 It's a pleasure to be here. I'll be discussing scattering amplitudes and my actual title will be the connection with inter-gubility. The sort of overarching question is, is how do massless particles interact? And the context in which I will be discussing this question is in flat space. And I will be considering this from the viewpoint of a perspective picture. So when we start from tree level and add loops and so on, a lot of ideas have been developed from this viewpoint because there's a huge program of people making precision calculation for LHC physics. So we benefit a lot from this extensive study. Massless particles, so most particles in the world are massive, but describing them as massless is relevant when you describe the sort of long range interaction, which for example in QCD means that distance much longer than inverse masses and also of course look at distance much smaller than 1 over lambda QCD when we apply, yeah. So there's a wide range of energy scales in QCD where the approximation particles are massless is a good approximation for maybe not for the top part but finding particles. So they're very interesting to study and these are the ones I will study in these lectures. But I will discuss the general case, non-super symmetric, but a lot of progress recently has occurred. Yeah, one distance smaller than lambda QCD. Oh, sorry. Thank you. Yeah, so the masses of the like quartz, yes. Thank you. Sorry, what's this equation? Yeah, yeah, this is short distance compared to mass and confinement. Okay, so there'll be a very interesting connection in a special case of N-quals-4, Super Youngness, very special theory where at least in the planar limit there's a connection with integrable system. So we'll be using this general theory that has been developed for practical calculations in QCD and so on to address to discuss how we see integrability arise in this setup, in this special theory. And at the same point, at the same time using integrability, I think the long-term goal of this program is that we learn how to calculate much more efficiently. Again, okay. Will that be okay? I will try to stick to that point. Sorry. Okay, so to start in perspiration theory we'll start beginning at three level and it will be very important, very useful to understand three level thoroughly. We'll be using a lot of simplification that occurs in three plus one space-time dimensions. So when we discuss massless particle for momenta, you can write as a, you can take the, you can construct the vial operator from it, which is a two by two matrix, which is of this type, P zero plus PZ, P zero minus PZ, PX plus I, PY, this is basically P slash for vial fermion. And that two by two matrix is the property that if P square is massless, then the depth of P alpha dot is equal to zero, and P alpha alpha dot can be written as a product to spinners. Again? Many small levels. Okay. So the important equation is that P alpha alpha dot is equal to a product of two spinners. And yeah, these spinners are the solution of the vial equation. This, yeah, so this is factorization in four dimension, which is very useful to simplify notation. So we're looking at the S matrix of one shell particle. The first case is one particle goes to one particle. That's just the identity, there's nothing to say. The next case is three point, three particle amplitude. And if you try to have three on-shell particles interact in four dimensions, one, two, three, the on-shell condition imposed at all moment are zero. And since P three is P one plus P two, it's imposed at P one dot P two is equal to zero. Here, these lambdas, all environments can be written in terms of these lambdas here. And P one dot P two is equal to what people write like this, where one two is equal to the anti-symmetric symbol of lambda one alpha and the two beta. So if you anti-symmetrize those two things, you make a Lorentz invariant product of the spinners. You have the similar Lorentz invariant product of the bar spinners. And all Lorentz invariant products can be written in terms of these invariants. And for the three point amplitude, this is equal to zero, which should use that at least one of them has to vanish. So there are two cases, which I would draw first like this. That's when, well, if the case when one two is equal to zero, one can show that one three and two three are equal to zero also. So there's a case where our angle brackets vanish. And that means that the lambda one is proportional to lambda two, it's proportional to, all the lambdas are proportional. And there's another vertex in the theory in which they are, in which the lambda two are proportional to each other. So there are the two basic vertices. Normally, you might say that if you have a three point vertex in Minkowski space, the normal conclusion, if you think about the real momenta, is that all the momenta have to be coordinated with each other. Here we're doing something slightly more general. If we allow the momenta to be complex, then these and these are independent cases. And now in Minkowski signature, you will have both of them simultaneously. But as we'll see later, it's