WEBVTT 00:00:00.000 --> 00:00:01.000 So we're very happy to have 00:00:01.000 --> 00:00:10.000 Very happy to have Bailey from Chinese Academy of Sciences. She's going to tell us about ADS 3 quantum gravity finite and expansions. 00:00:10.000 --> 00:00:17.000 And there's going to be dinner with the speakers. So if anyone's interested, please see Tom after. 00:00:17.000 --> 00:00:23.000 Okay, thank you very much for the introduction, and thank you for the opportunity to give 00:00:23.000 --> 00:00:29.000 to, uh, to present this work. So this is done with 00:00:29.000 --> 00:00:31.000 Chihoni from last, um… 00:00:31.000 --> 00:00:36.000 November, and also some work in progress. 00:00:36.000 --> 00:00:43.000 Uh, so the general context of this work is that we want to understand finite and quantum gravity. 00:00:43.000 --> 00:00:46.000 via ADSCFT. 00:00:46.000 --> 00:00:52.000 So remember, so take the ADSFT, and then there's three levels. 00:00:52.000 --> 00:00:54.000 Depends on how it appears. 00:00:54.000 --> 00:01:04.000 So the weakest form is that when n goes to infinity, so that would be the correspondence between the QFT in the planar limit and the classical gravity. 00:01:04.000 --> 00:01:12.000 And then the intermediate stage is the form is the — you still take the large N, but you want to include all the 00:01:12.000 --> 00:01:17.000 1 over n expansion. So that would be the corresponding. 00:01:17.000 --> 00:01:24.000 The correspondence between non. 00:01:24.000 --> 00:01:30.000 Non-planar QFD and then the perturbative gravity. 00:01:30.000 --> 00:01:38.000 So the case I want to discuss today is the strongest form when you really take the n to be finite, say even 1. 00:01:38.000 --> 00:01:40.000 So then, um… 00:01:40.000 --> 00:01:43.000 the correspondence, the finite, um, 00:01:43.000 --> 00:01:46.000 Kft, and then at the 00:01:46.000 --> 00:01:52.000 Quantum gravity side, you need to include all the non-perturbative effects. 00:01:52.000 --> 00:01:56.000 So, on the community side, of course, you can easily just set n to be finite. 00:01:56.000 --> 00:02:00.000 The question is, how do you reproduce this final statement? 00:02:00.000 --> 00:02:02.000 From the gravity side. 00:02:02.000 --> 00:02:09.000 So I want to, in this context, discuss a pretty old puzzle or 00:02:09.000 --> 00:02:12.000 code that streaming exclusion principle. 00:02:12.000 --> 00:02:16.000 Uh, so, so here's this statement. 00:02:16.000 --> 00:02:20.000 So on the boundary side, 00:02:20.000 --> 00:02:22.000 We compared the degrees of freedom 00:02:22.000 --> 00:02:26.000 of the final n relative to angles infinity. 00:02:26.000 --> 00:02:29.000 So, um… 00:02:29.000 --> 00:02:32.000 So then the degrees of freedom at Finan should be. 00:02:32.000 --> 00:02:35.000 You can easily see that there should be, uh, an… 00:02:35.000 --> 00:02:39.000 fewer, or should be fewer than the degrees of freedom at the. 00:02:39.000 --> 00:02:45.000 and goes to infinity. For example, for the gauge theory, for the, say, for the 00:02:45.000 --> 00:02:47.000 for the Ink 2 for Super Yamyu. 00:02:47.000 --> 00:02:54.000 And this reduction of the degrees of freedom from angles to infinity to finite n. 00:02:54.000 --> 00:02:57.000 is realized by the trace relation. 00:02:57.000 --> 00:03:00.000 Meaning that, uh… 00:03:00.000 --> 00:03:05.000 the trees… so… so say the 00:03:05.000 --> 00:03:07.000 They can, um… 00:03:07.000 --> 00:03:13.000 So you have an embedded matrix, and then you take the trace of x to the some power. 00:03:13.000 --> 00:03:21.000 If the power is larger than n, so then you can rewrite it, you can express it in terms of the trace of x to the power. 00:03:21.000 --> 00:03:23.000 When the power is less than N. 00:03:23.000 --> 00:03:32.000 So similarly, for the symmetric orbifold, so this reduction is realized by just the constraint that the total cycle length cannot be bigger than n. 00:03:32.000 --> 00:03:36.000 And on the other hand, you… now you do the same. 00:03:36.000 --> 00:03:43.000 On the gravity side. But from the gravity side, you know that when you include a non-perturbative effect. 00:03:43.000 --> 00:03:48.000 You have more stuff, right? I mean, now you have black holes and the ball brain and stuff. 00:03:48.000 --> 00:03:53.000 So naively, when you include a non-protective effect, you will have a more degree of rhythm. 00:03:53.000 --> 00:03:57.000 So then the strain inclusion principle just says that the 00:03:57.000 --> 00:04:05.000 From the gravity perspective, this bulk non-perturbed effect should somehow actually cut down the number of states. 00:04:05.000 --> 00:04:10.000 So today, I want to see how to actually realize this. 00:04:10.000 --> 00:04:17.000 How this principle is actually realized. 00:04:17.000 --> 00:04:22.000 So, right, so the goal is to see the bulb mechanism of the strain exclusion principle. 00:04:22.000 --> 00:04:32.000 So the strategy today is that I'm looking at the so-called finite expansion of the grand canonic ensemble partition function. 00:04:32.000 --> 00:04:39.000 So the starting points that suppose you can compute the guanon ensemble partition function. 00:04:39.000 --> 00:04:43.000 So this zeta is the fogacity that counts for N. 00:04:43.000 --> 00:04:50.000 And then this queue is a collective fugacity of everything else that linear fuel theory. 00:04:50.000 --> 00:04:57.000 So this is a starting point. So now, if you start with this, there is something N you've been talking about. 00:04:57.000 --> 00:05:00.000 Yes, the big N. 00:05:00.000 --> 00:05:04.000 So you're filling over different quantum field there is? 00:05:04.000 --> 00:05:07.000 Um, I consider them to be a family. 00:05:07.000 --> 00:05:09.000 Yes. 00:05:09.000 --> 00:05:18.000 Yes. They're adding together U1, U2, U2? Yes, yes. Yes. 00:05:18.000 --> 00:05:22.000 And reduce partition functions so that they Bps. 00:05:22.000 --> 00:05:26.000 So April, or no? 00:05:26.000 --> 00:05:31.000 A prior or not, but today I'm going to discuss the BPI sector. 00:05:31.000 --> 00:05:40.000 So suppose you can… I mean, first, it might be hard to get a closed formula if you… if you consider the non-BPS case. 00:05:40.000 --> 00:05:45.000 But suppose you have this guanony ensemble partition function in the closed form. 00:05:45.000 --> 00:05:51.000 Then there are two ways to obtain the canonical ensemble partition function. So this is what we're really interested in. 00:05:51.000 --> 00:05:55.000 So, you can just take the expanders and take the. 00:05:55.000 --> 00:05:58.000 Zn coefficient here. 00:05:58.000 --> 00:06:03.000 I mean, this is equivalent to you do a contour integral at z equal to 0. 00:06:03.000 --> 00:06:07.000 And, uh… expansion convergence. 00:06:07.000 --> 00:06:13.000 Uh, yes, yes, this is assuming if you converges. 00:06:13.000 --> 00:06:19.000 So now you can also try to deform this control at the zeta plane. 00:06:19.000 --> 00:06:21.000 And then you can look at the… 00:06:21.000 --> 00:06:23.000 Polls outside this contour. 00:06:23.000 --> 00:06:29.000 And then you can ask yourself whether this Zn can be rewritten in 00:06:29.000 --> 00:06:32.000 In another sum of residue. 00:06:32.000 --> 00:06:43.000 So a priorities might not work, right? I mean, your zeta plane might have a branch cut and the wall of singularity. So you might not have the residue theorem. 00:06:43.000 --> 00:06:49.000 So, uh, but there might be hope for the protective sector in the holographic QFT. 00:06:49.000 --> 00:06:52.000 Because there you might think that there should be a… 00:06:52.000 --> 00:06:58.000 Found the way of computing the this set n and then a bog way of computing it. 00:06:58.000 --> 00:07:00.000 So then sometimes the… 00:07:00.000 --> 00:07:02.000 They should match. 00:07:02.000 --> 00:07:06.000 So suppose you want to find such a formula. 00:07:06.000 --> 00:07:13.000 So, the first, you have to determine the set of contributing poles. 00:07:13.000 --> 00:07:18.000 And then the crucial part is that this might depends on the regime of this kiln. 00:07:18.000 --> 00:07:21.000 And, uh… 00:07:21.000 --> 00:07:29.000 And then because we want to understand this as a gravity statement, so for each contributing pole or a family contributing pole. 00:07:29.000 --> 00:07:32.000 You need to find the corresponding bulk interpretation. 00:07:32.000 --> 00:07:40.000 And then remember this puzzle of straining exclusion principle. So why do we say that this can help you understand the straining exclusion principle? 00:07:40.000 --> 00:07:45.000 So as you see in this game, you always have some negative molds. 00:07:45.000 --> 00:07:48.000 And then you need to do the some analytic continuation. 00:07:48.000 --> 00:07:55.000 So in this process, you'll pick up some signs that depends on some integer numbers. 00:07:55.000 --> 00:08:04.000 So this is crucial principle is just due to the science at the given order of Q, there will be cancellations. 00:08:04.000 --> 00:08:07.000 from different sectors. So that will cut down the… 00:08:07.000 --> 00:08:12.000 number of states. So this is a rough picture. 00:08:12.000 --> 00:08:20.000 Let me use the simplest example I know to explain how this actually, so you can see this? 00:08:20.000 --> 00:08:23.000 The hat? In the last line. 00:08:23.000 --> 00:08:26.000 Uh, yes, I haven't defined it. So, uh… 00:08:26.000 --> 00:08:34.000 So here you want to redefine. So this set height. 00:08:34.000 --> 00:08:39.000 will be defined using this residue. Yeah, I. 00:08:39.000 --> 00:08:45.000 I define this set later for different examples. 00:08:45.000 --> 00:08:47.000 Questions and queues. 00:08:47.000 --> 00:08:50.000 Yes, no, fugacities. 00:08:50.000 --> 00:08:53.000 It's a collective of fogacies. 