text:
From poo to food: Kenyan toilet waste key for new animal feed
NAIROBI (Reuters) - Kenyan farmer Victor Kyalo's chickens have doubled the number of eggs they are laying. The reason: Human excrement.
He is feeding them food from a Nairobi-based organics recycling company. Sanergy harvests waste from toilets it operates in a franchise network in Nairobi's sprawling slums and feeds it to fly larvae, which become high-quality animal feed.
Kyalo says his customers have noticed the difference in the past three weeks: yellower yolks and larger eggs.
"Before we were getting like five trays (of eggs) per day, but now we are getting 10," Kyalo said. "It's kind of perfect for me."
As the world looks to feed 10 billion mouths by 2050, businesses harvesting insects — either for human consumption or as animal feed — are growing. They promote themselves as a greener alternative to traditional feed such as soybeans, whose cultivation can lead to deforestation and the overuse of farm chemicals.
Fast food giant McDonald's and U.S. agricultural powerhouse Cargill Inc are among many large companies studying using insects for chicken feed to reduce reliance on soy protein in the $400 billion-a-year animal feed business.
By 2023 the global edible insect market could triple to $1.2 billion from current levels, market research firm Meticulous Research said last year.
In developing countries like Kenya, where the World Bank says nearly two-thirds of urbanites live in slums, feeding waste to fly larvae could solve both sanitation and nutrition problems.
Faeces from more than two-thirds of Nairobi's inhabitants go untreated because there are not enough toilets. Many others are not cleaned out regularly, Nairobi City Water and Sewerage Company said.
During the rains, they often overflow, polluting local waterways. That can make workers ill. Days off slow Kenya's economy by around 1% annually, its Health Ministry said.
FILE PHOTO: Black soldier flies are seen breeding inside netted enclosures at the Sanergy organics recycling facility near Nairobi, Kenya, March 10, 2019. Picture taken March 10, 2019. REUTERS/Baz RatnerFROM LOO TO FOOD
David Auerbach co-founded Sanergy eight years ago to deal with sanitation. The waste management franchise provides more than 2,500 toilets to 100,000 people daily.
Lilian Mbusia runs one of Sanergy's franchises, charging residents of Mukuru Kwa Ruben slum in the south of the city 5 Kenyan shillings (5 U.S. cents) to use her blue "Fresh Life" toilets.
Nestled beneath her squat-toilets are small blue barrels that, once full, are sealed and taken to an organics recycling factory in Machakos County, a bumpy 40-minute drive outside the city.
Beds of writhing black soldier fly larvae feast on a mix of excrement and food waste from hotels and agri-businesses.
That produces two products for farmers: fertilizer and animal feed.
In 10 days the larvae munch their way through 70% of the waste, leaving behind a manure laden with nitrogen and calcium, which becomes organic fertilizer.
Once the recycling plant is expanded later this year, Auerbach said it will provide 400 tonnes of fertilizer. Larvae production will ratchet up from 7 tonnes to 300 tonnes per month.
"Right now we are receiving equity debt, and grant investment to scale up operations," Auerbach said. "We're on track for profitability by the end of 2020."
The plump white larvae are boiled in hot water to kill off pathogens, Michael Lwoyelo, managing director of Sanergy, said.
The larvae are then sold to animal feed millers, who grind them into powder mixed with other ingredients to create a balanced diet for poultry, pigs and fish.
Frederick Wangombe, an animal nutritionist at Unifeed, a Kenyan animal feed miller that uses Sanergy's black soldier fly product, envisages it replacing fish meal from Lake Victoria, which can contain sand and other impurities, or expensive soy beans from Zambia.
Slideshow (18 Images)
"The egg farmer doesn't want to know what's in the feed, they want to know the performance," he said.
($1 = 101.1000 Kenyan shillings)
Reporting by Hereward Holland; Editing by Maggie Fick; Katharine Houreld and Alison Williams
Our Standards:The Thomson Reuters Trust Principles.Read More
http://feeds.reuters.com/~r/reuters/entertainment/~3/xcQGURI9rmQ/from-poo-to-food-kenyan-toilet-waste-key-for-new-animal-feed-idUSKCN1TI1BN
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redpajama_set_name: RedPajamaGithub


Beautiful and unique home! Homes in this close-knit neighborhood of Wyandotte do not come up for sale often. Walk to downtown Wyandotte in minutes and enjoy everything it has to offer. This home has plenty of charm including a tiled porch & sunroom, 4 level architectural deck, private garden with lights, bamboo & perennials, custom driveway with pendants, custom natural cedar siding, original stained glass, woodwork, doors, skylights & garden window.


text:
Невидљиви човек () је америчко-аустралијски научнофантастични психолошки хорор филм из 2020. године који је написао и режирао Леј Вонел. Прати жену која верује да је прогања насилни и богати дечко чак и након његовог наизгледног самоубиства. Она на крају закључује да је он стекао способност да постане невидљив. Главне улоге тумаче Елизабет Мос, Елдис Хоџ, Сторм Рид, Харијет Дајер, Мајкл Дорман и Оливер Џексон-Коен. Представља међународну копродукцију Сједињених Држава и Аустралије. Због пандемије ковида 19 филм је био доступан за дигитално рентирање три недеље након америчке биоскопске премијере, а завршио је на петом месту по заради у САД за 2020. годину.

Радња

Сесилија Кес, архитекткиња из Сан Франциска, заробљена је у насилној вези са имућним инжењером оптике и бизнисменом Адријаном Грифином, који је третира као своју имовину. Једне ноћи Сесилија надрогира Адријана својим диазепамом и побегне из његове добро обезбеђене куће уз асистенцију своје млађе сестре, Емили.

Сесилија се сакрије у кући свог пријатеља из детињства, детектива Џејмса Ланијера и његове ћерке тинејџерке, Сидни. Две недеље након Сесилијиног бекства, Адријан наизглед изврши самоубиство и остави јој пет милиона долара. Његов брат, адвокат Том Грифин, одради папирологију. Сесилија сумња на присуство још неког у кући након неколико чудних догађаја, али Џејмс је уверава да је она само истраумирана и параноична. Током интервјуа за посао, она примети да је садржај њеног пословног портфолија уклоњен, а потом се онесвести. Лекар јој саопшти да су јој у крвотоку пронађени високи нивои диазепама, што је био узрок њене несвестице. У свом купатилу она нађе ону исту окрвављену бочицу диазепама којим је надрогирала Адријана током свог бекства.

Сесилија, у друштву Џејмса, састане се са Томом. Она верује да је Адријан лажирао своју смрт и употребио своју оптичку експертизу да постане невидљив како би је малтретирао, али је та замисао одбачена. Касније Сидни удари нека невидљива сила док ова теши Сесилију; претпостављајући да је то учинила Сесилија и да постаје неуравнотежена, Сидни и Џејмс оду. Оставши сама у кући, Сесилија испроба разне тактике да ухвати фигуру за коју верује да је ударила Сидни. Она пронађе Адријанов мобилни телефон у поткровљу, на који одмах прими СМС поруку "изненађење". Сесилија баци фарбу низ таванске мердевине и она се размаже по претходно невидљивој фигури. Наступи насилни окршај, али она побегне. Она оде до Адријанове куће да прегледа његову лабораторију, где пронађе још једно одело за невидљивост, чиме су њене сумње потврђене. Одмах након што она сакрије одело у њихову бившу спаваћу собу, невидљива фигура је изнова нападне, те она побегне и назове Емили. Њих две се састану у ресорану, где невидљива фигура прекоље Емили и стави нож у Сесилијину руку, смештајући јој убиство.

Док чека суђење, Сесилија је притворена у психијатријској болници, где сазна да је трудна. Том јој понуди да јој издејствује ослобађање од оптужби уколико она пристане "да му се врати" и одгаја дете, имплицирајући да је Том помогао свом брату да инсценира самоубиство. Он открије да јој је Адријан заменио таблете за контрацепцију диазепамом да би затруднела. Сесилија одбије понуду и украде му пенкало. Те ноћи она кобајаги покуша самоубиство да би измамила невидљиву фигуру. Када фигура покуша да је заустави, она га убоде више пута пенкалом, услед чега се камуфлажно одело поквари. Обезбеђење стигне на лице места, али их фигура онеспособи пре него што побегне из болнице, уз Сесилију за петама. Штитећи њено нерођено дете, фигура запрети да ће, уместо њу, напасти оне које она воли.

Сесилија одјури до Џејмсове куће и затекне фигуру у нападу на њега и Сидни. Она га попрска пеном из апарата за гашење пожара да га учини видљивим, а потом га упуца на смрт, али у оделу затекне Тома. Полиција упадне у Адријанову кућу и нађе га живог, везаног у подруму, тврдећи да га је Том држао као затвореника. Они закључе да је Том убио и Емили, али Сесилија верује да је Адријан подметнуо свог брата као одговорног за сопствена недела.

Да би искамчила Адријаново признање, сада ослобођена Сесилија сретне се с њим у тајности озвучена, док Џејмс слуша разговор неколико блокова даље. Она пристане да му се врати, али само ако он призна убиство Емили. Адријан инсистира да је Том одговоран, али устврди да му је то искуство променило поглед на живот и начин на који је поступао према њој. Након што Адријан изговори реч "изненађење", Сесилија оде у купатило "да се среди". Неколико тренутака касније камера видео-надзора сними како Адријан наизглед самог себе прекоље. Сесилија се врати и позове 911, наизглед потресена. Ван видног угла камере, међутим, она буљи хладнокрвно у умирућег Адријана и зачикује га, из чега се да закључити да је она употребила резервно камуфлажно одело да га убије.

Када Џејмс стигне, Сесилија потврди оно што је ухватила камера. Он примети да она износи одело, али јој дозволи да оде.

Референце

Спољашње везе

Филмови 2020.
Амерички филмови
Аустралијски филмови
Филмови на енглеском језику
Научнофантастични филмови
Амерички хорор филмови
Филмови студија Universal Pictures
Филмови Taramount Film-а
Научнофантастични хорор филмови
Римејкови хорор филмова
Хорор филмови 2020-их

redpajama_set_name: RedPajamaWikipedia


One of the biggest investments a person can make in themselves is to invest in a coach. To help them dial up, progress or breakthrough to the next level and phase or stage of their lives. But it doesn't come cheap and why would it. Why would something that enables you to unlock your potential be cheap? If it did would you take it as seriously, would you be as committed, would you want whatever goal, ambition or dream as much? Probably not, and that's why it can't be free. If a person has "no skin in the game" so to speak they are not as likely to be committed, take their time or the coaches time as seriously, be more prone to cancelling, rescheduling or going months between sessions making it hard for coach or coachee to gain any real traction.
I never used to get why coaching was so extraordinarily expensive, but now standing on the other side as a coach I of course totally get it. It is so much more than just about being paid for your time and your worth. It's about being able to seek out those individuals that are ready to show up and do the work. That are ready to put their money where their mouth is so to speak.
An individual that has chosen to make the investment in themselves comes to the conversation with a totally different mindset and from a totally different place. Hungry and eager to get going but most importantly ready to do the work. That is very different from the mindset of someone who approaches coaching from a place of uncertainty, trepidation or expense.
Do you think that individual is going to get the most from coaching if at the back of their mind all they can think about is how much money they're spending? No. So my ask of you is that if you are going to invest in a coach do it wholeheartedly. Do it because you're ready to show up and do the work. Do it because you buy into the value that a coach brings to the table and because you understand the value that comes from unlocking your untapped potential. Do it because you believe you are worth the investment, because you are buying into yourself as much as you are buying into a coach.


