Happy International Women's Month. Today, we're gonna be talking about the best woman that has ever existed and that will ever exist. Today, we're gonna be talking about a gong girl. Happy International Women's Month to Amy Duncan. Oh! Amy Duncan is from Good Luck Charlie. This is the best woman that I've ever met in my entire life and I haven't even met her. Ooh! Gong girl, why, come on girl, I love you. This movie starts off with none other than Ben Affleck playing Nick Dunn, married to character Amy Dunn and it opens up with him basically in shambles. Like Ben Affleck really is one for the people because he does exude some sort of familiarity in all of his work. Basically, it starts off with him like walking into this bar that he owns that he like wanted to make. Like he like is like, I'm gonna, I'm gonna be a bar owner. I'm just having a bad day.   Then you cut to the Nick and Amy meet you. Amy. Amy, who are you? They're fun. They're sarcastic. They're different. They're not like other couples. They have fun little quirky jokes. She's different, he's different and that's how they find each other. Who are you? I'm the guy to save you from all this awesomeness.  What's the gift for fire? Wood. So would you get her? There's no good gift for wood. He like gets a call from someone that's like, I think like your cat's like running around. And so Nick is like, okay, like be right there. He starts walking to the house. He's like, the door is open. That's really weird. He starts walking through the house. He seems so fine.  And he's like, oh my gosh, wife kidnapped. Isn't that crazy? So then he calls the police and they come and they're walking through the house. And through the entire scene, these officers are suspicious. They're like, um, like what is up with this guy? They're suspicious of this guy because he's oddly like calm about his wife missing. Like he's like, yeah, I checked up here. Like everything's fine. Should I be concerned? There's a dress laid out and they're like, oh is it date nine? And he's like, yeah, it's our anniversary today. Like I feel like if my partner went missing the first, I wouldn't be so excited that it was like our anniversary today.  Like just like a little bit more somber due to the fact that like your partner could potentially be missing. I don't know about you.  Probably already in tears, already sobbing. Like the table was broken.  What do you mean your comment collected?  They're like, sure. The officers are like putting like post-it notes on things and that she's finding like, obviously you put post-it notes for like the evidence. She's finding little splatters of blood everywhere. And the other guy that's like her coworker is like looking at Nick and is like, oh, you slipped up there. You forgot to clean up the, you forgot to clean up that blood over there, huh? Oh, that's crazy that that one splatter of blood got there, huh? That's crazy. And Nick is like, and then it cuts to Nick Dunn being DNA swabbed and investigated by the police for the disappearance of his wife, Amy Dunn.  He later on gets more stressed out by like people thinking that it's him. But even in the first like clips of him being like investigated or whatever, he's very much just like, I gotta get investigated by the police later. Like it's really annoying. And as he's getting investigated by the police, he basically knows nothing about his wife. He like has no idea about her schedule, about anything that she does and answers questions completely incorrectly. So through all of Amy's journal entries, you're also seeing the, you know, the life that they were living together and how, you know, like when they were first together, they had a very happy marriage. They knew each other like the back of their own hand. You know, they had inside jokes. They were spontaneous.  They were like getting each other the same gifts because they knew each other so well. It was like all a lot of fun. It was the perfect relationship ever. And they, you know, they got married and they were having a blast. They were literally two peas in a pod. Cut back to the investigation. They find out about another character named Noelle. And Noelle is Amy's best friend, which according to Nick Dunn earlier on, he said that Amy had no friends and that she didn't hang out with anyone. Turns out she has a Southern little housewife best friend that is super concerned about where she has been. So one of the most iconic scenes from the movie is the press junket, the like press release for Amy Dunn's disappearance. So Amy's parents finally come down to the police station to make the statement with Nick and there's all these cameras and journalists being like, what happened? It does a very big deal. You know, a housewife has gone missing in the middle of the day in broad daylight. Is this like, should we be concerned about like, you know, something coming up? It's a very big deal in suburbia. And so Nick Dunn goes up and he's like, hi, my name is Nick Dunn. I'm the husband of Amy Dunn. I'm here to bring awareness to her disappearance. She disappeared from my house two days ago. If anyone has any information on her whereabouts, please contact me immediately.  And then the parents go on this long spiel. She's a scholar, she's kind, she's beautiful. Please, we're putting together a search party for her. Please go to www.findamie.com. Everyone's like, that's how normal people would respond to someone going missing. So they finish it, they're like, Amy, come home. And then they like stand next to the missing poster of Amy. And this is the moment one of the most iconic shots of I've ever seen in my entire life. And Ben Affleck like nails it. Like, I seriously don't think there could have been a better person to play Nick Dunn. I think Ben Affleck was perfection. Like, I seriously don't know how they got such the perfect person to play Nick Dunn. He goes in front of the poster.   Like it's so cheeky, it's so like chagrin eating manner. Like it's, it's like, like I wouldn't be surprised if he like put up a peace sign. He was like, and even his sister who's watching him at this press junk and it's like, ooh. Even like, she's weird now. She's like, sir, everyone is slightly suspicious of Nick. Like, without saying it, they're like, what is up with this guy? Like, what is wrong with him? Even his own blood relative is like, what is wrong with my brother? The next part of this movie is a scavenger hunt. Amy leaves scavenger hunt notes, presumably for their anniversary. And the detectives on the case says we should follow these clues to see what her last moves were and what her thought process was before she went missing. I know this one. They go on this little thing, they like find the first clue, they find the second clue.  Nick goes to his father's house where he finds the third clue, but then the police show up and they're like, I thought this was gonna be like the house. And he was like, no, like it's like not the brown house. Then he starts taking on initiative with the scavenger hunt by his lonesome. He's like, I wanna figure out this scavenger and all on my own. The thing is that like a lot of this stuff ends up just being, cause he's weird. Like it's not incriminating, but the fact that he does these things are just because he's a freak. So finally getting into some of the nitty gritty parts of Amy and Nick's relationship, we go back in time to kind of the downfalls of their relationship. We see that Nick has been laid off on his writing job. That's one of the common interests that they both had was that they were both writers. And he's gotten laid off and he's not working. He's on the couch. He's gaming. He's literally a gamer boy.  And Amy says something that I dig up within this freaking movie. I dig it up. She goes, I don't mind. I just wish she asked first. And you expect me to hate her?  You're asking me to hate her during international women's history month? And if you couldn't already hate this character more, Nick is revealed to have a secret lover, Emily Radikowski. Emily Radikowski is his secret lover. Emily Radikowski, my body. She's an author. She's also a writer. All these people are writers. And she is his lover. And it's like, that's enough. That's all we needed to know. That's all we needed to know. I don't need to know much. I just need to know that. You're cheating on her. You're cheating on your missing wife. Your wife that just disappeared. That is the breadwinner of this household. She's the breadwinner of this household. Everything is under her name. The house, your bar, the cars. And you're cheating on her. And then she disappeared. And then she disappeared. What did you expect me to think? You're the worst.  Cause how are you going to have your wife that pays for everything and then cheat on her? How are you gonna have a wife that provides all that for you and then cheat on her? And with author of my body, Emily Radikowski. With her of all people. So through seeing the downfalls of their marriage, we have a flashback from Amy's journal of, you know, the realities of their relationship once they're in Missouri and that she ends up confiding in him that she thinks that it's time for them to start having children. He makes it into a fight. He's like, we won't be the type of people that have a baby to save our marriage.  And then she pushes him and then he pushes her and she falls into the railing of the stairs.  I was like, whoa. What do we have here according to Amy's journals and according to the police investigation that we're currently working on? We have this guy, presumably, doesn't know anything about his wife. You don't know if she has friends.  Sure, y'all are married. His wife went missing and he seemed to be very nonchalant about it. He's smiling in the pictures next to his wife's missing poster. And he just seems to be tiptoeing around everything and then we find out that he's cheating and then throughout the journal entries, we also see that he was abusive towards his wife.  