00:08:53.000 --> 00:09:01.000 Uh, so, so this is the simplest example I know. So you look at the 40 into 4 superyum, and you look at the high Bps partition function. 00:09:01.000 --> 00:09:06.000 So there is a very easy. So this is the result of the guan ensemble position function. 00:09:06.000 --> 00:09:09.000 So there's only one queue. 00:09:09.000 --> 00:09:13.000 And this, um… 00:09:13.000 --> 00:09:17.000 This is that type, so you can see that there's only simple poles here. 00:09:17.000 --> 00:09:19.000 So, um… 00:09:19.000 --> 00:09:21.000 Okay, so you can first compute the 00:09:21.000 --> 00:09:28.000 Find the end result by doing the expansion. And this is the result. So in this case, it's happened to be 00:09:28.000 --> 00:09:30.000 just that you replace this, so… 00:09:30.000 --> 00:09:33.000 Infinity by this n cut up. 00:09:33.000 --> 00:09:44.000 And then you do this trick of the finite N-span. Partition function of what? S3 times s1? Yes, so you put this on the — 00:09:44.000 --> 00:09:50.000 S. Cross is one. But you only looking at the high BPI sector. 00:09:50.000 --> 00:09:53.000 So, um… 00:09:53.000 --> 00:09:56.000 respond to just kidding. 00:09:56.000 --> 00:10:04.000 So, you spend this, and then this just measure the I will also explain later. It's measured the single trace. 00:10:04.000 --> 00:10:11.000 operators, I mean, the multi-park link, the single… there's just one matrix. It's a fugacity for what charge? 00:10:11.000 --> 00:10:13.000 Little cube. 00:10:13.000 --> 00:10:19.000 For the, uh, for the charge of this high BPS states in this, uh, 00:10:19.000 --> 00:10:21.000 So it's a new one inside the… 00:10:21.000 --> 00:10:27.000 Marcia? Yes, yes, it's basically the arsometry. 00:10:27.000 --> 00:10:35.000 CO2 asymmetry, I think. 00:10:35.000 --> 00:10:38.000 Right, so now you apply the same tree. 00:10:38.000 --> 00:10:43.000 So I apply this finite expansion. So this would be the, uh, um, 00:10:43.000 --> 00:10:46.000 This result. So this is what you get. 00:10:46.000 --> 00:10:56.000 By deforming this contour, and then because this has only simple folds, so you can apply the residue theorem, you just pick out all the all the 00:10:56.000 --> 00:11:02.000 Post outside z equal to 0. So this is the result. 00:11:02.000 --> 00:11:11.000 So now I want to match the left-hand side. So the left-hand side, the fugacity, the Q has to be inside the unit disk. 00:11:11.000 --> 00:11:15.000 So to match this, I have to expand the right-hand side. 00:11:15.000 --> 00:11:19.000 So to recalculate using the… 00:11:19.000 --> 00:11:21.000 Young Military, you calculate Z. 00:11:21.000 --> 00:11:23.000 the grand canonical and so on. 00:11:23.000 --> 00:11:27.000 Right. Correct. Yes. Okay. 00:11:27.000 --> 00:11:33.000 So that's fine. And then you're defining the non-perturbative gravity as the 00:11:33.000 --> 00:11:38.000 Counter-integral? 00:11:38.000 --> 00:11:48.000 So, right now, I'm going to show that this is what gravity result gives you. But are you calculating it? Or is this a prediction for what gravity should be? 00:11:48.000 --> 00:11:52.000 I'm going to… so, okay, so let's go ahead. 00:11:52.000 --> 00:11:54.000 So first, I want to 00:11:54.000 --> 00:12:01.000 Uh, so the first there are two parts. First, you want to rewrite this. Rewrite this by a sum of residue. 00:12:01.000 --> 00:12:08.000 The second part is that you want to find the interpretation of this residue in terms of the bulk. 00:12:08.000 --> 00:12:19.000 So, right now, I don't know what they correspond to in the bulk yet. I just see that because this has only simple posts, I'm picking out all the poles outside the small dig. So you're aiming to… 00:12:19.000 --> 00:12:28.000 interpret the principles of gravity, but we don't know that interpretation yet. Not yet. I mean, so in the next page or two. 00:12:28.000 --> 00:12:35.000 So, um, I mean, if you want to compare this, you also have to… so this is just a result coming from this contour integral. 00:12:35.000 --> 00:12:42.000 So you have to expand this in the regime of the left-hand side of the queue. The queue is inside unit disk. 00:12:42.000 --> 00:12:47.000 So you see that. So here, this is you have a negative modes. 00:12:47.000 --> 00:12:53.000 So if you want to spend this in the unit, this has this expansion. 00:12:53.000 --> 00:12:56.000 So you already see that there's a, um… 00:12:56.000 --> 00:13:01.000 minus 1 to the Q, and then you have an upshift of the Q of the energy. 00:13:01.000 --> 00:13:06.000 So this is a result, and I want to interpret this both 00:13:06.000 --> 00:13:09.000 Understand this both at the level of the 00:13:09.000 --> 00:13:12.000 a QFT and the… 00:13:12.000 --> 00:13:19.000 Gravity. So okay, so this is a result of this infinite expansion or the so-called giant graviton expansion. 00:13:19.000 --> 00:13:24.000 So let's look at the meaning. So maybe I should have explained this. What is Zn earlier. 00:13:24.000 --> 00:13:32.000 So this comes the number of independent gate invariant operators. So it's a product of the trace. So it's a single matrix. 00:13:32.000 --> 00:13:36.000 The embedded matrix and the derivatives. 00:13:36.000 --> 00:13:45.000 So, as said infinity. So this is when n goes to infinity. So then all the traces there. 00:13:45.000 --> 00:13:48.000 Independent. 00:13:48.000 --> 00:13:55.000 So now, uh, from the boundary perspective, how do you see that the… from when you go from infinity to finite n? 00:13:55.000 --> 00:14:00.000 how the degrees of freedom get reduced. So that's just because the trace relation. 00:14:00.000 --> 00:14:11.000 So suppose your X is just a number. So then the one by one matrix. So then the trace of a square equal to trace of x. The whole thing squared. 00:14:11.000 --> 00:14:14.000 And then this is an equal 2 trace relation. 00:14:14.000 --> 00:14:19.000 So on and so forth. And then, so now, what's the meaning of the, um, 00:14:19.000 --> 00:14:23.000 Uh, right-hand side. So, so this set infinity is just this guy. 00:14:23.000 --> 00:14:26.000 And then this thing here. 00:14:26.000 --> 00:14:31.000 You see this that hat k here has this minus 00:14:31.000 --> 00:14:37.000 Minus 1 to the K. So the leading one is 1 0. So that means the 00:14:37.000 --> 00:14:39.000 So this is a counting… 00:14:39.000 --> 00:14:45.000 So the leading term is just one. That's when you have no relation. 00:14:45.000 --> 00:14:47.000 And then the k equal to 1, so it's negative. 00:14:47.000 --> 00:14:50.000 So it's counting the trace relation. 00:14:50.000 --> 00:14:56.000 And then the… so you already see that the trace duration starts at the q to the N. 00:14:56.000 --> 00:15:04.000 And then, the fact that you have a higher k, that means that your trace relations are not independent. So you have relations of relations. 00:15:04.000 --> 00:15:07.000 So this is the alternating sign. Yes. 00:15:07.000 --> 00:15:12.000 Yes. Can you understand that precisely the various terms that he had from synergies? 00:15:12.000 --> 00:15:18.000 Yes, yes. I mean, in this case, you can. So it's basically you're counting the. 00:15:18.000 --> 00:15:23.000 the resolution of this V, the vector 00:15:23.000 --> 00:15:33.000 at infinity, and then more modified. You literally have a projective module on these curves in the resolution. Yes, yes, yes. 00:15:33.000 --> 00:15:35.000 Yeah, this is not my work. This is a… 00:15:35.000 --> 00:15:37.000 Uh, work by 00:15:37.000 --> 00:15:42.000 by Ji Hong, I think 2022. That's cool. 00:15:42.000 --> 00:15:44.000 In this case, it's very clean. 00:15:44.000 --> 00:15:50.000 I mean, when you have, like, quarter Bps, when you have more than one. 00:15:50.000 --> 00:15:59.000 While matrix or two matrices are already not that clean. But I mean, that's a general picture. 00:15:59.000 --> 00:16:05.000 So now we want to understand the meaning of the same. 00:16:05.000 --> 00:16:08.000 formula in terms of gravity. 00:16:08.000 --> 00:16:12.000 So, so now that infinity, um, 00:16:12.000 --> 00:16:16.000 Compute the hot BPS partition function. 00:16:16.000 --> 00:16:21.000 taking notes. And then this, so Z. 00:16:21.000 --> 00:16:26.000 It's, uh, the HubS open string citation on the Kjang graviton. 00:16:26.000 --> 00:16:31.000 So, um, the fact… so 00:16:31.000 --> 00:16:40.000 Uh, here, so you look at this one, so you see that there is exactly the same, almost the same as the Z n. 00:16:40.000 --> 00:16:45.000 If you replace this k by n, and then the Q inverse by Q. 00:16:45.000 --> 00:16:51.000 So that just means that this excitation is first form is the same because it's 00:16:51.000 --> 00:16:55.000 partition function on also the D3 brain. 00:16:55.000 --> 00:16:59.000 So the BPS partition function, the D3 brain. 00:16:59.000 --> 00:17:05.000 Uh, so that's why it's the same. And then the fact that it's a Q inverse is because it, uh, 00:17:05.000 --> 00:17:11.000 This configuration of this giant graviton is this asteroid inside S5. So it's a maximal giant. 00:17:11.000 --> 00:17:15.000 So any excitation can only shrink the S3. 00:17:15.000 --> 00:17:19.000 So that's why it's a negative mode. That's why it's a. 00:17:19.000 --> 00:17:24.000 I mean, this business says that there is a negative mold. 00:17:24.000 --> 00:17:31.000 But so you do this negative modes, and then you do this analytic. 00:17:31.000 --> 00:17:34.000 Can you remind me why this? 00:17:34.000 --> 00:17:38.000 These three frames are stapled. They're wrapped on S3. 00:17:38.000 --> 00:17:42.000 Yes, I mean, they're kind of stable because this is the Meyer effect. 00:17:42.000 --> 00:17:47.000 They're puffed up by the fluxes. By the flux. Yes. 00:17:47.000 --> 00:17:53.000 But I feel excited, then it's still a lowered energy. 