Step up your keychain game with this stunner. This leather lightning bolt in silver leather will have all eyes on you. Use it for your keys or use it as a charm for your favorite handbag. This is a way to accessorize your accessories.


text:
Q: shell script blocking process I wrote a one small shell script(demo_script) and the content are following
exec > /tmp/log 2>&1
   sudo apt-get install -y telnet
i used "bash demo_script &" to run this script as daemon. But it doesn't install software and keep on running in background.
After captured process details used "ps -ef"
root     28882 28881  0 13:35 pts/0    00:00:00 sudo apt-get install -y telnet
root     28883 28882  1 13:35 pts/0    00:00:00 apt-get install -y telnet
root     28901 28883  0 13:35 pts/0    00:00:00 apt-get install -y telnet
i don't know why its created like this.
and log message also
Reading package lists...
Building dependency tree...
Reading state information...
The following NEW packages will be installed:
  telnet
0 upgraded, 1 newly installed, 0 to remove and 71 not upgraded.
Need to get 0 B/67.1 kB of archives.
After this operation, 167 kB of additional disk space will be used.

A: It's waiting for user input.  use "--yes" or something:  http://linux.die.net/man/8/apt-get

redpajama_set_name: RedPajamaStackExchange


text:
namespace But
{
namespace Log
{
namespace Destination
{
namespace Common
{

class VerboseTextStream: public TextStream
{
public:
  struct Depth
  {
    size_t value_;
  };

protected:
  VerboseTextStream(std::ostream& os, Depth depth): TextStream{os}, depth_{depth} { }

private:
  void toStreamFormat(std::ostream& os, Backend::FieldInfo const& fi) override
  {
    detail::VerboseStreamAndTrimVisitor satv{&trim(), &os, depth_.value_};
    fi.visit(satv);
    os << endline();
  }

  const Depth depth_;
};

}
}
}
}


We have a F550 that currently has super singles but the tires that were installed don't meet the GVW. We have acquired new tires that meet the GVW but are narrower than the original tires - so we need new custom rims built. Does anyone have any recommendations on companies that build custom rims for super singles?


text:
Sex Work Made Me Value My Body More
How Sex Work Made Me Value My Body More
Lily Yin
Illustrated by: Anna Sudit
Imagine you're in a dungeon dressed in thigh-high stockings clipped into lace garters. You're wearing a tiny black thong under a black leather corset that's pulled so tight you can barely breathe — but it gives you the perfect hourglass shape. When you catch a glimpse of yourself in the mirror, you think I can't believe this is my body, and you feel a rush of confidence. You're not wearing the high heels you walked in with, because there's a handsome naked man kneeling at the base of your bed, kissing your feet, rubbing them, and telling you that you're the most beautiful woman he's ever seen. He addresses you on his knees, and wouldn't dare do anything to upset you. He has paid you £250 for this privilege. To him, you are a goddess.
Now, imagine you're at home wearing sweatpants and browsing through Tinder. The introductory messages you're getting are commands from men: "take off my pants," "give me your number," "let's fuck," "send me a nude." These men make lewd comments about your breasts and describe the rough sex they'd like to have with you. And if you do decide to meet them, many try to pressure you into sex or ask, "Don't I at least get a kiss?" These men think they're entitled to your body, simply because you exist.
You finally meet a somewhat normal guy, and he asks you out to drinks. He's a lawyer, and you're not sure you guys have much in common, aside from the fact that you both live in New York and enjoy beverages. This guy will be taking up your time, which is now worth at least £2500 an hour. You won't be compensated for this time, and you're actually missing out on money by not booking sessions. You know you'll feel compelled to entertain or please the guy in some way or another, and wonder what will happen when you explain to him what you do for a living. You can already picture his judgmental smirk as he reduces you to a sexual object — if, by some chance, he hadn't already. At the end of the date, he'll assess whether or not he wants to see you again as you do the same. But the power dynamic isn't in your favour, and the boundaries aren't well-defined.
At least when you're in the dungeon, you know you have the upper hand, the boundaries are clear, and there's an understanding that everything is fantasy.
These scenarios aren't hypothetical for me: I worked as a dominatrix in a New York City dungeon for two years. It's how I earned a living as I completed my master's. When I first started domming, I stopped dating. It wasn't a conscious decision, but I think it was my attempt to protect myself from the crude men I kept meeting on dating apps and reclaim control over my body and I how I perceived it.
But I wasn't always so attuned to my needs. Growing up, I discovered early on that women tended to receive compliments based on physical attractiveness and niceness, so I formed the belief that my value was dependent on those two attributes. Although I was always shy and uncomfortable with my body, I adhered to the implicit social pressure to dress pretty during the day and sexy at night. By my teens, I was dependent on men's approval. I was reading countless magazine articles about "how to entice a man" or "how to keep a man," which left me emotionally needy, overly giving, and nice to a fault in my relationships. I'd succumb to the pressure to have sex with guys I was only lukewarm about. I felt that if I listened to my gut and got to know them before hooking up, they'd lose interest. Unsurprisingly, sex left me feeling used and devalued.
Sex work was never on my radar, but when I first heard about working as a dominatrix, it seemed like the ultimate validation, and I wondered if men would consider me attractive enough to pay for the pleasure of being around me. I sent in my photos to a nearby dungeon, and I was immediately called in for an interview and offered a job as a dominatrix. I was definitely flattered about landing the gig, but I also began to think of it as a means to self-discovery. I hated how docile and submissive I had become, especially when it came to my body, and I thought playing the role of a domme might help me actively rework my wiring and my relationship to societal expectations.
For me, entering into the sex industry was a highly personal decision. I hoped it would give me the opportunity to suss out all of my complicated feelings about my body and my relationships with men. And it did, but not in the way I expected. (Contrary to popular assumptions, and just like many women in the industry, my decision to be a sex worker was not fuelled by desperation or victimisation.)
When I started the job, it blew my mind that men were willing to pay hundreds of dollars to be in a room with me. As I watched clients enter the dungeon, their faces bright red with nervousness, I began to understand my power for the first time in my life. These men were willing to do whatever I commanded them; they were so eager to please me and earn my favour. All of them knew that they'd never possess me — and none would ever imagine they'd be able to penetrate me. But of course, I could choose to penetrate them.
I began to learn about the concept of "body capital," a term that describes the assessment of the worth of a body. As a dominatrix, I held the majority of the body capital, since I wouldn't have sex with my clients, no matter how much they paid me. My power was in my denial and withholding, and that elevated me in that dynamic. And when I later left the dungeon and went independent, I had control over the price they'd pay, and I had the power to adjust the rate if I particularly liked a client. The funny thing is that the majority of my clients, most of whom were good-looking white men, either held high-power jobs in their daily lives or had amassed certain socioeconomic privileges — the cost of coming to the dungeon alone required clients have a mix of disposable income, privacy, and control over their own time. This flipping of the script — from me feeling obligated to please men to me having men at my complete will — dramatically changed my relationship with myself and my body.
For the first time in my life, I didn't sleep with someone unless he made me feel both emotionally and physically safe.
In the two years I dommed, I never had a serious relationship or dated much at all. Before, I would talk to men on dating apps and meet them for a drink and maybe kiss or hook up with them. But after domming, I stopped being comfortable participating in the laissez-faire hookup culture, since it didn't feel fulfilling to me. Though I stayed on the apps as a voyeur, my accounts remained inactive and my messages unanswered. The idea of purposely going to meet a man who might expect sex from me seemed unpleasant at best and dangerous at worst. I may have been in a relatively progressive profession, but I became much more conservative in my dating life. By the time I finished my first year of domming, I noticed that I had stopped wearing tight-fitting clothes and instead wore baggier outfits. The last thing I wanted was any sort of sexual attention when I wasn't working.
At one point, I thought I'd finally made the progress I was waiting for. For the first time in my life, I didn't sleep with someone unless he made me feel both emotionally and physically safe. Instead of jumping into sex like I did before, I first needed to know that the person was willing to compensate me emotionally; I needed to spend time with him and be courted. I'd unconsciously perform calculations in my head to determine "payment" was received before I'd cede my body capital, even though I wanted to sleep with all of these men. Oddly enough, I found that valuing myself more — by listening to my personal needs — made men value me more, too.
Even still, none of these experiences turned into relationships, because there was always a part of me that wasn't able to give over my entire self. Looking back, I think I was compensating for how vulnerable sex made me feel, which led to even more mental calculations of value and worth — and caused me to become emotionally unavailable to men. I also think I purposely picked men who were either also emotionally unavailable or somehow incompatible with me (because of an age difference or physical distance). Even though these guys treated me better than the men I had in my life before I knew my "value," I'd keep these interactions one-dimensional. I was too busy trying to keep up with the changes happening internally to allow myself to be vulnerable in the dating world beyond sex.
It took me a long time to realise that relationships aren't always an even exchange. They demand vulnerability and compromise, not a constant maintenance of power dynamics. But my internal power struggle with men has always directly related to how I treated and regarded my body, and so the evolution of my body image has been inextricable from my relationships with men.
But I soon realised that the newfound value I was finding in myself was still about traditional beauty standards. As I became more connected to my body and its needs, I started to be more conscious of how it felt and how it moved, and I took better care of it — it was now the source of my income, after all. Although this definitely had its benefits, I also became a little too fixated on my physical appearance.
By the end of my two years of domming, I had returned to my old habits, and I noticed I was unconsciously conforming to conventional standards of beauty. I grew out my once-short, unnaturally coloured hair; I became more stringent with my makeup routine; I took birth control pills to make my breasts larger (it's well-known that larger breasts can allow sex workers to earn more money). And I was even more perceptive of whether others were sexually attracted to me. I started domming because I wanted to better understand my relationship with beauty and attractiveness, but after seeing the power my body held, I ended up internalising these ideals and allowing them to govern me even more — just from a different vantage point.
This all led to an important realisation: Knowing the power my body holds over others is separate from understanding and appreciating its real worth on a deeper level.
I decided to take a break from domming a few months ago so that I could really think about what this realisation means to me, and how I can use what I've learned in the past two years to grow as a person. As addictive as being worshipped and desired is, I want to figure out how to love myself without the outside world's validation. It turns out, two years of being coddled by clients has made my ego even more fragile. The girl I used to be, the one who would freely give herself over to men and who was constantly unsure about her attractiveness, is no more. That insecure, naive person has been replace by a guarded woman who is more sure of her aesthetic value, but dubious of men's intentions and sensitive to being objectified or praised. When men tell me that I'm beautiful, it's no longer validating for me — it just makes me wonder what it is they want.
For now, I've found some sort of equilibrium. I'm starting to understand that my body is truly worth something, and that no man's approval or dollar value should be able to change that. Working as a domme has allowed me to see that I alone have control over my worth; it kickstarted a new relationship with my body. I'm still trying to reconcile what this all means, and I know I have a lot more work to do. But at the very least, sex work has empowered me to be selective about who I share my body with. And for the first time in my life, I finally realise that I'm entitled to that.
*The writer's name has been changed to protect her identity.
Dominatrix Body Image - Sex Work Made Me Value My Body
written by Lily Yin
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redpajama_set_name: RedPajamaCommonCrawl


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\newcommand{\rrr}{\big\rangle\big\rangle}
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\begin{document}
{}~ \hfill\vbox{\hbox{CTP-SCU/2018005}}\break
\vskip 3.0cm
\centerline{\Large \bf  Derive three dimensional geometries from entanglement entropies of CFT$_2$}

\vspace*{10.0ex}
\centerline{\large Peng Wang, Houwen Wu and Haitang Yang  }
\vspace*{7.0ex}
\vspace*{4.0ex}
\centerline{\large \it Center for theoretical physics}
\centerline{\large \it Sichuan University}
\centerline{\large \it Chengdu, 610064, China} \vspace*{1.0ex}
\vspace*{4.0ex}

\centerline{pengw@scu.edu.cn, iverwu@scu.edu.cn, hyanga@scu.edu.cn}
\vspace*{10.0ex}
\centerline{\bf Abstract} \bigskip \smallskip
In this paper, we propose an approach to derive three dimensional geometries from the dual CFT$_2$ entanglement entropies. We only employ the holographic principle and do not use any assumption about the AdS/CFT correspondence and bulk geometry. Our strategy is to use both UV and IR-like CFT$_2$ entanglement entropies to fix the bulk geodesics. By a simple trick, then the metric can be extracted from the geodesics. As examples, we derive the pure AdS$_3$ metric from the entanglement entropies of free CFT$_2$, and more importantly the BTZ black hole from the entanglement entropies of finite temperature CFT$_2$. Consequently, CFT$_2$ with finite size or topological defects can be worked out by simple transformations. Following the same steps, in principle, it is possible to derive  all three dimensional (topologically distinct)  holographic classical geometries   from the  dual CFT$_2$ entanglement entropies.

