OK thank you very much. Can you hear me?So first of all I'd like to thank the organizers for the opportunity to do this nice counter-workshop.And so I'm always happy to be back here.So yes, and my talk is based on the collaboration with J-1 Son in UC San Diego and also these two papers.And also another work with Pralita Agawar and J-1 Son 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 around the RG flow.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 examples.So the example 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 quantum theory with SUM gauge group.Then we have gauge coupling G with also three adjoint chiral multiplets.And with this superpotential phi 1 phi 2 phi 3 trace of phi 1 phi 2 phi 3 and we have some coupling H.So it is known that this theory in the IR flow to n equal to 4 super amuse 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 other 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 conformant field theory with E6 global symmetry.So this is interesting example.In this talk I want to present an infinitely many example of the infrared supersymmetry enhancement from n equal to 1 supersymmetry in 4D to n equal to supersymmetry in 4D.I'm always staying in the 4 dimension.Okay so yes.Then let me in this introduction let me show you the simplest model and simplest RG flow which we found which is the sum n equal to n equal to 2 Lagrangian 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 to RG S Lagrangian theory.So I prepare one slide to review the RG S Lagrangian theory.So what is the RG S Lagrangian theory?So this was originally found at the special point on the Coulomb branch of n equal to 2 SU3 pure superion mill theory.So there is a Coulomb branch of n equal to 2 theory.Then there is a special point where the massless particle appears.And so in this SU3 theory there is a very special point where the mutually non-local 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 to SCFC.So this theory is called RG S Lagrangian theory.And this theory is strongly coupled and we know the central charges which is this rational numbers.So the central charges are some number which count the degrees of freedom.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 six fifths.So we have just only one Coulomb branch parameter, Coulomb branch operator with dimension this one.So this theory is sort of the minimal non-trivial n equal to SCFT.Why?Because these people show that very recently that this theory...So there is a central charge bound for n equal to theory, n equal to super conformal field theory.And this theory saturates this bound.So I mean the central charge C saturates the bound.So it's not by A but I mean this is some simplest theory or minimal theory in n equal to supersymmetric gauge theory in four dimensions.So it's interesting to study this kind of theory.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.So I want to explain what this theory is.So this is which we found at the first time.So this is the theory.So we consider n equal to one supersymmetric theory with SU2 gauge group.And the following chiral multiblets.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 multiblets.And also we have phi.This is in the joint representation of SU2.Up to here the matter content itself is something like n equal to SU2 gauge theory with one flavor.But we add a few singlets.This one means the singlets.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.So the first term is like n equal to supersymmetric.But another term which breaks n equal to n equal to one.We have some coupling with m1, m3, m5, m3 prime to other fields.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.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 what's the gauging variant operator in this theory?We have many, of course.But 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 squared operator.This is again the gauging variant.So of course there are many, but we focus on these operators.Then we want to know what's the IR theory of this simple, rather simple, gauging theory.To know the IR theory, the most fastest way to do is to know what the central charge is in the IR theory.So there is a work by Anselmi Friedman-Grissau Johansenthat the central charges A and C of some super conformal field theorycan be written in terms of the anomalies of the IRR symmetry, something like this.So if you know the cubic anomalies and trace anomalies of the IRR symmetry,we can compute A and C by using this formula.So the problem is to find what's the IRR symmetry.So in our problem, the IRR symmetry 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.So if you fix this epsilon, we know what's the IRR 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 an 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.So this epsilon tells you what's the true IRR symmetry in the IRR.I guess it's clear.So we can do this maximization in our model.So let's do that.