All right, wonderful. All right, good morning, everyone. So I thought about what I should do, given as I said that I was going somewhat more slowly than I wanted to, but I decided that it would not be good to leave you dangling, having you sort of promise some cool things. So I want to explain, and in any case, this will again be relevant, I hope in the end for you to appreciate even more deeply what's special about the Higgs. I want to explain, at least in the beginning of this lecture today, from a deeper point of view, what's why we get such incredible rigid structure associated with possible interactions of massless elementary particles. So I alluded to it a few times that the second we say that we have massless particles, then all of a sudden there's an humongous constraint on our imaginations as theoretical physicists, and there's only a tiny menu of possible spins and kinds of interactions that we're allowed to have, that any consistent theory of the world would have to have particles of spin zero, one half, one, three halves and two, that the spin two would be unique and would actually have to correspond to general relativity at long enough distances, that the spin one would have to have the Yang-Mill structure, and that the spin three halves would have to go along with the supersymmetry and supergravity. So the fact that, and almost all of these we've seen in nature now, we of course knew about spin a half and spin one and spin two for a long time, and what's very special is that we finally discovered our first example of an elementary particle with spin zero, and that's the most zero-authoritative thing about what makes the Higgs special and we'll again talk more about why that's so perplexing from many points of view. But I want to explain where this comes from in the most direct possible way, where the sort of basic fact of the architecture of the way the world works comes from the most basic possible way. And as I said already in the first lecture, this is something that just given the principles of relativity and quantum mechanics, basically from pure thought, theorists can figure out this amazing fact about the world. And then the only choices you have is a choice for what precise group representation and matter content you might have in the Ethereum, whether you do or don't have supersymmetry, but that's it. There aren't a million other ways the world could be and we've just found one way to describe it using our formalism that we learn in field theory books. This is completely rigid. There's no other way it could have been. And so I want to explain this along the way we'll understand that the, let me just give you the structure of what we're going to talk about. We're first going to talk about, as I already reviewed last time, in the standard way of thinking about things, let's say we're talking about particles of spin one, we really imagine that there's a Lorentz tensor for photon scattering, what you can call the Feynman amplitude, that we then dot into polarization vectors in order to get something that transforms, you see the actual amplitude just depends on holicities and momenta, and it should transform such that M of lambda P and H goes like the little group rescaling e to the i theta M of P and H. So the amplitude is something that transforms on this side under the Lorentz group, but on this side, bad notation on the Lorentz group, on this side under the little group, they're both LGs, but okay. And so that's really what the transformation property of amplitudes is. This object just transforms like a Lorentz tensor. And so we try to invent these polarization vectors to be something that transforms on the mu index by the Lorentz group and somehow on the holicity index by the little group. But as we said, no such objects exist. These polarization vectors don't exist uniquely, okay? Polar vectors epsilon mu H don't exist, and only these gauge redundant equivalence classes exist. Epsilon mu goes to epsilon mu plus alpha P mu, okay? And therefore, every time we describe physics with polarization vectors and momenta, we are giving a redundant description. Now it's fine, it's perfectly okay, we'll get correct answers from it, but we're dragging around things in this description of the physics that aren't there in the actual answer, okay? So what we really want is something that directly transforms like that, okay? So the first thing we're going to do is introduce variables. So amplitudes are not, they're not functions of polarization vectors and momenta, okay? Because the polarization vectors don't actually exist. They're only redundantly functions of polarization vectors and momenta. So we want to find what variables the amplitudes actually depend on. So step one is going to be to introduce the so-called spinor-holicity variables. And if you've seen this, any part of this story before, this is often used as a trick, well, say what this means in a second, but it's introduced as a trick for representing null momenta for massless particles in a nicer way, but it's not a trick. It's really that we're going to introduce some new variables which do precisely have the property that they transform under Lorentz and little group in the correct way. Okay, so amplitudes will directly be functions of these new variables, these spinor-holicity variables, not momenta and polarization vectors. So this is a good first start because we now actually see what amplitudes are really functions of rather than giving a redundant representation for it. So after we learn that, the second step is to look at the very simplest possible amplitudes you can think of which are three particle amplitudes. And three particle amplitudes are something that you normally don't sort of talk about in the courses so much, but they are the most fundamental amplitudes that we can have, as I'll describe, that there's a reason we don't, well, we'll get there. And an amazing fact is that if you have massless particles, if you specify the holicities of the massless particles and that's it, this amplitude is fully fixed up to the strength, up to overall strength. So up to the sort of overall coefficient out in front, it's fully fixed by symmetries. So it's fully fixed just if you specify the holicities, the form is totally fixed. And then we'll talk about what that totally fixed form is. And it doesn't matter if the spins are zero, one-half, 33 million, okay? There's an amplitude, there's a consistent three particle amplitude for any possible spin. This is already fascinating. And we'll expose all kinds of hidden relationships between theories that we know and love, okay? For instance, we'll see that if we look at the three particle amplitude for gravitons and the three particle amplitude for gluons, the three particle amplitude for gravitons is literally the square of the three particle amplitude for gluons. That's an amazing connection between two things that seem to have nothing to do with each other. If you look at the Lagrangian for general activity and you expand it out around flat space, the three particle amplitude looks like a horror with like a hundred terms, looks absolutely nothing whatsoever like the square of the gluon amplitude. And yet all of that complexity is in our head. All of that complexity is a consequence of the Lagrangian way of writing things. And actually, there not only are these things all extremely simple and we'll write them in one line, but there's even hidden relationships between them that become exposed in this way of thinking about it. But the three particle amplitudes are essentially completely determined by kinematics and are there for any particle. Already at this point, we will see that one of the things that I mentioned, which is very special, which is that if you have a single particle, the single kind of particle, A, A, A, that this three particle amplitude can only exist for spin zero, which is the case of the Higgs, or for spin two. And for spin two, it's some very high dimension operator involving gravitons that we'll never see. But that's one of the extremely interesting things given the importance of the Higgs self interaction and the three point coupling. It's to see that in fact, there's the only elementary particle that can enjoy this property of interacting with itself in this way is actually the Higgs. So we'll come back to that point in the end. But that's a parenthetical comment. Then after all of this, after we see that the three particle amplitudes are fully fixed by symmetries, then the dynamics comes in, the physics comes in, when we start trying to write down four particle amplitudes. So when we start trying to write down a four particle sample, dude, okay, there's a question there. What's important here is that I am demanding that all of these three quantum numbers are the same. So we'll see that for example, gluons, the usual interaction for gluons looks like minus, minus plus for the holicities. And not only that, it has colors, A, B and C, and these colors have to be different from each other. It's actually given by, there's a structure constant there. So the interaction vanishes. First, it changes the holicities of the particles. And secondly, we even have to change the color labels. So if you want to have something where it's literally the same particle interacting with itself, then these are the only possibilities. And we'll see why that, why that, okay. But now coming back to the physics, we're gonna look at a four particle amplitude. And the four particle amplitude is gonna have to satisfy a consistency condition coming from locality and uniterity. Okay, and locality tells us that the only poles for the four particle amplitude, if we imagine working in some tree approximation, the only poles of the four particle amplitude will take place when S, T and U goes to zero, where these are the usual Mandel-Siemen variants.  