For Valentine's Day, I thought I'd buy a gun. Amy Dunn decides she needs to buy a gun to protect herself from her husband. She doesn't know what to do. She can't go back to her parents because if she went back to her parents, she would have to tell them the truth. Amy Dunn is scared for her life in the presence of her husband according to everything that she writes in her journal. So we have the visual, vigil, vigil. So the town has a vigil for Amy and they have their candle lit. They're hoping to gather the town so that they can still find Amy. And Nick does a speech where he's like, I love my wife so much. I want you, if anyone can help us, please give your help if you know anything about my wife's disappearance. Please come to us, help us find Amy. And for anyone that may be too scared to ask, I have nothing to do with my wife's disappearance. I have been cooperative with the police. I have not hired a lawyer. I am not guilty of anything. I love my wife so much and I just want her to come back home. And then Amy's BFF Noelle shouts. She goes, what did you do to your pregnant wife, Nick? You tell Mammy we're six weeks pregnant. What did you do to your pregnant wife? Did you tell them that she was six weeks pregnant? I thought this was a classy party. Um, Gag, what do you mean six weeks pregnant and missing? And everyone is like gag. They're like, oh my God, mom. The equivalent of like throwing tomatoes at a person. They're like, get him, mom him, get him. You made a pregnant woman's, you made her disappear. And they don't even know that he's cheating on her. So whoa, whoa, whoa. You're saying that Amy was pregnant. And so then he's at his house with the detective and the detective takes this opportunity to start asking Nick some questions. So Nick is like, I didn't know that she was pregnant. I don't even know who this Noel chick is.  Like they barely know each other. And the detective is like, they look like pretty good friends to me. Boom, picture evidence.  Hikes, book club, babies, eating. They were together. They were literally, they were kind of, they were kind of together. And while we're at it, let's talk about the scene of the crime. Let's dig a little deeper while we're here. The scene of the crime was kind of a little bit sketchy, right? So the police detective starts going on to the investigation on the crime scene. She showed that it was obviously stays because for someone to struggle, there was not a lot of things in disorder surrounding the scene. And then they did a loomis test of, you know, the chemicals within near the blood spatter, which was in the kitchen. And they found that a lot of blood had been mopped from the kitchen. Amy lost a lot of blood in there, Nick. A lot. Oh my God. Nick was like, what the freak? Why would someone mop something up and then stage something else? And they were like, well, cleaning up and staging suggest homicide, which would be people coming from inside the house, people that know Amy, kidnapping and messy crime scenes lead us to think of people from outside the house, like homeless people, which Nick had suggested earlier on the film that there was a high homeless population within their town. You know, we have a pretty serious homeless problem in our neighborhood. You maybe you guys should check that out. So they're like, what the freak is the truth? And then they go, what are you and your wife fight about? What are you and your wife argue about? What pisses you off? Oh, money, lack thereof. Me and my ex, the same. And then they show him that he's in $100,000 within credit card debt. And he's like, well, what the freak? I did not, I didn't make any of those purchases. I don't even golf. Why would I need all these golf clubs? And they reveal that Nick has also bumped up Amy's life insurance to 1.2 million. And Nick said that Amy wanted him to do that. Gagged. Messy crime scene cleaned up and then staged as a kidnapping or breaking credit card debt that he didn't even know about and bumped up life insurance. I don't know about you guys. I don't know about you guys. I'm just giving you the facts. I'm just giving you the facts of what we're watching here. I'm not trying to insert an opinion. I'm not trying to persuade you in any way. I'm not trying to make you tilt one side or the other. I'm not trying to do that to you. But for me, someone's a little fishy. Drum roll, please. Finally, we are now onto Amy's perspective. Amy's perspective, day one of presumably missing the morning of her disappearance. I am so much happier now that I'm dead. Amy starts the narration finally talking to the audience of how she makes herself disappear to frame her husband for the murder of herself. We did it. We did it, Joe. Yes, guys. Amy Dunn was behind the entire thing. She's kind of like really mother for that. Like mother, mama, mummy, dearest. Like that's a true serve of the century.  He took everything from me and that's basically murder to me. So I'm gonna frame him for actual murder. He embarrassed me and no one embarrasses me. No one embarrasses mummy. So you will go to jail for killing me. Yeah.  She took out a bunch of money to make it seem like they had money issues and a lot of credit card debt, which they like just didn't. She just made that happen. She waits for him to start his day so she can finally stage her disappearance on the morning of July 5th. So Amy goes into the details of how she faked her disappearance, how she extracted the blood from her arm, how she made this crime scene to have just the right amount of mistakes to be suspicious of anyone or to look further into it. All the journal entries she wrote prior in hopes that the police would find the journal that she had within the beginning parts of her story being true and then as things got more serious, money troubles, abuse and the need to get away becoming fake. She creates an elaborate story to frame Nick for being a horrible husband, someone who she was scared of, someone who she needed to escape from, who she was scared something was gonna happen to her if he got mad enough at her. And she does this all in hopes that one day she will die, she will go out with a bottle of pills and throw herself in the river where they will find her body and truly and finally incriminate Nick Dunn for the murder of Amy Dunn and hopefully he will die one day too. And as she's telling this, she finally enters one of the best portions of a monologue I've ever heard. Bind you, this entire monologue of her describing her incriminating Nick and faking the crime scene is magnificent. The way that Rosemond speaks, the way that she has this monotone yet typed voice, it's clean, it's crisp, but it's a monotone and frigid at the same time where there's just that little lack of actual emotion that makes it scary. Nick loved the girl that I was pretending to be. Cool girl. Men always use that, don't they? That's their defining compliment. She's a cool girl. Cool girl is hot, cool girl is gay, cool girl is fun, cool girl never gets angry at her man. She only smiles in a chagrin loving manner and then presents her mouth for faking. I gotta eat that.  That's how deeply she devoured this scene. It was impeccable. It was amazing. It was life changing. No life should ever be the same after witnessing this scene.  She was the perfect image for him. She stooped to his level to be the girl of his dreams but for him, she forged the man of her dreams. She made him rise to her level. She inspired. She made him get back on his ass, get back on his game. Who would have thought millions of girls would watch Gone Girl and love the so-called villain of the story because I do. Let me ask the audience. So things back by Nick Dunn are getting a little bit more crazy. They're getting a little bit more suspicious because Nick has caught on that Amy is framing him for the murder of her. They find the shed of all the debt that she had created for them, all the stuff that she had bought as the ending anniversary gift.  So Nick decides to lawyer up.  I'm coming back for everything. So he decides to get his lawyer, Tyler Perry. Well, right now it's a he said, she said. And all the clues that she left on her anniversary scavenger hunt actually have clues to incriminate him for not only his infidelity, but also the murder of Amy. And all while this is going on, Amy is making some acquaintances at the place that she is staying at. So the friends that she's making, they basically had her her story because she said that her ex-husband cheated on her and she ran away because he was abusive towards her. And they basically were like, I don't really think that's true.  So she's like, fuck, what am I gonna do?  So Amy's new plan has to do with her ex-boyfriend. This ex-boyfriend was mentioned on earlier in the film for someone that had been sending letters to Amy. Amy uses this as her backup plan. She calls up her ex-boothing. She's like, hey, I need your help. I ran away from home. He forced me to leave. He said that he'd find me and kill me if I didn't do what he wanted. I lost the baby and they meet up and he's like, oh my God, Amy, like you need to tell the police. And she's like, no, I can't tell them. Like if I go to them, everyone will hate me. Everyone will like, no one's gonna believe me that this happened. Everyone thinks I may have gone missing. And he's like, I'm gonna say you want my lake house so you can stay and be safe, Boogie. This is also Neil Patrick Harris. You saved me. And then as Amy's running off with her ex-boothing and then Nick is getting ready for his press interview on clearing his name, a press release happens from none other than Emily Radikowski. She comes to the stand and she says, I was involved with a married man and I apologize for everything that I've done to hurt everyone in this situation. The parents caught her and they said, we believe that Nick Dunn has something to do with the disappearance of our amazing Amy. We hate him. He's lied to us again and again and he cheated on his pregnant wife. And he's the reason why she's gone. And like everyone is gagged. Everyone is gasping. Everyone is like literally like they need a second. Everyone's pausing because what do you mean?  