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 unitary bound.So in the four dimensional theory,there is a uniformity bound.And for the scalar field,the minimum 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 unitary.So we have to check that.There is no,all the operator should be greater than one.And then if I found some operator,which has dimension less than one,I interpret that.So we start from some UB theory,and some operator has dimension greater than one.But after a maximization,we found some operator,which has dimension less than one.So it means that the dimension will be decreased.And at some point of the RG flow,that operator hits the unitary 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 oneis 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 M1will hit the unitary bound.So I subtract this contributionfrom the central charge.Then I redo the maximization.So in the second trial,M3 and M3 prime will hit the unitary bound.And then I subtract this contribution.Again do the maximization.In the third trial,it is okay.So there is only one M5 operator.Then I get some epsilon,which is this number.And it is surprising thatI get the rational number in this trial.So usually if you do the maximization,you 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.And also the dimension of M5 operatorwill be six-fifth.So these numbersyou saw in the previous slideSo I think that this theory,at least by looking these dimensionsand central charges,this theory is address Douglas.So this is just a briefreview of our theory,which is our simplest case.We have in the introduction.That's possible.Operator doesn't hit the unitary bound.Okay,so there is 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 carewhether operator hits the unitary boundis these operators.So maybe more largerscaling dimensions.This exists.Some operator will exist.Okay,so we don't havecomplete understandingof these operators.Where these operators will come tothe n equal to multiplied or not.But of course this,for example,this M5or one chiral match plate.So this is not enoughto get n equal tocool-on branch operators.So we should have something extra.This is what we wanted to do,but there is some small discrepancyabout this consideration.So we don't have complete understanding of it.Yes,okay.Okay,this is alsowe 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 descriptionin this way,in the super potential,because later I want to relatethis theory to some otherUB theory.This is I will show you later.So in order to do this,I keep these termsand also M1,M3,M5,M3 prime.Okay,other questions?You are fine?Okay,so we saw thatsome,this simplest model.We have some Lagrangian descriptionwhich flow to the,which may be flow toN equal to Algester Grass theory.So how useful of this theory?And if you are a generationof Gaillot's class S theory,you may not worry aboutwhether the theory hasLagrangian or Lagrangian description.But the two have Lagrangianis very useful in the sensethat of course,to studytheongly coupleinteracting theories,for example,non-bps sector or somethingmore thanbps sector.And more concretely,we can computethe partial functionif you have the Lagrangianby using localization technique.So for example,we haveN equal 1 theory.The simplest one isthe super conformal index.We can compute the indexin the full generality.We have three,so we can havethree fugacities in this index.So thissuper conformal indexwe can computeor some other partial functionscan be computed.I'm not so much familiarwith theN equal 1localization,but I thinkthere are morewhich can be computed.Okay,so indeedthe super conformal indiceswhich we compute in this wayby using Lagrangian description.We can recoverthe known result,previous resultin the special limits byVikan,Nishinaka andKordoba and Shao.So these people studyby using other methodto get the indexin the special limitof fugacities.So we can recoverfrom this super conformal indexthis result.So in this sensethis Lagrangian descriptionis I guess usefuland I want to ask you to useour Lagrangian description.Okay,sobutwe don't know whythere is this kind ofN equal 2 enhancementin the IR.So this is the big problemand I want tounderstand what's going onand even we don't knowyet we don'tprove that the IR theoryhas N equal 2 supersymmetry.We just look at thethe central charges and dimensionof the operators andwe thought that this is N equal 2theory by matching.Okay,right.Okay,so indexis more.Yes.Yes.By using indexwe canit's more likemore likelyto have.I'm gettingN equal 2 theory.Yes.Thank you.That's true.Yes.So okay.But we want toknow whyI getN equal 2 theoryaftergetting some decoupling fieldsand throw away this,we getN equal 2.Also theinteresting thing is thatwe start from somewhereN equal 2 like theory.So we startfrom SU2 gate theorywith some adjoint.And this adjointthis theory shouldhas cool on branchby parameterized bytrace pi squared.Okay.But this trace pi squaredwill be gone.And thiswill be decoupled.Then M5 will bethe cool on branch operatorin the IR theory.But thiswas not cool on branch operatorin the UV.Right.Sothe cool on branchwill appear in the