very, you will get a lot of information if we distinguish those two cases. So what are these vertices? Let's consider Yang-Mills theory, for example. So in Yang-Mills theory, the external states are labeled by polarization, which can be gluonous two-elicity polarization, plus or minus. Yes. OK, yeah, square bracket is the same where, well, actually I should write it down there. Let me be very correct. So it's the symmetric product of two spinners. And that defines a Lorentz-Envariant product of these spinners. What's important about this product is that it's Lorentz-Envariant. So if we look at per Yang-Mills, so when we scatter, we have polarization vectors. A good thing about this formalism is that it allows us to get rid of the polarization vector. So the M2 doesn't depend on polarization explicitly. So what are the polarization expressed in terms of lambdas? If we have a plus polarization, you can write it as denotation. The simplest guess for a polarization vector will be something like lambda, lambda, alpha dot. That's the simplest thing you can write in terms of lambdas. But that's proportional to p, p, alpha dot. And that decouples by gauge invariance. So how do we write polarization vector? We write lambda for actually, we write it for plus. It would be proportional to one lambda. But for the other guy, we can choose lambda. We choose some arbitrary reference. And we divide by something which somehow removes the dependence on this reference. So that would be the definition of polarization vector. So we have to introduce a gauge choice to write this polarization. However, it turns out that this gauge choice doesn't matter at all, and the amplitude does not depend on it. And one way to check this is you take epsilon 1 of mu 1. You take the difference between different polarization vectors. The difference is proportional to mu 1 alpha over mu 1 lambda minus mu 2 alpha over mu 2 lambda times lambda tilde alpha dot. And this combination, the numerator is proportional to mu 1, mu 2 lambda, antisymmetrized in 1, 2. And there's a very useful identity, which comes from the fact that the mu lives in the two-dimensional vector space. If you're antisymmetrized with respect to 2, that's equal to lambda to 1. So that way, you see that if you change the gauge choice, you get something proportional to p, and it decouples. So the absolute doesn't depend on, it never depends on this choice of mu, and it only depends on the lambda. It's very convenient. So what is the three-point vertex of the theory? Well, if we plug in this epsilon in the expression for the three-point vertex, we can compute, we get an answer, just some Feynman diagram. But we don't have to compute to know what the answer will be, because these lambdas have a very nice property that p is invariant under the following, what's called a little group ambiguity. If you take lambda goes to e to the i theta lambda, and lambda tilde goes to e to the minus i theta lambda, and the tilde, then p is unchanged. p goes to p. And under, actually, let me put over two here. Under this phase ambiguity, and this is just saying there's a phase ambiguity in choosing your spinors. Under this phase ambiguity, epsilon plus goes to e to the, let's see, epsilon plus goes to e to the minus i theta epsilon plus, and epsilon minus goes to e to the plus i theta epsilon minus. So this phase ambiguity track what is called the elasticity of the particle. The exponents are always related to the elasticity of the particle. As general, it's true for fermions also, for scalars, there's no phase ambiguity, of course. And just because of this phase ambiguity, so what do we know about this vertex here? If we take, for example, I will take this plus plus minus vertex. Because the lambdas are aligned, it can only depend on lambda tilde's. It has to have these phase ambiguities, so that it needs to have two lambdas, so this is two, three, one. It needs to have lambda tilde two, lambda tilde three, lambda tilde one, upstairs, square. And we have to contract this into Lorentz-Envarn quantity. And one can see that there's only one way, actually, to contract these things together, which is to write the following combination. Sorry, something is wrong. One, two, three. So this vertex is proportional to square bracket of two, three, cube divided by two, one, two, three, one. Let's see, oh, I got it backwards, sorry. Of course, yeah, lambda tilde's on top, okay, good. Lambda tilde two, lambda tilde three, over lambda tilde one. And you see, indeed, that this few kind of weights in two, there's three on top, one down, so there's two on top, two threes on top, and one down downstairs. And it's the only contraction, which has this homogeneity. So without doing any calculation, the vertex has to be proportional to a number times this. And the calculation tells you what the number is. It's FABC, the structure constant. Other minus sign? Sorry, I'm very, yeah, so it's for the lecture, I'll be sloppy with signs. I will give reference to the literature. So they are very good reviews and lectures notes on this on the internet, and for the signs I would refer to the