\vfill
\eject
\baselineskip=16pt
\vspace*{10.0ex}
\tableofcontents

\section{Introduction}

As a manifestation of the non-local property of quantum mechanics,
quantum entanglement attracts considerable attention in recent years.
The entanglement entropy (EE) measures the correlation between subsystems
and is one of the most distinct features of quantum systems. Considering
the simplest configuration, a quantum system is divided into two subsystems:
$A$ and $B$. The total Hilbert space is accordingly decomposed into
$\mathcal{H}=\mathcal{H}_{A}\otimes\mathcal{H}_{B}$. Tracing out
the degrees of freedom of the region $B$, one obtains the reduced
density matrix of the region $A$: $\rho_{A}=\mathrm{Tr}_{\mathcal{H}_{B}}\rho$.
The entanglement entropy of the region $A$ is evaluated by the von
Neumann entropy $S_{A}=-\mathrm{Tr}_{\mathcal{H}_{A}}\left(\rho_{A}\log\rho_{A}\right)$.
It is clear that $S_{A}=S_{B}$.

Motivated by the AdS/CFT correspondence \cite{Maldacena:1997re},
and the Bekenstein-Hawking entropy of black holes, Ryu and Takayanagi
(RT) proposed to identify the minimal surface area ending on the $d$
dimensional boundary of AdS$_{d+1}$ with the entanglement entropy
of the CFT$_{d}$ living on the boundary of AdS$_{d+1}$ \cite{Ryu:2006bv, Ryu:2006ef,Hubeny:2007xt}.
In the case of $d=2$, the minimal surfaces are geodesics and the
RT formula was verified extensively. For follow-ups and references,
one refers to a recent review \cite{Rangamani:2016dms}. The success
of the RT formula leads to an inspiring conjecture that gravity could
be interpreted as emergent structures, determined by the quantum entanglement
of the dual CFT \cite{VanRaamsdonk:2009ar,VanRaamsdonk:2010pw}. This
idea was further developed by Maldacena and Susskind to conjecture
an equivalence of Einstein-Rosen bridge (ER) and Einstein-Podolsky-Rosen
(EPR) experiment \cite{Maldacena:2013xja}, namely ER=EPR. To justify
these conjectures, one question has to answered: Can the dual bulk
geometry, specifically the metric, be rebuilt from given CFT entanglement
entropies?

Although, everyone believes that the dual geometries must be asymptotic
AdS, there is still no perfect method to derive the dual geometries
from entanglement entropies of CFTs. One might think it is trivial
to derive the dual geometry from a   CFT$_{d}$, since they
 are supposed to possess the same $SO(2,d)$ symmetry.  This is not true since all
CFTs share the same symmetry and so do the dual geometries. The symmetry
argument is a necessary but not a sufficient condition to determine the dual
geometry of a CFT. But what matters much more is precisely the sufficient condition, i.e.
deriving  the dual geometries from CFTs. The bottom line is that the simple symmetry argument
cannot carry on beyond the vacuum configuration. Therefore, it is
necessary to find some other systematic method applicable to all excited
states. Of course, since we hope to \emph{derive} the dual geometries,
one can NOT assume the geometries satisfying any dynamic equation, say, the
Einstein equation. The dynamic equation should also be derived from
the CFT information. In literature, there are several attempts, but
none of them can really rebuild the dual geometries unambiguously. The
tensor network is only able to construct the discrete AdS \cite{Swingle:2009bg}.
Another major method resorts to integral geometry. The concept of
kinematic space is introduced and it is argued that the kinematic
space of AdS$_{3}$ is dS$_{2}$, which can be read off from the Crofton
form defined as the second derivatives with respect to two different
points of the given entanglement entropy of CFT$_{2}$ \cite{Czech:2015qta}.
But it was not proved that AdS$_{3}$ is the only choice to have dS$_{2}$
as the kinematic space. Moreover, this method can only apply to the
static scenario naturally.

It is not hard to foresee our journey to discover geometry reconstruction
would be blocked by two difficulties:
\begin{itemize}
\item It is a standard homework to calculate the minimal surface from the
metric. Now that the CFT entanglement entropy is identified with the
minimal surface, in order to get the geometry, one needs to extract
the metric from the minimal surfaces, which looks forbidden.
\item It is usually not hard to reduce a higher dimensional theory to a
lower dimensional one after setting some limits or boundary conditions
to eliminate extra degrees of freedom. However, since the CFT$_{d}$
entanglement entropy is identified with the minimal surface attached
on the boundary of the dual $d+1$ dimensional geometry, in order
to reconstruct the $d+1$ dimensional bulk geometry, we have to find
some way to uniquely fix the extra degrees of freedom, namely the
bulk geodesic when $d=2$.
\end{itemize}
In a previous work \cite{Wang:2017bym}, we proposed an approach to
solve these two difficulties for $d=2$. It turns out there is a simple
way to extract the metric from given geodesics, the minimal surfaces
for $d=2$. Considering a single geodesic $x=x\left(\tau\right)$
, $\tau\in\left[0,t\right]$ which connects points $x$ and $x'$
on a manifold $M$, such that $x\left(0\right)=x$ and $x\left(t\right)=x^{\prime}$.
$L(x,x')$ is the length of the geodesic. Then the metric proves to
be

\begin{equation}
g_{ij}=-\:\underset{x\rightarrow x'}{\lim}\partial_{x^{i}}\partial_{x^{j^{\prime}}}\left[\frac{1}{2}L_{\mathrm{}}^{2}\left(x,x^{\prime}\right)\right].\label{eq:derive metric}
\end{equation}
For more details, one refers to a comprehensive review \cite{Poisson:2011nh}.
As an illustration, noting that along a geodesic, the norm of the
tangent vector $g_{ij}\dot{X}^{i}\dot{X}^{j}$ is constant. So for
very small distance $t\to0$, we have

\begin{equation}
\frac{1}{2}L^{2}(x,x')=\frac{1}{2}\left[\int_{0}^{t}d\tau\:\sqrt{g_{ij}\dot{X}^{i}\dot{X}^{j}}\right]^{2}\approx\frac{1}{2}\,\lim_{t\to0}\,g_{ij}\frac{\Delta x^{i}}{t}\frac{\Delta x^{j}}{t}t^{2}\approx\frac{1}{2}g_{ij}\Delta x^{i}\Delta x^{j}.
\end{equation}
Since the quantity $\sigma\left(x,x^{\prime}\right)\equiv\frac{1}{2}L^{2}\left(x,x^{\prime}\right)$
plays a central role to address the radiation back reaction of particles
moving in a curved spacetime, it has a specific name: Synge\textquoteright s
world function.

Therefore, what is left is to generalize the geodesics located on
the boundary to generic geodesics in the bulk. In order to fix the
expression of bulk geodesics, we find that in addition to the usually
used UV entanglement entropy, the IR-like entanglement entropy of
the CFT is a must.

In the paper \cite{Wang:2017bym}, we only addressed the vacuum configuration,
i.e. the free CFT$_{2}$ with zero temperature and infinite length.
We showed explicitly the dual geometry must be AdS$_{3}$, as expected.
The purpose of this paper is to demonstrate that this approach also
works for excited states, specifically, the CFT$_{2}$ with finite
temperature, whose dual geometry is supposed to be the BTZ black hole.
It is well known that $3D$ gravity is topological as a consequence
of the general relativity. Out of general geometries, the Einstein
equation selects those having no local degrees of freedom to describe
gravity. However, since we hope to derive the dual geometry, we can
not use any results from the general relativity. So, the local agreement
of the dual geometries is supposed to be unknown, and should be revealed
by the derived metrics from the entanglement entropies. Since the
entanglement entropy is a non-local quantity, it is easy to understand
that one can not directly transform the entanglement entropy of free
CFT to that of the finite temperature CFT, for they having different
topologies. In contrast, once we derived the BTZ black hole from the
finite temperature CFT, one can easily extend the result to the finite
size CFT under transformation $\beta=-iL$ or CFT with topological
defects under transformation $\beta=-iL/\gamma_{\mathrm{con}}$, since
they have the same topology, a cylinder. More importantly, the BTZ
derivation indicates that, with our approach, it is possible to derive
all $3D$ classical (topologically distinct) geometries from the entanglement
entropies of the dual CFT$_{2}$.

The reminder of this paper is outlined as follows. In section 2, we
briefly review some useful results of CFT entanglement entropies,
which help us to determine the geodesic length in the bulk geometry.
In section 3, we summarize how to get the asymptotic AdS$_{3}$ from
the entanglement entropies of a pure CFT$_{2}$. We show how to derive
the BTZ spacetime from the entanglement entropies of the finite temperature
CFT in section 4. In section 5, we present some inspirations and conjectures.

\section{Some useful results of CFT entanglement entropy}

In this section, based on refs. \cite{Calabrese:2004eu,Calabrese:2009qy},
we briefly summarize some results of CFT$_{2}$ entanglement entropies
that we are going to use in the rest of the paper. For a quantum system
consisting of two parts $A$ and $B$, the entanglement entropy (EE)
for the subsystem $A$ is defined by the Von Neumann entropy:

\begin{equation}
S_{EE}=-\mathrm{Tr}_{\mathcal{H}_{A}}\left(\rho_{A}\log\rho_{A}\right),
\end{equation}

\noindent where the reduced density matrix $\rho_{A}=\mathrm{Tr}_{\mathcal{H}_{B}}\rho$
and $\rho=\left|\Psi\right\rangle \left\langle \Psi\right|$. Usually
in QFT, it is difficult to calculate the Von Neumann entropy directly.
Alternatively, one uses the ``replica trick'' to calculate $\mathrm{Tr}\rho_{A}^{n}$,
then

\begin{equation}
S_{EE}=-\underset{n\rightarrow1}{\lim}\frac{\partial}{\partial n}\mathrm{log\:Tr}\rho_{A}^{n}.\label{eq:density to S}
\end{equation}

\noindent Considering an 1+1 dimensional Euclidean QFT with a local
field $\phi\left(t_{E},x\right)$, it turns out

\begin{equation}
\mathrm{Tr}\rho_{A}^{n}=\frac{1}{\left(Z_{1}\right)^{n}}\int_{\left(t_{E},x\right)\in\mathcal{R}_{n}}\mathcal{D}\phi e^{-S_{E}\left(\phi\right)}\equiv\frac{Z_{n}\left(A\right)}{\left(Z_{1}\right)^{n}},
\end{equation}

\noindent where $Z_{n}\left(A\right)$ is the partition function on
an $n$-sheeted Riemann surface $\mathcal{R}_{n}$, $Z_{1}$ is the
vacuum partition function on $\mathcal{R}_{2}$. The partition function
is given by the two-point function of twist operators $\mathcal{T}$
and $\tilde{\mathcal{T}}$. For an infinitely long system, when fixing
$t_{E}=it=0$,

\begin{equation}
Z_{n}\left(A\right)=\left\langle \mathcal{T}_{n}\left(u,0\right)\tilde{\mathcal{T}}_{n}\left(v,0\right)\right\rangle _{\mathbb{C}}=\frac{1}{\left|u-v\right|^{2\Delta}},\label{eq:twist}
\end{equation}

\noindent where $\triangle=\frac{c}{12}\left(n-\frac{1}{n}\right)$
is the scaling dimension and $c$ is the central charge. So, the entanglement
entropy is:

\begin{eqnarray}
S_{EE} & = & -\underset{n\rightarrow1}{\lim}\frac{\partial}{\partial n}\mathrm{\log\:Tr}\rho_{A}^{n}=\frac{c}{3}\log\frac{\triangle x}{a}+c_{1}^{\prime},\label{eq:EE infinite}
\end{eqnarray}

\noindent where $c_{n}^{\prime}\equiv\log c_{n}/\left(1-n\right)$
and $a$ is an energy cut-off which ensures the factor inside the
$log$ dimensionless. $\Delta x=x'-x$ is the size of the entangling
region $A$.