IRsomehow.Sothis is some verypeculiar things in thispro,in this model.Soand wedon't understand why this kindthing will happen in this model.So any commentsand suggestions are helpful.And I cannot talk about whywe getN equal 2 enhancementin this talk.Sothenhere in the rest of my talkI want to saysomething more abouthow I construct this kindof Lagrangian theory.Okay.Iguess so.Soin the IR theory thereshould be a cool on branch.Andthe theory in the cool on branchshould be u1 gauge theory,some mattermultiple.This u1is not the u1in the UV theory.So maybeyeah,butactually I don't know what happened.might be.Okay.SoSo later I want to show you howI get this kindof Lagrangiantheories.Andactually this methodisbyokay.So afterwe have,okay.So by thismethod can be applied to anyN equal 2 supacon-Humafil serieswith non-Aberianglobal symmetry.Sookay.So we can getmany Lagrangian descriptionswhichprote to n equal to fix point in the IR.Sothis is the first point.Andif I have a time,Iwant to say something aboutthe supacon-Humafil index.Okay.Okay.I have 30 minutes.Nowokay.Let me go on.So what'sthe deformation?So let'sfirstforget aboutwhat I said in theintroduction,the Lagrangiantheory,and please concentrateon this deformation.Let'ssuppose that we haveN equal to supacon-Humafil which I call t.Andwith thenon-Aberian flavor symmetryf.Soyou haveany theory withflavor symmetryf.Okay.So this flavor symmetrycan be a subgroup offull flavor symmetry of thistheory t.Thenokay.Sojust a convention thatsince this is N equal to 2 theorywe have SU2R symmetryand U1R symmetry.Herethis I3,twiceI3 oftwice I3 is theculturn of the SU2R.Okay.And this Ris thegenerator of U1R.Sowe use a convention thatthis 2 iswritten as j plus and j minus.just a matter ofconvention.Okay.Sothento this theorylet us couplethe N equal to 1 chiralmultiplet Min the adjoint representationof F.Okay.Weadd one chiralmultipletor chiralmultiplet in theadjoint representation of Fwith this superpotential.So heremu is the operatorin this N equal toSCFT whichappear in the army high stockyesterday.Whichthis mu has theSU2R spin onein the spin onerepresentation of this SU2R.So okay.So this muis in thisN equal to SCFTT.And Icouple thismultiplet N equal to1multiplet to musomething like this.This is thesuperpotential which we have.Since the mufield,mu operatorhas spin onein the of SU2Rwhich means thatOkay.So which means thatI3,this isCultando SU2R isCharge 1.So this hasJ plus J minus Charge20.So thenthe M should have02Charge.Okay.So since J plus and Jminus should be theRCharges,R likeCharges.So it shouldhave 22RCharges in thesuperpotential.Okay.So thenOkay.So butso far at this pointwe don't get anything newbecauseif you just couplethe chiral march bet in the joint representation and tosee what happened in the IRyou just get the coupled theorywith N equal tosuperconquon field theory andthe chiral march bet.So thisis nothing interesting.But if you dosomething moreyou get somethinginteresting.So andthis kind of thing is alsorelated to the Amihae story yesterday.But okay.So we givesome nilpotent webto the chiralmarch bet M.Sookay.So this nilpotentweb is specified by theRow which from SU2to F where Fis the flavor symmetry.So in the caseokay.So this web break ofcourse breaks the flavor symmetryto the small amount.Thenso in the case ofif the flavor symmetryF is SUnor A typethis webis classified by thepartition of Nor Young Diagram.Andfor other groupsthere might be the classification.So this one.Thesetwo deformationsdefine the theory in the IRand which is leveledby the UBN equalto theory and alsothe way of the embedding.So this IR theory hastwo levels.And so since Ido,I addedthe N equal to one chiralmarch bet,genericallythis theory is N equal to one.Okay.There is no reason thatalthough we start,Istart it from the N equal to twotheory,but there is no,butthis theory purely N equal to one.So there is no reason that thistheory is going toN equal to,but we seein the many examples thatwe get N equal tosuper symmetry in the IR.Okay.But let me justdo some a bitrather complicated thingswhich Ineeded.Sowe had herewe had here mu andm.So here muandm are both in thejoint representation of flavor symmetry.Okay.So thenafter giving aweb,this jointrepresentationbroken or decomposedinto the SU2 representationwhere SU2 isthis one of the embedding.So thisjoint will be broken to SU2representations.Okay.SoandOkay.So thenby giving a web,assowe label the SU2representation something like thiswhere jlabel the spinrepresentation and j3 is acomponent of the inside thej-spin representation.Okay.So j3 isj2-jwhich is integer or half integerofcourse.ThenI label m and mu somethinglike this.Thenby giving a webthe super potential is somehowperformed something like thisbecause I give you a web to mso we get this termonly the viewonly the view.So this isproblematic because somehowthis term breaks the originalU1 symmetryU1R symmetryU1R symmetrybut if I considersome combination of theseU1 with this rowwe can preserve these2U1s.SoOkay.So let's consider thisj plus prime and j minusprime and this