So the only poles happen when S, T and U has zero and uniterity tells us that near those poles, let's say as S goes to zero, we have to be able to interpret this as one over S. So the residue of the pole has to be one over S times the sum of the exchange of certain particles here. So we have some particles, maybe they have some label A and some spins. And if the helicity here is plus and the helicity on the other side has to be minus, but we have to be able to interpret the four particle amplitude as arising from the residue has to be interpreted as the product of three particle amplitudes for production and decay roughly of the intermediate state, okay. This is going to turn out to be a huge constraint, an enormous constraint. It's this constraint, these two constraints that are going to kill almost all the possibilities and leave us with the extremely restricted menu of possibilities that we're talking about before and we'll discover from this rule why gravity has to have universal interactions, why we have to have the Yang-Mills structure, why we have to have supersymmetry when we have spin three halves and so on and so forth. It's all going to come from a single algebraic condition which is demanding that any function that we write down here has poles in the correct spot and this property of uniterity is sometimes called factorization for obvious reasons, that's a four particle amplitude factorized properly, okay. So that's our plan, I want to go through these points and I won't go through every aspect of the last thing. I'll do a few examples just so you can see how it works. You can then work out the rest for yourself or unfortunately I don't remember the archive number off the top of my head but this is all described in a section of a paper I wrote a number of years ago. I mean, it's summarizing earlier work as well and in this paper, all of this stuff is extended to particles with general mass and spin, not just massless particles, but the story is most interesting in the case of or most constrained in the case of massless particles as I've emphasized. There's a paper that I wrote the title scattering amplitudes for all masses and spins back in 2017 with Yutin Huang and Jimmy Huang. So if you look at, I think it's a section four of this paper, all of this stuff that I'm talking about is actually described in more detail. Also in this paper, you will learn how you can talk about the sort of production and decay of any massive particle of any spin with any properties. If you're interested in doing understanding the production of possible mass three resonances of some blue balls or some crazy stuff at a collider you don't need to open anyone's book or learn any Lagrangian or rely on anyone. You can do the calculation in a few lines yourself using the kind of formalism that I'm talking about here as well as what's described in more detail in this paper. So this is actually potentially a relevance for just thinking about collider physics from the bottom up if you want to play with it. All right, so let's begin with the plan. So part one is to explain the correct kinematical variables. So the correct kinematical variables and remember that when we're describing our particles we tacitly think in terms of giving them a momentum but that it satisfies an on-shell condition. So let's say M is zero, then we have to say that P is equal to what's the data? I mean, there's only three degrees of freedom that specifies the momentum, right? If I give you the real, the spatial momentum the energy is fixed.  Because I have to say that the time component is constrained in this way. So step one is to express the momentum in such a way that there's no constraints on it at all. And so we're gonna do this by a usual trick that perhaps many of you have seen where you take a four momentum and you dot it into the Pauli matrices where sigma zero is defined to be one and sigma i is sigma i or just the usual Pauli i matrices.   So when, so look at this matrix P, what do you notice about this matrix P? It's the most general, it's the most general Hermitian two by two matrix, okay? And notice that any Hermitian two by two matrix would look like some A, B, C, C star. And so I can get every A, B, C, C star in this way, okay? So on the one hand, it's the most general Hermitian two by two matrix. Now, notice also another very cool thing. What is the determinants of P?   Okay, so now what happens, what happens then if I look at this matrix P, suppose I take P and I do an arbitrary similarity transformation on, okay? So let's say I take P goes to L dagger P L and here L is just any two by two matrix. Actually, I'm gonna ask that it has determinant one, okay? So technically we'll say that L is an SL two C and every time you see this L, it means that the determinant is one, but other than that, it's a general linear transformation, okay? So it's a completely general two by two linear transformation, okay? Well, clearly if I look at L dagger P L is also Hermitian. So note, I'm not demanding that L is unitary, okay? So L is a totally random two by two linear transformation just with unit determinant. So therefore, this is some P prime, which is also Hermitian. And therefore, since it's Hermitian, I can write this as some P prime mu, sigma mu. So if I say this again, so I have P is sigma mu P mu and L dagger P L is some sigma prime mu, some sigma mu P prime mu, but notice that the determinant of, if this guy I call it P prime, the determinant of P prime is equal to the determinant of L dagger determinant of P, determinant of L. And since I assume that these are one, determinant of P prime is equal to determinant of P. And so I learned something cool. I learned that P prime squared, which is the determinant of P prime is equal to P squared, which was the original momentum that I started with. So that means that P prime, whatever it is, is a Lorentz transformation on P. And in this way, we discover the most fundamental spin-a-half representation of the Lorentz group. You may have seen this kind of trick before just for the rotational group. That's also the nicest way of discovering this spin-a-half representation just for ordinary rotations. And that follows from this argument if you just drop the P zero part, and then you notice that the matrix is the most general traceless formation matrix. And then you have to restrict yourself to unitary transformations that preserve that traceless property, okay? But the most general guy is the one that I said, all right? So okay, so that's how we discover that the Lorentz group is the same as SL2C. Now, let's say that we are talking about the complexified Lorentz group. Well, in other words, here I mean that those momentum, P mu could be complex, could be complex numbers. Well, I would still define this P as in just the way I did before P zero plus P three, P one minus I P two. Now these things can be complex. So these I's don't matter as much. I can just reabsorb them into redefining what I mean by the P's. But anyway, let me keep writing it this way. And so now if I took P goes to just some completely two totally general matrices and L and an R. So I don't have to make this one the dagger of the other. Then this would also be a complex two by two matrix. And so long as L and R both have determinant equal to one, then exactly the same argument would tell me that P prime is some Lorentz transformation on P. Now a complex Lorentz transformation. So what we've learned is that the real Lorentz group is SL2C and it acts like P goes to L dagger PL. The complex Lorentz group is just two copies of SL2C and it acts like P goes like totally general L, P, R. And there's a final cool point to discuss. What if I just looked at a general two by two real matrix? So I'm gonna write this, let me write this a general real matrix. And I'll just keep writing it this way.  Of course, I don't have to write it this way. It's a general matrix. I could just write this as A, B, C, D as well. Okay, totally general two by two real matrix. I'm writing it this way only because now if we look at the determinant of P, now it looks like P0 squared minus P3 squared minus P1 squared plus P2 squared. So you'll notice that this is two comma two signature. Okay, so if I think of this as P squared, then it's a signature with two pluses and two minuses. All right, so if we're in two comma two signature rather than three comma one signature, then we could also treat the Lorentz group in this way. But then what we would have is in two two signature, P would be real and we would again have P goes to L, P, R except now P is real and L and R are real two by two matrices. So these are sort of three closely related ways of the closely related aspects of the Lorentz group. The real Lorentz group is one copy of SL2C. The complex Lorentz group is two copies of SL2C. The Lorentz group in two two signature is two copies of SL2R. All right, good. So now let's go back to our program. So let's say we're trying to represent the momenta of the particles. So now I'm gonna talk about this P mu sigma mu and often this is written as P alpha, alpha dot, okay? This is just to emphasize that we have a two by two matrix. And if you've never seen this notation before, do not be scared, it's just literally a two by two matrix. So this would be alpha equals one, alpha equals two, no alpha dot there, sorry. So this is just telling you how to read the entries of this matrix. So alpha equals two and then we'd have alpha dot equals one and alpha dot equals two.  