All Units, unless otherwise stated, will come with a Yellow Return to Service Tag and a 30 Day Warranty. 8130 available upon request.
Listing DOES NOT include ANTENNA, MOUNTING TRAY WITH CONNECTORS, WIRING DIAGRAMS or anything else. Listing is for the instrument shown only, exactly as seen in pictures. Antennas, mounting trays, connectors, and wiring diagrams may be available upon request.
Due to to unreliable online shipping quotes, please do not pay for item until you have received an invoice from us. We require international buyers to use PayPal. International buyers are responsible for any duties and/or taxes charged by their country upon delivery. Shipping costs and transit times may be significantly higher or longer than domestic shipments. Sometimes no tracking feature is available. Your county's customs inspection procedures may slow the estimated delivery time provided by the carrier. All customs forms are marked 'merchandise' or 'used merchandise' accordingly.


Listen to Travis Scott's new song "Skeletons" featuring Tame Impala, The Weeknd & Pharrell.
It's been a couple years in the making, but after numerous setbacks & delays, Travis Scott's highly anticipated Astroworld album has finally arrived. The 17-track project features guest appearances from a slew of house hold names, including Migos, The Weekend, Nav, Swae Lee, Frank Ocean, Juice WRLD, Gunna, Kid Cudi, and Pharrell, among others.
Take a listen to the lengthy, New Song (below) and let us know what you think!
The post Travis Scott – Skeletons Ft. Tame Impala, The Weeknd & Pharrell [Zippyshare + 320kbps] appeared first on HipHopSongs.


I'm a strong believer in sharing a good feeling. The customer who was responsible for the 1000th sale will receive our Online Teddy Bear Making course worth $ 65 for free. Avril from the UK assures me that she is delighted to be that person!
I would love to be counted in your wonderful giveaway.
One question - I am a teddy bear, may I join in the fun?
Being a bear, you are probably highly qualified to participate. Come to think of it, I am going to disqualify all badgers and skunks. Will be holding thumbs that Prudence's number gets pulled out of the hat!
Congratulations on a 1000 sales!! Wishing much success and joy for many more! May it always bring you happiness!!!!
Anyone who wins your On Line Course will be a very lucky person!
Congratulations on your first 1,000.
I would love to enter the comp. for the on line class.
The online class could not go to a nicer person. I will be holding thumbs for your number!
Congratulations Megan on reaching your 1000th sale, may the next 1000 come quick!
Much as I like Teddy Bears, I just don't have any spare time to make one, the miniature business keeps me too busy, so please don't include me in the draw.
You are already a wonderful miniature maker, Elga. I would be surprised if you had time for teddy bears too! Thank you for your kind words, soon I will have to offer a miniature give away.
Gwyneth tells me that she is blocked from viewing blogs on her computer at work, sensible employer ;-) I'm posting on her behalf to enter her name in the draw for online classes.
A few other ladies have also contacted me through FB and personal messages to ask to be entered in the draw. I am making a list of names. If you are unable to comment on the blog for some reason, contact me and I will add your name.
Congratulations for your 1000 sales! Just now I was looking for an online teddy bear making course and here you happen! Pls enter me for the giveaway, I do not know anything about teddy bear making and this is such a great chance! thank you so much!
Well done Megan! You are making the world a better place. You are an inspiration, and I know bear-making will be my most favourite hobby forever. I just wish there was more time in a day, or that I could quit my job and only make bears all day ha ha ha. Other alternative is to work VERY HARD now so I can retire early and make lost of bears! Please put me in the draw! I would love to sharpen up on my bear making skills and techniques, and learn to do it perfect like you!
You're welcome, Eddy the Teddy! Had a look at your social media. Nice work!


Research shows good teaching makes a big difference in how much kids learn. But the United States lacks an effective system for training new teachers or helping them get better once they're on the job. This documentary examines why, and asks what it would take to improve American teaching on a wide scale. We meet researchers who are trying to understand what makes teaching complex, and how to determine whether someone is ready to be a teacher. We also visit U.S. schools that are taking a page from Japan and radically rethinking the way they approach the idea of teacher improvement.
In the United States, we tend to think that improving education is about improving teachers - recruiting better ones, firing bad ones. But the Japanese think about improving teaching. It's a very different idea.
The American RadioWorks team includes Samara Freemark and Sasha Aslanian. Special thanks to the Hatcher Group and Kohnstamm Communications.
Support for "Teaching Teachers" comes from the William and Flora Hewlett Foundation, Lumina Foundation, the Corporation for Public Broadcasting and the Spencer Foundation. A note of disclosure: Deborah Ball is chair of the Spencer Foundation board and the Spencer Foundation has given money in the past to support research by Akihiko Takahashi. The foundation has no editorial influence on ARW's coverage.


I absolutely LOVE children's books. I love simple yet clever stories and whimsical illustrations.
Iggy Peck, Architect, is about a boy that has a passion for buildings and architecture. It all started when he was two. Iggy used all sorts of things to build towers: diapers, peaches, sod, chalk. When he was in second grade, Iggy's teacher tried to discourage this passion. Miss Greer didn't think grade two was the place to talk about architecture. Of course, in the end, Iggy and his architecture saves the day.
This book is so fun to read out loud and the illustrations (some on graph paper, my favorite) are wonderful. I love books that are as fun for grown ups to read as it is for kids.


11, 12th November 2016 Current Affairs MCQs, Quiz, Questions: Current Affairs for November 2016, Daily Multiple Choice Questions (MCQs) for India GK, World GK and current affairs with Collection of daily objective type Question by www.Indiagk.net based on General Knowledge (GK) and General Science (GS) Questions for UPSC, State PSC, SSC, Police exam, Railway exam, SBI, Bank PO, IBPC, SSC, LDC, UDC, Army, Airforce, Navy, Coast Guard, Bank Clerk, TET and all entrance examination with current affairs News, Multiple Choice Questions (MCQs) available on Website www.IndiaGK.net and Android Apps with daily updates:.
Who became the first Indian American to be elected to US House of Representatives?
Who has been appointed as the Interim Chairman of TCS?
China's e-commerce website Alibaba, set a new world record with sales figures of $20 Billion on which day?
Public Service Broadcasting Day observed is observed on which date ?


text:
New Bills in MI Legislature Add to Narrative of Refugee Exclusion
A Syrian migrant carries a child as she walks along a railway track after crossing into Hungary from the border with Serbia, 29 Aug 2015. Photo courtesy of Reuters/Bernadett Szabo
Media coverage in the past week has been a cascade of reactions to the horrific Islamic State (IS) attacks in Paris. The acts of terror, which left 130 dead (per NPR), were unforgivable, as is the general religious fanaticism that serves as a crucible for such unspeakable acts as the ongoing civil war in Syria, the university attacks in Kenya, and mostly recently, the hostage crisis in Mali, among many others.
The genesis of the Islamic State is nuanced, as is its expansive, hateful ideology, but one thing is certain: its very survival, as has been artfully outlined by former Secretary of State Madeline Albright and others, is dependent upon the fragmenting, isolating, divisive reactions to those who fear them. When these reactions happen at a policymaking level, where victims are villainized, doors are shut, and help is withdrawn, the Islamic State has achieved its desired effect.
Governor Snyder in 2011. Photo courtesy of Michigan Municipal League.
Enter Michigan Governor Rick Snyder and thirty other governors, all of whom have declared Syrian refugees unwelcome in their states by calling for "reviews" of the already-vetted screening processes currently in use. But it's not just state-level executive branches getting involved in constructing an unwelcome environment for refugees; in Michigan, state legislators are already drafting legislation that, like Governor Snyder's decree, holds no water legally in excluding refugees.
Jim Runestad, a Republican out of White Lake, Michigan, says he is drafting two bills in the State House related to refugee resettlement. The first seeks to prohibit the expenditure of state funds to assist the federal Refugee Resettlement Program except in cases in which the state legislature and relevant county governments approve the resettlement. In an interesting departure from reality, Runestad noted, "Since states are sovereign and cannot be forced to comply with this voluntary program, I think a full review of the state's role in the resettlement program is the smart thing to do." As we already know, the matter of refugees and immigration in general is a federal law issue. States cannot legally bar refugees from entering, nor can they engage in foreign policymaking by specifying that refugees from certain countries of origin are permitted, while others are not.
Runestad's second bill would place a complete ban on the resettlement of refugees from failed states. Runestad justified this measure, alleging, "We do not have access to government databases from failed states, so we don't have access to the backgrounds of those seeking to relocate in Michigan." However, White House officials outright reject the claim that too little data exists on Syrian refugees, saying, "There is intelligence on refugee populations, including Syrian refugees…Iraqis and Syrians tend to be a very, very heavily documented population. And members of families tend to have passports and family registries and military books, and they have a lot of information in most instances."
Finally, and perhaps most ironically, such proclamations about stymying resettlement in the wake of the Paris attacks is nothing short of declaring support for what is already common practice for refugee resettlement. According to Vox: "The sad truth is that the US's insistence on screening Syrian refugees carefully, and its almost paranoid aversion to admitting anyone whose family might have had any form of contact with any extremist group at any point, created a bottleneck that for years prevented nearly any Syrian refugee from coming to the US. The federal government has just, in 2015, started devoting enough resources to screening refugees who've fled Syria to start allowing them to come into the country in any numbers at all." So while outright exclusion satisfies our most banal, fearful inclinations, the reality is, we're hardly letting any Syrians in anyway.
A US Coast Guard patrol boat around Liberty Island in New York. Photo courtesy of Cristo Vlahos.
Earlier this week, the House voted to set impossibly high hurdles meant to halt acceptance of Syrian and Iraqi refugees, or, as they are otherwise known, those that need help most urgently. These include new FBI background checks, and personal reviews from the Director of National Intelligence, the Director of the FBI, and the Department Homeland Security Secretary of each and every individual application before admission. And while this transparently knee-jerk policy reaction to the attacks in Paris is exclusionary and has been condemned by President Obama and other immigrant advocates and human rights groups, it does have one thing going for it – it is actually in the jurisdiction of the U.S. House of Representatives to make such a move, as is not the case for Mr. Runestad and the State House.
So why have state lawmakers who knowingly are not positioned to challenge federal law taking such stands? This brings us to the ugliest thing about the policy positions and unwelcoming statements from these elected officials – they don't even matter. Statements of this kind from elected leaders about policy moves outside of their realm of influence mean to foster a hostile environment for refugees. They accomplish nothing except to lend legitimacy to xenophobia, and make the direct victims of terrorism feel unwelcome and unsafe in a country whose very sum and substance is supposed to be a haven for those fleeing persecution.
While the resettlement issue remains polarizing in civil society, unlike many of our elected officials, there are advocates who still work on behalf of refugees and attempt to create safe and welcoming communities in response to the hostility witnessed on a policy level. A MoveOn.org petition calling on Governor Snyder to restore the Syrian refugee program in Michigan has generated 5,253 signatures as of 4:30 pm on Friday. Religious organizations are working tirelessly as well – St. Vincent Catholic Charities in mid-Michigan still wants to resettle at least 100 Syrian refugees in the area. And finally, in Pontiac, where plans for the construction of housing developments for Syrian refugees and a community center are in the works, supporters of the project are not folding under pressure from county official L. Brooks Patterson, who condemned the continuation of the project in inflammatory and politicized terms. In response, the Oakland county treasurer affirmed, "…Oakland County is made up of many thousands of Americans born in other countries, including many where terrible things happened, and that they are contributing greatly to Oakland County's prosperity and security every day."
Paris was attacked last week in an appalling display of the boundless capacity of human hatred and mercilessness and the ghastly consequences of religious extremism and zealotry. As prominent leaders both at home, and in the international community, American lawmakers were handed an opportunity to react with compassion for the victims and with condemnation for the attackers. Instead, our elected officials are painting all with a broad brush, and reacting only with the latter.
UPDATE (April 8, 2016): Rep. Runestad has officially introduced House Bills 5528-5529, meant to require more stringent reporting by resettlement agencies to the state and include specifics on the policies outlined above. Runestad's office argues that the pair of bills provide "an outline for communication between state departments and local governments in order to maximize placement success and minimize security risks," but refugee resettlement officials and immigrant advocates take issue with the proposals. The managing attorney for the Michigan Immigrant Rights Center says that not only are many of the provisions outlined in the bill already being done, but the bills do damage in contributing to the culture of suspicion of Muslim and Syrian refugees, and they "suggest that refugee resettlement may have a negative impact on our state." The bills have been referred to the House Committee on Oversight and Ethics.
asylum, Immigration, Iraqi, ISIS, refugee crisis, Syria, Syrians
Another Thick Green Line: Militarization and Our Borders
The Usual Caveats Apply: Amendments to the Visa Waiver Program
Free Migration as an American Principle

redpajama_set_name: RedPajamaCommonCrawl


text:
Most Popular TV Networks & Services, December 2014 [TABLE]
By David Erickson | December 30, 2014
Netflix is among the top-3 "cool" networks or services for a leading 26% of TV viewers aged 18-49.
Top Must-Have Pay TV Channels [CHART]
ABC tops the list of must-have channels for a pay TV service, with 68.4% of 2,000 US broadband households surveyed by The Diffusion Group.
Defiance Will Pair TV Series With Simlutaneous Real-Time Video Game
By David Erickson | April 24, 2012
via adage.com Syfy is already pitching advertisers on integrations into "Defiance," the coordinated TV series and multiplayer online game that it plans to introduce in April 2013. The complicated nature of the project, in which plot points will appear on the TV show and in the game at about the same time, requires bringing in…


text:
\section{Introduction} \label{int}

\bigskip

D. Ruelle in  \cite{Reullenoncommutative} considered  a general setting (which includes hyperbolic diffeomorphisms on manifolds)  where he is able to describe a formulation  of the concept of Gibbs state based on {\bf conjugating homeomorphism} in the so called Smale spaces.
On this setting he shows a relation of KMS states of $C^*$-algebras with H\"older equilibrium probabilities of Thermodynamic Formalism. Part of the formulation of this  relation  requires the use of a non trivial  result by N. Haydn (see \cite{HaydnGibbs}). Later, the paper \cite{HD} by N. Haydn and D. Ruelle  presents a shorter proof of the equivalence.

 Here we consider similar problems but now on the symbolic space and the dynamics will be  given by the shift. We will present
a simplified  proof of the equivalence mentioned above. The main result of the paper is Theorem \ref{mai} on section \ref{E}. One can get a characterization of the equilibrium probability for a potential defined on the lattice $\{1,2,...,d\}^{\mathbb{Z} - \{ 0 \}  } $ without using the Ruelle operator (which acts on the lattice $\{1,2,...,d\}^\mathbb{N}  $). The probability we get is invariant for the action of the shift $\tau$ acting on $\{1,2,...,d\}^{\mathbb{Z} - \{ 0 \}  } $.

The proof of this result will take several subsequent sections.

In section \ref{C} we show the relation of these probabilities with the KMS dynamical $C^*$-state  on the $C^*$-Algebra associated to the groupoid defined by the homoclinic equivalence relation.

In the previous sections we introduce several results which are necessary for the simplification of the final argument on section \ref{C}.

We present several examples  helping the reader on the understanding of the main concepts.

On \cite{RuelleBook} and also on the beginning of the book \cite{BLL} it is explained the relation of equilibrium states of Thermodynamic Formalism
with the corresponding concept in Statistical Physics. The role of KMS $C^*$-dynamical states on Quantum Statistical Physics is described on \cite{Bra}.  KMS $C^*$-dynamical states correspond to the DLR probabilities (see \cite{CL} for definition) in Statistical Mechanics.

Working on the symbolic space avoid several technicalities which are necessary for hyperbolic diffeomorphisms on manifolds (where one have to use stable foliation, the local product structure, etc...).

Our proof consider mainly potentials    $A:  \{1,2,...,d\}^{\mathbb{Z} - \{ 0 \}  } \to \mathbb{R}$ which depend on a finite number of coordinates. The case of a general H\"older potential (more technical) can be obtained by adapting our reasoning but we will not address this question here.

On the papers \cite{LM1} and \cite{LO} the authors consider among other things a relation of KMS probabilities with eigenprobabilities for the dual of the Ruelle operator (which are not necessarily invariant for the shift). This problem is analyzed on the lattice $\{1,2,..,d\}^\mathbb{N}$ which is a different setting that the one we consider here. The equivalence relations are also not related.  Despite  some similarities that can be perceived in the statements of the main results obtained in the two settings we point out that the reasoning on the respective proofs are quite different.

Lecture 9 in \cite{GiaII} presents a brief introduction to $C^*$-Algebras and  the KMS condition.

