I hope I pronounced that right. And he's going to tell us.
about strings on Ads 3 times s 3 times s 3 times S1, which has the most symmetry and turns out to be the hardest case. Well, thank you very much. Thank you very much for the invitation to come here.
I think I've come here once before, but it's so many years ago that I wouldn't have found a place.
But I'm not that often in the US. So indeed, as Craig was mentioning.
Oops, click somewhere to click somewhere. Somewhere on your signage.
That's good. Okay. So as… So as Greg was indicating, the CFT dual of strings on Ads 3 equals 3 equals T4 has been studied in great detail, and everybody agrees what the what the CFT dual is. It's the symmetric orbital fold of T form.
The situation for the seemingly simpler because more symmetrical case of ADS 3 equals 3 equals 3 equals this one has been less clear. So let me first remind you of what people know or believe they know, and then I'll try to explain to you what recent progress we've made in identifying the CFT tools.
So the dual of string theory in Ads 3 constants 3 constants 3 consists one. It's maximally supersymmetric, so it will have n equals to 4 super conformant symmetry. So there will be 4 super currents. But the R symmetry is bigger than for the.
more familiar small n equals to 4 superconformant geometry. And this is simply due to the fact that you have 2 S3s in the target space. Each S. 3 has an SO4 and So 4 has an SU 2 plus SU. 2. So you have an SU 2 plus SU 2 on the first S3 and SU 2 plus SU 2 on the second S. 3.
So you would expect SU2 plus SU2 acting on the left and SU2 plus SU2 acting on the right, and therefore you would expect that the dual CFT as Verasero, which comes from ADS3, and then two SU2s with levels that.
corresponding to the size of the corresponding S3s and for supercharges. And there is, in fact, an algebra you can write down that closes and satisfies Jacobi. It's what's called the large n equals to 4 superconformal algebra. I spare you writing down.
the full set of computation relations. But you can believe me that it exists, and it has a very specific central charge which is expressed in terms of the levels of the 2 SU tubes. So for the small n equals to 4, the corresponding statement is that you just have one level, and the central charge is 6 times that level.
For the large N equals to form, you have two levels and the central charge is the product, 6 times the product divided by the sum.
So this is the what you sort of know based on symmetries. And then there is sort of the the most naive proposal you would think is that if you are, if you are simple minded, you would say Ads 3 crosses 3 cross T 4 is the symmetric orbifold of T4. Ads. 3 cross s 3 cos K. 3 is the symmetric orbit full of K. 3. So ads 3 cross s3 cross.
History S1 should be the symmetric orbifold of S3 cross s1. And what does S3 cross s1 mean? It's basically the so-called s-kappa theory. It's an SU2 level K theory describing the S3, and the U1 theory describing the S1.
And it's a super n equals to one super conformal version of it. So it's really SU 2 plus U1 plus 4 free fermions. And that was believed to be a reasonable proposal for what the CFT dual of this background should be.
Now, more recently, written has looked into it. There's another proposed natural proposal, which is that it's the sigma model on the instant on noduli space, and what witness recently shown. This has indeed larger than equals to 4 super conformal symmetry. So that's another natural candidate.
But. be that as it may. So these are candidates based on symmetries, or based on naive expectations from what you know from ADS 3, because S3 cross T43.
Now, as Greg in this famous paper with Martineck, and Strominger has shown this is unlikely to be true, and the reason was that the Boer and collaborators had done the supergravity analysis on this background.
And I classified all the supergravity BPS states, and what Greg and collaborators pointed out is that the symmetric orbit S kappa has far fewer Bps states than those predicted by supergravity. So the BPF spectrum of the two.
description doesn't match, and then you would believe that means they can't possibly be dual, because at least the BPS states with Hot.
Okay, so this is sort of a part of that. Now there is oops. Now there is a.
There is actually a small wrinkle to that, which is that.
Yes, I did show it some years ago with Lawrence About. In fact, that was Lawrence's master thesis at ETH was to calculate the… to redo the calculation of the Boer honestly. And what we found is that.
the argument that he had given, which was purely based on group theory, is not actually capturing the entire physics, and there are actually far fewer Bps states than he predicted.
Because some of them get really lifted once you really solve the — I mean, you look for the solutions of the Laplace equation, the eigenfunctions, and you determine the eigenvalues, and you see that the supergravity states are actually far fewer, and the BPS spectrum actually matches correctly the symmetric orbital of Skappa.
Provided at least that K. One of the cases, one and the other one is kappa plus 1.
Again, it's only checking the BPS spectrum. It's not showing that this is the correct answer, but at least the reason for why Greg and collaborators argue that it couldn't possibly be true isn't quite correct, because the supergravity spectrum was wrong. I mean, their analysis was correct, but the supergravity spectrum was.
However, this is only a very special case, and you could wonder whether we can find a more general answer, and whether we can say more.
Now, there are 2 prongs with which I'm going to approach this problem. So you believe it's a symmetric. In this case, you believe it's a symmetric product. The actual answer is the symmetric product.
I don't see any reason why it's not. Spectra now match. Vps spectrum matches. Yes. Vps spectrum matches whether I have a strong opinion whether it is exactly that or not is a different matter, because I'm… But at least on the level of the Bps spectrum, that's true, and that's one candidate, but it's, I think, only a candidate if one of the levels is high. I thought there was this…
analogs, we elliptic genus that also one could compare. Well, but the analog of the elliptic gene is basically, again, captures non-homorphic index. No, no, no, the non. There's a non-holomorphic.
Right. But I think it, again, checks basically just the aspects of the BPS spectrum. It's not going beyond the.
Okay, anyway. So okay, so I'm not. I mean, this is this is not what I'm going to talk about today. So so there are 2 recent ways in which we have been trying to make progress on this, and one is coming from the fact that we have recently made progress in understanding the situation for ADS 3 crosses 3 cross T4.
And that reliant on understanding the pure the Schwartz sector background, because for the pure Schwartz sector background, there's a resume in a written model description. And for the case at hand, the relevant resume in a written model is SL2R level K plus SU2 level K plus SU2 level k plus u1.
And what you could hope is that you could sort of follow the same line of argument of the papers with Lawrence and Rogers, where we derive the symmetrical overfold of T4 more or less from first principles from this virtual description. One could hope that one should be able to do that.