will betheU1 symmetry in theIR theory also.Then someanother argument shows thatfor each so okay sowe saw that mum and mu which decomposeinto the SU2representationsand for each SU2representations onlythe lowest component survivesI mean the m willgoes tothe componentsin the mjust only this componentwhere for eachspin j representationthe component which hasminus j3 equals to minus jis the only onewhich survive after the deformationso after allwe get this super potentialokaythis is justsmall computationokay yeahso as a okay as asomeone will combine to theto thecombine to the flavorcurrent andit makes the flavor currentto non-bps because flavor symmetryis brokenor some other willdick up.Then okayso this is the small thingokay so thenwe can dofor each modelwe can consider thecentral charges by using a maximizationlike in the Lagrangian which we saw in the introductionif we know thecentral charges of tthe uv theoryif you know the n equal tocentral charges t and alsoflavor central chargef of fsymmetry f we can do the maximizationthe formula for suwhen f is su nand if Ichoose the embeddingwhich called principle embeddingwhich breaks the flavor symmetry completelythe formula is something like thisokay sothese are trace ofsome u1 global symmetryand here a and tis thecentral charge of n equal to theorywhich we have in the uvthen k is theflavor central chargeand other is nothingwe have just nand if we start from su nso this is the formulaand we can do the maximizationso the inputis the n equal tocentral chargesand flavor central chargeand also thehow you embed su2 inside fso we can dothatand this isthe general prescriptionto deformn equal to theorywith anyflavor symmetryso let's see the examplethe first example is very simplewe haveas the n equal to theorysqfttthe su2with 4 flavorsthis is somehowsimple Lagrangian theoryLagrangiansuper conformal field theoryand this theory has the flavor symmetryso8okaythen we consider the principle embeddingof thisso8okay so what we do isstart from this theoryso8adjointand give a verbto break the so8flavor symmetry completelyokaynear potent verb to break so8completelyso the adjoint representation of so828 representationis broken tothree dimensionalone three dimensional and twoseven dimensional and onedimensional representation of su2andfor each representationwe have just one componentsurvivewhich i call m1-1and m3-3m3-3and m5-5this one threethree five denotesdenotes the spin representationthis is a spin onerepresentation which has three dimensionalspin threerepresentation which is seven dimensionalokayso these fouronly surviving componentand thiswill be the singletwhich appear in the Lagrangian descriptionwei show you in the introductionokayso then of course we give a verbto theto this mso this is the Lagrangian theoryso we can doeverything in the Lagrangian levelso thenby giving a verb to mwill be the mass termfor the quarks of su2with four flavorsso if you do integrate outintegrating out the massive fieldafterallwe get su2 theorywith one flavorand some adjoint and these four fieldsthesuperpotential in the introductionintroduction can be obtainedafterintegrating outso thisokay so thisthe previous modelLagrangian model can be obtained from su2with four flavor byso8 principle embeddingso this is howwe get the Lagrangian theoryin the IRand if you do the maximizationyou get n equal to theoryAlgesta Glass theoryokayso this is one thingokay soI do theI consider the embedding which breaksso8 completelyso why youdon't consider any other embeddingso we didso we have many embeddingsfor so8so maybeyou are not familiar with about thisbut okay sojust thisthis is a labelso for exampleother choice of embeddingswhich is something like thiswhich preserves only su2 flavorssymmetryif I consider these embeddingsin the IRagain we get n equal to enhancementbut this is not Algesta Glass theorybut some generalizationof Algesta Glass theorywhich is called H1theoryso originalAlgesta Glass theory is called H0theory andnext Algesta Glass theoryis called H1and this is somehow in theKodaira classification of rank 1SCFTokay so we can get this oneor for another embeddingwith preserveU1 times U1 global symmetryin the IR I get H2 theorywhich has su3global symmetryso okay so in this examplethethe supersymmetry is enhancedin the IR and also the globalsymmetry is enhancedin the IR to su3so many enhancementyes I think soyes I think soyesokay soI'm not sureI'm not sure we didthe computationcompletely okayyeah this is what we aredoing right now soyeah I'myeah I'm not sureso far yes but should becan be seenby computing the indexokay so anothersutter example is this5.3and this case I get somerational number a equals tosomesomething this strange numbersand see something strangeand we are not sure whether thisis n equal 1 or n equal 2and so rationalitydoesn't tell you this is n equal2 theory because for n equal1 theory we can get rationalnumber or irrational numberso we don't knowwhat the theory isyesyeah it's not dimension 1yeahyeah sowe are not sure aboutthis theory isbut