 Ok. Ok, benvenuti a tutti a questo secondo giorno. Vogliamo dire, prima di iniziare, che ci sarà la foto della conferenza a 10. Quindi, non lasciate il ruolo dopo il lecture, perché avremo una piccola di tutti noi qui. Siamo andati. Ok, benvenuti a tutti e dobbiamo andare al secondo lecture. Ovaninda, su The Conformal Bootstrap in Melly Space. Ok, quindi... L'intero giorno ho avuto un grande avvio di questo programma, dove questo è andato. E, senza che il programma sia tecnico, pensavo che io avessi scelto le dettagli, le omettere le tecniche, e poi avrei messo come potete imparare le dettagli tecniche. E, spero, in questa ora, alla fine di questa ora, vedete come avremo questi mesi, come i mesi di fine-men di cui ho detto di yesterday. Quindi l'ultimo ingrediente è melanespace e in un senso melanespace è naturalmente anche in un modo in cui pensate di conformi blocchi. Allora, facciamo un'occhio. Quindi questo è conformi blocchi e mac polinomiali. Quindi conformi blocchi e melanespace se vuoi. So there's an integral representation of the conformal blocks. So the conformal blocks yesterday I told you that they satisfy a differential equation. But there's also an integral representation of the conformal block. So I don't have the time to review the integral representation, but if you want to learn about the integral representation, it's very well covered in Dolan and Osborne in 2011. So the idea is that when we write down the integral representation in position space, you need to do a certain class of integrals which look like this. So there is some normalization factor here, but there is an integral of a position that you need to do. There's a product of these factors x minus xi raised to minus 2 Li such that the summation over Li is equal to D, the dimensionality. And you have to do an integral over x. Now these are conformal integrals and there is a famous formula in conformal field theory that allows us to do that, which is called the Simon's star formula. So essentially the Simon's star formula converts this into a complex integral. And the result of that is, incidentally, so let me just make a note of that, so there is a formula called the Simon's star formula. It's called a star because you have x and that x is joined to various points x1, x2, x3, x4 and you have to carry out the integral over x. So roughly speaking it looks like a star. And if you want to know the derivation of the Simon's star formula, it's given very well, at least the way that I learnt it, the most accessible one instead of Simon's paper was in an appendix of one of the early papers that Dolan and Osborne wrote. I think it's the super conformal field theory paper in 2002. But you can look at one of the first three papers that they co-authored and it's in one of the appendices there. So once you apply the Simon's star formula, all you have to get is this. So there's an integral over some set of variables which I'm labeling, so these ij's are labels. There's a bunch of gamma functions. And then finally some powers. So the x integral has been removed and you've traded that for an integral over these complex variables. And in this formula the sij's are restricted to satisfy these kind of constraints. So for a four-point function, so this is actually true for any n-point function, but for a four-point function the number of independent sij's are two. So this formula that I've written, I can run from one to n. It's completely general. But for a four-point function there are two independent sij's and these correspond to the fact that there are two independent cross ratios. Four-point function, two independent sij's. And it's nicer to make a transformation. You can solve for them. So let's solve for S12 and S13. And then nicer to refine the coordinates in this manner. So you introduce S and T instead of S sub 1, 2 and S sub 1, 3. You introduce S and T. In terms of these variables, the conformal block, GUV, in melon space looks like this. So there is a function of S and T times a measure factor. It's conventional to pull this measure factor out. This measure factor that I'm pulling out essentially comes from these product of gamma functions. It's just convention. You may pull it out or you may not pull it out, but it's convention to pull this factor out times gamma of S plus T squared times u to the power S, v to the power T. So in terms of the u variable and v variable, this complex integral is nothing but the melon transform with respect to these variables. The contour that you have is basically both these S and T contours run from minus i infinity to plus i infinity with possibly a shift. And you close the contour on the right or the left depending on which channel you're considering. So for example, because this is... If you were in the S channel, you would expect u to be small and so you need to pick up smaller powers of u. So you have to close the contour on the right. This is the melon transform of the conformal block. And explicitly it takes this form. Loads and loads of gamma functions and it takes, especially writing on a board, writing out everything, takes a lot of time. So I will just write out the key pieces and times a projector, I'll explain what this is. So the explicit poles are in these gamma functions. So there are poles in S equals to delta minus L by 2 plus N and S equals to D minus delta minus L by 2 plus N. So there are two sets of poles. These don't introduce poles, they will introduce zeros. And these zeros are extremely important in the representation of the conformal blocks. This is a polynomial. This is what is called the Mach polynomial. And it's a polynomial of degree L. So if it is a spin