 Well, no, thank you for inviting me here Because it's nice to see so many friends. It's also sad, you know, the Think about the occasion for the talk I Met Vadim several times in attack You know over we had these fantastic meetings to land on Odita meetings And I think it was not there the first time but he probably was there. I mean in 84 And I remember one episode from 1984. We had one of these Marvelous parties in someone's apartment and we were I don't know 30 people 40 people in a two-room apartment Yeah, I think that was that time it was that's true because it was not such as it was no no That was in 81. No, it was any how So it was completely crammed and there were two sober people and that was Vadim and Paolo de Vecche And they were sitting, you know, like that and they were discussing and they mentioned the roti paper. So that was Then the last time I saw Vadim was when we were in in Japan in October of 1987 and he was We were a few people Dan was there also the two sashes and And Vadim took a great responsibility there I would say he gave several talks and he was really on top of everything Then just I guess it was a month later. I got this letter from from Sasha Telling me about what had happened which was Extremely sad But I'm really glad that we can honor him here by this time I I have to tell you now that what I'm going to talk about is Essentially what I talked about in Paris almost a year ago. I have been completely out of business For good reasons as you know for the last six months or so and I've not been able to do and this is That this is not, you know Leading to any great conclusion of what I'm gonna say, but it's it's something that I've been Thinking about I mean the the origin of what I'm gonna say goes back actually to what I was talking about, you know in 1984 in this land of Norita meeting So but it's It's coming back to you know all these beauties that we see in an equals four and in and we say it's super gravity and I Think my message will be that there is more to it than Where we have seen before So, let me see if I can get some I wanted to Do it a bit interactively so I will have some formula here and then I will have my extra blackboard up there So I'm going to do something very simple Namely to start with it. Just think about young mills So I will write So you I start with that and then I will um essentially just Consider the light con gauge, which is that A plus well, let me put a gauge in X is equal to zero and I'm going to Do use a notation where I'll write X plus minus and I will use X plus as the time and then del minus is the conjugate That will mean also that Del plus it's just a space derivative Okay, and also I will I will also use the notation that X is So if I introduce this here and then In the functional interior I can add and subtract and then what I will find is that I will get that so del plus a minus a will be something like D Well, actually I can now what I'll do is that I will now switch on the next slide. So you will see So I will have This expression that is di a And then there is a term which is g f a b c One of a del plus a b I del plus a c I Well, actually I will eventually also go over to an expression have a and a bar and the Here are one over del plus Which is of course a non locality. So this is a non locality along the icon, but it's It's nothing really dangerous about that. So I will have lots of expressions with like that So if I just do this and I introduce it back into the Into the action I get an expression like the one upstairs and So you recognize that It's an ordinary kinetic term, but you know, we have to interpret as I said del plus is the time This is a three-point coupling and complex connect and a four-point coupling like that so But I you know, this is really a Formalism which is working on shell. So what I will do is that I will Really look at equations of motion and See if you look at this equation of motion, so Minus I del minus This is really like P minus well one on a P minus on a So this but of course this thing This is really the The momentum operator acting on that which means that this is one of the generators of the Poincare Algebra and we will interpret this this really as a right here as the variation with respect to this generator of the field and If I if I write that I wonder why there's this Reflection down here. You will see an expression like that. So so this operator this generator Is then non-linearly realized on the field, so this is the kinetic term up there and so I have a Which the three-point coupling gives this term and I write complex conjugate there It's not the complex conjugate of that, but if you go back One step You see that here here is the term which has a a bar a and here is another one Which has a bar a a and when I take the the equation of motion with respect to a I very respect a bar so This term is is we know it which means that I'm not gonna bother about this term. I know I know what it is Once I know this term, I know that term not directly. I have to go via the Lagrangian, but So now I can go back and you can find all the Poincare Generators acting on the field and you get a completely sort of group theoretical approach And every generator that has a minus component will be dynamical and hence non-linear So it's a non-linear realization of the Poincare algebra. For example, the this this is the spin Operator it has this term minus one. This is just telling you that it's a spin one field So that's the only Only thing you see from the spin part when you when you act here in this way and So what we did At this time 30 years ago was that we somehow abstracted out from this formalism How to write so the any higher spin? You just find these representations you find this generator and of course we also have to look at the other generators and the So I will write there is these are the dynamical ones or what I call this And this so I've complexified this also this is into the transverse direction, but it's a minus direction This you see it's it will have x plus del minus minus x minus del plus but we can always Put x plus equal to zero we do the algebra to x plus equal to zero So then this is trivial. So we only have to worry about this guy this guy and this What I call P minus So to find any field theory in a sense, you know massless field theory to start with you only have to find these generators and Make sure that they transform correctly with the