And what I want to explain to you in the first half of the talk is that this is indeed true, and we've more or less, I believe, derived what's the duality, what's the dual CFT? For the pure Neverschwarz background, where the levels of the two SU2s are 2 and the level of the SO2 are S1. And I'll explain that to you.
And it's not the symmetric orbit float of. as couple. But it's very closely related to it.
Okay, so this is one approach, and the other approach is to look at a problem in terms of microscopic deep brain configuration. So if so one road towards understanding that a symmetric orbit hold of T4 is the CFT dual of the ADS 3 cross 3 cross T. 4 came from the D1 D 5 brain system, and looking at the open string degrees of freedom and understanding the structure of it.
So you could attempt to do something similar here. So what the other idea is that we are looking at this D brain configurations where you start with two stacks of five brains that are orthogonal to one another. So the the check marks indicate the Neumann directions of the.
the the tangential directions of the corresponding brains. So they are orthogonal to one another. So they preserve supersymmetry in the D. One brain is along the intersection of the 2. And what you could try to do is to analyze the open string degrees of freedom and deduce from there.
What that you will see if T is, and that's the other half, what I'm going to try to explain to you. Progress with understanding this. So this was attempted by David Tong about 10 years ago. I think he gave convincing arguments that in the infrared, this requires large and equal to four superconformal symmetry.
But he couldn't identify the CFT to which it would flow, and we are going to make a proposal for what the CFT to which this will flow to, and I'll explain to you the evidence we have for it and the ingredients that go into this analysis.
So this is… this is the plan of the talk. I want to describe progress on two fronts. So the first half, which will be briefer, will be to explain the tensionless Schwarz-Lever-Schwarz background version, where I think we can be very concrete. This is based on work with I wrote with a PhD student who is Vitrikul.
was now at Weizmann, and then an important follow-up paper with Lawrence last year. And then the other thing is something that I'm currently doing with Rajeshkopakuma, where we are looking at this deep brain system in a very specific limit, which you should think of as being sort of the analog of the.
0 truth limit of the usual D1 define point system, and I'll explain to you the consequences of that.
So that's that's the plan of the talk. So I shall first explain to you a detention listener with Schwarzen Schwartz background, and I have a brief section explaining you to you where in the bigger model space of supergravity.
This version sits in and where this version will sit into, and then I'll explain to you our evidence and our description for the CFT dual for the deep brain system. And then there are many open questions. So this is work in progress, so it's not that we have done the second half completely, but I think there's.
I'm quite excited about it, so I'll explain to you what remains to be done.
Okay, so please interrupt me if you have any questions.
Okay, so let's start with the tensionlessness with Schwartz over Schwartz background. And this is a really relatively well, it hinges on one important observation. So the so if you are in the background with Nobel Schwartz flux, then you can describe Ads. 3 in terms of an Sl. 2r resume in a written model.
But because we are in the world of superstring theory, it's the end equals to one superconformal version of SL2R. So what this means is this is the affine D algebra together with fermions in the adjoint representation and the corresponding commutation relations. So you have the SL2R factor for ADS3, then you have an SU2 factor for one S3, another SU2 factor for the other S3.
And the U1 factor for the S1. And demanding that the whole thing as C equals to 15 gives you this relation between the levels of the three algebras, which translates into the condition of the central charges I was writing down earlier.
And what we've learned in this early analysis is that the tensionless limit arises if the SL2R theory is at level one. So one idea you could have is that you can ask, okay, when can this k be equal to one?
But since K plus and k minus have to be integers, positive integers, the only way this can be run is if k plus and k minus are both equal to 2.
Well, there's 2 solutions, either K plus is equal to one and k minus is equal to infinity. That's basically ads 3 equals s 3 cross flat space. And the other solution is that both of them are equal to 2, a half plus a half is 1.
Now, why is this interesting? Well, this is interesting because that actually has a very simple world sheet description. You see, the supersymmetric level 2 SU 2 theory is nothing but 3 free fermions.
Because when you decouple the fermions, you reduce the level by two for SU2, so once you have decoupled the fermions, there is no bosonic degrees of freedom left. So the supersymmetric SU2 theory at level 2 is just 3 free fermions.
So therefore, we have three free fermions from the 1S23, 3 free fermions from the other S3. From the S1, we get one boson and one fermion, and the rest is this SL2R level 1 theory, which, by the same token, I can write in terms of 3 free fermions and the decoupled SO2R.
But for SO2R, the shift goes the opposite direction. So this gives you SL2R level 3.
And so this is the world sheet theory. It's basically lots of free fields, namely altogether 3 plus 3 or 10 fermions, 7 plus 3 fermions in 10 fermions. One goes on from here. The Sl. 2 r Level 3 theory. But you're going to have to impose a physical state condition which is going to kill two bosons and 2 fermions.
So so the hero of this story is So on level 3. So what can you say about Sl 2 on level 3? In general, the affine, the Sl. 2 R affine theories are very complicated. But SL2R level 3 seems to have a remarkably simple.
Modular invariant spectrum and consistent set of correlators. That is really the simplest affine So to r theory I've ever seen. So the proposal is that what we should take as a to our level 3 to be is a specific spectrum where you only include the continuous representations.
And all their spectrally flowed images, and you integrate over this P parameter that characterizes the continuous representations. But you think of P as an integral going from minus infinity to plus infinity. I… you're introduced basically 2 copies for each representation, and you think of them as a representation, and it's conjugate representation.
And it is formally, this is a modular invariant spectrum. I mean, obviously, it's a little bit delicate because all of these representations are infinite dimensional. So you have to regularize it.
But to the extent that you can believe. formal arguments, you would think this is a modular invariant partition function describing strings matrices for these things, right? Right. So but I think they normally looked at the discrete representations, and they didn't normally do the spectral flow.
Yeah, so this is only continuous representation. So let's see. So this is the.
You're you're taking the continuous representation of SL2R, and then you're turning that into a representation of the affine. Right? So so this is what I mean here. Have you checked the modular curvance? Yes, yes. So so these characters, I mean, you can write down the characters, and you formally it's I mean, it's integrals and so on, but formally, this is multilinevariant.
Now, okay, so this looks very formal. But what's really striking is that this model with this spectrum has a remarkably simple set of correlations, correlation functions that we checked to be crossing symmetric and satisfying the conditioning homological equation.
