OK thank you very much. Can you hear me?Yes.OK.So first of all, I'd like to thank the organizers for the opportunity to do this nice kind of workshop.And so I'm always happy to be back here.So yes, and my talk is based on the collaboration with J-One Song in UC San Diego and also these two papers.And also another work with Pralit Agawar and J-One Song and another paper will appear maybe soon.And OK, so I should first apologize that in my talk there is no geometry up here.Although the workshop is on geometric correspondence of the gauge series, so I should apologize.But I think there should be, there exist the geometric understanding of the phenomena which I have in my talk.And also I think that this geometric understanding should be helpful to understand to know what the mechanism of the phenomena which I have.So if you have any comments or any complaints, any suggestions, please let me know and please stop me.So what's the phenomena?The phenomena is the supersymmetry enhancement in 4D quantum field theory.So here we consider more precisely the enhancement of supersymmetry in the 4D quantum field theory.Along the RG flow.OK, so of course this is some enhancement.So we have some supersymmetry in the UV fixed point or some higher energy scale.And we want to see the enhancement of the supersymmetry along the RG flow more precisely in the IR.So of course it's conceptually possible to have some accidental symmetry in the IR.So we sometimes see that symmetry will be enhanced in the IR.There are some many examples.But for supersymmetry it is not so many, there is not so many examples.So the example of which I know is something like this.If you know something more please let me know.The first one is very somehow trivial.So let's consider n equal to 1 4D dimensional theory with SU and gauge group.Then we have gauge coupling g with also 3 adjoint chiral multiplets.And with this superpotential 515253, trace of 515253.And we have some coupling H.So it is known that this theory in the IR flow to n equal to 4 supereme theory.So in other words this theory is somehow the deformation.Just the deformation of n equal to 4 theory by changing the coupling constant of gauge coupling or this coupling to the superpotential.So in a sense this is somehow just the deformation of n equal to 4 theory.And in the IR this flow back to the n equal to 4 theory.So this is somehow trivial example of the supersymmetry enhancement in the IR.So yeah I'm skeptical about this.So this is somehow trivial.So as a recent example is by Gatte,Lazamart and Willett.They consider some n equal to 1 Lagrangian theory where some coupling constant is said to be infinite.So we cannot do some perturbation theory because the coupling constant is infinity.So they show that or they argue that this theory flow to in the IR to n equal to E6 theory.The super conformal field theory with E6 global symmetry.So this is interesting example but in this talk I want to present an infinitely many example of the infrared supersymmetry enhancement.From n equal 1 supersymmetry in 4D to n equal 2 supersymmetry in 4D.I'm always staying in the 4 dimension.Okay so then the let me in this introduction let me show you the simplest model and simplest RG flow which we found.It is the sum n equal and n equal 1 SU2 gauge theory to n equal 2 RG Douglas theory.So this is simplest one which we have.So let me show you first this example.Okay so first of all if you are not familiar with this n equal 2 RG Douglas theory.So I prepare one slide to review the RG Douglas theory.So what's RG Douglas theory?So this was originally found at the special point on the Coulomb branch of n equal 2 SU3 pure supian mill theory.Okay so there is a Coulomb branch of n equal 2 theory.Then there is a special point where the massless particle appears.And so in this SU3 theory there is very special point.There is a very special point where the mutually non-local massless particle appears.Massless particle appears.So mutually non-local means that we have some field with electric