L operator that you're getting exchanged. If O is a spin L operator, then this is a Mach polynomial. S and T both have a degree L. So there are two sets of poles here. So these are the usual physical pieces that you would pick out. But there's also these pieces which are not physical because these are what are called the shadow pieces. So these are shadow poles. Shadow poles are, whenever you have an operator whose dimension is delta, the shadow operator has dimension D minus delta. So you naturally get these shadow poles as well. And of course you don't want the contribution of the shadow poles. So you require to add a projector which will project out the contributions of the shadow poles. So basically the projector has zeros precisely where you have the poles here. So for example you could have a trigonometric function like a sine function which will get rid of these poles without changing the residue at S equal to delta minus L by 2 plus N. So this is what it is. Now one thing that I should point out that the only poles that are there come from this factor and you could add additional factors to this which will leave the residue unchanged and also the location of the poles unchanged. So for example if you added, see the, here the poles are at S equals to delta minus L by 2 plus N. So if you were looking at the sum over the residues at these poles so you could write this as sum over some function of T over S minus delta minus L by 2 minus N and you sum it over N. So there is a label here as well, these residues. Then actually you could add additional pieces like S over delta minus L by 2 plus N raised to whatever power that you want such that the residues and the location of the poles are unchanged. Is that clear? So for example suppose I added a square then if I evaluate the residue at delta minus L by 2 plus N this factor is going to become 1. So the residues are unchanged. It does not introduce any additional poles. And the effect is simply that of adding a polynomial because you could have added and subtracted 1. So this piece is going to have precisely this factor and S plus delta minus L by 2 plus N. So you will be, so this additional piece that you're adding is essentially a polynomial in S. Okay, so there are these kind of ambiguities that are there in these representations, yes. Yeah, so we will, so his question is that unless and until we are closing the contour on the right and using the residue theorem to pick up the poles what does this project actually mean? So if you just need, if you can't close the contour on the right that's what he's asking. So we are going to assume that we will be able to close the contour on the right and it falls off fast enough which will allow us to close the contour on the right and you might have to use the fact that u is less than 1 or something along those lines so that you get an exponential suppression coming from these kind of factors. So that's an important point, yes. You cannot add arbitrary factors. The projector has to be such that it allows you to close the contour, that's correct. So we want to require the explicit representation of the Mach polynomials. The Mach polynomials actually look pretty complicated. There are multiple sums, multiple finite sums but there is no compact nice expressions that are known for the Mach polynomials unfortunately. This explicit representation of the Mach polynomials can be found in that paper by Dolan and Ausman in 2011. So motivated by this structure, Mach in 2009 and I suppose in 2009 he actually proposed that the Mellon transform, he proposed a Mellon transform of a CFT amplitude, a Mellon transform of CFT correlator. He proposed that it's natural to think about the Mellon transform of a CFT correlator leading to the definition of what is referred to as a Mellon amplitude. This is your CFT correlator in position space, identical scalars for simplicity. Pull out this position dependent factor and then a generalization of a 4-point function of this kind you can define a Mellon amplitude in this way. It's conventional to pull this factor out and this factor satisfies some nice properties and we will make use of that. So this is something that I will refer to as some measure factor. As I said, it's convention. This MST here is what is referred to as a Mellon amplitude and it satisfies certain nice properties. It's supposed to be a Mellomorphic function so this Mellon amplitude MST is a Mellomorphic function. It only has simple poles, Mellomorphic function. It has simple poles, no branch cuts. It has poles in different channels. These poles correspond, are located at twist by 2 plus N. So twist was delta minus L and that's what where it is. The poles are delta minus L by 2 so this is twist by 2 plus N. The residues at these poles are related to three point functions to three point correlators and exhibit factorization. Going from one channel to other so if you wanted to start off with an S channel expression of that kind if this is an S channel expression by that I mean that you're taking the first two operators and doing the operator product expansion and you're taking the last two and doing it. So that is what I was calling the S channel yesterday. Then to go from the S channel to the T channel so this was the S channel so 1, 2, 3, 4 the T channel is this expression when I'm taking 1 and 4 close together and 2 and 3 close together and the U channel is the usual way that we would think of the U channel 1 and 3, 2 and 4 so this is S, this is T and this is U so if you start off with the S channel expression to go to the T channel S to T you need to make the following replacement and this I'll leave as