In \cite{EL1} and \cite{EL2} a relation of KMS states in a certain $C^*$-Algebra and eigenprobabilities of the dual of the Ruelle operator is considered.

In a different setting the paper \cite{BK} also consider the homoclinic equivalence relation.

\section{Conjugating homeomorphisms}

In order to simplify notation in this section $\Omega = \{1,2,...,d\}^{\mathbb{Z} - \{ 0 \}  }$, and the dynamic $\tau: \Omega \to \Omega $ is such that
$$\tau (..., x_{-n},...,x_{-2}  , x_{-1} \,| \,x_1,x_2,...,x_n,...)=(..., x_{-n},...,x_{-2},   x_{-1} , x_1\,|\, x_2,...,x_n,...)$$
And the usual metric $d$, is defined in such way that for $x,y\in \Omega$ we set
$$ d(x,y) = 2^{-N},$$
$N\geq 0$, where for
$$x=
(..., x_{-n}, ... ,x_{-1}\,|\,x_1,..,x_n,..\,)\,\,\,,y=
(..., y_{-n}, ... ,y_{-1}\,|\,y_1,..,y_n,..\,),$$ we have $x_j=y_j$, for all $j$, such that, $-N\leq j\leq N$
and, moreover $x_{N+1}\neq y_{N+1}$, or $x_{-N-1}\neq y_{-N-1}$. Given $x,y$ we denote $\vartheta(x,y)=N$, therefore $\vartheta(x,y) = - \log_{2} (d(x,y))$.

Given $x,y \in \Omega$, we say that $x\sim y$ if
$$ \lim_{k \to + \infty} d(\tau^k x, \tau^k y)=0$$
$$\text{and}$$
\begin{equation} \label{equivalencee}
\lim_{k \to - \infty} d(\tau^k x, \tau^k y)=0.
\end{equation}

This means there exists an $M\geq 0$ such that $x_j=y_j$ for $j> M$ and $j< -M$. In other words, there are only a finite number of $i$'s such that $x_{i} \neq y_{i}$. In this case  we say that $x$ and $y$ are homoclinic.

$\sim$ is an equivalence relation and defines the groupoid $G\subset \Omega  \times \Omega$ of pairs $(x,y)$ of elements which are related (see for instance \cite{Ren0}, \cite{Ren2}, \cite{LM1} or \cite{LO}).

 Let $\kappa(x,y)$ be the minimum  $M$ as above. Therefore $x_{\kappa(x,y)} \neq y_{\kappa(x,y)}$ or $x_{-\kappa(x,y)} \neq y_{-\kappa(x,y)}$. Note that $\vartheta(x,y) \leq \kappa(x,y)$ and could be strictly less. Note that $\kappa(x,y)$ is defined just when $x \sim y$.
\medskip

\begin{example}
For example in $\Omega=\{1,2\}^{\mathbb{Z}-\{0\}}$ take
$$ x= (...,x_{-n},...,x_{-7},1,2,2,1,2,2\,|\, 1,2,1,2,1,1,x_7,...x_{n},..)$$
and
$$ y= (...,y_{-n},...,y_{-7},1,2,2,1,2,2\,|\, 1,2,1,1,1,2,y_7,...y_{n},..)$$
where $x_j=y_j$ for $|j| > 6=\kappa(x,y).$
In this case $d(x,y) = 2^{-3}$ and $N=\vartheta(x,y) = 3.$
\end{example}

 Given a  H\"older function $U:\Omega \to \mathbb{R}$ it is easy to see that if $x$ and $y$ are homoclinic, then the following function is well
 defined
 \begin{equation} \label{Vq} V(x,y) = \sum_{n =\,-\,\infty}^\infty (U (\tau^n (x)) -  U (\tau^n (y)) ).\end{equation}

 This function satisfies the property
 $$ V(x,y) + V(y,z) = V(x,z)$$
 when $x \sim y\sim z$.

 A function $V $ with this property will play an important role in some parts of  our reasoning. We will not assume on the first part of this work that $V$ was obtained from a $U$ as above.

\medskip

\vspace{0.5cm}

Now we will describe a certain class of {\bf  conjugating homeomorphism} for the relation $\sim$ (see \eqref{equivalencee}) described above.

Given a fixed $x$ and $y$ ($y$ in the class of $x$)  we  define the open set $\mathcal{O}_{(x,y)} =B_{\frac{1}{2^{\kappa(x,y)}}}(x)= \{ z \in \Omega : d(x,z) < 2^{-\kappa(x,y)+1}  \} $.

 We will define for each such pair $(x,y)$ a conjugating homeomorphisms $\varphi_{(x,y)}$ which has domain on
 $\mathcal{O}_{(x,y)}.$

We denote for $m,n \in \mathbb{N}$
 $$\overline{ x_{-m}x_{-m+1}...x_{-1}\,|\, x_{1}...x_{n-1}x_{n} }=$$
$$ \{z \in \Omega \,|\, z_j= x_j, \, j=-m, -m+1,...,-1,1,2,...,n-1,n   \},$$

 and call it the   cylinder determined by the finite string
 $$x_{-m}x_{-m+1}...x_{-1}\,|\, x_{1}...x_{n-1}x_{n}.$$

 We will say that a cylinder, or a string, is \textbf{symmetric} if $n=m$.

Note that given   $x\sim y$
$$\mathcal{O}_{(x,y)}=   \overline{ x_{-\kappa(x,y)}\,\, x_{-\kappa(x,y) + 1}\, ... \,x_{-1}\, \vert\,  x_{1}\, ... \,x_{\kappa(x,y) -1}\,\, x_{\kappa(x,y)}}  $$
$\mathcal{O}_{(x,y)}$ is a symmetric cilinder.

Now we shall define the main kind of {\bf conjugating homeomorphisms} that we will be using. Given $(x,y) \in G$, let $n = \kappa(x,y)$, we define a conjugating $\varphi=\varphi_{(x,y)}$ with domain
$$\mathcal{O}_{(x,y)}=B_{\frac{1}{2^{n}}}(x) = \{ z \in \Omega : d(x,z) < 2^{-n+1}  \}=  \overline{ x_{-n}x_{-n+1}...x_{-1}\,|\, x_{1}...x_{n-1}x_{n} },$$
 where $\varphi_{(x,y)}: \mathcal{O}_{(x,y)} \to B_{\frac{1}{2^{n-1}}}(y)$ is defined by the expression:
 $z$ of the form
 $$ z= ( ... z_{-n-2} z_{-n-1} \,{\bf \,x_{-n} x_{-n + 1} ... x_{-1} }\vert {\bf x_{1} ... x_{n} \,} \,z_{n+1} z_{n+2} ...)$$
 goes to
\begin{equation}\label{varphicorreto}
\varphi_{(x,y)}(z) = ... z_{-n-2} z_{-n-1}\,{\bf  y_{-n} y_{-n + 1} ... y_{-1}
}\vert {\bf y_{1} ... y_{n} }\, z_{n+1} z_{n+2} ...
\end{equation}
We shall call these transformations the family of \textbf{symmetric conjugating homeomorphisms}. We shall denote by $S$ the set of symmetric conjugating homeomorphisms obtained by considering all pairs of related points $x$ and $y$.

\medskip
 Note that  the homeomorphism $\varphi_{(x,y)}$ transforms the cylinder $O_{(x,y)} = \overline{ x_{-n}x_{-n+1}...x_{-1}\,|\,x_1 ... x_{n-1}x_{n} }$ in the cylinder $\overline{ y_{-n}y_{-n+1}...y_{-1}\,|\, y_1...y_{n-1}y_{n}}.$

 The graph of $\varphi_{(x,y)}$ is on $G$.

 A more explicit formulation of the concept of symmetric conjugating homeomorphism  will be presented on next section via expressions  (\ref{kre}) and  (\ref{varphicorreto7}).

 \medskip

\medskip
\begin{example}
Consider
$$  x= (... 1\,1\,2\,1\,1\,2\,2\,2\,2\,1\,\,\, 1\,1 \vert 2\, 1\, \,\,2\,1\,2\,2\,1\,2\,2\,2\,1\,1...) $$
and
$$y =( ... 1\,1\,2\,1\,1\,2\,2\,2\,2\,1\,\,\,1\,2 \vert 1\,2 \, \,\,2\,1\,2\,2\,1\,2\,2\,2\,1\,1 ...) $$
in this case $\kappa(x,y)=2, $ and for $z$ of the form
$$ z = (... z_{-4}\,z_{-3}\,\,\, 1\,1 \vert 2\, 1\, \,\,z_3\,z_4\,z_5...)$$
we get

$$\varphi_{(x,y)}(z) =   (... z_{-4}\,z_{-3}\,\,\, 1\,2 \vert 1\, 2\, \,\,z_3\,z_4\,z_5...).$$

\end{example}

\vspace{0.5cm}

It is easy to see that the family of symmetric conjugating homeomorphisms we define above has the following properties: given $x \sim y$

a)  $\varphi_{(x,y)} :\mathcal{O}_{(x,y)}\subset \Omega \to \Omega$ is an homeomorphism over its image

b)  $\varphi_{(x,y)}(x)=y$, and

c) $\lim_{k\to \infty} d(\tau^k (z), \tau^k ( \varphi_{(x,y)}(z))=0$ and $\lim_{k\to -\infty} d( \tau^k (z), \tau^k ( \varphi_{(x,y)}(z))=0$.

Item c) implies that $z$ and $\varphi_{(x,y)}(z)$ are on the same homoclinic class.

\medskip

\section{$C^{*}$-Gibbs states and Radon-Nikodym derivative}

We consider the groupoid $G\subset \Omega \times \Omega$ of all pair of points which are related by the homoclinic equivalence relation.

We consider on $G$ the topology generated by sets of the form
$$ \{\,\, (z, \varphi_{(x,y)}(z))\,\,\text{where}\,\,x \in \mathcal{O}_{(x,y)}\,\,\}.$$

This topology is Hausdorff (see \cite{Reullenoncommutative}).

 Now consider a continuous  function $V: G \to \mathbb{R}$ such that
 $$V(x,y) + V(y,z) = V(x,z),$$
 for all related $x,y,z$. Note that this implies that  $V(x,x) = 0$ and $V(x,y) = - V(y,x)$.

Here we call $V$ a modular function.

Under some other notation  the function $\delta(x,y)= e^{ V(x,y)}$  is  called  a modular function (or, a cocycle).

\medskip

\begin{definition} Given a function $V:G \to \mathbb{R}$ as above
we say that a probability measure $\alpha$ on $\Omega$  is a \textbf{$C^{*}$-Gibbs probability} with respect to the parameter $\beta \in \mathbb{R}$ and $V$, if for any $x\sim y$
\begin{equation} \label{gibbsR}
\int_{O_{(x,y)}} \exp(- \beta V(z, \varphi_{(x,y)}(z))) f(\varphi_{(x,y)}(z)) d \alpha(z) =
\int_{\varphi_{(x,y)}(O_{(x,y)})} f(z) d \alpha (z),
\end{equation}
for every every continuous function $f : \Omega \to \mathbb{C}$ (and conjugated homeomorphism $(O_{(x,y)}, \varphi_{(x,y)})$).

\end{definition}
\medskip

We will show on section \ref{C} a natural relation of this probability $\alpha$ with the $C^*$-dynamical state on a certain $C^*$-algebra. This is the reason for such terminology.

The above definition was taken from \cite{Reullenoncommutative}. This is a version of  the Renault-Radon-Nikodym condition (Def. 1.3.15 in \cite{Ren0}).

\medskip

It is easy to see that the above definition is  equivalent  to say that: given
any pair of finite strings
$$x_{-n} x_{-n+1}...x_{-1}, x_1 ...\,x_{n-1} x_{n}\,\,\text{ and}\,\, y_{-n}y_{-n+1}...y_{-1}\,y_1 ... y_{n-1}y_{n},$$
$n \in \mathbb{N}$, the transformation

\begin{equation} \label{kre} \varphi: \overline{ x_{-n}x_{-n+1}...x_{-1}\,|\, x_1...x_{n-1}x_{n} } \to \overline{ y_{-n}y_{-n+1}...y_{-1}\,|\,y_1...y_{n-1}y_{n}}
\end{equation}
defined by the expression:
\begin{equation}\label{varphicorreto7}
\varphi(z) = (... z_{-n-2} z_{-n-1} \,\,y_{-n} y_{-n + 1} ... y_{-1} \,\vert\, y_{1} ... y_{n}\,\, z_{n+1} z_{n+2} ...),
\end{equation}
where
$$ z=  (... z_{-n-2} z_{-n-1} \,\,z_{-n} z_{-n + 1} ... z_{-1} \,\vert\, z_{1} ... z_{n}\,\, z_{n+1} z_{n+2} ...),$$
is such that for any continuous function $f:  \overline{ y_{-n}y_{-n+1}...y_{n-1}y_{n}} \to \mathbb{R}$
\begin{equation} \label{gibbsR1}
\int_{\overline{ x_{-n}x_{-n+1}...\,|\,...x_{n-1}x_{n} } } e^{- \beta \,V(\,z\,,\, \varphi (z)\,)}  f(\varphi (z)) d \alpha(z) =
\int_{ \overline{ y_{-n}y_{-n+1}...\,|\,...y_{n-1}y_{n}}  } f(z) d \alpha(z).
\end{equation}

\medskip
Note in particulary that by taking $f=1$ we get
\begin{equation} \label{gibbsR2}
\int_{\overline{ x_{-n}x_{-n+1}...\,|\,...x_{n-1}x_{n} } } e^{- \beta \,V(\,z\,,\, \varphi (z)\,)}   d \alpha(z) =
\int_{ \overline{ y_{-n}y_{-n+1}...\,|\,...y_{n-1} y_{n}}}    d \alpha(z).
\end{equation}

In the moment we only  consider symmetric conjugating homeomorphisms of the form $\eqref{varphicorreto7}$.

We will show on section \ref{E} a relation of the \textbf{$C^{*}$-Gibbs probabilities}  $\alpha$ with the {\bf Gibbs (equilibrium) probabilities} of Thermodynamic Formalism.

 In a more explicit formulation  $\alpha$ is  such that given any conjugating $(O_{(x,y)}, \varphi_{(x,y)})$ of the form \eqref{kre}, and continuous function $f : \Omega \to \mathbb{C}$
$$
\int_{O_{(x,y)}} e^{-\beta V( z, \varphi_{(x,y)}(z))} f( \varphi_{(x,y)}(z)) d \alpha(z) =$$
$$\int_{O_{(x,y)}} e^{-\beta V( ( ...  z_{-n}, ..., z_{-1} \vert z_{1} ,... ,z_{n},  ...    ), ( ...  z_{-n-1} y_{-n}, ..., y_{-1}  \vert y_{1} ,... ,y_{n} ,z_{n+1},  ...    ))} f(\varphi_{(x,y)}(z)) d \alpha(z) =$$
\begin{equation} \label{jk1} \int_{\varphi_{(x,y)}(O_{(x,y)})} f(z) d \alpha(z).
\end{equation}
In this case, clearly the Radon-Nikodym derivative of the change of coordinates $\varphi $ is
$$  e^{-\beta V( ( ...  z_{-n}, ..., z_{-1} \vert z_{1} ,... ,z_{n},  ...    ), ( ...  z_{-n-1} y_{-n}, ..., y_{-1} \vert y_{1} ,... ,y_{n} ,z_{n+1},  ...    ))}.$$

In order to simplify the notation sometimes on the text we will consider the value $\beta=1$.