charge and another field which have magnetic charge or dionic charge.So both two fields will be massless.So in this case we don't have the local Lagrangian description.So they say that this is strongly coupled and this will be the super conformal field theory.So now we know much about this strongly coupled n equal 2 SCSE.Okay so this theory is called RG Douglas theory.And this theory is strongly coupled and we know the central charges.Which is this rational numbers.Okay so the central charges are some number which count the degrees of freedom.Okay so A and C is something like this.And also this theory has one dimensional Coulomb branch parameter which we call u with scaling dimension 6 fifths.Okay so we have just only one Coulomb branch parameter Coulomb branch operator with dimension this one.And okay so this theory is sort of the minimal non-trivial n equal 2 SCFT.Why?Because these people show that very recently that this theory so there is a central charge bound for n equal 2 theory.n equal 2 super conformal field theory and this theory saturate this bound.So okay so I mean the central charge C saturate the bound.So it's not by A but I mean this is some simplest theory or minimal theory in n equal 2 supersymmetric gauge theory.In four dimensions.So it's interesting to study this kind of theory.Okay so this is the one which we want to obtain in the IR fixed point.So now I want to say what okay so let me go back to the previous slide.And so I want to explain what this theory.So this is which we found at the first time.This is the theory.So we consider n equal 1 supersymmetric theory with SU2 gauge group.And the following chiral multiplex.So we have q and so this first line denotes the SU2 representation.SU2 is a gauge SU2.So we have q and q prime which is in the fundamental of SU2.So we have two fundamental chiral multiplex.And also we have phi.This is in the joint representation of SU2.So up to here the matter content itself is something like n equal 2 SU2 gauge theory with one flavor.So but we add a few singlets.This one means the singlet.So we have m1 m3 m5 m3 prime.We have four singlets in this theory.This is rather simple.So then we have a superpotential something like this.This is some tedious but we have to add this one.Okay so the first term is like n equal 2 supersymmetric.But another term which breaks n equal 2 to n equal 1.Okay we have some coupling with m1 m3 m5 m3 prime to other fields.Okay so and then one can check that in this theory with this superpotential.There are only two u1 symmetries.The one is the r symmetry.The symmetry like r symmetry which is something like this.Okay so we have this.And another symmetry is which I call u1f.And this is just global symmetry.So these two are only non-anomalous global symmetry in this theory.So then okay so what's the gauge invariant operator in this theory.We have many of course.But okay so let me focus on these operators.Of course m1 m3 m3 prime m5.This was singlet of the SU2.So it is gauging variant operator.And also teres pi square operator.This is again the gauging variant.So okay so of course there are many but we focus on these operators.Then okay so we want to know what's the IR theory of this simple rather simple theory.To know the IR theory the most fastest way to do is to know what the central charge is in the IR theory.Then so the okay so there is a work by Anselmi Friedman Grissel Johansen.That the central charges A and C of some super conformal field theory can be written in terms of the anomalies of the IR r symmetry.Something like this.So if you know the cubic anomalies and trace anomaly of the IR r symmetry.We can compute A and C by using this formula.Okay so the problem is to find what's the IR r symmetry.So in our problem the IR r symmetry is the should be the combination of two u1 symmetry which we have in the scale of the Lagrangian.So this should be a combination something like this.So here we have epsilon which is undetermined.And we want to so if you fix this epsilon we know what's