reference. Yeah, pleasing your all signs in this lecture. And similarly, this other vertex is FABC, and now two, three to be minus, minus, plus one. Okay, so what about other possible vertices? For example, plus, plus, plus, if we have this guy here and plus, plus, plus. Now it will have to be proportional to all lambda tilde's on top. That will have to be proportional to, and one can see that the only thing consistent to this homogeneity is this quantity. Times some coefficients, C, one, two, three. However, in that case, the dimension of this is different. So each of these brackets is dimension one, and the two point vertex needs to have dimension one. So C will have dimension minus two, mass dimension minus two. So this coupling here will be irrelevant in the inference. So, and I started this lecture discussing, I will discuss the long distance, so I will ignore such term. So that would be a higher dimensional operator either through the theory. So you could generate this term only by adding higher dimensional operator. And for the Yang-Mills Lagrangian, it will not appear, because the coupling is dimensionless. And another case you could consider is what about minus, minus, minus. Then you get one over all that. Now the coupling constant as dimension plus four, which is really bad. And as we'll see shortly, this will lead to, it's not consistent. You cannot find, it's inconsistent with four-particle scattering. So I will discuss four-particle in a second, but this does not correspond to any self-consistent theory. So it will not consider it. So now that we've discussed three point, let's move on to four-particle. So we can imagine, again, doing the Feynman diagram calculation. Many of you have sort of done it in many, many classes as students to do it. All the permutations, this diagram, this diagram. We could imagine doing this calculation and plugging in the explicit expression for this epsilon, and we would get an answer. But there's no need, again, to do it. And this is one thing I want to emphasize in this lecture, is that self-consistency of the theory basically fixes the answer, again. So, and what makes this possible is the fact that we're scattering particles with spin, and the spin comes with this little group ambiguity. Because of this little group ambiguity, we can say, for example, that the scattering amplitude for four-particle, two of them, negative elicity and two of them, plus elicity, must be equal to something which has the correct phase ambiguity, and that something would be one, two, I just write, can write anything. So that's an example of some dimensionless thing which has the correct phase ambiguity. And then some function which doesn't have phase ambiguity, so it depends only on S and T, where S and T are the Mandelstam invariants. Well, that's the case. And at three level, this function here would be a rational function. What can we say about it? Well, it's a rational function and we never get double poles. So, can only have single poles, one over S, one over T, okay? It would be dimensionless, four-particle scattering is dimensionless, and yeah, in general the dimension, yeah, anyway. So it could be, I could write S over T, S over U, but the one for S here is actually pretty bad because S, which I define as P one plus P two square is equal to one, two, let me write bigger, sorry. One, two, one, two is also equal to three, four, three, four. This has a cubic pole at three, four goes to zero. That's very bad. The physical amplitude never has anything stronger than single pole. So not only this is ruled out, but actually we need an S on numerator or both terms. And we cannot even have a constant because it's not proportional to S. So the amplitude must be of this form for two constants, C one and C two. So we don't have to do a calculation, we know the amplitude has to be like this. And this is actually very constraining because the amplitude has three factorization channels on which it must factorize. But it only has two coefficients, C one. So somehow there's some identity that's required for the theory to make sense. And consider for example this, what's the simplest way to fix this C one is to look at what happens when T goes to zero. Then this term becomes irrelevant. If you take T goes to zero, the amplitude must factorize on a product of three point vertices. And let's say that T is a, let's say that two, three goes to zero more precisely. And as I mentioned, it's useful to consider limits in which not only particles become collinear but consider complex momenta where I'm allowed to take two, three goes to zero but two, three fixed. If you insist the momenta are real, both of them will have to go to zero at the same time. But we get much more information by considering these two limits separately and we have the right to do that because amplitude is defined as a complex function. And if we take two, three goes to zero, then the only vertex which is non-vanishing when these are zero is this vertex. The amplitude has to proportional to product of this vertex times one of it. And the coefficient of course is the structure factor air. Let's call this