 Okay, so for this last lecture, we'd like to cover two topics, two short topics. One of them is momentum twisters, which are very nice variables, which in many of these calculations make your life nice, and the momentum twisters were invented in the context of this dual conformal symmetry in order to make it manifest. So they are very closely related to the old twisters of pin rows. So I recall that the dual conformal symmetry becomes visible when we look at the amplitudes in the dual momentum space. If you hold the momenta coming into an amplitude, like this, we're going to do region momenta yi, y plus 1, such that yi minus yi minus 1 is equal to pi. And then we have a conformal symmetry reacting on these y's. This is a very good, this is a standard trick to linearize the action of conformal symmetry, which is to introduce six vector, because the conformal symmetry is SO4 comma 2, let us just call it SO6, to be not to formal. To introduce the following six vectors, which is components yi mu, so four of the components are the one you start with. And the two extra components are two null coordinates, yi plus, which I would set to 1, and yi minus, which I would set to yi mu square. And the metric is such that this vector is a null six dimensional vector. And then conformal symmetry become, conformal transformation become rotations on this space. This notation is very useful, and people use it in the CFT bootstrap, for example, to do calculations in conformal field theory. So it makes conformal invariance easy to see, they're just rotations of this vector. So SO4 comma 2 rotation, conformal transformation are just these rotations. So we're already halfway to define the momentum twisters. Moment twisters are just the spinors of these guys. So if you have a null vector in six dimension, it's a mathematical fact that it can be written as a product of two spinors. So y is equal to, I'll call it two spinors, ZA and ZB. And what are the indices here? The way it works is that SO6 is like SU4, and the spinors are in the four. So these guys are four spinors. And if you take the anti-symmetric product of two such spinors, you get a vector. So a null vector can be written as an anti-symmetric product of these four, two of these spinors, and these are called twisters. You get two twisters, you make a point. In the context of amplitudes, the way this is used is that for a specific context, you define the following object. So we have all of these points. And to each of these points, each point corresponds to two twisters. But the twisters will be shared among neighbors. So we really just need n twisters, n momentum twisters. And they're defined by the following formula. The first two components are the lambda spinor we've seen before. And the bottom two components are lambda i, alpha. They are related to these y-coordinates, so the alpha index is contract, it says alpha and alpha dot index. And one thing that can be shown is that if you take Zi wedge Zi plus one, that gives a matrix, so that gives a two by four matrix, which is lambda i, lambda t, lambda i plus one dot, dot, dot, here. But the wedge product, you can shift, you can change the basis as much as you want. And this becomes proportional to one to one, up to just the reshuffling of the columns. This is proportional to yi alpha dot. So this is the way that the vector yi is encoded in this product. So given z's, you extract lambdas, extract lambdas, y's, and lambda tildes, so we can extract all the data given the z's, and you can go both ways because this map is explicit. So there's a one to one correspondence in twisters and on-shell kinematics. And momentum conservation is automatically solved because defining the momentum as differences like that, they automatically add up to zero. So it's a very nice variable. They solve both momentum conservation and on-shell conditions for you. Yes, there's a cyclic identification. So yn plus one is equal to y1. But it's not a constraint here. This constraint is automatically assigned. So I give you a set of y's, just have to give you y1 to yn, and n plus one you define as y1. There's no constraint there. So instead of having four constraints, now we have four symmetry because you can shift the y's as much as you want, and this equation is unchanged. So you solve the four constraints and replace them by four symmetries. I don't have so much time. So what I want to show you is some application of these twisters. So they're useful to rewrite the amplitudes in a way which makes the symmetry manifest. So the basic step, so recall that the amplitudes are proportional to, we can always pull out this supersymmetry delta function. This is like delta eight. We can always pull out this factor in front of the amplitude and define a rescale amplitude, which doesn't have this factor. The super amplitude has weight, so it receives some phase if we rotate the lambdas, and a very natural way to absorb this phase is to divide by this denominator. So we define objects like this, and then this Mn, we can write Mn depending on the amplitude. Because we have a recursion relation for A, one can