Isn't it? Isn't it Nick? So Nick does his interview and we'll get to that in a second. But Amy is with her ex-boo and Neil Patrick Harris. He's weird as well. The thing is that like all the men that she frames are like also already weird. Like they're not like good guys. Granted, they didn't do the crime but like they're also kind of freaks at the end of the day. So with Nick's press release and everyone on his side suddenly and Amy living with her ex-boo and Neil Patrick Harris. Everything seems to be going better for Nick.   Amy puts in a hit note. She calls the line and says there's some strange activity outside of Margo Dunn's house in the back shed. And so they come up. They say we have a search warrant for this property Nick Dunn. We're searching this motherfucker. And so they search the shed and they find the shed that has all of the debt that Amy took out in Nick's name. I'm talking the golf clubs at TV, a robot dog. And they're like, mm-hmm. So you're hiding, you're hiding stuff from us. You just keep lying to us, Nick.    The murder weapon to a doll. There was a doll earlier on and I'm not gonna get into the details. There was a doll earlier on.  And they start reading him the journal entries from Amy Dunn that they had earlier on but they didn't tell him about. And they start reading it off to him that the last note was that she was fearful for her life that her husband may kill her. Nick Dunn, you are under the rest of the murder of Amy Dunn. It's now 26 days after the appearance of Amy Dunn. And Amy Dunn is living up the perfect housewife life for another man, Neil Patrick Harris. And this is where Amy starts her new framing technique. She's going to frame her ex-bookie Neil Patrick Harris for a new crime. And while he was at work, while he was doing stuff, she had forcefully inserted something into her body. So if she had been tested, it would be consistent with wounds of a sexual assault.  Like I want you bad. And then they proceed to have intercourse. And as they're having intercourse, she waits till he completes himself and slits his throat as he does. So there is evidence within her and it looks like an act of defense. And then she comes up to her house.  She has returned home. She is making sure that Nick Dunn is not going to jail for the murder of Amy Dunn. So Amy Dunn comes back and says that she was kidnapped by her ex-lover and taken, held hostage, and she was being and sexually assaulted before she could kill him and make a way out. Nick Dunn is not responsible for the murder of Amy Dunn. She was kidnapped by an ex-lover.   And she goes, checkmate, motherfucker. Not only did I frame you for the murder of me, I'm also clearing your name with a Bob. New Bob, who this?  Watch out for girls that just got a Bob because they're about to frame you for their murder. So in the most magnificent ending, Amy Dunn survives. Amy Dunn survives being kidnapped. She was kidnapped guys. This crazy ex-guy came and kidnapped her and she survives.  I want to leave you. I don't want to be with you.  I did all this to get back to you. When I saw you on TV telling me that you loved me and you needed me back home, that's the Nick guy marry. That's the Nick I wanted. She doesn't want much from him. She just wants him to be the guy that she married.  That is such a good line.   Like I don't like using like explicit words like that because like I get it. Like I don't know why it doesn't like flow off of my like American lingo as well because like it's more of like a European thing.  Yeah, that's something serious. It's something serious. It's something that we should all take notes of.  And you guys are also having a baby together. Congratulations, congratulations. You're having a baby together. This has got to be one of the best movies I've ever seen. This has got to be one of the most lively movies I've ever seen. It's just so empowering. If you don't watch Girl and Girl and you don't feel empowered to leave whatever shipback husband you have, then I don't know what it is made for. I think that a lot of the times people like to look at Gone Girl as like, you guys know that like Amy Dunn was the villain, right? Oh my God, boring, boring, sloppy, lazy. If you just say that in response to anything that I talk about with Gone Girl, then you can go get out of here. You can get out of here because you're missing the whole best part of the movie. There's a big, grand part of the movie that is so fascinating about a woman who was trapped in a loveless marriage who decides to have the audacity to cheat on her after she sacrificed so much for him. She's literally pays for the house, the house, cars, and everything is in her name. She moved to freaking Missouri, Missouri. She moved to Missouri from New York City, concrete jungle for him, and he decides to cheat on her.    That's what she wanted to happen to him. Can we blame her? I don't know.  I think she is the cunt that we all married.  Stand up to all the married cunts. Cause I applaud you. I love Amy Dunn, and I think this is the best movie that's ever created. I love David Fincher and not in a way like other people do cause I'm different and I'm better and I'm special. I'm a special David Fincher fan. Let me hear it for all the hot David Fincher fans out there. I wanna hear you scream. There's a certain side of David Fincher fans that are just not the same as us.  Let me hear all the hot girl David Fincher fans fan. Stand up. Young girl, more like come on girl. Let's go party. Bye guys.


So today we want to understand how the correlators match between the two descriptions. So remember, we have on the one side string theory and ADS3 cross S3 cross 4, and that we described in terms of an SL2R Resomino-Witton model at level 1 plus an SDN equals to 1 version, say. I mean, I'm a bit, I'm writing this again in the NSR formulation. I had the impression that the hybrid formalism didn't reach enormous level of enthusiasm. So let's just stick with this. And you can understand this as well in that description as in the hybrid description. So this is a string theory on this, so this is the world sheet. And this is due to the symmetric orbit fold of T4, which is the CFT. And what we want to do is we want to calculate correlation functions in that theory. And the boundary of ADS3, you should think of as being a cylinder, and if you compactify it, it will be a sphere. So these correlators will always live on the sphere. So this will be correlators on the sphere. So this will be things like you take your space time sphere and you insert various points where certain fields from the symmetric orbit fold theory sit. And throughout today, the coordinates in this space will always be denoted by X. So X will always be a coordinate of the correlation functions of the symmetric orbit fold of T4. And this is to be distinguished from the parameter on the world sheet. Now the world sheet will not always be a sphere, although for most of today I'll concentrate on the case where the world sheet is a sphere. So in general here I will have a world sheet of whatever genus is appropriate, but for most part I'll look at G equal to zero. And the coordinates here I will denote by Z. So just to be clear, whenever you see a Z, it will be a coordinate on the world sheet. Whenever you see an X, it will be a coordinate in the symmetric orbit fold theory. And I'm looking at sphere correlators of twisted sector states in the symmetric orbit fold theory. So this will be something like, if you wish, so the simplest case would be a correlator of this sort of type, say a three point function, X1, X2, X3 are three points on the sphere and this is say the twisted sector ground state in the W1, W2 and W3 cycle twisted sector. So that's the sort of thing we are going to calculate in the X space, in the symmetric orbit fold space. And I first of all want to explain to you how you calculate this correlator in the symmetric orbit fold irrespective of whether it comes from ADS or not.  And then I want to relate it to a calculation in the world sheet theory and I'll explain to you how this correlation function of X will be reproduced from the string theory perspective. You know, from the string theory perspective life is a bit complicated because as you know in general, we'll have to integrate over the Z variables. The insertion points of the vertex operators on the world sheet are modular. You have to integrate over, say for a four point function, you would have to integrate over the cross ratio of the Z's and what we should manage to achieve is that after we've done this integral, we should reproduce the correlator as we calculate it from the dual CFT. So that's the roadmap we have in mind. So how do you calculate this correlator? I mean, there are different ways of doing it but there's one smart way of doing it that will fit very nicely with what we have in mind. So how do you calculate that? Now remember what a sigma w field looks like. Remember this has the property that if I start, let's take w to be the cycle from 1 to w and then if I start, remember in this simple case of a single boson, if we take dx i around this field, we ended up with dx i plus 1. So that was the effect of this twist field. The twist field makes sure the first one goes to the second. The second one goes to the third. The third, the w1 goes to the first one.  That's the structure of this twist detector field. So now what's the smart way of calculating it? Well, the smart way of calculating it is to change coordinates and to go to the covering space. So we want to find the covering space locally. And what do I mean by the covering space? Well, the covering space will be a map gamma and I'll write it in a map of z. So this set at this stage, it's not obvious that it will have anything to do with this. At this stage, it's just the parameter I call z to distinguish it from x. But I call it z because ultimately it will be connected to that set. But that's far from clear at this stage. So gamma of z will be a map from some neighborhood of gamma to the minus 1 of x to a neighborhood of x. So let's call this x0, which is the point where the twist fields it. And this map should be such that z gets mapped to x0 plus z times xz minus, and let's call this point z0, z0 to the w. So what you're looking for is a map that sort of w fold covers the point x. So why is this a smart thing to do? Well, think about in the covering, so here's the covering space. This is the covering space. In the covering space, we have the point z0. This gets mapped down in our real space. This gets mapped to x0. And this map has the property that if I go once around up here, if I multiply z minus z0 by e to the 2 pi i z minus z0, then I go w many times downstairs, right, because of the power of w. So if I circle once around here, I circle w many times around here. And you see after going around w many times, it's become single valued. So this is the map that locally maps the individual coordinates dx i single valued, right, because as you go once around upstairs in the covering map, you go w times around downstairs. So you go from x1 to x2 to x3 to x4, up to xw, back to x1, x2, each of them gets mapped to itself. So from the point of view of the covering space, this map will be single valued in terms of the individual dx i's. So in some sense, you've removed the effect of this twist by going to this covering space. Now locally, it's clear how you do it.  