F, let's say a B, C, D. Let's say the colors are B, C, D, A. So this is F, B, C, E, F, A, D, E. So this product of color factors times one over T. I will not evaluate the vertices explicitly but basically the vertices as required as you will expect give you exactly this structure so that C one is equal to this. So you fix C one from this factorization. Similarly, you fix C two and that would be basically F, A, B, sorry, F, B, D, E, F, A, C, E over U. I'm sorry, the one where T is, is, is, is not, yeah. So the C is just this numerator. And C two is just this, but now, so that's the U channel. So we've used these two channels and we'll fix all two coefficients. So now what do we get from the third channel? For the third channel, now we know C one and C two so we predict what the amplitude is and the S channel, the amplitude, of course, must be proportional to F, A, B, E, C, D, E. And it actually works because of the Jacobi identity. If you take C one minus C two, you get that. Again, I'm missing things, but the point is the theory will not make sense. You could not construct a consistent four point amplitude if the Jacobi identity was not satisfied. And the reason this, you get this kind of constraints for particles with spin is that we're forced to pull out these phases. And these phases always come with some pull. So the fact that we have to pull out these factors leads to these kind of constraints. And yeah. So yeah, for spin one, the Jacobi identity can derive it from, consistency of the four-point function. This analysis has been done later for IR spin. For spin two, we derive in this way the equivalence principle so that all particles must couple in the same way to the graviton, and there can be only one graviton and so on, massless graviton. And for massless IR spins in flat space, you get that there's no solution. And the reason these constraints get so much stronger at IR spin is that you get these factors with IR and IR power. So if you have graviton, four graviton, you have something like that. There's a quadrupled pole, and you need to cook up something which cancel three powers of it for the amplitude to make sense. And it's very difficult to have this cancellation without introducing poles in the other channels. So the higher the spin, the more constraints you get. For massless scattering. So that's an important lesson. So move on to, so that's the five-point amplitude. We obtained it in some form in terms of FABCs. At IR points, IR points, we could keep going and just make an ansatz and fix all the coefficients using factorization. Because we already have four points, it will somehow be useful as a building block, but there's a better way. And yeah, the reason it would work at IR point is that, and you just say that if you have a five-point amplitude, the dimension is minus one. In general, the dimension of the amplitude is equal to, because the S matrix is dimensionless, the S matrix is a delta function times the T matrix. So the dimension of the T matrix, the momentum considering the delta function, so the dimension of the T matrix is four minus number of particles. Because four comes from cancelling this delta four, delta function, and each particle, each unshelt state is dimension one. So as you go for, so this is an example for n equal three, the three-point vertex as dimension one, which is indeed what we add here, this is dimension one. The four-particle amplitude is dimensionless, that we just saw. And the five-particle amplitude as dimension of one over mass. So basically, this analysis becomes more and more constraining at IR points because, just by dimension analysis, you need more and more denominators at IR points. So it's just impossible to write something at five points which doesn't have a denominator. And then if it has a denominator, it has a pole somewhere and can constrain this pole by factorization. So once you have the three-point particle, you construct the four-particle amplitude, and then all our points are fixed by factorization. And the efficient way to solve this factorization constraint, there's at least one efficient way is recursion relation. So that's a way to solve factorization constraints. Systematically, that's much, that IR points becomes more practical than just listing all the possible structures you can have. But conceptually, it's solving, it's doing exactly what we did by hands here for four-particle. And the idea of the recursion is to introduce, so you might think that to use solve factorization can be difficult because amplitude depends on many, many variables. If you have a five-particle amplitude, if you scatter five particles or two particles, move to three particles, there's a lot of angles and relative energy fractions. There's a lot of variables that these amplitudes depend on. And if you try to solve, if you try to think of it as a function of, I don't know, five complex variables, it's very difficult analysis. The trick is to restrict it, project it into some plane so that you look at it as a function of just one variable at a time. And so you introduce one parameter, the formation of the amplitude that