two years, and now everybody is over 6'0?


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Mercedes Class S
The class S is the flag ship of the Mercedes Line. Launched for the first time in 1972, the Mercedes class S combines the most refined level of comfort and leading edge technology into a luxury, high performance sedan. Among its many features are independent air conditioning. leather interior and tinted windows to ensure your ride in ultimate comfort. This Mercedes Class S 320 will provide you a special touch of class to your stay in Romania.
Mercedes Class E
The E-Class has never been the biggest, fastest nor most technically sophisticated of the Mercedes-Benz cars, but it is surely the most recognizable Mercedes in the world. The appeal of this mid-size luxury sedan lies in a combination of safety, luxury, practicality, sport and status that no other category can match. Noise, vibration and harshness control is first rate. Inside, the E Class is an exercise in comfort and safety from top to bottom. This top of the range vehicle offers the passenger a full compliment of luxury extras, including automatic air-conditioning and full leather upholstery.
You can pay for our services using : VISA-VISA ELECTRON-MASTERCARD-MAESTRO
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Big Battery Low Vision App for Low Vision users.
Big Battery is a Large Print Battery App for Visual Impairment Aid.
Big Battery displays your battery level in Large Print. It's a lot easier to read than the normal Battery level for partial sight.
This App works for Ipad and Iphone. In portrait mode, you have more details such as time remaining for Wifi, 3G etc. If you just want to hide the details and just see the Battery level, rotate the device to landscape and the details will be hidden.
There are 4 high contrast modes to choose from, just tap this button ❖, your favorite mode will be saved even if you close the app and will remain until you change it. There is also a low glare invert mode to match the Device invert mode settings.
To allow for maximum size print, the word 'hours' has been removed to save space. All numbers represent the estimated hours of time left for that activity.
Your Battery Level will be easy to read in all lighting conditions. You do not have to increase your screen brightness. Use it conveniently in Outdoor light, Indoor light and very low lighting. Even with patchy screen reflection, there is clarity.


Smooth fresh greenish tasting liquor is characteristic of this unique green tea.The tea imparts the fresh green character you would expect to be a part of any springtime tea experience!
Nitrogen keeps the tea fresh for up to 8 years in an unopened pack. Small batch packed in a HACCP facility in Canada. Luxury Loose Leaf Tea.


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Now Reading Late Fan-Favorite Ben's Cat Named Vox Populi Winner
Late Fan-Favorite Ben's Cat Named Vox Populi Winner
Content provided by BloodHorse
Ben's Cat after winning the 2015 Jim McKay Turf Sprint at Pimlico Race Course. (Eclipse Sportswire)
The late Ben's Cat, four-time Maryland Horse of the Year, has been awarded the 2017 Secretariat Vox Populi Award, or "Voice of the People," with record-breaking voter participation.
The award, created by Secretariat's late owner Penny Chenery, annually recognizes the horse whose popularity and racing excellence best resounded with the American public and gained recognition for Thoroughbred racing.
Ben's Cat (Eclipse Sportswire)
During Ben's Cat's eight-year career, which included three races in 2017, the 11-year-old multiple stakes winner finished in the money in 48 of his 63 starts, earned $2.6 million and widespread fan devotion that grew from his Mid-Atlantic base and spread internationally.
Hall of Fame trainer King Leatherbury bred, owned, and trained Ben's Cat.
While many of Thoroughbred racing's stars emerge from the Triple Crown trail and other noteworthy stakes races, the beginnings for Ben's Cat were humble. Leatherbury entered the then 4-year-old in a $16,000 maiden-claiming race at Pimlico Race Course in 2010. The dark bay gelding, who was of modest breeding and had broken his pelvis a couple of years prior, won that race — and he kept winning year after year, on the turf and the dirt, during a long and accomplished career that culminated this past June and boasted 26 stakes victories. He died just a few weeks after his retirement following complications from colic.
"My mother created this award to honor the horses whose stories draw people to Thoroughbred racing and convert them to fans," said Kate Tweedy, daughter of the late Penny Chenery, who established the Vox Populi Award in 2010. "She certainly admired the journey of Ben's Cat and would have been thrilled with the outcome of this year's voting. Our family looks forward to meeting Mr. Leatherbury and sharing this special celebration with racing fans."
Leatherbury, 84, who races under The Jim Stable, continues to train on the Mid-Atlantic circuit and has a career total of nearly 6,500 wins.
"It is quite an honor for Ben's Cat to receive this impressive award," Leatherbury said. "I never would have guessed his fan base reached that far. Ben's Cat won a lot of races, and I sure did enjoy racing him. This is really something."
The Tweedy family will present the custom-made Vox Populi trophy to Leatherbury Jan. 20 at Santa Anita Park. In conjunction with the presentation, a variety of festivities will be offered to the public.
Ben's Cat was one of a select group of nominees submitted by the Vox Populi committee. In the online poll, thousands of Vox Populi voters from more than 50 countries showed their support by choosing Ben's Cat. He joins other Vox Populi Award winners: California Chrome (2016 and 2014), American Pharoah (2015), Mucho Macho Man (2013), Paynter (2012), Rapid Redux (2011), and Zenyatta (inaugural 2010).
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performing arts (symphony, ballets & plays).
Below I have listed some of my favorite restaurants.
you to view their menus online.
Hope you enjoy my selections.
Excellent food for the price.
Owned by the same individuals who run Mustard Seed.


