 Okay, oh boy. Yes. Okay, so this makes the momentum conservation manifest But now we also want to make manifest that the sides of the polygon are null Right, and if you remember last time to make momentum null We could make that manifest with these spinor-holicity variables Writing p is lambda lambda tilde, but then it wasn't obvious that the momentum add up to zero So now what we're going to do is take everything that we've learned From our discussion of twister space yesterday and now apply it to this picture. Okay, so let's imagine What does this polygon look like? So this is a picture in spacetime. Of course, it's not our spacetime, right? This is this funny spacetime in which the coordinates of units of momenta Okay, so we'll call that we can call this the dual spacetime It's a spacetime with coordinates of units of momenta, but let's ask What does this picture look like of the bunch of points null separated in the corresponding twister space? Okay, well, what does any point look like in twister space? A point if you remember looks like a line in twister space So this is in spacetime the dual spacetime now does it look like in the twister space now? This twister space because it's associated with the spacetime in momentum space is called momentum twister space so momentum Twister space Well, let's say what it looks like. So that x1 is a line There it is. So this is the line one X2 is another line But it's not a random line because x2 is null separated from x1 So what does that mean about the second line? It intersects the first one, right? That's what null separation means So the second line is another line. There it is two, but it has to intersect one Okay, so this is a line two The third one is a line three, four, five, and so what we see is In twister space in momentum twister space this picture of the null polygon Corresponds to a whole bunch of lines one of which intersects the next Okay Okay, great. Now, how can I describe that picture? How can I give the data to totally specify that picture of a bunch of lines one intersecting the next? I just give you the intersection points Yeah, the last comes back and intersects the first. Yes, that's right. That's why we have a closed closed polygon, right? So all I have to do is hand you just the intersection points now now This is still in momentum twister space. I have to have to send z1 and some z2 z3 Up to zn and now I get to specify these z's completely freely, right? I guess totally free unconstrained bunch of data z's Out of which I can build this line z1 z2 z2 z3 z3 z4 up to zn z1 okay so the point xa is Associated with the line so the point in space time in this dual space time xa is associated with a line za za plus one in momentum twister space and Then so who is this edge then who is this null ray that corresponds to p1? That's the z's directly Okay, so remember that the that the lines in twister space are points in space time and points in twister space Are the null rays in space time and that exactly corresponds to the momentum? So now now we're done Because we've accomplished what we wanted now I just have to hand you a bunch of completely unconstrained data z1 through zn. They're just beautiful four-dimensional vectors Totally unconstrained and out of them. I can build a bunch of Null momenta that add up to zero Okay, which is exactly the data for the scattering process now Let's just do it very explicitly remember I told you that if you have a za and a zb in general that the xab associated with it is Mu a lambda b minus Let's say I have a z1 and a z2 Okay, then the x a a dot is mu 1 lambda 2 minus mu 2 lambda 1 over lambda 1 lambda 2 a dot a a dot a okay, so so so With this line z1 z2 is this point x x 1 2 associated with that line And so it's just a little exercise for you to work out What is the x a that's associated with these z's so let me write each z a as lambda a and some mu a And the exactly the way that we were doing before it's a four-dimensional vector. So so exercise is You can work out what is x a? Associated with z a z a plus one and Then after you work out x a you can work out. What is p a which is x a plus one? minus x a And and verify that p a is null Which means that you can write the p a as? lambda a lambda tilde a Where the lambda is just the lambda that we see here and there's an explicit formula for lambda tilde And let me just give you what the answer is lambda tilde a turns out to be Mu a minus one sorry lambda a minus one lambda a Mu a plus one plus cichlata. So plus a lambda a plus one lambda a minus one mu a plus lambda a lambda a minus justification a lambda a plus one mu a minus one over a minus one lambda a A log a plus one Okay, so it's a completely concrete formula now, right? You give me a bunch of, you give me, so give me Z a, which is lambda a and mu a, and from here I'm going to build a bunch of p a equals lambda a, lambda tilde a. The lambda a are just the lambdas here. The lambda tilde's are given by this formula. And you can check, so it's clear that the p squared is zero, and you can check that the sum of p a over a is equal to zero. So that's a beautiful exercise for you to see how that turns out with those little brackets there, upstairs, that the sum of p a is equal to zero. In order to show that, you need a small identity about these brackets, which is very simple to, very simple to approve. If we have those lambdas, all these things are two-dimensional vectors. So if you have any two-dimensional vector, you can write it as a linear combination of two other two-dimensional vectors, right? And that simple fact is the following identity, a b lambda c plus cyclic plus c a lambda b, so if I have some index here, plus b c lambda a is equal to zero. You can verify that that's just the identity that tells you how do you expand the two-dimensional vector. So how do I expand the two-dimensional vector lambda a as a linear combination of lambda b and lambda c, which are two other two-dimensional vectors? So this identity, many has many names, it's also Cramer's rule, it has many, many names. If you use that basic identity, then it's a beautiful exercise to show directly that this definition for the lambda tilde satisfies momentum conservation. So after this journey into twister space and all these spin or hostity variables and so on, we're finally done. I can hand you completely unconstrained data, a bunch of four-dimensional vectors, and from that I can specify the non-momenta that correspond to a scattering process. Now this problem of just the free particles, just the labeling of the free