Brito-Cachezzo-Feng-Witten introduced. And this, the simplest deformation is the one they used is you take, it's likely, the simplest thing you would try is just take lambda one, sorry, let's take lambda n, let's define the shifted one as lambda n plus z lambda one. So that changes lambda. That doesn't work because total momentum is not conserved. And you have to respect momentum observation. But you can fix it in a simple way by shifting lambda tilde one in the following way, minus z lambda tilde n. And so I'm just, the only modifying I went up p1 and pn, but in such a way that pn of z plus p1 of z, which by division is lambda, so if I put alpha, alpha dot in this case, this is lambda n alpha of z, lambda tilde n alpha dot plus the one alpha and then lambda tilde one alpha dot at. And it's only these two guys depend on z. So this guy, this guy are not changed. And if you compute this quantity, so the lambda n, lambda tilde n term and lambda one and the tilde one term are there. And then the z term, you get a cross term which exactly cancels in those two terms. So total momentum remains conserved for all values of z. So it's a smart, it's really the simplest deformation that you can do, which is consistent with momentum conservation and for which your particles remain on shell plus two. So if we look at the M2 as a function, so now we look at the M2 as a function. Actually, yeah, this is what I will not need. Yeah, so we're looking at this M2 as a function of z. So what can we say about this amplitude? There's a bunch of poles. And the poles come from when, in some channel, all the poles in an absolute comes from factorization. So we will have some diagram on the left where n at, sorry, let me put one at, two bunch of labels here. And we call these labels I. Can you see what's to I? Let me call this set of label I. And then there's some amplitude on the other side where n at shifted momentum and other stuff here. So all these poles occur when the momentum in that propagator go to zero. So the poles are p1 at of z plus pI total momentum in that channel. Is equal to zero. And that is a linear polynomial in z. Because yeah, p1, this is basically p1 plus pI square plus two. The cross term here gives you two z times what I call what you can write like this. Where this by definition is the vector which corresponds to lambda one, lambda tilde n of alpha dot, dotted into pI. So that notation should be, I hope it's sensitive. If anyone's confused by it, just please stop. But the point is that, yeah, that is what this notation is. So the pole condition is that this is zero. And the point is that this is linear equation in z. So it's easy to solve. That deformation is very convenient. So in the complex plane, we have a bunch of poles because in general the solutions are complex. There's a bunch of poles. And there's one pole for each channel in which one is on one side and then the other side. And the residue of each pole we know, at least recursively because it's a lower point amplitude. It's a product of two things that are lower points. So we know all the poles. So if we're lucky, if az goes to infinity, if az goes to zero, at z goes to infinity, then from all its poles, we can reconstruct what a is. And the question which demonstrates that is the question, if this limit is true, then this is true. And that is equal to a of zero plus sum over poles, one over z on the pole times a, times the residue. So from the residues, which are product of amplitudes, you can reconstruct the function. What we want, which is the function, the undeformed amplitude, the function here at z equals zero. Sorry, I'm tilting down. So the solution to that can be written very much, can be written in a compact form. Yeah, let me erase this. And the solution, basically, yeah, the statement is that a n is equal to sum over all the channels i of an amplitude on the left, which would depend on particle one, all the particles in i, and some intermediate pole label, which call n, some amplitude on the right, right here. And then, let's call this j, and then n at. So very related to the shifted momenta. And in front, there's just the pole, p one plus, but in that channel, square, which is just this propagator. So that's a recursive formula for three amplitudes. So I will keep that thing. Just need to keep the vertices here. Keep the four point amplitude. So, yeah, so I have 15 minutes left, so I will briefly discuss the simplest solution of this equation. And yeah, and there are, for the simplest elicities. And at the same time, I will introduce a useful concept, which is that of color order amplitude. So if we look at the four particle amplitude, it involves these gauge theory structure constants. And a good way to deal with them is to, so for example, if we have this structure constant, it's to write it as a trace of generators. This can be written as trace, if we, yeah, if we write that e and f and delta e f, you can write this like that. And using that delta e f is trace of t e, t f, you can write that as t a, t b, t c, t b. So it's a trace of product of two commutators. That gives you this environment up to a