And now the idea is, if you calculate a correlator of that kind, what you should be doing is you should patch together the local covering maps of that type into a global covering map, where you have some remand surface on top that has the property that near each of these insertion points, it's doing exactly that. And it's therefore undoing the twisting that happened downstairs. Now, at first you may think, is this possible? Okay, so let's give an example. So the simplest example is you take three points with w equals to three. So let's take w equals to three at x1, w equals to three at x2, and w equals to three at x3. And for simplicity, set x1 equal to 1, x2, 0, 1 and infinity. And let's choose also the sets. The pre-image is to sit at 0, 1 and infinity. That is without loss of generality because you see by the Möbius symmetry, I can always move three points to three points. So then I claim the corresponding map, the corresponding covering map in this case is of the form z to the four minus two times z to the three divided by one minus two z. Okay, I've learned this by heart. I wouldn't be able to spot it as I'm not that good a calculator. But what you can do, you can either be smart and see it in your head or you can type it into Mathematica. But if you check what this, how this behaves, how does this behave near zero? Well, near zero you see the first term goes like minus two times z to the cube. So this is exactly of that form that map zero to zero plus some coefficient which turns out to be minus two in that case times the third power of z minus z zero. And then what happens near one? So if I calculate gamma of z minus one, but let's write it gamma of z is of the form if I expand it around one, it goes like one minus two times z to the minus one to the power three for z near one. And then at infinity what I have to look at is gamma to the minus one of one over u. So this is the coordinate chart around infinity and you map it back to run zero. And then you can read off that it goes like minus two u cubed, which describes it for u equal to zero, which is the same as z goes to infinity. So this is an example of such a map. I mean, it's holomorphic. It's in fact the map from the sphere to the sphere. It's fourfold. I mean, every generic point has four pre-images because it's a fourth-order polynomial divided by a fourth-order polynomial. Well, the fourth-order polynomial in the denominator has been degenerate, it's fourth-order over fourth-order. So it has four pre-images and it maps zero to zero, one to one, infinity into infinity. And at each of the point, it has a threefold covering. So it will undo this correlator. And the idea is that instead of calculating this correlator, I applied this covering map, which is a holomorphic map. Correlation functions transform nicely under coordinate transformation under holomorphic maps. So I lift it up to a map of the covering space. And then on the covering space, this map is, this correlator is effectively trivial because, you see, once you've moved it up to the covering space, then everything is single valued. Any trace of this funny sort of insertions where things got sort of a twisted have disappeared. And because I'm looking at this twisted sector ground state, the idea is that once I've lifted it up to the covering space, it's just the vacuum correlator. Every single trace of this twist field has disappeared. The only purpose of the twist field in life was to impose this twisting. But by this local coordinate transformation, I've undone the twist things or nothing is left. So therefore, this correlation function is just a conformal factor associated to this conformal transformation. So I just have to calculate a conformal anomaly that comes from this coordinate transformation and then I've calculated this correlation function. That's the idea. That's the smart way of defining, of calculating the correlation functions of symmetric or before theories. Now- Question? Yes. Shouldn't we expect some remnant of the parameter capital N? Oh, sorry. What I'm saying here is, well, no. I mean, sorry, yes and no. Yeah, I'll come to that in a second. You're right, but let me explain one thing and then I'll explain it. Okay, so the point is, so in this example, the covering map I've written down is a sphere. So in this example, this example is a map from the sphere to the sphere. But in general, the covering surface needn't be a sphere. So the covering surface, so the covering map, so if you ask what are the possible covering maps and you allow yourself your covering space to have arbitrary topology, then you get different covering maps. There's not a unique covering map. There's a finite number of covering maps. And in general, the covering space that covers, say, this configuration needn't be a sphere. In fact, there is the Riemann-Hurwitz formula. So the Riemann-Hurwitz formula tells you exactly what the genus of the covering surface is in terms of the parameters w and the number of covering. So the Riemann-Hurwitz formula is i is equal to one to n. So if on endpoint function, w i minus one over two plus one minus g, where g is the genus of the covering surface is equal to m and m is the number of pre-images. Number of pre-images at a generic point. So for example, for this example, you see w is equal to three. So three minus one over two is equal to one. There are three points. So this gives you three plus one is four. It's equal to four. So the genus is equal to zero. So this is an example of a map that's from the sphere to the sphere, a covering map and it's fourfold. It's a fourfold cover. Each point in the downstairs sphere has four pre-images in the upstairs sphere. But in fact, there's also a torus covering. There is in fact also a torus covering where you choose the number of pre-images to be three and then you find out that the genus has to be equal to one. So there are a number of different coverings in particular for this configuration as at least a sphere covering and a torus covering. Now coming back to Francesco's questions, what is the role of the genus of this covering map and what you discover is that the correlation functions in their large and expansion, the one over N dependence of the correlators is controlled by the genus of the covering space. So the way you should think about it is that if you calculate these correlators, what you have to do, you have to sum over all possible covering maps. You get some conformal factor associated to the covering map gamma. And if you expand this in a one over N expansion, what you realize, and this was an observation of Rastelli and Razumat, is that the correlation function, if you ask, so the diagram for a given gamma, it contributes as N to the one minus G minus N over two. But this N is the N of the symmetric orbit fold. So this N, so I'm thinking of taking N large. I'm calculating this correlation function in a larger expansion. And what I'm saying is that the prescription, this is sum over all possible covering maps from the point of view of conformal feature that's summing over the different conformal blocks. And a given covering map contributes, if you think about it in the one over N expansion, in a way that depends on the genus of the covering map. And so that's a fact that you can just read off from the structure of the symmetric orbit fold correlators. I mean, there's a theorem that you can calculate them in this fashion. And then if you look at each term there, you can say each term goes as N to the one minus G minus N over two, where G is the genus of the covering map of the covering surface that I'm considering. Now remember that N is to be identified with, or one over N is to be identified with the genus. So we know from the point of view of the dual string theory that the genus should go like, that the string coupling constant should go like one over square root of N. So if I write this in terms of the genus, this will go like the genus of the string, sorry, the string coupling constant to the minus two G. And therefore it'll suggest that the genus, but that is exactly how the genus expansion of a string theory should appear. So what this suggests is that somehow the contribution that comes from a specific genus here, if I interpret it via the ADS CFT duality, that should be the contribution that from a world sheet perspective comes from a world sheet of the corresponding genus. So the genus expansion that's natural here that comes from just expanding out the one over N expansion of the symmetric orbital correlators should match with the genus expansion as you write it down in conformal feed theory because the one over N expansion behaves exactly like the G string expansion under the usual correspondence between the string coupling constant and one over N. I think there's a question there. Sorry, when we do an orbit fold usually in 2D, we get some symmetry that we usually call quantum symmetry corresponding to this orbit fold. And usually when we have a symmetry, the correlators will fit in some, we will follow some selection of rules or something related to this symmetry. And here we can understand something of this, what we get from this or... Well, first of all, I think the quantum symmetry is usually you can