\medskip

\medskip

We will consider a larger class of conjugating homeomorphisms.

\begin{definition}
Given $n$ and $m$ and
pair of finite strings

\begin{equation} \label{pre}x_{-n} x_{-n+1}...x_{-1}, x_1 ...\,x_{m-1} x_{m}\,\,\text{ and}\,\, y_{-n}y_{-n+1}...y_{-1}\,y_1 ... y_{n-1}y_{m},
\end{equation}
$n,m \in \mathbb{N}$, the transformation

\begin{equation} \label{kre} \varphi: \overline{ x_{-n}x_{-n+1}...x_{m-1}x_{m} } \to \overline{ y_{-n}y_{-n+1}...y_{m-1}y_{m}}
\end{equation}
defined by the expression:
\begin{equation}\label{varphicorreto7}
\varphi(z) = (... z_{-n-2} z_{-n-1} \,{\bf \,y_{-n} y_{-n + 1} ... y_{-1} \,\vert\, y_{1} ... y_{m}\,}\, z_{m+1} z_{m+2} ...),
\end{equation}
where
$$ z=  (... z_{-n-2} z_{-n-1} \,{\bf \,x_{-n} x_{-n + 1} ... x_{-1} \,\vert\, x_{1} ... x_{m}\,}\, z_{m+1} z_{m+2} ...),$$
is called a {\bf non-symmetric conjugating homeomorphism} associated to the pair (\ref{pre}).
\end{definition}

 Proposition \ref{krt} claims that if $\alpha$ is a $C^{*}$-Gibbs probability, then the relation \eqref{jk1} is satisfied for a bigger class of $\varphi$ transformations, i.e. not necessarily symmetric. Before that  we shall provide the reader with an example of idea of the proof.

\begin{example} \label{wer}
Consider the non-symmetric conjugating homeomorphism $\varphi : \overline{0 \vert 11} \to \overline{1 \vert 10}$ given by
$$
\varphi(... z_{-3} z_{-2} 0 \vert 1 1 z_{3} ...) = ... z_{-3} z_{-2} 1 \vert 1 0 z_{3} ...
$$
we shall prove that if $\alpha$ is a $C^{*}$-Gibbs measure then relation \eqref{jk1} is valid for $\varphi$. This is actually straightforward, first divide the domain and image of the function into symmetric cylinders,  and in these cylinders apply relation \eqref{jk1}. So in this case consider $\varphi_{0} : \overline{00 \vert 11} \to \overline{01 \vert 10}$, and $\varphi_{1} : \overline{10 \vert 11} \to \overline{11 \vert 10}$ such that
$$
\varphi_{a} (... z_{-3} a 0 \vert 1 1 z_{3} ...) =
(... z_{-3} a 1 \vert 1 0 z_{3} ...)
$$
for $a = 0$ or $a = 1$. Now notice that
$$
\int_{\overline{0 \vert 11}} e^{-\beta V(x, \varphi(x))} f(\varphi(x)) d \alpha(x) =
$$
$$
\int_{\overline{00 \vert 11}} e^{-\beta V(x, \varphi(x))} f(\varphi(x)) d \alpha(x) + \int_{\overline{10 \vert 11}} e^{-\beta V(x, \varphi(x))} f(\varphi(x)) d \alpha(x) =
$$
$$
\int_{\overline{00 \vert 11}} e^{-\beta V(x, \varphi_{0}(x))} f(\varphi(x)) d \alpha(x) + \int_{\overline{10 \vert 11}} e^{-\beta V(x, \varphi_{1}(x))} f(\varphi(x)) d \alpha(x) =
$$
$$
\int_{\overline{01 \vert 10}} f(x) d \alpha(x) + \int_{\overline{11 \vert 10}} f(\varphi(x)) d \alpha(x) =
\int_{\overline{1 \vert 10}} f(x) d \alpha(x) .
$$
What proves that relation \eqref{jk1} is valid for this conjugating.
\end{example}

\begin{proposition} \label{krt} Assume $\alpha$ is $C^*$-Gibbs for $V$ as in (\ref{jk1}), then for any
non-simmetric homeomorphism $(\varphi,\mathcal{O})$, as defined on (\ref{varphicorreto7}), we have that for
$n,m \in \mathbb{N}$, the transformation

$$
\int_{ \overline{ x_{-n}x_{-n+1}...x_{-1}\,|\, x_1\,...x_{m-1}x_{m} } } e^{-\beta V( z, \varphi(z))} f( \varphi(z)) d \alpha(z) =$$
$$\int_{O} e^{-\beta V( ( ... z_{-n-1} z_{-n}, ..., z_{-1} \vert z_{1} ,... ,z_{m},z_{m+1}  ...    ), ( ...  z_{-n-1} y_{-n}, ..., y_{-1}  \vert y_{1} ,... ,y_{m} ,z_{m+1},  ...    ))} f(\varphi(z)) d \alpha(z) =$$
\begin{equation} \label{jk3} \int_{\overline{ y_{-n}y_{-n+1}...y_{-1}\,|\, y_1\,...y_{m-1}y_{m}}} f(z) d \alpha(z).
\end{equation}

\end{proposition}

We leave the proof (which is similar to the reasoning of example \ref{wer}) for the reader.

As a particular case we get

\begin{equation} \label{jk09} \int_{ \overline{|\, x_1\,...x_{m} } } e^{-\beta V( z, \varphi(z))} f( \varphi(z)) d \alpha(z) = \int_{\overline{|\, y_1\,...y_{m}}} f(z) d \alpha(z).
\end{equation}
for given  $\overline{|\, x_1\,...\,x_{m} }$, $  \overline{|\, y_1\,...\,y_{m}}$ and the corresponding conjugating homeomorphism $\varphi$.

\medskip
It is possible to consider more general forms of conjugating homeomorphisms as described on the next example.

\begin{example} \label{poi} Consider the homeomorphism $\varphi : \overline{112 \vert 2} \to \overline{1 \vert 122}$ given by
$$
\varphi(... z_{-4} 112 \vert 2 z_{2} z_{3} z_{4} ...) =
(... z_{-4} z_2 z_3 1 \vert 1 2 2 z_4 ...).
$$

Note that $\overline{112 \vert 2}  $ is translation by $\tau^{-2}$ of the set $\overline{1 \vert 122}.$

As in the previous example we will prove that if $\alpha$ is a $C^{*}$-Gibbs probability then relation \eqref{jk1} is also valid for such $\varphi$ and $\mathcal{O} = \overline{112 \vert 2}$. First consider the conjugating homeomorphisms, $\varphi_1$, $\varphi_2$, $\varphi_3$ and $\varphi_4$, given by
$$
\varphi_1 (... z_{-4} 112 \vert 2\, {\bf 11} z_{4} ...) =
(... z_{-4}\,{\bf 11} 1 \vert 1 2 2 z_4 ...),
$$
$$
\varphi_2 (... z_{-4} 112 \vert 2\, {\bf 12}\, z_{4} ...) =
(... z_{-4}\,{\bf  12}\, 1 \vert 1 2 2 z_4 ...),
$$
$$
\varphi_3 (... z_{-4} 112 \vert 2\,{\bf 21}\, z_{4} ...) =
(... z_{-4} \,{\bf 21}\, 1 \vert 1 2 2 z_4 ...),
$$
$$
\varphi_4 (... z_{-4} 112 \vert 2 \,{\bf 22} z_{4} ...) =
(... z_{-4} \,{\bf }\,{\bf 22}\, 1 \vert 1 2 2 z_4 ...).
$$

Therefore we have that
$$
\int_{\overline{112 \vert 2}} e^{V(x, \varphi(x))} f( \varphi(x)) d \alpha (x) =
$$
$$
\int_{\overline {112 \vert 211}} e^{V(x, \varphi(x))} f(\varphi(x)) +
\int_{\overline {112 \vert 212}} e^{V(x, \varphi(x))} f(\varphi(x)) +
$$
$$
\int_{\overline {112 \vert 221}} e^{V(x, \varphi(x))} f(\varphi(x)) +
\int_{\overline {112 \vert 222}} e^{V(x, \varphi(x))} f(\varphi(x))  =
$$
$$
\int_{\overline {112 \vert 211}} e^{V(x, \varphi_1(x))} f(\varphi_1(x)) +
\int_{\overline {112 \vert 212}} e^{V(x, \varphi_2(x))} f(\varphi_2(x)) +
$$
$$
\int_{\overline {112 \vert 221}} e^{V(x, \varphi_3(x))} f(\varphi_3(x)) +
\int_{\overline {112 \vert 222}} e^{V(x, \varphi_4(x))} f(\varphi_4(x)) =
$$
$$
\int_{\overline {111 \vert 122}} f(x) +
\int_{\overline {121 \vert 122}} f(x) +
\int_{\overline {211 \vert 122}} f(x) +
\int_{\overline {221 \vert 122}} f(x) =
$$
$$
\int_{\overline{1 \vert 122}} f d \alpha (x)
$$
where some of the $d \alpha$ where omitted. Since we proved that
$$
\int_{\overline{112 \vert 2}} e^{V(x, \varphi(x))} f( \varphi(x)) d \alpha (x) =
\int_{\overline{1 \vert 122}} f d \alpha (x)
$$
for any measurable function $f$ then we have that relation (\ref{jk1}) is satisfied.
\end{example}

\vspace{0.5cm}

In analogous way as in last example one can define a conjugating $\varphi$ such that

$\varphi :  \overline{ x_{-n}...{\bf x_{-r}\,...\,x_{-1}}\,|\, x_1\,...x_{m} }\to  \overline{ x_{-n}x_{-n+1}... x_{-r-1}\,|\, {\bf x_{-r}\,... x_{-1}}\,x_1\,...x_{m}}.$

We will consider such transformation $\varphi$  in the next result.
\medskip

\begin{proposition} \label{krtt} Assume $\alpha$ is $C^*$-Gibbs for $V$ as in (\ref{jk1}), then for
$n,m \in \mathbb{N}$, and $0<r$, such that, $ r\leq n$, we get

$$
\int_{ \overline{ x_{-n}x_{-n+1}...x_{-r-1}\,{\bf \, x_{-r}\,x_{-r+1}...x_{-1}}\,|\, x_1\,...x_{m-1}x_{m} } } e^{-\beta V( z, \varphi(z))} f( \varphi(z)) d \alpha(z) =$$
\begin{equation} \label{jk15} \int_{\overline{ x_{-n}x_{-n+1}... x_{-r-1}\,|\, {\bf x_{-r}\,x_{-r+1}... x_{-1}\,x_1\,}\,...x_{m-1}x_{m}}} f(z) d \alpha(z).
\end{equation}

\end{proposition}

{\bf Proof:} The proof is similar to the reasoning of example \ref{poi}. One have just to consider the homeomorphisms
$$\varphi(...z_{-n-r-1}\,  z_{-n-r} ...z_{-n-1}  x_{-n} x_{-n+1}...x_{-1}\,|\, x_1\,...x_{m-1} x_{m} {\bf  z_{m+1}... z_{n+r}}\, z_{n+r+1} ...) =$$
$$
(...  z_{-n-r}\, {\bf z_{m+1}... z_{n+r}\,} x_{-n} x_{-n+1} ... x_{-r-1}\,|\, x_{-r}\,x_{-r+1}......x_{-1} x_{1}...x_{m-1} x_{m}  \, z_{n+r+1}  ...).$$

Note that
$$\hat{\tau}^{-r} \,(\overline{ x_{-n}x_{-n+1}...x_{-1}\,|\, x_1\,...x_{m-1}x_{m} } \,)=$$
$$ \overline{ x_{-n}x_{-n+1}... x_{-r-1}\,|\, x_{-r}\,x_{-r+1}... x_{-1}\,x_1\,...x_{m-1}x_{m}}.$$
\qed

We want to show that is $\alpha$ is $C^*$-Gibbs for $V$, then, the pull back $\rho=\tau^* (\alpha)$ is also $C^*$-Gibbs for $V$.

The next example will help to understand the main reasoning for the proof of the above claim.

\begin{example} Suppose
$
V(x,y)
$ is defined
when $x \sim y$.

Given $\alpha$ consider  the pull back $\rho=\tau^* (\alpha)$.

Consider
$$\varphi : \overline{11 \vert 2 1}\to \overline{21 \vert 12},$$
where
$$ \varphi ( ... x_{-4} x_{-3}\, 1 1 \vert 2 1\, x_3\, x_4...\,) =( ... x_{-4} x_{-3}\, 2 1 \vert 1 2 \, x_3\, x_4...\,),
$$
and
$$\varphi_1 : \overline{112 \vert 1}\to \overline{211 \vert 2},$$
where
$$ \varphi_1 ( ... x_{-5} x_{-4}\, 1 1 2 \vert  1\, x_2\, x_3...\,) =( ... x_{-5} x_{-4}\, 2 1 1 \vert  2 \, x_2\, x_3...\,).
$$
If for any continuous function $g$ we have that
$$
\int_{\overline{11 \vert 21}} e^{V(x, \varphi(x))} g( \varphi(x)) d \alpha (x) =
\int_{\overline{21 \vert 12}} g (x) d \alpha (x),
$$
then, for any continuous function $f$ we have that
$$
\int_{\overline{1 1 2 \vert 1}} e^{V(x, \varphi_1(x))} f( \varphi_1(x)) d \rho (x) =
\int_{\overline{2 1 1 \vert 2}} f (x) d \rho (x).
$$

In fact both properties are equivalent.

Note first that $ \varphi_1 \circ \tau = \tau \circ \varphi.$

Therefore,
$$
\int_{\overline{1 1 2 \vert 1}} e^{V(x, \varphi_1(x))} f( \varphi_1(x)) d \rho (x) =$$
$$\int  I_{\overline{1 1 2 \vert 1}}(x)\,  e^{V(x, \varphi_1(x ))} f( \varphi_1(x)) d \rho (x)=$$
$$
\int I_{\overline{1 1 2 \vert 1}}(\tau(x)) e^{V(\tau (x), \varphi_1(\tau(x)))} f( \varphi_1(\tau (x))) d \alpha (x) =
$$
$$\int I_{\overline{1 1 2 \vert 1}}(\tau(x)) e^{V(x, \varphi_1(x ))} f( \varphi_1(\tau (x))) d \alpha (x) =
$$
$$  \int  I_{\overline{1 1 2 \vert 1}}(\tau(x))\,  e^{V(x, \varphi_1(x ))} f(\tau (\varphi(x)))) d \alpha (x)=
$$
$$  \int  I_{\overline{1 1 \vert 2 1}}(x)\,  e^{V(x, \varphi_1(x ))} f(\tau (\varphi(x)))) d \alpha (x)=
$$
$$  \int_{\overline{1 1 \vert 2 1}}\,  e^{V(x, \varphi_1(x ))} f(\tau (\varphi(x)))) d \alpha (x)=
$$
$$  \int_{\overline{2 1 \vert 1 2}}\,   f(\tau (x)) d \alpha (x)=
$$
$$  \int I_{\overline{2 1 \vert 1 2}}\, (x)  f(\tau (x)) d \alpha (x)=
$$
$$  \int I_{\overline{2 1 \vert 1 2}}\, (\tau^{-1} \circ \tau ) (x)  f(\tau (x)) d \alpha (x)=
$$
$$  \int I_{\overline{2 1 \vert 1 2}}\, (\tau^{-1} (x))  f(x) d \rho (x)=
$$
$$  \int_{\overline{2 1 1  \vert 2 }}\,   f(x) d \rho (x).
$$

Above we took $g = f \circ \tau.$

From the above reasoning we get that  both properties are equivalent.

\end{example}

\begin{proposition} \label{kttt}
 If $\alpha$ is $C^*$-Gibbs for $V$, then, the pull back $\rho=\tau^* (\alpha)$ is also $C^*$-Gibbs for $V$.

\end{proposition}

{\bf Proof:} Suppose  $\alpha$ is $C^*$-Gibbs for $V$.

The reasoning of the proof is just a generalization of the argument used on last example.

Consider for $r,s>0$
$$\varphi : \overline{a_{-r}...a_{-1} \vert a_1 a_2...a_s}\to \overline{b_{-r}...b_{-1} \vert b_1 b_2...b_s},$$
where
$$ \varphi ( ... x_{-r+2} x_{-r+1}\, a_{-r}...a_{-1} \vert a_1 a_2...a_s\, x_{s+1}\, x_{s+2}...\,) = $$
$$( ... x_{-r+2} x_{-r+1}\, b_{-r}...b_{-1} \vert b_1 b_2...b_s\, x_{s+1}\, x_{s+2}...\,) ,
$$
and
$$\varphi_1 : \overline{a_{-r}...a_{-1}  a_1 \vert a_2...a_s}\to \overline{b_{-r}...b_{-1}  b_1 \vert b_2...b_s},$$
where
$$ \varphi ( ... x_{-r+2} x_{-r+1}\, a_{-r}...a_{-1}  a_1 \vert a_2...a_s\, x_{s+1}\, x_{s+2}...\,) = $$
$$( ... x_{-r+2} x_{-r+1}\, b_{-r}...b_{-1} b_1 \vert  b_2...b_s\, x_{s+1}\, x_{s+2}...\,) ,
$$

Adapting the argument of last example one can easily show that  if for any continuous function $g$ we have that
\begin{equation} \label{outr}
\int_{\overline{a_{-r}...a_{-1} \vert a_1 a_2...a_s}  } e^{V(x, \varphi(x))} g( \varphi(x)) d \alpha (x) =
\int_{\overline{b_{-r}...b_{-1} \vert b_1 b_2...b_s} } g (x) d \alpha (x),
\end{equation}
then, for any continuous function $f$ we have that
\begin{equation} \label{outr1}
\int_{ \overline{a_{-r}...a_{-1}  a_1 \vert a_2...a_s}} e^{V(x, \varphi_1(x))} f( \varphi_1(x)) d \rho (x) =
\int_{ \overline{b_{-r}...b_{-1}  b_1 \vert b_2...b_s}} f (x) d \rho (x).
\end{equation}

As $\alpha$ is $C^*$-Gibbs for $V$, then (\ref{outr1}) is true for any $f$. From (\ref{outr1}) it follows that
$\rho$ is $C^*$-Gibbs for $V$.