We can easily develop this result to other geometric background by
utilizing conformal transformations $z^{\prime}\rightarrow z=z\left(z^{\prime}\right)$
on two-point functions:

\begin{equation}
\left\langle \mathcal{T}_{n}\left(z_{1}^{\prime},\bar{z}_{1}^{\prime}\right)\tilde{\mathcal{T}}_{n}\left(z_{2}^{\prime},\bar{z}_{2}^{\prime}\right)\right\rangle =\left(\frac{\partial z_{1}}{\partial z_{1}^{\prime}}\frac{\partial z_{2}}{\partial z_{2}^{\prime}}\right)^{\triangle}\left(\frac{\partial\bar{z}_{1}}{\partial\bar{z}_{1}^{\prime}}\frac{\partial\bar{z}_{2}}{\partial\bar{z}_{2}^{\prime}}\right)^{\triangle}\left\langle \mathcal{T}_{n}\left(z_{1},\bar{z}_{1}\right)\tilde{\mathcal{T}}_{n}\left(z_{2},\bar{z}_{2}\right)\right\rangle .\label{eq:CFT trans}
\end{equation}

\noindent For example, to calculate the entanglement entropy of a
CFT$_{2}$ at finite temperature $2\pi\beta^{-1}$, one maps the infinitely
long cylinder $z^{\prime}$ to the plane $z\left(z^{\prime}\right)$
by the following transformation:

\begin{equation}
z^{\prime}\rightarrow z\left(z^{\prime}\right)=e^{\frac{2\pi z^{\prime}}{\beta}}.\label{eq:finite trans}
\end{equation}

\noindent The two-point function of the finite temperature CFT$_{2}$
is

\begin{eqnarray}
\left\langle \mathcal{T}_{n}\left(u^{\prime},0\right)\tilde{\mathcal{T}}_{n}\left(v^{\prime},0\right)\right\rangle _{\mathbb{C}} & = & \left(\frac{\partial}{\partial u^{\prime}}e^{\frac{2\pi u^{\prime}}{\beta}}\right)^{\triangle}\left(\frac{\partial}{\partial v^{\prime}}e^{\frac{2\pi v^{\prime}}{\beta}}\right)^{\triangle}\left\langle \mathcal{T}_{n}\left(u,0\right)\tilde{\mathcal{T}}_{n}\left(v,0\right)\right\rangle _{\mathbb{C}}\nonumber \\
 & = & \left|\frac{\beta}{\pi}\sinh\frac{\pi\triangle x}{\beta}\right|^{-\frac{c}{6}\left(n-\frac{1}{n}\right)}.
\end{eqnarray}

\noindent Therefore, the entanglement entropy is given by

\begin{equation}
S_{EE}=\frac{c}{3}\log\left(\frac{\beta}{\pi a}\sinh\frac{\pi\triangle x}{\beta}\right)+c_{1}^{\prime}.
\end{equation}

\noindent It should be stressed that though the two-point function
of the finite temperature CFT$_{2}$ can be obtained from that of
the free CFT$_{2}$ by a conformal map, there is no coordinate transformation
to connect their entanglement entropies. The reason is that entanglement
entropy is a global quantity associated with the topology, while these
two systems clearly have different topologies. In contrast, since
a finite size system has the same topology as a finite temperature
system, the entanglement entropy for a finite size system is obtained
by replacing $\beta\rightarrow L_{S}$ and imposing the periodic boundary
condition,

\begin{equation}
S_{EE}=\frac{c}{3}\log\left(\frac{L_{S}}{\pi a}\sin\left(\frac{\pi\triangle x}{L_{S}}\right)\right)+c_{1}^{\prime},
\end{equation}

\noindent where $L_{S}$ is the circumference of the given system.

It turns out that, when deriving the BTZ geometry, we also need the
entanglement entropy of boundary conformal field theory (BCFT). BCFT
is a CFT whose boundary satisfies conformally invariant boundary conditions.
Considering an one dimensional semi-infinite long system $x\in\left[0,\infty\right)$.
The boundary obviously locates at $x=0$. The $n$-sheeted Riemann
surface $\mathcal{R}_{n}$ now consists of $n$ copies in the region
of $x\geq0$. The transformation from complex coordinates on $\mathcal{R}_{n}$
to $\mathbb{C}$ is $w\rightarrow z\left(w\right)=\left(\left(w-il\right)/\left(w+il\right)\right)^{1/n}$.
The partition function $Z_{n}\left(A\right)$ on the $n$-sheeted
Riemann surface $\mathcal{R}_{n}$ becomes the one-point function
of twist operator $\mathcal{T}$ . For any primary operator $\mathcal{O}$,
the one-point function is

\begin{equation}
\left\langle \mathcal{O}\left(z\right)\right\rangle =\frac{1}{\left(2\mathrm{Im}\:z\right)^{\triangle}}.
\end{equation}

\noindent The scaling dimension of $\mathcal{T}$ still equals $\triangle=\frac{c}{12}\left(n-\frac{1}{n}\right)$.
So, we have

\begin{equation}
\left\langle \mathcal{\mathcal{T}}\left(il\right)\right\rangle =\frac{1}{\left(2l\right)^{\frac{c}{12}\left(n-\frac{1}{n}\right)}}.\label{eq:one-point of BCFT}
\end{equation}

\noindent It is then straightforward to find

\begin{eqnarray*}
S_{EE} & = & -\underset{n\rightarrow1}{\lim}\frac{\partial}{\partial n}\mathrm{\log\:Tr}\rho_{A}^{n}=\frac{c}{6}\log\frac{2\triangle x}{a}+\tilde{c}_{1}^{\prime}.
\end{eqnarray*}

\noindent Applying the transformations (\ref{eq:finite trans}) onto
the one-point function (\ref{eq:one-point of BCFT}), we get

\begin{equation}
\left\langle \mathcal{\mathcal{T}}\left(il^{\prime}\right)\right\rangle =\left|\frac{\beta}{\pi}\sinh\frac{2\pi\triangle x}{\beta}\right|^{-\frac{c}{12}\left(n-\frac{1}{n}\right)}.
\end{equation}

\noindent Therefore, the entanglement entropy of BCFT at finite temperature
is

\begin{equation}
S_{EE}=\frac{c}{6}\log\left(\frac{\beta}{\pi a}\sinh\frac{2\pi\triangle x}{\beta}\right)+\tilde{c}_{1}^{\prime}.\label{eq:BCFT}
\end{equation}

\noindent Note the $\Delta x$ here is the entanglement length of
BCFT, which is one half of the entanglement length of the corresponding
CFT.

In most of the cases, when one mentions the entanglement entropy,
he/she really refers to the UV entanglement entropy, which is precisely
what we have discussed thus far. However, when a free CFT is perturbed
by a relevant operator, the correlation length $\xi$ (IR cut-off)
takes a finite value. In the IR region $\Delta x\gg\xi$, the UV entanglement
entropy (\ref{eq:EE infinite}) is no longer valid and there exists
an IR entanglement entropy. The simplest way to calculate the IR entanglement
entropy is to consider the action

\begin{equation}
S=\int d^{2}x\left(\frac{1}{2}\left(\partial_{\mu}\varphi\right)^{2}+\frac{1}{2}m^{2}\varphi^{2}\right),
\end{equation}

\noindent where $m\rightarrow0$. The partition function $Z_{n}\left(A\right)$
on $n$-sheeted Riemann surface $\mathcal{R}_{n}$, can be calculated
with the identity \cite{Calabrese:2009qy}

\begin{equation}
\frac{\partial}{\partial m^{2}}\log Z_{n}\left(A\right)=-\frac{1}{2}\int d^{2}xG_{n}\left(\mathbf{x},\mathbf{x}\right),
\end{equation}

\noindent where $G_{n}\left(\mathbf{x},\mathbf{x}\right)$ is the
two-point function on $\mathcal{R}_{n}$, satisfying the equation
of motion: $\left(-\nabla^{2}+m^{2}\right)G_{n}\left(\mathbf{x},\mathbf{x}^{\prime}\right)=\delta^{2}\left(\mathbf{x}-\mathbf{x}^{\prime}\right)$.
It is easy to find

\begin{equation}
\frac{\partial}{\partial m^{2}}\log\frac{Z_{n}\left(A\right)}{\left(Z_{1}\right)^{n}}=\frac{1}{24m^{2}}\left(n-\frac{1}{n}\right).
\end{equation}

\noindent Integrating $m^{2}$ on both sides, we get

\begin{equation}
\log\frac{Z_{n}\left(A\right)}{\left(Z_{1}\right)^{n}}=\frac{\log a^{2}m^{2}}{24}\left(n-\frac{1}{n}\right)\rightarrow\frac{Z_{n}\left(A\right)}{\left(Z_{1}\right)^{n}}=\left(ma\right)^{\frac{1}{12}\left(n-\frac{1}{n}\right)}.
\end{equation}

\noindent Therefore, the IR entanglement entropy is

\begin{eqnarray}
S_{EE}^{IR} & = & -\underset{n\rightarrow1}{\lim}\frac{\partial}{\partial n}\log\:\frac{Z_{n}\left(A\right)}{\left(Z_{1}\right)^{n}}\nonumber \\
 & = & -\underset{n\rightarrow1}{\lim}\frac{\partial}{\partial n}\left[\left(ma\right)^{\frac{1}{12}\left(n-\frac{1}{n}\right)}\right]\nonumber \\
 & = & \frac{1}{6}\log\frac{\xi}{a},\label{eq:EE IR}
\end{eqnarray}

\noindent where $c=1$ for one field $\varphi$ and we introduced
the IR cut-off $\xi=m^{-1}$.

The time dependent entanglement entropies can be calculated by completely
the same procedure. At each step, one simply includes the time-like
variable to get

\begin{eqnarray}
\mathrm{infinite\quad system:} &  & S_{EE}\left(t\right)=\frac{c}{3}\log\frac{\sqrt{\left(\triangle x\right)^{2}-\left(\triangle t\right)^{2}}}{a},\label{eq:time uv}\\
\mathrm{finite\quad temperature:} &  & S_{EE}\left(t\right)=\frac{c}{3}\log\left(\frac{\beta}{2\pi a}\left[\sqrt{2\cosh\left(\frac{2\pi\triangle x}{\beta}\right)-2\cosh\left(\frac{2\pi\triangle t}{\beta}\right)}\right]\right),
\end{eqnarray}

\noindent which are well-defined when two-points are space-like separated.