So P one, one dot would be P zero plus P three and so on, okay? So that's all we mean by this notation. And so we see that under the Lorentz group, P alpha, alpha dot goes like something on the left, alpha, beta and something on the right, alpha dot, beta dot on P beta, beta dot. And depending on whether we're Lorentz or two-two signature are complex, either if we're Lorentz then R has to be the complex conjugate of L. If it's complex, they're independent and complex. If it's two-two signature, they're independent and real, but we can sort of talk about them all in this uniform language. Okay, okay, so now let's say we want to impose that P squared equals zero. So now we wanna talk about massless particles. And so we wanna impose that P squared is equal to zero. Okay, then what does that mean? What does that mean? That means that the determinant of this matrix, P alpha, alpha dot is equal to zero. And so what does that mean? That means that the matrix has to have rank one. It's either zero or it has to have rank one, okay? And so if it has rank one, it means that P alpha, alpha dot, I have to be able to write as the outer product of two-two dimensional vectors. So I have to be able to write it as lambda alpha times lambda tilde alpha dot, okay? So this is what follows from masslessness is that I can write my two-by-two momentum matrix as the outer product of two-two dimensional vector. For example, suppose that the P mu was equal to E E zero zero or sorry, this is the z direction. So let's say P was equal to E zero zero E, then P mu, then my matrix P alpha, alpha dot would just be two E zero zero zero. And my lambda alpha would just be square root of two E zero and lambda tilde alpha dot would be square root of two E zero, okay? That's a choice I could make. So then clearly if I take the lambda lambda tilde, I get this matrix, okay? So you see, this looks very beautiful because now instead of giving you some constrained four vector, if I just freely give, freely give lambda and lambda tilde, from this, I construct a P alpha alpha dot, lambda lambda tilde, which is null. So this seems wonderful that now with no constraints at all, I give you freely two independent lambda and lambda tilde and from them I build something which is null. Now notice that if I'm in the real Lorentz group to real momenta, then since P alpha alpha dot has to be Hermitian, then in this case, lambda tilde alpha dot has to equal literally the complex conjugate of lambda, okay? That's the only way this matrix would be Hermitian. This is to make this matrix Hermitian. For complex Lorentz, lambda and lambda tilde are independent complex two vectors and in two-two signature, lambda, lambda tilde are independent real two vectors. Now I keep saying all of these things because we're going to want in a moment to imagine all of these amplitudes that we're writing down as in fact, analytic functions of the momenta. And so we're going to want to think about them in general as the functions of complex momentum. That's going to be a very important zero-author part of this story. That's why I keep emphasizing these different things. So for complex momenta, lambda and lambda tilde are independent. For if you want to think about them as independent and real, then you're really doing it in two-two signature. And there's something special about three-one signature that it forces one of them to be the complex conjugate of the other one. So they're complex, but they're forced to be complex conjugate of each other. Okay, now comes the crucial point, which is that, remember, we know that a normal momentum has three degrees of freedom, but it looks like I have written it, if I give you lambda and lambda tilde, it looks like this is a free two vector. This is a free two vector. So it looks like I still have four degrees of freedom here. So what's going on? And what's going on is really beautiful. What's going on is that given a P, if you give me a P, I do not uniquely associate with it a lambda. Lambda, lambda tilde are not uniquely defined. And you can see why, because the formula involves a product. And so I can always rescale lambda goes to T lambda for any T and lambda tilde goes to T inverse lambda tilde. And this leaves P invariant. If I'm in three one signature, so if I'm in three one signature real momentum, lambda tilde has to equal the complex conjugate of lambda, as we said, and therefore T inverse has to equal T star. So that means that T is a phase. So what have we learned? We've learned that P alpha, alpha dot does not give me uniquely a lambda alpha and lambda alpha dot, because whatever it is, in three one signature, I can rotate this lambda by a phase and P is left invariant. Now, what does that remind you of? That's exactly the little group. That's precisely the little group, okay? And in fact, you can follow an argument, essentially completely paralleling the arguments about Wigner that we talked about before. You can say, okay, I'm gonna choose a reference K, alpha, alpha dot, I'll write the reference K alpha, alpha dot as some lambda K, alpha, lambda K, alpha dot, okay? So this is now, I'm locking, I'm making a choice here, just like we're doing before, I'm making a choice. Okay, so I'm gonna make a choice to write this in a particular way. I'm going to define the lambda for a general momentum P to be some specific Lorentz transformation that depends on P for the lambda of K. Similarly, for lambda tilde alpha dot P is gonna equal the thing on the right, alpha dot beta dot of P, lambda tilde beta dot of K. Okay, so this is a particular way of defining what I mean by lambda and lambda tilde for all momenta, starting from a reference momentum. And then after you do all of that, you discover that the lambda that you associate with some Lorentz transformation on P is not equal to the alpha beta associated with that Lorentz transformation on lambda beta of P, which is what you might have thought, but is equal to that up to this little group transformation, exactly the little group transformation, okay? So that's what's beautiful about these variables. That's what's conceptually important about these variables. The lambda alpha variables are exactly things that transform under Lorentz and little groups correctly. So we've discovered not with polarization vectors, we can't do with polarization vectors, but any object that we build out of these lambda's and lambda tilde's is guaranteed to be something that transforms like an amplitude transforms under Lorentz transformations and the little group. All right, let's see, there's a few questions. Let me see if I can look at them. Oh, let's see, are there any... Model comments, yes. Okay, there's some comments. Okay, very good. All right, okay, so maybe I'll do one more thing and then we'll take a little break. Is that okay? Okay, but actually maybe I can actually, are there any questions? All right, so let me, okay, so this means that if I want to talk about an amplitude for, Fabio, can you just give me five minutes and then we'll come to your question. All right, so that if I want to talk about an amplitude, what is an amplitude now sort of from this point of view, it's going to be labeled by the lambdas and the lambda tilde's for my eighth particle. Remember, A is going to run from one through N, if I have N particle scattering, and the helicities, but it has to be an object that satisfies that if I take lambda A and I rescale it by TA and lambda tilde by TA inverse, lambda tilde, that this needs to pick up the little group phase associated with the particle A. So if I do this on all the T's, there'd be two A, TA to the power of negative two H times M of lambda, lambda tilde and the helicities, okay? And this negative sign is purely conventional. The two is there because remember, these lambdas are the spin of half representation. So when I say lambda goes to E to the I theta lambda, this is for spin one half or helicity one half. And therefore, so this is what I'm calling T. So T corresponds to helicity minus one half. And therefore, in general, this would be T to the negative two for general helicity. All right, so this is the, so we've now seen that amplitudes are actually functions of these spin or helicity variables. All right? Not polarization vectors and momentum, but they're directly functions of these spin or helicity variables. And maybe I'll just say one more thing before we stop. What are the Lorentz invariance that I can build out of lambda or lambda tilde? So remember that since L, lambda is gonna just transform according to one SL2C, lambda tilde is according to transform according to another SL2C. So the only invariant tensors that we have are the epsilon symbol, okay? So epsilon alpha beta and epsilon alpha dot beta dot. Therefore, the only invariance that I can have are to contract things with the epsilon symbol. That's it. So for instance, if I have lambda one and lambda two alpha, I can define what you can call lambda one, lambda two with this angle, with this angle bracket. Sometimes this will be abbreviated just as one, two when there's no confusion for what we mean by it. But this is nothing other than epsilon alpha beta, lambda one, alpha, lambda two beta, okay? So if I write it again here, so one, two which would be the same as lambda one, lambda two is just contracting with an epsilon symbol, okay? So epsilon lambda one, alpha, lambda two beta. And I can also do the other kind of contraction which is a square bracket, lambda tilde one, lambda tilde two and this would be a contraction like so. And how are these related to familiar things that we know and love? Like let's say I have two momenta p one dot p two. Okay, what is p one dot p two? Well, okay, we can write it out explicitly the first time we're doing it. So it's the lambda one alpha, lambda tilde one alpha dot and then lambda two beta, lambda two beta dot tilde. And of course the only way the adopt product can work well, we just have to contract with the epsilon alpha beta epsilon alpha dot beta dot. And so we see that p one dot p two is actually equal to the angle bracket one two times the square bracket one two. And that makes sense because the momenta remember are invariant under the little group action and therefore any expression that's only a function of the momenta has to have no weight when you rescale lambda and lambda tilde. So if we rescale the lambdas then one two would pick up a weight but the angle brackets would pick up a weight but these guys would pick up the opposite weight. So this would transform like t one, t two times one two but this would transform like t one inverse t two inverse times one two. And so the factors of t one and t two cancel. Okay, so these would just go back to itself, okay? So anything that's built out of momenta alone is going to be invariant under this action of the little group, but the amplitude itself is not invariant, the amplitude itself has to pick up some of these phase factors to reflect the little group property. So the amplitude is not invariant. Let me actually give you a quick example just so you see what some amplitude looks like. So here's a famous amplitude for gluons. I won't be too careful about the, I