We point out that it is equivalent to ask the $C^*$-Gibbs property for $V$ taking symmetric cylinders or taking not symmetric cylinders (this is implicit on the proof of Proposition \ref{krtt}).

\qed

\bigskip

\section{Modular functions and potentials} \label{mod}

 As we mentioned before given a H\"older function $U:\Omega \to \mathbb{R}$ there is a natural way (described by (\ref{Vq})) to get
 a continuous function $V$ satisfying the above property.

We  suppose now that $V$ is such that $ V(x,y) = \sum_{k=-\infty}^\infty [U(\tau^k (x) - U(\tau^k (y)]$, when $x\sim y$, where $U:\Omega \to \mathbb{R}$ is H\"older (see (\ref{Vq})). The function $U$ will sometimes be called a \textbf{potential}. We shall also suppose that $U$ is a finite range potential, or equivalently that it depends on a finite number of coordinates, that is,  there is $k\in \mathbb{N}$ and a function $f: \{1,...,d\}^{k} \to \mathbb{R}$, such that, for all $x\in \Omega$ we get
$$U(x)= U \,( ... x_{-n} x_{-n+1}...x_{-2} x_{-1}\,|\, x_1\,x_2...x_{m-1} x_{m} ...)= f(x_1,x_2,...,x_k),$$
$n,m>0,$ for this fixed $f$ and $k>0$, where $U:\Omega \to \mathbb{R}$. In this case we say that $U$ depends on $k$ coordinates.

\medskip

{\bf Remark 1:} By abuse of language we can write $U: \{1,2,..,d\}^\mathbb{N} \to \mathbb{R}.$
\medskip

If $x \sim y$ isn't hard to see that there is a finite $M>0$, such that,
$$
V(x,y) =
\sum_{k=-\infty}^\infty [U(\tau^k (x)) - U(\tau^k (y))]=
\sum_{k=-M}^M [U(\tau^k (x)) - U(\tau^k (y))].
$$

In this way, if $z\sim \varphi(z)$, then,
$$ V(z,\varphi(z)) = \sum_{k=-M}^M U(\tau^k (z)) -  \sum_{k=-M}^M U(\tau^k (\varphi(z))= \sum_{k=-M}^M [U(\tau^k (z) - U(\tau^k (\varphi(z))]. $$
Therefore, in this case, equation (\ref{jk1}) means
$$
\int_{\overline{ x_{-n},...,x_{-1}\,|\,x_1,x_2,...,y_n   }} e^{\sum_{k=-M}^M U(\tau^k (\varphi(z))]-\sum_{k=-M}^M U(\tau^k (z)) } f(\varphi(z)) d \alpha(z) =$$
\begin{equation} \label{jk2}\int_{\overline{ y_{-n},...,y_{-1}\,|\,y_1,y_2,...,y_n   }} f(z) d \alpha(z)
\end{equation}
\medskip
If $\alpha$ is $C^{*}$-Gibbs for $V$, and  $ V(z, \varphi(z)) =\sum_{k = - \infty}^{+ \infty} U(z) - U(\varphi(z)) $ we also say {\bf by abuse of language that $\alpha$ is $C^{*}$-Gibbs for $U:\Omega \to \mathbb{R}$}.

\medskip

\vspace{0.5cm}

\begin{definition} Given a function $V:G \to \mathbb{R}$, $
V(x,y) =
\sum_{k=-\infty}^\infty [U(\tau^k (x)) - U(\tau^k (y))]$, with $U$ of H\"older class,
we say that a probability measure $\alpha$ on $\Omega$  is  \textbf{ quasi $C^{*}$-Gibbs probability} with respect to the parameter $\beta \in \mathbb{R}$ and $U$, if there exists constants $d_1>0$ and $d_2>0$, such that, for any $x\sim y$ and any $O_{x,y}$,
$$ d_1 \,\int_{O_{(x,y)}} \exp(- \beta V(z, \varphi_{(x,y)}(z))) g(\varphi_{(x,y)}(z)) d \alpha(z)\,\leq$$
\begin{equation} \label{qgibbsR}
\int_{\varphi_{(x,y)}(O_{(x,y)})} g(z) d \alpha (z) \leq  d_2 \,\int_{O_{(x,y)}} \exp(- \beta V(z, \varphi_{(x,y)}(z))) g(\varphi_{(x,y)}(z)) d \alpha(z)
\end{equation}
for every every continuous function $g : \Omega \to \mathbb{C}$ (and conjugated homeomorphism $(O_{(x,y)}, \varphi_{(x,y)})$).

\end{definition}
\medskip

A $C^*$-Gibbs probability is a quasi $C^*$-Gibbs probability.
\medskip
\medskip

We say that a potential
$\tilde{U}:\{1,2..,d\}^\mathbb{N} \to \mathbb{R}$ - which depends on a finite number of coordinates - is {\bf normalized}, if for $k$ large enough and for any $(x_1,x_2,..,x_k)$ we get $\sum_{j=1}^d e^{\tilde{U}(j,x_1...,x_{k-1})}=1$ - in particularly,
we get $e^{\tilde{U}(x)} =e^{\tilde{U}(x_1...,x_{k})}<1$ for all $x=(x_1,x_2,...)\in \{1,2,...,d\}^\mathbb{N}$.

From this follows that for any $w= (w_1,w_2,...,w_m,...)  \in \{1,2,...,d\}^\mathbb{N}$ and $n\in \mathbb{N}$,
 $$\sum_{z_1, z_2,..,z_n =1  }^d e^{ \sum_{j=0}^{n-1} \tilde{U} (\sigma^j(z_1, z_2,..,z_n,w_1,w_2,w_3,...,w_m,...)}=1 $$
where $\sigma$ is the shift acting on $\{1,2,..,d\}^\mathbb{N}.$

Suppose  for such $U$ that $\alpha$ is quasi $C^*$-Gibbs for $U$ (satisfies the equation
(\ref{qgibbsR}) for any continuous $g$). This implies in particular that  for any cylinders of the form $ \overline{|\,x_1^0,x_2^0,...x_s^0}$ and $\overline{|\,y_1^0,y_2^0,...y_s^0}$, there exist $d_1,d_2>0$, such that,
$$d_1
\int_{\overline{|\,x_1^0,x_2^0,...x_s^0} } e^{-\beta V( z, \varphi_{(x,y)}(z))} g( \varphi_{(x,y)}(z)) d \alpha(z) \leq$$
\begin{equation} \label{jk9}  \int_{\overline{|\,y_1^0,y_2^0,...y_s^0}} g(z) d \alpha(z) \leq d_2
\int_{\overline{|\,x_1^0,x_2^0,...x_s^0} } e^{-\beta V( z, \varphi_{(x,y)}(z))} g( \varphi_{(x,y)}(z)) d \alpha(z) ,
\end{equation}
where  $\varphi$ is the associated conjugating  homeomorphism, such that,
$$\varphi: (\overline{|\,x_1^0,x_2^0,...x_s^0} ) \to \overline{|\,y_1^0,y_2^0,...y_s^0} $$

\medskip

\begin{example} \label{poise} Consider the homeomorphism $\varphi : \overline{112 \vert 2} \to \overline{\vert 1122}$ given by
$$
\varphi(... z_{-4} 112 \vert 2 z_{2} z_{3} z_{4} ...) =
(... z_{-4} z_2 z_3 z_4 \vert 1 1 2 2 z_5 ...).
$$

Note that $\overline{112 \vert 2}  $ is translation by $\tau^{-3}$ of the set $\overline{ \vert 1122}.$

Consider the conjugating homeomorphisms, $\varphi_1$, $\varphi_2$, $\varphi_3$ and $\varphi_4$, given by
$$
\varphi_1 (... z_{-4} 112 \vert 2\, {\bf 11} z_{4} ...) =
(... z_{-4}\,{\bf 11}  \vert 1 1 2 2 z_5 ...),
$$
$$
\varphi_2 (... z_{-4} 112 \vert 2\, {\bf 12}\, z_{4} ...) =
(... z_{-4}\,{\bf  12}\,  \vert 1 1 2 2 z_5 ...),
$$
$$
\varphi_3 (... z_{-4} 112 \vert 2\,{\bf 21}\, z_{4} ...) =
(... z_{-4} \,{\bf 21}\,  \vert 1 1 2 2 z_5 ...),
$$
$$
\varphi_4 (... z_{-4} 112 \vert 2 \,{\bf 22} z_{4} ...) =
(... z_{-4} \,{\bf }\,{\bf 22}\,  \vert 1 1 2 2 z_5 ...).
$$

Suppose $\alpha$ is quasi-$C^*$ Gibbs and satisfies (\ref{qgibbsR}).

Therefore,
$$
\int_{\overline{112 \vert 2}} e^{V(x, \varphi(x))} f( \varphi(x)) d \alpha (x) =
$$
$$
\int_{\overline {112 \vert 211}} e^{V(x, \varphi(x))} f(\varphi(x)) +
\int_{\overline {112 \vert 212}} e^{V(x, \varphi(x))} f(\varphi(x)) +
$$
$$
\int_{\overline {112 \vert 221}} e^{V(x, \varphi(x))} f(\varphi(x)) +
\int_{\overline {112 \vert 222}} e^{V(x, \varphi(x))} f(\varphi(x))  =
$$
$$
\int_{\overline {112 \vert 211}} e^{V(x, \varphi_1(x))} f(\varphi_1(x)) +
\int_{\overline {112 \vert 212}} e^{V(x, \varphi_2(x))} f(\varphi_2(x)) +
$$
$$
\int_{\overline {112 \vert 221}} e^{V(x, \varphi_3(x))} f(\varphi_3(x)) +
\int_{\overline {112 \vert 222}} e^{V(x, \varphi_4(x))} f(\varphi_4(x)) \leq
$$
$$
\frac{1}{d_1}\,\,[\,\int_{\overline {111 \vert 122}} f(x) +
\int_{\overline {12 \vert 1122}} f(x) +
\int_{\overline {21 \vert 1122}} f(x) +
\int_{\overline {22 \vert 1122}} f(x)\,] =
$$
$$
\frac{1}{d_1}\,\int_{\overline{ \vert 1122}} f d \alpha (x),
$$
where some of the $d \alpha$ where omitted. We proved that
$$
\int_{\overline{112 \vert 2}} e^{V(x, \varphi(x))} f( \varphi(x)) d \alpha (x) \leq \frac{1}{d_1}
\int_{\overline{\vert 1122}} f d \alpha (x),
$$
for any measurable function $f$.

Taking $f=1$, we get that
$$
\int_{\overline{112 \vert 2}} e^{V(x, \varphi(x))}  d \alpha (x) \leq \frac{1}{d_1}
\int_{\overline{\vert 1122}}  d \alpha (x).
$$

As $e^{V(x, \varphi(x))}$ is strictly positive we get that if $\alpha(\overline{\vert 1122})=0$, then, $\alpha(\overline{112 \vert 2})=0.$

Using the inequality for $d_2$ in  (\ref{qgibbsR}) we get in a similar way that  if $\alpha(\overline{112 \vert 2})=0$, then, $\alpha(\overline{\vert 1122})=0$.

One can also show that
$$\int_{\overline{\vert 1122}}  d \alpha (x)\leq d_2
\int_{\overline{112 \vert 2}} e^{V(x, \varphi(x))}  d \alpha (x) .
$$

\end{example}

\medskip

\begin{proposition} \label{oup} Suppose $\alpha$ is quasi-$C^*$-Gibbs, then
$$\alpha( \overline{a_{-r}...a_{-1} \vert a_1 a_2...a_s})>0,$$
if and only if,
$$\alpha( \vert \overline{a_{-r}...a_{-1}  a_1 a_2...a_s})>0.$$

Moreover, there exist $b_1>0,b_2>0$, such that, for any  cylinder set of the form  $\overline{a_{-r}...a_{-1} \vert a_1 a_2...a_s}$ we get
$$   b_1\,\,\,\alpha(\,\overline{a_{-r}...a_{-1} \vert a_1 a_2...a_s}\,) \leq  \alpha(\,\overline{|\,a_{-r}...a_{-1}  a_1 a_2...a_s}\,)\leq$$
\begin{equation} \label{tutu}
b_2\,\,\alpha(\,\overline{a_{-r}...a_{-1} \vert a_1 a_2...a_s}\,) .
\end{equation}

\end{proposition}

{\bf Proof:} We left the proof for the reader which is an adaptation of the reasoning of Example \ref{poise}.

Note that for $x\sim y$ the values $e^{V(x, y)}$ are uniformly bounded above and below.

\qed

\medskip
 The next result shows that we can always consider normalized potentials (see \cite{PP} for general results) on the definition of quasi $C^*$-Gibbs  probability.

\begin{theorem} \label{opp} Suppose the probability $\alpha$ on $\Omega$ is $C^{*}$-Gibbs for H\"older potential $U$. Assume, $X:\Omega \to \mathbb{R}$ is such that  $X=U+ g - g \circ \tau+ \lambda$, where $g: \Omega \to \mathbb{R}$ is a continuous function, then $\alpha$ is quasi $C^{*}$-Gibbs for $X$.
\end{theorem}

\textbf{Proof:} Suppose that for any continuous $f$ we have
$$
\int_{O_{x,y}} e^{\beta \sum_{k=-\infty }^\infty U(\tau^k (\varphi(z)))-\beta \sum_{k=-\infty}^\infty U(\tau^k (z)) } f(\varphi(z)) d \alpha(z) =$$
\begin{equation} \label{jk7}\int_{\varphi(O_{(x,y)})} f(z) d \alpha(z)
\end{equation}

Note that
$$ \sum_{k=-\infty }^\infty g(\tau^k (\varphi(z)) -\sum_{k=-\infty}^\infty g(\tau^{k+1} (\varphi(z)))$$
and
$$ \sum_{k=-\infty }^\infty g(\tau^k (z)) -\sum_{k=-\infty}^\infty g(\tau^{k+1} (z))$$
are both bounded by constants related to the supremum of $g$.

Therefore,
$$ \sum_{k=-\infty }^\infty X(\tau^k (\varphi(z)))-\sum_{k=-\infty}^\infty X(\tau^k (z))=$$
$$ [\,\,\sum_{k=-\infty }^\infty U(\tau^k (\varphi(z))]-\sum_{k=-\infty}^\infty U(\tau^k (z))\,]+$$
$$ [\,\sum_{k=-\infty }^\infty g(\tau^k (\varphi(z)) -\sum_{k=-\infty}^\infty g(\tau^{k+1} (\varphi(z)))\,]-$$
$$ [\,\sum_{k=-\infty }^\infty g(\tau^k (z)) -\sum_{k=-\infty}^\infty g(\tau^{k+1} (z))\,]$$
is bounded above and below by constants which do not depend on $x\sim y$, $O_{x,y}$ and corresponding $\varphi_{x,y}$.

Then, $\alpha$ is quasi $C^{*}$-Gibbs for $X$.

\qed

By Proposition 1.2 in \cite{PP} given a H\"older potential $U: \Omega \to \mathbb{R}$, one can find $W$ depending on positive coordinates $(1,2,3,..,n,...) =\mathbb{N}$ and a continuous function $v:\Omega \to \mathbb{R}$,  such that,  $W=U+ v - v \circ \tau.$

From the structure of the proof one can see that if $U$ depends on a finite number of coordinates, then,   both $W$ and $v$ also depend on  a finite number of coordinates.

More precisely, there exist $\tilde{W}:\{1,2,...,d\}^\mathbb{N} \to \mathbb{R}$ an $r$, such that,
$$ W  (...  x_{-n-1} \, \,x_{-n} x_{-n + 1} ... x_{-1} \,\vert\, x_{1} ... x_{m}\, x_{m+1}  ...)=$$
$$ \tilde{W}  ( x_{1} ... x_{m}\, x_{m+1}  ...)=K( x_{1} ... x_{r}\,),$$
for a certain function $K:\{1,2,...,d\}^r \to \mathbb{R}.$

Now, by Theorem 2.2 in \cite{PP}, given such $\tilde{W}$ one can find, $u$ and positive constant $\lambda$, such that,
 $\tilde{W}=\,\tilde{U}+ u - u \circ \tau+ \lambda$. Moreover,  $\tilde{U}: \{1,2,...,d\}^\mathbb{N} \to \mathbb{R}$ and  $u: \{1,2,...,d\}^\mathbb{N} \to \mathbb{R}$ both  depend on a finite number of coordinates.