And this is given by this formula. So what Lawrence and I observed is that, so this is not the normal SL2R level K.
correlate us analytically continue to k equal to 3. This is the second set of correlation functions that only exists for SL2R and level 3. And so the vertex operators are labeled by the spectral flow parameter. P is the parameter.
So P is this parameter. W is the spectral flow parameter. And then inside this representation, the state is labeled by the quantum number with respect to J3 and J 3 bar 0. What is the spectral flow go from minus infinity to infinity?
Yeah, so this is always so I think on the in spectrum, it only goes from 0 to infinity, because when you rotate.
I mean, under… the out states will then go from… have the negative spectral flow, because as you move these states in in the X space in the target space, you're basically also rotating the spectral flow direction.
So the spectrum is always for the if you write on the spectrum as for the in states, then you only have positive numbers of spectral flow. W's are positive. Yes.
So the statement is the correlation functions are given in terms of sums over covering maps. So this vertex operators depend on where you insert the vertex operator on the world sheet, and they depend on where you insert the vertex operator in the dual CFT.
So X is a parameter that lives on the dual, on the sphere of the boundary sphere of ADS3. And you rotate different values of X are rotated into one another by the translation operator in the dual CFT.
And that is the normal coordinate on the worksheet. And what we propose is that these correlation functions are mostly 0, and they're only non-zero if there is a covering map that maps the Zs to the x's, such that.
Near the point X, there is a WI fold. Near Xi, there's a Wi-fold cover.
And so this answer is a sum over this possible covering maps with this delta functions, because these covering maps don't generically exist in moduli space times parameter. They depend on the covering map times something that looks like a free boson. Yeah, I didn't understand what you meant by instakes and out states.
So it's your Ads. First of all, your ads is Lorentzian. No, it's Euclidean. So what do you mean by? Well, so what I mean by so from the from the CFT perspective by in states, I mean evaluating the vertex operators for Z equal to 0. So letting them act on the.
So, so this is, first of all, a statement about a Westminor written model.
A priori, it has nothing to do with ADS3. What I'm claiming there is a consistent CFT. This consistent CFT is generated by these vertex operators acting on the vacuum and the states are identified by saying I take z goes to zero and apply it to the vacuum, that is the state field correspondence.
And for that, if I said Z and X to 0, that's the state field correspondence. And what I'm in response to Greg's question.
Because I'm introducing this X parameter. The X parameter will act on the vertex operators by conjugation with the target space. So the vertex operator corresponding to any state evaluated at x and z.
is equal to the vertex operator evaluated at x equal to 0 in Z by e to the x l minus 1 spacetime.
and X minus one space time is just e to the xj plus 0 on the world sheet. So from the point of view of this.
So what I'm doing is at x equal to Z equal to 0, I only include states with W positive, because if I take x to infinity, that will affect also the spectral flow, and it'll effectively return change the spectral flow. The positive spectral flow by negative to negative spectral flows. So I would be over counting.
If I included both W positive and negative, and I include the X parameter as a way of parameterizing value states that you will see.
But on the face of it, this is just a pure Cft formula. What I'm claiming is there is a set of… So so the X dependence is implicit in the covering map gamma. Yes, correct. And what's the last factor that you circled? The pi hat and the AI right? So so so so what I'm looking at are covering maps, gamma.
of that that have the property that they're equal to Xi plus AI. Yes. But yes, so here I'm only looking at sphere covers over sphere covers. So near that.
equal to Zi. It has to map for Xi plus a wi branching around it. And there's a parameter that characterizes the sort of scaling behavior near this point. And these are these A parameters. And pi.
So discovering maps must have poles. And the pi is the product of the residues of the poles. So it's data associated to the covering map that is a little bit baroque. But it's something that's very explicitly known. So so this is just a very explicit formula. I'm simply saying this correlator is 0 unless the X's and Zs are such that I can find a holomorphic covering map that maps the zi's to the Xis with this branching behavior.
In which case. So this is this delta function. Obviously, this delta functionality depends on n minus 4 points, because 3 points by Sl. 2 are on the world sheet, and 3 points by Sl. 2 are on the on the space time. I can move anywhere I want. So 3 points I can always map.
So once I have more than 3 points, I get delta function constraints. And for every additional point, I get one additional delta function.
And it depends on some combinatorical data that comes from the covering map. So this is the covering map data. And then it's the other piece is basically telling you that p behaves like the momentum of a free bow.
Now, this is a bizarre formula. You would think this formula has no no looks complicated. But what we've checked is that that this formula satisfies the is crossing symmetric and solves the condition examiner logic of equation.
at least Y equals to 4, and we believe this to be a consistent formula of a consistent set of correlation functions for that theory. So again, this is kind of orthogonal to what Teschner did. This is orthogonal to what Tesh. So this is a proposal. Now, what's neat about this proposal? You see, if you think about this in the context of string theory in ads 3.
If you think of the Z as being the world sheet variable, then you have to integrate the z's over the entire moduli space. If you do this on the sphere, they're going to be n minus 4 free integrals, and there are n minus 4 delta functions. So what happens is when you do this modular integral.
The correlation function will just pick up contributions from finitely many points in moduli space. Maybe from those points where a covering map exists, and the answer is just the sum over covering maps times some finding combinatorial factors. And that's exactly the structure.
of the correlation functions of the symmetric orbitals, because in the symmetric orbit fold, one way of calculating symmetric orbit correlators is to write it in terms of sums of covering maps.
where you go to the covering surface to undo the orbi folding, the twisting, and therefore, if that is the right answer, the world sheet modular integral will reproduce more or less by construction, the symmetric orbits for correlation.
So whenever you have this formula, you know that that you will see if T will be a symmetric orbital, because the correlation functions on the world sheet will just repackage for you the way you calculate the symmetric orbit full correlators.
And at the same time you can calculate the spectrum. You can solve the physical state condition, and you can ask the symmetric orbit of what are you going to get? And that's relatively simple in this example, you see. So this is what we have. We have 10 fermions. We have Sl. 2r level 3, and we have one boson. So what's left over, we get 8 fermions because 2 fermions get eaten.