won't be too careful about the, telling you exactly what it is just that it's a bunch of spin one particles but I want to emphasize a point here, okay? So let's say a four particle amplitude and this is called the Park-Taylor formula and it looks like one three to the fourth over one, two, two, three, three, four, four, one. It's very exceptional that it's so simple that it only involves angle brackets and so on. That's very special to this case, but that's not what I wanna illustrate. So first you see it's shockingly simple, okay? If you write this expression with polarization vectors it's some kind of complicated numerator, lots of polarization vectors and you have no idea what the answer actually looks like. In fact, most of you who've done courses and you've seen results written in terms of polarization vectors, you've never actually seen the amplitude. What you've seen is some redundant representation of the amplitude and you've never seen the amplitude. Maybe you've taken the mod squared to calculate a cross section and averaged over spins or over polarization but you rarely see an actual amplitude in the course because you can't, you only see this redundant representation of it, okay? So this is what an actual amplitude looks like. So if you've never seen it before this is your introduction to what an actual amplitude looks like and notice something very cool here. I didn't even have to tell you, well, so first let's check that it satisfies the rule. It should satisfy that it's lambda one goes to t one lambda one that this amplitude should pick up a factor of t one to the negative two times the holicity of the particle one which is claimed to be negative one, okay? So this should go like t one squared M, does it? Yes, it does. There are four ones upstairs and there's two ones downstairs, okay? So when I do this shift lambda one goes to t one lambda one I indeed pick up a factor of t one squared, okay?  Just by reading off the weight of the expression when you rescale lambda and lambda tilde that by itself tells you what the holicity information is on the left. Notice also that here's an expression that I'm not allowed to write down, okay? So let's say I write down plus two four to the fourth. This is a no-no. This makes, does not make any sense because these things have different weights, okay? So this has a weight that would correspond to this has a weight that would correspond to a holicity is one minus two plus three minus four plus. Now what would this one correspond to? That would be something that would have exactly the opposite would be one plus two minus three plus four minus, okay? So I can't add expressions that have different weights under the rescaling of the lambdas and the lambda tilde. This freedom to rescale lambdas and lambda tilde is the little group. And so the phase that you pick up or the power of T that you pick up when you do the rescaling is just locked and reflects the holicities of the particles, okay? So maybe that's all I'll say for now and we'll take a break and come back and start doing some dynamical thing. But yeah, let me pause now and ask if there are any questions. Okay, so there's a question by Maria and probably there was a question from Fabio. So we want to hear from Fabio first. Yeah, hi. Yes, my question was about the change in the, in the signature of the metric. I didn't understand why there was a change. Oh, no, I was just doing a few examples. So if you want to ignore it, you're free to ignore that. But it's useful to think, I mean, it's going to be important for us to think about the amplitude of the general function of general complex momentum. The reason for that, I mean, the deep reason for that, which we won't take, we won't talk about a lot, but the deep reason for that, which I briefly alluded to before, is that the fact that the amplitudes have, the way that causality is ultimately reflected in the scattering amplitudes is that they're appropriately analytic functions of the momentum. So that means that whether we like it or not, and of course you see it in books, even if you didn't know any fancy thing, you get functions of momentum that are like, they don't look like some crazy magnitude of P or step function of P. They're like simple functions, like powers of P, logarithm of P, and so on. They're functions that have analytic continuations for complex momentum. That's not an accident. It's not an accident that the amplitudes need to be, you need to be able to analytically continue them to complex momentum. And so it's useful from the get-go to sort of learn how to think about the momenta, even if they were complex. So that's why, even though we begin with the real Lorentz group, where this matrix P is Hermitian, it's useful to just immediately go to the complex case, okay? So to the complex case where this is just a general complex two-by-two matrix, okay? And the only reason I talked about two-two signature is that later it might be useful to visualize. So if this is the case, then when we write P is lambda, lambda tilde, then these are both complex, okay? These are both complex two-by-two, complex two vectors. It's gonna be useful for us in a moment to visualize them. So it'll be useful to think about the case where P alpha alpha dot is lambda alpha, lambda tilde alpha dot, where these are real two vectors, independent real two vectors. And the only reason to say that thing about the two-two signature is that when they're independent real two vectors, this corresponds to a situation where the momentum is actually in two-two signature, okay? Okay, so if I were to take the determinant of the matrix, I would find that, yeah, okay, okay. Yeah, that's right. And that's the exercise we did, right? So let me not write it even the way I did before. Let me do it just like this. So this is the matrix P, okay? Now A, B, C, D are just general, but let's say real. And so the determinant is A, D minus B, C, okay?  Yeah, all right. Pluses and two minus. Okay, thank you very much, yeah. Okay, all right, Maria, go ahead. Okay, hi Nima, I have two questions. First of them is that I understand that the goal of this derivation was to find a matrix element that under Lorentz transformations just picks up a phase from the leader. That's right, that's right. So at some point I got lost because as we were discussing just the day, the little group was SU3 for massive particles or a Euclidean group in two dimensions for massless particles, but at some point you started calling a little group transformations to this phase rescaling. Yes, yes, yeah, that was the whole business yesterday that for the general Euclidean group, the representations that we would have when the TX and TY were non-zero would force you to have an infinite number of particles, okay, and so we restricted our attention to those ones where the TX and TY were zero and that's how we got this dropped. That's how we went from the general Euclidean group down to the ones that leave just the, where the TX and TY are at the origin and those guys just are acted on by the phase. So that's the sort of drama is that there are these strange continuous spin representation. So the thing that you get out of the box is these funny continuous spin representation. We haven't seen them in nature and maybe they're just not consistent. So if we have massless particles, we have to further make the assumption that sort of T is at the origin. And then the only thing that acts on the state, then TX and TY just give you zero when they act on the state. And so the only information you have is the rotation by Z, which is the holicity. And in general, in D space-time dimensions, we have SOD minus one rotations for massive particles. But because of this strange, same strange funny phenomenon, we would have the Euclidean group in D minus two dimensions which would solve this continuous part, but we have to sort of go down to the origin and then we're left with rotations in D minus two dimensions. So the little group for massless particles is SOD minus two. So that's the sort of discontinuity in going from massive to a massless, but it has to do with our just, our throwing out of these strange, continuous spin representation. My second question was this amplitude that you wrote in the last page, it somehow reminds me some conformal invariant. Oh yes, we'll be seeing that in just a moment. In fact, if you know something about conformal field theory, you know famously in conformal field theory that three-point functions, three-point functions of conformal primaries are completely determined by conformal symmetric and up to the strength. And exactly, it's mathematically, I mean the words are different, but it's mathematically essentially exactly the same argument for why three-particle amplitudes for massless particles are completely determined by symmetries as well. Okay, so yeah, so if you just hang on, you'll be very familiar with the argument that we're going to see. And even the formula that we write down is virtually identical to the corresponding formula in CFTs. Okay, thank you. Max, Chu. Yeah, hi. So this representation for the amplitudes which you introduced here, I can see how it simplifies life enormously. If you deal with massless, spend one and spend two particles because you don't have to deal with just redundancies anymore. But if you talk about spend zero particles, it seems as if this doesn't bring us anything new or any simplification. Is this true or can we even learn something about spend zero particles through this mechanism? I think like purely from this, just from this rewriting of the variables, it's not especially, you're right, just the rewriting of the variables doesn't help. But in fact, the kind of point of view that's associated with this that we're only gonna care, we're not gonna talk about polarization vectors, we're not gonna care about the physical unshell states and so on, that point of view is universally useful. If you like what's going on, the whole business of the difficulty with polarization vectors when you have massless particles with spin is the sort of beginning of there, there are many sort of qualitative realizations early on. There's something funny. There's