 \medskip

 {\bf Remark 2:} Therefore, from Theorem \ref{opp} if $\alpha$ is $C^*$-Gibbs for a H\"older potential $U:\Omega \to \mathbb{R}$, which depends on a finite number of coordinates, we can assume that $\alpha$ is quasi-$C^*$-Gibbs for {\bf another} potential, denoted $\tilde{U}$, which  is normalized and depending on a finite number of coordinates.

By abuse of language one can write  $\tilde{U}: \{1,2,...,d\}^\mathbb{Z} \to \mathbb{R}$.
 \medskip

\medskip
\section{Equivalence between equilibrium measures and $C^{*}$-Gibbs measures} \label{E}

 First we present two important and well known theorems (see theorems 1.2 and 1.22 in \cite{bowenbook} and also \cite{RuelleBook}).

 We will consider without lost of generality that $\beta=1$.

$ \mathcal{M}_{\tau} (\Omega)$ denotes the set on invariant probabilities for $\tau$ acting on $\Omega$.

\begin{theorem} \label{Bowen1.2} (see Theorem 1.2 in \cite{bowenbook})
Suppose $U : \Omega \to \mathbb{R}$  is of H\"older class.  Then, there is a unique $\tau$-invariant $\rho \in \mathcal{M}_{\tau} (\Omega)$, for which one can find constants $C_{1} > 0$, $C_{2} > 0$, and P such that, for all cylinder $\overline{|\,y_1^0,y_2^0,...y_s^0}$ we have
\begin{equation} \label{defbowen}
C_{1} \leq
\frac{\rho (\overline{|\,y_1^0,y_2^0,...y_s^0})   }
{\exp \left( -P\, s + \sum_{k=0}^{s-1} U(\tau^{k}x) \right) }
\leq C_{2},
\end{equation}
where
$$x = (...x_{-k}, x_{-k+1},...,x_{-1}\,|\,x_{1},...,x_m,x_{m+1},...) \in \overline{|\,y_1^0,y_2^0,...y_s^0} \subset \Omega ,$$

\end{theorem}

\medskip

We call (\ref{defbowen}) the Bowen's inequalities.

\medskip

\begin{definition} The probability $\rho=\rho_U$ of Theorem \ref{Bowen1.2} is called \textbf{equilibrium probability} for the potential $U$.
\end{definition}

\begin{theorem}
 Given $U$ as above and $\rho_{U}$ the equilibrium measure for $U$, then $\rho_{U}$ is the unique probability  on $\mathcal{M}_{\tau} (\Omega)$, for which
$$
h(\rho_U) + \int U d \rho_U = P(U) =\sup_{\nu \in \mathcal{M}_{\tau}} \{h(\nu) + \int U d \nu\},
$$
where $h(\nu)$ is the entropy of $\nu$.
\end{theorem}

For a proof see \cite{PP} or \cite{bowenbook}.

 $P(U)$ is called the pressure of  $U$. One can show that the $P$ of (\ref{defbowen}) is equal to such  $P(U).$

\medskip
Remember that if $\alpha$ is $C^{*}$-Gibbs for $V$, and  $ V(z, \varphi(z)) =\sum_{k = - \infty}^{+ \infty} U(z) - U(\varphi(z)) $ we also say by abuse of language that $\alpha$ is $C^{*}$-Gibbs for $U:\Omega \to \mathbb{R}$.

\medskip

Note that if $\alpha$ is an equilibrium probability for a H\"older potential $U$, then, it is also an equilibrium probability for
$U + (g \circ \tau)  - g + c$, where $c$ is constant and $g: \Omega \to \mathbb{R}$ is  H\"older continuous (see \cite{PP}). In this way we can assume without lost of generality that $\alpha_U$ is an equilibrium probability for a normalized potential $U$. If $U$ is normalized then $P(U)=0$.

If $\alpha$ on $\Omega$ is $C^{*}$-Gibbs for $U$, then, from Remark 2  we have that $\alpha$ is quasi-$C^*$-Gibbs for {\bf another} potential $U$ which is normalized.

Note that given $U$ we are dealing with two  definitions:  $C^{*}$-Gibbs and Equilibrium.  From the above comments  we can assume in either case that $U$ is normalized.

 The bottom line is: we can assume (see \cite{PP}) that the H\"older potential $\tilde{U} = U + (g \circ \tau)  - g + c$ is normalized, depends just on future coordinates
$\tilde{U}: \{1,2,...,d\}^\mathbb{N} \to \mathbb{R}$ and has pressure zero.

 We will work here (due to Theorem \ref {opp} and the above comments)  with the case where the probability $\alpha$ - which is $C^*$-Gibbs for the potential $U$ - is also a quasi-$C^*$-Gibbs probability for the potential $\tilde{U}$ satisfying Pressure $P=0$.  In this case, if we want to prove
 expression (\ref{defbowen}) for such probability $\alpha$ over $\Omega$, this can be simplified just showing that there exist $c_1,c_2>0$, such that,
 \begin{equation} \label{defbowen23}
c_{1} \leq
\frac{\alpha(\overline{|\,y_1^0,y_2^0,...y_s^0}) }
{\exp \left( \sum_{k=0}^{s-1} \tilde{U}(\sigma^{k}x) \right) }
\leq c_{2}.
\end{equation}
where $\sigma$ is the shift acting on $\{1,2,..,d\}^\mathbb{N}$ and
where $x$ is of the form
$$x = (y_1^0,y_2^0,...y_s^0,x_{s+1},...,x_m,x_{m+1},...) \in \{1,2,..,d\}^\mathbb{N} .$$

{\bf Remark 3:} Indeed, due to Remark 2 we get that $\tilde{U} = U + (g \circ \tau)  - g + c$, where $g$ depends on finite coordinates. Therefore,  to show (\ref{defbowen23}) - for {\bf $\alpha$ which is $C^*$-Gibbs for $U:\Omega \to \mathbb{R}$} - is equivalent to prove (see details on the proof of Theorem \ref{opp}) that there exists $C_1,C_2>0$, such that,
 \begin{equation} \label{defbowen25}
C_{1} \leq
\frac{\alpha(\overline{|\,y_1^0,y_2^0,...y_s^0}) }
{\exp \left( \sum_{k=0}^{s-1} U(\tau^{k}x) \right) }
\leq C_{2}.
\end{equation}
where $\tau$ is the shift acting on $\{1,2,..,d\}^\mathbb{Z}$ and
where
$$x = ( ... x_{-2},x_{-1}\,|\,y_1^0,y_2^0,...y_s^0,x_{s+1},...,x_m,x_{m+1},...) \in \{1,2,..,d\}^\mathbb{Z} .$$

\medskip

It's important to note that main equivalence (equilibrium and $C^*$-Gibbs)  is still valid in a more general settings of H\"older potential in a general Smale Space. D. Ruelle proved on the setting of hyperbolic diffeomorphisms  that Equilibrium implies $C^{*}$-Gibbs in his book \cite{RuelleBook}, see theorems 7.17(b), 7.13(b) and section 7.18). On other hand Haydn proved in the paper \cite{HaydnGibbs} that $C^{*}$-Gibbs implies Equilibrium. Later, the paper \cite{HD} presents a shorter proof of the equivalence.

On the two next sections we will present the proof of the following theorem.

\begin{theorem} \label{mai}

 Given a potential $U$ depending on a finite number of coordinates, then,  $\alpha$ is the equilibrium measure for $U$, if and only if, $\alpha$ is
 $C^*$-Gibbs for $U$. As the equilibrium probability is unique we get that the $C^*$-Gibbs probability for $U$ is unique.

 \end{theorem}

\vspace{7pt}
\section{Equilibrium implies $C^{*}$-Gibbs}

The fact that Equilibrium state  implies $C^{*}$-Gibbs
 was proved by Ruelle in a  general setting. The proof is in  the  book  \cite{RuelleBook} (see theorems 7.17(b), 7.13(b) and section 7.18).

 For completeness we will explain the proof on our setting.

 We drop the $(x,y)$ on
$ \varphi_{(x,y)  }$ and $\mathcal{O}_{(x,y)  }$.

\begin{lemma} \label{7.17Ruelle}
Let $(\Omega, \tau)$ be the shift on the Bernoulli space $\Omega =\{1,2,...,d\}^{\mathbb{Z}-\{0\}}$ and $\rho_{0}$ be the $\tau$-invariant probability measure which realizes the maximum of the entropy, or, simply the equilibrium state for $U = 0$. If $(\mathcal{O}, \varphi)$ is a conjugating homeomorphism, then
$$
\int_{\mathcal{O}} f(\varphi(x)) d \rho_{0}(x) =
\int_{\mathcal{\varphi(\mathcal{O})}} f(x) d \rho_{0} (x)
$$
\end{lemma}

{\bf Proof:} Given
$$\mathcal{O}= \overline{ x_{-n}x_{-n+1}...x_{-1}\,|\, x_1\,...x_{m-1}x_{m} } $$ and
$$ \varphi(\mathcal{O})= \overline{ y_{-n}y_{-n+1}...y_{-1}\,|\, y_1\,...y_{m-1}y_{m} },$$
we have that for any $r<m$ and $k<n$
$$ \rho_{0} (\overline{ x_{-k}x_{-k+1}...x_{-1}\,|\, x_1\,...x_{r-1}x_{r} }  )  =d^{ -\,(\,r+k\,)}= $$
$$ \rho_{0} (\overline{ y_{-k}y_{-k+1}...y_{-1}\,|\, y_1\,...y_{r-1}y_{r} }).$$

Therefore,

$$ \int_\mathcal{O} I_{  \overline{ x_{-k}x_{-k+1}...x_{-1}\,|\, x_1\,...x_{r-1}x_{r} }  }(\varphi(x)) d \rho_0(x)=$$
$$     \int_{ \varphi(\mathcal{O}) } I_{  \overline{ y_{-k}y_{-k+1}...y_{-1}\,|\, y_1\,...y_{r-1}y_{r} }  }(y) d \rho_0(y).             $$

From this follows the claim.

\qed

\medskip

We denote by $C^{\alpha}(\Omega)$ the set of $\alpha$ H\"older functions on $\Omega$.

\begin{lemma} \label{7.13Ruelle} (see corollary 7.13 in \cite{RuelleBook})
	Consider the shift space $(\Omega, \tau)$ and $A, B \in C^{\alpha}(\Omega)$. Write for $a<0$ and $b>0$
	$$
	Z_{[a,b]} =
	\int\,\,\, e^{ \sum_{k=a}^{b-1} B \circ \tau^{k}   }\,\,d \,\, \rho_{A}
	$$
	Then, $Z^{-1}_{[a,b]} \,\,(\exp \sum_{k=a}^{b-1} B \circ \tau^{k} )\,\, \rho_{A}$ tends to $\rho_{A+B}$ in the vague topology, when $a \to - \infty$ and $b \to + \infty$.

In particular, taking $A=0$, when $a \to - \infty$ and $b \to + \infty$, we get that
$$Z^{-1}_{[a,b]}\,\,\, e^{ \sum_{k=a}^{b-1} B \,\circ\, \tau^{k} }\,\,\, \rho_{0}\,\, \to\,\, \rho_B,$$
where
$$
Z_{[a,b]} =
	\int\,\,\, e^{ \sum_{k=a}^{b-1} B \circ \tau^{k}   }\,\,d \,\, \rho_{0}
$$

\end{lemma}

\begin{theorem} \label{rtw9}

If $\rho_{B}$ is an equilibrium state for $B$ then it is a $C^{*}$-Gibbs state for $B$.
\end{theorem}

\textbf{Proof:}  The statement holds for $B = 0$ by Lemma \ref{7.17Ruelle}. Moreover, Lemma \ref{7.13Ruelle} allow us to extend this result for all $B \in C^{\alpha} (\Sigma_{N})$ in the following manner:
$$
\int_{\varphi (\mathcal{O})}  g(x) d \rho_{B} (x) =
\lim_{\substack{a \to - \infty \\ b \to  \infty} } Z^{-1}_{[a,b]} \int_{\varphi(\mathcal{O})} \exp \left( \sum_{k=0}^{b-1} B \circ \tau^{k} (x) \right)  g(x) d \rho_{0} (x) =
$$
$$
\lim_{\substack{a \to - \infty \\ b \to  \infty}} Z^{-1}_{[a,b]} \int_{\mathcal{O}} \exp \left( \sum_{k=0}^{b-1} B \circ \tau^{k} \circ \varphi (x) \right) g \circ \varphi(x) d \rho_{0} (x) =
$$
$$
\lim_{\substack{a \to - \infty \\ b \to  \infty}} Z^{-1}_{[a,b]}  \int_{\mathcal{O}} \exp \left( \sum_{k=0}^{b-1} B \circ \tau^{k} \circ \varphi (x) - \sum_{k=0}^{b-1} B \circ \tau^{k}(x) \right)
$$
$$
\quad \quad \quad \quad \quad \quad \quad \quad \exp \left( \sum_{k=0}^{b-1} B \circ \tau^{k}(x) \right) g \circ \varphi(x) d \rho_{0} (x) =
$$
$$
\lim_{\substack{a \to - \infty \\ b \to  \infty}} Z^{-1}_{[a,b]} \int_{\mathcal{O}} e^{\left( -V(x, \varphi(x)) \right)} g \circ \varphi(x) \exp \left( \sum_{k=0}^{b-1} B \circ \tau^{k} (x) \right) d \rho_{0} (x) =
$$
$$
\int_{\mathcal{O}} e^{\left( -V(x, \varphi(x)) \right)} g \circ \varphi(x) d \rho_{B} (x).
$$

Since the equality
$$
\int_{\varphi (\mathcal{O})}  g(x) d \rho_{B} (x) =
\int_{\mathcal{O}} e^{\left( -V(x, \varphi(x)) \right)} g \circ \varphi(x) d \rho_{B} (x)
$$
was verified for any conjugating homeomorphism $\varphi$ and any $g$, then it follows that $\rho_{B}$  is an $C^{*}$-Gibbs state for $B$.

\qed

\vspace{7pt}
\section{$C^{*}$-Gibbs implies Equilibrium}

Given a $C^*$-Gibbs probability $\alpha$ for the H\"older potential $U$ we want to show that $\alpha$ is the equilibrium probability for $U$.

In order to simplify the reasoning we assume that $U$ depends on $r$ coordinates.

Therefore, we can assume from section \ref{E} and Remark 2 on the end of section \ref{mod} that $\alpha$ is a quasi-$C^*$-Gibbs probability for {\bf another} normalized potential {\bf which we also denote by $U$}.

More precisely, $U:\Omega \to \mathbb{R}$ is such that there exist $ f: \{1,2,..,d\}^r \to \mathbb{R}$ and $\tilde{U}$ satisfying
$$ U (..., x_{-2}, x_{-1}   \,|\, x_1, x_2,..,x_r, x_{r+1}, x_{r+2},...  ) = \tilde{U} \,( x_1, x_2,..,x_r, x_{r+1}, x_{r+2},...  ) =$$
$$ f(x_1, x_2,..,x_r).$$

We also assume that for any $w= (w_1,w_2,...,w_m,...)  \in \{1,2,...,d\}^\mathbb{N}$ and $n\in \mathbb{N}$,
\begin{equation} \label{sav0} \sum_{z_1, z_2,..,z_n =1  }^d e^{ \sum_{j=0}^{n-1} \tilde{U} (\sigma^j(z_1, z_2,..,z_n,w_1,w_2,w_3,...,w_m,...)}=1,\end{equation}
where $\sigma$ is the shift acting on $\{1,2,..,d\}^\mathbb{N}.$

If we know that $\alpha$ is $\tau$-invariant and  also satisfies the Bowen's inequalities for $U$, then, it will follow that $\alpha$ is the equilibrium probability for $U$ by Theorem \ref{Bowen1.2}.

We will show first that a quasi $C^*$-Gibbs probability $\alpha$ for $U$ satisfies the Bowen's inequalities (\ref{defbowen25}) for $U$.

Later we will show that a $C^*$-Gibbs probability $\alpha$ is invariant for $\tau$ (see Proposition \ref{outr}). This will finally show (see Theorem \ref{rwt3}) that "$C^{*}$-Gibbs implies Equilibrium".

Note that we want to show (\ref{defbowen25}) but due to Remark  3 we just have to show (\ref{defbowen23}).