undo the orbit fold by taking the orbit fold with respect to the quantum symmetry. That only works in the Abelian case. You can't undo non-Abelian orbit folds. But obviously you still have... So the obvious quantum symmetry just works for Abelian orbit folds because the twisted sector just... You can characterize the twisted sector by some phase and then the phases give you a selection rule and if you orbit fold by that, you go back to the original theory. For the non-Abelian orbit fold, you can't do it. This has to do with this complication that the twisted sectors are now labeled by conjugacy classes rather than group elements. So there's no easy way to reconstruct the full original theory from taking the orbit fold of the orbit fold. At least that's as far as I know, that's not known. There is no... The story about non-invertible symmetry in this case. Well, maybe there's something like that, but certainly none of the conventional orbit folds will do the job. Now, maybe there is some more general construction, but I'm not aware of a general construction. And actually I talked to Yifang about this and I think that would be interesting to see what one could use non-invertible symmetries here, but that's certainly not known. Now, obviously there are selection rules. So for example, this correlator can only be non-zero if you can find three cycle permutations that multiply to give you the identity. So, but that you can do, right? I mean, you could, for example, take three times the same one, but that would give you the torus covering. Here you have to take one that involves four different numbers, but there are different ways of multiplying three cycle permutations together to give you the identity. But beyond that, I don't think there's any global symmetry for a non-invertible orbit fold like the symmetric orbit fold. Thank you. There's another question over there. Sorry, just to, so the cover coordinate you call Z. Yes. And then the same Z as the word should you connect. So this is anticipating what I just explained. I mean, at a covering space at the moment, I shouldn't have called Z. I should have called babysitter because it has at this stage nothing to do with the Z. But over there I explained because of the one over N dependence, you will expect this Z to be the Z and the upshot of my talk today is that this Z is the Z. So I'm sort of suggesting to you already where I'm going, but I'm using a notation that will go down. But at the moment, this has nothing to do with the world. This is just some abstract covering space. Hi. So I want to be sure I understood this sum over gammas. So for three point function, it seems that there are possibly a unique, so are you referring to higher point function where you want to take all these? No, but for the three point function, I mean the question is how many channels do you have, right? Yeah, exactly. So you have mined the four or higher point function, I guess. So you have mined an OPE where you have an infinite number of three point function then an infinite number of gammas, something like that. No, no, I mean for a fixed three point function, there are different channels. So I'm probably going to screw this up, but so there's one obvious channel which is, so you pick the permutation one, two, three, one, two, three and one, two, three, right? That would be one solution, right? But remember in the symmetric orbit fold, you have to pick representatives. So when I say a w-cycle tested sector, what I mean is I take any w-cycle, but I haven't specified that it's necessarily one, two, three, right, because I'm labeling the states by elements in the conjugacy class. So this will be one configuration that will contribute to this three point function, but there's also another and if I can find my notes, I can tell you what it is. Maybe I can't, here it is, there's one, there's another one, three, four, one, three, two, four and one, two, three, that will also do the job, right? So one goes to two, two goes to four, four goes to one, two goes to three, three goes to two, three goes to one, one goes to three, four goes to three, three goes to four. So that's also another solution. So there are two conformal blocks for the three point function of a three-cycle twisted sector operator because that is a different channel than that channel, right? Because you see that involves three elements in the covering space, that involves a four-fold cover. So that's the genus one contribution and that's the genus zero contribution. And when you calculate this correlation function, you have to sum over the conformal block coming from here and from here and what I'm saying is in the larger limit, this conformal block dominates over this conformal block because this conformal block will be down by a power of one over n because it has genus one relative to genus zero. And then if you look at more complicated correlators, you get finite number of contributions from different genus and their one over n dependence is controlled by the genus of the corresponding covering surface. Okay, thanks. So another mild question is there are other correlators, right? This twist feels, why are you focusing on this specific set of correlators matching and not? Well, first of all, I'm looking at a single cycle. I'm looking at a single particle states from the string theory perspective. So I'm only going to look at states that live in a single cycle twisted sector. So they'll always live in a cycle of one length because otherwise there would be multi-particle states and I wouldn't see in perturbative string theory. Now, obviously I don't necessarily have to look at the ground state. I could look at some excited state and the punchline is if you look at some excited state, then you don't get a trivial correlator on the covering surface, but you get a correlation function of the seed theory on the covering surface. So you can relate it to something relatively simple and the interesting bits that I'm trying to explain to you really hinges on which cycle it comes from rather than exactly which state you look at. The covering map is always the same covering map. Any further questions? Okay, good. So now, I've written here a sum. Now, people know that when I write sums, I don't necessarily mean sums. Maybe sometimes I mean integrals because I've been sloppy when I wrote down the spectrum of the SL2R, the seminal written model. But the point is here I actually mean a sum. Covering maps are rare. Generically, they don't exist. Covering maps are something that's... They're really a finite number of covering maps. There's not a continuum of covering maps. And in order to explain that, and that will be crucial for what we are going to see, let's look at a case where you look at a gene as zero covering. So let's look at a gene as zero case. So let's use the Riemann-Hurwitz formula for the example of G equal to zero. So then this term isn't there. Okay, so if you have on the sphere, what does the most general holomorphic map look like? Well, my function gamma of Z will be of the form minus PmM of Z divided B by P plus M of Z, where P minus plus of M of Z is a polynomial of degree M. Right, so this is the most general M to one map from a sphere to the sphere. I mean, it's just the ratio of polynomials. Now, what is the condition for this to be a covering map? Well, the condition has to be that gamma of Z should be equal to xi plus something of order Z minus Zi to the wi. So I have in mind, I'm looking at an endpoint function. So I'm having an endpoint function. And the points sit at coordinate xi and they have winding wi. And I want the pre-images to sit at position Zi. So I'm asking, so what this means is that near Z equal to Zi, it has to map to xi. And then it has to be order Z minus Zi to the wi. That's the covering map condition. But now, you see, we can just plug this answer into this formula. So what does it say? But now maybe people can't see anymore. So maybe I'll continue over them. So what this means is that minus PmM of Z divided by P plus M of Z is equal to xi plus order Z minus Zi to the wi. So now multiply by P plus M of Z. This is some generic polynomial. And then bring this to the other side. And what do we learn? We get the equation Pm minus of Z plus xi P plus M of Z must be of order Z minus Zi to the wi. That's the constraint I have to satisfy in order for this to be a covering map, right? I mean, if I multiply this through, this is a generic polynomial. It will generically not have any zero or anything at Z equal to Zi. So therefore, you can just multiply it through. So xi multiplies P plus, you bring P minus to the other side and then it has to be of order Z minus Zi to the wi. So let's count how many parameters we have in our answers and how many constraints we have to satisfy. Well, how many parameters do we have? Well, this is polynomial of order M. So it has M plus one parameters. This is a polynomial of order M. So it has another M plus one parameter. So I have two M plus two, three parameters. But one parameter drops out because I'm looking at the ratio, right? The overall scale drops out. So the number of parameters in my answers is equal to two M plus one. On the other hand, how many constraints do I have? So the number of constraints I have to satisfy is equal to, well, at each point xi, so at sum for each i, so sum i is equal to one to N. I get how many constraint? I get wi constraint because I have to make the order zero term vanish, the order one term vanish, up to the order w minus one term vanish. So I get wi many constraints. But now let's look at the Riemann-Hurwitz formula. You see the Riemann-Hurwitz formula tells you that the sum over N of wi is equal to two M minus two. So this is equal to two M minus two plus N. That's the number of constraints that you have to solve. If you want, if you, so what we are specifying here is we are specifying the coordinates xi. We specify the w i's and we specify the z i's. So if you specify all of that information, we have two M minus two plus N constraints, but we have only two M plus one free parameters. So the number of parameters minus constraints is equal to two M minus, plus one minus two M minus two plus N. And what