00:17:53.000 --> 00:17:56.000 This is also a general feature in ODIS Game. 00:17:56.000 --> 00:18:01.000 So, you see the negative modes does two things. So, one, 00:18:01.000 --> 00:18:08.000 It gives you these alternating signs. So this realized this is a mechanism that they realized Schuny's corrosion principle. 00:18:08.000 --> 00:18:14.000 And then there's this another upshift that will be also important later. 00:18:14.000 --> 00:18:18.000 Okay, so we're… I don't understand the energy. 00:18:18.000 --> 00:18:24.000 How do I see it from the formula? 00:18:24.000 --> 00:18:27.000 What do you mean by the energy option. 00:18:27.000 --> 00:18:29.000 Maybe I, um… 00:18:29.000 --> 00:18:38.000 So you have this one, you want to expand. So this is a an elite function, and you want to do a power series expansion. 00:18:38.000 --> 00:18:42.000 So this expansion depends on 00:18:42.000 --> 00:18:45.000 Whether your ex is inside the unit disk, 00:18:45.000 --> 00:18:48.000 We're outside. 00:18:48.000 --> 00:18:53.000 So if we insert is document1 and plus x squared, right? 00:18:53.000 --> 00:18:59.000 This is also start from minus X. 00:18:59.000 --> 00:19:05.000 And then and then one chance first. 00:19:05.000 --> 00:19:13.000 I mean, so this is the upshift. 00:19:13.000 --> 00:19:18.000 Okay. Why is it energy? And what what? 00:19:18.000 --> 00:19:20.000 Because I… 00:19:20.000 --> 00:19:28.000 Because, I mean, this is my kill, right? So now this is just one factor. I have many factors. The charge. 00:19:28.000 --> 00:19:38.000 Is it hard? Is it hard? 00:19:38.000 --> 00:19:48.000 I'm lost. And so if you don't expand, this is just an identity. I mean, you can write what's inside here as 00:19:48.000 --> 00:19:53.000 Some offer, um… 00:19:53.000 --> 00:20:03.000 I'm sorry. 00:20:03.000 --> 00:20:10.000 And if I don't respond, it's an identity. But the whole point is that the, uh… 00:20:10.000 --> 00:20:17.000 Depends on whether it's a negative mode or not. You have different way of spending. 00:20:17.000 --> 00:20:23.000 But, I mean, this is it, uh, this is the basically. 00:20:23.000 --> 00:20:30.000 How it works in all the examples. So you do this expansion and then you see the 00:20:30.000 --> 00:20:36.000 negative moles, and then the when you have negative modes, you have to respond in a different way. 00:20:36.000 --> 00:20:43.000 And then this pick up some alternating signs, and then you have negative mode, you mean negative power of Q. 00:20:43.000 --> 00:20:46.000 Uh… 00:20:46.000 --> 00:20:51.000 I could say a negative vote. I think you just mean the negative power of Q. 00:20:51.000 --> 00:20:55.000 No, but it depends on what… where… I mean, in this case, yes, but uh… 00:20:55.000 --> 00:21:03.000 By negative modes, I mean that when I have some stuff here. By negative modes, it just means that whatever is inside here. 00:21:03.000 --> 00:21:06.000 This is the last, I'm sorry. 00:21:06.000 --> 00:21:09.000 Um, later that month. 00:21:09.000 --> 00:21:11.000 So in this case is a move. 00:21:11.000 --> 00:21:15.000 Q to the inverse. 00:21:15.000 --> 00:21:24.000 But later, you will see that it's slightly different. 00:21:24.000 --> 00:21:28.000 So, uh, and so this is the 00:21:28.000 --> 00:21:35.000 everything about the PPI sector for the case. So, any questions? 00:21:35.000 --> 00:21:37.000 And, uh, so… 00:21:37.000 --> 00:21:40.000 Um, so we want to apply this technique. 00:21:40.000 --> 00:21:49.000 Yes. You had an interpretation of the leading terms. KK votes from S. 5, and then the D. 3 brains. 00:21:49.000 --> 00:21:51.000 I have interviewed me for everything. 00:21:51.000 --> 00:21:56.000 I have interpretation of everything. 00:21:56.000 --> 00:22:03.000 Both from the boundary perspective? No, but in terms of non-perturbative… yeah, no, we already said that. 00:22:03.000 --> 00:22:05.000 He told me it was the projective resolution. 00:22:05.000 --> 00:22:13.000 That's the module of these operators trace x to the k. Yes. For finite n. 00:22:13.000 --> 00:22:15.000 That's great. That's beautiful. But now, 00:22:15.000 --> 00:22:18.000 What about the gravity interpretation? 00:22:18.000 --> 00:22:21.000 So you gave one. You have Z infinity. 00:22:21.000 --> 00:22:24.000 As the KK votes. 00:22:24.000 --> 00:22:28.000 Um… 00:22:28.000 --> 00:22:33.000 Oh, I see. That's an interpretation for all… that's a bulk interpretation for all K. Yes. 00:22:33.000 --> 00:22:36.000 Yes. Cake giant grammatized. Yes. 00:22:36.000 --> 00:22:39.000 And so now no other non-perturbative effects. 00:22:39.000 --> 00:22:43.000 matter here, then. 00:22:43.000 --> 00:22:49.000 But, uh, I mean, the fact that you have this Q to the Kn, that means that the 00:22:49.000 --> 00:22:52.000 This non-perturbulous correction. I understand that it's a non-protector. 00:22:52.000 --> 00:23:01.000 perturbative effect, and just there are other non-perturbative effects in string theory. Right, but in this… I mean… You have an interpretation. Yes. 00:23:01.000 --> 00:23:04.000 In terms of non-perturbative effects for every k in the sub. 00:23:04.000 --> 00:23:06.000 Yes, yes. 00:23:06.000 --> 00:23:10.000 Right, this is the simplest case. 00:23:10.000 --> 00:23:15.000 So I want to apply the same technique for ADS3 CFT2. 00:23:15.000 --> 00:23:22.000 So we want to have these three questions in mind when we apply this technique. 00:23:22.000 --> 00:23:24.000 So first is the 00:23:24.000 --> 00:23:26.000 Do all posts still contribute? 00:23:26.000 --> 00:23:38.000 And second, so what will be the meaning of this contributing pose? As we will see that in this case, there won't be any giant graviton configuration, so will be the analog of giant graviton for ADS 3. 00:23:38.000 --> 00:23:40.000 And also, uh… 00:23:40.000 --> 00:23:50.000 It's a student exclusion principle still implemented by the… I already told you the punchline, but yeah, this is another question. You basically want to check that the. 00:23:50.000 --> 00:23:58.000 Whether the technique that applied for this simple example still holds for this more complicated case. 00:23:58.000 --> 00:24:06.000 So this is the ADS3CFT2 setup. It's a classic setup. 00:24:06.000 --> 00:24:10.000 Uh, is this T1T5 system in the type 2B. 00:24:10.000 --> 00:24:14.000 And the bulk is a type 2B supergravity in this background. 00:24:14.000 --> 00:24:18.000 So the M4, we choose to be T4 or K3. 00:24:18.000 --> 00:24:20.000 Because we want to preserve the 00:24:20.000 --> 00:24:24.000 small superconformal symmetry. 00:24:24.000 --> 00:24:33.000 So you start from the d1d file, and then you're in this R background. So Q1 is a number for d1 and q phi is number d5. 00:24:33.000 --> 00:24:41.000 So I'm going to look at the BPS sector so we can move to a point where there's a fairground. 00:24:41.000 --> 00:24:43.000 And the 00:24:43.000 --> 00:24:51.000 So this is going to be useful later when I need a worksheet technique to do some computation. 00:24:51.000 --> 00:25:01.000 So the boundary is symmetric orbit for CFT. So is this sigma model of M4, and then this is the symmetric orbifold 00:25:01.000 --> 00:25:07.000 So this is the standard setup. And so the so here this I will say something more about the symmetric orbifold. 00:25:07.000 --> 00:25:11.000 So… 00:25:11.000 --> 00:25:15.000 So this is… so this is sigma model is either T4, you have a 00:25:15.000 --> 00:25:20.000 For scalars and for fermions, and the case rate is 00:25:20.000 --> 00:25:24.000 a resolution of the T4 mods at 2. 00:25:24.000 --> 00:25:31.000 So, uh, the symmetric orbital sector is labeled by the contrasting classes of symmetric goods. 00:25:31.000 --> 00:25:36.000 And in particular, for the single particle spectrum. So you just have one 00:25:36.000 --> 00:25:43.000 cycle. So if it's the length is W, it's the WTC sector. 00:25:43.000 --> 00:25:47.000 So, and so this symmetric orbit fault has a 00:25:47.000 --> 00:25:54.000 A small ink to 4, 4, superconformal symmetry. 00:25:54.000 --> 00:26:02.000 So let's first talk about how BPI spectrum. So let's look at the NLS sector. So there. 00:26:02.000 --> 00:26:08.000 The high PPI sector. Happy PI spectrum is given by the primary on the left on the right. 00:26:08.000 --> 00:26:14.000 So this is the conformal weight, and this is the SU2 charge. 00:26:14.000 --> 00:26:17.000 So this is the canonical ensemble partition function. 00:26:17.000 --> 00:26:22.000 So the Y will be the fugacity. So there is this tool here. 00:26:22.000 --> 00:26:26.000 And so you're tracing over the current spectrum. 00:26:26.000 --> 00:26:30.000 Um, so this is the and so the 00:26:30.000 --> 00:26:34.000 The Guang Canon, for example, partition function is given by this. 00:26:34.000 --> 00:26:39.000 So this is for both T4 and K3. So the only information 00:26:39.000 --> 00:26:46.000 That's the goal seen here is this Hodge diamond here. 00:26:46.000 --> 00:26:49.000 So now we have the, um… 00:26:49.000 --> 00:26:52.000 All we need to start this again. 00:26:52.000 --> 00:26:55.000 So for… 00:26:55.000 --> 00:27:03.000 You get a quantum potential functioning. You do this expansion, you pick out the… Could you go back to the ground color? 00:27:03.000 --> 00:27:06.000 So again, if you have… 00:27:06.000 --> 00:27:08.000 What, simple polls? 00:27:08.000 --> 00:27:12.000 I assure related to more than just simple pose. 00:27:12.000 --> 00:27:17.000 You have to wait, wait a little bit. 00:27:17.000 --> 00:27:29.000 So, so this is Zm. And so, okay, so I first want to, before I look at all this poles, before I want to apply this procedure, I want to first see what I want to reproduce. 00:27:29.000 --> 00:27:33.000 by these techniques. So I first want to show you what, um… 00:27:33.000 --> 00:27:36.000 The result is from the boundary perspective. 00:27:36.000 --> 00:27:43.000 So this is from the boundary perspective, you're just expanding this and you pick up the Z coefficient. 