derive a recursion relation for M. It's just a simple algebraic step to carry out what this factor does. So I give the answer. So the BCF recursion took the following form. We add that the amplitude for n particle, if we shifted n and 1, we did this shift. The amplitude was expressed as a sum of product of lower point amplitude, where shifted 1 was on one side, shifted n is on one side, and the sum run over all channels in which you can cut the planar amplitude. So it's a BCF recursion, and one can write an equation. One can write this equation at the level of the Mn's, and the answer is simply Mn, so Z1 goes to Zm. One term is just m2, one less twisters. That takes care of one channel, where you have three point vertex here. And then the rest, there's a sum over j, which runs from 2 to n minus 3, of, there's a pre-factor which replaced this propagator, which turns out to involve this funny bracket, which I will explain in a second, of five twisters, 1j plus 1, n minus 1, times the product of m on the left, which involves twisters 1, 2j, and the exchange particle, from two twisters here to this guy, and some m on the right, of mj plus 1, so that's the answer. So let's look at some cases. So the basic property of this bracket is that it's proportional to four fermions, so for mHv, all the fermions are 0. For mHv amplitude, which have all but two negativity gluons, these two negativity gluons are absorbed by this factor, and the recursion, we can ignore this term, and the recursion is simply m, mn, mn minus 1, and they're all equal to 1. So in the first lecture, we struggled to derive the mHv amplitude by the recursion, but now we're building on this, on the fact that we've already done this, and now the mHv amplitude is trivial. We've just pulled out this factor, that we work so hard to derive in the first place. For nMHv, now we have to work to first order in this bracket. So in this sum, these m's are just 1. So nMHv, the recursion is very simple to solve, mn is equal to sum over ij, 1i i plus 1. Some examples, m5 is 1, 2, 3, 4, 5, m6 is 1, 2, 3, 4, 5, which is m5 term, plus 1, 2, 3, 5, 6, plus 1, 3, 4, 5, 6. So that's nMHv. And you can keep going, then squareMHv gets more complicated, but you get some sum of products of two of these brackets. So this gives a very, very compact way of solving the recursion. And this object, essentially, with this compact notation, you can essentially fit the answer in your head, which is very striking. And what is this bracket? It's made out of the simplest SU4, and of the simplest SU4 slash 4 environment. One thing I haven't told you is that we need to supersymmetrize these twisters here. You can these twisters are explained in a good way, a good reference, which is still a good place to read about this. What was the original reference? It's a paper by August, 0905.47, 1472, these status. So this bracket is the basic environment. It's the basic SU4 slash 4 environment, and it's built out of the simplest SU4 environment, which is a determinant of Zi. You can make a supersymmetric environment, Ijklm is equal to, you made it out of these simpler environments, Ijkl, Ijklm, all the cyclic products up to M, I, J, K, and on top, the fermionic delta function, Ijkl, chi, or chi is this guy, chimm, some r-symmetry index, plus cyclic. So it's a rather compact object, and to evaluate it, it's just computing a bunch of determinants. So the amplitude at any number of points is expressed in terms of products of such determinants. So what is the upshot of this formula? So the following facts, so first the dual conformal symmetry is manifest. So for all, what, it manifested for all trees, any number of points, and this is just the SU4 slash 4 environment. They can write it algebraically as JAB equal to zero, where JAB is somewhere particles of Zi V over D, Zi. It's just the generator of rotation. There's a something which is not obvious, but is a true fact, is that there's actually a yang-yang symmetry, which holds for each term of the expression. It's true actually term-wise in this recursion. So there's something like a level one generator, which is equal to zero, and this level one guy is defined as sum over i smaller than j of essentially commutator of Zi D over Zi, Zj D over Zj, with AB in this case. And the simple way to see that this has to be there is that this contains in particular some elements of the ordinary conformal transformation. So by this variable, we made the dual conformal symmetry of the theory manifest, but the theory also is a conformal symmetry. And when you express it in terms of these variables, it takes this more complicated form. And when you commute these level zero and level one guy and commute level one guy with each other, you get an infinite algebra. So this was completely unclear. If you just compute Feynman diagrams or even from the recursion in the first lecture, it was completely unclear that the amplitude of this symmetry, but in this form, using these variables, the symmetry becomes another fact is that the recursion can be extended the level of the loop integral. And essentially, at the loop