Two bosons get eaten, so we keep that boson, and we keep the radial direction of SL2R, which is the one that is labeled by momentum P. This P parameter of the continuous representation becomes the momentum of that boson, because that's exactly how it appeared here.
in the in the correlation function as defined on the world sheet. And then the rest of the Sl. To r is just there to make sure that you get a sum over covering maps, i.e. You reproduce the structure of the symmetric order.
So therefore, you can basically read off from this formula that the dual CFT is the symmetric orbifold of one compact boson coming from the S1, one non-compact boson, which is the radial direction from Ads. 3 plus 8 fermions. And it's if you.
I that this is the correct description for SL2R level 3. This is something you can simply read off from the sound, so that there is a remarkable world description for SL2R such that you can read off the dual Cft answer, and if you do, that's the answer. So this is.
Literally a derivation of the ADS-CFT correspondence. Right, so this is the analog of what we did for ADS3, because S3 equals T4, except here we can do it pure NSR.
For ADS 3, cross S3, cross T4, we had to go to this hybrid description because the SU2s were at level 1 at 1, and if you were to decouple the fermions, you would get SU2s at level minus one, and you couldn't do that.
But here, this is an honest derivation. Yes. What happens to the string loop corrections to this? Very good. So so here we've honestly only done it for the sphere coverings. But in the for the hybrid string.
What you can check is that this localization property is not just true on the sphere, but it's also true for hierogenous world sheet.
And what you furthermore know is, if you write down symmetric orbit, so symmetric orbital correlation functions you can write in terms of sum over covering maps. And the one over n dependence of that expansion is connected to the genus of the covering surface.
The higher genus covering surfaces are subleading in 1 over N because they involve fewer active colors, and then the combinatorical factors are down by powers of 1 over n.
And because this higher gene is covering surfaces come from hygienus world sheets, it basically tells you that this analysis is true not just to leading order in 1 over n, but to all orders in 1 over N. In fact, you can read off from that that the correct description is not the symmetric orbifold.
of T4, with N being proportional to G to the minus a half, but it's the grand canonical ensemble.
It's the it's the chemical potential in the grand canonical ensemble that becomes the coupling constant in the dual CFT. And you can see this very explicitly from the way the coupling constant dependent of the higher genus surface is rearranged themselves to reproduce the correct.
extremely expensive. No, no, I don't understand. So usually we think there's a sigma model.
describes the tool. Right. But now you're talking about ensembles of theories, right? So what I'm saying is so.
for the usual D1, D5 brain system. People believe it's a symmetric orbital of that number for the fun for the NS sector.
But you would have to do is you look at NS5 brands and fundamental strengths. And what I'm saying is the number of fundamental strings is not a good super selection sector.
You can't really restrict n, because in string theory, you will continuously produce additional fundamental strings, and therefore in the NS version of this duality, it's the grand canonical ensemble of the symmetric orbital that appears. And it's very clear from the.
Hygieneous corrections to our analysis that that must be the answer. Then the combinatorical factors come out right, and if you try to simply identify G with 1 over N, G with 1 over square root of n, the higher the subleading 1 over n corrections wouldn't come out correct.
But is what you're saying relevant to the current debate about whether ad involves ensembles. Well, yeah, I don't think so. I think this is a sample in a different sense. It's really, it's really just a way of keeping track of what the parameters are.
I mean, I think of this as one theory. I'm just thinking of it that a correct dictionary between the coupling constant and the parameter in the symmetric orbifold is not n, but it's the chemical potential in the grand canonic ensemble conjugate train. Isn't there some story about long strings going off the… This is what it is. I mean, I mean, effective, you see here we are in the tensionless regime, and therefore everybody is long strings, because there is no space for short strings. That's the reason why you only have continuous representation.
That's basically the picture. But the reason why you have to go to the Grand Canonical ensemble is, I think, because the number of fundamental strings is not a good quantum number to keep phi.
For the D. One, the 5 point system, you would expect it to be a fixed number of them, because they're really super selection sectors parameterized by Roman Roman, whereas for the pure Neville Schwartz case, that's not the case.
Anyway, so this is, I think, what we're showing not only obviously, as Craig will have probably observed. You see, this looks basically like a 0 squared, because these are 8 fermions and 2 bosons. 0 is one boson and 4 fermions. So this is basically a 0 square.
So what you learn is that the dual Cft is a 0 squared, except that one boson is compact and one is non-compact.
But this is more or less a derivation in that little corner. So.
How this generalizes to other values? I don't know. But at that corner, that's what we understand.
Okay, so this is this is the evidence number one that this is the correct Cft dual at one point in a very funny point in what's a tensionless point in evenness.
And our attempt. The second half will be to try to do the analogous thing for the tensionless Vermont Ramont.
But let me first comment a little bit about where we sit in this modulized space. So as was explained in this paper of Greg and collaborators, you can relate the supergravity, the geometry to the fluxes, and it depends on which background you're looking at it.
If you write it in pure NSNS flux, that's the formulas they wrote down. So the radius of ADS and the radius of the three spheres are related to the number of fundamental strings and NS5 brands. So they think of.
two sets of NS5 prints, or two sets of NSNS flux, and the radius of the circle is of that form.
And the point we've looked at so far is basically the point where we take these to be 2 and disproportional to N. So then we see this is the tensionless point, because this is a tensionless regime, because the radius of the ADS space and the 3 spheres are are string scale.
and the radius of the circle has the usual G squared n behavior. So if G h did so the hatted parameter, the Novi-Schwarz parameters, if that goes like n to the minus a half, then this will have a finite size.
So this just tells you that this is the usually expected dictionary between the coupling constant string theory and the end parameter from the point of view of the symmetric orbit. So this is one corner of this problem.
and I want to contrast this with the with the pure Vermond-Dramont case. Now, in the pure Vermont-Dramont cases, they also wrote down. This is the relation of the parameters. So it looks similar. I mean, it's obtained by.
by some usuality. And what we want to analyze here is a somewhat different regime. So we want to look at the analog of the tensionless regime. So we want these these sizes again to be small. So we want G times Q. 5 plus and minus to be small.
which we want to think of as basically being the two parameter of the analog of the dual Cft. And we are going to take the regime where Q. 1 is equal to one. The reason why we take Q, one is equal to one is that only then do we have a plausible.
proposal for what the CFT do is. But what we really want to be is in this tensionless regime where Q g times Q. 5 plus goes to 0. So we take G to 0, and Q. 5 plus can be finite or large, but this ratio should go to 0. So G must go to 0 faster.