an exercise it's very easy to actually do this you have to go from S channel to the T channel you have to make the following replacement S goes to T plus delta phi and T goes to S minus delta phi in my conventions and if you want to go from the S channel to the U channel then you have to make the replacement S goes to delta phi minus S minus T leaving T untouched so this is how you go from one channel to the other oh sorry yeah yeah yeah thank you we don't change the position yeah thanks if I put in an arbitrary amplitude it might not satisfy these properties I mean you want them to satisfy if it satisfied then obviously they would be crossing symmetric on their own I mean when because the anomalous dimension goes to 0 or something ok yeah right but that's in the limit of L goes to infinity so even if L is large and finite you have again you have to repeat the question sorry so his question is that M is not strictly a metamorphic function because when you have these large twist operators sorry large spin operators then they will accumulate because they'll have the same so for example twist 2 large spin will all have land up having the same anomalous dimension approximately because the anomalous dimension goes to 0 so they have this roughly speaking the same dimension so the poles in that limit would sit on top of each other so all the poles will coalesce and it will become a branch cut that's what you're saying or a branch cut or something that is in the strict L going to infinity limit yeah that is true so I'm pretending that L is large but finite so that they are never on top of each other if you want I said that for the four point function I use the fact that it is just a function of the reduced co-relator just a function of u and v and as a result of which I just had the uv space corresponding to this ST space you could define for anything I mean you find my diagrams you do use melon bonds techniques any other questions ok so this M of ST actually satisfies some other properties I'll make a note of these here and these are also trivial to derive so I'll just make a note of these and leave them as an exercise for you to derive so suppose you start off with the S channel so this is the S channel 1, 2, 3, 4 now of course it does not make any difference if I interchanged 3 and 4 or 1 and 2 that should not affect the representation for the co-relator and so that actually tells you that M of ST and I'll put a suffix S to denote it by denote the channel so M of ST is going to be equal to M of S S of minus S minus T so that's 1 and this is actually a property that is respected by the Mach polynomials so this is the symmetry of the Mach polynomials for identical scalars but that you know where this comes from now and using this and this is actually important you can show that if I write down an expression of this kind use this property and this is again an exercise you can show that M of ST plus M of T plus delta phi S minus delta phi plus M of delta phi minus S minus T which is basically the sum of so if this is the S channel then this would be the T channel and this would be the U channel from that transformation that I've written right in the end this expression times this measure u to the power S v to the power T dS dt divided by 2 pi i squared this expression is going to be manifestly crossing symmetric in the sense that if I use this expression go to the make this transformation to the T channel or I use that expression make the transformation to the U channel it will remain invariant and this this is actually important to show that this is true this measure factor that I've pulled out which is this I've written it in several places delta phi minus S squared gamma of minus T squared gamma of S plus T squared is actually invariant under all three transformations both of these transformations is that clear so M is a mailing amplitude and it's automatically crossing symmetric right if I add up all three channels it's crossing symmetric but suppose I give you some arbitrary function of S and T M of ST doesn't have to be yes so what is the definition of M? no it's some function ok let's call this f sorry sorry I should call it f and f is such a function that satisfies this property then you can show that this new function that I've constructed is crossing symmetric sorry for confusing the notation is that point clear? now suppose I started off by saying that f of ST was this the mailing amplitude corresponding to the usual conformal block and I tried to make a crossing symmetric correlator by just adding the various channels so can you do that the first observation that you have is that the S channel here has this set of zeros at delta phi minus S squared gamma of delta phi minus S squared so S equals to delta phi plus N those are zeros those are double zeros now gamma of delta phi minus S squared when you do that replacement to go to the T channel this factor just simply becomes gamma of minus T squared so that the T channel expression which is this so this will be proportional to one over gamma of delta phi minus S squared so these double poles that comes from the measure at S equal to delta phi plus N but this guy under the transformation is going to become gamma of minus T squared so the double poles that this measure has at S equal to delta phi plus N will remain similarly for this this will become gamma of delta phi minus S minus this is S plus T squared this factor when you do this transformation to the U channel is going to become this factor so both the T and U channels will have double poles that comes from this measure factor because this will have gamma of delta phi minus S squared