level, one adds a term. So at loop level, in the planar limit, you have all sorts of loops that you to answer that are labeled by extra variable y zero, y zero prime, y zero prime prime. And in the recursion, you get terms where these loops just get opened apart, just cut in the middle. And there's an extra term where you cut open one of these loops. There's a term that looks like schematically like that. One and go like that. That opens. It's not meant to represent any kind of insect, but yeah. So there's a generalization to loop level. And that recursion also makes this symmetry manifest to our loop order. So that's one way we know for certain that the symmetry is there. In addition to the T duality argument from a DSCFT that I mentioned previously. Yeah. So that's a good question. So at three level, it's linearly realized. At loop level, the loop integral is covariant under the symmetry, but that's the integral. When you do the integrals, funny things happen. And the statement is that the amplitude is invariant under a deformed yang-yang, which is not linearly realized. So it changes the number of particles. Loop integral, so I would just write. So then we're in a modified yang-yang, which is not linear. So we wrote it down to our, well, wrote down what was the modification in the paper by myself and building on previous work by Beiser and others. And the fact I want to mention is that this kind of structure, ii plus one, jj plus one occurs naturally when you solve this recursion. These kind of structures occurs naturally when you solve certain geometry problem. And an example is you have a convex polygon. I just tell you it's convex. It's a bit of a trailer example, but you have a convex polygon with points i, i plus one, i plus two, blah, blah, blah. The boundaries are i plus one. They involve consecutive guys. So in general, when you have geometry problems, which involve complex convex objects, occurs in convex geometry problem. Whenever you have some kind of convex problems, this kind of thing occur naturally. And this was generalized to the simpler problem was generalized to a more complicated geometry problem, which reproduced all the structure by Neymar Kenyamir and Erslav Tranka. So Neymar may or may not talk about it lecture soon, but yeah, so I would just mention it that this leads to an interpretation of the amplitude, the volume of a certain convex object, which was called the amplitude right now, whose faces somehow of the correct combinatorics that they reproduce all this i plus one, jj plus one pattern. It's unfortunate that we really don't have more time to discuss this, but I wanted to mention it. The last thing I want to mention about trees is that I hope I didn't over-emphasize the trees during these lectures. So I've this, you shouldn't get the impression that this business is about computing trees. This business is really about computing loops, but people, me, some people here, you do all this complicated computation, loop computations, you get some answer, you realize there's some amazing structure, and then you try to understand it. And then ultimately it always boils down to an amazing property of the trees that were there and you were sitting in your face for years and you just am not recognized it. So the trees are really important. Even though at the end really the business is about computing loops. Yeah, last thing I want to say about the trees, you don't really have to evaluate these things by hand anymore because there are computer packages which do that for you. So if you need to compute some tree amplitude for some process, so you would just want to stare at what the expression looks like. I mentioned this one package by my colleague, Jake Borgerly, bcfw.m. It's in 1011.2447, so you can get an analytic expression for any amplitude you fancy. Yeah, 2447. So that's what I wanted to see about about trees, three amplitudes. The second thing I would like to do to wrap up these lectures is to discuss the cost-pandamless dimension in N4. So what is this object? So scattering amplitudes have entire divergences. They actually have soft and corneal divergences. So the scattering amplitude is a well-developed theory for these divergences. Essentially they exponentiate. That's the main thing you should know about these divergences is that they exponentiate. And it's essentially due because of Wilson's pictures that there is a freedom at different length scales, they couple from each other. So very infrared radiation, they couple from hard processes that lead to exponentiation. In the case of a planar, the exponent in general, it's exponentiate, but you still have to compute the exponent. The exponent is particularly simpler in the planar conformal feed theories, but there's a completely general story. But for planar CFTs, the exponent takes the form of some coefficient in the anomalous dimension. For some reason there's a factor of 8. Some are particles. Then some log square pi dot pi plus 1 over infrared cutoff times some constant. And this pi dot pi plus 1 arises from sub-gluons emitted between two adjacent legs. If you want to know