if the dimension is 1if the dimension is 1this will be excludedif the dimension is equal to branchand in order to have such a bigyou need to have I guess1,000 or moreno for surethis isfor sure this is not dimension 1but it's not like1,000so dimension 2yeahmaybe somebody know aboutyeah okay sofor other embedding we getthe irrational numberso it means thatthe theory is n equal 1in the IRokay so thesetwo cases at leastwe get n equal 2 enhancementand for this onecase we don't knowand the other is n equal 1SCFTin the IRokay so this is the case for SU2with four flavorsso why not wetry to generalize this SUNcan flavorsthis onesorry 7-1maximalmaximal yesmaximal is justnothingwe don't do any deformationwe start from the nextmaximal oneyes and otherrather smallerembeddingdoesn't give you any n equalso thetop linethey all produce the samebecause of SO entirelyyes yesso there iswhich oneyesyesyour hasty diagramyou start frommaximal5,3 is the next oneand you have threeand which is5,1 cube and 4,4and combine to3,2,1 squaredyes that's correctyeahso then okayso this is the exampleokay so then let's start toSUN with do I have timeoh 10 minutesso okay let's tryto SUN with two n flavorsthis has a global symmetrySU2N times U1soI don't thinkof the U1Arberian part and just considerSU2N partand let's consider the principleembedding for SU2Nwhich okay so this breaksthe SU2N toSU2N times U1to just U1okay so by thisI'm not sure whether I havetimelet me skip the detailsbut I getthis dimensionsand also thiscentral chargesso this is the onewhich call A1 toN-1 theory bySergeowhich got this classificationand the central charges are obtainedby maybe first bySharpier and Tachikawaso okay so againwe get start fromsome SUN toN flavorsand we get in the IRthe N equal to theoryyes soor we can consideras an example with SPNwith 2N plus 2 flavorsand this caseit's flavor symmetry is SONso then wewe sorrynot SON, SO2N plussorry yesthen the principle embeddingwhich breaks completelywill get I getA1, A2N theorywhich is some generalization ofAzurestagrass theoryokay soby this examplewe hope thatall the super conformalthe theory can be obtainedby this kind of deformation in the IRso this might beto try to considerto considerother all N equal toSCFTs andwhat result in the IRand but it isvery difficult tosweep all the casesand what we trywhat we haveso faris okay so thisflow we already seethis is the tablefor the principle embeddingwhich breaks the flavor symmetrycompletelyso this one alreadythis one is already solvedand then other caseswhich is rank 1we can start from rank 1 SCFTwith N equal to super symmetrywhich is called H1H2D4E6E7E8okay so this hasSU2SU3SO8this one already we thoughtand E6E7E8 Global Symmetryand we can considerthe breakingcompletely break the embeddingwhich break completely this flavor symmetryand all the caseswe get H0theorythe original RGSTAGRASorwe can start from A1DKtheorywhich at least has SU2 Global Symmetrywhich we considerand in this casein the IRby breaking SU2 Global Symmetrywe get A1AK-1theoryso many enhancementswill happenbut there is a casethat there is no IR enhancementandthe one is otherrank 1theoriesby the classification byArziles and his collaboratorsand sowe try otherrank 1theoriesbut there is no enhancementthe IR theory is n equal 1maybe we get newn equal 1theory but I don'tand alsoif you are familiar with class S theorythere is so calledTN theoryand we can do forTN theory has SU3SUN cubed Global Symmetryand we canconsider the breaking of these threeSUNby this we get n equal 1theoryin the IRor the otherthing is the n equal 4supermusecan be seen as n equal 2theoryif you look at n equal 4theorywith SU2 gauge groupas n equal 2theorythis has SU2 Global Symmetryso this can be usedas our deformationand if you look at IR theorythe n equal 1ok so there is somemaybe we can try moreat least we have this kind of tableok sothe first question iswhat is the patternof thiswhy these twoclasses are dividedwhy thesehas n equal 2 enhancementin the IRthe hintwill bethis is justwe notice thatbutthe first onethis onehas 2D chiral algebrawith Sugawa constructionthis is byBeam and Rasteryand his collaboratorsand this should be okso the first this oneall of them hasok soif you have n equal 2theorythis correspond to2D chiral algebrais associated tobut the chiral algebrahas Sugawa constructionis not the alwaysso thenthe chiral algebraassociated to thishave the Sugawa constructionand for thisit does not have itso this is something to dowith 2D chiral algebrabut only exceptionis n equal 4SU2this hasthis has chiral algebrawith Sugawa constructionthere is one exceptionbut somethingrelated tothe chiral algebra businessok sobutI guess this case is chiral algebra with this flavor groupSU2,SU3,SU8yeah so yesand this isSU2,at leastSU2andSU2 timesU1but the rank is somethingminus the negative valuethen otherok sothis listwas only for thethe flavor symmetrycompletelyand so we can doother embeddingsok soI think thisif you dothethe nilpotentdeformationinside n equal 2SU2you may getthis isthis deformation