particles, is of course conformally invariant, right? Now it's conformally invariant in this dual space. This is a different conformal symmetry than the conformal symmetry in ordinary spacetime. This is a funny conformal symmetry in this dual space where the coordinates have units of momenta. So it's called the dual conformal symmetry. And what we just talked about now is pure kinematics, right? It's just labeling massless particles, free massless particles, nothing else. The data is nicely acted on by this dual conformal symmetry simply because conformal transformations preserve light rays. It's a big surprise and sort of miracle of the planar n equals 4 super Yang-Mills theory that that dual conformal symmetry is actually a symmetry of the scattering amplitude. And it's associated with this picture that the scattering amplitudes have a, which are amplitudes for gluons in the original spacetime, gluons scattering the original spacetime, can be thought of in some appropriate sense as the expectation value of a Wilson loop in this spacetime. This Wilson loop with these null polygonal edges in this spacetime. So because of that fact that there is this conformal symmetry in the actual answer, these variables have even more significance. They're not just beautiful variables that lets us talk about the external kinematical data in this free unconstrained way. They're also the variables on which this hidden dual conformal symmetry acts as simply as possible. The dual conformal symmetry acts on the momentum twister variables just as 4 by 4 linear transformations, SL4, just as we talked about yesterday. Okay? All right. Now, one last bit of kinematics before we move on to a geometry is this was all bosonic and now I want to tell you how to think about the supersymmetry. And I'm going to be a little brief about this partially because in the picture of the amputahedron, the supersymmetry is going to get completely bosonized, okay? So that's part of the beauty of the story, but anyway, but let me at least describe the variable so we know precisely what we're talking about. So now let's just talk about supersymmetry in scattering amplitudes. And here there's a lovely fact. So you all know that if you want to describe a theory using a Lagrangian, already if it has n equals 1 supersymmetry, you have to be smart, right? You have to be Wesson's Amino to figure out how to write down supersymmetric Lagrangians. And it should always make you suspicious in physics where you have to be smart. You shouldn't have to be smart in physics. Physics is smart. Human beings aren't as smart as physics is. So every time human beings have to be smart, it's because there's something you haven't understood about the physics. The point of life in theoretical physics especially is not to be clever and ingenious, it is to be simple and deep. And when you have to be too clever and ingenious, it means that you haven't understood the picture simply and deeply enough. And supersymmetry is actually a perfect example of this because if someone hands you a Lagrangian, if someone hands you a free Lagrangian, okay, it's already beautiful to realize that it has this amazing symmetry between bosons and fermions, but it's not so totally crazy. You have a free theory, okay, you can maybe discover a symmetry like that. But then at the interacting level, you all know that there's something weird about supersymmetry. If someone hands you the component Lagrangian for the Wesson Amino model, it's not obvious that it's supersymmetric. You have to go check these nonlinear variations of, right, there are these nonlinear corrections of the super transformations. And then, of course, people teach you about superspace and, okay, then you sort of feel that it's all been made simple, but it's a little funny. You introduce all these auxiliary fields that aren't really there in order to make the action of the supersymmetry manifest. And then, so you see, supersymmetry is already not like other symmetries. If someone hands you a global U1 symmetry, you know ahead of time, what does a global U1 symmetry mean? It's independent of what the theory is. You know what a global U1 symmetry does to a scalar field. You know what its variation is, independent of what the Lagrangian is. But supersymmetry is not like that. The symmetry depends on what the theory is. Depends on the Lagrangian, okay? And of course, it's strange that when you have more and more supersymmetry, it's harder and harder to write down a Lagrangian that makes them manifest, right? So there is this interesting tension between supersymmetry and locality. And with n equals 1 supersymmetry, we can have Lagrangians with some work that we have to learn about in courses. n equals 2, you have to be crazier and more Russian in order to learn how to build these harmonic superspaces, you know, and n equals 4, there's no Lagrangian, right? There's, so, which is manifestly supersymmetric. How weird is that, that the more beautiful the theory becomes, the harder it is to make the symmetry manifest at the level of the Lagrangian. So again, this is one of many hints that there's something wrong with the Lagrangians, but what I want to stress is that the story is diametrically opposite on-shelf, because when you're talking about what supersymmetry does to physical particles, you know exactly what it does, right? You don't have to be smart, you don't have to be clever. You know exactly what it does to the on-shelf particles. It takes a glue on and it converts it to a gluino, okay? So the action of supersymmetry on the on-shelf states is totally manifest. And that's why we can, when we have maximal supersymmetry in fact, it makes your life when we write down amplitudes as simple as possible, okay? Unlike with the Lagrangian where it makes your life impossibly complicated, okay? So let's actually jump immediately to theories with maximal supersymmetry. So we're talking about n equals 4 in four dimensions. And we know we have the Suzy algebra and this is going to be some p alpha alpha dot delta ij, okay? So i runs from 1 to 4 for the four supersymmetries. And up to now, we were labeling our scattering amplitudes for gluons, let's say, by their holicities, pluses and minuses, right? And that's slightly ugly, it has this funny discrete label on it. So if you want to talk about n particles, you have two to the n different scattering amplitudes for all the different choices of pluses minus for any given leg. But when we have an extra symmetry, we should diagonalize the symmetries as much as we possibly can, right? So it's