00:27:43.000 --> 00:27:49.000 And then also the Z infinity is the Kk spectrum. So this you can compute by. 00:27:49.000 --> 00:27:54.000 Taking the residue at the closest to the. 00:27:54.000 --> 00:28:00.000 to the origin. So this is the harsh diamond of the various symmetric products of K. 3. 00:28:00.000 --> 00:28:03.000 Uh, it's a harsh diamond of the case itself. 00:28:03.000 --> 00:28:05.000 No, on the next slide. 00:28:05.000 --> 00:28:10.000 On the other side… 00:28:10.000 --> 00:28:13.000 Those matrices, that's the arch diamond of this. 00:28:13.000 --> 00:28:17.000 Uh, yes. Symmetric products of K3. Yes, yes. 00:28:17.000 --> 00:28:19.000 Okay. Yes. So… 00:28:19.000 --> 00:28:28.000 Now, yeah, right, so to present this, it's more convenient to just specialize in y and y bar, and then just show you the 00:28:28.000 --> 00:28:30.000 And… 00:28:30.000 --> 00:28:32.000 this power series. 00:28:32.000 --> 00:28:34.000 Uh, with just one variable. 00:28:34.000 --> 00:28:40.000 So now angles infinity, this is the KK spectrum. 00:28:40.000 --> 00:28:42.000 So I want to compare. 00:28:42.000 --> 00:28:51.000 So with this benchmark. 00:28:51.000 --> 00:28:57.000 So you can see that suppose I want to start — so this I can compute the n degrees infinity. 00:28:57.000 --> 00:29:06.000 So I want to start from the gravity and then from the gravity, this is what I know already. I want to start from this angle to infinity and reproduce all this polynomials. 00:29:06.000 --> 00:29:09.000 So this is an infinite series. 00:29:09.000 --> 00:29:11.000 So you can see that the… 00:29:11.000 --> 00:29:16.000 What I can… I can match. So the blue one is the… for each N. 00:29:16.000 --> 00:29:21.000 So the blue one is what the, um… 00:29:21.000 --> 00:29:23.000 The part that. 00:29:23.000 --> 00:29:26.000 The mesh, they can reproduce by the KK spectrum. 00:29:26.000 --> 00:29:33.000 So you can see that the right, so maybe I explained, this is the. 00:29:33.000 --> 00:29:39.000 The energy sector brought of the 00:29:39.000 --> 00:29:42.000 The SU2 charge, and then the… 00:29:42.000 --> 00:29:46.000 conformal weight. So this this blue line is the chiral primary. 00:29:46.000 --> 00:29:48.000 And then this, uh, uh… 00:29:48.000 --> 00:29:53.000 This curve here is a black hole bond. 00:29:53.000 --> 00:29:56.000 So, um… 00:29:56.000 --> 00:29:58.000 So, uh… 00:29:58.000 --> 00:30:04.000 So the current primary only touched this bond by one point. 00:30:04.000 --> 00:30:07.000 So once you touch the. 00:30:07.000 --> 00:30:09.000 this black hole bond. 00:30:09.000 --> 00:30:14.000 You can see that the after this point is a heavy regime. 00:30:14.000 --> 00:30:19.000 And then this blue one, you can see that that's all less than 00:30:19.000 --> 00:30:26.000 lesson. So if you look at the one side, that's L0 equal to J naught less than quarter n. 00:30:26.000 --> 00:30:35.000 So the blue part, so the blue part is the live spectrum, and then the 00:30:35.000 --> 00:30:37.000 And then, look at this polynomial. 00:30:37.000 --> 00:30:40.000 The polynomial, this, uh… 00:30:40.000 --> 00:30:44.000 So here I'm just plotting 00:30:44.000 --> 00:30:46.000 this number, so this is the log scale. 00:30:46.000 --> 00:30:49.000 So you see that the rich. 00:30:49.000 --> 00:30:55.000 This intersecting point, I mean, you can think of it as the degeneracy of the small black hole. 00:30:55.000 --> 00:30:59.000 And there's not really a black hole. 00:30:59.000 --> 00:31:01.000 in the box. But 00:31:01.000 --> 00:31:08.000 Because it touched this black hole bond, so you can think of it as a degeneracy of the smallest small black hole. 00:31:08.000 --> 00:31:13.000 So that's so so this curve is… 00:31:13.000 --> 00:31:19.000 So this is the one because I'm talking all overlappingly, so it's hard to see. 00:31:19.000 --> 00:31:23.000 So this is a 122 one. 00:31:23.000 --> 00:31:26.000 And then in between, so 123, 00:31:26.000 --> 00:31:30.000 267, so it goes like this. 00:31:30.000 --> 00:31:32.000 So it's all these triangles. 00:31:32.000 --> 00:31:35.000 So that means that the, um… 00:31:35.000 --> 00:31:40.000 In this case, this non-perturbative effect is so strong that 00:31:40.000 --> 00:31:43.000 Hmm. Um. 00:31:43.000 --> 00:31:51.000 So it's so strong that it turns starting from this PK mode, it turned back. 00:31:51.000 --> 00:31:55.000 An H1 case, the whole series. 00:31:55.000 --> 00:32:04.000 So the non-protective effect can start with an infinite series and then not truncate and then make it into a polynomial. 00:32:04.000 --> 00:32:08.000 So that's what I want to, um, 00:32:08.000 --> 00:32:11.000 Uh, reproduced by, by applying this, uh, 00:32:11.000 --> 00:32:21.000 Financial expansion to see how, starting from this infinite series, you can get a polynomial. 00:32:21.000 --> 00:32:27.000 Yeah, so yeah, this is what I just said, that after you hit this midpoint. 00:32:27.000 --> 00:32:32.000 I mean, this non-protective effect dominates. 00:32:32.000 --> 00:32:40.000 So I want to apply this… So the Y, the fugacity, Y is the fugacity for angular momentum, if I remember correctly. 00:32:40.000 --> 00:32:45.000 So, um, I'm looking at the chiral chiral spectrum. 00:32:45.000 --> 00:32:52.000 So, in the chiropractor spectrum, this is the definition. Yes. It's the same as energy. Yes. 00:32:52.000 --> 00:32:56.000 Yes. Yes, yes. So that's this straight line here. 00:32:56.000 --> 00:33:00.000 They're all just lined up here. I have a finer graph here. 00:33:00.000 --> 00:33:03.000 But, um… 00:33:03.000 --> 00:33:08.000 So, uh, so you see, we know we understand the blue one. 00:33:08.000 --> 00:33:14.000 But from the gravity, the blue one reproduced by the gravitational KK spectrum. 00:33:14.000 --> 00:33:20.000 So what we don't understand this black one. Black one is in the middle range. And then this. 00:33:20.000 --> 00:33:25.000 The red one is even harder. The red one is this heavy part. 00:33:25.000 --> 00:33:28.000 So I want to understand both the right one. 00:33:28.000 --> 00:33:32.000 And the medium part and the red one. 00:33:32.000 --> 00:33:36.000 By this final expansion. 00:33:36.000 --> 00:33:38.000 So as you go to higher values of A, 00:33:38.000 --> 00:33:41.000 You expect that for a fixed value of… 00:33:41.000 --> 00:33:45.000 fixed power is why the difference in the coefficients. 00:33:45.000 --> 00:33:49.000 is going to be converge to 0, right? At any for any y. 00:33:49.000 --> 00:33:51.000 Yes, it's a polynomial. 00:33:51.000 --> 00:33:53.000 Right, but… 00:33:53.000 --> 00:33:57.000 So for L equals 5, the power of Y to the 12. 00:33:57.000 --> 00:33:59.000 So yeah, just too long. 00:33:59.000 --> 00:34:03.000 I mean, sorry, just relying — I mean, they're all polynomials. 00:34:03.000 --> 00:34:07.000 For any finite… it's polynomial. I understand, but is there. 00:34:07.000 --> 00:34:09.000 Is it, is it interesting to study the 00:34:09.000 --> 00:34:12.000 The rate at which they fall by difference decays. 00:34:12.000 --> 00:34:15.000 Eventually, as you take large n. 00:34:15.000 --> 00:34:18.000 on fixed value of Y, for the fixed power of Y. 00:34:18.000 --> 00:34:23.000 The difference in the numerically coefficient between the n equals infinity result. 00:34:23.000 --> 00:34:27.000 And the finite end result, does that have some kind of interesting role or 00:34:27.000 --> 00:34:38.000 Maybe. Maybe. 00:34:38.000 --> 00:34:45.000 Right, so, uh, right, so going back to this, um, learning ensemble partition. 00:34:45.000 --> 00:34:47.000 function. So this is the thing that 00:34:47.000 --> 00:34:53.000 So this is the alignment of the single T4 or single case. 00:34:53.000 --> 00:34:55.000 Right, so, uh… 00:34:55.000 --> 00:34:59.000 Uh, so you see that the 00:34:59.000 --> 00:35:02.000 There's four corners, they are shared by T4 and K3. 00:35:02.000 --> 00:35:08.000 So for both T4 and T3, you have four family of simple poles that correspond to these four corners. 00:35:08.000 --> 00:35:13.000 So because of this zeta to the n here, 00:35:13.000 --> 00:35:17.000 So, I mean, so basically in this case, this N, 00:35:17.000 --> 00:35:22.000 So, for each K, you have a 00:35:22.000 --> 00:35:32.000 So, okay, so you have plus minus, plus, minus. That's described as four corners. And then you have this K that basically correspond to this n here. 00:35:32.000 --> 00:35:40.000 And then for each K, because you have a set to the n, so you have a root of unity factor here. 00:35:40.000 --> 00:35:43.000 So this is the old simple pose. 00:35:43.000 --> 00:35:49.000 And then you also have a wall of essential singularities from this center of the Hodge diamond. 00:35:49.000 --> 00:35:51.000 So this is the… 00:35:51.000 --> 00:35:54.000 Uh, the post structures. 00:35:54.000 --> 00:35:56.000 So let me plot this. 00:35:56.000 --> 00:35:59.000 This, uh… 00:35:59.000 --> 00:36:04.000 The family of a post relative to this wall of essential singularity. 00:36:04.000 --> 00:36:10.000 So this part will depend on the — because it's 00:36:10.000 --> 00:36:16.000 Because it's a polynomial, so you can your Y and Y bar can. 00:36:16.000 --> 00:36:18.000 go from zero to infinity. 00:36:18.000 --> 00:36:20.000 So, yeah, y and y bar. 00:36:20.000 --> 00:36:23.000 So the boundary would be the… 00:36:23.000 --> 00:36:30.000 Whether… so this is one, this is one. And whether the Y and Y bar is inside or outside the unit disk. 00:36:30.000 --> 00:36:33.000 And then the relative size of them. 00:36:33.000 --> 00:36:37.000 So this, I start from this regime. 00:36:37.000 --> 00:36:42.000 And then so this is I'm plotting the 00:36:42.000 --> 00:36:44.000 I'm prouding all the 00:36:44.000 --> 00:36:48.000 These four family of poles relative to the wall of essential singularity. 