the general story as well as a sort of modern proof of this exponentiation, I recommend a recent paper by Fayage Solis. It's a well understood theory. And at the end of the day, when you compute physical scattering process, as usual, what matters is this cutoff scale becomes a renormalization scale. It becomes finite. And for example, if you have a process involving narrow jets with energy e, the exponent will look something like the cross-section will have an exponent, something like gamma cos 4. It's a factor of 2 because you square the amplitude to get the cross-section. And one of the log is collinear. So it will involve this cone angle. And the other log is soft. So if you have two jets and you have energy smarter than this, if you instead of the energy outside your jet is smarter than certain cutoff, so there's a soft log and there's a corneal log. In general, I'm too bad for jets of this form. And yeah, because much of LHG physics is about jets, these logs, and this coefficient is an important observable in quantum field theory. In n equals 4, it turns out that this gamma cosp can be calculated to all orders using intergability. And so to relate gamma cosp to a spin chain, for example, as I mentioned yesterday, the way to go is that this gamma cosp involves collinear physics. So it involves what is called twist operators, operators of low twist. Which, and it turns out that the soft divergence just turns out that they're related to large spin. So one has to look at operators of low twist and large spin. Just turns out. So for example, if one can calculate the anomalous dimension of operator trace of scalar field, a derivative to the s, and another scalar field where v is a null vector, that's a covariant derivative. If you know the anomalous dimension of these operators, which have twist 2 because the vector is null, the spin, if you know these at large spin, you know gamma cosp. So the dimension of O minus, sorry, dimension for s minus a spin goes to gamma cosp. There may be a factor, probably a factor of 2, log s plus a constant. So at large spin, there's a growth and gamma cosp is a coefficient of this dog. So in that way, this coefficient which occurs in amplitude is related to something can calculate from these local operators, which is mapped to this integrable spin chain. The fact that you can map this problem to an integrable spin chain doesn't mean that it's easy. There are many heart problems involving spin chains. It turns out, but this turns out to be a problem that can be solved. I wish I could tell you something about how it was solved and so on, but I think it's even more interesting to discuss the answer. And the answer is the following. Beisert, Eddin, and Stadecker around 16.7, and the answer is the following. Define a matrix, k i j, whose matrix elements are 2 j minus 1, some integral, d t over t, Bessel function of i, 2 g t, Bessel function of j, 2 g t, divided by e t minus 1. Then gamma cosp is equal to 4 g square times the n-verse of 1 plus k, the top left component of that big matrix. So k is an infinite dimensional matrix. So this is a bit formal, but let me explain. So in practice, the coefficient, the matrix element of k, let's compute the first one, for example. k 11, let's assume that the coupling is small, and g square is, this equation is lambda over 16 by square. If the coupling is small, you see that because of this exponential factor, we care only about t over order 1, and then the argument of the Bessel function is small. So we can, so k 11 involves Bessel function 1, and which is just linear in its argument. So k 11 is approximately 2 g, sorry, 2 j, this is 1, time integral n zero times infinity, d t over t, and the Bessel function gives g t square, divided by e t minus 1, and integral d t t over this is that of, remains that of function, 2 j, that's 1. So that gives 2 times z of function of 2, which is 5 square root 3 times g square. So the matrix elements of k are small in persuasion theory, and in general, k i j is over j g, g to the i plus j. So they get smaller and smaller. So if you want to compute this inverse, for example, to 20 loop orders, you just need to keep terms of other j g to the 40. So if you just keep, if you want 20 loops, just keep 40 by 40 matrix, invert it. I cannot do that, but the computer can do that. So the problem is essentially solved. And even when the coupling is strong, the matrix elements go to zero at large entries. So even if the coupling is 10 or whatever, you can still get numerics out of that matrix. And there are other ways to solve this at strong coupling. So what does the solution look like? It's really the first quantity which was successfully interpolated between weak and strong coupling. So on this axis, I put g square, which is lambda over 16 pi square. On this axis, I put gamma cos. At weak coupling, so we basically computed the two loop order here. Gamma cos is 4g square times 1 minus pi square over 3g square plus order g4. So it starts like 4g square. So it starts then only. At strong coupling, it goes, it damps and down. So it doesn't flatten up, but it damps down. But