We assumed $\alpha$ is such that (\ref{jk9}) is true, that is, there exists $d_1,d_2>0$, such that, for any continuous function $g$
$$
 \,d_1 \int_{\overline{|\,x_1^0,x_2^0,...x_s^0} } e^{- V( z, \varphi_{(x,y)}(z))} g( \varphi_{(x,y)}(z)) d \alpha(z) \leq $$
\begin{equation} \label{jk11}  \int_{\overline{|\,y_1^0,y_2^0,...y_s^0}} g(z) d \alpha(z) \leq d_2 \int_{\overline{|\,x_1^0,x_2^0,...x_s^0} } e^{- V( z, \varphi_{(x,y)}(z))} g( \varphi_{(x,y)}(z)) d \alpha(z).
\end{equation}

As $U:\Omega \to \mathbb{R}$ is $\gamma$-H\"older we get that it is bounded
$$\sup |\,\sum_{k = - \infty}^{\infty} U(\tau^{k}(x)) - U(\tau^{k} y)| \,$$
for all $x \sim y.$

Moreover, there exists $K_1>0$, such that, for any   $   x \sim y$ we have that $|\,\sum_{k = - \infty}^{\infty} U(\tau^{k}(x)) - U(\tau^{k} y)|\leq K_1 $.

We denote $\mathcal{U}= \sup_{x \in \Omega} U(x) -  \inf_{x \in \Omega} U(x)$.

\begin{lemma} \label{rwt1}

Given the normalized H\"older potential $U(x) = f(x_1, x_2,..,x_r)$, consider $x_1^0\,...x_s^0$ and  $y_1^0\,...y_s^0$ fixed, and also  $a,b \in \{1,2...,d\}$ fixed.

Now, taking $s>2r$ we  fixed

$$x=(...x_{-m}x_{-m+1}...x_{-1}\,|\,  x_1^0\,...x_s^0\, x_{s+1},  x_{s+2},...x_{m-1}x_{m} ...)\in \overline{|\,x_1^0,x_2^0,...x_s^0}  $$

$$y=(...x_{-m}x_{-m+1}...x_{-1}\,|\, y_1^0\,...y_s^0\,x_{s+1}, x_{s+2},...x_{m-1}x_{m} ...) \in \overline{|\,y_1^0,y_2^0,...y_s^0},$$

and also

$$x_a=(...x_{-m}x_{-m+1}...x_{-1}\,|\,  x_1^0\,...x_s^0\, {\bf a}\, x_{s+2}...x_{m-1}x_{m} ...)\in \overline{|\,x_1^0,x_2^0,...x_s^0}  $$

$$y_b=(...x_{-m}x_{-m+1}...x_{-1}\,|\, y_1^0\,...y_s^0\, {\bf b}\, x_{s+2}...x_{m-1}x_{m} ...) \in \overline{|\,y_1^0,y_2^0,...y_s^0}.$$

Then,

$$|\sum_{k = - \infty}^{\infty} U(\tau^{k}(x_a)) - U(\tau^{k} (y_b))|\,\leq $$
$$2\,r\, \mathcal{U}\,+\,\,|\sum_{k = - \infty}^{\infty} U(\tau^{k}(x)) - U(\tau^{k} (y))|\,.$$

\end{lemma}
{\bf Proof:} Note that as $s>2\, r$

$$|\sum_{k = - \infty}^{\infty} U(\tau^{k}(x_a)) - U(\tau^{k} (y_b))|=$$
$$|\sum_{k = -r}^{\infty} U(\tau^{k}(x_a)) - U(\tau^{k} (y_b))|=$$
$$|\sum_{k = -r}^{s+ 1}  U(\tau^{k}(x_a)) - U(\tau^{k} (y_b))| =
$$
$$|\,[\, U (...\,x_{-m}\,x_{-m+1}...x_{-r-1}  x_{-r} \,\,|\,\overbrace{x_{-r+1}...x_1^0}^r,...  x_s^0 \,{\bf a}\,x_{s+2}...x_{m}x_{m+1}   ...) -$$
$$\, U (...x_{-m}x_{-m+1}... x_{-r-1}  x_{-r}\,\,|\,\overbrace{x_{-r+1}...y_1^0}^r, ...,y_s^0 {\bf b} \,x_{s+2}...x_{m}x_{m+1}  ...) \,]\,+$$
$$\,[\, U (...\,x_{-m}\,x_{-m+1}...x_{-r}  x_{-r+1} \,\,|\,x_{-r+2}...x_1^0, x_2^0...  x_s^0 \,{\bf a}\,x_{s+2}...x_{m}x_{m+1}   ...) -$$
$$\, U (...x_{-m}x_{-m+1}... x_{-r}  x_{-r+1}\,\,|\,x_{-r+2}...y_1^0,y_2^0 ...,y_s^0 {\bf b} \,x_{s+2}...x_{m}x_{m+1}  ...) \,]\,+$$
$$...$$
$$\,[\, U (...\,x_{-m}\,x_{-m+1}...x_{-1} \,\,|\,\overbrace{ x_1^0 x_2^0\,...x_r^0}^r,...  x_s^0 \,{\bf a}\,x_{s+2}...x_{m}x_{m+1}   ...) -$$
$$ U (...x_{-m}x_{-m+1}...x_{-2} x_{-1}\,\,|\, \overbrace{y_1^0 y_2^0\,...y_r^0}^r,...y_s^0 {\bf b} \,x_{s+2}...x_{m}x_{m+1}  ...) \,]\,+$$
$$ ...+$$
$$\,[\, U (...\,x_{-m}\,x_{-m+1}...x_{-1} x_1^0 x_2^0\,... \,x_{s-r-1}^0\,|\,\overbrace{x_{s-r}^0...  x_s^0 \,{\bf a}}^r\,x_{s+2}...x_{m}x_{m+1}   ...) -$$
$$\, U (...x_{-m}x_{-m+1}...x_{-2} x_{-1} y_1^0 y_2^0\,...\,y_{s-r-1}^0\,|\,\overbrace{y_{s-r}^0...y_s^0 {\bf b}}^r \,x_{s+2}...x_{m}x_{m+1}  ...) \,]\,+$$
$$\,[\, U (...\,x_{-m}\,x_{-m+1}...x_{-1} x_1^0 x_2^0\,... \,x_{s-r}^0\,|\,x_{s-r+1}^0...  x_s^0 \,{\bf a}\,x_{s+2}...x_{m}x_{m+1}   ...) -$$
$$\, U (...x_{-m}x_{-m+1}...x_{-2} x_{-1} y_1^0 y_2^0\,...\,y_{s-r}^0\,|\,y_{s-r+1}^0...y_s^0 {\bf b} \,x_{s+2}...x_{m}x_{m+1}  ...) \,]\,+$$
$$\,[\, U (...\,x_{-m}\,x_{-m+1}...x_{-1} x_1^0 x_2^0\,... \,x_{s-r+1}^0\,|\,x_{s-r+2}^0...  x_s^0 \,{\bf a}\,x_{s+2}...x_{m}x_{m+1}   ...) -$$
$$\, U (...x_{-m}x_{-m+1}...x_{-2} x_{-1} y_1^0 y_2^0\,...\,y_{s-r+1}^0\,|\,y_{s-r+2}^0...y_s^0 {\bf b} \,x_{s+2}...x_{m}x_{m+1}  ...) \,]\,+$$
$$...+$$
$$[\, U (...x_{-m}x_{-m+1}...x_{-2}x_{-1}\, x_1^0\,...x_s^0 \,|\,\,{\bf a}\,x_{s+2}...x_{m}x_{m+1}   ...) -$$
$$\, U (...x_{-m}x_{-m+1}...x_{-2} x_{-1}\, y_1^0\,...y_s^0 \,|\,{\bf b}\,x_{s+2}...x_{m}x_{m+1}    ...) \,]\,\,|\,\,=$$
$$|\,\,\,[\, U (...\,x_{-m}\,x_{-m+1}...x_{-r-1}  x_{-r} \,\,|\,\overbrace{ x_{-r+1}...x_1^0}^r...  x_s^0 \,{\bf x_{s+1}} \,x_{s+2}...x_{m}x_{m+1}   ...) -$$
$$ U (...x_{-m}x_{-m+1}... x_{-r-1}  x_{-r}\,\,|\,\overbrace{x_{-r+1}...y_1^0}^r, ...,y_s^0 {\bf x_{s+1}} \,x_{s+2}...x_{m}x_{m+1}  ...) \,]\,+$$
$$\,[\, U (...\,x_{-m}\,x_{-m+1}...x_{-r}  x_{-r+1} \,\,|\,x_{-r+2}...x_1^0, x_2^0...  x_s^0 \,{\bf x_{s+1}}\,x_{s+2}...x_{m}x_{m+1}   ...) -$$
$$ U (...x_{-m}x_{-m+1}... x_{-r}  x_{-r+1}\,\,|\,x_{-r+2}...y_1^0,y_2^0 ...,y_s^0 {\bf x_{s+1}} \,x_{s+2}...x_{m}x_{m+1}  ...) \,]\,+$$
$$...$$
$$\,[\, U (...\,x_{-m}\,x_{-m+1}...x_{-1} \,\,|\, \overbrace{x_1^0 x_2^0\,...x_r^0}^r,...  x_s^0 \,{\bf x_{s+1}}\,x_{s+2}...x_{m}x_{m+1}   ...) -$$
$$ U (...x_{-m}x_{-m+1}...x_{-2} x_{-1}\,\,|\, \overbrace{y_1^0 y_2^0\,...y_r^0}^r,...y_s^0 {\bf x_{s+1}} \,x_{s+2}...x_{m}x_{m+1}  ...) \,]\,+$$
$$ ...+$$
$$\,[\, U (...\,x_{-m}\,x_{-m+1}...x_{-1} x_1^0 x_2^0\,... \,x_{s-r-1}^0\,|\,\overbrace{x_{s-r}^0...  x_s^0 \,{\bf a}}^r\,x_{s+2}...x_{m}x_{m+1}   ...) -$$
$$ U (...x_{-m}x_{-m+1}...x_{-2} x_{-1} y_1^0 y_2^0\,...\,y_{s-r-1}^0\,|\,\overbrace{y_{s-r}^0...y_s^0 {\bf b}}^r \,x_{s+2}...x_{m}x_{m+1}  ...) \,]\,+$$
$$\,[\, U (...\,x_{-m}\,x_{-m+1}...x_{-1} x_1^0 x_2^0\,... \,x_{s-r}^0\,|\,\overbrace{x_{s-r+1}^0...  x_s^0 \,{\bf a}\,x_{s+2}}^r,...x_{m}x_{m+1}   ...) -$$
$$\, U (...x_{-m}x_{-m+1}...x_{-2} x_{-1} y_1^0 y_2^0\,...\,y_{s-r}^0\,|\,\overbrace{y_{s-r+1}^0...y_s^0 {\bf b} \,x_{s+2}}^r ,...x_{m}x_{m+1}  ...) \,]\,+$$
$$\,[\, U (...\,x_{-m}\,x_{-m+1}...x_{-1} x_1^0 x_2^0\,... \,x_{s-r+1}^0\,|\,x_{s-r+2}^0...  x_s^0 \,{\bf a}\,x_{s+2}...x_{m}x_{m+1}   ...) -$$
$$\, U (...x_{-m}x_{-m+1}...x_{-2} x_{-1} y_1^0 y_2^0\,...\,y_{s-r+1}^0\,|\,y_{s-r+2}^0...y_s^0 {\bf b} \,x_{s+2}...x_{m}x_{m+1}  ...) \,]\,+$$
$$...+$$
$$[\, U (...x_{-m}x_{-m+1}...x_{-2}x_{-1}\, x_1^0\,...x_s^0 \,|\,\,{\bf a}\,x_{s+2}...x_{m}x_{m+1}   ...) -$$
$$\, U (...x_{-m}x_{-m+1}...x_{-2} x_{-1}\, y_1^0\,...y_s^0 \,|\,{\bf b}\,x_{s+2}...x_{m}x_{m+1}    ...) \,]\,\,|\,\,\leq $$
$$2\, r\, \mathcal{U} \,\,+\,\, |\,\,\,[\, U (...\,x_{-m}\,x_{-m+1}...x_{-r-1}  x_{-r} \,\,|\,x_{-r+1}...x_1^0...  x_s^0 \,{\bf x_{s+1}}\,x_{s+2}...x_{m}x_{m+1}   ...) -$$
$$\, U (...x_{-m}x_{-m+1}... x_{-r-1}  x_{-r}\,\,|\,x_{-r+1}...y_1^0, ...,y_s^0 {\bf x_{s+1}} \,x_{s+2}...x_{m}x_{m+1}  ...) \,]\,+$$
$$\,[\, U (...\,x_{-m}\,x_{-m+1}...x_{-r}  x_{-r+1} \,\,|\,x_{-r+2}...x_1^0, x_2^0...  x_s^0 \,{\bf x_{s+1}}\,x_{s+2}...x_{m}x_{m+1}   ...) -$$
$$ U (...x_{-m}x_{-m+1}... x_{-r}  x_{-r+1}\,\,|\,x_{-r+2}...y_1^0,y_2^0 ...,y_s^0 {\bf x_{s+1}} \,x_{s+2}...x_{m}x_{m+1}  ...) \,]\,+$$
$$...$$
$$\,[\, U (...\,x_{-m}\,x_{-m+1}...x_{-1} \,\,|\, \overbrace{x_1^0 x_2^0\,... x_r^0}^r,..., x_s^0 \,{\bf x_{s+1}}\,x_{s+2}...x_{m}x_{m+1}   ...) -$$
$$\, U (...x_{-m}x_{-m+1}...x_{-2} x_{-1}\,\,|\,  \overbrace{y_1^0 y_2^0\,...y_r^0}^r,...y_s^0 {\bf x_{s+1}} \,x_{s+2}...x_{m}x_{m+1}  ...) \,]\,+$$
$$ ...+$$
$$\,[\, U (...\,x_{-m}\,x_{-m+1}...x_{-1} x_1^0 x_2^0\,... \,x_{s-r-1}^0\,|\,\overbrace{x_{s-r}^0...  x_s^0 \,{\bf x_{s+1}}}^r\,x_{s+2}...x_{m}x_{m+1}   ...) -$$
$$\, U (...x_{-m}x_{-m+1}...x_{-2} x_{-1} y_1^0 y_2^0\,...\,y_{s-r-1}^0\,|\,\overbrace{y_{s-r}^0...y_s^0 {\bf x_{s+1}}}^r \,x_{s+2}...x_{m}x_{m+1}  ...) \,]\,+$$
$$\,[\, U (...\,x_{-m}\,x_{-m+1}...x_{-1} x_1^0 x_2^0\,... \,x_{s-r}^0\,|\,\overbrace{x_{s-r+1}^0...  x_s^0 \,{\bf x_{s+1}}\,x_{s+2}}^r...x_{m}x_{m+1}   ...) -$$
$$\, U (...x_{-m}x_{-m+1}...x_{-2} x_{-1} y_1^0 y_2^0\,...\,y_{s-r}^0\,|\,\overbrace{y_{s-r+1}^0...y_s^0 {\bf x_{s+1}} \,x_{s+2}}^r...x_{m}x_{m+1}  ...) \,]\,+$$
$$\,[\, U (...\,x_{-m}\,x_{-m+1}...x_{-1} x_1^0 x_2^0\,... \,x_{s-r+1}^0\,|\,x_{s-r+2}^0...  x_s^0 \,{\bf x_{s+1}}\,x_{s+2}...x_{m}x_{m+1}   ...) -$$
$$\, U (...x_{-m}x_{-m+1}...x_{-2} x_{-1} y_1^0 y_2^0\,...\,y_{s-r+1}^0\,|\,y_{s-r+2}^0...y_s^0 {\bf x_{s+1}} \,x_{s+2}...x_{m}x_{m+1}  ...) \,]\,+$$
$$...+$$
$$[\, U (...x_{-m}x_{-m+1}...x_{-2}x_{-1}\, x_1^0\,...x_s^0 \,|\,\,{\bf x_{s+1}}\,x_{s+2}...x_{m}x_{m+1}   ...) -$$
$$\, U (...x_{-m}x_{-m+1}...x_{-2} x_{-1}\, y_1^0\,...y_s^0 \,|\,{\bf x_{s+1}}\,x_{s+2}...x_{m}x_{m+1}    ...) \,]\,\,|\,= $$
$$2\, r\, \mathcal{U}\,\,+\,\,|\sum_{k = -r}^{s+ 1}  U(\tau^{k}(x)) - U(\tau^{k} (y))|.
$$

\qed
\medskip

We will adapt the formulation of Proposition 2.1 in \cite{HaydnGibbs} to the present situation.