00:36:48.000 --> 00:36:58.000 So this is a plus minus, plus minus. I'm writing, I'm using this circle and triangle or the hollow one. It's the plus and the minus refer to? 00:36:58.000 --> 00:37:02.000 Uh, this, uh, 4 diamonds. 00:37:02.000 --> 00:37:10.000 This is the result, and this… So that way of labeling extremal points on the assignment? Yes, yes. 00:37:10.000 --> 00:37:16.000 So, I mean, because, I mean, so originally you have our R goes through from 0, 1, 2. 00:37:16.000 --> 00:37:23.000 And because they have a minus one, so you change the R minus one to plus minus. 00:37:23.000 --> 00:37:27.000 natural from the left cell too. Yes. 00:37:27.000 --> 00:37:33.000 Right, so it's just easier to call them plus minus than 0, 2. 00:37:33.000 --> 00:37:35.000 Actually, our paper will probably use the 02. 00:37:35.000 --> 00:37:39.000 Uh, so this is k equal to 1, so this is here. 00:37:39.000 --> 00:37:41.000 And click 2. 00:37:41.000 --> 00:37:43.000 So you have this tool. 00:37:43.000 --> 00:37:47.000 And sweet, you have this, uh, uh, uh, 3. 00:37:47.000 --> 00:37:50.000 Finally? 00:37:50.000 --> 00:37:53.000 4, 5, 6, 7, 8… 00:37:53.000 --> 00:37:55.000 So, you can see that the… oops, sorry. 00:37:55.000 --> 00:37:59.000 For the hierarchy, they also accumulate. 00:37:59.000 --> 00:38:05.000 like hugging this essential singularity. 00:38:05.000 --> 00:38:07.000 Uh, so here. 00:38:07.000 --> 00:38:15.000 We won't… because you have a central singularity, so you don't have a residue theorem. So you have to decide what are the contributing posts. 00:38:15.000 --> 00:38:20.000 So to decide this, we have to actually look at this, um, 00:38:20.000 --> 00:38:25.000 Uh, so this is the… All for what? So you're starting with a thing. 00:38:25.000 --> 00:38:28.000 You're starting with a contour risk overall. 00:38:28.000 --> 00:38:30.000 The finite n partition function. 00:38:30.000 --> 00:38:35.000 was given by a small contour of origin. Yes, his own origin, yes. 00:38:35.000 --> 00:38:41.000 So now I'm looking… You're trying to deform that contour around the origin. I'm just looking at the outside. 00:38:41.000 --> 00:38:48.000 Because of this war of singularity, there is no guarantee that what's inside this small circuit can be reproduced. 00:38:48.000 --> 00:38:51.000 by the post onset at all. 00:38:51.000 --> 00:38:56.000 But I'm just trying to see what I'm just trying to understand what contour deformation you have in mind. 00:38:56.000 --> 00:39:00.000 What I have in mind is that whether I. 00:39:00.000 --> 00:39:02.000 I'm looking at everything outside this, uh… 00:39:02.000 --> 00:39:06.000 Also z equal to 0. 00:39:06.000 --> 00:39:09.000 I mean, so everything outside is the potent. 00:39:09.000 --> 00:39:14.000 This is the pose that potentially might contribute. 00:39:14.000 --> 00:39:16.000 I'm trying to see whether I can. 00:39:16.000 --> 00:39:20.000 Pick a subset of this pose and then reproduce it at N. 00:39:20.000 --> 00:39:25.000 contribute to what I mean, you could just, you know, formally look at the sum over 00:39:25.000 --> 00:39:28.000 of residues over some set of poles, but you're 00:39:28.000 --> 00:39:33.000 Your whole motivation was to start from a contour reticle with a small contour around the origin. 00:39:33.000 --> 00:39:38.000 Yes, I want to reproduce that as a sum of residues. 00:39:38.000 --> 00:39:42.000 is that you can do some kind of contour deformation at small. 00:39:42.000 --> 00:39:44.000 But I hope it's because of your 00:39:44.000 --> 00:39:48.000 line of essential singularity. Yes. 00:39:48.000 --> 00:39:51.000 By the way, happens in, you know, the Ising model. 00:39:51.000 --> 00:39:55.000 I don't quite see why. 00:39:55.000 --> 00:39:59.000 What concert deprecation you even have in mind. 00:39:59.000 --> 00:40:02.000 So… 00:40:02.000 --> 00:40:09.000 Maybe I just look at here. So this is the set of plane, right? I mean, I didn't write it here. Originally, the country is inside. Right. 00:40:09.000 --> 00:40:14.000 Now I'm looking at everything outside, and then ask whether uh 00:40:14.000 --> 00:40:21.000 by any chance, send me over the residue of the stuff outside my reproduce that N. 00:40:21.000 --> 00:40:24.000 Then I know how to compute, then from the boundary, just, uh, 00:40:24.000 --> 00:40:32.000 expanding this grand counting ensemble partition function by looking at the 00:40:32.000 --> 00:40:35.000 satitude and coefficient. So there's no ambiguity there. 00:40:35.000 --> 00:40:37.000 I'm now, I want to reproduce that. 00:40:37.000 --> 00:40:46.000 By a sum of a residue of the — Oh, it might work or it might. Yeah, yeah, exactly. It might not. 00:40:46.000 --> 00:40:51.000 Right, because exactly, so there's no residue theorem. I mean, that's why this is a, uh, 00:40:51.000 --> 00:40:53.000 That's why I'm giving a talk. 00:40:53.000 --> 00:40:57.000 Uh, about this, because it's not as trivial as the hot BPS sector. 00:40:57.000 --> 00:41:08.000 Right. So exactly, there's no guarantee that it will work precisely because of this essentially singularity. Otherwise, you just sum over all of them. 00:41:08.000 --> 00:41:14.000 Yeah, that's precisely the question the point. 00:41:14.000 --> 00:41:19.000 So now, uh, I know that I have this 00:41:19.000 --> 00:41:21.000 Uh, figures, and uh… 00:41:21.000 --> 00:41:26.000 So this is just one region like Europe. 00:41:26.000 --> 00:41:31.000 And so in this region, and I want. 00:41:31.000 --> 00:41:40.000 And I'm plotting the location of this four family of pose relative to this wall of central singularity. 00:41:40.000 --> 00:41:49.000 So I asked myself, do they all contribute? I mean, you might have already guessed that probably was inside the wall of singularity contribute. 00:41:49.000 --> 00:41:51.000 But you want to have a proof of Y, right? 00:41:51.000 --> 00:41:57.000 So to do that, so I probably should put a question on. 00:41:57.000 --> 00:42:00.000 Question mark here, because it's… 00:42:00.000 --> 00:42:02.000 This is what you hope. 00:42:02.000 --> 00:42:10.000 Right? So now you can ask yourself, is it possible to choose a subset of this. 00:42:10.000 --> 00:42:13.000 Pose, such that. 00:42:13.000 --> 00:42:16.000 The sum of all this residue will reproduce at N. 00:42:16.000 --> 00:42:19.000 So that's the the logic here. 00:42:19.000 --> 00:42:22.000 So for that, you need to evaluate this residue. 00:42:22.000 --> 00:42:30.000 So you already see that the k is this orbital, I mean, the k is this Km for each K, you have a 00:42:30.000 --> 00:42:38.000 that K of the set K image. So it's natural to for given K, it's natural to sum over all the M. 00:42:38.000 --> 00:42:42.000 So now, for each K, you have a k 00:42:42.000 --> 00:42:47.000 basic for families. So this is my definition of this, uh, the Z hat. 00:42:47.000 --> 00:42:52.000 So that hat is for fixed k, you summing over all the zk image. 00:42:52.000 --> 00:42:55.000 So that will have some geometric meaning later. 00:42:55.000 --> 00:42:58.000 So this is just a result. 00:42:58.000 --> 00:43:01.000 And so now you just have some… 00:43:01.000 --> 00:43:04.000 analytic function here. 00:43:04.000 --> 00:43:06.000 So, but to… 00:43:06.000 --> 00:43:10.000 To say something, to say anything physical. 00:43:10.000 --> 00:43:12.000 Um, with the… 00:43:12.000 --> 00:43:21.000 Like, in terms of the boundary variable, we have to expand this. So this then precisely the exactly the same game as the high BPS sector earlier. 00:43:21.000 --> 00:43:32.000 So except that now this suspension, how do you expand this depends on which chamber you are in this cell y and y bar. 00:43:32.000 --> 00:43:41.000 So, so yeah, this is already I wrote earlier, so basically you have so you have lots of factors like this, and then this X is a 00:43:41.000 --> 00:43:43.000 basically look at them. 00:43:43.000 --> 00:43:51.000 look like something like here. And then this zero expansion depends on whether this, uh, uh, that this X is inside or outside. 00:43:51.000 --> 00:43:58.000 of the unit disk. And then so what we observe is that for a given 00:43:58.000 --> 00:44:07.000 timber. So the timber, I mean this. So there are eight chambers, and also, also, you have this the 00:44:07.000 --> 00:44:13.000 the boundaries between them. So they're seeing different cases for each of sitting cases, you need to, um. 00:44:13.000 --> 00:44:16.000 discuss this. 00:44:16.000 --> 00:44:19.000 So, the observation is that. 00:44:19.000 --> 00:44:23.000 This poll, if the pole is inside this wall of singularity. 00:44:23.000 --> 00:44:27.000 Then, this set has a finite number for negative modes. 00:44:27.000 --> 00:44:30.000 And it's also an infinite number of negative modes. 00:44:30.000 --> 00:44:32.000 So what does it mean here? So, 00:44:32.000 --> 00:44:43.000 Uh, so this is also similar to earlier. So the suspension, if you have negative modes, so this different way of analytic so this analytic continuation or different way of expanding it. 00:44:43.000 --> 00:44:47.000 have to expect. So one is that you get an oscillating factor. 00:44:47.000 --> 00:44:52.000 And also, when you do this, you have energy upshift. 00:44:52.000 --> 00:45:01.000 So now, suppose you have only finite number of negative modes. So that's fine. You have a sign, and then you have a finite upkeep for your energy. 00:45:01.000 --> 00:45:04.000 But suppose you have an infinite number of negative moles. 00:45:04.000 --> 00:45:08.000 Then basically your energy is so high that you just don't see it. 00:45:08.000 --> 00:45:10.000 So that's the… so then they don't contribute. 00:45:10.000 --> 00:45:17.000 do have a way of understanding this in terms of homological algebra, like in the FPPS states? Not yet. 00:45:17.000 --> 00:45:20.000 I think, yeah, not yet. 00:45:20.000 --> 00:45:25.000 I mean, that one, even for the quarter BPS case is not that, uh… 00:45:25.000 --> 00:45:32.000 Uh, easy. We can discuss later. 