not a good idea to label the states by the holicities. Now what do we have? What is the particle content we have? We have a positive holicity gluon, we have one positive holicity gluon, say, we have four positive holicity. We have four holicity plus a half fermions. We have six scalars with zero holicity. We have four fermions of a holicity minus a half. And we have one gluon of a holicity minus one. And I'll explain why I'm using these indices on them for a second. Okay, well, why am I using these indices? Because I can build all these states starting from one of them and applying the queues, okay? I have four queues, so I use this and I apply, maybe I should have put this downstairs and upstairs. Doesn't really matter. I apply the queues and I build these guys and I apply them twice and I build those three times, I build those four times, I build those and they're anti-symmetric and these are symmetry indices, okay? So this is the spectrum of states that I have. But then why should I label them by these holicity states? That's dumb, right? I should find some linear combination of them that's acted on as nicely as possible by supersymmetry. In other words, I should try to diagonalize as many of the symmetries as I can. That's why we label the states by momenta because we're diagonalizing translations. So let's try to also diagonalize the queues as much as we can. We can't diagonalize both because they anti-commute. So I can choose to diagonalize one of them. And I'm going to choose to diagonalize q tilde, okay? So we're going to say q tilde alpha dot j on some state. And it's going to be a Grassman coherent state, okay? So these a to tilde is going to be a Grassman variable. And this is going to be a to tilde j. And what's gonna, so this is a state that's labeled by its momentum, which is a lambda and lambda tilde, and some Grassman variables a to tilde. And so this state is gonna satisfy that q on the state is a to lambda tilde back on the state. So these are the nicest possible states that we can use. And concretely, that's just a particular linear combination of all these states. Which linear combination, the state a to tilde is just e to the, is roughly e to the a to tilde q on, let's say, plus. So this would be, concretely, it would be plus plus a to tilde i plus a half i. Plus one over two factorial, a to tilde i, a to tilde j, zero ij. Plus one over three factorial, a to tilde i, a to tilde j, a to tilde k. Minus a half plus one over four factorial, a to tilde, just abbreviate to the fourth, minus one, okay? A to ijkl, ijkl. All right, so it's just this particular linear combination of all of the holicity states. So that's the good way to label the external data when we have supersymmetry. And so what we're left with in the end is a very beautiful object. You see, when we have the actual holicities, even though we stripped off these color factors, and the momenta have this cyclic structure acting on them when we stripped off the color factors, the actual amplitudes aren't cyclically invariant. Because the choice of holicities breaks the symmetry between the different particles. But now when I label all the states by these Grassman coherent states, then the cyclic symmetry is manifest on everybody. So the super amplitude, this is a super amplitude, is just a function of lambda, lambda tilde, and a to tilde, which is completely cyclically invariant. And you can actually show that the correct weight under all the holicities means that if you rescale these by t and t inverse and the a to tilde, the tilde is there to remind you that it's like a lambda tilde as far as the scaling goes. So if you scale this by t, a inverse, that you should just pick up a factor t, a to the minus 2 times m of lambda, lambda tilde, and a to tilde, okay? And so these super amplitudes are functions of the lambda and the lambda tilde, and they're polynomials in a to tilde, right? They're Grassman variables, they're polynomials in a to tilde. And so because they're polynomials in a to tilde, we can expand them as a sum of something I'll call, it's for n particles, m of n and k hat of lambda and lambda tilde, and a to tilde, where this guy has four k hat a to tilde's in it. So it's a polynomial in the a to tilde's, there's a piece with no a to tilde's, four a to tilde's, eight a to tilde's, four k a to tilde's in general. Why is it a multiple of four? Because there's this r symmetry index acting on the a to tilde's, and so everything has got to be contracted with the r symmetry epsilon symbol. And so they come in fours, the a to tilde's come in fours. All right, so now I've introduced this variable k hat, a little exercise you can do is, so it's just literally polynomial. So if you expand it on the a to tilde's, it'll have a piece that has like a to tilde's, one to the fourth, a to tilde's, three to the fourth, and then multiplied by something. And then a piece, a to one, a to two, a to five, a to seven, times something else and so on. And the component amplitude, you just pick off by those coefficients, okay? So for example, if you want an amplitude with two gluons being particle one and particle five having negative helicity and everyone else having positive helicity, what do you do? To pick out the one with positive helicity, you put the a to tilde for those particles to zero. So a to tilde for everyone to zero, except for particle one and particle five. You want them to have negative helicity, so you extract the piece of the super amplitude that looks like a to tilde one to the fourth, a to tilde five to the fourth, okay? And just that component of the super amplitude is the corresponding physical helicity amplitude, okay? So there's no magic for how you extract, this is just a packaging. It's just a generating function to combine all of the helicity amplitudes into one object under which supersymmetry acts nicely. And in fact, this k hat corresponds, the component amplitudes in this piece of the super amplitude is the one that will correspond to k hat negative helicity gluons. But now there's a very nice thing, because momentum is conserved, there's a delta function for momentum conservation. And similarly, there's a super partner for the delta function of momentum conservation that we always have to have because of the supersymmetry, okay? So that means that just like normally when we write an amplitude, we pull off a delta function for momentum conservation that would just be lambda lambda tilde. There's also an analog of that, which is this guy, okay? So that's the super analog of momentum conservation. And one