is something differentbecause I add n equal 1 chiral much bit and giving a verb tothis chiral much bitso your case is giving a verb tomuso yeahbut I'm not suremaybe we can tryok sothenso what's the other embeddingsso we don't knowcompletely but we knewexperimentally what happenedso very experimentallywe observed that if there is no enhancementin for the principalembedding there is no enhancementin the other embeddingsso the principal embedding isis mostcase which have mostwhich have enhancementin the IRyeah thenokso then quite very experimentallytheembeddings with smallerflavor symmetriestend to have enhancementok in this sensethe principal embedding which breaksthe flavor symmetry completelyis theembedding which havelikely havethe enhancementbut we don't know completely yetwhy this kind of embedding happenedso ok soI'm running out of timeok so I havetwo slides forsuper confirm my indexso maybe I havetwo minutes ok let melet merun now runningso okok this is for the expertI completely skip the detailso theok so we haveso I should mention thatthein the exampleswe only try the Lagrangian theorythe starting point isLagrangian theory andwe deform to Lagrangian theoryand to see what the IR theorybut as I said beforewe can choose thisT at any theoryany n equal to theorywe can start fromwe can start from theorywithout Lagrangianok soso in this sense we havewe can't have some Lagrangiantheory whose IR theoryis n equal to or somenon Lagrangian theory whose IR theoryis n equal to it's up to youso but if I chooseLagrangian theory we cancompute the index or we can computethe partition functionso and but in this waywe can compute the superconformal index if you know the matter contentof the Lagrangianso then this is somehowthe definition of the indexand I don't want toyes I thinkthis is the case withthe theory has conformalsymmetryfirst of all I'm not surewhether there is theRsymmetry in the non-conformal caseso we shouldhave SU2R timesU1Rand I think non-conformalcase U1R will be broken byanomaly soin this case I don't get enough U1sin the IRyeahyeah but we didn't tryso maybesomething interesting happenedfor the non-conformal casebut I justfocus on the conformal theoryokayso thenokay so this isthat's right so in theH0 theoryokay so H0 is the originalRGS Douglas theory which Iconsider in the introductionand I show youwhat's the matter contentand here we haveintegralbecause of theSU2gauge symmetryand this part comes fromthevictor multiplyand twofundamentalsand this is a jointandwe havethewe haveso we have four singletsbut three will be decoupledwe justthrow away these threecontributionthen I just multiplythis contributionthis is for the remaininggauge singletmultiple this oneso then finally alsotrace phi square will bedecoupled this isonly one subtle thing to computethe trace phi squareis the trace phi squareand it's not likejust remove the contribution from phiso there should be trace phiqq primeor qq or q prime q primeoperator soto remove the contribution fromphi is not goodbut I just remove the contributiontrace phi squareconsider that some chiral algebraso this is onewhich appear in the denominatorso this is the indexwhich I haveso thenwhat I should dois just compute the integralbefore thatwe should considerthe irr symmetry is differentfrom the ubr symmetryso in order to take into accountwe shouldsubtract this onexi into t timesp and qso by this we getthis prescription is to choosethe correctr symmetry in the irafter the maximizationok sobasically one can do the integralsorry integralthis integral and we getthe someseries in terms of p and qand pq and pthen theok so we can try thelimit for examplethe simplest one iscoolon index limitwhich is set pq over tequals to u and pqt goes to0 the simple oneand I get very simple oneand you can see that there isthe one operator which hasdimension 6 fifthso this shows thatthere is some coolon branchwith dimension this oneand also we can do somemcdonald limitwhich agrees with the index bycooled over and showby order by orderso it's very difficult to showthat thisto take the limit in thisin this levelso after doing theintegration and wecompare by taking a limitand compare to theprevious indexfor full generalityand this can be done forother theories I guessin principleyes so ok so this isthe last slide andso we have many questionsso first of allwe are doing fornon-principle embeddingsthe previous paperswe try to consider principle embeddingand we do something moresomething complicatedof course it's notit's not easy to complete the listwe don't have any completelist of any called two theoryso it's impossiblenow to complete the listthe deformationso as I saidwhat's the condition of UB theoryT for the enhancementthis might be interestingand why there is enhancementmaybe this is related tostring theory constructionif we consider some geometricconstructionthis might beansweredok sothis is related to some stringand some theoryrealization or howcool-on branch appear in the IRmight be the interesting thingsok I stop here thank you very much