Then the Q5 plus goes to infinity. Now, this is a somewhat singular limit. So this is a singular limit, and that is, I think, always the case. If you look at the the analog of the tensionless regime for pure Vermont Vermont case.
And the reason is, so first of all, it is the tensionless limit because by construction, these parameters get small. So this is GQ times GQ over the sum of two gqs. So this goes like one, they all go to zero in the same in the same way.
This actually will go to infinity, because you see, you have lots of Gqs down here, and one upstairs. So this goes like one over this parameter cubed.
So two to three halves. So this will go to infinity. So this tells you that the radius, the additional radius will decompactify.
But more importantly, you also know from the Bmn limit that this will be a somewhat singular background. So if you look at the BMN limit.
The BMN description of strings on ADS3, cos3, cos3, cross S1, then you know that you're going to get expressions of that form. This is the sort of Ramond-Vamond version of the dispersion relation, but you know that for ADS3, cos3, cos3, cos 1, you have one complex flat direction.
And that means one of these mu i parameters is equal to 0.
This is obvious for the S1, but the because it's complex, there's actually a second direction that is also flat that comes from the combination of the radial direction to the 2 S3s.
And therefore, for one of these terms, this square root will just become n times gq over j. And if GQ goes to 0, what this means, you're going to get a large number of very, very low-lying states, because the.
As G2 goes to zero, then small numbers of n, they will all be degener. So there will be a pileup of lots of low-lying states, and that will happen, I think, whenever you look at this… at the tensionless limit in the Ramon-Ramond case, it'll be the same for ADS3, because S3 cross T4.
Except for T4, you have two complex flat direction. So you have even more noise states. So here we have fewer, so maybe this is a slightly better corner to understand.
And finally, if you think of it from the point of view of the D1, D5, D1, D5 prime perspective, then G q 5 going to 0 basically means that if a Europolis parameters go to 0, because that's the relation between the Fayo and Oopoulos parameters.
And the charges of the number of D5 brains. And this is, in some sense, necessary so that you can sit at a combined Higgs branch of the 2d1, d5, 1D5 prime systems. But again, this is a somewhat singular point, and you'll see the singularity also appearing naturally.
From the perspective of the dual CFT. But you'll see this in a second.
Okay, so this is what we are aiming for. We are now going to try to understand this background of this D1 D 5 brains, and we are trying to read off what's the CFT do.
So, to start with, you would think you do the most naive thing. So you have the five brands.
Then you have a set of orthogonald 5 brands, and you have the one brands going along the intersection. So, as a good physicist, what you do is, let's 1st ignore one set of the 5 brains and analyze the original D1, the 5 system. And then we analyze the other D one the 5 system, and then we try to put them together.
So, if you look at the D1, D5 system, then that's sort of old folklore. So what you do is you the D one effect on hypermultiplex in the adjoint device for vector and hypermultiles in the adjoint, and d1 and D5 have a hypermultiplets in the bifundamental.
And the… if you see that the Higgs branch and the vector multiplex have vanishing expectation value, the FIA localist term fixes the adjoint hypermultiplets, and then the degrees of freedom are the remaining degrees of freedom are the hypermology, the bifundamental hypermultiplines, and that's where the usual.
Q1, Q5 symmetric orbital degrees of freedom come from.
Now, there's a residual gauge symmetry, and this is the Weil group, and that's the reason why you get the symmetric orbifold of this hypermultipleptic degrees of freedom. So this is the usual law about how you argue why the CFT dual of the D1 D5 brain system should be the symmetric orbifold of four.
a four-dimensional degrees of freedom, I mean, for Q1, Q5 degrees of freedom, and they describe the T4.
the end copies of the T4 that sit inside the symmetric. We could also think of this as instant on modulizer. We can also think of it in this way with instant on number of Q. Absolutely. So I'm I'm taking here the perspective of this sort of microscopic brain system, because.
For the combined system, that is something I have to.
Okay, so so so how does the geometry look like? So so there's a very from a point of view the affine parameter. That's the blow up. That's that's making it non-commutative sometimes, right? But your Fi is going to 0.
Right. So we want to think about sort of starting a slightly non-zero for your local associate ourselves of the Higgs branch, but we are going to describe the limit where you come from a non-zimar for your Lopez parameter and send it to zero. This is the regime we are trying to describe.
Okay, so I'm thinking in terms of instant on moduli space, I'm taking the commutative limit of yes.
Now nowadays, one simple problem with this idea, which is I mean, people have tried this before. So what's what's the bottleneck in this construction? The bottleneck in this construction, you can understand geometrically as follows. So so let's 1st con look at the D one system.
So did you mind if I system has open strings from the D1 and the D 5, and the fermions in the D1 D. 5 system come from the Raman sector of this open string spectrum, and the Ramon sector will have 0 modes in the direction where you will have a dd direction.
Right? Because ND, you would have half integer mode. So the zero modes and the Ramon sector sit in the direction 5, 6, 7, 8. And therefore the ground states of the Raman sector sit in form a representation of the Clifford algebra coming from these fermions.
And therefore, it'll be charged under the SO4 associated to these four directions. These are the… this will therefore be the asymmetry of the D1 brain system, because that's the symmetry under which the fermions in the hypermultiplets are charged.
Now, obviously, for the d1, d5 primes is still the same statement is true, except for the d1, d5 prime system, the 0 ram 1, 0 modes will come from the direction where you have DD with respect to D1 D5 primes, and this will be associated to the other 4. It'll be associated to the so 4 coming from 1, 2, 3, 4.
because they are talking to they are orthogonal to one another.
And then the problem, as already long ago observed by Witten, is that the demand system obviously has also hypermultiplate bosons. They will come from the Noveschwarz sector, and then the Nova-Schwarz sector, you have 0 modes for the fermions.
in the ND directions. So the hypermultiple bosons will be charged under will transform on the rotations of the first SO4, and therefore will be charged under the asymmetry of the D1, D5 prime system.
Now, they're obviously not charged under the asymmetry of the D1, D5 system, but they are charged under the asymmetry of the D1, D5 prime system, and as a consequence, you can't just put them together, because you can't have an R-Symmetry, which is an affine symmetry. You can't have bosons charged under an R symmetry, because bosons have zero modes and the zero modes are shared between left movers and right movers.