\section{Asymptotic AdS$_{3}$ from entanglement }

For the sake of completeness and convenience, we summarize how to
derive the asymptotic AdS$_{3}$ from free CFT$_{2}$ entanglement
entropies. For more details, one refers to our previous work \cite{Wang:2017bym}.
We consider the static scenario first. The entanglement entropy of
an infinite size CFT$_{2}$ at zero temperature is given by eqn. (\ref{eq:EE infinite}).
This entanglement entropy is identified with the length of the geodesic
attached on the boundary of the dual geometry.

\begin{equation}
L_{\mathrm{boundary}}=4G_{N}^{(3)}S_{EE}=R\log\frac{\left(x-x^{\prime}\right)^{2}}{a^{2}},\qquad R=\frac{2G_{N}^{(3)}c}{3},\label{eq:UV L}
\end{equation}

\noindent where we neglect the irrelevant constant $c_{1}^{\prime}$.
The subscript of $L_{\mathrm{boundary}}$ indicates the geodesic ends
on the boundary of the dual geometry. From holographic principle \cite{tHooft:1993dmi,Susskind:1994vu, Susskind:1998dq},
energy cut-off generates an extra dimension, so we have $a\to y$.
With two coordinates $x$ and $y$, The most general extension of
(\ref{eq:UV L}) is

\begin{equation}
L_{\mathrm{boundary}}=R\log\left[\frac{\left(x-x^{\prime}\right)^{2}f\left(x,x^{\prime};y,y^{\prime}\right)+\left(y-y^{\prime}\right)^{2}}{g\left(x,x^{\prime};y,y^{\prime}\right)}\right],\label{eq:eta define}
\end{equation}

\noindent where $f(x,x';y,y')$ and $g(x,x';y,y')$ are regular functions,
invariant under exchanging two endpoints $\left(x^{\prime},y^{\prime}\right)\leftrightarrow\left(x,y\right)$.
When the endpoints coincide, the geodesic length must vanish. Therefore,
the geodesic length can be expanded in a power series for nearby endpoints
and we have:

\begin{eqnarray}
\eta\to\infty: &  & L_{\mathrm{boundary}}=R\log\left(\eta^{2}\right),\nonumber \\
\eta\to0: &  & L_{\mathrm{bulk}}=R\log\left(1+C_{1}\eta^{\alpha}+C_{2}\eta^{2\alpha}+\ldots\right),
\end{eqnarray}

\noindent where we defined

\begin{equation}
\eta^{2}\equiv\frac{\left(x-x^{\prime}\right)^{2}f\left(x,x^{\prime},y,y^{\prime}\right)+\left(y-y^{\prime}\right)^{2}}{g\left(x,x^{\prime},y,y^{\prime}\right)}.\label{eq:simple eta}
\end{equation}

\noindent The subscript of $L_{\mathrm{bulk}}$ indicates the geodesic
is a generic one in the dual geometry. All geodesic lengths read from
the entanglement entropies are $L_{\mathrm{boundary}}$ and satisfy
$\eta\to\infty$. What we want to do is to determine $f(x,x';y,y')$
and $g(x,x';y,y')$ through $L_{\mathrm{boundary}}$, and then substitute
$\eta$ into $L_{\mathrm{bulk}}$ to get the bulk geodesic.

\vspace{2em}

\noindent \textbf{Step 1:} We first consider the entanglement entropy
with IR cut-off given by eqn. (\ref{eq:EE IR}):

\begin{equation}
S_{EE}^{IR}=\frac{c}{6}\log\frac{\xi}{a},\qquad L_{\mathrm{boundary}}^{IR}=R\log\frac{\xi}{a}.
\end{equation}

\noindent Since both $a$ and $\xi$ are varying energy cut-offs,
they are simply two points on the coordinate axis $y$. Looking at
the general expression (\ref{eq:eta define}), under the condition
$\xi\gg a$, it is ready to see that this must be the case $y=a$,
$y^{\prime}=\xi$ and $\triangle x=0$, as illustrated by Fig. (\ref{fig:The-entanglement-entropy}).

\begin{figure}[H]
\begin{centering}
\includegraphics[scale=0.3]{fig_IR}
\par\end{centering}
\centering{}\caption{\label{fig:The-entanglement-entropy}The entanglement entropy with
IR cut-off gives the geodesic connecting $y=a$, $y^{\prime}=\xi$
and $\Delta x=0$.}
\end{figure}

\noindent We immediately obtain

\begin{eqnarray}
L_{\mathrm{boundary}} & \rightarrow & R\log\left[\frac{1}{g\left(x,x;a,\xi\right)}\left(a-\xi\right)^{2}\right]\nonumber \\
 & \rightarrow & L_{\mathrm{boundary}}^{IR}=R\log\frac{\xi}{a}.
\end{eqnarray}

\noindent and thus

\begin{equation}
g\left(x,x';y,y^{\prime}\right)=yy^{\prime}\left(1+\mathcal{O}\left(x-x^{\prime}\right)^{2}\right).\label{eq:g function}
\end{equation}

\vspace{2em}

\noindent \textbf{Step 2:} We next come to the entanglement entropy
with UV cut-off given by eqn. (\ref{eq:EE infinite}):

\begin{equation}
S_{EE}^{UV}=\frac{c}{3}\log\frac{\triangle x}{a},\qquad L_{\mathrm{boundary}}^{UV}=R\log\frac{\left(x-x^{\prime}\right)^{2}}{a^{2}}.
\end{equation}

\noindent It is easy to see this picture corresponds to $y=y'=a$
in the general expression (\ref{eq:eta define}), as illustrated by
Fig (\ref{fig:The-entanglement-entropy-1}).

\begin{figure}[H]
\begin{centering}
\includegraphics[scale=0.35]{fig_UV}
\par\end{centering}
\centering{}\caption{The entanglement entropy with UV cut-off supports the geodesic connecting
$y=y^{\prime}=a$ with arbitrary $\Delta x$.\label{fig:The-entanglement-entropy-1}}
\end{figure}

\noindent We get

\begin{eqnarray}
L_{\mathrm{boundary}} & \rightarrow & R\log\left[\frac{f\left(x,x^{\prime},a,a\right)}{g\left(x,x^{\prime},a,a\right)}\left(x-x^{\prime}\right)^{2}\right]\nonumber \\
 & \rightarrow & L_{\mathrm{boundary}}^{UV}=R\log\frac{\left(x-x^{\prime}\right)^{2}}{a^{2}}.
\end{eqnarray}

\noindent Therefore, combining eqn. (\ref{eq:g function}), the expression
of $g\left(x,x^{\prime},a,a\right)$, we get

\begin{equation}
f(x,x';0,0)=1+\mathcal{O}\left(x-x^{\prime}\right)^{2},
\end{equation}
and then

\begin{equation}
f(x,x';y,y')=1+\mu_{1}\left(\frac{y}{\xi}+\frac{y^{\prime}}{\xi}\right)+\ldots
\end{equation}

\noindent where $\mu_{i}$ are constants. Finally, we substitute the
determined function $\eta$ back into $L_{\mathrm{bulk}}$:
\begin{equation}
L_{\mathrm{bulk}}=R\log\left[1+C_{1}\left(\frac{\left(x-x^{\prime}\right)^{2}\left(1+\mu_{1}\left(\frac{y}{\xi}+\frac{y^{\prime}}{\xi}\right)+\ldots\right)+\left(y-y^{\prime}\right)^{2}}{yy^{\prime}\left(1+\mathcal{O}\left(x-x^{\prime}\right)^{2}\right)}\right)^{\alpha/2}+\ldots\right].
\end{equation}

\noindent As we emphasized, this expression is valid only for $\Delta x\to0$,
$\Delta y\to0$, precisely the conditions for applying eqn. (\ref{eq:derive metric}),
as follows

\begin{equation}
g_{ij}=-\underset{x\rightarrow x'}{\lim}\partial_{x^{i}}\partial_{x^{j^{\prime}}}\left[\frac{1}{2}L_{\gamma_{A}}^{2}\left(x,x^{\prime}\right)\right].
\end{equation}

\noindent Moreover, it is remarkable that when using this equation
to get the metric, one finds only $\alpha=1$ is physical. We thus
obtain

\begin{eqnarray}
g_{xx} & = & -\frac{1}{2}\:\underset{\left(x^{\prime},y^{\prime}\right)\rightarrow\left(x,y\right)}{\lim}\:\partial_{x}\partial_{x^{\prime}}L_{\mathrm{bulk}}^{2}=C_{1}^{2}\left(1+\mu_{1}\left(\frac{y}{\xi}\right)+\ldots\right)\frac{R^{2}}{y^{2}},\nonumber \\
g_{yy} & = & -\frac{1}{2}\:\underset{\left(x^{\prime},y^{\prime}\right)\rightarrow\left(x,y\right)}{\lim}\:\partial_{y}\partial_{y^{\prime}}L_{\mathrm{bulk}}^{2}=C_{1}^{2}\frac{R^{2}}{y^{2}}.
\end{eqnarray}

\noindent We can set $C_{1}=1$, $R=R_{AdS}$. The metric is

\begin{equation}
ds^{2}=\frac{R_{AdS}^{2}}{y^{2}}\left[\left(1+\mu_{1}\left(\frac{y}{\xi}\right)+\mu_{2}\left(\frac{y}{\xi}\right)^{2}+\ldots\right)dx^{2}+dy^{2}\right].
\end{equation}

\noindent Including the time-like direction is straightforward by
using equation (\ref{eq:time uv}). The metric is

\begin{equation}
ds^{2}=\frac{R_{AdS}^{2}}{y^{2}}\left[-G\left(y\right)dt^{2}+F\left(y\right)dx^{2}+dy^{2}\right],\label{eq:asym AdS}
\end{equation}

\noindent where

\begin{eqnarray}
G\left(y\right) & = & 1+\rho_{1}\left(\frac{y}{\xi}\right)+\rho_{1}\left(\frac{y}{\xi}\right)^{2}+\ldots,\nonumber \\
F\left(y\right) & = & 1+\mu_{1}\left(\frac{y}{\xi}\right)+\mu_{1}\left(\frac{y}{\xi}\right)^{2}+\ldots.
\end{eqnarray}

\noindent It is exactly the asymptotic AdS$_{3}$ in the Poincare
coordinates. When removing the relevant perturbation by taking the
limit $\xi\rightarrow\infty$, it becomes the pure AdS$_{3}$

\begin{equation}
ds^{2}=\frac{R_{AdS}^{2}}{y^{2}}\left(-dt^{2}+dx^{2}+dy^{2}\right).
\end{equation}

\section{BTZ spacetime from entanglement}

We now consider a finite temperature CFT$_{2}$. There are obviously
two energy scales: UV cut-off $a$ and the temperature: $T^{-1}=\frac{\beta}{2\pi}\equiv\beta_{H}$.
We use the notation $\beta_{H}\equiv\frac{\beta}{2\pi}$ for simplicity
in the rest of the paper. The temperature introduces a natural upper
bound for the energy generated extra dimension: $y\le\beta_{H}$.