the the kinetic generators and In this way you can you know, it's just algebraically you you construct any of the known field theories and I Even I think I gave a talk when a Poincare was 150 years old I gave a talk called, you know field theory as representations of the Poincare algebra So so these are all just representations of its algebra so it looks, you know completely group theoretic and So this is a different way of looking at field theory And this is what I thought all the time that This was say completely gauge fixed version and you can imagine that on on another planet, you know somewhere in the Milky Way Where you have some creatures that this is might have been the way that they constructed field theory They might not have worried about gauge invariance and that they have just because what we are dealing with here I just the physical physical degrees of freedom. So we have an a in a bar with two degrees of freedom and Of course This was the starting point for when we did n equals 4 and by doing n equals 4 like that We could prove that it was finite in perturbation theory. This is what I discussed, you know in 84 and in in Moscow And But then you can ask what happens when you go to gravity. What does gravity look like in this way? And you see Again What you will do is that we just have to find you start with finding this three-point coupling And in fact, I will show it to you Up there. It's a very Simple expression the three-point coupling. You see it's really binomial coefficients in a sense here And that's why it's very easy to to generalize this to a higher spins because you You know the number of derivatives here just tells you what the illicit is essentially So this is the three-point coupling then of course, there's a four-point coupling which is much more complicated But you can get a lot of information from the three-point coupling and and so this is what I'm going to Look at I Will now make things a bit more complicated. I will introduce some much more general formalism Where I write this expression essential like that I introduce this operator. It's an exponential in this del bar over del plus With factor a so what I'm doing here is that I just expand this to power a square take the derivatives and Then you also wrote me more general expressions here for the del pluses, but Don't look too much into that Then what you'll find then is that? This is a very compact way of doing things because we can do it for any spin at the same time And it's also a very good way of doing it because all these Kinetic generators which we have which gives you know tells you about the illicit or whatever They are automatically satisfied. So you only have to find once you write this expression like that You have to write these two generators in a certain form and then you you you just check the algebra that it satisfies the Poincare algebra Okay There are no index on H. Sorry Yeah, so this is sort of an in a sense an index free formulation and Okay, so And when you do it like this it looks very unique you can so you can in principle construct ordinary gravity this way It's hard to to to guess if you would like if you had to guess this four-point coupling here It's hard and it's a lot of calculations to do to check that it works But this is a way of constructing it and it looks unique as I said, but The question you can ask is that since we have a dimensionful coupling constant here We can also allow for terms with more derivatives So what happens if we look at the corresponding terms with four derivatives? And then Of course you can see immediately that what you have to do is that you have to add in one Dell bar and one Dell in order to have the right illicit in so the form of the expression is very natural Now I in What I do is that this was this exponential at all. I'm now introduced both and an exponent here Coefficient with four Dell bar and one for Dell So I write this expression now you see a three means that there are three Dell bars and one Dell So this is the Expression we start and then Since we have done this many times we know this is the expression for the dynamical part so to speak in these two generators They look at this one. So there are two coefficients here, which are in principle unknown But this is the form they have to be in order to satisfy all the other generators So the the only generators we have to check is how these three Commute with each other So we do that and Then We got a big a bit of a surprise I would say because we got solutions for all values of N and M So we got infinitely many terms possible here For This so this was sort of against what I was saying before that if this was a fully fixed gauge theory Fully fixed gauge. Yeah, and that that the algebra should give you everything Somehow it doesn't because somehow why We had thought you know naively that we would get Two terms here because in principle, you know in gravity that that should be two counter terms at the one loop level You know, this is this famous calculation of Tufte Weltman You essentially have an R squared and you have an arm you knew arm you knew You don't have the arm you knew a row square because of the Gauss-Bonnier terms. You should that get two terms So we were very mystified by this this Check this calculation hundred times, you know, whether we had done any mistake, but no, we couldn't find them And in fact Another virtue of this formula is that this generalized to to any Then a higher order counter term You know with the only thing we have to do is that we have to add more derivatives But it's the same calculation because you know it we do the calculations regardless of what is what it says here So to speak so we can do all counter terms all higher order counter terms in the same calculation and We only have to have to vary this m and n according to these rules, so this is the the spin of the Or the loop border I should say so if you take a 137 loop You just have to put L equals to 137 and this is the type of term you have there Okay, and you will have you know a fantastic number of derivatives, but it's still the algebra is working So That was a problem, so we have to go back to the drawing board again to see What are we missing I? Mean I've I've been telling you that you could do field theory by just doing group theory and something is missing here So let's go back. Well. No, I have to tell you one more thing Because we also know that These terms