you see is that this is equal to three minus N. So what does this tells you is that you see generically if N is bigger than three, you have more constraints than parameters. Which is another way of saying that generically the covering map doesn't exist. The covering map will only exist if the z i's satisfy N minus three constraints. I mean, the z i's must lie on a hypersurface of co-dimension N minus three in order for you, at least generically, to have a chance to find the covering map because the number of parameters is simply too small to satisfy all the constraints that you have to satisfy. But this number is a very suggestive number, right? Because if we are doing world sheet theory on the sphere, if we're calculating a string diagram, we have to calculate, we have to integrate over the z's. Three of z integrals are for free because of the möbius symmetry. So we always have to do N minus three z integrals. We have to integrate N minus three cross ratios. But you see here that there are exactly N minus three constraints that allow us for the existence of a covering map. So this has the structure. So this suggests the following way in which this wants to work. So suppose I can write these constraints in terms of the variables z four up to z N, then I would expect that from the world sheet point of view. So far I've only done symmetric orbital analysis. Now I make a leap of faith. Now I say, how could I possibly get it from the string theory side? And there's a very, very natural way in which you can get it from the string theory side. And the natural way is the following. The vertex operators on the string theory, well they're going to depend on x i and z, x one and z one and up to w n, x n and z n. So they should be a sum over all the covering maps labeled by J and for each covering map, I'm going to get an n minus three delta functions. And the delta functions will be of the form gamma J to the minus one of x four, x i minus z i times some conformal factor. So I propose that is a natural way in which the world sheet theory should look like. And why is this natural? Well this is natural because once you do the string theory, remember in string theory we have to do the integral over dz four up to dz n of this correlation function, right? Because if you calculate an endpoint function on the sphere, there are n minus three moduli we have to integrate over and here for simplicity I've chosen z one, z two and z three to be fixed. So then I have to integrate over z four up to z n.  It will pick out exactly the configuration of z for which the covering map exists. And then I will end up with the sum of our covering maps with the conformal factor of the covering map which is exactly what the symmetric obfault answer is. But what I'm saying is if you stare at the symmetric obfault answer given the fact that the covering map only exists on co-dimension n minus three, what it suggests is that the world sheet correlators are actually delta function localized on the low chi, low psi, where the covering map exists. And then the integral over the moduli of my world sheet will automatically produce the structure of the correlation function that's calculated by the symmetric obfault theory. It's a bit of a, I mean, probably takes a moment to get used to it. I mean it's very unusual to think about a correlation function to be delta function localized. If you open a yellow book on a random page, that's not what CFT correlators look like. CFT correlators are rational functions. So this looks very far-fetched, but if you think about what would the world sheet theory have to be like in order to reproduce the symmetric obfault answer, in some sense that's the only thing that can really work, right?  Or at least if this was true, then it would manifestly reproduce the symmetric obfault answer because the integrals over the free parameters in my string theory just kill the delta functions that appear here, and I'm left with some over-covering maps as calculated from the symmetric obfault perspective. Okay, so that's our claim. Our claim is that this world sheet theory has exactly this property. Now this, as I just said, this is a bold claim because this is not what normal conformer features look like, but our conformer feature isn't entirely normal. It has this funny spectrally float sectors, it has all this funny stuff, and unfortunately I don't have that much time left, but what I want to do in the rest of my lecture is to convince you that we have actually proven that this is true for our world sheet theory. Our world sheet theory has exactly that structure, and thereby it manifestly reproduces this symmetric obfault answer, and this does not only work at genus zero, it also work at higher genus. So if you do the corresponding analysis here, you get a minus G term from here counting for the genus, the moduli of the genus, and they are always exactly getting fixed by these constraints, and it reproduces exactly the structure that I was outlining here, namely that the world sheet gives you exactly the contributions. The world sheet of genus G gives you the contributions that come from the covering surface of genus G. That's exactly how it works out. Now, this is a little bit, so what I want to spend the rest of my, so unless there are questions, what I want to spend the rest of my lecture is to explain in how you can prove that this identity is true in our specific world sheet theory. And obviously I won't give the full proof because I don't have enough time, but I want to sketch how this works. But please ask me anything if this is unclear what I'm trying to do. I'm not sure whether that means that I've lost all of you or whether it's abundantly clear. But if you don't ask me, there's nothing I can do. Okay, so how do we try to prove this from the symmetric, from the world sheet perspective? Now from the world sheet perspective, so there's one key insight, which in retrospect is a pretty obvious key insight, but maybe it's something that people didn't quite appreciate in its full significance. So in the first three lectures, what I explained to you was how the states of the symmetric orbital theory can be reintroduced from a physical state on the world sheet theory. So I established a map from a vector space of states here to the vector space of states here. Now, as you're probably familiar in two-dimensional conformal filter, in fact in higher dimension conformal filter as well, there's what's called the operator state correspondence. To every state, you can construct an operator. But here we have to be a little bit careful exactly what we mean by that. So suppose I pick a state in the symmetric orbital theory. So this is a state now. And in the symmetric orbital theory, I do the usual operator state correspondence, which means I identified states with the corresponding vertex operator at x equal to zero. That's the usual way in which you do it in conformal feed theory. At x equal to zero, the vertex operator creates the state. But this is at x equal to zero because this is the symmetric orbital T4 that lives in x space. Now, what I explained to you is that this state corresponds to a state that which I call phi hat in the world sheet theory. And in the world sheet theory, I also have the operator state correspondence. So to that state, I can associate it with a vertex operator, which I'll denote by phi hat of z. And that's the vertex operator that's associated to the state. And again, I make the association as is conventional to say that a vertex operator at zero, that light on the vacuum produces for me the state. Just like you always do in 2D CFT. Now, let me remind you how you can characterize the z dependence of such a vertex operator. The z dependence of that vertex operator is actually totally obvious because you see we know what the translation operator is on the world sheet. The translation operator is e to the z times l minus one. And therefore, this vertex operator is really necessarily of the form e to the z l minus one. The vertex operator associated to z equal to zero e to the minus z l minus one, right? The z dependence is always trivial. It just comes from conjugation with the translation operator and the translation operator on the world sheet is l minus one. Where l minus one is the we're a sorrow generator on the world sheet. But so far, this describes the vertex operator that corresponds to the state being inserted at x equal to zero. So I should have really written here, this is the vertex operator corresponding to x equal to zero and z and this is the vertex operator associated to x equal to zero and z, z equal to zero because you see I've identified it with a state and the state is really associated to the vertex operator in the dual CFT being at x equal to zero. So now I can ask what is the vertex operator associated to this symmetric orbital state if I insert it at an arbitrary position x and an arbitrary position x in the dual CFT and an arbitrary position z on my world sheet. Now, given that I've just explained to you how it works for the world sheet CFT, it's now obvious what I have to do. You see also in space time, there is a translation operator, there is a Möbius group and therefore the x dependence of the space time vertex operator comes from conjugation by the translation operator in space time. What's the translation operator in space time? Well, it's l minus one space time but l minus one space time from the point of view of my world sheet theory is simply j plus zero. So obviously this is equal to e to the x j plus zero e to the z l minus one, the vertex operator associated to the state x equal to zero at coordinate zero and then it's just conjugated by the corresponding and what I'm using here is that j plus zero is the same as l minus one in space time. So that's how my vertex operators on the world sheet will depend on x and on z. And there is no ambiguity in how to define this once I've identified the states, it's clear how it will depend on x and z independently because I know what the translation operator is on the world sheet and I know what the translation operator is in space time. And note, there is no ordering ambiguity here because j plus zero commutes