The time-dependent entanglement entropy of the finite temperature
CFT$_{2}$ was given in section 2, which is

\begin{equation}
S_{EE}\left(t\right)=\frac{c}{3}\log\left(\frac{\beta_{H}}{a}\left[\sqrt{2\cosh\left(\frac{\triangle x}{\beta_{H}}\right)-2\cosh\left(\frac{\triangle t}{\beta_{H}}\right)}\right]\right).
\end{equation}

\noindent As $\beta_{H}\rightarrow\infty$ ($T\to0$), it reduces
to the entanglement entropy of the free CFT. The static case is achieved
by taking $\triangle t=0$. The corresponding geodesic ends on the
boundary and the length is

\begin{eqnarray}
\frac{L_{\mathrm{boundary}}}{R} & = & \log\left(\frac{\beta_{H}^{2}}{a^{2}}\left[2\cosh\left(\frac{\triangle x}{\beta_{H}}\right)-2\cosh\left(\frac{\triangle t}{\beta_{H}}\right)\right]\right),\label{eq:BTZ boundary L}
\end{eqnarray}

\noindent Based on the holographic principle, the energy cut-off generates
the extra dimension $a\to y$. In last section, we have shown the
dual geometry of infinite size CFT$_{2}$ at zero temperature is asymptotic
AdS$_{3}$. For pure AdS$_{3}$ ($\xi\rightarrow\infty$), the geodesic
length between $\left(x,y,t\right)$ and $\left(x^{\prime},y^{\prime},t^{\prime}\right)$
turns out to be

\begin{equation}
\cosh\left(\frac{L_{\mathrm{bulk}}}{R}\right)=\frac{\left(\triangle x\right)^{2}-\left(\triangle t\right)^{2}+y^{2}+y^{\prime2}}{2yy^{\prime}}.\label{eq:pure AdS L}
\end{equation}

\noindent In order to match eqn. (\ref{eq:BTZ boundary L}) on the
boundary and eqn. (\ref{eq:pure AdS L}) as $\beta\to\infty$ simultaneously,
the most general expression of the bulk geodesic of the dual geometry
for the finite temperature CFT$_{2}$ can only take the form

\begin{equation}
\cosh\left(\frac{L_{\mathrm{bulk}}}{R}\right)=\frac{\beta_{H}^{2}}{yy^{\prime}}\left[f\left(x,x';y,y^{\prime};t,t'\right)\cosh\left(\frac{\triangle x}{\beta_{H}}\right)-g\left(x,x';y,y^{\prime};t,t'\right)\cosh\left(\frac{\triangle t}{\beta_{H}}\right)\right].\label{eq:BTZ cosh}
\end{equation}

\noindent where $f\left(x,x';y,y^{\prime};t,t'\right)$ and $g\left(x,x';y,y^{\prime};t,t'\right)$
are regular functions to be determined and they must be invariant
under $\left(x^{\prime},y^{\prime},t^{\prime}\right)\leftrightarrow\left(x,y,t\right)$.
The $cosh$ on the LHS of eqn. (\ref{eq:BTZ cosh}) is determined
by eqn. (\ref{eq:pure AdS L}), and the function form on the RHS of
eqn. (\ref{eq:BTZ cosh}) is determined by eqn. (\ref{eq:BTZ boundary L}).
We do not put the term of $\cosh\left(\frac{\triangle y}{\beta_{H}}\right)$
in eqn. (\ref{eq:BTZ cosh}) because its existence, if any, can be
absorbed in the undetermined functions $f\left(x,x';y,y^{\prime};t,t'\right)$
and $g\left(x,x';y,y^{\prime};t,t'\right)$. Similarly, the factor
$\frac{1}{yy'}$ outside the bracket is simply for convenience. Therefore,
the aim is the same as the free CFT situation: we apply various constraints
to determine the functions $f$ and $g$, and then use $L_{\mathrm{bulk}}$
to get the metric.

\vspace{2em}

\noindent \textbf{Step 1:} When $\beta_{H}\gg y=y^{\prime}=a$, $L_{\mathrm{bulk}}$
must reduce to $L_{\mathrm{boundary}}$, given by eqn. (\ref{eq:BTZ boundary L}),

\begin{eqnarray}
L_{\mathrm{bulk}} & = & R\log\left(\frac{\beta_{H}^{2}}{yy^{\prime}}\left[f\left(x,x';y,y^{\prime};t,t'\right)2\cosh\left(\frac{\triangle x}{\beta_{H}}\right)-g\left(x,x';y,y^{\prime};t,t'\right)2\cosh\left(\frac{\triangle t}{\beta_{H}}\right)\right]\right)\nonumber \\
 & \rightarrow & R\log\left(\frac{\beta_{H}^{2}}{a^{2}}\left[2\cosh\left(\frac{\triangle x}{\beta_{H}}\right)-2\cosh\left(\frac{\triangle t}{\beta_{H}}\right)\right]\right).
\end{eqnarray}

\noindent So as $\beta_{H}\gg y$ and $y^{\prime}$, we have

\begin{eqnarray}
f\left(x,x';y,y^{\prime};t,t'\right) & = & 1+\mu_{1}\left(x,x';t,t'\right)\left(\frac{y}{\beta_{H}}+\frac{y^{\prime}}{\beta_{H}}\right)+\mu_{2}\left(x,x';t,t'\right)\left(\frac{y}{\beta_{H}}+\frac{y^{\prime}}{\beta_{H}}\right)^{2}+\ldots\nonumber \\
 &  & +\rho_{1}\left(x,x';t,t'\right)\left(\frac{yy^{\prime}}{\beta_{H}^{2}}\right)+\rho_{2}\left(x,x';t,t'\right)\left(\frac{yy^{\prime}}{\beta_{H}^{2}}\right)^{2}+\ldots,\nonumber \\
\nonumber \\
g\left(x,x';y,y^{\prime};t,t'\right) & = & 1+\bar{\mu}_{1}\left(x,x';t,t'\right)\left(\frac{y}{\beta_{H}}+\frac{y^{\prime}}{\beta_{H}}\right)+\bar{\mu}_{2}\left(x,x';t,t'\right)\left(\frac{y}{\beta_{H}}+\frac{y^{\prime}}{\beta_{H}}\right)^{2}+\ldots\nonumber \\
 &  & +\bar{\rho}_{1}\left(x,x';t,t'\right)\left(\frac{yy^{\prime}}{\beta_{H}^{2}}\right)+\bar{\rho}_{2}\left(x,x';t,t'\right)\left(\frac{yy^{\prime}}{\beta_{H}^{2}}\right)^{2}+\ldots,
\end{eqnarray}

\noindent where $\mu_{i}$, $\rho_{i}$, $\nu_{i}$, $\bar{\mu}_{i}$,
$\bar{\rho}_{i}$ and $\bar{\nu}_{i}$ are regular and bounded functions
no matter what the values of $\triangle x$ and $\triangle t$ are.

\vspace{2em}

\noindent \textbf{Step 2:} As $\beta_{H}\rightarrow\infty$, or $\beta_{H}\gg\triangle x$,
$\triangle t$, $y$ and $y^{\prime}$, the general expression (\ref{eq:BTZ cosh})
must match the pure AdS$_{3}$ background (\ref{eq:pure AdS L}).
From step 1, we know the leading term of $f$ and $g$ is the unit.
So, we have

\begin{eqnarray}
 &  & \cosh\left(\frac{L_{\mathrm{bulk}}}{R}\right)\nonumber \\
 & \simeq & \frac{\beta_{H}^{2}}{yy^{\prime}}\left[f\left(x,x';y,y^{\prime};t,t'\right)\left(1+\frac{\left(\triangle x\right)^{2}}{2\beta_{H}^{2}}+\ldots\right)-g\left(x,x';y,y^{\prime};t,t'\right)\left(1+\frac{\left(\triangle t\right)^{2}}{2\beta_{H}^{2}}+\ldots\right)\right]\nonumber \\
 & = & \frac{1}{2yy^{\prime}}\left[f\left(x,x';y,y^{\prime};t,t'\right)\left(\triangle x\right)^{2}-g\left(x,x';y,y^{\prime};t,t'\right)\left(\triangle t\right)^{2}\right.\nonumber \\
 &  & \left.+2\beta_{H}^{2}\left(f\left(x,x';y,y^{\prime};t,t'\right)-g\left(x,x';y,y^{\prime};t,t'\right)\right)+\ldots\right]\nonumber \\
 & \rightarrow & \frac{\left(\triangle x\right)^{2}-\left(\triangle t\right)^{2}+y^{2}+y^{\prime2}}{2yy^{\prime}}.\label{eq:cond.3}
\end{eqnarray}

\noindent On the other hand, when calculating the metric by using
eqn. (\ref{eq:derive metric}), it is easy to see that $f\left(x,x';y,y^{\prime};t,t'\right)$
and $g\left(x,x';y,y^{\prime};t,t'\right)$ enter $g_{xx}$ and $g_{tt}$.
But we know that for large $\beta_{H}$, it must reduce to the asymptotic
AdS in the Poincare coordinates, we thus conclude that $f\left(x,x';y,y^{\prime};t,t'\right)$
and $g\left(x,x';y,y^{\prime};t,t'\right)$ are independent of $x,x'$
and $t,t'$. Note $f$ and $g$ are dimensionless, we therefore rewrite
the general expression of the geodesic length as

\vspace{1em}

\noindent\fbox{\begin{minipage}[t]{1\columnwidth - 2\fboxsep - 2\fboxrule
\begin{equation}
\cosh\left(\frac{L_{\mathrm{bulk}}}{R}\right)=\frac{\beta_{H}^{2}}{yy^{\prime}}\left[f\left(\frac{y}{\beta_{H}},\frac{y'}{\beta_{H}}\right)\cosh\left(\frac{\triangle x}{\beta_{H}}\right)-g\left(\frac{y}{\beta_{H}},\frac{y'}{\beta_{H}}\right)\cosh\left(\frac{\triangle t}{\beta_{H}}\right)\right],\label{eq:general expression of L}
\end{equation}
\end{minipage}}

\vspace{1em}

\noindent with

\begin{eqnarray}
f\left(\frac{y}{\beta_{H}},\frac{y^{\prime}}{\beta_{H}}\right) & = & 1+\mu_{1}\left(\frac{y}{\beta_{H}}+\frac{y^{\prime}}{\beta_{H}}\right)+\mu_{2}\left(\frac{y}{\beta_{H}}+\frac{y^{\prime}}{\beta_{H}}\right)^{2}+\ldots\nonumber \\
 &  & +\rho_{1}\left(\frac{yy^{\prime}}{\beta_{H}^{2}}\right)+\rho_{2}\left(\frac{yy^{\prime}}{\beta_{H}^{2}}\right)^{2}+\ldots,\nonumber \\
\nonumber \\
g\left(\frac{y}{\beta_{H}},\frac{y^{\prime}}{\beta_{H}}\right) & = & 1+\bar{\mu}_{1}\left(\frac{y}{\beta_{H}}+\frac{y^{\prime}}{\beta_{H}}\right)+\bar{\mu}_{2}\left(\frac{y}{\beta_{H}}+\frac{y^{\prime}}{\beta_{H}}\right)^{2}+\ldots\nonumber \\
 &  & +\bar{\rho}_{1}\left(\frac{yy^{\prime}}{\beta_{H}^{2}}\right)+\bar{\rho}_{2}\left(\frac{yy^{\prime}}{\beta_{H}^{2}}\right)^{2}+\ldots
\end{eqnarray}

\noindent Moreover, from (\ref{eq:cond.3}), matching the $y$ direction
of the pure AdS$_{3}$ gives an important constraint

\begin{equation}
f\left(\frac{y}{\beta_{H}},\frac{y^{\prime}}{\beta_{H}}\right)-g\left(\frac{y}{\beta_{H}},\frac{y^{\prime}}{\beta_{H}}\right)=\frac{1}{2\beta_{H}^{2}}\left(y^{2}+y^{\prime2}\right)+\mathcal{O}\left(\frac{1}{\beta_{H}^{4}}\right).
\end{equation}