the IR r symmetry but we don't know what the epsilon is so far.And then there is a technique to know what's the epsilon which is called a maximization by a interrogator Wecht.So this is basically the method to know what the mixing is.And this epsilon is determined by maximizing the trial central charge which computed by this formula.So if I maximize this function we get epsilon.This epsilon tells you what's the true r symmetry in the IR.Okay I guess it's clear.So we can do this maximization in our model.So let's do that.Okay so going back we can do and we can fix epsilon but there is one thing one subtlety in this maximization.Which is that if you do the maximization sometimes some operator hits the unitality bound.So I mean okay so in the four dimensional theory there is a unitality four dimensional conformal theory.There is a unitality bound and for the scalar field the minimal dimension scaling dimension of the scalar field is just one.And this is the case for the free field.And if it's not free field it should be greater than one.If the dimension is less than one it's not unitality.So we have to check that there is no there is all the operator it should be greater than one.And then if I found some operator which has dimension less than one I interpret that okay.So we start from some UB theory and some operator has dimension less than greater than one.But after a maximization we found some operator which has dimension less than one.So it means that dimension will be decreased and at some point of the RG flow we hit that operator hit the unitality bound.And this becomes dimension one.So at this point I interpret that this operator becomes free and decoupled.So the prescription to implement this one is to subtract the contribution of this operator.Whose dimension is less than one.Okay so let's do that.And for the first trial of the maximization I found that trace phi squared and m1 will hit the unitality bound.So I subtract this contribution from the from the central charge.Then I redo the maximization.So in the second trial m3 and m3 prime will hit the unitality bound.And then I subtract this contribution.Then I again do the maximization.In the third trial it is okay.So only there is only one and five operator.Then I get some epsilon which is this number.And it is surprising that I get the rational number in this trial.So usually if you do the maximization we get some irrational number.Not rational number.But this time I get some rational number and A and C is this value.So maybe you saw this value in the previous slide.Also the dimension of m5 operator will be six-fifth.So these numbers you saw in the previous slide in the RGSTAGRAS theory.So I think that this theory at least by looking these dimensions and central charges this theory is RGSTAGRAS.So this is just a brief review of our simplest case.We have in the introduction.Can you make each environment out of Q1?Yeah, that's possible.Yes, but this operator doesn't hit the unitality bound.But then it will be additional.Okay, so there is an additional one.Okay, I didn't write in this slide.Of course there are many additional operators.Yes, but only the operator which I have to care whether the operator hits the unitality bound is these operators.Other operator will have more larger scaling image.This exists.Some operator will exist.Okay, so we don't have a complete understanding of these operators.Where these operators will come to the n equal to multiplied or not.But of course this, for example, this m5 is n equal to one chiral multiplied.So this is not enough to get n equal to Coulomb branch operators.So we should have something extra.This is what we wanted to do.But there is some small discrepancy about this consideration.So we don't have complete understanding of it.Okay, this is also we found after this work.Actually we can throw away m1, m3, m1 and m3 prime in the beginning.Okay, so we can do but I keep this in the Lagrangian description in this way in the super potential.Because later I want to relate