with l minus one. l minus one and j plus zero commute so there's no funny ordering I have to choose whether I first translate in space time, first on the world sheet and then in space time or first on the world sheet and then in space time doesn't matter. They commute with one another and that's how the vertex operator will depend on x and z. So this is what I meant with this symbol here. This is the vertex operator that will depend on where I insert a state in the dual CFT. Where this crosses is in x space and it depends on z that is where the corresponding vertex operator is inserted on my world sheet surface and the z's is the stuff I have to integrate over when I do a strength theory so that at the end of the day I'm only left with the function of the x's. So what I have to do is I have to calculate the correlation functions of these vertex operators. Okay, so now I at least know which problem I have to solve and now I can go about trying to solve that problem. Now the key trick, so you may think what's the big deal about this funny exponential but this funny exponential has a very significant impact and what's the significant impact of that exponential? So I was planning to, it's most easily explained in terms of, if I can find my notes, it's most easily explained in terms of introducing the free field realization of SL2R. How should I do it? Can I have maybe a few more minutes? Yes, yes, sure. Okay, so I have to, so at this stage I have to introduce a free field. This is for the purpose of making this a little bit more pedagogical, otherwise it becomes a little bit complicated. What I claim is that this has a free field realization in terms of what I like to call symplectic bosons. I think string theory people tend to call this beta gamma system. So these are bosonic coordinates and they're characterized by the following commutation relations, psi alpha r eta beta s is equal to epsilon alpha beta delta r minus s. So alpha and beta take values in plus minus, epsilon plus minus is equal to plus one and epsilon minus plus is equal to minus one and the other epsilon's are zero. So these are, this is like a beta gamma system. And then I claim this beta gamma system realizes the SL2R, F and cut smoothie algebra at level one. If I write the plus, the j plus and j minus generators there, the bilinears of the form. So j plus is eta plus psi plus, j minus is eta minus psi minus understood as fields. And then j three is equal to a minus a half times eta plus psi minus, always normal ordered, minus a half times eta minus psi plus. Okay, so I hope that this is not scary for you. I mean, these are free fields. This is like a beta gamma system. So this is not so different from stuff you've seen. And what I claim is if you look at these combinations, they're all spin a half field. So these guys are spin one fields. And when you calculate the OPEs, you realize SL2R at level one. That's what I claim. For example, in string, sorry. Is it exactly the beta gamma system or further bosonized like in string theory, we have this eta psi system? Well, I mean, it's bosonic field to have a simple pole OPE, right? That I thought is called beta gamma system, right? I mean, the corresponding field psi alpha eta beta of Z and W goes like epsilon alpha beta over Z minus W. That's the OPE, right? I thought that's what you call a beta gamma system. Yeah, maybe I'm assuming that's what I mean. That's what it is, right? And if I write it in terms of mode, that's what it looks like. And then if I look at the normal order products, that gives me SL2R at level one. Thanks. Okay, so that is, that is, now I've lost my notes again. Okay, so this is something you can check and it's not a rocket science to check it. And it's true. It requires maybe a little bit care, but it's true. So we need that. And then remember, in the world sheet theory, we don't have just the SL, so I'm looking at just the bosonic piece of it. I'm just looking at the SL2R. I'm ignoring the fermions. So the SL2R bosonic algebra can be generated by these fields. And then the other thing we need to know is what spectral flow does. I explained to you a spectral flow does to these fields, but actually this spectral flow comes from a natural spectral flow that you can formulate in terms of the free fields.  So this is how spectral flow acts on the free fields and that just generates the spectral flow as we've seen in the generator. As you see, the plus components get shifted downwards, the minus components get shifted upwards. So the plus component gets shifted downwards by two minus a halves, so it gets shifted down by minus one. The minus component gets shifted, moved up by two plus one, so it gets shifted up by plus one and then the zero mode stays the same and the normal ordering term gives you the shift in the zero mode term. Anyway, I mean it's, if you don't trust me on that, then I probably have no chance to explain you the rest. This is not rocket science, this is something you do. Okay, so now why is this funny formula, why am I so excited about this formula about the correct definition of the vertex operators? So why is this such an important formula? You see, what we want to do in order to understand these correlators, we want to understand the word identities with respect to these three fields. So what we're going to do is we're going to look at the correlators where we insert one of the symplectic bosons and see what the poles are of the symplectic bosons with respect to everybody else. So what we're interested in is the pole, the OPE of xi plus of zeta with a vertex operator of this kind. Now, what's important is you see, j plus is proportional to eta plus and xi plus and eta plus and xi plus commute both with xi plus because the only commutator that's not trivial is between xi plus and eta minus, but there's no eta minus inside. So this xi plus doesn't care about the fact that there's an e to the xj plus zero standing here because it just goes through. So its OPE will behave exactly as it originally behaved and because it gets shifted downwards, what you find is that this OPE you can calculate. So this goes like the sum over r of the mode numbers and it'll be, so let's specify the spectrally flowed sector by this upper index and then what you find is that this is going to be xi plus r minus w over two in this w fold spectrally flowed sector, you shift this down by w times minus a half acting on phi hat and then x and z.  That's just the usual way the OPE works and this is independent of the x variable because it commutes with the e to the xj plus zero term and then if you look at it, what's the first term that survives? If phi hat is the highest rate state, then this will survive is r is a less than w over two. So the leading term will go like zeta minus z to the minus w plus one over two and then it'll go like this term. So that's the structure of this OPE. So the xi plus OPE with this is this form. Okay, so that's how xi plus works. But now what about xi minus? Well, when you try to do the same calculation with xi minus, you see previously we didn't have this factor of x here. We didn't have this factor of e to the xj plus. So it's like the OPE we would have calculated previously except we have to ask what happens when I move xi minus past e to the xj plus and that you can easily calculate what is xi minus over zeta with e to the xj plus zero. Well, if you think about it, you can expand this out, j plus zero has a term that matters because the xi minus is going to e to the eta plus and spit out xi plus and what you're finding is that this is equal to e to the xj plus zero times xi minus of zeta. That's the term where nothing has happened and then you get a correction term and the correction term is goes like minus x times xi plus of zeta.  And it's purely algebraic, right? I mean, there's no guesswork involved here. This just follows from the commutation relations of these modes with one another. So why the xi plus is invisible to the e to the xj plus, the xi minus isn't and therefore when I calculate this OPE, if I move it past the e to the xj plus zero, I have to calculate the OPE of minus x times the OPE I've just calculated and then I have to calculate it. So this will go like minus x times this OPE except I sort of moved out the, well actually there it doesn't matter because that commutes with x and then it has a term that goes like plus the OPE I would get from xi minus and the OPE I get from xi minus is actually much more regular than normal because under spectral flow, the xi minus modes get shifted the opposite way and if you calculate it, what you find is it goes like, the correction term goes like plus zeta minus z to the power w minus one over two times vw of xi minus of zero on phi x and z. So the important fact is that xi plus and xi minus have a different OPE when you bring them near these vertex operators and more importantly, this OPE depends in an intricate way on the coordinate x where you've inserted the vertex operator in the dual CFT and algebraically it comes from this identity. This is the key identity. The key identity is that the translation operator in space-time has an impact on xi minus. Geometrically what this means is you see we've picked the spectral flow direction but under translation in space-time, the spectral flow direction isn't really invariant and you pick up correction terms and the correction term you pick up is this correction term. Obviously when you set x equal to zero, this goes away, but for x not equal to zero, you get this correction term. And that is, I mean it really comes all out of just doing things properly. I mean there was no ambiguity in how to define this operator. There's no ambiguity in calculating that. There's no ambiguity in calculating these OPEs. This is just dictated to you once I've defined the theory. There's no freedom here. Okay, so now there's one small additional fact I have to mention and then we'll see what happens. So unfortunately this is a super string theory and super string theory correlators are awkward because there are things like picture changing and stuff. So we have to incorporate picture changing and the correct way of incorporating picture changing in this context is that you, is that you, that the real correlator has some additional picture changing operators inserted. And what this means is that the real correlator that you have to calculate is not just the product of all of these vertex operators, but the real correlator we have to calculate is the product that has alpha equals to n minus two of these sort of picture changing type operators. And then you have this product of i is equal to one to n v w i of phi hat i x i and z i. So that's the physical correlator. That's the physical correlator.  