\vspace{2em}

\noindent \textbf{Step 3:} When two endpoints of a geodesic coincide,
the geodesic length vanishes exactly. Plugging $x=x^{\prime}$, $y=y^{\prime}$
and $t=t^{\prime}$ into eqn. (\ref{eq:general expression of L}),
we get
\begin{eqnarray}
\cosh\left(\frac{L_{\mathrm{bulk}}}{R}\right) & = & \frac{\beta_{H}^{2}}{y^{2}}\left[f\left(\frac{y}{\beta_{H}},\frac{y}{\beta_{H}}\right)-g\left(\frac{y}{\beta_{H}},\frac{y}{\beta_{H}}\right)\right]\nonumber \\
 & \rightarrow & 1,
\end{eqnarray}

\noindent which leads to

\begin{equation}
f\left(\frac{y}{\beta_{H}},\frac{y}{\beta_{H}}\right)-g\left(\frac{y}{\beta_{H}},\frac{y}{\beta_{H}}\right)=\frac{y^{2}}{\beta_{H}^{2}}.
\end{equation}

\vspace{2em}

\noindent \textbf{Step 4:} From eqn. (\ref{eq:BCFT}), the BCFT gives
the entanglement entropy of the half line. But we should replace $\Delta x$
by $\Delta x/2$ here since we use $\Delta x$ to represent the total
size of the entangled region,

\begin{equation}
S_{EE}=\frac{c}{6}\log\left(\frac{2\beta_{H}}{a}\sinh\frac{\triangle x}{2\beta_{H}}\right)\Longrightarrow L_{\mathrm{BCFT}}=R\log\left(\frac{2\beta_{H}}{a}\sinh\left(\frac{\triangle x}{2\beta_{H}}\right)\right).
\end{equation}
When $\triangle x\rightarrow\infty$, it becomes

\begin{equation}
L_{\mathrm{BCFT}}=R\log\left[\frac{\beta_{H}}{a}\exp\left(\frac{\triangle x}{2\beta_{H}}\right)\right].\label{eq:half 1}
\end{equation}

\noindent As illustrated in Fig. (\ref{fig:The-left-hand}), the geodesic
corresponding to this entanglement entropy connects $y=a$, $y'=\beta_{H}$.

\begin{figure}[H]
\begin{centering}
\includegraphics[scale=0.4]{fig_CAB}
\par\end{centering}
\centering{}\caption{The left hand side picture shows the geodesic identified from CFT
entanglement entropy. The endpoints are fixed at $y=y^{\prime}=a$.
The solid curve in the right hand side picture represents the geodesic
identified from BCFT entanglement entropy. The endpoints are fixed
at $y=a$, $y'=\beta_{H}$ and the entangling length is $\Delta x/2$.
\label{fig:The-left-hand}}
\end{figure}

\noindent On the other hand, by using the general expression of the
geodesic length (\ref{eq:general expression of L}), we have two other
ways to calculate the length of this geodesic. The first way is to
straightforwardly substitute $y=a$, $y'=\beta_{H}$, $\Delta x/2\to\infty$
into (\ref{eq:general expression of L}) to get

\begin{equation}
L_{\mathrm{bulk}}\rightarrow L_{\mathrm{half1}}=R\log\left(\frac{\beta_{H}}{a}\,f\left(\frac{a}{\beta_{H}},\frac{\beta_{H}}{\beta_{H}}\right)\exp\left(\frac{\triangle x}{2\beta}\right)\right).\label{eq:half 2}
\end{equation}

\noindent It is easy to understand that this length is one half of
the geodesic length connecting $y=y^{\prime}=a$, $\Delta x\to\infty$.
So the second way is

\begin{equation}
L_{\mathrm{bulk}}\rightarrow L_{\mathrm{half2}}=\frac{1}{2}\,R\log\left(\frac{\beta_{H}^{2}}{a^{2}}\,f\left(\frac{a}{\beta_{H}},\frac{a}{\beta_{H}}\right)\exp\left(\frac{\triangle x}{\beta_{H}}\right)\right).\label{eq:half 3}
\end{equation}

\noindent These three lengths (\ref{eq:half 1}), (\ref{eq:half 2})
and (\ref{eq:half 3}) ought to be identical, as illustrated in Fig. (\ref{fig:The-left-picture}).
We thus obtain

\begin{equation}
f\left(\frac{a}{\beta_{H}},\frac{\beta_{H}}{\beta_{H}}\right)^{2}=f\left(\frac{a}{\beta_{H}},\frac{a}{\beta_{H}}\right)=1.
\end{equation}

\begin{figure}[H]
\begin{centering}
\includegraphics[scale=0.4]{fig_half}
\par\end{centering}
\centering{}\caption{The left picture is given by BCFT. The middle picture is obtained
from $L_{\mathrm{bulk}}$ by setting $y=a$, $y^{\prime}=\beta_{H}$
and $\triangle x/2\rightarrow\infty$. The right picture is also given
by $L_{\mathrm{bulk}}$ from a different point of view, by setting
$y=y^{\prime}=a$ and $\Delta x\to\infty$. The solid lines in all
the three pictures describe the same object. \label{fig:The-left-picture}}
\end{figure}

\noindent The derivation of this constraint does not require $\beta_{H}\gg a$.
As long as $\beta_{H}$ is the upper bound of $y$, the derivation
is justified. Since $a$ is a varying cut-off not beyond $\beta_{H}$,
satisfying $0<a/\beta_{H}\le1,$ we can safely replace $\frac{a}{\beta_{H}}$
by $\frac{y}{\beta_{H}}$ to get:

\begin{equation}
f\left(\frac{y}{\beta_{H}},1\right)^{2}=f\left(\frac{y}{\beta_{H}},\frac{y}{\beta_{H}}\right)=1.
\end{equation}

\vspace{2em}

\noindent \textbf{Step 5:} An important lesson we learned from the
free CFT$_{2}$ case is that, in order to completely determine the
dual geometry, we need to know the geodesic length between $a$ and
$\beta_{H}$ with $\triangle x=0$, i.e. the vertical geodesic. To
be consistent, this particular geodesic length must be provided by
the CFT$_{2}$ entanglement entropy. In the free CFT$_{2}$, the IR
entanglement entropy precisely fits the requirement. In the finite
temperature CFT$_{2}$, there is no such IR entanglement entropy.
To solve this problem, we map the finite temperature system to a finite
size ($L_{S}$) system by replacing $\beta\rightarrow L_{S}$ and
impose the periodic boundary condition\footnote{To make the discussion simpler, we do not choose the equivalent replacement $L_S=i\beta_H$.}

\begin{equation}
\beta_{H}=\frac{\beta}{2\pi}\rightarrow\frac{L_{S}}{2\pi}.
\end{equation}

\noindent Therefore, the geodesic length between $a$ and $\beta_{H}=\frac{\beta}{2\pi}$
in the finite temperature system equals that connects $a$ and $\frac{L}{2\pi}$
in the finite size system,

\begin{eqnarray}
\mathrm{finite\;temperature} &  & \mathrm{finite\;size}\nonumber \\
L_{\mathrm{geodesic}}\left(a,\frac{\beta}{2\pi}\right) & = & L_{\mathrm{geodesic}}\left(a,\frac{L_{S}}{2\pi}\right).
\end{eqnarray}

\noindent Noting that $\frac{L_{S}}{2\pi}$ is the radius of the finite
size system with the circumference $L_{S}$, therefore, this geodesic
goes from boundary to the center of the circle, as illustrated in
Fig. (\ref{fig:The-geodesic-between}).

\begin{figure}[H]
\begin{centering}
\includegraphics[scale=0.4]{fig_map}
\par\end{centering}
\centering{}\caption{\label{fig:The-geodesic-between}The geodesic between $y=a$ and $y^{\prime}=\beta_{H}$
at a finite temperature system is mapped to a finite size system,
corresponding the radius of a circle from $y=a$ to $y=\frac{L}{2\pi}$.}
\end{figure}

\noindent We know that the entanglement entropy of a finite size system
is

\begin{equation}
S_{EE}=\frac{c}{3}\log\left(\frac{L_{S}}{\pi a}\sin\left(\frac{\pi\triangle x}{L_{S}}\right)\right),
\end{equation}

\noindent The maximal entanglement entropy is achieved by splitting
the circle into two equal regions, $\Delta x=L_{S}/2$. The corresponding
geodesic is nothing but a diameter

\begin{equation}
S_{EE}=\frac{c}{3}\log\left(\frac{L_{S}}{\pi a}\right),\qquad L_{\mathrm{boundary}}=2R\log\left(\frac{L_{S}}{\pi a}\right).
\end{equation}

\noindent It is then easy to get what we want

\begin{equation}
L_{\mathrm{radius}}=\frac{1}{2}L_{\mathrm{boundary}}=R\log\left(\frac{L_{S}}{\pi a}\right).
\end{equation}

\noindent We now map $L_{S}\to\beta$ to get the geodesic length between
$a$ and $\beta_{H}=\frac{\beta}{2\pi}$ in the finite temperature
system:

\begin{equation}
L=R\log\left(\frac{\beta}{\pi a}\right)=R\log\left(\frac{2\beta_{H}}{a}\right).
\end{equation}

\noindent Therefore, from the general expression (\ref{eq:general expression of L}),
as $x=x^{\prime}$, $t=t^{\prime}$, $y=a$ and $y^{\prime}=\beta_{H}$
, we have

\begin{eqnarray}
L_{\mathrm{boundary}} & = & R\log\left(\frac{2\beta_{H}}{a}\left[f\left(\frac{a}{\beta_{H}},\frac{\beta_{H}}{\beta_{H}}\right)-g\left(\frac{a}{\beta_{H}},\frac{\beta_{H}}{\beta_{H}}\right)\right]\right)\nonumber \\
 & \rightarrow & R\log\frac{2\beta_{H}}{a}.
\end{eqnarray}

\noindent We thus obtain

\begin{equation}
f\left(\frac{a}{\beta_{H}},\frac{\beta_{H}}{\beta_{H}}\right)-g\left(\frac{a}{\beta_{H}},\frac{\beta_{H}}{\beta_{H}}\right)=1.
\end{equation}

\vspace{2em}

\noindent For convenience, we summarize all the constraints we have
obtained for the general expression (\ref{eq:general expression of L})
of geodesic length:

\begin{eqnarray}
 &  & f\left(\frac{y}{\beta_{H}},\frac{y^{\prime}}{\beta_{H}}\right)-g\left(\frac{y}{\beta_{H}},\frac{y^{\prime}}{\beta_{H}}\right)=\frac{1}{2\beta_{H}^{2}}\left(y^{2}+y^{\prime2}\right)+\mathcal{O}\left(\frac{y^{4}}{\beta_{H}^{4}}\right),\hspace{1em}\beta_{H}\gg y,y',\label{eq:R3}\\
\nonumber \\
 &  & f\left(\frac{y}{\beta_{H}},\frac{\beta_{H}}{\beta_{H}}\right)^{2}=f\left(\frac{y}{\beta_{H}},\frac{y}{\beta_{H}}\right)=1,\label{eq:R4}\\
 &  & f\left(\frac{y}{\beta_{H}},\frac{y}{\beta_{H}}\right)-g\left(\frac{y}{\beta_{H}},\frac{y}{\beta_{H}}\right)=\frac{y^{2}}{\beta_{H}^{2}},\label{eq:R5}\\
 &  & f\left(\frac{a}{\beta_{H}},\frac{\beta_{H}}{\beta_{H}}\right)-g\left(\frac{a}{\beta_{H}},\frac{\beta_{H}}{\beta_{H}}\right)=1.\label{eq:R6}
\end{eqnarray}