very quick way of seeing that that corresponds to a supersymmetry is that half of the supersymmetries correspond to translations of eta tilde, okay? And they correspond to translations of eta tilde by something proportional to lambda tilde, okay? So that supersymmetry is a translation in superspace. And the half of the supersymmetries that we're making manifest are just the translations eta tilde goes to eta tilde plus something times lambda tilde. And so under that variation, because the sum of lambda lambda tilde is zero, that delta function is invariant, okay? So that's the super partner of momentum conservation. So you see from here, this thing already has eight eta tildes in it. So you see without doing anything that every amplitude has to have at least k hat equals two, okay? Because we have to have these guys in it. Now, I'm going to pull out a factor here, which is just lambda one, lambda two. Well, let me just write it right as one, two, two, three, up to n one. And what does that do? Well, once I have that factor, this takes into account all these weights. Every particle appears downstairs twice, so everything has perfectly good weights. And so what's left is a function of, well, it's still a function of lambda, lambda tilde, and eta tilde. Except now this has 4k eta tildes in it, where k is equal to k hat minus two, okay? Okay, now, this thing out in front has a beautiful interpretation and meaning. This is the famous Park-Taylor amplitude that I told you about. And in fact, if you have an amplitude just for negative helicity gluons, well, two negative helicity gluons in everyone else positive helicity, or if we want the sector that has k equals zero or k hat equals two, k equals zero, that expression amazingly turns out to be the exact tree scattering amplitude, okay? So hundreds of pages of Feynman diagrams actually collapse to that single beautiful simple term that, as I told you yesterday, was written down exactly 30 years ago yesterday, okay? So that's already amazing that all the tree amplitudes for any n with k hat equals zero or in components with two negative helicity gluons is equal to that guy. It's so important to us now that we take it out. It's almost kinematical now. It's like the delta function for momentum conservation. It all comes out, okay? This thing which is leftover, let me call this r for leftover, remainder, okay? This thing which is leftover, now it has weight zero, weight zero under rescaling, under lambda, lambda tilde under, well, it has weight zero under lambda goes to t lambda, etc. And so that's the guy that we're going to be interested in computing. But now here is a wonderful surprise. So I told you that, so we talked a moment ago about how to express the lambdas and lambda tildes in terms of these momentum twisted variables, right? That makes the momentum conservation manifest. And that's fine now because everything is making sense on the support of this delta function for momentum conservation. It turns out that, just like I wrote down, so we have a z which is lambda and mu, now we can extend this to a super z. Some curly super z which is lambda, mu, and then underneath there's a bunch of Grasmann variables, eta, okay? There are four of these, and there's four of these. And the relation between a to tilde, so just like we have lambda tilde is a minus 1 a mu a plus 1 plus dot dot over a minus 1 a, a plus 1. Similarly, we have that a to tilde is a minus 1 a little a to plus 1 plus cyclic over a minus 1 a, a a plus 1, okay? And if we do that, the super momentum conservation is made obvious, just like ordinary momentum conservation was made obvious by the other variables. But the amazing upshot is that, so this function r can be written of lambda, lambda tilde, and eta tilde can be written as a function of the super z variables, z a, which is invariant under just invariant zero weight under the z a goes to t a, z a, given super symmetrically. So that's just, that's just the fact that we're using these variables to express. These are the good variables, right? As we said, unconstrained variables, which give us the external kinematical data, now also even super symmetrically. But the amazing fact, the non-kinematical fact is that r is actually invariant under a dual super conformal symmetry, which is SL 4 slash 4 acting as linear transformations, these super z. All right, so that's this famous hidden dual conformal symmetry of the n equals 4 scattering amplitudes. Okay, but now, now we're done. So, so this is the, we've set up the problem. What we'd like to do is understand how do we compute this function? How do we compute r? And of course we can compute it with finding diagrams in principle. And if we do it, we get all the horrendously complicated mess. But we'd like to know, is there a different starting point for computing them? And can we understand from this different starting point why the final answer looks like it comes from local evolution in space time and all the other things that we talked about before. Okay, now, this would be an entire lecture series by itself to tell you the methods that have been developed to learn to calculate these scattering amplitudes without Feynman diagrams. And these go under the names of various recursion relations, BCFW recursion relations, lots of other interesting ideas. And to actually explain how they work, as I said, would be a whole lecture series in itself. But let me tell you philosophically how it works. Is that, instead of using Feynman diagrams to do the calculation, that of course makes it manifest every step of the way that the answers you get are local and unitary. What you do is you use the fact that the answer is local and unitary to make an ansatz for the answer and then use a locality and unitary to determine it. So you're exploiting locality and unitary to determine the structure of the answer. For example, what happens, let me just very quickly sketch it. What happens for BCFW recursion is you say, let's say you have an amplitude. And something that you know about the amplitude is that if you take, if some subset of the momenta add up to be on shell, then the amplitude has to have a pole and has to factorize on the pole. So let's say the momenta P1 plus P2 plus P3 go on shell. So P1 plus P2 plus P3 squared goes to zero. Then this amplitude has got to factorize, has to go like 1 over P1 plus P2 plus P3 squared times a little four particle amplitude on the left and a four particle amplitude on the right with 1, 2, 3, 4, 5, 6 and some intermediate line which is going on shell. And you have to