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