But the current symmetry split left movers and right movers apart. So so this is the reason why you can't just put the two sets of hypermultiplets together and say, this is the C of T that describes this cyst. And this was… explained long ago by David. So that was the puzzle. How do I describe this combined system?
Now, already in the original paper of Wittens, and then subsequently by Aroni and Bercuse, and then by Sleipkenwitten, they pointed out that there's a potential way in which this could work, namely, the idea is that if you integrate out the hypermultiplets, you can get an effective description of what you think of as the throat region to connect the Higgs and the Coulomb branch.
and that is described by an N equals to 4c Lewis theory, which in effect is the same as the S Kappa theory. The S. Kappa theory is really S. 2 level copper plus 4 free fermions and one free boson.
And it contains two small n equals to 4 sub-algebras, namely either the one corresponding — sorry, the fourth free fermions form SU2 level 1 plus SU2 level 1. You add one SU2 to this, you get an SU2 level copper plus 1, and you get one SU2 level 1.
And there are two small n equals to four algebras inside the system. And you can think of this either as a small n equals to 4 with this central charge or a small n equals to 4 with this central charge. If you switch on the appropriate background charge for the boson. And the idea was that somehow this describes the sort of combined Higgs and Coulomb.
degree of freedom, where depending on which of the two n equals the force you look at, it captures the degrees of freedom that come from the Coulomb branch of the Higgs branch.
So our proposal is that we should treat one set of D1, D5 brains in terms of the conventional hypermultiplets, and the other set of the 1d5 brands we are going to describe in terms of this as Kappa theory. And we want to ask whether there is a way of sort of putting these 2.
These 2 separate descriptions together. So one is sort of the direct P1D5 hypermultiplex, and we take Q1 is equal to one. So we just have.
for Q5 hypermultiplets coming from 1D1 and Q5 D5. And the other set of D1, D5 prime brains we want to describe in terms of this as kappa theory that comes from sort of an effective description of this approach. So where did you use Q1 equals 1?
This is for Q1 equals 1. This is for Q1 equal Q five and q five. Sure. So at this stage, we could do it for general Q1 and we could write down a similar formula for Q1 and Q5, but for what I'm about to propose on the next formula, this will only match if Q1 is equal to normal.
So so you see a priori, it's not clear that you can combine these things. This is a small and equals to 4 theory. This is a large N equals to 4 theory. You want to combine it into a large N equals to 4 theory. It's not clear how you do this. It's not clear whether there is a CFT that will contain this and form a large n equals to 4 superconformal field theory.
And I think this is where I think stood until recently. I mean, people thought that this may be a way out, but nobody had an idea of how you would… what Cft would fit the bill of somehow containing this and defining a large N equals to force super conformal.
Now, what he proposes is that there is a very natural candidate that will do this. And this is the Wolfsay space coset. So this requires a little bit of explanation.
So it's been long known. I should actually, I should add some references about the Wolf Space Coast. It's not something we invented. This is a theory that's already, as Greg pointed out earlier to me already mentioned in that paper, it goes back to the 80s of people analyzing the super conformal field theories.
It's SUN plus 2, the supersymmetric version of SUN plus 2 divided by SUN, the supersymmetric version. And if you decouple the fermions, this becomes Sun plus 2 level K plus SO4n plus 4 level 1. This is just a fancy way of saying there are 4 and plus 4 free fermions.
divided by S2N level K plus. And so on the regular side, the S2n's are just bosonic WCW3. So this is bosonic. So whenever there's a 1, I mean the super affine theory with the free fermions, once I decouple the free fermions.
The level gets reduced by n plus 2, so this level K, and so you get n plus 2 squared minus 1 fermions from the numerator. You subtract out n squared minus 1 fermions from the denominator, so you retain 4n plus 4 of free fermions, and this one, if you decouple it, just becomes SUN level K plus 2.
And what has been checked is that that is a theory that has large n equals to force super conformal symmetry. And in fact, these are the levels. It has levels K plus one, n plus 1, and then the central charge is n plus 1 times k plus 1 over n is k plus 2, which is the.
central charge of the larger infrastructure. Now, it's maybe not obvious to you, but this is the correct thing to do. And in fact, you have to do some small modification.
But what you have to do is you have to divide out by an additional yuan, because you want the answer ultimately to be level rank due, because it has to be symmetrical under the exchange of n and k, and it is not if you don't divide out by the by the U. 1.
But you see, once you do that, there's another funny thing that happens. You see, this 4n plus 4 free fermions, actually, only 4N of them are charged under the SUN, and 4 are singlets. It's really very geometrical. You see, you have the n plus 2 times N plus 2 matrix.
So this is N plus N. This is 2 cross 2. This is n, comma, 2, and this is 2, N.
So this 4 m plus 4 free fermions are the come from here and from from here. I mean so you define all those. So the 4n plus 4 free fermions is you divide all the the strand fermions. So you retain these 2 N.
These 2n and these 4. So there are 4 free fermions that are singlets. So whether you write them in the numerator, or whether you write them separately doesn't matter, because the Sun actually doesn't.
Act on it. So what you do is you separate out the four free fermions that are unaffected by the code set. You write them as a separate four free fermions, and then you divide out by U1 and you add in a U1, so as to retain the central charge.
And what I now claim is that this is level rank dual.
And what I furthermore claim is that it precisely has the form of the four n hypermultiplex plus an S kappa vector. And this is sort of the extra factor, just like for the symmetric orbifold of T4, you have one extra T4.
That's the symmetric lobby fold of one tug. So this is like the ectopus degree. This is some sort of center of mass degrees of freedom.
So how how do you see that this matches? How do you see that this fits the bill? But actually, it's you can see this very explicitly. So what you do is you take the S2n plus 2 level K. This is really this. You write it as soon plus SU 2 plus u. 1.
plus n comma 2 plus N comma 2. So this is this is the numerator SUN. So this is this SUN level SN plus 2 level K in the numerator.
and the 4 and 3 fermions are just n comma 2 plus n bar comma 2. So there are 4 entry fermions, 2 n of them are in the fundamental, 2 n of them are in the antifundamental. But there are actually.