sum over all the helicities of whatever the particles are, H and negative H, that can run through the loop. So as I said, if you compute Feynman diagrams, this is just obvious because that's what the diagrams are doing. You're summing over all, but you have these virtual things going on in the middle, virtual particles going in the middle. But it's clear that if you go on shell, those virtual particles become real. And the whole apparatus of having the gauge redundancy and all the rest of it then guarantees that this happens. So that's the purpose in life of the usual formalism to make this factorization property manifest. And that's also how you can check if someone said I did the calculation, you can check if they're lying or not by whether this factorization happens. So but conversely, this can awaken you to the possibility that maybe you can forget about that prescription and just find a way to present an answer that makes that fact manifest, right? What you can do as an intermediate though is use this fact to determine the amplitude in a different way. This is a little bit like what you're familiar with, with functions of a single complex variable. When you have functions of one variable and you know what all the poles are, and the residues of the poles, then using Cauchy's theorem, you can just determine what the function is everywhere, okay? So here we know, now it looks superficially a lot more complicated, because we have a function of many, many variables. And in this multi-dimensional space, we know where the poles are, and how it has to factorize on the poles. But the beautiful idea of VC F and W, Brito Cachazzo, Fang and Witten, was to explore this multi-dimensional space just in one-dimensional directions, okay? So what you do, you see, generically if I hand you this thing, of course these momenta are not on shell, right? But all you do is you take two of the particles, let's say three and four. You take P3 and P4, and you deform them. You add P3 plus Z times some momentum Q, and then you have to take that momentum Q away from somewhere, so you take it away from the next guy, right? You choose Q so that everything is null. So you choose Q such that Q squared is equal to zero, and Q dot P3 and P4 is equal to zero. And so now what you have is a function of just Z. Of all the other momentum and a function of Z. But now as you move around in Z, somewhere in the Z plane, you're going to expose all of these singularities, all these poles, right? As you change Z, then somewhere you'll put these intermediate particles on shell. So as a function of this single variable, Z, you have a bunch of poles, but the poles correspond, each pole precisely corresponds. You know what the residue is, it precisely corresponds to the factorization of the amplitude to lower pieces. A crucial thing is to be able to control the pole at infinity, and that's where the real miracle happens, that there are no poles at infinity. That's not at all obvious from Feynman diagrams. If you draw them, you would naively think that things blow up at infinity, you have problems at infinity, but quite remarkably there are no poles at infinity. So you can completely reconstruct what the object looks like, just from the knowledge of these poles at finite locations, that just correspond to lower amplitudes on factorization channels. And so on this way, you can build a higher amplitude by gluing together lower amplitudes, and I want to stress how important that is. It's not like Feynman diagrams, we have these virtual particles in the middle. There's no virtual particles in this picture. You take on shell lower point amplitudes, and you smash them together in a way that keeps everything on shell, and you build higher amplitudes in this way, okay? So that's the basic idea of the BCFW recursion, and using this method, you can take any higher tree amplitude and compute it recursively in terms of lower ones that you've computed already, all the way down to three particle amplitudes. And the three particle amplitude is something that turns out to be completely determined by symmetries, by Poincare symmetry, okay? So you don't even have to think a structure is just nailed. So everything is just beautifully built up in this way. But as I said, many lectures to explain how that works properly, but that's the spirit. But something I want you to see from here is that in this way of writing, in this way of doing the calculation, you're first of all, it's highly non-unique. There's millions of ways of doing this recursion. Who do you choose to deform? And you can make different choices as you recurse down, okay? So you never get the answer in a unique form. Another remarkable feature of the answer is that each term in the answer, each term cannot correspond to a local scattering process in space-time. And the reason is that each term has poles that don't look like just a sum of a subset of the momenta. That's because each term is what you get from taking lower point amplitudes, glued together, which do have poles there, but then you shift and you deform away from them, okay? So each term has some good poles that correspond to local space-time processes and some spurious bad poles that can't possibly correspond to local processes in space-time, okay? And only the sum over all these collection of objects is free of those spurious poles and gives you the right answer. So these are all very evocative of some kind of picture, you know? And then when you study these individual building blocks more, the individual BCFW terms, you start discovering they have all sorts of connections, well, they have all sorts of remarkable properties that satisfy lots of highly non-trivial identities between each other. Each term by itself turns out to be, you see, in this picture, I've only talked about ordinary momenta. I didn't tell you anything about the dual space, dual conformal symmetry, where it just went all the way back to ordinary amplitudes. But each one of these terms, when you finally do the recursion all the way to the end, you get a sum of a bunch of terms. Each one of the terms by themselves turn out to be in varying terms of the dual conformal symmetry, okay? Each term sees the full infinite dimensional symmetry of the theory. There's an infinite dimensional symmetry. There's the dual conformal symmetry which was hidden. There's the obvious conformal symmetry in the usual space time which is manifest. And those two symmetries together commute to an infinite dimensional symmetry algebra known as the Yangian of SL4 or SL4 slash 4. This is a hidden