00:45:32.000 --> 00:45:40.000 Uh, so this, the statement of, uh, which family are inside the war and which is out depends on the chamber. So this is this, uh, 00:45:40.000 --> 00:45:45.000 8 chambers, and then also the boundary. 00:45:45.000 --> 00:45:50.000 So, let's start from this. 00:45:50.000 --> 00:45:53.000 It's like Green Chamber. 00:45:53.000 --> 00:45:56.000 And then for the lightning camber, 00:45:56.000 --> 00:45:58.000 You see that there are two. 00:45:58.000 --> 00:46:03.000 These two families inside and these two hollow ones are outside. 00:46:03.000 --> 00:46:05.000 And then you can change the chamber. 00:46:05.000 --> 00:46:09.000 So this is the borderline. You see, there's only one contributing. 00:46:09.000 --> 00:46:18.000 And so these two triangle, they basically just overlap, and then they land on on the wall. 00:46:18.000 --> 00:46:21.000 And then so it just, uh… 00:46:21.000 --> 00:46:28.000 You just run you just come counterclockwise, and you cover all the 00:46:28.000 --> 00:46:33.000 quoters, all the chambers, and then the also the borders. 00:46:33.000 --> 00:46:35.000 And they, um… 00:46:35.000 --> 00:46:43.000 So you, you, you basically can write down what's the plus minus, plus minus family contribute for each chamber. 00:46:43.000 --> 00:46:45.000 So this is the result. 00:46:45.000 --> 00:46:48.000 And then we also have a proof that the 00:46:48.000 --> 00:46:53.000 They are all you need. Then we'll sum them up, they reproduce the Zn. 00:46:53.000 --> 00:46:56.000 So, so… 00:46:56.000 --> 00:46:59.000 Means that, um, 00:46:59.000 --> 00:47:04.000 This is the equal sign. What's the equal sign? 00:47:04.000 --> 00:47:10.000 20 slides here. 00:47:10.000 --> 00:47:14.000 This is an equal sign. 00:47:14.000 --> 00:47:19.000 So, um, so I have a proof that simple pulse? 00:47:19.000 --> 00:47:23.000 Uh, just from the simple pose inside wall of singularity. 00:47:23.000 --> 00:47:26.000 So the, um… 00:47:26.000 --> 00:47:30.000 Um, right, so the proof is that, uh, 00:47:30.000 --> 00:47:33.000 Hey, what's outside, they don't contribute. 00:47:33.000 --> 00:47:35.000 be, uh… 00:47:35.000 --> 00:47:41.000 What's inside there? No, there's no residue theorem, amazing. So those residues actually get… Yes, exactly. 00:47:41.000 --> 00:47:49.000 Right, right. So the proof is that we have a cutoff, and that's um. 00:47:49.000 --> 00:47:51.000 Um… 00:47:51.000 --> 00:47:54.000 We have a cutoff here, and then we have cut off, you have residue theorem. 00:47:54.000 --> 00:48:04.000 And then you increase the cutoff and you see that the destabilize. 00:48:04.000 --> 00:48:10.000 So, um, I can either show you the proof later, or I want to demonstrate. 00:48:10.000 --> 00:48:16.000 I want to reproduce this, so what we looked at earlier, this character spectrum. 00:48:16.000 --> 00:48:22.000 So I want to just show you this polynomial with one variable. 00:48:22.000 --> 00:48:25.000 So this is the y equal to y bar, so this is this line. 00:48:25.000 --> 00:48:31.000 So we already see that there only is a minus minus family contribute. So let me just demonstrate this. 00:48:31.000 --> 00:48:38.000 So, yeah, this is the plot of for this y equal to y bar less than 1. 00:48:38.000 --> 00:48:44.000 Is this only the minus minus is inside this wall? 00:48:44.000 --> 00:48:53.000 So this is a cutoff. Why don't the higher order poles, besides the simple poles contribute? 00:48:53.000 --> 00:49:02.000 Yeah, no, no, you're right, exactly, exactly. Yes, yes, Will. So you said, put a cut off. Yes. That's how you prove this. 00:49:02.000 --> 00:49:12.000 But they potentially contribute, but they have a… the number of negative moles grows, I think, quadrantly with the cutoff. 00:49:12.000 --> 00:49:14.000 So when you increase the… 00:49:14.000 --> 00:49:19.000 The cutoff and then you have like the number of negative modes also. 00:49:19.000 --> 00:49:22.000 Growth. 00:49:22.000 --> 00:49:27.000 So I want to demonstrate with the most quantum case when n equal to one. 00:49:27.000 --> 00:49:31.000 So this, the target, this we want to reproduce. 00:49:31.000 --> 00:49:33.000 by this. 00:49:33.000 --> 00:49:35.000 some from the bulk perspective. 00:49:35.000 --> 00:49:42.000 So remember, I have a sum of k from 1 to infinity. So I want to. 00:49:42.000 --> 00:49:47.000 impose a… so this kind of is different from this other cutoff I just mentioned. 00:49:47.000 --> 00:49:49.000 So, uh… 00:49:49.000 --> 00:49:53.000 So you want to eventually, so this kind of infinity, 00:49:53.000 --> 00:49:59.000 So I want to show how when you increase this cutoff approach this. 00:49:59.000 --> 00:50:07.000 This final end result. So this k equal to 1, this is the KK spectrum. So then this is just one. 00:50:07.000 --> 00:50:11.000 And then when you increase it, you see that the match better and better. 00:50:11.000 --> 00:50:14.000 Actually, I have some better picture here. 00:50:14.000 --> 00:50:18.000 So I'm drawing this in the science log. 00:50:18.000 --> 00:50:20.000 Um, crot. 00:50:20.000 --> 00:50:25.000 So this is the this horizontal direction is the energy. 00:50:25.000 --> 00:50:31.000 And this is the degeneracy. So we want to match 122 and 1. 00:50:31.000 --> 00:50:34.000 So this line is the 00:50:34.000 --> 00:50:37.000 KK spectrum is basically the cutoff is 1. 00:50:37.000 --> 00:50:42.000 So you see that the… so when they match, I put the yellow circle there. 00:50:42.000 --> 00:50:48.000 So you see that if you only if you cut off is one, then you only match the leading one, not surprisingly. 00:50:48.000 --> 00:50:54.000 And then, so then the k equal to 2k, you can already see that it's negative. 00:50:54.000 --> 00:50:58.000 So then there's a hope that if we bring this. 00:50:58.000 --> 00:51:02.000 This degeneracy down. So you put this, you include this cake to 00:51:02.000 --> 00:51:10.000 tool in your sum, so that means the cutoff is 2. So you see that now you manage to match also the second term. 00:51:10.000 --> 00:51:15.000 So then if this kick to sweeping back up positive again. 00:51:15.000 --> 00:51:21.000 And then you include it, you match the first three. 00:51:21.000 --> 00:51:24.000 But, so then you repeat it then. 00:51:24.000 --> 00:51:28.000 You see that once as you move, 00:51:28.000 --> 00:51:30.000 You match… 00:51:30.000 --> 00:51:32.000 One more. 00:51:32.000 --> 00:51:34.000 And you can also see that the 00:51:34.000 --> 00:51:45.000 For hierarchy, they start to have the effect at higher and higher energy. 00:51:45.000 --> 00:51:50.000 So, I think it's already quite striking. This one, you reproduce the 00:51:50.000 --> 00:51:56.000 to include one result. So in some sense, so this is sometimes it's just a repeat of the that one. 00:51:56.000 --> 00:52:00.000 But it's not as striking, but it's just… it looks. 00:52:00.000 --> 00:52:08.000 You work in the exact same way. 00:52:08.000 --> 00:52:12.000 Alright, so… 00:52:12.000 --> 00:52:14.000 I have the picture for… 00:52:14.000 --> 00:52:22.000 So I just showed you how it works for this line, and then you can try it for all these other chambers almost the same way. 00:52:22.000 --> 00:52:24.000 So we conclude that 00:52:24.000 --> 00:52:32.000 Now we want to conclude that this final expansion in terms of this Z hat can reproduce a final result. 00:52:32.000 --> 00:52:37.000 So now we want to understand what are the bulk interpretation. 00:52:37.000 --> 00:52:44.000 Of all this stuff had. So the intuition is that there should be some kind of a zirky orbital geometry. 00:52:44.000 --> 00:52:48.000 So the first thing is that the… 00:52:48.000 --> 00:52:54.000 From the earlier from the earlier computation, you see that there's a tree level part. 00:52:54.000 --> 00:52:59.000 So the energy is immediate orbifold of the BTZ black hole? 00:52:59.000 --> 00:53:03.000 Uh, no, it's more like the orbifold of the 00:53:03.000 --> 00:53:05.000 ADS rule. 00:53:05.000 --> 00:53:15.000 Which is a special case. Yes. So this is below the black hole. Just, oh, I see. Okay. 00:53:15.000 --> 00:53:17.000 ADS 3. 00:53:17.000 --> 00:53:27.000 Uh, right, so you have to, I think, I mean, so so in some normalization, the Btc is 0. And this is all the black hole state is above 0 and then the 00:53:27.000 --> 00:53:32.000 Global ADS is minus one, so all the 00:53:32.000 --> 00:53:39.000 All the forces here, I mean, there are like minus k squared. 00:53:39.000 --> 00:53:42.000 Right, that's why it's called median range. 00:53:42.000 --> 00:53:49.000 So this is also from the analog of the giant graviton. So the giant graviton is also that the 00:53:49.000 --> 00:53:53.000 The giant graviton is a medium energy. 00:53:53.000 --> 00:53:55.000 And then because it's negative modes, 00:53:55.000 --> 00:53:59.000 You have this upshift. It takes you from the. 00:53:59.000 --> 00:54:03.000 medium to the, uh, to the hive. 00:54:03.000 --> 00:54:05.000 Uh, oh, so, so here… 00:54:05.000 --> 00:54:09.000 The difference is that at this medium range energy, 00:54:09.000 --> 00:54:14.000 So, so this n is scale at C. 00:54:14.000 --> 00:54:19.000 And then so you already can backreact. So this is different from the between the ADS 3 and ADS5. 00:54:19.000 --> 00:54:28.000 So for the ADS-5 case, the right-hand side of this giant graviton of this finite expansion is interpreted as a giant graviton brain. 00:54:28.000 --> 00:54:32.000 So here, because of this. 00:54:32.000 --> 00:54:35.000 The scaling at ADS3 case. 00:54:35.000 --> 00:54:39.000 They backrect, so you get backgrounded geometry. 00:54:39.000 --> 00:54:45.000 So also from the polls, you already see exactly the structure. 00:54:45.000 --> 00:54:50.000 And also from the spectrum, you see this 1 over case fractional spectrum. 