this theory to some other UB theory.This is I will show you later.So in order to do this, I keep these terms and also m1, m3, m5, m3 prime.Okay, other questions?You are fine?Okay, so we saw that some this simplest model.We have some Lagrangian description which flow to the,which may be flow to n equal to algesta glass theory.So how useful of this theory?And if you are generation of Gaillotto's class S theory,you may not worry about whether the theory has Lagrangian or Lagrangian description.But the two have Lagrangian is very useful in a sense that of course to study the strongly coupled interacting theories.For example, non-BPS sector or something more than BPS sector.And more concretely, we can compute the partial function if you have the Lagrangian by using localization technique.So for example, we have n equal 1 theory.The simplest one is the super conformal index.So here we can compute the index in the full generality.So we can have three fugacities in this index.So this super conformal index, we can compute.Or some other partial functions can be computed.I'm not so much familiar with n equal 1 localization.But I think there are more which can be computed.So indeed, the super conformal indices, which we compute in this way by using Lagrangian description.We can recover the known result, previous result in the special limits by Vikan Mishinaka and Koldoba and Shao.So these people study by using other method to get the index in the special limit of the fugacities.So we can recover from this super conformal index these results.So in this sense, this Lagrangian description is I guess useful.And I want to ask you to use our Lagrangian description.Okay, but we don't know why there is this kind of n equal 2 enhancement in the IR.So this is the big problem.And I want to understand what's going on.And even we don't know yet, we don't prove that the IR theory has n equal 2 supersymmetry.We just look at the central charges and dimensions of the operators.And we thought that this is n equal 2 theory by matching.Okay, right.Okay, so index is more, yes, yes.By using index, we can, yeah, it's more like, more likely to have, I'm getting n equal 2 theory.Yes, thank you.That's true, yes.So okay, but we want to know why I get n equal 2 theory.After getting some decoupling fields and throw away these, we get n equal 2.Also, the interesting thing is that we start from somewhere n equal 2 like theory.So we start from SU2 gate theory with some adjoint.And this adjoint, this theory should has Coulomb branch by parameterized by trace pi squared.Okay, but this trace pi squared will be gone.Then this will be decoupled.Then M5 will be the Coulomb branch operator in the IR theory.But this was not Coulomb branch operator in the UV.So the Coulomb branch will appear in the IR somehow.So this is some very peculiar things in this, in this flow, in this model.And we don't understand why these kind of things will happen in this model.So any comments and suggestions are helpful.And I cannot talk about why we get n equal to enhancement in this talk.So then here in the rest of my talk, I want to say something more about how I construct this kind of Lagrangian theory.Okay, I guess so.So in the IR theory, there should be a Coulomb branch.And the theory in the Coulomb branch should be U1 gauge theory, some matter much.But this U1 is not the U1 in the UV theory.So maybe, yeah, but actually I don't know what happened.Okay, so later I want to show you how I get this kind of Lagrangian theory.And actually this method is, by, okay, so after we had, okay, so by this method can be applied to any n equal to super Coulomb field theories with non-Aberian global symmetry.So, okay, so we can get many Lagrangian descriptions, which brought n equal to fixed point in the IR.So this is the first point.And if I have a time, I want to say something about the super Coulomb index, 30 minutes.Now, okay, let me go on.So what's the deformation?So let's first forget about what I said in the introduction, the Lagrangian theory.And