infinite dimensional symmetry of planar n equals 4 super Yang-Mills. But each term of the BCFW recursion has it term by term, okay? So there's something obviously important about these building blocks. They know about all the symmetries, conformal symmetry, dual conformal symmetry. They don't care much about locality in space time, because they're not. They're piece by piece of spurious poles. The amplitude wants to be built out of them, but in not a unique way, you can do it in lots and lots of different ways to get the final answer. They have spurious pieces that cancel miraculously amongst them to get the final answer. What's going on? Okay, so what's the structure which is controlling this? In other words, of course, one answer is, well, we just derived it from quantum field theory, stupid. And so that's what it is. But it had a very strong sense that these objects were associated with another world of ideas, another set of structures, with their own logic for why they were being put together in this particular way. And that's the story of positive geometry, positive Resmonian and the Amphitheidron that I want to tell you about in the rest of the lectures. But yes. Well, for triumplitudes, it doesn't matter so much, because you color, order all of them, and everything is, there are no multi-trace pieces there. But at loop level, it matters. So at loop level, you can extend these recursion ideas also to loop level. Then there the clarity matters a lot. Okay, yes. Yes, well, many of these things are true for gravity as well. But maybe I'll make some comments about gravity at the end of the lectures. These things about triumplitudes are very universal. They're also true for gravity. And they have exactly the same flavor. That you have pieces that are built out of on-shell objects. They have spurious poles. They cancel miraculously some. All these things are very universally true. It's in the context of planarine equals four that we've taken the story as far as we can to the direction of finding the second starting point that it all comes from. But just to give you a flavor, let me tell you what the amplitudes look like. The simplest amplitudes with has k hat equals three or k equals one. Well, the very simplest one, well, the very simplest, the very simplest are k hat equals two and k equals zero. And at tree level, this function r that I just talked about before is just one. So, and that's what I just told you, that the tree amplitude for k hat equals two is this famous part Taylor amplitude that we just factored out. So that's the absolute simplest case. It's just been factored out to be one. So the next simplest case has that k hat equals three. So three negative helicity gluons and k equals one. Again, tree level. And these are called the, so these are called mhv amplitudes, maximally helicity violating for obscure historical reasons. And these, imaginatively, are next to maximally helicity violating, it's nmhv. OK, now there's a much more interesting formula for r that you can get from BCFW recursion relations, OK? And now I'm going to write it, and as I told you, it's not unique at all. You can get it in many, many different ways. But let me give you one particular way of doing it recursively. And this is going to be a function of these super momentum twist variables, as I told you. And you just have to take it on faith, but this is what it looks like. It looks like the sum of a bunch of objects, one i, i plus one, one i, i plus one, j, j plus one, sum over i and j. These are particle labels between one and n. So what is this funny, brackety object? Well, this is what it is explicitly. So in general, one, two, three, four, five, for any five labels, one, two, three, four, five is the following. So I'll explain all these things in a second. Bracket one, two, three, four, eight of five plus cyclic divided by one, two, three, four, two, three, four, five up to five, one, two, three. OK. Now, these ades are the Grassman variables that we're talking about. And this delta four is in the sense of delta functions for Grassman variables. So it's really the fourth power of that object in the Grassman variables. In the sense of delta of eta is equal to eta. But what are these brackets? Well, remember, the only symmetry that I have is sl4, sl4 slash 4. But there's the bosonic part of sl4. It's just z goes to any four by four linear transformation times z. So zi goes to Lij, zj. And so that means the only symmetric tensor that I have is the epsilon symbol, epsilon ijkl. And so the only invariants that I can build are things like zA, zB, zC, zD. And this bracket is just defined to be epsilon ijkl times all the z's contracted together. So those are the only invariants that can ever appear. These four brackets. And every time in the rest of this lecture, in the next, I have an object in brackets, it means it's contracted with the epsilon symbol, because we don't have any metrics. All we have is the epsilon symbol in this entire story. Very good. So it's a very simple object. Now, let's just talk a little bit about the kind of poles that we expect are going to appear in our answer. Because as we said, the only poles we expect to appear are when, for tree amplitudes, are when a subset of the momenta, let's say from PA all the way up to PB, go on shell. So when I have PA plus up to PB squared goes to zero, is the only time I expect to find poles. And in this picture of the polygon, of the light-light polygon, this just means if this is PA, if this is PA, and this is all the way up to PB, this just corresponds to, this is xA, and that's xB plus 1. The sum of all these P's is just xB plus 1 minus xA. And so we expect to have a pole when xB plus 1 minus xA squared is going to zero. When the corners of this polygon become not separated from each other, maybe I should have put a P B minus 1 here so that this is xB minus xA squared goes to zero. And what does that correspond to in this twister space? What does it correspond to when two points become not separated? Well, this means that the line zB plus 1 has got to intersect the line zA, zA plus 1. And so that pole means that we should have poles when zB, zB plus 1, zA, zA plus 1 goes to zero. When that four bracket goes to zero, it means that those four z's are linearly dependent, so they all lie in a plane, and therefore those lines intersect each other. So ahead of time, we know that the only poles of the amplitude should correspond to, let me write it as zI, zI plus 1, zJ, zJ plus 1 goes to zero. So somehow, even though the amplitude is given as a function of