strong coupling, gamma is approximately, at weak coupling, gamma approximately 4g square. The strong coupling result is an agreement with strings in the mirror right here. What else can we say about this perseverance expansion? So this is a rather explicit result. One thing that people have extract from it is that the weak coupling expansion converge. So this is an agreement with the general argument that Marcos gave in his lectures. And it converts because we're doing large Nc. In general, the tear-off expansion at large Nc is expected to converge in some radius. And here we can actually compute what the radius is. And it turns out to be that g is equal to 1 quarter, which can also be written as lambda is equal to pi square. Or if you like alpha strong types quantity, alpha strong would be g square, young miss over 4 pi. That would be equal to pi over 4 Nc. So that's the radius of convergence of the series. And when you pass this point, you can see that you've entered a strong coupling regime. That's where the transition between weak and strong coupling. You can look at the strong coupling expansion. The expansion is the alpha prime expansion of string theory. This expansion turns out to be asymptotic. It turns out to be non-boral summable. The problem is that this is what Marcos described as a renormal line, which occurs because this string theory contains a asymptotically free SO6 model in it, which has a non-protective scale. Sort of like lambda QCD, which goes like m over e to the one over g at some, the coupling at some scale. So there is a scale which is similar to lambda QCD in which the theory becomes strongly coupled. And the scale is non-protective. And that causes the series to diverge. We can actually see more. The weak coupling expansion diverges because the function as a pole at g squared, lambda equal minus this. But it's a pole. Because the divergence is caused by a pole, you can rearrange the series to resum past it. So you can resum past the first pole. So if you do normal preservation theory, diverge and air, if you try to resum it, you get garbage past the reasons of convergence. But you can resum it all the way up to here. So that's resum, and that's the next pole. At strong coupling, the series is asymptotic and it goes bad very quickly. However, people, so Kaczynski and Basso, have analyzed the first non-protective correction. And when you, which as Marcos explains, only will define after you do the Borel resumption of the asymptotic series. So they've analyzed, so they did the Borel resumption including the, so you can do Borel resumption of the asymptotic series. Plus including leading non-protective effect. And then you get a curve which goes all the way up to here. So you can beautifully, you have beautiful agreement over some overlapping regime of validity. So quite a beautiful story. And all of these, all of these ingredients that Marcos described, you can see them in action. And that, this is pretty much what I wanted to say. My early, oh, I'm not good at improvising though. I wanted to be sort of finished on time for the lecture at the center. So one thing I could mention is that, so this is a beautiful story. And actually let me talk a bit about the story here. So it took a while for people to derive this, this solution. One of the difficult things is, so the basic ingredient in this spin chain is, so this one ingredient is what is called the scattering phase. So remember yesterday I mentioned this Eisenberg spin chain. We have operators of this type. We have, it's like a qubit. We have two different possible states on each site. And if you take two of these x's and you scatter them across each other, what you get is a phase. And you can calculate this phase order by order in the coupling, but in general from the integrality viewpoint, this phase, which depends on the momenta, from the viewpoint of integrality, this phase is an input. We can imagine calculating order by order, but you also can imagine being more clever and somehow knowing this phase. And the zero for the business of these guys was to try to guess these phase. So they made the first guess. As far as I understand the story, try the first guess and then they turn the crank and try to extrapolate the equation to strong coupling and it didn't work. It didn't agree with ADS shift. So then they took harder and they realized that there was some consistency condition which fixed this phase that they had not been careful enough with. So by imposing consistency condition, then they got exactly two solutions. And one of which they were able to show numerically that it seemed to be in agreement with the ADS shift. These two solutions differed at for loop. So differed by loops, one of which seemed to agree with the ADS safety and they had some other things they found nice about the solution. And somebody did the for loop calculation of this gamma cusp. So that was Bern Dixon and collaborators and they found exact agreement with it. And now, okay, now the story is settled and there are ways to derive this phase from consistency principles. It's been checked through five