is this point clear is this yeah the measure is this factor that I have pulled out that is mu of S t that is the definition of the measure F of S t is just this capital B so imagine that we are doing it for the usual S channel conformal block the usual S channel conformal block has this one over gamma of delta phi minus S squared and that is expected because you don't want there to be any poles except the physical ones so these are completely expected so what I'm saying is that suppose you started off by considering this little f to be the S channel conformal block the usual conformal block trying to make it crossing symmetric then the crossed pieces which will be the T and U channels will have these double poles corresponding to this factor ok ah so the point that I'm trying to make is that and this is something that you can convince yourself that regardless of whatever expression that you start off with here so suppose you thought that you would be able to add these or some factors that will get rid of these double poles that are there in the measure you will always land up with a crossing symmetric expression which will have some poles that comes from the measure factor so that's the point that I'm trying to make here so let me just emphasize that so you might have thought that ok instead of this gamma of delta phi minus S squared suppose I started off with 1 over gamma of delta phi minus S squared times gamma of minus T squared times gamma of S plus T squared that will get rid of all the double poles that are there in the measure but you can't do that because the S channel would then have the wrong residues they'll have the wrong residues corresponding to the physical poles so you can't add arbitrary things ah so ah there is no way of getting a crossing symmetric expression which is free of any and I'll start calling them spurious poles any spurious poles that comes from this measure factor is that clear? is spurious poles because you don't see them in the operator product expansion so you can do an explicit calculation suppose in this expression I started off by saying that let's suppose this was the usual conformal block and I said that you will start having these spurious pole contribution in the cross channels so you won't be able to build this crossing symmetric expression of this kind but suppose you did this calculation for using Witton diagrams for simplicity let's consider just a scalar exchange Witton diagram so this is just a scalar exchange you write down the bulk to boundary boundary to bulk bulk to bulk these are all the bulk to boundary propagators and you do the calculation in the usual way then the expression, the Mellon amplitude for this diagram has been worked out by several people for example you can find this expression in a paper by Penedones Paulos this expression has been worked out and you can the expression takes on the following form so in terms of the reduced amplitude in terms of U and V the Mellon amplitude so I've pulled out this measure factor it's just a 3F2 hypergeometric function whose arguments are delta minus, delta by 2 minus S, 1 plus delta by 2 minus delta phi same factor 1 plus delta by 2 minus S delta minus d by 2 plus 1 and all evaluated at 1 at the unit argument so 3F2 has 3 entries so these are the AIs and these are the BIs in a hypergeometric function so this is a 3F2 hypergeometric function and also there is all this is divided by delta minus S over 2 so you see the point here is that if I look at the S channel V10 diagram which is the S channel exchange then the Mellon representation of the V10 diagram is such that it no longer has these zeros does not have this factor oh yeah sorry is that point here so the usual conformal block has this factor uh which gets rid of these additional S dependent poles that comes from this measure you do the V10 diagram calculation and you will find that that particular factor is gone it's no longer there that's actually in some sense it's desirable in the sense that now if you go from the S channel to the T channel and the U channel if you now add the in the usual way all three channels will now have these double poles that come from the measure factor is that clear it's a real statement so V10 diagrams, double poles now in the language of ADS CFT you are you're familiar with this because whatever poles are coming from the measure where are they located so the poles are located when comes from this factor in response to S equals to delta phi plus N and the physical dimension of the operators because the physical operators are located at delta minus L by 2 plus N so the physical dimension of the operators would be S equals to 2 delta phi because of this factor plus L plus 2N so these operators are what in the ADS CFT language are double trace operators and so from the ADS CFT in the large N limit the measure factor actually takes into account the contribution from these double trace operators so that's why from the melon amplitude point of view don't require to add the contribution of the double trace operators in the melon amplitude they are taken into account by this measure now in the large N limit that is ok but here I am really not talking about a large N limit I have just made an observation that is independent of any large N limit or anything of that kind now in usual CFTs you don't have operators with exactly these dimensions there is always an anomalous dimension maybe small but then there is anomalous dimension there are no physical operators unless and until there is an extra symmetry if there is a supersymmetry or something that protects the dimension of the operator the dimensions of this kind