please concentrate on this deformation.Let's suppose that we have n equal to super Coulomb field theory, which I call t.And with the non-Aberian flavor symmetry f.So you can, you have any theory with flavor symmetry f.Okay, so this flavor symmetry can be a subgroup of full flavor symmetry of this theory t.Then, okay, so just a convention that, so since this is n equal to two theory, we have SU2R symmetry and U1R symmetry.Here, this I3, twice I3, I3 of twice I3 is the cartoon of the SU2R.Okay, and this R, small R is the generator of U1R.So we use a convention that these two is written as J plus and J minus.Just a matter of convention.Okay, so then to this theory, let us couple the n equal to one chiral multiple m in the adjoint representation of f.Okay, we add one chiral multiple or chiral multiple in the adjoint representation of f with this superpotential.So here, mu is the operator in this n equal to SCFT, which, which appeared in the army high stock yesterday.Which, this mu has the, okay, SU2R spin one, in the spin one representation of this SU2R.Okay, so this mu is in this n equal to SCFTT and I couple this, this, this multiplied n equal to one multiplied to mu, something like this.This is the superpotential, which we have.Since the mu field, mu operator has spin one in the, of SU2R, which, which means that, okay,so which means that I3, this is Cartando VSU2R is, is charge one.So this has J plus J minus charge two zero.So then the m should have zero two charge.Okay, so since J plus and J minus should be the, should be the R charges, R like charges.So it should have two two R symmetry, R charges in the superpotential.Okay, so then, okay, so, but so far, at this point, we don't get anything new.Because if you just coupled the chiral match bet in the joint representation, and, and to see what happened in the IR,you just get the coupled theory with n equal to super compound field theory and chiral match bet.So this is nothing interesting.But if you do something more, you get something interesting.So, and this, this kind of thing is also related to the army high stock yesterday.But, okay, so we give some nilpotent web to, to the, to the chiral match bet m.So, okay, so this nilpotent web is specified by the embedding,low, which from SU2 to F, where F is the flavor symmetry.So in the case, okay, so this web break of course breaks the flavor symmetry to the smaller amount.Then, so in the case of, in, if the flavor symmetry F is SUn or A type,this web is classified by the partition or partition of n or a young diagram.And for other groups, it's, there might be the classification.Okay, so this one, these two, two deformations define the theory in the IR.And which is labeled by the UV n equal to theory and also the way of the embedding.Okay, so this IR theory has two, two levels.And so, since I do, I added the n equal one chiral match bet,generically this theory is n equal one.Okay, there is no reason that although we start, I started from the n equal two theory,but there is no, but this theory purely n equal one.So, there is no reason that this theory is, is going to n equal two.But we see in the many examples that we get n equal two super symmetry in the IR.Okay, so, but let me just do some a bit rather complicated things which I need it.So, we had here, we had here mu and m.So, here mu and m are both in the adjoint representation of flavor symmetry.So, then after giving a verb, these adjoint representation are broken or decomposed into the SU2 representation,where SU2 is this one of the embedding.So, this adjoint will be broken to SU2 representation.Okay, so, and, okay.So, then the by giving a verb,so we level the SU2 representation something like this,where j level the spin representationand j3 is a component of the inside the spin, j spin representation.Okay, so j3 is j to minus j,which is integer or half integer.Then, I double m and mu something like this.Then, by giving a verb,the superpotential is somehow deformed something like this,because I give you a verb