are zero on shell so How is that implemented here? Well, it's implemented in them Well it this looks a Bit difficult, but remember that the equations to motion for this field is this in my way of writing it so Considered if I look at this box acting on on an expression like that Let me do it over here. So Essentially what I'm doing is that I have a box on something with an H and an H so this is Del plus del minus minus del del bar Acting on but and I'm going to use the equations to motion because Of course, I when I use the equations to motion I get higher order terms, but I'm not worrying about those terms so When so I have to do like this rule here But I would every time that this guy acts here. I will get the corresponding term there So that will not give it so the only type of expressions that we have to worry about is this term like that But for that we use the equations of motion. So we said that that is del del bar or del plus and then there are terms whether it's a del H and Del bar H Like that and If you put all these terms together You'll see that it's again So I do it here a more general tab put it It's again of the same form as I have just have to change the del pluses and Which derivative I take so to speak so if I put this back I would say that If I have any this is the the starting in Equations motion here. I have a typical counter term if it's of this form I can just use this formula To rewrite this guy as long with the box up there And Then I make a field read definition like this and then we absorb it So by a field read definition, we can absorb all these counter terms that I've shown which of course we we should Because I mean among those that should be the correct ones and they are zero on shell. So everything is zero on shell So Then you might ask how or not would you ever be able to get the counter term which is It's non-zero on shell because you know at the at the next level at the two loop. There should be a term which goes like Essentially cubed so to speak your Well, if you go back and look at this expression here here you see it In order for the a term like that to be It So you could be you would be able to write like that you have to have both of the derivatives suppose I don't have This derivative say then I cannot do it because I cannot have it to do q-1 So the only way to have a counter term is that if it's only written in terms of one type of derivative I'm sorry Okay, sorry, I got I blew it I Missed that slide somewhere. I don't know where it's gone. So the point is that I can have a term a counter term which If it's of the following form I can have a term which goes essentially like six of those Wait a minute. No, I can have a term like that that will have illicit it to and that has There it is only one term one type So such a term I can write down I Can check it and when I do that then in fact, you know, we find a unique result there So we can find the tube the two-point coupling term we can find and that's unique and it's zero It's non-zero on shell, but let's go back to this problem that we had because that's So let Checking young bills. This is what I said we did with which was this gauge and we eliminated This fields through this equation so The question is then how much have we fixed the gauge? Well, so this is the gauge events that we start with of course in order to still be in that gauge We have to demand that this Should satisfy this condition We also have to demand, you know that any variation should Be invariant under well this should be invariant on the one that means that we have to To solve for this but this means, you know that there is sort of a remaining gauge invariance in the theory Even though we it's fully fixed and in young means we never use it. So so so the lesson is that There must be some some kind of of invariance in the system which Point career didn't know about so to speak and how would it how would we do that? Yeah Well before I say that there is one thing So if we still have some kind of gauge invariance here once you're able to write this in a gauge invariant way So if if you go back to this expression that I had up here One can check, you know this invariance But can we write it in a covariant form well in fact what we we can because what we can do is that we can introduce this Sort of covariant derivative which looks like this and it has to term like that and then you know since this the transformations we do Satisfy this condition it doesn't really matter if it's a del plus there or the one of a del plus So so this is certainly gauge covariant under this remaining gauge invariance and writing like that we find an expression for the For this P minus which was called the Hamilton as a quadratic form like that So this is in this way This is the whole expression for for young mills in this way of doing it, which is I Think very nice of course we have not been able to really use it. We've been trying but Still I You can also you know imagine that perhaps someone can now make a complicated field redefinition And it really it would really look like a few a free theory in a sense Then you would put every all the dynamics into the measuring the functional integral Doesn't really help but But it's it's good to know that for For young mills you have this expression And then if we check the gravity case it can also be written in this form So there must be some kind of invariance left there this And then what is that so one can go that what one can do is one can start With the covariant form and try to really check look at all the equations look at all the the fields that you that we have eliminated and try then to to find That events that's a very complicated way, but what can make a shortcut? But you because you can guess that part of the remaining gauge invariance what it would look like and It's rather natural that because of helicity this is essentially the only form it can have so I introduce a parameter here XI which has elicit the one And and Then this is the form you can write And I use this condition and it also introduced that So this XI is XI of X bar So this is I thought I changed it This is sort of one Well, I'm just saying that this is a remaining gauge invariance because it's not the whole but we only need to know enough, you know to to be able to to To select the right terms But because there's a problem with this symmetry which and it It does not close. So these are infinitesimal