00:54:50.000 --> 00:54:54.000 So they all suggest that you probably should. 00:54:54.000 --> 00:54:57.000 Look for some ZK Orbifold. 00:54:57.000 --> 00:55:04.000 So we want to — so we have the result of the Z hat k. 00:55:04.000 --> 00:55:08.000 Start obtained by doing this finite expansion. 00:55:08.000 --> 00:55:17.000 Now we want to actually reproduce it from some real computation that you have the interpretation of the bulk. 00:55:17.000 --> 00:55:20.000 It's this model… 00:55:20.000 --> 00:55:23.000 This doesn't have any debates there, right? 00:55:23.000 --> 00:55:30.000 They do have deep brain, but they don't… Yeah, but I thought you shifted it to. 00:55:30.000 --> 00:55:33.000 And the Bush-Schwartz hotline. 00:55:33.000 --> 00:55:39.000 Yes, but I can also look… I can also do the same for the arrow sector. 00:55:39.000 --> 00:55:43.000 I mean, it's just easier to see it in the sector. 00:55:43.000 --> 00:55:46.000 I mean, there are boundary states. 00:55:46.000 --> 00:55:48.000 in this model, but they don't… 00:55:48.000 --> 00:55:51.000 Uh, they don't fit the bill. 00:55:51.000 --> 00:55:54.000 They don't. I mean… 00:55:54.000 --> 00:55:58.000 We don't have a boundary state description for this yet. 00:55:58.000 --> 00:56:02.000 I mean, that's also what we're trying to look for, whether there's some deep brain that can. 00:56:02.000 --> 00:56:06.000 After backreaction, maybe some kicking monopole. 00:56:06.000 --> 00:56:10.000 I mean, right now we don't… we don't have a boundary state description for this. 00:56:10.000 --> 00:56:13.000 These are not open strings. 00:56:13.000 --> 00:56:15.000 scripts. 00:56:15.000 --> 00:56:20.000 Hmm. 00:56:20.000 --> 00:56:24.000 I think it's a bit… maybe you can. 00:56:24.000 --> 00:56:28.000 Ask this question again when I show you this computation in terms of worksheet. 00:56:28.000 --> 00:56:31.000 It's, uh… 00:56:31.000 --> 00:56:33.000 It's kind of also… 00:56:33.000 --> 00:56:36.000 is, maybe you can see later. 00:56:36.000 --> 00:56:44.000 So this suggested that you want to try to reproduce this in terms of some Orbit for geometry. 00:56:44.000 --> 00:56:48.000 So, uh, this is the, uh, punchline. 00:56:48.000 --> 00:56:50.000 That we're going to show that the Z hat 00:56:50.000 --> 00:56:54.000 that had Kate is the character partition function. 00:56:54.000 --> 00:56:56.000 on this orbifold background. 00:56:56.000 --> 00:57:01.000 So this is the old game that. 00:57:01.000 --> 00:57:07.000 Because it's a carcass sector, so it doesn't matter where you are at, so you can move to this. 00:57:07.000 --> 00:57:11.000 Last background with just one Q5. 00:57:11.000 --> 00:57:15.000 So in this background, there's a exact worksheet description. 00:57:15.000 --> 00:57:21.000 So we're going to use that worksheet description to compute this. 00:57:21.000 --> 00:57:23.000 So that turns out to be, uh, 00:57:23.000 --> 00:57:29.000 much easier to do than a true supergravity computation. 00:57:29.000 --> 00:57:34.000 So, so this is the orbital geometry. So write down this ADS 3 quasi. 00:57:34.000 --> 00:57:38.000 So, this is before the awful. This is the identification. 00:57:38.000 --> 00:57:43.000 So now you can do the orbit folding. So before identification. 00:57:43.000 --> 00:57:48.000 There's actually a two parameter family of different titsaki orbifold. 00:57:48.000 --> 00:57:51.000 So this labeled by this S bar. 00:57:51.000 --> 00:57:59.000 So when SS bar is the plus minus that correspond to our earlier plus minus that's appeared in the finite expansion. 00:57:59.000 --> 00:58:08.000 Asymmetric means different on the left and right movers? Yes. As in a heterotic. Yes. 00:58:08.000 --> 00:58:16.000 Right, so let's demonstrate with this. So yeah, so this is just… so before you do the orbit for this. 00:58:16.000 --> 00:58:18.000 This physical goal all the way back. 00:58:18.000 --> 00:58:29.000 And before I identify, I mean, so now for the k equal to 3, you just cover 2 pi over 3, and then you identify. 00:58:29.000 --> 00:58:35.000 Um, you can also do a cornea transmission to show that they are indeed asymptotic ADI3 cross S3. 00:58:35.000 --> 00:58:48.000 But with the deficit angle. So you know that they fit in with this same asymptotic. So you kind of want to include them in your gravitational path integral. 00:58:48.000 --> 00:58:51.000 FPPS. 00:58:51.000 --> 00:58:56.000 This is just, you're, right, so this is the orbifold that, um, 00:58:56.000 --> 00:59:02.000 preserve the supersymmetry. So, I mean, so that's why you want to. 00:59:02.000 --> 00:59:05.000 That's why… 00:59:05.000 --> 00:59:11.000 You want to offer both the ADS resources, both ADS 3 factor and S3 factor at the same time. 00:59:11.000 --> 00:59:13.000 Otherwise, then preserve the supersymmetry. 00:59:13.000 --> 00:59:16.000 So the fact that you want to 00:59:16.000 --> 00:59:21.000 Or before both sides in this coordinated fashion, because the supersmature. 00:59:21.000 --> 00:59:25.000 And then on top of that, we want to consider just the, um, 00:59:25.000 --> 00:59:35.000 Um, happy PS partition function on top of this. So this is just a tree level part. 00:59:35.000 --> 00:59:39.000 So we're going to use this worksheets. 00:59:39.000 --> 00:59:41.000 Uh, description. 00:59:41.000 --> 00:59:44.000 of the symmetric orbifold, the 00:59:44.000 --> 00:59:47.000 Here, given by Eberhard, Gabidi and Copa Kumar. 00:59:47.000 --> 00:59:50.000 So this is some. 00:59:50.000 --> 00:59:52.000 Uh, hybrid, uh… 00:59:52.000 --> 00:59:55.000 high performerism. 00:59:55.000 --> 01:00:03.000 So basically it computes a single particle spectrum of this space-time CFT, the symmetric orbit fold at n goes infinity. 01:00:03.000 --> 01:00:14.000 So you don't need to know the details, you just need to know that the physical spectrum here is summing over all the spectra flow sector given by labeled by integer. 01:00:14.000 --> 01:00:22.000 So the spectral flow here basically correspond to the — so this is labeled by W. W is integer. 01:00:22.000 --> 01:00:28.000 Here it correspond to the W3C sector of the symmetric orbit fold. 01:00:28.000 --> 01:00:33.000 So, so the vacuum is when W equal to one. So means means that the 01:00:33.000 --> 01:00:36.000 From the… 01:00:36.000 --> 01:00:40.000 Uh, a… 01:00:40.000 --> 01:00:48.000 from the… this symmetric group conjugate class is the one over the 11. 01:00:48.000 --> 01:00:51.000 So now the W2 is that the 01:00:51.000 --> 01:00:56.000 you're twisting your running twice, and then from the. 01:00:56.000 --> 01:01:01.000 Worship perspective, you have a string that one twice. 01:01:01.000 --> 01:01:03.000 So that's all standard. 01:01:03.000 --> 01:01:05.000 So now, how about for the orgifos? 01:01:05.000 --> 01:01:08.000 So this gentleman here, 01:01:08.000 --> 01:01:10.000 Uh, they… 01:01:10.000 --> 01:01:12.000 They, they, um… 01:01:12.000 --> 01:01:17.000 extend this warship description also for this orbifold case. 01:01:17.000 --> 01:01:22.000 So the result is that it's the same spectrum for the each. 01:01:22.000 --> 01:01:25.000 Before the spectral flow, 01:01:25.000 --> 01:01:27.000 It's the same spectrum. 01:01:27.000 --> 01:01:30.000 And then they sum over all the 01:01:30.000 --> 01:01:36.000 spectral flow sector, but then now the spectral flow is integer. So it's fractional. 01:01:36.000 --> 01:01:43.000 It's the n over k, and this case correspond to this Zk. 01:01:43.000 --> 01:01:45.000 So now, uh… 01:01:45.000 --> 01:01:51.000 Let me demonstrate with the k equal to 3. So for the vacuum, so their proposal is that, uh, 01:01:51.000 --> 01:01:58.000 For the vacuum of this — so this is related to whether this is a open, whether it's open. 01:01:58.000 --> 01:02:01.000 So, uh… 01:02:01.000 --> 01:02:04.000 This is the worksheet series that, uh, on… 01:02:04.000 --> 01:02:07.000 This orbital geometry, but it turned out. 01:02:07.000 --> 01:02:14.000 It describes a sub-sector of the original asymmetric order for CFT that correspond to the n-orifold geometry. 01:02:14.000 --> 01:02:16.000 So the difference is that now. 01:02:16.000 --> 01:02:18.000 you change your vacuum. 01:02:18.000 --> 01:02:24.000 So the vacuum is changed to a highly non-perturbative vacuum. 01:02:24.000 --> 01:02:26.000 So that's the proposal. 01:02:26.000 --> 01:02:31.000 So this is the vacuum states. The vacuum state is already three winded. 01:02:31.000 --> 01:02:33.000 phone, uh, from this, uh… 01:02:33.000 --> 01:02:35.000 Um… 01:02:35.000 --> 01:02:40.000 From this system. 01:02:40.000 --> 01:02:45.000 Yes, but he's so highly excited. It's like. 01:02:45.000 --> 01:02:49.000 I mean, this is n divided by K. 01:02:49.000 --> 01:02:52.000 So before you're… 01:02:52.000 --> 01:02:54.000 Your vacuum is like one. 01:02:54.000 --> 01:02:56.000 Yeah, and there it is oh… 01:02:56.000 --> 01:03:02.000 There's an audience. 01:03:02.000 --> 01:03:07.000 Uh, sorry, sorry, this is… this is not exciting yet. You're exciting on top of this. 01:03:07.000 --> 01:03:15.000 Yeah, yeah. But so, so now this is okay, like many pay sector, and so the n divided by K. This is much higher than 01:03:15.000 --> 01:03:21.000 That's why it's already non-perturbative. 01:03:21.000 --> 01:03:26.000 So this is a vacuum, but you see that because this. 01:03:26.000 --> 01:03:33.000 Is this W equal to n over k. So the smallest one is actually the one with the lowest energy is not this vacuum. 01:03:33.000 --> 01:03:38.000 So, actually, you have a state that's actually lower than the vacuum. 01:03:38.000 --> 01:03:44.000 So you can actually lower your energy. So that's where the negative modes come from. 01:03:44.000 --> 01:03:50.000 So, for the K equals to 3, you have actually two negative modes. One correspond to the 01:03:50.000 --> 01:03:52.000 W equals 1 over 3. 