nothing wrong. I mean, it's perfectly correct, but there's something a little off about the usual sort of Lagrangian based textbook way of thinking about these things. For example, when you're learning things in field theory courses, you start with a scalar theory, and then that's very simple. Lagrangian's very simple. Then you go on to spin one, maybe. Then already it's a little bit more complicated. You have to have gauge invariance and the formulas look more complicated. And then maybe you'll talk about gravity. I mean, even perturbations around flat space and anyone who's stared at gravity and Lagrangian, it's a horror, right? You take G mu nu is A mu plus H mu nu, already the three point vertex is 60 terms. Someone told you calculate the two to two amplitude of gravitons in gravity at tree level, you'd say thanks a lot for the 10 week homework exercise. It looks like a horror. It's like a thousand terms, just a total mess. So in other words, in the standard way of thinking about things, what is the ordering of the simplicity of amplitude? Scalar, the simplest of all. Yang-Mills, a little bit more complicated and gravity, a horrible mess. The amazing fact is that the actual amplitude, the final answer, the actual amplitude, the simplicity both at the most sort of visceral level of just seeing what the expressions look like. And even at a more formal level of the sort of amount of computation for a computer that's needed to get them is ordered in exactly the opposite way. And it's just extraordinary that the gravity amplitude is the simplest of all. And then the Yang-Mills is still pretty simple and the five to the four amplitude is very complicated, okay? So when you sum over all the Feynman diagrams, there's no special cancellation, nothing, you get some kind of mess for a five cube, but humongous number of cancellations for Yang-Mills and even more cancellations for gravity, but you don't see that with the polarization vector, right? You just see a bigger and bigger and more horrible mess. So this is kind of so that so, and the fact that when we start adding spin in the usual point of view, we have to add these more redundancies to describe the physics and yet the answer again is simply the order of the other way around. It's just some invitation to try to think about the physics from an alternative point of view, that's all. But strictly speaking, you're right that just using these variables doesn't give you, I mean, you can use them, but they'll always show up in these combinations that you can call dot products and you. So it's not especially helping you with anything for a scale. Okay, and can I ask a follow up and quick question on what you just said? So you said that actually there's some notorious reverse that for spin two, then we do this easier. It looks simpler than for spin one, it's one, but from what you just showed us, how can we see that this is the case that it is actually a reverse? Well, you're not gonna, not from what I've shown you, but what we're going to do right now. So for example, we're gonna write down these three particle amplitudes and then we're gonna do this consistency check on four particle scatter, okay? And as I told you, when we do this check, sorry, as I told you when we do this check, we're gonna find very few theories work, okay? But in the process of doing them, not only will we confirm that they're consistent, but we'll actually write down the answer. So we'll actually be computing them. So we'll be doing a sort of a, one-minute calculation of the two-to-two gluon amplitude and a one-minute calculation of the two-to-two gravity amplitude. And I promise you that those are not one-minute calculations from Feynman diagram, okay? For gluon amplitude is what you do in a course, maybe on a problem set. And on a graviton amplitude, this was first done by Bryce DeWitt in a tour de force paper in the early 1970s.  Okay, thank you very much. Okay, any other questions or was that it? Since that's it for now. Okay. Should we continue now or do you wanna have a little break or how do you wanna do it? We can have a little break. A short break perhaps. Okay. Okay, five minutes. Five minutes is perfect. All right, see you guys soon, all right. Great. We're gonna start, we're done with kinematics and we're going to start talking about interactions. And as we'll see, these three particle amplitudes are almost kinematics as well because as I've mentioned already, a few times everything is determined by symmetries. Let's talk for a second why we normally don't talk about three particle amplitudes for massless particles. First, let's remind ourselves what we do normally talk about. What we do normally talk about are four particle amplitudes, one, two, two, three, four. And here the amplitudes, maybe we think of it's leaving let's say there are scalars. They're a function of the Mandelstein variance.  So we can have relations between these guys. We have S plus T plus U equals zero since we're talking about massless particles. But anyway, we have a function of two independent variables, S and T. Okay, so that's fine. But let's say we're talking about the three particle amplitudes like what could it be a function of? Let's say you can add scalars, nothing fancy. Let's say I just had scalars. What could this be a function of? I have one, two, and three. Well, what is P one, dot P two? P one, dot P two is the same as two P one, dot P two is P one plus P two squared. Remember all the P squared are equal to zero. But P two, P one plus P two squared is P three squared is equal to zero. Okay, so all of the Mandelstein variance for this three particle amplitude because they have momentum conservation. So crucially, of course, because that P one plus P two plus P three equals zero. And if the particles are all massless, all of the kinematical invariance equals zero. Okay, so there's nothing for the amplitude to be functions of, apparently, right? There's no dependence at all. There's all the PI, dot PJs are zero. So, well, okay, so what could it be? What could this three particle amplitude be if they're just scalars? Well, all it could be is a constant. So let me call the constant mu, okay? And that's all it could be, all right? Now, of course, this is what you get from a Lagrangian, which is mu phi cubed, right? And so if someone handed you the Lagrangian mu phi cubed, you would write down the Feynman rule mu, and that's what the amplitude would be. It would just be mu. And so from here, we learned something important as well, that the units of the three particle amplitude is mass dimension one, okay? So the units is mass dimension one, it's the same as the units of mu in the pi cubed theory, but there's nothing else that it could be. And notice that if you're sort of very naive, you would think, oh gosh, but there's millions of cubic interactions that I could write down. I could write down pi cubed in the Lagrangian. I could write down, you know, phi d mu phi d mu phi, I could write down box phi d phi d phi, I could sprinkle derivatives all over the place here, okay? And yet I'm telling you there's only one amplitude you ever get, one, two, three. You only ever get this one amplitude, all it can be is mu. So what's going on, of course, is that in this case, this amplitude gives you something non-zero. The rest of them all give you something zero, right? Because when you go on shell, for example, let's say this was my choice in the Lagrangian, then what I'd get is like p two dot p three for the amplitude and that's equal to zero, all right? More generally, they're not always zero, but we're going to see that the three particle amplitude is completely locked, okay? And therefore even when there appears to be millions of terms you can write down in the Lagrangian, when you put the, when you turn them into an amplitude by dotting with polarization vectors, they'll all give you the same answer, okay? Up to the overall, up to the overall scale or this finite number of answers up to the overall scale. Okay, so in general then, if we have general helicities, what's going to happen is that the amplitude, the only things that it can depend on are sort of polarization and helicity. And if just for one moment, I step back and show you how this works with ordinary polarization vectors, maybe you'll get a little more intuition before we do it sort of properly with these spin and helicity variables. But let's say I have a, let's say I have a, I don't know, I have a spin zero, spin zero and a spin one. And I want to describe this. So this is a particles one, two and three, but I want to write down an amplitude for this guy. Then, well, there's no dependence on p one dot p two or anything like that, but I have a polarization vector for particle three. I have epsilon three mu and this can multiply some linear combination. And remember that epsilon three dot p three is equal to zero, which also means that epsilon three dot p one plus p two is equal to zero. So this term vanishes and the combination of these two terms, which is plus will vanish. Therefore, the unique amplitude that I could have, the absolutely unique amplitude I could write down here is just epsilon mu up to some g times p one mu minus p two mu. So this shows you that the amplitudes have pure polarization. There's no dot products of momenta. They're purely determined by polarization. And, all right, so good. So let's now go back and see how we can think about these things in terms of the spinor-holicity variables. And first, let's see what is the condition. So remember, we have momentum conservation that now says that lambda one, lambda one tilde plus lambda two, lambda two tilde plus lambda three, lambda three tilde is equal to zero. If I had longer, I would explain how to think about this in a lovely geometric way, but we don't really need it right now. All we need to know is that we've already seen that p one dot p two, et cetera, that these are all equal to zero. So for instance, we know this means that angle bracket one two, one two is equal to zero, one three, one three equals zero, and two three, two three equals zero, okay? So let's say I take the top, top formula. So here I have a choice, either