They're all regular. Okay, so now I've accumulated all the data that I need in order to explain to you the water identities. Yes. My question is about the correlators that do not necessarily preserve the winding number. So the spectral flow variable. So... They are totally singular. These are the vertex operators where you've moved the points here all into one position. These are non-zero, right? So you can compute them. But they are singular. They are not the right correlators to consider. These are the, so I mean if you ask about it from the point of view of the symmetric orbit fold the winding number will only add if the two points sit on top of each other. Then the winding number obviously adds. If they don't sit on top of each other, it never adds. But what would be the string theory interpretation? Because from the string theory side we know that the winding numbers can be in principle violated in this case. Right, but so these are correlated. So previously people have calculated these correlators but they didn't introduce the X variable. If you don't introduce the X variable what this means is that all your points are bunched up on one point. So then from the point of view of the dual CFT you're calculating not the correlation function you're calculating the limit of all the points coming together. Then obviously winding number adds. If you're interested in the honest correlator of the symmetric orbit fold then you have to evaluate them at different points and then winding number will not add as we will see. And it's all characterized in terms of covering maps. But that's the data that really controls the correlators of the symmetric orbit fold that controls the one over N dependence and so on. So the correlators that were looked at previously are sort of the funny singular limit. And it's not the correct, I mean in some sense you must be able to retrieve them from there but they are very, very singular. That's the limit where the covering surface degenerates. The regular correlators where they sit at different points winding number can never add. And you get the structure of diffusion rules that are appropriate for the symmetric orbit fold. That's exactly how people calculate symmetric orbit fold correlators. Okay so now our plan is that we are looking at these correlators and we want to analyze the word identities and in order to analyze the word identities what we have to do we have to insert psi plus and minus and now I've managed to mislay my crucial piece of paper but maybe I find it here. Yes I find it, it's the final one. So the end is near. So I propose to define the following functions. So okay let me just write it down. It's a product of one to n of zeta minus x i to the w i plus one over two divided by the product of alpha is equal to n minus two of zeta minus u alpha and then it's the correlation function where I insert psi plus of zeta and then I put the product of the w and the product of the v's. So it's basically this correlator and what I do is I insert psi plus and minus inside the correlator here and then I multiply it with this funny prefect. Okay so now what are these functions? Well let's think about it. So let's ask where do these functions have a pole? Well originally you would think these functions have a pole when psi plus and minus go close to one of the vertex operators. But if psi plus comes close to a vertex operator the pole is of order minus w plus one over two and if psi minus comes to the vertex operator the pole is of the same order because of this term. There's also a pole of that order but in those types I've exactly removed that pole by multiplying it by zeta. Sorry this is zeta minus zi to the wi plus one over two. So this factor is designed to kill this pole, right? I mean psi plus near each of these vertex operator has a pole of order w plus one over two and I multiply it by a pre-factor by hand that is just a zero of that corresponding order. So this has no poles at zeta equal to the zi's and at zeta equal to w it has a simple zero and I've cut divided by the simple zero so it still doesn't have a pole. So this function doesn't have a pole in zeta and a function that doesn't have a pole in zeta and is holomorphic is called the polynomial. So p plus and minus of zeta are polynomials because they don't have any zeros and they don't have any poles. Now what's the degree of the polynomial? Well the degree of the polynomial is what? Well I have to calculate the degree of the pre-factor so I get the sum from i is equal to one to n wi plus one over two. Then I get minus, so this is the pre-factor coming from here that obviously adds to the degree and then this subtracts to the degree so then I have minus the sum of alpha is equal to one over n minus two times one. These are the things I divide out through and what is the degree of this? Well the degree of this you can determine by asking how does it behave when zeta goes to infinity but you know how it behaves when zeta goes to infinity because zeta is a spin and a half field so it'll go like one over zeta. So there'll be a minus one from the behavior. That's the zeta behave, the large zeta behavior of this correlation function which simply follows from the fact what zeta does, xi plus does out at infinity. Now if I find my help sheet then I will realize that this expression here is exactly equal to, so now I can't find my sheet anymore but this expression is exactly equal to m so this goes exactly like, I can rewrite this, this goes like, this goes like the sum over i is equal to one to n of w i minus one over two because I can subtract these things out and then plus one because I'm only subtracting out n minus two ones and I have to add in another one so this is the degree of this map and this is exactly m as coming of the Riemann Horowitz formula that I described before. So these are polynomials of degree m, right? I mean they have no poles, that's the degree and the degree is exactly the degree as predicted by the Riemann Horowitz formula. Now what about the characteristic identity that characterizes the covering map? Remember what we have to look at is p minus of zeta minus x plus xi of p plus of zeta but look at this, if you look at this combination then xi minus of zeta plus x times p plus, this kills exactly this term. So what's left behind is it goes like something to the order, so this goes like order zeta minus zi to the power w i minus one over two but then remember there's an additional pre-factor standing outside that goes like zeta minus z to the power w i plus one over two plus w i plus one over two. So you see that this combination goes exactly like zeta minus z i to the w i because w i minus two plus w i plus one over two is w i. So therefore these functions are polynomials of degree m that satisfy the characteristic equation to be defined in the covering map so I conclude that the covering map gamma of zeta is actually equal to p minus divided by p plus. I see the other way around. There's a minus sign, I've forgotten.  So this is minus of zeta divided by p plus of zeta. So what this tells you is that the world sheet theory allows you to recover the covering map. The covering map arises naturally out of the world sheet theory by virtue of looking at these correlation functions and just dressing them up so that they become polynomials of the right degree and the construction guarantees that these correlation functions of the world sheet theory produce for you exactly the covering map. But remember the covering map doesn't generically exist. But if the correlators produce you the covering map then something with this construction must go wrong for those configurations where the covering map doesn't exist because we know the covering map generically doesn't exist. Now if you trace through this argument carefully what you see is that the thing that has to go wrong is that p plus has to go to zero. This construction must break down. And then you can show that if p plus goes to zero then the correlator without the insertion of psi plus must also go to zero. So you prove from that that because you can reconstruct the covering map from the correlators by inserting an additional field and you know that the covering map will sometimes not exist. You can prove that whenever the covering map doesn't exist the correlators even without the insertion of the psi fields must be equal to zero. I, the correlators on the world sheet can only be non-zero at the location where the covering map exists. And if you do a little bit more work you can show that it's actually a delta function type localization property so that when you do the integral over the world sheet modular you really recover the sum of covering maps. So that's the reason why this world sheet theory knows about covering maps and what it really tells you is you see this identity, this identity tells you that psi plus and psi minus want to be the twister variables and this is like the incidence relation that recovers the spacetime coordinate from the twister variables. So what this strongly suggests is that the symplectic bosons play the role of the spacetime twisters and the localization, the incidence relation of the spacetime theory is encoded by the world identities of these symplectic boson theories. And this is the idea that has then subsequently motivated us to try to imitate this also for ADS-5 and but obviously I don't have any time anymore to sketch that, but here in the ADS-3 case we can really nail this down and it matches exactly with what you would expect from the symmetric orbital answer. So my time is truly up so I'll stop here and thank you for your attention. Thank you.