Andrea Russett is an American actress and YouTuber, who runs the popular YouTube channel, 'GETTOxFABxFOREVER'. Check out this biography to know about her childhood, family life, achievements and fun facts about her life.
Andrea Russett is an American actress and YouTuber, who runs the popular YouTube channel, 'GETTOxFABxFOREVER'. Apart from her YouTube channel, she is also active on other social media platforms, such as Instagram, Vine and Snapchat. Andrea has been a diehard Justin Bieber fan since her teens and he was also the very reason for her first video uploaded on YouTube. As a 14 year old girl, Andrea made her first video and posted it on YouTube to participate in a contest that could have given her the chance to meet Justin Bieber in person. This was the beginning of the internet career of this high school dropout who firmly believes that "everybody learns differently" and for her doing something is the only way to learn. She first appeared as an actress in 2014 in the teen comedy film, 'Expelled', where she acted alongside Cameron Dallas. In 2013, she started her own clothing line named 'FLAWD Clothing'.
By the age of 14, Andrea Russett was already a YouTube Star owing to her stop motion video named "JUSTIN BIEBER ONE TIME MUSIC VIDEO" that went viral within few days. However, the rapidly increasing number of views on her video did not win her the chance to meet Justin Bieber but instead she became an internet celebrity. She worked as a radio jockey in a local radio station before she was pushed to stardom by her music and lifestyle based videos. In 2013, Russet signed a contract with 'Fullscreen, Inc.', a global network of content creators and brands on YouTube, and was relocated to California. In the same year, she started designing for her own clothing line and launched her own brand FLAWD Clothing.
Soon after moving to California, Andrea also started focusing on her acting career. In 2014, she first landed her role as an actress in the feature length comedy film, 'Expelled'; the video of which was released by 20th Century Fox Home Entertainment. Thereafter, Andrea appeared in different small film roles and film shorts in the upcoming years. She also made appearances as guest in television shows. Her career took the next step in 2016 when she co-produced and starred in 'Sickhouse', a thriller film that was initially released in brief installments on Andrea's Snapchat account. The film received praises from the viewers as well as from the critics due to its "different" storytelling technique. The online popularity of the film also fueled the making of a sequel to the film, which is yet to be released.
In 2016, she appeared in two short films, 'Lovesick' and 'A Peculiar Tale'. In the same year, she did a cameo in the film 'Mike and Dave Need Wedding Dates'. She also appeared on the reality television series 'Catfish: The TV Show' aired by MTV. 2016 was also the year in which Andrea Russett released her first compilation "Mind Medicine" with the motto to give more exposure to talented but underrated EDM artists like Coleman Hell and Robert DeLong. In 2017, she is back as the anchor of the weekly talk series 'Apologies In Advance'. Andrea Russett enjoys over 2 million followers on both YouTube and Instagram.
She loves animals and also has come up with her initiative 'dollarsforcollars' for raising money to be used in preventing cruelty on animals. She was dating fellow YouTube personality, Kian Lawley, but things seem not to turn out well between the two. Currently, she is in a relationship with the singer, Kenny Holland.


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Watch Official Music Videos In Facebook Watch
Back in the day if you wanted to watch an official music video by your favorite singer or band, you would have to rely on TV channels like MTV. These days, a lot of musicians post their videos directly to YouTube where users can seek it out and watch it on demand. Now it looks like musicians will have a new avenue and it comes in the form of Facebook Watch.
According to Facebook, "Today, we're adding a new way for people to come together around music by bringing official music videos to Facebook in the US. Starting this weekend, you'll be able to discover, watch and share music videos from today's top artists to up-and-coming bands and classics across various music genres on Facebook."
If you head on over to Facebook Watch, you will notice a new Music section where you'll be able to discover artists based on genre or mood. According to Facebook, they actually started building on this back in 2019 with partners in India and Thailand to lay the foundation. It was also previously reported that Facebook was negotiating with various record labels to get the rights to stream music through its platform.
As to why you would choose Facebook Watch over YouTube, we suppose it boils down to personal preference, but with Facebook, you will be able to tag your friends more easily and share your reactions to videos.
Filed in General. Read more about Entertainment, Facebook and Music. Source: about.fb
This AI Playing A Non-Stop Bass Solo Is All Kinds Of Impressive