normalization, which I will not bother with, not to lose too much time. And this you can expand basically as this trace, t a, t b, t c, t d, minus plus and some other guys. And what is called the color order amplitude is the coefficient of this guy in the expansion of the amplitude. So there is n minus one factorial color order amplitude. So in other words, you write the amplitude as sum over trace, but sum over permutation of n minus one particles. This tree, it's called t a one, t sigma a two, t sigma a n times the amplitude times the amplitude for one sigma a two, sigma a n. And it's n minus permutation of n minus one guy because the trace is cyclic, you can always put one guy first. And actually the trace is a flip symmetry, it's really, it is this really number, but yeah. But basically each of these cyclic amplitude is defined and tested symmetry on the, on the reversing the order. So these amplitude here are contrary to the reason there are, the main reason they are useful is that they are unambiguous because if you look at this amplitude here, you can define c one is the coefficient of this and c two is the coefficient of this in a basis where there's all your two building blocks. But you could use the Jacobi density to rewrite this as this minus this. And then this quotient c one will be different. So the quotient of these f's is not uniquely defined because of the Jacobi identity. But this coefficient here are unambiguous. So they are, they are, and one, one simple way to see that they are unambiguous is to imagine you have SUN gauge theory where n is big enough, SUN c, if nc is greater than number of particles you're scattering. And if you consider the scattering amplitude where particle one is just above the diagonal like that, ta two is equal to the next guy just above the diagonal and so on. Then out of all these products, the only products which survive would be ta one, ta two, ta three and so on. So just by evaluating the amplitude on that specific configuration of color, you extract this. So that shows that it's completely unambiguous. Yeah, in string theory these are what would correspond to an amplitude with the topology of the disk. And the ordering is the ordering of the vertex operators on the outside of the disk. So this was inspired in the 80s by this. But yeah, so useful notation gauge theory. And for the four particle amplitude, if you compute this guy and write it in terms of angle brackets, turns out to have a very simple expression which is just one, two to the four divided by one, two, two, three, three, four, four, one. This is basically what this c one, yeah, it's basically what this s over t times this is. It doesn't come deeper than that. This is compact and elegant expression. And for solving the recursion, it's also relatively, this notation also helps because you have much fewer factorization channels which preserve given ordering. So if you want to compute, basically you have to write down factorizations which preserve the psychic ordering. So for, if you deform in N one, you only have three channel for the five particle amplitude. One of them is one, actually just two channels. One, two, three, four. And the other is one, two, three, four, to be five at. So in the planar limit, the number of factorization channel is very small. It's just number of particles minus three. And if you consider a partially simple elicity structure, the simplest, actually I don't need this anymore. Now that I've written this. So just as a one illustration of the recursion relation, discuss the amplitude where you have all, but two particles are plus elicity when everybody has the same elicity, but two guys. This will turn out to be a simple generalization of this formula. And what happens in this special case is that, let's write, let's put the elicities here. This is one, two. So one is minus, two is minus, three is plus, and these are plus. And this is minus, minus, plus, plus, plus again. What you see is that this is a four point amplitude with a minus, with three plus and one minus. And this, I haven't discussed it so far, but it's actually zero, and you can basically show by showing that it cannot have any pole. It's a, I leave that as an exercise. Give it a different argument, but yeah. This is zero, so this amplitude is zero. There's just one diagram, this one. And this is non-zero because we have a plus, plus, minus three point vertex, which is non-zero, and this is a minus, minus, plus, plus vertex, which we just computed given by this guy. So this amplitude, so let's evaluate this graph. So we get one over the total momentum in that channel, which is P four plus P five square, evaluating this factor in front. And one over P four plus P five square is one over four five, A of five pluses. Then there's a factor which is the amplitude on the left. So just write A four, one at two, three. Let's call this guy, intermediate guy, middle guy M. And then there's this amplitude on the right, which is the three point vertex given here, which is plus, plus. So the two plus, the two same go on top. Five is shifted downstairs, this four M, M. This we know, so basically