\medskip

For fixed
$\overline{|\,x_1^0,x_2^0,...x_s^0} $ denote
$$ U_a = \overline{|\,x_1^0,x_2^0,...,x_s^0, a }$$
$a=1,2..,d.$

Note that
\begin{equation} \label{sav1} \sum_a \alpha (U_a) \leq d \,\,\,\alpha (\overline{|\,x_1^0,x_2^0,...x_s^0}). \end{equation}

Consider now a fixed $\overline{|\,y_1^0,y_2^0,...y_s^0}$ and $\varphi_{a,b}$, $a=1,2,...,d$,  $b=1,2,...,d$, denotes the
conjugating homeomorphism from $U_a$ to  $\overline{|\,y_1^0,y_2^0,...y_s^0\,b}= \varphi_{a,b} (U_a).$

Note also that for each $a$
\begin{equation} \label{sav2}  \alpha (\,\overline{|\,y_1^0,y_2^0,...y_s^0}\,)= \sum_{b=1}^d \alpha (\, \varphi_{a,b}\, (U_a)\,).
\end{equation}

Denote
$$ K = \sup_{m\in \mathbb{N}} \{ \sum_{k=0}^{m-1}    [\tilde{U} \tau^k (u) - \tilde{U} \tau^k (v)]\,\,,\, \text{where}\,\, u,v\in \overline{|\,a_1,a_2,...,a_m}\,\,,$$
$$ \text{and}\,\, (a_1,a_2,...,a_m)\in \{1,2,..,d\}^m\,\}.$$

Note that on the above expression we do not ask that $u \sim v$.

Note that if $\alpha$ is $C^*$-Gibbs and satisfies (\ref{jk11}) we get in particular that
$$
 \,d_1 \int_{\overline{|\,x_1^0,x_2^0,...x_s^0} } e^{- V( z, \varphi_{(x,y)}(z))}  d \alpha(z) \leq $$
\begin{equation} \label{jk19}  \int_{\overline{|\,y_1^0,y_2^0,...y_s^0}}  d \alpha(z) \leq d_2 \int_{\overline{|\,x_1^0,x_2^0,...x_s^0} } e^{- V( z, \varphi_{(x,y)}(z))} d \alpha(z).
\end{equation}

\medskip
\begin{proposition} \label{este1} Suppose $\alpha$ is quasi-$C^*$-Gibbs for $U$ as above. Then, there exists a constat $c_1>0$, such that,

 $$ c_1\leq  e^{ - \sum_{k=0}^{s-1}  \tilde{U} \tau^k (x)}\,\, \alpha(\,\overline{|\,x_1^0,x_2^0,...x_s^0}\,)$$
 for any cylinder $\overline{|\,x_1^0,x_2^0,...x_s^0}$ and any $x$ on the cylinder.

 The $\alpha$-probability of any cylinder is positive.

\end{proposition}

{\bf Proof:} We assume that (\ref{jk19}) is true.

 Assume first that $s>2r$.

From (\ref{jk2}) for a fixed $x \in \overline{|\,x_1^0,x_2^0,...x_s^0}$ and $y \in \overline{|\,y_1^0,y_2^0,...y_s^0}$,
where (as in  Lemma     \ref{rwt1})

$$x=(...x_{-m}x_{-m+1}...x_{-1}\,|\,  x_1^0\,...x_s^0\, {\bf x_{s+1}},  x_{s+2},...x_{m-1}x_{m} ...)\in \overline{|\,x_1^0,x_2^0,...x_s^0}  $$

$$y=(...x_{-m}x_{-m+1}...x_{-1}\,|\, y_1^0\,...y_s^0\,{\bf x_{s+1}}, x_{s+2},...x_{m-1}x_{m} ...) \in \overline{|\,y_1^0,y_2^0,...y_s^0},$$

we get from Lemma \ref{rwt1} that
$$\alpha (\, \varphi_{a,b}\, (U_a)\,) \leq d_2 \, e^{ 4\,K } e^{ \sum_{k=0}^{s-1} [\tilde{U} \tau^k (y_b) - \tilde{U} \tau^k (x_a)]}\,\, \alpha(U_a)\,\leq\, $$
$$d_2\, e^{ 4\,K + 2\,r\, V} e^{ \sum_{k=0}^{s-1} [\tilde{U} \tau^k (y) - \tilde{U} \tau^k (x)]}\,\, \alpha(U_a).$$

Then, from (\ref{sav2})

$$ \alpha ( \overline{|\,y_1^0,y_2^0,...y_s^0} )  =  \, \sum_{b=1}^d  \alpha (\, \varphi_{a,b}\, (U_a)\,) \leq d_2\, d\, e^{ 4\,K + 2 r V} e^{ \sum_{k=0}^{s-1} [\tilde{U} \tau^k (y) - \tilde{U} \tau^k (x)]}\,\, \alpha(U_a) .$$

From this and  from (\ref{sav0}) we get

$$1= \, \sum_{y_1^0,y_2^0,...y_s^0=1 }^d  \alpha ( \overline{|\,y_1^0,y_2^0,...y_s^0} )   \leq d_2\, d\, e^{ 4\,K + 2r V} e^{ - \sum_{k=0}^{s-1}  \tilde{U} \tau^k (x)}\,\, \alpha(U_a) ,$$
and, finally, for $x =(  ... ,x_{-t},..., x_{-2}, x_{-1}   \,|\, x_1, x_2,..,x_t,...) \in  \overline{|\,x_1^0,x_2^0,...x_s^0}$
$$ d = \, \sum_{a=1}^d \,\sum_{y_1^0,y_2^0,...y_s^0=1 }^d  \alpha ( \overline{|\,y_1^0,y_2^0,...y_s^0} )   \leq \sum_{a=1}^d  d_2\, d\, e^{ 4\,K + 2 r V} e^{ - \sum_{k=0}^{s-1}  \tilde{U} \tau^k (x)}\,\, \alpha(U_a) =$$
 $$ d_2\, d\, e^{ 4\,K + 2 r V} e^{ - \sum_{k=0}^{s-1}  \tilde{U} \tau^k (x)}\,\, \alpha(\overline{|\,x_1^0,x_2^0,...x_s^0}).   $$

This shows one of the inequalities in (\ref{defbowen23}) for the case $s>2r$.

As there exist just a finite number of distinct cylinders $\overline{|\,x_1^0,x_2^0,...x_s^0}$ of size smaller than $2 r$ we get the claim of the proposition.

This also shows that  the $\alpha$-probability of any cylinder $\overline{|\,x_1^0,x_2^0,...x_s^0}$ is positive when $\alpha$ is quasi-$C^*$-Gibbs.

By Proposition \ref{oup} we get that any cylinder of the form $\overline{x_{-m}...x_{-1} \vert x_1 x_2...x_s}$ has positive $\alpha$-probability.

\qed

\medskip
\begin{proposition} \label{este2} There exists a constat $c_2>0$, such that,

 $$  e^{ - \sum_{k=0}^{s-1}  \tilde{U} \tau^k (x)}\,\, \alpha(\,\overline{|\,x_1^0,x_2^0,...x_s^0}\,) \leq  c_2,$$
 for any cylinder $\overline{|\,x_1^0,x_2^0,...x_s^0}$ and any $x$ on the cylinder.

 The $\alpha$-probability of any cylinder is positive.

\end{proposition}

{\bf Proof:} We assume that (\ref{jk19}) is true.

On a similar way as before we assume first that $s>2r$.

Consider  fixed $x \in \overline{|\,x_1^0,x_2^0,...x_s^0}$ and $y \in \overline{|\,y_1^0,y_2^0,...y_s^0}$ as in  Lemma \ref{rwt1}:

$$x=(...x_{-m}x_{-m+1}...x_{-1}\,|\,  x_1^0\,...x_s^0\, {\bf x_{s+1}},  x_{s+2},...x_{m-1}x_{m} ...)\in \overline{|\,x_1^0,x_2^0,...x_s^0}  $$

$$y=(...x_{-m}x_{-m+1}...x_{-1}\,|\, y_1^0\,...y_s^0\,{\bf x_{s+1}}, x_{s+2},...x_{m-1}x_{m} ...) \in \overline{|\,y_1^0,y_2^0,...y_s^0},$$

Using an analogous reasoning as in proposition \ref{este2} we get that
$$\alpha(U_a)  \leq \frac{1}{d_1}  \,e^{ 4\,K } e^{ \sum_{k=0}^{s-1} [\tilde{U} \tau^k (x_a) - \tilde{U} \tau^k (y_b)]}\,\, \alpha( \varphi_{a,b}(U_a)\,)\leq\, $$
$$\frac{1}{d_1}\,e^{ 4\,K + 2\,r\, V} e^{ \sum_{k=0}^{s-1} [\tilde{U} \tau^k (x) - \tilde{U} \tau^k (y)]}\,\, \alpha (\, \varphi_{a,b}\, (U_a)\,) .$$

Therefore,

$$ e^{ \sum_{k=0}^{s-1}  \tilde{U} \tau^k (y)} \alpha(\overline{|\,x_1^0,x_2^0,...x_s^0})= e^{ \sum_{k=0}^{s-1}  \tilde{U} \tau^k (y)]} \sum_{a=1}^d \alpha(U_a) \leq $$
\begin{equation} \label{aci1} \frac{1}{d_1}\,e^{ 4\,K + 2\,r\, V} e^{ \sum_{k=0}^{s-1} \tilde{U} \tau^k (x) }\,\, \sum_{a=1}^d \alpha (\, \varphi_{a,b}\, (U_a)\,). \end{equation}

Finally, as $  \overline{|\,y_1^0,y_2^0,...y_s^0\,b}= \varphi_{a,b} (U_a)$ we get  from (\ref{sav0}) and  (\ref{aci1})
$$ d\, \alpha(\overline{|\,x_1^0,x_2^0,...x_s^0})= \sum_{b=1}^d \sum_{y_1^0,y_2^0,...y_s^0=1 }^d  e^{ \sum_{k=0}^{s-1}  \tilde{U} \tau^k (y)} \alpha(\overline{|\,x_1^0,x_2^0,...x_s^0}) \leq $$
$$  \frac{e^{ 4K + 2r V}}{d_1} e^{ \sum_{k=0}^{s-1} \tilde{U} \tau^k (x) }\, \sum_{a=1}^d \,\sum_{b=1}^d \sum_{y_1^0,y_2^0,...y_s^0=1 }^d \alpha (\varphi_{a,b}\, (U_a))= \frac{d\, e^{4K + 2r V}}{d_1}\, e^{ \sum_{k=0}^{s-1} \tilde{U} \tau^k (x) }. $$

This shows the claim of the proposition.

\qed

\medskip
Now we have to show that $\alpha$ is invariant by $\tau$.

\begin{corollary} \label{rwt15}
 If $\alpha_1$ and $\alpha_2$ are  quasi $C^*$-Gibbs for $U$,  where
 $$ U (  ... ,x_{-n} ,... ,x_{-2} ,x_{-1} \, | \,x_1,x_2,...,x_r, x_{r+1},...  x_{m} ...   ) = f(x_1,x_2,...,x_r)$$
 for some fixed $r$ and function $f:\{1,2,...,d\}^r\to \mathbb{R}$,
 then $\alpha_1$ is absolutely continuous with respect to $\alpha_2$.

\end{corollary}

{\bf Proof:}
We assume that $U$ is normalized. Suppose $\alpha_1$ and $\alpha_2$ are  quasi $C^*$-Gibbs for $U$.

Expression (\ref{defbowen23}) for $\alpha_1$ and $\alpha_2$  will determine, respectively, constants $d_1^1,d_2^1$ and
$d_1^2,d_2^2$.

From last Propositions there exist constants $Y_1>0$ and $Y_2>0$, such that,  given for any cylinder $\overline{|\, x_1,x_2,..,x_n}$ we get
$$    \frac{\alpha_1 (\overline{|\, x_1,x_2,..,x_n})}{e^{ \sum_{k=0}^{n-1} U(\tau^k (... ,z_{-1}^0\,|\,x_1,...,x_n, z_{n+1}^0,..  ))}  }      \leq Y_1   ,$$
and
$$   Y_2 \leq  \frac{\alpha_2 (\overline{|\, x_1,x_2,..,x_n})}{e^{ \sum_{k=0}^{n-1} U(\tau^k (... ,z_{-1}^0\,|\,x_1,...,x_n, z_{n+1}^0,..  ))}  }        ,$$
for some $(... ,z_{-1}^0\,|\,x_1,...,x_n, z_{n+1}^0,..  )$.

Therefore,
$$ \frac{Y_2}{Y_1} \alpha_1 (\overline{|\, x_1,x_2,..,x_n})\leq  \alpha_2 (\overline{|\, x_1,x_2,..,x_n}).$$

Now consider a cylinder set of the form
$$(\overline{  x_{-m},...x_{-1}|\, x_1,x_2,..,x_n}).$$

Expression (\ref{tutu}) for $\alpha_1$ and $\alpha_2$ will determine, respectively, constants $b_1^1,b_2^1$ and $b_1^2,b_2^2$.

Then, by  Proposition \ref{oup} we get that

$$b_1^1 \,\alpha_1 (\overline{  x_{-m},...x_{-1}|\, x_1,x_2,..,x_n})\leq
\alpha_1 (\overline{|\,  x_{-m},...x_{-1} x_1,x_2,..,x_n})\leq$$
 \begin{equation} \label{uaua} \frac{Y_1}{Y_2} \alpha_2 (\overline{|\,  x_{-m},...x_{-1} x_1,x_2,..,x_n})\leq \frac{Y_1}{Y_2} b^2_2 \alpha_2 (\overline{  x_{-m},...x_{-1}|\, x_1,x_2,..,x_n}).\end{equation}

The Borel sigma-algebra over $\Omega$ is generated by the set of cylinders of the form $\overline{  x_{-m},...x_{-1}|\, x_1,x_2,..,x_n}$.

As the probability $\alpha_j(B)$, $j=1,2$, of a Borel set $B$ is obtained, respectively, as an exterior probability using probabilities of the generators we finally get that the analogous inequalities as in (\ref{uaua}) are true with the same same constants, that is,
\begin{equation} \label{uaua1}    b_1^1 \alpha_1(B) \,\,\,\leq \,\,\,\frac{Y_1}{Y_2} b^2_2 \alpha_2 (B ).\end{equation}

Therefore, $\alpha_1$ is absolutely continuous with respect to $\alpha_2$.

\qed

\begin{proposition} \label{outr} Assume $\alpha$ is $C^*$-Gibbs for $U$, then,
 $\alpha$ is invariant for $\tau$.

\end{proposition}

{\bf Proof:} From Corollary \ref{rwt15}  we get that any two $C^*$-Gibbs probabilities for $U$ are absolutely continuous with respect to each other.

Suppose $\alpha$ is $C^*$-Gibbs, then, $\alpha_1=\tau^* (\alpha)$ is also $C^*$-Gibbs by Proposition \ref{kttt}. If $\alpha \neq  \tau^* (\alpha)$
then, following Theorem 2.5 in \cite{HD} we get that $\rho_1 = |\alpha_1-\alpha | + \alpha_1-\alpha$ and $\rho_2 = |\alpha_1-\alpha | - \alpha_1 +\alpha$ are also  $C^*$-Gibbs. But $\rho_1$ and $\rho_2$ are singular with respect to each other and this is a contradiction.

¨Therefore, $\alpha=\tau^* (\alpha)$.

\qed

\begin{theorem} \label{rwt3}

 Suppose $U:\Omega \to \mathbb{R}$ is of the form
 $$ U (  ... ,x_{-n} ,... ,x_{-2} ,x_{-1} \, | \,x_1,x_2,...,x_r, x_{r+1},...  x_{m} ...   ) = f(x_1,x_2,...,x_r),$$
 for some fixed $r$ and fixed function $f:\{1,2,...,d\}^r\to \mathbb{R}$.

 If $\alpha$ is  $C^*$-Gibbs for the potential $U$ then $\alpha$ is the equilibrium state for $U$.

\end{theorem}

{\bf Proof:} As we know by Proposition \ref{outr} that $\alpha$ is $\tau$ invariant and, moreover, we also know that $\alpha$ is quasi-$C^*$ invariant for another normalized potential, it follows from  Proposition \ref{este1},  Proposition \ref{este2} and Theorem \ref{Bowen1.2} that $\alpha$ is the equilibrium probability for $U$

\qed

Another conclusion one can get  from the above reasoning is that the concepts of quasi $C^*$-Gibbs and $C^*$-Gibbs are equivalent on the lattice $\mathbb{Z}.$

\bigskip

\section{Construction of the $C^{*}$-Algebra} \label{C}

Remember that we consider the groupoid $G\subset \Omega \times \Omega$ of all pair of points which are related by the homoclinic equivalence relation.