\noindent From eqn. (\ref{eq:R4}) and (\ref{eq:R6}), we get

\begin{equation}
g\left(\frac{a}{\beta_{H}},\frac{\beta_{H}}{\beta_{H}}\right)=0.
\end{equation}

\noindent Since $a$ is a varying quantity, $y$ or $y'=\beta_{H}$
must be a zero of $g(y/\beta_{H},y'/\beta_{H})$ . Moreover, $g$
must be symmetric for $y$ and $y'$. So, the function form must be

\begin{equation}
g\left(\frac{y}{\beta_{H}},\frac{y'}{\beta_{H}}\right)\propto\left(1-\frac{y^{n}}{\beta_{H}^{n}}\right)^{\kappa}\left(1-\frac{y'^{n}}{\beta_{H}^{n}}\right)^{\kappa}\left(\cdots\right)
\end{equation}

\noindent On the other hand, from eqn. (\ref{eq:R4}) and (\ref{eq:R5}),
one gets

\begin{equation}
g\left(\frac{y}{\beta_{H}},\frac{y}{\beta_{H}}\right)=1-\frac{y^{2}}{\beta_{H}^{2}}
\end{equation}

\noindent It then easy to fix $n=2$ and $\kappa=1/2$ and

\begin{equation}
g\left(\frac{y}{\beta_{H}},\frac{y'}{\beta_{H}}\right)=\sqrt{\left(1-\left(\frac{y}{\beta_{H}}\right)^{2}\right)\left(1-\left(\frac{y^{\prime}}{\beta_{H}}\right)^{2}\right)}\left[1+\left(\frac{\Delta y}{\beta_{H}}\right)^{2}\left(\sigma_{1}+{\cal O}\left(\frac{y}{\beta}\right)\right)\right].\label{eq:g expansion}
\end{equation}

\noindent Similarly, from eqn. (\ref{eq:R4}), we get

\begin{equation}
f\left(\frac{y}{\beta_{H}},\frac{y'}{\beta_{H}}\right)=1+\left(\frac{\Delta y}{\beta_{H}}\right)^{2}\left(1-\frac{y^{m}}{\beta_{H}^{m}}\right)^{\delta}\left(1-\frac{y'^{m}}{\beta_{H}^{m}}\right)^{\delta}\left[\theta_{1}+{\cal O}\left(\frac{y}{\beta}\right)+\cdots\right],\label{eq:f expansion}
\end{equation}

\noindent where $m$, $\delta>0$ are some numbers.

\noindent The story is not over yet. In order to match eqn. (\ref{eq:R3}),
there must be $\sigma_{1}=\theta_{1}$. When applying eqn. (\ref{eq:derive metric})
to calculate the metric, noting a limit $\Delta x,\Delta y,\Delta t\to0$
is going to be imposed after making the derivatives, it is easy to
see that terms proportional to $\left(\frac{\Delta y}{\beta_{H}}\right)^{2}$
only contribute to $g_{yy}$, but not to $g_{xx}$ and $g_{tt}$.
So, looking at eqn. (\ref{eq:g expansion}) and (\ref{eq:f expansion}),
without altering the derived metric, equivalently, we are free to
pack all the corrections into $g(y,y')$ and simply set $f(y,y')=1$.
Finally, we get

\begin{equation}
\cosh\left(\frac{L_{\mathrm{bulk}}}{R}\right)=\frac{\beta_{H}^{2}}{yy^{\prime}}\left[\cosh\left(\frac{\triangle x}{\beta_{H}}\right)-\sqrt{\left(1-\left(\frac{y}{\beta_{H}}\right)^{2}\right)\left(1-\left(\frac{y^{\prime}}{\beta_{H}}\right)^{2}\right)}\left(1+\left(\frac{\Delta y}{\beta_{H}}\right)^{2}\cdot{\cal O}\left(\frac{y}{\beta}\right)\right)\cosh\left(\frac{\triangle t}{\beta_{H}}\right)\right].
\end{equation}

\noindent Applying (\ref{eq:derive metric}), we obtain the metric
of the BTZ black hole,

\begin{equation}
ds^{2}=\frac{R_{AdS}^{2}}{y^{2}}\left(-\left(1-\frac{y^{2}}{\beta_{H}^{2}}\right)dt^{2}+dx^{2}+\left(1-\frac{y^{2}}{\beta_{H}^{2}}\right)^{-1}\left(1+{\cal O}\left(\frac{y}{\beta_{H}}\right)\right)dy^{2}\right).
\end{equation}
Since the finite size system has the same topology as the finite temperature
one, a simple way to derive the dual geometry of the finite size system
is to use the transformation $\beta=-iL$, which leads to the pure
AdS$_{3}$ in the global coordinate. Similarly, one can also derive
the dual geometries of CFT$_{2}$ with topological defects under transformation
$\beta=-iL/\gamma_{\mathrm{con}}$.

\section{Discussions and conclusion}

In this final section, we discuss our results and give several inspirations
as well as conjectures. In summary, we demonstrated an approach to
derive three dimensional dual geometries, such as asymptotic AdS$_{3}$
and BTZ black hole, from the entanglement entropies of $\mathrm{CFT}_{2}$.
Our derivation only relies on the holographic principle without any
assumptions about AdS/CFT and bulk geometry. We list the steps as
follows
\begin{enumerate}
\item Identify the energy cut-off as an extra dimension.
\item Identify the entanglement entropy with geodesics length of the unknown
dual geometry. The geodesics are attached on the boundary.
\item Write down the bulk geodesic length by making the most general extension
of the geodesic ending on boundary to include the extra dimension.
\item Use properties a geodesics must respect, say, zero length for coincide
endpoints, to impose constraints on the bulk geodesic function form.
\item Use the IR-like entanglement entropy representing a geodesic whose
one endpoint stands on the boundary and another stretches into the
bulk of the unknown geometry, to restrict the bulk geodesic function
form.
\item Apply eqn. (\ref{eq:derive metric}) to derive the metric.
\end{enumerate}
It is convincible that our approach, from the derivation of the BTZ
black hole, works for all three dimensional geometries. As we explained,
since they have the same topology as the finite temperature CFT$_{2}$,
the finite size CFT$_{2}$ and CFT$_{2}$ with topological defects
can be easily worked out by simple transformations, though an independent
parallel derivation is desirable. Basically, we only need to consider
one representative for each topology. So, the next non-trivial steps
are to investigate the chiral CFT$_{2}$ or finite size thermal CFT$_{2}$.
Probably, the only obstacle is to find the IR-like entanglement entropies
from the CFT side. The existence of the IR-like entanglement entropies
is unquestionable, but it might be very hard to calculate from the
CFT side. A compromise is to borrow the IR-like entanglement entropies
from holographic calculation, if not so strict. For CFT living on
surfaces beyond torus, we believe the derivation still could be done,
but one has to know the entanglement entropies, both UV and IR-like,
which are very hard to obtain from CFT side. In addition, it would
be of great interest to consider the CFT$_{2}$ which are dual to
sourced gravity. We may learn more non-trivial things from these examples.

From our derivation, it is obvious that it is impossible to determine
the bulk geometry by UV entanglement entropy only. For higher dimensional
case $d+1>3$, the situation is complicated. One reason is that there
is no available method to calculate the metric from minimal surfaces.
Another issue is that even for one single topology, there are many
inequivalent gravitational structures in higher dimensions. We are
not sure if there will be other subtleties arise.

One interesting question is: \emph{Is it really necessary to identify
the entanglement entropy with the geodesic to derive the metric?}
The answer seems negative. As we know, the arguments of the entanglement
entropy include both spacetime directions $t$, $x$ as well as energy
cut-offs such as $a$, $\xi$, $\beta$....After identifying the energy
scale as an extra dimension $y$, we could introduce a generalized
entanglement entropy $\mathcal{S}_{EE}\left(t,t';x,x';y,y'\right)$
as follows:
\begin{itemize}
\item We denote the energy generated dimension as $y$. So the energy cut-offs
are different values on the dimension $y$.
\item Since $y$ is on the same footing as the ordinary spacetime directions
$x$, $t$, it is natural to generalize $S_{EE}(t,t';x,x';a,\beta,\cdots)$
to $\mathcal{S}_{EE}\left(t,t';x,x';y,y'\right)$ in the most generic
manner.\\$\Longrightarrow$This step corresponds to extend the boundary
attached geodesic to the bulk geodesic.
\item Under various limits, the generalized entanglement entropy $\mathcal{S}_{EE}\left(t,t';x,x';y,y'\right)$
is supposed to reproduce all the entanglement entropies of a specified
CFT, such as the UV or IR-like entanglement entropies . \\$\Longrightarrow$This
step corresponds to use various entanglement entropies to determine
the behaviors of the regular functions in our approach.
\item The generalized entanglement entropy $\mathcal{S}_{EE}\left(t,t';x,x';y,y'\right)$
should be renormalized, since the infinities of QFT are caused by
energy, which is now a new dimension. \\$\Longrightarrow$ This step
corresponds to demand vanished geodesic length for coincident endpoints,
and other consistencies.
\end{itemize}
Therefore, it is easy to see that our previous calculations naturally
fit the procedure. Now, we immediately have an interesting equation:

\begin{equation}
\frac{1}{2}\mathcal{S}_{EE}^{2}\left(x;x+dx\right)=g_{ij}\left(x\right)dx^{i}dx^{j}+\mathcal{O}\left(dx^{2}\right).\label{eq:new interpretation}
\end{equation}

\noindent All the derivations in this paper can be put into this pattern.
Consequently all GR quantities, such as the connection, Riemann tensor
etc. can be constructed subsequently. This equation indicates some
new interpretations:
\begin{itemize}
\item Spacetime is not an emergent structure from quantum entanglement,
but is quantum entanglement itself, simply looked from a different
angle.
\item Quantum entanglement with different lengths knits the spacetime.
\end{itemize}
There is another point deserves a special stress. Our results demonstrated
that when we treat the energy scale as a usual space-like dimension,
the CFT contains almost all the classical information of the dual
geometry, at least for $d=2$. In AdS/CFT correspondence, to compare
the correlation functions of the dual theories, limits are taken to
push the AdS$_{d+1}$ bulk-to-bulk correlation function onto the boundary,
and then match the CFT$_{d}$ correlation function \cite{Witten:1998qj}.
But how to lift the CFT$_{d}$ correlation function into the bulk
directly is still an open question. From our derivations, however,
once treat the energy scale as an extra dimension, after imposing
some consistent constraints, it might be possible to derive the bulk-to-bulk
correlation function from the boundary-to-boundary one. Think it over,
there are two equations to govern the dynamics of operators in QFT:
the Callan-Symanzik (RG) equation and the equation of motion (EOM).
The Callan-Symanzik equation tells us how the operators evolve with
respect to energy scales. The EOM determines the evolution of the
operators with respect to spacetime coordinates. Therefore, logically,
it is very nature to conjecture

\vspace{2em}

\noindent\fbox{\begin{minipage}[t]{1\columnwidth - 2\fboxsep - 2\fboxrule
\[
\mathrm{Callan-Symanzik\;(RG)\;equation\qquad+\qquad EOM\;on\;flat\qquad=\qquad EOM\;in\;the\;bulk},
\]
\end{minipage}}

\vspace{2em}

which implies a unification of the RG equation and the field EOM.

\vspace{5mm}

\noindent {\bf Acknowledgements}
We are deeply indebt to Bo Ning  for many illuminating discussions and suggestions.  We are also very grateful to Q. Gan, S. Kim, J. Lu,  H. Nakajima, S. Ying and S. He for very helpful discussions and suggestions. This work is supported in part by the NSFC (Grant No. 11875196, 11375121 and 11005016).

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