transformations that we can do Which the the theory should be invariant under if you if we try, you know to look at the closure of it We will get back to into some mess But in order to just check it's enough to know this It I shouldn't say this this is We don't know exactly what this means because since Since it's not a closing symmetry. We cannot really use it sort of as a geometry, but What happens is that it will really select the right type of terms for us So when we check, you know all these one loop countertimes we have It turns out that there's just one left and now you see So these were the expressions and it's only one combination of N and M namely this one That survives So now we're down to just one term, but we were supposed to have two and So that's could be a bit of a headache, but it's a simple solution to that I Said that it should be two. However You know, this is just at the three-point level If you go to the four-point level you would see two but at the three point those two expressions would give the same so But it means that we have it works that that's the way to do it and Let me see So now here I have this expression for the for the two loop it should be of this form Coverently and as I've said there is only one Type or term we can have someone which we are called b6 which means that it's exactly of this form So you see, you know, that's an H with an H bar and H bar and then and the Poincare Poincare algebra gives you this unique Solution and if you check this gauge emery, it works so So it's so somehow I mean we've understood what was the problem and and the The new lesson is that is you know, it's not enough just to close Poincare So Poincare didn't I mean Poincare didn't know about the counter terms. I mean into about so it's not enough You just use group theory in order to select say a General theory for spin to and higher so So what are then this other invariant terms? So what this gauge? This remaining gauge invariance. What does it do? It doesn't take away, you know, unphysical degrees of freedom But what it does is that it's selecting us selecting the spin to from higher spins so So that's what it has a different Role here it's it's assuring you that that we get just spin to Because the other ones are some kind of invaders from higher spin those terms that we wrote Could be terms coming from from higher spins And we have to somehow get rid of them and that's what we So so in that sense You might say that In a sense you can go back to to Poincare that Poincare would have sold this because the the idea is so close that the way we do it It's you cannot really you cannot check the immediate that you have spin to fields in those terms So this is the sort of the criteria for keeping that But it looks like you use this remaining gauge invariance. You have to use some kind of local invariance In order to do it in order to select that so that comes on top of all the the Poincare generators So When people are now interested in higher spins, we are we were really trying to get rid of the highest pins So this is the way of getting rid of the highest pins Okay, so But you know this remaining Immers looks like if you're a sore algebra. So at some stage I thought but it does not close We have not been able to really find a two-dimensional Local algebra that works It's not clear that it should exist. It would be nice, you know, and I Been discussing with Sasha some state that there might be some there is something in in also in this gravity theory with that we don't see by the eye which is there, you know and We we get it from from This local invariance But it's I haven't really presented it in the best possible form But as I said, I've been out of business for some time. So I've not been able to think about it but this the message in in the in the gravity case is that that there is something Doing it in this form, which is very much stripped down to the bare bones. You'd only use the Real degrees of freedom You you have to to do something else and it was this This was just a simple exercise though. We started just because we wanted to see what is the corresponding expressions in any crusade supergravity And then I'm not gonna be able to say so much about but In any crusade supergravity is Described by a field which no indices is just an X and you have eight status And if you expand it, you have all the supermultiple in it So it's a beautiful way, you know, you cannot you just have to to write the whole expression in terms of this five This five sets for sort of a chirality constraint and what we would call an inside out so you see that if That is sort of symmetric You take the complex conegate and you turn it around and you get it back. You have H bother your age You have cyber there psi So it it sets a rather Strict constraints and this chirality constraint for example shows Will show immediately that We will not be able To take this counterterm over to supergravity because then it would mean That it would on this side We will have a delta phi and here we will have Phi But this is satisfying a chirality condition that this has the opposite chirality So it's not gonna work So so of course this people know from the beginning of supergravity that there are no two loop counter terms in supergravity Dole also works for any cruise one, but here it's obvious. So you will see that it's much more Constrained here I was also saying that In gravity and in in young males you have this nice way of writing the Hamiltonian this p-minus as a quadratic form in in any cruise for and any Young males and a goose eight it works the same and there it sort of has a meaning If you look at the supersymmetry algebra Then I will write it in the following way I write one which I Divide up the spinner into one plus and one minus. This is what I call the kinematical one This is the one which just gives you a p-plus, which is the number This is a guy which gives you a p-minus, which is the Hamiltonian and then the mixed is like that So this expression is again, you know like so the square root of the Hamiltonian is like the supersymmetry So you can use this if you use this in the integral form One can rewrite this Hamiltonian in the following way again as a quadratic form You see this is an expression, which is just this supersymmetry this non-linear supersymmetry transmission on that And the it's complex conjugate and there is some kind of measure here So you can write the