They have the same central charge as SU2 level N plus SUN level 2 plus U1 plus one U. I mean, as you easily check, if you add the conformal dimensions of these specimen models, they add up exactly to 2F.
and the Sun level K plus 2 of the denominator is just a diagonal subalgebra of this S unk from the numerator and that Sun from the 3 fermiuls. So there's an SUN sitting inside here. There's an SUN sitting inside here, and this SUN is the diagonal SUN.
sitting in the tuition. So what does this mean for the degrees of freedom that survive?
Well, what it means is that you have the fermions.
And then you have to construction. Say again. When you say survive. After you take the coset. Now you have to take the singlets with respect to S to n and U1.
So what you have to do is you can think of this as — so the numerator, you have an SUN, that will get killed by the… This will survive because it's neutral under the SGN. This U1 boson that gets basically killed by this U. 1. And then you have the n comma 2 plus m bar comma 2 bosonic degrees of freedom coming from the SUN plus 2.
And from the fermions, you just have n comma 2 plus n bar comma 2. So you think of this as being the fermions and the bosons of the hypermultiplets.
Subject to the SUN condition, and what's left over is the SU2 level K from here, plus the U1 plus the four free fermions, and that gives you the S-cappa theory. So this theory contains.
Exactly the hypers, namely the SUN singlets coming from the N comma 2 from bosons and fermions, together with an S kappa factor. That's just the leftover piece after you've assembled the Wolf space coset.
I don't understand what happened to the superchargers. You know, all of this mixing up of the degrees of freedom. That's a very good point. But what you have checked, but one can check is that this cursor is actually larger than equals to four super conformal. Yes. But this is also n equals to force superstitions. Sorry, no, the combined. I mean, if you retake out this SU 2 into here, then they are separately N equals to 4.
But the n equals to the supercharges are basically the bilinears of a boson from here and a fermion from here. So you have to take an S2N thing. Let's really take this guy.
together with this guy that forms a singlet and it transforms as the 2,2 with respect to the two sq2s. So this gives you the 4… super current. Now, this coset actually has a higher spin symmetry, so it has more than large and equals to 4, but in particular it has large and equals to 4. So the large and equals to 4 generators you can find inside there.
Here, I'm just counting degrees of freedom to match with the hypers and the rest. But it is, in fact, super confusing. The whole space also has WN symmetry. Yes, it has W infinity symmetry. Oh, wow. So it is which is again what you would expect at the tensionless point, because an attention this point you would expect to see the.
Higher spin symmetry emerging in the Wolf space coset actually has a W infinity symmetry.
So so so this means that this basically matches the degrees of freedom you get from this simple minded description, provided you identify the parameters with the number of 5 grains and 5 prime brains. And what's neat is that this description, although it's not manifest the level rank dual.
It is actually a level ranked keyword. So for every representation of that coset, if you replace the numerator and the denominator young tableaus and you transpose them, you get a representation of the level rank dual coset that has exactly the same conformal image.
And that you can check. So this spectrum is in fact level rank dual, despite the fact that it's not manifestly so, because you've treated one set of D5 brain one way and the other set of D5 brains the other way. But once it's reassembled itself, it actually is totally symmetrical.
under this description. So we regard this as sort of good evidence for the fact that that is, I mean, it's a large and equals to force super conformal field theory, and it reproduces these degrees of freedom.
Now, it has a few other interesting features. So there is this. There's this old paper by they looked at D 5 D5 prime system, and they analyzed the U1s.
that sort of the transfers you once. And in fact, the transverse U1s can be identified.
Essentially, literally, with the formulas given in that paper, with the denominator you want and the U1 that survives in the coset and the denominator you want is what they call the flat x2 direction. So what they observed is that this system looks like it has a three-dimensional symmetry rather than a two-dimensional symmetry. So there's one additional direction.
that emerges, this direction is exactly the U direction by which we divide.
So this is just a non-compact direction once we've divided it out because of what I explained earlier.
Whereas the R direction is a specific linear combination of the, so you see there's one combination of this U1 boson and this U1 fermion that survives this coset.
And this U1 is what we call this R current here. It can be written as a certain linear combination of the bosonic, the correctly normalized bosonic and fermionic U1 generators. And it is it plays the role of the radial ADS3 or linear dilotron direction from the point of view of Kutazov.
in cyberc and and Itsaki. And so this is this is like the flat ads 3 direction this we saw in the tensionless NS regime in the tensionless NS regime, we saw this one non-compact boson, which came from the radial direction of ADS 3.
And also here, this boson is a boson without background charge, and in fact, it's the lowest higher spin current of the W infinity symmetry, which this background is. And I think this is probably a lesson.
at the tensionless point, I think this linear dile transformer has to be 0, otherwise you wouldn't have a W infinity symmetry, because the W infinity symmetry comes exactly from this additional U1 current that is present. What do you mean the linear delta? You mean the background charge for the linear? Yeah. So the background charge for the linear adult one is 0. So it's really just a regular flat boson, just like the R direction in the NSNS spectrum.
We're also wondering whether it's the second flat direction you see in the BMN limit, because in the BMN limit from the Bmn perspective, you also see 2. There's a complex flat direction, one corresponding to the S1 of the additional direction, and one from a certain linear combination of the.
of the transverse directions to the 5 brains, and I think this is probably what this R current is as well, but that we are not 100% sure yet.
The other thing is that this is a I said, this is a somewhat singular background, and this is also visible in the dual CFT. The dual CFT has lots of light light states, because there are the states. If you label the cursor representations in this sort of language in terms of the young tablets of SUN plus 2 and SUN.
Then if you take them to be the same, the conformal dimension is just the number of boxes divided by n plus k plus 2. So when n and k are large, and you take small young diagrams, you get gazillions of very, very low-lying states.
And I think this is probably the reflection of the fact that you have this pileup of low-lying states from the DMN perspective.
and it probably also means that this theory isn't just vector light, because you have more degrees of freedom than just the bilinears of the of what you would have thought of as the vector degrees of freedom which come from F0 and 0, F. But again, this is something we haven't.
completely understood so far. And finally, the worst-based curse that has many, many Bps states that are potential larger equals to form modular. In fact, there are these combinations of young diagrams that give rise to exacto Bps states with a plus equals minus equals to a half, which.