one two equals zero or the square bracket one two equals zero. And then if a tiny exercise I'll leave for you, very, very tiny one line to show that once you made the choice that one two is equal to zero, then momentum conservation actually forces that two three equals zero and one three equals zero as well. Or once you've made this choice, the other ones have to be zero as well. In other words, and when these brackets are zero, what does it mean? When lambda one, lambda two, epsilon lambda one, lambda two is equal to zero, that means that as two-dimensional vectors, lambda one is proportional to lambda two, okay? So that tells us something interesting that just momentum conservation for three particles that either lambda one is parallel to lambda two is parallel to lambda three, or lambda tilde one is parallel to lambda tilde two is parallel to lambda tilde three, okay? So that's something kind of peculiar about the configuration of lambda, that either the lambdas are all parallel and the lambda tildes are generic or the lambdas tildes are parallel and the lambdas are generic. Now, this actually explains something about why we don't normally talk about three-particle amplitudes in courses. Like intuitively, let's forget about all this for a second. Why don't we talk about three-particle amplitudes because we say, look, if we have a gluon or a photon, let's say going in this direction, then the only way it can conserve momentum is to split into two other photons that are going in exactly the same direction, right? So this is the only picture we have for the conservation of momentum for an energy for massless particles. And so it's a very singular configuration that's exactly collinear configuration where the photon goes and splits into two other photons that are going in exactly the same direction. And in fact, if you work out the amplitude for this, if you literally plug in the polarization vectors and you shove in the momenta and stuff, you'll find that this amplitude is equal to zero. So literally this amplitude is equal to zero for Minkowski and momentum. And we can see why from our slightly fancier point of view now because we just saw that for the three-point amplitude, we either have to have lambda one parallel to lambda two, lambda three, as I said, or lambda tilde one parallel to lambda tilde two, parallel to lambda tilde three. But in three-one signature for a normal real momenta, lambda tilde has to equal lambda star. So if we're really talking about real momenta, then if we have lambda one proportional to lambda two proportional to lambda three, then we must also have the lambda tilde one is proportional to lambda tilde two is proportional to lambda tilde three, just because they're the complex conjugates of each other. Okay, so that's why Minkowski and momentum are a little too degenerate in order to actually see the full structure of the three-part of my amplitude. Because what's happening for complex momenta is just that the lambdas are forced to be parallel or the lambdas are forced to be parallel. But because three-one signature forces one to be the complex conjugate of the other, it forces everything to be parallel and that's too degenerate in order to actually see the actual amplitude. So that's why it was useful and important that we were thinking sort of to begin with slightly more generally about the momentum. Okay, so there's a configuration then, either when the, so let's say that the lambdas are, let's say that the lambda tildes are parallel. Okay, so if the lambdas are all parallel, then the only thing that I have at my disposal are these brackets, one, two, two, three and three, one. Because the other brackets are all zero, right? One, two, the other square brackets are all equal to zero. So these are the only things that I have. And so now let's try to write down an expression, okay? So here I have holicity one, holicity two, holicity three and any expression that I write down will look like some one, two to the A, two, three to the B, three, one to the C. Maybe with some strength of a coupling constant up in front. Okay? But now we have to impose that this transforms correctly, that it has the correct sort of a holicity way. So what is the formula? If I take this expression and I send lambda one to T one, lambda one, I have to pick up a factor of T one to the negative two H one and so on. So let's see what that tells me. When I send T one to, when I send, when I rescale T one, I pick up a power of A and a power of C. So from here, I learned that A plus C has to equal negative two H one. For T two, similarly, I pick a power of A and B so that the A plus B is equal to negative two H two. And finally for three, I have that B plus C is equal to negative two H three. And so these are three equations for three unknowns and I can solve for them easily. And A is equal to H three minus H one minus H two and so on. Okay, so let me, it's easier to sort of just write it and the answer to just remember it. So I have one, two, three, and this can be G one, two, to the H three minus H one minus H two, two, three to the H one minus H two minus H three, three, one to the H two minus H three minus H one, all right? And similarly, this was when all the, this was when all the, remember this was when all the lambda tildes are parallel. When all the lambdas are parallel, going back to the same expression, one, two, three, by exactly the same argument, we get G of the square bracket and now it's reversed. It's H one plus H two minus H three, two, three to the H two plus H three minus H one, three, one to the H three plus H one minus H two, all right? Now I just want you to pause for a second and I hope you appreciate how sort of amazing this is. We didn't have to, you know, there's no Lagrangian. We don't have to make any guess. We don't have to do any work. You might imagine you have a spin 17, it's very complicated, lots of indices. Doesn't matter. We just wrote down the answer, no thought, no choice, up to the strength of the overall coupling concept, all right? So let me, let's take a look at an interesting example. So let's say that the, the helicities were minus, minus then plus, okay? Just for a fun, this one ended up being relevant for a massless spin one of course. So these are all helicity minus one, minus one and plus one, okay? And so these are particles one, two and three. Well, if the lambda tilde's are parallel, the formula we get would be one, two cubed over one, three, two, three, G, and if the lambda's are parallel, it would be the other way around. It would be one, three, two, three over one, two cubed. Maybe it was some G prime, okay? We don't know. These are different configurations. So there's no reason why these two coupling constants even would have to be equal a priori. But I want you to notice something here. Between these two expressions, one of them is preferred. And we're gonna prefer this one and we're gonna set this G prime equal to zero. Okay, that's because we have another requirement is that we want that the limit of the amplitude, the limit in Minkowski's space should be smooth, should not bull up. Remember in Minkowski's space, if the lambda tilde's are parallel, then the lambda's also have to be parallel. And so here you see that this is going like one, two, all of these things are vanishing, but this is vanishing with a cubic power upstairs. There's only two powers downstairs. And so this goes to zero in Minkowski's space. Whereas this would actually go to infinity in Minkowski's space, okay? So if we want something to not blow up in Minkowski's space, then we have to put this structure to zero and we only have that one. And so the summary is that the amplitudes when h1, h2, h3 are equal to this expression that we wrote down, one, two to the h3 minus h1 minus h2, two, three to the h2 plus h3 minus h1, three, one to the h2 minus h3 minus h1 with some g. This is when the sum of all the holicities is positive and some other g prime or whatever, I'll just keep calling g the other way around. One, two to the h1 plus h2 minus h3, two, three to the h2 plus h3 minus h1, three, one to the h3 plus h1 minus h2 when the sum of all these holicities is negative. And in the case where the sum is equal to zero, in principle, we have both, but it actually turns out that these are going to be examples of things that don't give rise to consistently factorizing amplitudes unless h1, if the sum is zero. So these will all end up being inconsistent. I won't show you why, but you can discover it yourself by the methods that we'll talk about. The all end up being inconsistent unless all the holicities are zero, in fact. So that's another sort of special thing about the spin zero scattering. The spin zero scattering is the only scattering where the amplitude remains finite even in Mankowski spin, whereas there's not examples. And okay, so all the ones that are inconsistent other than the case of where all the spins are equal. Okay, now someone had a question. Was there a question? It disappeared in the meanwhile. Okay, I hope it would answer rather than just intimidate it away. Okay, good, so now let's go back and just stare at a few of these examples. So I wrote one of them down already. So this is all the velocity one. So we're gonna have minus one, minus one plus one. And this is G12 cubed over 1323. Okay, so that's what we're supposed to write down. Let's notice, and what about another one? What about if we had minus one, minus one, minus one? Well, this is a one, two times two, three times three, one. Times some other coupling constant, I'll call it kappa. But let's just stare at these things for a few minutes. So first of all, let's just verify. So in this expression, there's three ones upstairs and there's one, one downstairs. So this goes like t one squared and that's indeed compatible. This two is equal to negative two times negative one. And that's indeed compatible with it having calicity minus one, okay? So these are particles one, two, and three. So you can check in that same way that about the rest of the calicities. What about something else? Remember, the units of the three particle amplitude was supposed to be the units of