these z's, it should care about these funny i plus 1, jj plus 1 combinations. That's how it starts knowing that it has something to do with a spacetime, that despite the fact that in twister space, things are functions of points that it cares about the lines, too. It cares about these consecutive lines, and it cares about when the consecutive lines intersect each other in order to find the singularities in the right spot. But now let's look at this answer and see how many of these things are obvious in the answer. Well, first I hope you see it's an incredibly simple. We just wrote it down on one line. This would be thousands of pages of Feynman diagrams, got collapsed to this single, simple expression. But it has many funny features. First of all, the final answer is supposed to be psychically invariant. When you cycle all the labels over by one, but it isn't manifestly so, because particle one is special. I pick up particle one, the one i plus 1, not obviously cyclically invariant. Next is it obviously local. Locality means that the only poles are when i plus 1, jj plus 1 goes to 0. But it is not that either. If I look at the poles here from that expression, the denominator of the expression has pieces that look like, for example, one i, i plus 1, j. Then the next one, i, i plus 1, jj plus 1. This one is good. But then there's a bunch of other ones, three other ones, which are one i plus 1, jj plus 1. And then the other two. So this is bad. This is not a physical pole in general. That's not a physical pole. It has five poles, four of which are spurious. They're not physical. They're not local. But only one of which is physical. So this little example already shows you the kind of structure that we get when we stare at these nice representations of amplitudes that I alluded to. They're not given in a unique way. They're given by breaking some of the symmetries. There's many different ways you can do it. The pieces are incredibly simple, but the pieces aren't manifestly local. They have these spurious poles, which amazingly cancel on the sum. Now, in order for it to be true that all these things happen, these little objects have to satisfy identities. And in fact, there is a remarkable sort of six term identity that these five brackets satisfy, which you can use to prove all these things. For instance, if you could prove that this object is cyclically invariant, that would immediately tell you that it's free of these spurious poles. Because what the poles are, it depends on one here. And so if when you've cycled everything over one turned into two, then the location of those spurious poles would change. But so if the final answer is cyclically invariant, then it has to be free of those spurious poles. But the identity that tells you this is cyclically invariant turns out to be some very simple but sort of quite non-trivial to verify six term identity involving these basic brackets. So now what we're going to do in the third lecture, but before I get to the third lecture, I want to just do a tiny bit of projective geometry review. But what we're going to do in the next lecture is explain where this comes from. And just to give you a highlight, we're going to see that this answer is actually the volume of some geometric object. It's the amplitude, but if you never knew about the amplitude, we're going to associate. You give me the external momentum twister data, extended in some specific way, but you give me the data. And from that data, I'm going to construct a certain geometric object that doesn't even live in twister space. It lives in a slightly more abstract space. And the volume of that object is going to be the scattering amplitude. But in order to compute the volume, if we've never heard of the scattering amplitude, and we just want to calculate the volume of this object, you want to break it up into little pieces. In this case, it's a polyhedral object. It's literally a polytope. So you want to break it up into simplices, the same way that if you want to calculate the area of a polygon, you'd break it up into triangles. If you study the object, you discover there's a certain set of simplices that fit into it in a nice way, which are labeled by 1 i i plus 1 j j plus 1. And though they just fit into the object, and when you compute the volumes, it's exactly this formula. And the fact that the answer doesn't depend on the way you do the recursion is simply that you can triangulate this one underlying object in many ways. So that's what we're going to talk about. And that's the beginning of the story of the amplitude. But before getting there, I just want to spend the last few minutes here just giving you a very simple crash course on some aspects of projective geometry. Actually, these aren't going to be super critical for the necessarily brief exposition I have to give you. But just to make you a little bit more familiar, it's extremely, extremely simple. But it's sort of ironic that we are more familiar with much fancier geometries. We're more familiar with Euclidean geometry, or Monin geometry. These are much, much fancier geometries than projective geometry, which has almost no structure in it. And these fancier geometries were understood in the 1800s while projective geometry was understood in the 1500s when people were trying to learn how to draw in perspective. So if you look at really old drawings, people socked the drawing things. And they didn't look like the world you see. And then people had the great idea, what if I just draw what I see, literally? So you imagine everywhere you see a light ray comes into your eye. You just draw what you see there, and so on. And that gives you a picture. And depending on the plane that you project, all those light rays coming into your eye, you get a different picture, and you can draw from perspective in different directions. And that's the beginning of the idea that you should think of the geometry as a bunch of rays passing through an origin. And that's exactly what projective geometry is. So people like Des Alves and others in the 1500s, partially motivated by trying to learn how to draw more nicely, described projective geometry. But just a few concepts, just a little bit of projective geometry 001, OK? So let's first imagine that we want to, the basic point is that anything you do in geometry, any plane, any geometric questions that don't involve distance, are best thought of projectively. And again, the idea, let's say I'm talking about P2, I want to think of the space of all lines that go through the space of all lines that pass through