we just have to evaluate this. So what is the momentum five on the solution? So, sorry, yeah. What is the momentum? We need to know, we need to know five at, and we need to know four. It seems like I miss one information here. Let me just check. One thing which we can see on the pole, actually very simple, is that since we've deformed lambda, what is zero on the pole on that white vertex here? We have that four five at, sorry, M five at is equal to zero on the pole. So that tells us that M, remember that M is defined after rescaling is, sorry, wait a second. Just, just, just wait, yeah. It's equal to zero and the five five at is not, sorry. Five at is deformed. Yeah, that's what I was missing, okay. So this deformation I deformed five, but not five at. Not five tilde, sorry. So this angle bracket with five is on deformed. The only thing I need to compute is this M. This five at is such that four times five at is equal to zero because I'm on this factorization pole. And that basically tell me, and also four M is equal to zero. So I can take M, well, I think I can take M equal to, up to a choice, up to an irrelevant phase because M lambda, this is a short answer for lambda, M is equal to lambda four alpha. That's true because they're proportional to each other and the overall normalization doesn't matter. The overall normalization cancels between, the overall phase ambiguity cancels between these two factors. So I can take this and what can I say about M? Well, the vector M is P four plus P five at. And it's a null vector. So to extract its lambda tilde, I can project it on anything. And the convenient thing to projecting on is P one, sorry, P four plus P five at. If I project it on one and that's convenient because then the shift disappears because the shift of P five is proportional to lambda one. I can just write it like that. And now I'm done. And okay, I actually have to normalize it such that this product is equal to this. And basically normalization, you can work it out but just whatever, the simplest thing you will think which cancel the weight with respect to one and four. I'm saying this because you can do this calculation but after you've done it once, you can basically guess the answer. So telling you how to guess it, I think it's more useful. So M is this. And if we plug this into this factor, a brief algebra shows that, well, when we compute four M, let's actually do it. Four five cube, four M, for four M, the four cancels here. So we have four five, one five. Four five, one five. And for the other guy, we also get four five, but one four. And now we have one four squared from this guy two times. That cancels this. That's one. The four five, two of them cancels. And this four five cancels this. So in the end, if I write there, you will not see. So I will write here. In the end, we just get that a five is equal to one over four five. This and this cancels. Then there's one five, four five, five one. And then there's four one on top. Times a four of one minus two minus three plus four plus. And again, this only lambda tilde one is shifted, but a four depends only on lambda one. So you can ignore the at here. So the solution of this recursion relation is very simple. Well, so a five is, and yeah, let's write it explicitly so that you remember the pattern. So basically the four one cancels this and the product gives you one two to the fourth times one two, two three, three four, four five, five one. So the cyclic product of denominators. And obviously this recurses and you can keep going. So that's for the simplicity, this class of, this family of amplitude is given by this one line calculator credited to parameters who guessed it from a error six point calculation in 1980s. It's very, they really have to pull all the tricks and you use extensive computer power to manage to get this result by famine diagrams. But the end result is rather simple. But you can apply this recursion relation not just to this illicit, but actually at five point they're more or less all simple, but even for six point you can have more complicated things. And for all of these illicit, this formalism give you rather compact formulas. And which just are incredibly simpler to what you would get from something famine diagram. So this is the efficient way to think about, about three answers. So not only you get answers quickly, but you get answers which are compact enough that you can actually stare at them. Okay, so I was planning to discuss some supersymmetry, but I would conclude here. So yeah, so all I've said today is there's no supersymmetry. It's very, it's very, very quite general. Discussed amplitude in three level Yang-Mills theory. And the reason trees are so important, well first of all, you see that they're quite simple. They're much simpler than what you would think from famine diagrams. So you want to use them as a building block to compute loops. So what I will discuss in next lectures is how you go from trees to loops and trying to explore this simplicity. And I will make contact with intergability in a special case of, so that's roughly the plan of these lectures. Thank you.