to m.So, we get this term only mu.So, this is problematic because somehow this term breaks the original u1 symmetry, u1 r symmetry, u1 r symmetry.But, if I consider some combination of these u1 with this row,we can preserve these u1s, two u1s.So, okay.So, let's consider this j plus prime and j minus prime.And, this will be the u1 symmetry in the IR theory also.Then, some another argument shows that for e.So, okay.So, we saw that mu m and mu,which decompose into the SU2 representation.And for each SU2 representation,only the lowest component survives.I mean, the m will goes to the component in the m,just only this component,where for each spin j representation,the component which has minus j3 equals to minus jis the only one which survives after the deformation.So, after all, we get this superpotential.Okay.This is just a small computation.Okay.So, as someone will combine to the flavor currentand it makes the flavor current to non-VPSbecause flavor symmetry is broken.Or, some other will decouple.Then, okay.So, this is the small thing.Okay.So, then we can do for each model,we can consider the central chargesby using a maximization,like in the Lagrangian theory,which we saw in the introduction.If we know the central charges of T,the UV theory,if you know the n equal to two central charges,T and also flavor central charge,F of F,symmetry F,we can do the maximization.The formula for SU,when F is SUnand if I choose the embeddingwhich called principle embeddingwhich breaks the flavor symmetry completely,the formula is something like this.Okay.So, these trace of someU1global symmetriesand here,A and T is thecentral charge of n equal to theorywhich we have in the UV.Then,K is the flavor central chargeand other is nothing.We have just n.N is we start from SUn.So, this is the formulaand we can do the maximization.So, the input isn equal to central chargesand flavor central chargeand also theand also thehow you embed SU2 inside F.So, we can do that.And this is the general prescriptionto deformn equal to theoryflavor symmetry.So, let's see the example.The first example is very simple.We haveas a n equal to theoryScftTthe SU2 is four flavors.This is somehowsimple Lagrangian theoryLagrangian super conformalfuse theory.And this theory hasthe flavor symmetry SO8.Okay.We consider the principle embeddingof this SO8.Okay.So, what we do is start from this theoryand addSO8 adjointn equal to one chiral mass petand give a webto break the SO8flavor symmetry completely.Okay.Nilpotent webto break SO8 completely.So, the adjoint representationof SO8,28 representationis broken tothree-dimensional,one-three-dimensional,and two-seven-dimensional,and one-eleven-dimensionalrepresentation of SU2.Andfor each representationwe have just one componentsurvivedwhich I call M1minus 1and M3minus 3M3 primeminus 3and M5 minus 5.This one-three-three-fivedenotesdenotesthe spin representation.This isspin one representationwhich has three-dimensional,spin three representationwhich is seven-dimensional.Okay.So, these fourthese foursthe only survivingcomponents.And thiswill bethe singletwhich appearin the Lagrangian description.We...We...I show you in the introduction.Okay.Then, of course,we give a web toto the...to this M.Some field will...Okay.So, this is a Lagrangian theory.So, we can do it.Everything in the Lagrangian level.So, then,by giving a web to M,will be the mass termfor the quarks ofSU2 with four flavors.So, if you do integrate out,integrating out themassive field,after all, we getSU2 theorywith one flavorand some adjointand these four fields.The superpotentialin the introductioncan be obtainedafter...afterintegrating out this.So, this...Okay.So, this...the...theprevious model,Lagrangian modelcan be obtainedfromSU2with four flavorby SOAto principle embedding.So, this is howwe getthe Lagrangian theoryin the IRand if you do themaximization,you getN equal toSU2Algrestagrass theory.Okay.So, this is one thing.Okay.So, I do theembed...I consider theembeddingwhich breaksSOAcompletely.So, why youdon't considerany