but in usual theories you do not have operators of this kind so these cannot be physical poles and that's why we call them spurious poles contributions gamma of delta phi minus s square contributions have to cancel in physical amplitude and this is the key point a bunch of consistency conditions by looking at these cancellation conditions ok so this is an important point so if you have any questions to ask what do you mean yes yeah in the strict N going to infinity limit the dimensions of the double trace operators is literally that there are no physical there are no physical operators these dimensions because there are always anomalous dimensions which take you away from this I mean so but depends on so in the usual way that you write down written diagrams you're doing in 1 over N expansion even at the level of the measure ok so in 1 over N expansion the leading piece that you have is the N going to infinity piece where these are actually the double trace operators and the sub leading pieces will take you away from that yeah I mean it's a different way that we set up the area safety calculation here I'm just saying that that if I try to construct a crossing symmetric amplitude so what was the logic we said that we are going to try to construct a basis that is crossing symmetric by adding the S, T and U and you could have started by trying to take the S channel expression to be the usual conformal block the logic was that if you did that then in the T and U channels you'll still have these spurious poles coming from the contribution from the measure factor so that is something that is not desirable because only the cross channel will have these spurious poles now if I started off with the what I call the written diagrams then I said that look here this channel written diagram does not have these zeros so this channel written diagram will have these so-called spurious poles in the large n limit or n going to infinity limit are the double trace operators so that's why in the large n limit if you expanded everything in 1 over n expansion then the leading contribution would have these operators but in general you do not expect to see operators of this dimension precisely this dimension because here I'm not doing any large n here I'm not expanding anything here I'm setting up the exact expressions there's no expansion so there are no operators with precisely these dimensions so that's an assumption that I'm making typically there are no operators of that kind unless and until there is some symmetry that protects the dimensions so that's the assumption and that's why we are calling the poles that come from this gamma of delta phi minus s square as spurious poles any other questions? so if there are BPS theories where you have dimensions that are protected then you have to handle it differently so that's a good point because that's an important point so there are for example n equal to 4 you would need to handle it a little differently so now what we do is that we expand so suppose we have an arbitrary line and should do some things so suppose m of s t corresponds to the melon amplitude of some correlator that you want to consider so what we are going to do is we are going to expand this in the basis of tree level exchange Wheaton diagrams for this particular case the tree level so here I've just considered the delta comma zero but then you can do the analysis for delta comma l the expressions are known so what happens in the usual in the actual cases that you have some piece that looks like the 3F2 but in addition you also have a contribution so if I have delta comma l here then there will be a piece that comes from the mac polynomial so there will be some piece that looks like that so this is a contribution these deltas get shifted so there will be some l dependent shifts so these arguments get shifted that's not how they get shifted is not important but what is important is that when we expand in the basis of the tree level exchange Wheaton diagrams you have spurious poles s equals to delta phi plus n then goes from zero to infinity and these are double poles so each spurious pole because it's a gamma of it's a double pole gamma of delta phi minus s square times u to the power s when you complete the s integral then you will get u to the power delta phi plus n log u and u to the power delta phi plus n both of these are incompatible incompatible with OPE with the s channel OPE these have to cancel and so you will get a whole set of consistency conditions that arise from these cancellations yes, the what sorry computation using Wheaton diagrams said of doing a just mean field theory of generalized field theories I mean I don't see why you need an holographic description you don't require it it's a convenient way to study generalized field theories correct I wouldn't call them even generalized free fields it's just a convenient way that I mean if you want I could have just avoided referring to Wheaton diagrams at all I could have just said that look if I did the usual case then I just have these spurious poles from both these channels and as I argued that it is impossible spurious poles so it's better to actually introduce spurious poles in all three channels and then try to make sure that they cancel when you say Wheaton diagram I guess you are referring to a scalar in ADS that's it with ADS I'm not specified any bulk interaction no dynamical gravity no dynamical gravity it's just kinematical correct so I could have avoided this notion of Wheaton diagrams except that what we are doing is exactly Wheaton diagrams this is what you would call Wheaton diagrams but then if you wanted I could have