whole suit any crusade supergravity in this very compact form Which and You know also know what this guy is namely It is a part of the algebra even so it's a very constrained form You also know that it should be e7 invariant and In this level you need in in in this formulation e7 is not just sort of a gauge Duality symmetry it's a real symmetry of of the of the Hamiltonian It all the fields are transformed even the the fermions are transformed under e7 and So it's Very constrained form and then you might ask What would this remain engage in variance do in any crusade this is where I am you know, I'm just gonna Tell you that This field I'm just writing this to impress you because don't look at the details It has to be invariant under first of all there is this What is left from the repairment station? This is what is left from the from the local supersymmetry This is what's left from the from the gauge invariance and and then of course it has to be a chiral field It's this is a very complicated thing and the question is This algebra, you know, it looks Like any any crusade supervirus or if you really look at the details Unfortunately, as I said, it doesn't read it. It does not Close so it's it's not a good algebra. I don't know how to do it, but We were never able to do any crusade supervirus those in the old days But here we have you know, here we are building a two-dimensional theory into a four-dimensional one and we somehow have Parameters from the four-dimensional world which can help us do these things So this is Sort of group theoretically it's interesting if one can if we can Do something like that. I don't know if I can do it and and as I say I've not been able to look at this for the last eight months or so But so I will stop here, but and I will not give you any Nice results apart from that. This is my own Deliberations in a sense, but it really it tells me I think that there is this very Huge symmetry which is underlying these tears. It's the same You know with n equals 4 that any in these this form less N equals 8 is very much like the square of any n equals 4 the super field is the square of it And and you know that you get all these things that people have been trying to which you also seen in amplitudes You see here already at the level of the construction I Think there is a lot of things here which I have not understood which Perhaps will tell the something and I Unfortunately, I've not been able to do more than this. So let me stop there Question please Did you try to what if you apply this methods and ideas to what's in it? What would happen that you try to avoid? No, no, you can do it. Yeah, but but this is The this is Minkowski. I mean he's using and that is it here, but but But it's also interesting to look at the desires means because you What you really find is that you cannot When you once you go above spin to you really have to have all the the higher spins One can write very nice formula like the ones I had, you know for for these terms symmetry for for higher spins and I Don't know what to do with it, but it's Maybe it will close Um Well, you can but you you know It's very much streamed up for four dimensions, but I mean we have no what we have done we have in fact Done it in in also in 11 dimensions by what people used to call oxidations actually this is Actually, it's quite interesting, you know because somehow What is it that's Really telling you that the eniquus 8 that it's an 11 dimension 3. It's really the spinners because you know All the information is in the spinners and because when we When we did the eniquus 4 we started in ten dimensions and we just decomposed the spinners and we got four spinners and And and and the way to do it is to do you can take over essentially the expressions and Every time we have a derivative like that, which is it in two We have to add to it some terms here, you know with for example in well in in the young middle in the young midst case we write something like that and we have some They or did it 8 or something well I j's nothing So what one can construct in a new derivative so essentially what you do when you you go to higher dimensions is to add in the coordinates and write a new derivative and So then you can write also the 10 dimension case and the 11 dimension case in a very compact form I'm not sure that this quadratic form is working though. We have not been able to show that Yeah When you say the algebra does not close what does it give it give some things that could have some meaning or it's just really Strongly non-zero and you serve what they did when you say the algebra does not close with the well, you know Of course did I was in a sense I was not hiding it, but usually when you have conformal invariance you would make you say that you you have Delta X is and Delta X bar, you know, it's one with for us. It's the opposite that and of course that then it's clear that That's You're gonna mix up X with an X bar and then you're out in in the whole and then you're gone and I Don't at this level, you know, it works just like we're a sore it But once you get to try to close it, it doesn't which means that We don't know anything about C numbers as we in this algebra. I'm just using this as an infinitesimal Transformation which the theory should be invariant under And then it it does it works in the way that it gives us the right terms, but I cannot use it as Some underlying geometry, which I would have liked to do But but somehow I feel there is one but I haven't been able to see it That's right. The algebra does not close itself. I don't see how it can be a symmetry Usually what you have is a algebra close But when you write the algebra on the field only close up to the equations of question and then it's open But you know, can you say that when the algebra itself does not do? Well, it no the algebra closes in but but it doesn't what I'm saying is that Then you have to introduce new transformations, you know It's the algebra will eventually close but it will close into more as transformations than the one I'm writing here But I don't need them, you know in order to just check this but in order to understand that the full symmetry I Have to do that. So so what I would what I would have liked to do is to find some Sub-sub-algebra, which was closing of the whole remaining, but this I've not been able to find Well, yeah Okay, I've been able to to really get anything interesting out of that Okay