01:03:52.000 --> 01:03:54.000 is this guy here. 01:03:54.000 --> 01:03:57.000 And one correspond to the this 2 over 3. 01:03:57.000 --> 01:03:59.000 So that's where the… 01:03:59.000 --> 01:04:03.000 Um, this negative vote comes from. 01:04:03.000 --> 01:04:08.000 Right, so, so, so then, uh, we can just use this worksheet description. 01:04:08.000 --> 01:04:12.000 Uh, so we can reproduce this Zk. 01:04:12.000 --> 01:04:19.000 that we computed earlier from this residue computation. 01:04:19.000 --> 01:04:24.000 Uh, so, okay, so this is the half PPS. How much time do we have? 01:04:24.000 --> 01:04:28.000 Let me just announce the result for the quota Bps. 01:04:28.000 --> 01:04:31.000 For sure, well, go. 01:04:31.000 --> 01:04:33.000 Okay, so… so… 01:04:33.000 --> 01:04:42.000 This is the… so what I discussed is the result from the last December, and then we have been trying to generalize to the quarter Bps. 01:04:42.000 --> 01:04:46.000 So the quarter BPS, we look at this EDP genus. 01:04:46.000 --> 01:04:49.000 So this is a standard case, we look at the K3. 01:04:49.000 --> 01:04:53.000 And we have this TMEV formula. And 01:04:53.000 --> 01:04:55.000 Turner. 01:04:55.000 --> 01:05:02.000 Uh, the boundary result we want to reproduce is this with Jacobi form of weight 0 and the index k. 01:05:02.000 --> 01:05:05.000 And then we also can… 01:05:05.000 --> 01:05:10.000 goes to the sector. So this is what we want to reproduce starting 01:05:10.000 --> 01:05:14.000 This DMVV formula is the Grand Cannonic ensemble. 01:05:14.000 --> 01:05:19.000 a result that we want to do the counter interval with. 01:05:19.000 --> 01:05:22.000 So, yes, so this is the… 01:05:22.000 --> 01:05:30.000 We have this black hole threshold. We want to basically right now we can look at everything below the threshold. So those are the polar states. 01:05:30.000 --> 01:05:35.000 So the polarities means that everything that's below the threshold. 01:05:35.000 --> 01:05:39.000 So in terms of the NS, that means that on the 01:05:39.000 --> 01:05:46.000 Uh, rice sector, you have a car primary on the left sector, you have a… you have everything. 01:05:46.000 --> 01:05:49.000 So this is some products that you're… 01:05:49.000 --> 01:05:52.000 probably familiar with. So this is the RS sector. 01:05:52.000 --> 01:05:55.000 And drawing the polar states. 01:05:55.000 --> 01:06:00.000 So these dots here, they are the happy PS. 01:06:00.000 --> 01:06:05.000 piece… I mean, the polar bed, they're happy already covered before. 01:06:05.000 --> 01:06:08.000 And this triangle, they are the 01:06:08.000 --> 01:06:12.000 really quarter BPS power state. 01:06:12.000 --> 01:06:18.000 So I have all this degeneracy that I want to reproduce. 01:06:18.000 --> 01:06:21.000 And yeah, so this is the… 01:06:21.000 --> 01:06:23.000 Another sector plot. 01:06:23.000 --> 01:06:25.000 So you see that the… 01:06:25.000 --> 01:06:27.000 This is a… 01:06:27.000 --> 01:06:31.000 What I showed earlier, this line, this character primary line. 01:06:31.000 --> 01:06:37.000 And starting from ink to Phi, you see one. 01:06:37.000 --> 01:06:40.000 For the BPS polar state. 01:06:40.000 --> 01:06:43.000 Right, so… 01:06:43.000 --> 01:06:48.000 Yes, I'm just saying what I want to reproduce. So remember for the high BPS sector. 01:06:48.000 --> 01:06:51.000 Uh, it is the result of the… 01:06:51.000 --> 01:06:55.000 expansion and the. 01:06:55.000 --> 01:06:56.000 So. 01:06:56.000 --> 01:07:03.000 This is the access bar, this plus minus. Remember this which set contribute depends on the chamber here. 01:07:03.000 --> 01:07:08.000 So now we can do the same. So right now we can only do the polar state. 01:07:08.000 --> 01:07:18.000 So it's a similar result. So right now, but we have, apart from this plus minus plus minus. 01:07:18.000 --> 01:07:21.000 Rs can be, um… 01:07:21.000 --> 01:07:25.000 larger and it's constrained by this level merging condition. 01:07:25.000 --> 01:07:28.000 And then what we can do is that, uh, 01:07:28.000 --> 01:07:35.000 So this polar state is when you say restricted polar state, we're already, in some sense, go to the Michael Canon ensemble 01:07:35.000 --> 01:07:38.000 So for each microcontinent ensemble charge, 01:07:38.000 --> 01:07:43.000 Oh, you can see that only a finance set up for KIS contribute. 01:07:43.000 --> 01:07:49.000 And this is the one loop how indigenous that we can again reproduce from the 01:07:49.000 --> 01:07:55.000 worship description. So this is just saying that we can generalize our result to the polar part of the. 01:07:55.000 --> 01:07:59.000 quarter Bps. 01:07:59.000 --> 01:08:01.000 So let me sum up. 01:08:01.000 --> 01:08:08.000 The pulse structure is going to be a lot more complicated. But again, are you only selling over the simple poles? Again, there going to be. 01:08:08.000 --> 01:08:13.000 Aligned essential singularities in the zeta plane. 01:08:13.000 --> 01:08:17.000 Uh, I mean, slightly… I mean. 01:08:17.000 --> 01:08:26.000 Yes, it's accepted that the war is just your… whether you're inside your single upper half plane or not. 01:08:26.000 --> 01:08:31.000 I mean, this is the old story that you start from your DMV chamber. 01:08:31.000 --> 01:08:36.000 And then you can rewrite it, you can move to another chamber, and then you'll 01:08:36.000 --> 01:08:41.000 And you see, when you move to the other chamber, how many poles you're crossing. 01:08:41.000 --> 01:08:43.000 So for each pole that you're crossing, 01:08:43.000 --> 01:08:47.000 It corresponds to an orbit code, basically. 01:08:47.000 --> 01:08:52.000 I can show you the detail later. 01:08:52.000 --> 01:08:56.000 Uh, right, so that's the, uh… 01:08:56.000 --> 01:08:59.000 For the ADS3-CFT correspondence. 01:08:59.000 --> 01:09:07.000 I apply this final expansion on the HubS and the quota Bps sector, and the polar part of the quarter BPS sector. 01:09:07.000 --> 01:09:11.000 And then we see that the box title correspond to this orbifold. 01:09:11.000 --> 01:09:13.000 And then we checked it and says, uh, 01:09:13.000 --> 01:09:17.000 It works essentially the same way as the ADS-5 case. 01:09:17.000 --> 01:09:23.000 So the… or before you can view this as an ADS-3 analog of the junkert. 01:09:23.000 --> 01:09:25.000 So, um. 01:09:25.000 --> 01:09:29.000 Now, this is… I already emphasized earlier, that. 01:09:29.000 --> 01:09:34.000 Importantly, you see the effect of the negative modes. So that negative modes. 01:09:34.000 --> 01:09:43.000 From the symmetric orbital viewpoint, it's just that because you start with the K cycle, and then this k cycle can break. 01:09:43.000 --> 01:09:48.000 That's how you get to select negative mode. 01:09:48.000 --> 01:09:53.000 So this is a comparison of the ADS-5 versus ADS-3. 01:09:53.000 --> 01:10:00.000 So you start from the light, and you want to reproduce the 01:10:00.000 --> 01:10:03.000 all the… all the range. So… 01:10:03.000 --> 01:10:08.000 The middle one from the boundary perspective, that's when the trace relation kicks in. 01:10:08.000 --> 01:10:11.000 And then we have a, like, uh… 01:10:11.000 --> 01:10:14.000 When you have many, um… 01:10:14.000 --> 01:10:21.000 Giant Graviton, then the trace relation start to dominate, and then you reach your heavy, heavy regime. 01:10:21.000 --> 01:10:25.000 So having in ADS5 case means that there is order n squared. 01:10:25.000 --> 01:10:31.000 And for the ADS 3, it's the same. The only difference is that what do you mean by light, medium, heavy. 01:10:31.000 --> 01:10:33.000 But essentially, so… 01:10:33.000 --> 01:10:39.000 It's a complete parallel at this level. 01:10:39.000 --> 01:10:44.000 Um, so the, uh, the interesting further question is, 01:10:44.000 --> 01:10:49.000 Whether you can reproduce this dead head with a direct supergravity computation. 01:10:49.000 --> 01:10:53.000 And whether you can have a… 01:10:53.000 --> 01:10:59.000 systematic analysis of the analog of the trace relation. That's what the Greg asked earlier. 01:10:59.000 --> 01:11:03.000 And whether you can actually extend to the non-polar. 01:11:03.000 --> 01:11:05.000 Rajit. Yeah, that's that's it. 01:11:05.000 --> 01:11:08.000 Sorry if we're running over time. 01:11:08.000 --> 01:11:14.000 Thanks. 01:11:14.000 --> 01:11:20.000 We have a lot of questions and more. 01:11:20.000 --> 01:11:25.000 Why aren't there other non-perturbative effects? Why is it only ZK orbifolds? 01:11:25.000 --> 01:11:29.000 I think because we're focusing on the PPS sectors, 01:11:29.000 --> 01:11:38.000 I mean, why could there be other singular geometries? 01:11:38.000 --> 01:11:40.000 Am I not able to take orbifolds by… 01:11:40.000 --> 01:11:43.000 Other subgroups of, uh… 01:11:43.000 --> 01:11:46.000 Jesse, too, why couldn't I take the… 01:11:46.000 --> 01:11:48.000 I don't know a GK Orbital. 01:11:48.000 --> 01:11:54.000 So, what… It's DK instead of ZK. 01:11:54.000 --> 01:12:02.000 I mean, maybe if you look into it, maybe it doesn't even preserve the supersymmetry that you want to. 01:12:02.000 --> 01:12:11.000 Well, I mean, that's why it says approval. But I mean, maybe not. 01:12:11.000 --> 01:12:15.000 And my feeling is that there are probably a lot of non-perturbative effects in string theory. 01:12:15.000 --> 01:12:18.000 Why are they leaving once that matter. 01:12:18.000 --> 01:12:21.000 But maybe you want to go to, maybe, uh… 01:12:21.000 --> 01:12:26.000 Uh, yeah, I'm not sure if you go to… I think… 01:12:26.000 --> 01:12:39.000 For the quarter, even for the non-polar part of the quarter, you might not see them. That's the whole non-polar part. Yeah. 01:12:39.000 --> 01:12:42.000 Yeah, we can have a bed. 01:12:42.000 --> 01:12:52.000 Okay, let's thank the speaker again, and let me remind you that there's a dinner, so if anybody's interested in coming, please tell me.