just avoided talking about any of these Wheaton diagrams at all I would have said that I want these spurious poles from each channel the best thing to do is to start off with a conformal block representation and get rid of that zeros that comes from these gamma functions that are there in the denominator ok just understand if you take the generalized variable for which we know everything by now and you do the Mellon transform I think you should get automatically the same ok it's probably already worked out I mean yeah I mean the disconnected piece which comes from this generalized free field that is, I mean that's what you probably have in mind you are saying that can we reproduce the generalized free field answer so what's your question no I'm saying that you should be able to I'm essentially repeating what I said that you should be able to never mention with I agree with that I agree with that yeah the disconnected pieces are added in by hand by introducing explicit poles at the appropriate location so what we are saying is that I mean suppose so you want to suppose the usual thing would be the generalized free field answer for example would be 1 plus u by v to the power delta phi plus u to the power delta phi times that factor of x12 to the 2 delta phi x34 to the 2 delta phi this just comes from the identity exchange in the t channel and this just comes from the identity exchange in the u channel so to reproduce this we will just add in by hand something that has poles at s equal to delta phi and t equals to minus delta phi so this is something that we are going to do by hand if you want the disconnected piece we are adding in by hand so again you understand this integral really as just a series of poles because it's a bit confusing when we write the smelling integral one of its power is that you can close the contour to the left or to the right but it seems that you cannot do this here anymore because as soon as you added this piece it will not one of the poles will go to the left and that's what you are asking for for example this does not converge this integral over ds and dt does not converge but there is a u to the power s v to the power t so depending on depending on what you are choosing for u and v for example if you put u equal to a quarter it will have an exponential suppressor but let me take u equal to 3 then it will not work ok so we really are just closing the contour we are just really closing the contour and we are actually saying that we are looking at the full expression from the perspective of the s channel ope so we are looking at closing the contour in only one of the directions yeah we are not literally picking up the physical contributions from each channel so I think that is what you are asking so you are not literally picking up the physical contributions from each channel that's not what you are doing we won't get three times the answer we literally get the answer only once sorry to to wrap you again but I think Matteo was pointing out to me correctly completely generated it will be completely disconnected so let me re-ask the question so what's that what's that that cubic coupling of the wheat and egg that we are choosing no so we are just using there's no cubic coupling there's no coupling constant that we are putting here this is just a bulk to bulk propagator this is just the bulk to boundary propagator nothing I mean we are not there's no explicit vertex factor that we are thinking of there so is there a real connected term or there are purely disconnected I'm a bit confused I mean this is just a connected part of the correlator if you want but the way that we calculate wheat and egg ramps is to use these bulk to bulk bulk to boundary propagators but if there was an explicit vertex that would give rise to a contact term as well so that is something that we are not considering because those will contribute polynomials those would be related to the polynomial ambiguity in the wheat and egg ramps so that is something so since they are ambiguous we are making a choice and we are making some choice and throwing them away yes does that answer your question? so if you for example literally looked at the wheat and egg ramp it has to have the appropriate poles with the appropriate residues that are compatible with conformal invariance so you know where the poles are supposed to be they have to be at s equal to 2s by 2 plus n and the residues are fixed by conformal invariance they will be related to the mac polynomials you cannot tamper with there are other things that you can tamper with which are these polynomial pieces and that's up to you to fix does that make sense? up to a point yeah please ask because I will just sketch out what we do in the next couple of minutes and how we solve in the Wilson for the Wilson Fisher case how we extract results ok so let me just sketch out some of the details as to how you would use these cancellation conditions and extract results so actually so this came from the u piece but there is also a v dependence so there is for e there is actually if you wanted to expand around v equal to 1 for example you would get u to the power delta phi plus n times 1 minus v to the power m times this factor and similarly here so actually you would get an infinite set of equations that are labeled by two integers n and m and for each of these pairs you get a cancellation condition and how do we make use of those cancellation conditions so what is convenient as I mentioned in my second lecture yesterday if you look at the mac polynomial the function of s and t if you look at the mac polynomial precisely at s equal to delta minus l by 2 and for any