mass, okay? At the mass dimension one. So the units of this, remember one, two has units of momentum. So this is upstairs like momentum two, downstairs is like momentum squared. So this all has units of momentum or mass already and therefore this G is dimensionless. And in fact, this is exactly, almost exactly the amplitude in Yang-Milster, okay? That's the three particle amplitude in Yang-Milster. Again, when you do it with Feynman rules this would be epsilon P minus P and then contracting of a bunch of ways and so on. But anyway, this is just what the answer is. Now, what about this expression? This has units of momentum cubed already, right? So that means that this guy has to have units of one over mass squared. So we learned something interesting that this amplitude is more important at low energies than that one is. And if this was coming from a Lagrangian this is coming from the usual F squared in Yang-Milster's theory whereas this would come from an F cubed higher dimension operator, okay? And F cubed divided by mass squared, okay? So we know that the non-nibelian part of the Yang-Milster is gonna give us something dimensionless and that's this top amplitude whereas this other one with all minus solicities corresponds to a higher dimension operator. Well, let's actually take one more look at both of these expressions minus, minus plus. This is one, two cubed over one, three, two, three. And we would expect that by a Bose symmetry if I interchange one with two that Bose symmetry would say that the amplitude goes to the amplitude. But what happens to this? Remember that this one, two is contracting with an epsilon symbol. So one, two is equal to negative two, one. So if I interchange one and two this goes to negative itself, okay? That's very peculiar. So this amplitude appears to violate Bose symmetry. And in fact, as we'll see in a moment even though we can write it down as a three particle amplitude literally this thing will not give us a consistently factoring four particle amplitude. And again, if I had a lot longer I would show you how spin statistics and everything comes out of the considerations that we're talking about. But this is already to show you this is an example of something that actually this would violate Bose statistics but we don't know about that yet. It's certainly there as a perfectly fine structure we can write down. But what could you do if you wanted something that had Bose statistics and this will be forced on us anyway but just so you see what these things look like is that you can't just have one species of this particle you have to have many of them with some labels and then the amplitude would be one, two cubed over one, three, two, three but then you'd multiply by some FABC. At the moment I don't know anything about FABC being structure constants or anything like that but FABC if it's totally anti-symmetric would make this thing be Bose symmetric, okay? Because then when you interchange one and two you'd pick up a minus sign from this and you'd also pick up a minus sign from that so the whole thing would be both symmetric. And similarly for this expression, okay? Minus, minus, minus it would be one, two, two, three, three, one again time some of something else anti-symmetric, okay? But it has to be something anti-symmetric. So that's cool. Photons cannot have this interaction that would violate both symmetry and as we'll see in a moment as I said we'll discover why that's inconsistent even if you've never heard of Bose symmetry but in particular photons cannot have this interaction. So an interaction involving exactly the same spin particles just inconsistent if you only have one species of a particle. Okay, so now let's keep on going. Let's say that we had a single particle but with these holicities where this was of general spin S, right? I'm gonna call the spin here a curly S. Then what this amplitude would be is just the same thing one, two cubed over one, three, two, three to the power of S. That's it. And so if we put S equals two this is now there's nothing wrong with it. It's perfectly okay. It's both symmetric. Everything is fine. And that's the three particle amplitude in gravity. So this is the three point amp in GR.  This thing now has units of, if I put S equals two this thing has units of mass. So this would be mass squared. And therefore, if I put S equals two here the coupling constant would have to have units of one over mass. And that's what we call one over M block, okay? And so similarly, the other guy with all minus would be one, two, two, three, three, one squared. And now this thing as a dimension six already, right? So it has a momentum cubed squared is the dimension six. And so we'd have to have some overall one over mass to the fifth to make it make sense. And this is what would come from a Riemann cubed term in the Lagrange, okay? So once again, it comes from a higher dimension option. But if I compare S equals two gravity to S equals to Yang-Mills for S equals one we see this sort of amazing fact that the three particle amplitude for gravity is equal to the three particle amplitude for Yang-Mills squared. And this is the beginning of a long story for how gravity and Yang-Mills amplitudes are connected to each other, which I won't say anything about. But again, I just wanna emphasize if someone told you to go and calculate the three point the amplitude using the Lagrangian for gravitons expanded around the flat space you would not in a million years expect this kind of relationship nor in a million years would you expect the answer is so simple. It does not look simple with like, you know 100 terms in the Dundergage with contracting polarization vectors in crazy ways with each other. Okay, okay. So any questions about this? So we've now learned how to write down all the possible three particle amplitudes. There's a question in the chat if you want to. Yeah, yeah. No, no, so there's a question that about the anti-symmetry telling you anything about whether the gauge groups are simple. No, I mean, you're allowed to have direct products of many gauge groups if you like, but in a moment, just a moment, we're gonna discover why these FABCs actually have to be the structure constants of a lead group. So the fact that the Yang-Mills structure is there we're going to discover in just a moment. Okay, any other questions? Just as a reference, we're 10 minutes over the... Oh, we're 10 minutes over at this point. Yes. All right, if we're 10 minutes over, then I think I should stop. All right. We'll finish the discussion tomorrow then. Okay, okay, great. Thank you very much. Then we can collect the questions and there's one from Anna, okay. Hi, Rana. Go ahead. Hi, Prof. Thank you for these lectures. They've been absolutely amazing. I just, I'm dying to know, could you please tell us a little bit more about how the Yang-Mills and the gravity are connected? Now that you have some... Yeah, yeah, yeah, yeah, yeah, that there's a... Yeah, I mean, I'll tell you a few qualitative things about it. So we've seen already, so let me first back up to a super simple-minded idea, which is obviously wrong, but very often super simple-minded ideas have some kernel of the truth in them, okay? So we know that gravitons have spin two, right? So here's a graviton going along and has velocity two, so that's velocity two. So you might think couldn't it be that a graviton is just kind of like two gluons on top of each other, right? Or two spin one particles on top of each other? After all, if I had a gluon here and a gluon here, then if I have two of them on top of each other, the composite thing looks like it has spin two, okay? So the most simple-minded thing that you could imagine is maybe the scattering processes involving gravitons are exactly the same as what you get involving gluons, where you just imagine that every graviton is replaced with two gluons that are collinear to each other, okay? That's at the most simple-minded level. That's what you might think is going on. Now, why is that obviously ridiculous? Quote-unquote ridiculous, because let's say I did it like this. So here's the gluon in, here's two other gluons in, and then the problem is that this gluon hits that gluon and has some amplitude for them to come out at these angles, and these hit and they have amplitude to come out in some other totally different angles. So even if you set the guides in being parallel to each other, there's no way they're gonna be parallel to each other when they're coming out, right? I mean, so maybe there would be, if you somehow thought of the graviton as a bound state of the two gluons or something like that, but then okay, then all the excitement is about whatever is binding them to each other, and you're definitely not gonna be able to sort of independently treat them like this. But if I just keep on going, there is of course one situation where this sounds less crazy, which is for the three particle amplitude, where as I said, at least in Minkowski's space, you would just draw this picture, they're forced to be collinear with each other anyway. Okay, so in the splitting process, they're forced to be collinear with each other. So maybe that would give you a reason why the gluon, the square of the gluon is related to the graviton, all right? Except of course, this is true in Minkowski's space, but it's also true because both of the amplitudes are zero. So zero squared is equal to zero. So, but this is just the very beginning of an intuition for why there might be a connection up between them. But what we've seen very precisely is that the level of the three point amplitude for GR is equal to the three point amplitude for Yang-Mills square. Okay, so there is that exact correspondence. Now, there was actually another setting where people first saw this connection between gravity and gauge theory, which was in string theory. And here it's for another simple reason that in string theory, there's open strings and there's closed strings. And people knew that gluons came from open strings and gravitons came from closed strings. But of course, you can think of this closed string as really two open strings with their endpoints glued together, okay? So