One may have thought should be exactly marginal moduli. But from our perspective it looks as though they are not exactly marginal motor line. In fact, we believe there's only a small number of moduli. We would hope there's exactly one moduli that survives, but that is something, again, which we haven't totally understood, but it's clear from this analysis that only a very small subset of them can be potentially exact.
Okay. Oh, yeah, okay, so there's a connection to our old analysis of the symmetric orbifold in this context, which also fits a reasonably nicely into this, but that probably takes me a little bit too far here. We also believe we can identify the individual value terms. If you switch on one of them, then you break the large N equals to four super conformal symmetry because they carry charge with respect to the other SU2.
And they're also relatively natural moduli that would seem to do exactly that. And in fact, these are so 10 years ago.
Rajesh, I noticed that the symmetric orbifold in some sense, the Wolf space coset.
is a it's a subalgebra of the chiral algebra of the symmetric or before, and you can write the symmetric Orbifold in terms of representations of the Wolf space coset in the limit k goes to infinity. And if you do this, you can analyze where do the various moduli of the symmetric orbital come up from?
And there's one two-cycle twisted sector modulus of the symmetric orbital that corresponds exactly to one of these multilay that we see here, and the other modulus of the symmetric orbital, the untwisted sector modulus comes exactly from the 0f degrees of freedom.
So we are not exactly sure how the symmetric orbit foot fits directly into that picture, but it's encouraging to see that the moduli of the symmetric orbital play also a special role in this context.
Anyway, so there are clearly many open questions. But let me let me conclude. So I want. I've described 2 aspects of strings on alias 3, because the 3 quas 3 equals S1.
I've tried to convince you that we can basically derive the duality for the tensionless NSNS background, and what we find is that it's the symmetric orbifold of a 0 squared from this very explicit world sheet description.
And as I mentioned, what's neat about it is that you can directly describe this in the NSR formalism. You don't have to go to the hybrid string.
And it hinges on this simple and explicit description of the Bosonic SL2R resume written model. But once you believe that, in fact, it implies there are many, many more dualities you can derive. In particular, you can derive bosonic dualities.
between the symmetric orbit fold of R cross any CFT with central charge 17 is exactly dual to Bosonic string theory in ADS 3 described by the SL2R1 level 3 resuminum written model times some CFT of c equals to 17.
And so this is the Bosonic case. The other case is maximally supersymmetric, and you would hope there is something interesting in between that doesn't quite go between no supersymmetry to full-blown supersymmetry. And that's something we are exploring. But I think it's quite intriguing that this.
seems to be a very robust machine to generate dualities you can effectively derive. And I think that. But there's probably a feature of this special NS background that symmetric orbital is probably very simple.
And then in the other half of the talk I try to convince you that the CFT dual of that brain setup in the limit where G times Q 5 goes to 0 is this Wolf space coset.
Where the parameters of the Wolf space coset are related to the number of five brains and 5 prime brains in this manner.
And in particular, what I have explained is that this is larger than equals to four superconformal symmetry. It has the expected hypermultiple degrees of freedom with respect duality, and it has a higher spin symmetry, as you may have expected from that sort of background.
It's a bit singular, but you see this also in the Cft. But in a controllable fashion. You see these light states appearing.
And it would be interesting to understand exactly where the actual tooth, there should be one exactly marginal operator that, at the end of the day really survives.
And it would be good to identify this. I mean, we have a guess, but I'm not sure whether it's the correct case. And then you would hope that the light states would be lifted the moment you switch on this tooth parameter. This should sort of give some sort of.
charged to this to these states, and this looks vaguely plausible, because these light states, they all carry non-trivial charge with respect to this R generator. And the moment you switch on this tough parameter, I think the U1 current gets lifted, the higher spin symmetry gets lifted, and my suspicion is that you produce.
non-negative contributions to the conformal dimensions of these light states, and you're beginning to sort of resolve this somewhat singular setup. But this remains to be seen. And finally, it would be good to understand in which sense this is a vector model, a higher spin type theory, or really a string theory.
What we want to claim is that it's a string theory. But originally we thought it's a higher spin theory. So it's a question of what are really the independent single particle degrees of freedom? And how how should you think about? Yes, yes, I mean on the.
Originally, we thought this Wolf space coaster is due to the Vasilyev theory, but it was really only one half of it. This was the half where you set the lambda plus parameter to zero and lambda minus parameter arbitrary. That's basically reproduced the Vasilyev theory. So now, putting in this additional states, in particular allowing for the light states.
We suspect is in effect, producing a stringy spectrum from something that used to be just a Vasilyev type vector-like. But that's, again, something that's subtle, that hinges on what you should you think of as being the single particle states, what should you think of as the multi-particle states.
and so on. So that is again something we haven't completely understood anyway. So that's where we are, and I thank you for your attention.
So, are there any questions?
I have a question. So for the dual for ads. 3 times s 3 times T4 and Ads 3 times s 3 times K. 3.
We can think of the targets. We can think of the dual CFT in terms of the moduli space of instant tons on T4 and K3, respectively.
So one of the old proposals was. take the moduli space of instantons on S. 3 times S1.
And maybe that's… the holographic tool. And so you mentioned at the beginning of your talk, Quentin.
discuss that in a paper like almost 2 years ago now. It's amazing. So okay, so you could take you and instant on S3 times S1.
And then in that… In that same model, there's a B field you can turn on. So that's another parameter. So it's N and K. And then there's the instant on number. So those are supposed to be the 3 quantum numbers of D 5 D 5 prime, and D 1.
So what do you think? I mean, is is his description going to be valid in one regime and and yours in another regime? Or could it be that the moduli space these instant times is a wolf space?
be the same as the conformal filtery on a Wolf space? Well, we have only… we've only really attempt… we have only a proposal for Q1 equals 1. We only ever proposed that… That would be instant on number 1. Yes, so we only have a… so our analysis.
We have no Cft proposal for what happens for Q1 bigger than one. So we only have a 2 parameter space rather than the 3 parameter space. So whether this agrees with the instant one moduli space for that specific setup.
Well, once you go down to low instanton number, funny things can happen, for sure. Yeah. So that's… but this is a restricted setup, if you wish, because our Q1 is just more very interesting. Okay.
Any other questions? Let's thank Matthias again.
It's