value of t then this is nothing but what is in the mathematical literature what is referred to as a continuous hand polynomial the continuous hand polynomial is proportional to a specific kind of a 3f2 and the orthonomality conditions are known so in particular if you there is a measure factor I will write out the measure factor it's gamma square s plus t gamma square of minus t and if you take two of these polynomials labeled by l comma 0 and it's convenient to instead of looking at it at looking at it as being labeled by delta yeah let's let's consider this to be labeled by s then when you do the integral over t then this is just proportional to a chronica delta the proportionality constant is all known it's worked out a good source for this is the book on special functions by ASCII Richard ASCII Andrews Andrews and Roy and so using this orthonormal polynomial what you can do is you can expand the t dependence in each of these cancellation conditions so what you can do is you can write the t dependence in the following way so that you don't have to worry about what value of t you are going to choose the t dependence is all in terms of these continuous hand polynomials so for each of these each l prime you will get a cancellation condition so we set the coefficient of q l prime comma 0 t which is just a number because it has no t no t dependence it has no s dependence the coefficient is just a number or just it has no s or t dependence anymore because s we are setting equal to delta phi plus n and t has been factored out so we will set that equal to 0 and get and look at these conditions ok is that idea clear so in the last few minutes I suppose I have another 5 minutes 3 minutes how do we make how do we make progress from here so if you look at these conditions and set d equals to 4 minus epsilon for the Wilson Fisher you start by writing delta phi as 1 and actually don't require to specify this to be 1 this works out to be 1 as a leading contribution you write it like this in an epsilon expansion and whatever double field operators that you have also you write it in the same way these numbers integers that I am writing here these actually pop out from the calculation you don't have to start by assuming that this is 2 1 and you do the same thing for the double field higher spin operators which gradients so you write them as in the same way you do the same thing for the OPE coefficients again the OPE coefficients also you don't require to specify the order at which they start it comes out from the analysis so OPE also you write so for example the C phi phi phi square write this as C0 plus C1 epsilon plus C2 epsilon square and so on C phi phi stress tensor also you write in the same way so what you start by looking at is that there is a so each equation now gets labeled by L prime so you look at the L prime equal to 0 and L prime equal to 2 equation you look at these together you demand that there's a conserved stress tensor a unique conserved stress tensor so the dimension of this is 4 minus epsilon the dimension is delta is 4 minus epsilon and that's it that's the input that you need to put in you look at it in the epsilon expansion and you get a bunch of algebraic equations for this delta phi 1, delta phi 2 and so on looking at L prime equal to 0 and L prime equal to 2 è sufficient to fix this to be equal to minus half this to be equal to 1 over 108 and also delta phi 3 epsilon cubed this number is known to be 109 over 11,664 all these numbers come out from these analysis and Atish if you are there in the audience and still awake this was the point that I was trying to make this is 1 over 108 109 over 11,664 so this is roughly speaking the same as this so we can determine only up to 3 loops the 6 loop corrections are known I don't remember the numbers, precise numbers but I think they are comparable yeah I have the top of my head I don't really know so that people consider this Pate Borel re-summation and somehow the numbers are not very different from what you get by setting epsilon equal to 1 at this level you also fix this guy this actually works out to be minus 2 third this works out to be 19 over 162 and that's as much as you can do for phi squared for the yeah and looking at just l prime equal to 0 and l prime equal to 2 you can also fix the OP coefficients the OP coefficients for the stress tensor phi phi stress tensor you can fix all the way up to epsilon cubed and the OP coefficient for phi phi phi squared you can fix up to epsilon squared what you land up getting is a relation between the epsilon cubed OP coefficient with the cubic anomalous cubic order anomalous dimension here so that's the relation that you derive and if you look at the known result from Feynman diagrams that relation is actually I mean yeah so the once you use that relation then you can fix C3 but there are consistency conditions that C3 satisfies which passes those to look to determine the double field operators you need to set l prime to be arbitrary so l prime equal to some even integer and you can specify l prime to be arbitrary and then the formulas that I showed in the end of my second lecture in terms of the harmonic numbers those actually just pop out it's not something that we do order by order and we fix it it just pops out in terms of the harmonic numbers and so that's roughly speaking how we do these calculations and I I'm sorry I had to rush but since this is my last lecture I wasn't able to give you as much details as you would have liked to but then if you if you're interested in knowing more I can discuss with you in person and I'll tell you how exactly things work so I'll stop there, thanks