loop calculation that's been explicitly done. But this story is basically settled now. But it took a lot of, basically, to get this answer, they really had to put everything they had at strong coupling and everything they had at the time at weak coupling. There's really a lot. Everything that was available at the time went that result. The other ingredient from the integrability perspective is what is called, so there's a scattering phase. And the other sort of big thing is the wrapping correction. And if you just know the scattering phase, that doesn't fix completely the theory because if you have a chain, you have a short chain, you can have, so let's draw Feynman diagrams which describe the evolution of states with four scalars. At one loop, you draw diagrams like that and somehow the rest of the chain doesn't matter. And you could imagine you were doing calculation in an infinite chain. But at some loop order, you can have planar diagrams which wrap around the cylinder. And these you cannot predict from the infinite length chain. So these are extra effects. And it took years to people to arrive at the concrete way to deal with these effects. So now this goes into what is called PMU system, which I will not have time to describe unfortunately, but there's a really very interesting development. And one thing I personally like about this system is that it's formulated as in terms of NLST property. It uses the NLST property of a basic object, but it has this function of a certain auxiliary complex variable. And I like it personally because you can learn this equation without learning anything about spin chains or beta ansatz. You can really start from the actual equation, which involves NLST property. And they can try to work with it directly without going through this lot of baggage. But if you know this baggage, of course, it builds. Another direction is the application to amplitudes. So here, this cost-maneuveless dimension, which governs the infrared divergence of amplitude, is one way to relate amplitudes to local operators. But there's a much more general story. It's actually one remarkable things which turn out to be true, which is very useful for this, is the duality between scattering amplitude and Wilson loop. And the duality, it's really almost unbelievable. So if you have a scattering amplitude, it's labeled by momenta. And if you want to construct an object, you're told that this amplitude is invariant under conformal transformations of these Y coordinates. And it turns out that you can make this invariance manifest by instead computing an amplitude in terms of the polygon whose point on this Y is. Again, there's a vacuum expectation value in N equals 4, so it's very ended, in the same theory. So each of the edge of this null polygon has momentum p. And it's a closed Wilson loop by momentum conservation. In a way, this is the simplest thing you could write that as the right symmetry. On the other hand, you have to be really bold to write that down. And it was discovered by this duality in ADSCFT. Probably I doubt people would have guessed it otherwise. But it's a true equality, and you can check it perturbatively. You can do one loop calculation here. You can do this calculation here using some diagrams. And they miraculously agree. It's been checked extensively. And the fact that it works to our order is something which is made clear by this recursion for the loop integral that I mentioned. So there's no surprise anymore that it works. If someone does a five-loop calculation tomorrow and it doesn't work, that calculation was done wrong. It's not because it will not... No calculation can disprove it anymore. And this duality is very useful. One thing, a simpler way, and not simple, but one way that a group added by Basso, Sever and Vieira used to describe complicated polygons is to start by a simpler one. So if you start by understanding the four-partagon absurd, which is an odd square, if you understand the square very well, you can think of our point absurd as a small excitation of a square. So an hexagon, at least in certain limits, is a square plus a small bit. And a small bit you can think of as a local operator inserted on the edge. It's very simple. If you deform a Wilson line, you insert a field strength. So in the limit, the six-point amplitude, in the certain colonial limit, becomes this. It's a four-point amplitude, but dressed by some field strength. And now you can apply integrability to this, because this field strength, what dominates this... Let's say that there's some parameter u goes to zero in this limit. The amplitude will have some interesting scaling behavior. And that scaling behavior is governed by the dimension of certain operators which govern this limit. And this dimension, you can relate to this spin chain. So by computing all this dimension and opposing crossing symmetry, you can even try to fix the coefficients. And that is a long-standing project. They've made a lot of headway on this. So that's the direction. But in all of these developments, I think they remain... Yeah, I think I should stop here. Thank you.