this is the sense in which closed equals open squared. And this was actually very, very concretely there in string theory. In fact, the scattering amplitudes for closed strings in a very precise sense are given by a kind of squaring of the scattering amplitudes for open strings. They're not literally taking the functions and squaring them, but there's some kernel that you have to integrate. And instead of integrating it over the real axis, you integrate the same thing mod squared over the complex plane. That's the sort of relationship between amplitudes of closed and open strings. So that showed that there was a kind of a deep connection between gravity and Yang-Mills, completely unobvious in the Lagrangian, as I said, but that the amplitudes in string theory are really somehow related. What useful to think about closed strings is being the square of open strings. And the last sort of 10, 15 years, people have realized this in a practical way already at the level of the, already in the field theory limit. So it goes by many names of the sort of color kinematics connection or the double copy connection, but there is a precise formula of the following form that says that a gravity amplitude is equal to, this is a loose formula, Yang-Mills amplitude squared, multiplied by some dressings. And so it's these dressings here that take into account that it's not literally the square, which would have been crazy, but these dressings are one for the three point amplitude, but there's a very precise way in which they are addressed to make the statement true. So there really is a connection between them. And as I said, this is kind of a subterranean connection. You don't see it at all in the Lagrangian. You would never dream of it in a million years if you stared at the gravity Lagrangian and the Yang-Mills Lagrangian, but it is true. And I should say that going back this sort of deep connection between open and closed that's seen in string theory has been extended in the gauge gravity duality. And so the sort of gauge gravity duality, the ADSCFT correspondence is actually one more example of the same basic string theory phenomenon that you can think of as closed strings as a square of open strings. And so there's this close relationship between them is seen in the ADSCFT where the open string side is the CFT and the closed string side is the ADS. Any other questions? Okay, as Anna says, I'm also thanking for the very nice lecture. I will have one follow up question about this one, the first one. I think I heard another talk by you that you were talking that the gauge theories have F menu as their curvature and the gravity has an actual curvature of the four dimensional manifold. So could you comment on this geometrical correspondence as well from this perspective, or is there nothing more to say? Well, no, I mean, again, let me just just to be super explicitly clear about this. There's not only there's absolutely nothing wrong with the Lagrangian formalism and all the stuff you learn with the curvatures and the spacetime curvature on the gauge curvature. It's beautiful, it's correct. It describes reality in at least all the approximations that we see, and I'm not trashing it at all. Okay, so it's totally, totally, totally correct. And in fact, it's some things, for example, that we know that there is a relationship between the spacetime curvature and the internal space curvature of gauge symmetry is revealed in colloquial line theory. So you can start with something that just has gravitational curvature in a higher dimensional space, but when you compactify on some dimensions, part of the gravitational curvature will look like an internal space curvature from the point of view of the long distance theory, because they don't see the little manifolds that you're compactifying on. So the part of the gravitational curvature that's curved in the internal dimensions will look like a Yang-Mills curvature from the long distance point of view. Okay, so there's a good reason why all those things are connected to each other. And it's just that, so there's nothing wrong with that way of thinking about things. Not only is there nothing wrong, it's powerful and it's deep and it's correct. The reason for emphasizing things the way I've been emphasizing here is that as powerful and deep as it is, you should always remember that it's a formalism and formalism is distinct from physics all the time, okay?  And that's what I hope you're gathering from what I've been telling you is that all of this machinery, let's say we're interested in the interactions of elementary particles, we're interested in their scattering processes, all of the machinery of the Lagrangian and the fields and the gauge symmetry and all of that is a formalism is a convenient tool that we have introduced as theoretical physicists in order to describe the interactions of particles in a way that manifests the locality of their interactions. It's a tool and it's just important to remember that something is a tool because for different purposes, other tools might be better. First of all, and secondly, if you remember that it's a formalism and not a fact about nature, you'll have a more flexible mind because it might be that in future developments that it's important to not pay so much attention to this formalism because it might obstruct you from discovering something else, which is true. For example, very concretely, when we talk about gauge symmetries, if you take the attitude towards gauge symmetries that was sort of prevalent in books in the 1980s and it's still there in many courses today, although much less than it used to be, then what is the point of view? Gauge symmetry is the most important thing about the world. The gauge group of the world is the most important thing. Isn't it amazing? We can rotate the phase here independently then over there in the faraway galaxy and incredibly when we do that, it forces all these interactions on us and that's the amazing thing. For some reason, the world has this gauge symmetry and that's the deepest thing there is and everything follows from that. Okay, good. It's true that the gauge symmetry does all those things, but psychologically in your mind, you think, okay, that's the most important thing in the world. So if you believe that, it makes it very difficult to discover things like dualities. Okay, so something amazing that we learned in the 90s, even before to begin with, was they could have two completely different gauge groups that have identical physics, okay? The physics is completely identical and sometimes it's because the gauge groups are actually strongly coupled or they confine or something like that. So that makes it more reasonable that the sort of bound states that you get from the two different confinements might end up looking exactly the same, but this ended up being going on and on so that you could see that two different gauge groups had identical physics of every scale in every way, just identical physics. Now, if you believe that gauge symmetry is the most important thing in the world and God decided on the gauge group before doing anything else and so on, you'd be pretty shocked by that statement, right? Okay, which one is more important? Which is the fundamental one? Is it the G1? Is it G2? And in fact, if for some reason we thought it was G1, then you would never even think that there might be a G2 that describes exactly the same thing. So if however, the whole time you knew or you appreciated that this was actually a useful redundancy, then you're more flexible. You're like, okay, cool. Someone else found a different useful redundancy for describing exactly the same thing. This is even more dramatic in the context of the duality between gauge theory and gravity. When you have the duality between gauge theory and gravity, that's really shocking, right? If you told someone, look, gauge theory is the same as gravity, then if you're a gauge theory lover, you'd say, but where is the gauge group in the gravity description of the physics? And if you love general covariance, you'd say, hmm, I don't see any general covariance there in the gauge group description of the physics. And so both people would be very upset and very shocked that the most important thing they thought about the world is not reflected in the other description. However, when you know that they're both just redundancies for describing the same physics, it's much less surprising. Something like that is a possibility. So coming back to earth and what I was saying, we're not talking about anything as fancy quite as that yet, although it's actually not unrelated to some of those things, but whenever you have some physical phenomenon, here we're talking about the scattering of particles. It's important to look at the actual observable. The actual observable, what does it depend on? It depends on the momenta of the incoming particles and the outgoing particles and the holistic. That's it. Everything else that we do when we draw Feynman diagrams, polarization vectors, virtual particles, all of the rest of this is not there in the actual observable. It's there in a particular formalism for doing the calculation, but it's not there in the actual observable. And it's important to realize that sometimes you only have one formalism for doing the calculation. In that case, okay, it becomes convenient, maybe even your mind to assign reality to the formalism, but it's always useful, even if you don't know any other way of doing the calculation to sort of understand what part of what you're doing is there, is reflected in an actual invariant property of the answer and what part is part of the formalism. And then some day you might discover yet another formalism for doing the calculation. That's what I'm beginning to show you in these lectures, is a totally different formalism for getting exactly the same answers where we don't see the fields and we don't see all of that stuff, but we begin from this point of view where the particles are primary to begin. Okay, thank you. And my second and last question will be about this formalism. I don't know if it's a technical or deep question, but it seems that this spinor...