other embedding.So, we did.So, we have manyfor SO...SO8.So, maybe youare not familiarwith about this, butokay.So, justthis is a label.So, for example,other choice of embeddings,which issomething like this,which preservesonlySU2 flavor symmetry.If I consider these embeddingsin the IR,again, we getN equal to enhancement.But this is notAlgrestagrass theory,but some generalizationofAlgrestagrass theory,which is calledH1 theory.So,originalAlgrestagrass theoryis called H0 theory.And nextAlgrestagrass theoryis called H1.And this issomehow in theKodaira classificationof rank 1 SCFT.Okay.So, we can getthis one.For another embeddingwith preservesjust U1times U1global symmetryin the IR,I get H2 theory,which hasSU3global symmetry.So,okay.So,in this example,thethe supersymmetryis enhancedin the IRand alsothe global symmetryis enhancedin the IRtoSU3.So,many enhancement.Yes,I think so.Okay.So,I'mnot sureI'm not surewe didcomputationcompletely.Okay.Yeah.This iswhat we aredoing right now.So,yeah.I'myeah.Yeah.I'mnot sureso far.Yes.But should becan be seenby computingthe index.Okay.So,what we doneisin this example,theまた、定規の実際にはないと知らない。これは、一番のリメンジョンではない。だから、この理由は確かめない。そうすると、一番のリメンジョンは、特にこのリメンジョンは1に出てきて、ブラッシュオッケー、イヤ、イヤ1,0001,0001,0001,0001,0001,0001,0001,0001,0001,0001,0001,0001,0001,0001,0001,0001,0001,0001,0001,0001,0001,0001,0001,0001,000このスユー&は2Nの味を試してみてくださいこの1の味を試してみてください5.3?7.1?7.1?マクシムマル?マクシムマルですマクシムマルは何かをしていませんデフォメーションをしていません次のマクシムマルの味を試してみてください他のスモーラーの味を試してみてください2Nの味を試してみてください上のラインはエソインタイアリティと同じですはいどこにあるの?1.3.4.4エソインタイアリティと同じです一番上のラインは5.3の次に3 spreads4は5.1 termを購vat2.32.11では、これが例えです。では、10分間の時間を絞ります。では、2nの味を絞ります。これは、Global Symmetry SU2n × U1です。私は、U1アーベリアンプラットを考えません。これは、SU2n × U1 × Just U1です。では、私は、時間を絞ります。では、私は、ディテースを切り替えます。しかし、私は、このディメンジョンを取り付けます。そして、このディメンジョンを取り付けます。そして、このディメンジョンを取り付けます。そして、このセントランチャージーです。このディメンジョンは、A1 × N-1で、セルジュアのクラスフィケーションを取り付けます。セントランチャージーは、シャッピーエンタチカワーで、シャッピーエンタチカワーで、そして、このディメンジョンを取り付けます。そして、このディメンジョンを取り付けます。そして、このディメンジョンを取り付けます。そして、このディメンジョンを取り付けます。そして、この場所は、最初に、 accident两面とnを取り付けます。説明してくれました。全く分かっている場所で、それは、過程筆剪をさせ、1a2nの理由はアジュレスタグラスの理由ですこの例で全スーパーコンフォーマルフィールの理由がこのようなデフォーマーションで得られますこれらの理由は全スーパーコンフォーマーションで得られたのですしかし全スーパーコンフォーマーションで得られたのですこのようにこのフローが見えますこれがプリンスパルエンベディングのターボですこれが完全に味わいを止めることができますこのターボはこのように見えます次にランク1のケースが見えますランク1のSTFTとN-2のスーパーセメントはH1,H2,D4,E6,E7,E8このターボはSU-2,SU-3,SO8このターボはこのように見えますそしてE6,E7,E8はグローバルセメントですそしてこのターボは完全に味わいを止めることができますそしてこのターボはH0のケースですオリジナルアリジレスタグラスのターボですまた、A1,DKのターボはSU-2のグローバルセメントですそしてこのターボはSU-2のグローバルセメントでA1,AK-1のターボです多くのエンハンスメントができますしかし、このターボは無いアイアルエンハンスメントですそしてこのターボはランクワンのターボですアルジレスタグラスのターボでクラスを使ってランクワンのターボを試してみたらそれは無いアイアルエンハンスメントですエンハンスメントは新たにエンハンスメントができますそしてクラスSのターボはTNターボですTNターボはSU-3のグローバルセメントですそしてこの3のグローバルセメントのブレーキができますしかし、このターボはN-1のアイアルエンハンスメントですまた、N-4のスーパイアンミュースはN-2のターボですN-4のターボはSU-2のガイジングループでN-2のスーパイアンミュースはSU-2のグローバルセメントですこのデフォメーションで使えますアイアルエンハンスメントはN-1のグローバルセメントですでは、もう少しも試してみてくださいしかし、このターボを試してみてくださいまずは、このターボのパターンは?この2のグローバルセメントはN-2のアイアルエンハンスメントですこの2のグローバルセメントはN-2のアイアルエンハンスメントですこのターボはこのターボを見てみてくださいまず、このターボは2Dカイラルアジェブラとスガワーコンストラクションですこのターボはビーム、ラステリー、スガワーコンストラクションですこのターボはこの2のグローバルセメントは2Dカイラルアジェブラと2Dカイラルアジェブラと2Dカイラルアジェブラとスガワーコンストラクションはないです3Dカイラルアジェブラと2Dカイラルアジェブラとスガワーコンストラクションは2Dカイラルアジェブラでイ assignment点とでは、2Dカイラルアジェブラと2Dカイラルアジェブラで其1つ不明と思っていますなしかし、このカイラル・アルジェブラのビジネスに関することはありません。このカイラル・アルジェブラは、SU-2、SU-3、SU-8、SU-3、SU-8、SU-8、SU-8、SU-8、SU-8、SU-8、SU-8、SU-8、SU-8、SU-8、SU-8、SU-8、SU-8。ああ、それは次になって And this is at least SU2、At least SU2、SU2 x SU1。しかし、リントルは放置していないよりノ岩に引退してます This list was only for principle embedding which break the flavor symmetry completelyそして、それから別のエンベリングを行うことができます。はい。このエンベリングは、多くのアルジェバーが得られ、そして、それからエンベリングを行うことができます。私は、これからこのクラスターを作ることができます。ああ、はい。これを見せると、ニルポテントのデフォーマーションを作ると、Nは2系の内側に、Wアジェブラを作ることができます。このデフォーマーションは、Nは1カイラルマージュベットと、このカイラルマージュベットを作ることができます。ウアの個性は、、ミュウの組み合わせに比較することができます。でも、私はこれだけ確認で可能に、試みます。どうして、別のコンプリティングがあるの?だから、以前のコンプリティング、新しい経験を 行ったことを知らないの。