We consider on $G$ the topology generated by sets of the form
$$ \{\,\, (z, \varphi_{(x,y)}(z))\,\,\text{where}\,\,z \in \mathcal{O}_{(x,y)}\,\,\}.$$

This topology is Hausdorff.

We denote by $[x]$ the class of $x\in \Omega$. For each $x$ the set of elements on the class $[x]$  is countable.

We now come to the construction of the noncommutative algebra. Let $\mathcal{C}_{c} (G)$ be the linear space of complex continuous functions with compact support on $G$. If $A, B \in \mathcal{C}_{c} (G)$ we define the product $A * B$ by
$$
(A*B)(x,y) = \sum_{z \in [x]} A(x,z)B(z,y).
$$
Note that if $(x,y) \in G$ then they are conjugated and so the sum is over all $z$ that are conjugated to $x$ and $y$.

Note that
there are only finitely many nonzero terms in the above sum.
because the functions $A,B$ have compact support.

Considering the above, $A*B \in \mathcal{C}_{c}(G)$ as one checks readily, so that $\mathcal{C}_{c}(G)$ becomes an associative complex algebra. An \textbf{involution} $A \to A^{*}$ is defined by
$$
A^{*}(x,y) = \overline{A(y,x)}
$$
where the bar denotes complex conjugation.

For each equivalence class $[x]$ of conjugated points of $\Omega$ there is a representation $\pi_{[x]} \to \mathbb{C}$ in the Hilbert space $l^{2} ([x])$ of square summable functions $[x] \to \mathbb{C}$, such that
$$
((\pi_{[x]}A) \xi (y) = \sum_{z \in [x]} A(y,z) \xi (z)
$$
for $\xi \in l^{2}([x])$. Denoting by $\| \pi_{[x]} A \|$ the operator norm, we write
\begin{equation} \label{supremumnorm}
\| A \| = \sup_{[x]} \| \pi_{[x]} A \|.
\end{equation}

$I_{D}$ (the indicator function of the diagonal $D$) is such that for any $A\in C^{*}(G)$ we get $I_D * A = A * I_D=A$.

The completion of $\mathcal{C}_{c} (G)$ with respect to this norm is
 separable. It is called the    reduced $C^{*}$-algebra which is   denoted by   $C^{*}_r(G)$. The unity element $I_{D}$ is contained in this $C^{*}$ algebra.

\begin{remark} \label{continouscstar}

If $A \in \mathcal{C}_{c} (G)$ and $t \in \mathbb{R}$, we write
\begin{equation} \label{invariantstate}
(\sigma^{t} A)(x,y) = e^{i V(x,y) t} A(x,y)
\end{equation}
defining a one-parameter group $(\sigma^{t})$ of $*$-automorphisms of $\mathcal{C}_{c}(G)$ and a unique extension to a one parameter group of $*$-automorphisms of $C^{*}_r(G)$.

We say that $A \in \mathcal{C}_{c} (G)$ is analytic (a classical terminology on $C^*$-algebras) if the real variable $t$ on the function $t \to \sigma^t A$
can be extended to the complex variable $z\in \mathbb{C}.$ Under our assumptions this will be always the case. Therefore, $\sigma^{-\beta i} A$ is well defined.
\end{remark}

\begin{definition}

A state $\omega$ on $C^{*}_r(G)$ is a linear functional $\omega: C^{*}_r(G) \to \mathbb{C}$, such that, $\omega(A*A^{*}) \geq 0$, and $\omega (I_D) = 1$ (see \cite{Bra}).
\end{definition}

Such state $\omega$ is sometimes called a dynamical $C^*$-state.

\begin{definition}
 A state $\omega$ is invariant if $\omega \circ \sigma^{t} = \omega$, for all $t\in \mathbb{R}$.
 \end{definition}

 \begin{definition} Given a modular function $V: G \to \mathbb{R}$ as above and the associated  $\sigma_t$, $t \in \mathbb{R},$ we say that
 an invariant state $\omega:C^{*}_r(G)\to \mathbb{C}$ satisfies the \textbf{KMS boundary condition} for $V$ and $\beta\in \mathbb{R}$,  if for all $A, B \in C^{*}_r(G)$, there is a continuous function $F$ on $\{ z \in \mathbb{C} : 0 \leq \text{ Im} (z) \leq 1 \}$, holomorphic in $\{ z \in \mathbb{C} : 0 < \text{ Im} (z) < 1 \}$, and such that
\begin{equation}\label{KMS}
\omega(\sigma^{t} A * B) = F(t), \quad \omega(B * \sigma^{t} A) = F(t+i \beta)
\end{equation}
\end{definition}

\qed

Note that using \eqref{KMS} we have that $F(0) = \omega(A \cdot B)$ and
$$
F(0) =
F(-\beta i + \beta i) =
\omega(B * \sigma^{-\beta i} A).
$$
Therefore, for any $A,B$ we get
$$\omega(A\,*\,B) = \omega(B *  e^{-i \beta} A)$$
which is the classical KMS condition for $\omega$ according to \cite{Bra}. This condition is equivalent to \textbf{KMS boundary condition}.

\begin{theorem}

If $\mu$ is a probability measure on $\Omega$ then a state $w=\hat{\mu}$ on $C^{*}_r (G)$ can be defined by

\begin{equation}
\hat{\mu}(A) = \int A(x,x) d \mu(x)
\label{defstate}
\end{equation}

\end{theorem}

{\bf Proof:}  First note that it's easy to verify that for any $A$ in the $C^{*}_r$-algebra $\hat{\mu}$ is linear, $\hat{\mu}(A*A^{*}) \geq 0$ and $\hat{\mu}(I_{D}) = 1$. Now, note that since the diagonal $D$ is a compact set, then any continuous function $A : G \to \mathbb{C}$ has a maximum at $D$, therefore \eqref{defstate} is well defined for continuous function. $\hat{\mu}$ is also well defined on the $C^{*}_r$-algebra.

\qed

\bigskip As mentioned in \cite{Reullenoncommutative} the associated $C^*$ dynamical state  $\hat{\rho_0}$ for the probability $\rho_0$   (the case where $U=0$)  of lemma  \ref{7.17Ruelle} is a trace on the $C^*$-algebra $C^*(G)$.

\bigskip

\begin{definition} A probability $\nu $ on $\Omega$  is called a {\bf KMS probability} for the modular function $V$ if the state
$\hat{\nu}$ on $C^{*}_r (G)$  defined by
\begin{equation}
\hat{\nu}(A) = \int A(x,x) \nu(dx)
\label{defstate1}
\end{equation}
satisfies the KMS condition for $V$.  Here $G$ is the groupoid given by the homoclinic equivalence relation.

\end{definition}

This probability is sometimes called quasi-stationary (see \cite{LM1}).

The next claim was proved on \cite{Reullenoncommutative}. For completeness we will present a proof of this claim with full details.

\begin{theorem} If the probability  $\alpha$ on $\Omega$  is a $C^{*}$-Gibbs probability with respect to $V$ and $\beta$, then,
$\alpha$  is a KMS probability for the modular function $\beta \,V$. That is, the associated  $\hat{\alpha}$ is a $C^*$ dynamical state for the $C^*_r (G)$ algebra given by  the groupoid obtained  by the homoclinic equivalence relation.

\end{theorem}

{\bf Proof:}
Suppose $\alpha$ is a $C^{*}$-Gibbs state with respect to $\beta V$. We assume $\beta=1$

 $\hat{\alpha}$ is $\sigma^{t}$ invariant if for all $t \in \mathbb{C}$ it's true that
$$
\int \sigma^{t} A(x,x) \alpha (dx) =
\int A(x,x) \alpha (dx)
$$
which by definition \eqref{invariantstate} it's equivalent to
$$
\int e^{iV(x,x)t} A(x,x) \alpha (dx) =
\int A(x,x) \alpha (dx)
$$
but since $V(x,x) = 0$ then the state have to be $\sigma^{t}$ invariant.

Now we will show that if $A, B \in \mathcal{C}_{c}(G)$, then
$$
\hat{\alpha}(\sigma^{t} A * B) =
\int \alpha(dx) \sum_{y \in [x]} e^{iV(x,y)t} A(x,y) \cdot B(y,x)
$$
extends to an entire function. For this purpose we will pick $t_{0} \in \mathbb{C}$ and show that
\begin{equation} \label{basiclimit}
\lim_{t \to t_{0}} \dfrac{\hat{\alpha}(\sigma^{t} A * B) - \hat{\alpha}(\sigma^{t_{o}} A * B)}{t - t_{0}}
\end{equation}
exist. Indeed, the limit \eqref{basiclimit} is equivalent to
$$
\lim_{t \to t_{0}} \dfrac{1}{t - t_{0}} \left( \int \alpha(dx) \sum_{y \in [x]} e^{iV(x,y)t} A(x,y) \cdot B(y,x) - \right.
$$

$$
\left. \int \alpha(ds) \sum_{y \in [s]} e^{iV(s,y)t_{0}} A(s,y) \cdot B(y,s) \right) =
$$
\begin{equation} \label{tempholo}
\lim_{t \to t_{0}} \left( \int \alpha(dx) \sum_{y \in [x]} \dfrac{(e^{iV(x,y)t } - e^{iV(x,y) t_{0}} )}{t - t_{0}} A(x,y) \cdot B(y,x) \right) .
\end{equation}
Please always have in mind that for each $x$ the summation is over finite terms.

Now in order to use the dominated convergence theorem we shall work as if we were inside the integral. Let $R$ be a closed ball of radius $1$ centered in $t_{0}$. So we can consider the continuous function $f_{t_{0}} : R \backslash\{t_{0}\} \times $ supp$(A) \to \mathbb{C}$
$$
f_{t_{0}} (t,x) = \sum_{y \in [x]} \dfrac{(e^{iV(x,y)t } - e^{iV(x,y) t_{0}} )}{t - t_{0}} A(x,y) \cdot B(y,x)
$$
To extend $f_{t_{0}}$ for the case $t = t_{0}$ we need to solve the limit
\begin{equation} \label{definitionofL}
L_{t_{0}}(x) =
\lim_{t \to t_{0}} \sum_{y \in [x]} \dfrac{(e^{iV(x,y)t } - e^{iV(x,y) t_{0}} )}{t - t_{0}} A(x,y) \cdot B(y,x) =
\end{equation}
$$
\sum_{y \in [x]} \lim_{t \to t_{0}}  \dfrac{(e^{iV(x,y)t } - e^{iV(x,y) t_{0}} )}{t - t_{0}} A(x,y) \cdot B(y,x) =
$$
$$
\sum_{y \in [x]} iV(x,y) e^{iV(x,y) t_{0}} A(x,y) \cdot B(y,x).
$$
So define $f_{t_{0}} (t_{0},x) = L_{t_0}(x)$.

In this way $f_{t_{0}}$ is a continuous function defined on a compact domain. Therefore we may assume that both it's real and imaginary parts are limited by a value $M$ in the domain. Since the limit in \eqref{definitionofL} exist, then we may choose a sequence of function indexed by the $t$ variable, $\{f_{t_0}(t_{n}, x)\}_{n \in \mathbb{N^{*}}}$ that converge to $L_{t_0}(x)$ when $n \to \infty$,  e.g. $f_{t_{0}} (t_{0} + (1+i)/n, x)$. In this way the dominated convergence theorem assures that the limit \eqref{basiclimit} is equal to the integral:
$$
\int \alpha(dx) L_{t_{0}}(x).
$$
Indeed formally what we have is,
$$
\int \alpha(dx) L_{t_{0}}(x) =
\int \alpha(dx) \sum_{y \in [x]} iV(x,y) e^{iV(x,y) t_{0}} A(x,y) \cdot B(y,x) =
$$
$$
\int \alpha(dx) \lim_{n \to \infty} \sum_{y \in [x]} \dfrac{(e^{iV(x,y) t_{n}} - e^{iV(x,y) t_{0}} )}{t_{n} - t_{0}} A(x,y) \cdot B(y,x) =
$$
$$
\lim_{n \to \infty} \int \alpha(dx) \sum_{y \in [x]} \dfrac{(e^{iV(x,y) t_{n}} - e^{iV(x,y) t_{0}} )}{t_{n} - t_{0}} A(x,y) \cdot B(y,x) =
$$
\begin{equation} \label{sequencef}
\lim_{n \to \infty} \dfrac{\hat{\alpha}(\sigma^{t_{n}} A * B) - \hat{\alpha}(\sigma^{t_{o}} A * B)}{t_{n} - t_{0}}
\end{equation}
Now since the choosing of the sequence was arbitrary we could remake these calculations to any desired convergent sequence with the same result, therefore \eqref{sequencef} is equal to
$$
\lim_{t \to t_{0}} \dfrac{\hat{\alpha}(\sigma^{t} A * B) - \hat{\alpha}(\sigma^{t_{o}} A * B)}{t - t_{0}},
$$
what proves existence of the limit in equation \eqref{basiclimit}. This allow us to conclude that $\hat{\alpha}(\sigma^{t} A * B)$ is an holomorphic function everywhere.

Let $F(t) = \hat{\alpha}(\sigma^{t} A * B)$. Using a partition of unity on supp $A$ we may write $A = \sum A_{j}$, where supp $A_{j} \subset W_{j} = \{ (z, \varphi_{j}(z)): z \in \mathcal{O}_{j}\}$, and $(\mathcal{O}_{j}, \varphi_{j})$ is a conjugating homeomorphism.  Since supp $A$ is a compact set then we may assume the summation to occur over a finite amount of elements. Thus

$$
F(t) =\int_{\Omega} \sum_{j} \alpha(dx) A_{j} (x, \varphi_{j} x) B(\varphi_{j} x, x) \exp (i V(x, \varphi_{j} x)t) =
$$
$$
\sum_{j} \int_{\mathcal{O}_{j}} \alpha(dx) A_{j} (x, \varphi_{j} x) B(\varphi_{j} x, x) \exp (i V(x, \varphi_{j} x)t)
$$
and therefore
$$
F(t+ \beta i) =
\sum_{j} \int_{\mathcal{O}_{j}} [ e^{-\beta V(x,\varphi_{j} x)} \alpha (dx) ]
A_{j} (x, \varphi_{j} x) B (\varphi_{j} x, x)  \exp(i V(x, \varphi_{j} x) t)
$$
If $\alpha$ is an $C^{*}$-Gibbs state by \eqref{gibbsR} we have that
$$
F(t+\beta i) =
\sum_{j} \int_{\varphi_{j}(\mathcal{O}_{j})} \alpha (dy) B(y, \varphi^{-1}_{j} y) A_{j} (\varphi^{-1}_{j} y,y) \exp (iV(\varphi^{-1}_{j}y,y)t) =
$$
$$
\sum_{j} \int_{\varphi_{j}(\mathcal{O}_{j})} \alpha (dy) B(y, \varphi^{-1}_{j} y) \sigma^{t} A_{j} (\varphi^{-1}_{j}y,y) =
$$
$$
\int_{\Omega} \sum_{j} \alpha (dy) B(y, \varphi^{-1}_{j} y) \sigma^{t} A_{j} (\varphi^{-1}_{j}y,y) =
$$
$$
\int_{\Omega} \alpha (dy) (B * \sigma^{t} A) (y,y) =
\hat{\alpha} (B*\sigma_{t} A)
$$
so that $\hat{\alpha}$ satisfies the KMS condition.

\qed

redpajama_set_name: RedPajamaArXiv


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text:
Italy uncovers massive load of hash in ship's fuel tanks
This content was published on August 9, 2018 11:20 AM Aug 9, 2018 - 11:20
ROME (Reuters) - Italian police said on Thursday they found 20 tonnes of hash worth as much as 200 million euros (£180 million) in the fuel tanks of a Panama-flagged ship that was stopped in international waters and escorted to Sicily.
The entire 11-person crew, all from Montenegro, was arrested for international drugs trafficking, Italy's finance police said in a statement.
The research and survey vessel Remus left the Canary Islands bound for Egypt and Turkey, but police surveillance showed it had turned off its position transmitter near the coast of North Africa, raising investigators' suspicions.
With the permission of Panamanian authorities, Italian finance police seized the vessel in international waters on July 31 and escorted it to Palermo. There the fuel tanks were drained and the hash was discovered wrapped in sealed brown bags.
"The Mediterranean Sea again is confirmed to be one of the world's largest arteries for illegal trafficking," the police statement said. "The results of the operation are the fruit of attentive intelligence gathering and analysis of the ship's routes."
The North African hashish -- compressed cannabis resin -- was probably destined for the European market, police said. Morocco is the world's largest producer of hashish, which is usually smuggled into Europe through Spain, according the United Nations Office on Drugs and Crime.
(Reporting by Steve Scherer, editing by Larry King)