そのため、必ずリアルに分かれるのが、プリンスパルエンベディングには エンベディングがないプリンスパルエンベディングは エンハンスメントの最も大きな場合そのため エンベディングが小さくて エンハンスメントの最も大きな場合はプリンスパルエンベディングは エンベディングが最も大きな場合はしかし エンベディングはまだ完全に知らない時間がかかっているスーパーコンフォーマーエンデックスには 2つのスライドを持っています2分の時間がかかっていますこのスーパーのために エンベディングは完全に知らないそのため エンベディングの最も大きな場合はエンベディングの最も大きな場合は エンベディングの最も大きな場合はエンベディングの最も大きな場合はそして エンベディングの最も大きな場合はエンベディングの excelsANK lookフォーマスタスサタス エンベディングの最も大きな場合はラグランジャンシューリーはN is equal to or some non-Ragrangian theory whose IR theory is N is equal to.It's up to you.But if I choose the Lagrangian theory, we can compute the index or we can compute the partition function.But in this way we can compute the super conformal index if you know the matter content of the Lagrangian.So then this is somehow the definition of the index and I don't want to...Yes.I think this is the case with the theory has conformal symmetry.First of all, I'm not sure whether there is the R symmetry in the non-conformal case.So we should have SU2R times U1R and I think non-conformal case U1R will be broken by anomaly.So in this case, I don't get enough U1s in the IR.Yes, but we didn't try.So maybe something interesting happened for the non-conformal case.But I just focus on the conformal theory.Okay, so this is the slide.So in the H0 theory, okay, so H0 is the original RGS Douglas theory, which I consider in the introduction.And I show you what's the matter content, which is SU2 gauge theory with two chiral multiplates, fundamental chiral multiplates and one joint and four singlets.And here we have integral because of the SU2 gauge symmetry.And this part comes from the vector multiplate.And two fundamentals and this is a joint.And we have four singlets, but three will be decoupled.We just throw away these three contributions.Then I just multiply this contribution.This is for the remaining gauge singlet multiplate, this one.So then finally, also, trace phi square will be decoupled.This is only one subtle thing to compute.The trace phi square is the trace phi square.And it's not like just remove the contribution from phi.So there should be trace phi qq prime or qq or qprime qprime operator.So to remove the contribution from phi is not good.But I just remove the contribution, trace phi square, consider that some chiral multiplate.So this is one which appears in the denominator.So this is the index which I have.So then what I should do is just compute the integral.Before that, we should consider the IRR symmetry is different from the UBR symmetry.So in order to take into account, we should subtract this one,xi into t times p and q.This prescription is to choose the correct R symmetry in the IR after the maximization.Basically one can do the integral.We get some series in terms of p and q and pq and p.We can try the limits.For example, the simplest one is Coulomb index limit,which is set pq over t equals to U and pqt equals to 0.The simple one.And I get very simple one.And you can see that there is the one operator which has dimension 6 fifths.So this shows that there is some Coulomb branch with dimension this one.And also we can do some McDonald limit,which agree with the index by Coulomb and Shaw by order by order.So it's very difficult to take the limit in this level.So after doing the integration,and we compare by taking the limit and compare to the previous index.So we get the index for full generality.And this can be done for other series,I guess,in principle.Yes?So this is the last slide.So we have many questions.So first of all,we are doing for non-principle embeddings.The previous papers,we try to consider principle embedding.And we do something more,something complicated.Of course,it's not easy to complete the list.We don't have any complete list of n equal to theory.So it's impossible now to complete the list.The deformation.So as I said,what's the condition of UB theory T for the enhancement.This might be interesting.And why there is enhancement.Maybe this is related to some string theory construction.Or if we consider some geometric construction,this might be answered.So this is related to some string and some theory realization.How Coulomb branch appear in the IR might be the interesting things.Okay,I stop here.Thank you very much.