an origin in a three-dimensional space, that if I intersect with some plane, give me a bunch of points in two dimensions, OK? So this would be the example of the two-dimensional projective space. But what does it mean algebraically? Concretely, it means that I can imagine that I have a bunch of z's, let's say, in this case, three-dimensional z's, if I'm doing plane geometry, zi that are identified with t times zi. Then in practice, it means that, as we said yesterday, if I have a z0 at the top and a z1 and a z2, then I can always rescale the top to 1. And then the rest of the variables I can call, let's say x1 and x2. And these will be, again, just points in the plane, OK? x1, x2, and so on, OK? But why is this useful? It's useful because it allows us to unify, in one way, all possible linear structures we can have on the plane. And in particular, it allows us to say that any two lines intersect. Any two lines intersect. They might have an intersection pointed infinity. Because the central novelty of projective geometry is just that we add all the pointed infinity are added to our plane. And accessing the pointed infinity just corresponds to choosing the pieces of this three-dimensional space that are not accessible in these coordinates, or where z0 is equal to 0. But that's just another chart, which you have to use to cover the other parts of the projective space. But what I just want to spend five minutes doing is showing you how practical and useful it is. Even if you forget about all the twisters and amplitude and things like this, whenever you're doing any problems where you have to solve simple geometric problems on the plane or in three dimensions, it's much better to think about them projectively and solve the problems projectively. Let me give you some examples of this. So a very important point here is that there is an SL3 symmetry that acts on these z's. And that SL3 symmetry is the largest symmetry there is that maps straight lines into straight lines. So this is the crucial point. If we went the other way around, if someone just handed you a bunch of points on the plane, there's a two-dimensional plane and asks you, what are all the symmetries that are that map straight lines into straight lines? Then you would naively think, well, there's translations and there's two-by-two linear transformations. And those are all the symmetries that map straight lines into straight lines. But it's true, those do map straight lines into straight lines, but they have the additional feature that they map parallel lines into parallel lines. And if you actually just do the responsible exercise of the most general variation you can allow and just ask it to preserve straight lines, you discover there's a larger symmetry than SL2 cross-translations. There's an SL3 symmetry which maps straight lines into straight lines. But it has the feature that it doesn't map parallel lines into parallel lines. It'll take parallel lines and map them into lines that intersect. So this SL3 symmetry is a little bit hidden. It's a little bit hidden symmetry of the plane, which is the biggest symmetry there is that maps straight lines into straight lines. And you make that hidden SL3 manifest by thinking projectively, thinking of these things as the points in your two-dimensional space as secretly lines through the origin in a three-dimensional space with this SL3 symmetry acting on them. In order to go from this big projective space to the affine space, you have to specify some line at infinity, well, never mind. We'll talk about it in a second. So the SL3 symmetry is very, very important. And it also means that the only invariant tensor that we have, let's say, is just epsilon i j k, nothing else. So there's no distances, there's no metrics. The only thing we can imagine using in this whole business is the epsilon symbol. But let me just give you just a few minutes, just some concrete examples for why this is useful. We talked about what points are. Now what is an equation for a straight line? A straight line, if you're in junior high school, a straight line you say is ax plus by plus c equals 0. That's a nice equation for a straight line. And so always when you do these things are the pieces that have the x and the y and then there's some constant pieces left over. Now this equation that defines a straight line has a very nice projective interpretation. If I now introduce z to be 1xy and this has a lower index and a wi which has an upper index to be cab, then you see that this equation is just wi zi equals 0. So everything with these constants can secretly be thought of as something that was projective and you just rescaled the top component of the vector to be 1. So this is the equation for a straight line. In other words, if the w's are fixed, all the z's that satisfy this equation line a straight line. So what specifies the line is if the z's are points with indices downstairs, a line is a point with the index upstairs under the SL3 transformations. Now, so far so good. Now let's just do a few little exercises so you see why this is so useful. Let's say I give you two points. Here's a point with coordinates x1, y1. Again, just normally in the plane, an x2, y2, and you want to know what is the line connecting them? What is the equation for the line connecting them? OK, well, this is simple. We know how to find the equation for the line connecting them. But how would we think of it projectively? Well, what do I need to do projectively? Projectively, I have some z1, i, and a z2, i. Now what does it mean? What does it mean to build a line? I want to have an object which has an upstairs index. Sorry, these were downstairs. I want to build a w with an upstairs index which satisfies that w z1 is equal to 0 and w z2 is equal to 0. How can I build an object with one upstairs index out of something with two downstairs indices? I use the epsilon symbol. So wi is just epsilon ijk z1j z2k. And it just practically tells you exactly what the line is. And believe me, these things are trivial, but it's much easier to do it like this than to stare and draw a slope and an intercept. Especially if you're going to a three or four dimensions, well, it's exactly the same. If you want the plane containing three points, it's just epsilon ijk l, z1i, z1j, z2k, z3l, and so on. OK, let's go backwards. What if someone hands you two lines? Here's a line, w1, and here's a line, w2. And they intersect at a point. How do I find the intersection point? That's exactly the same. The intersection point for 1, 2 would just be downstairs. It would be epsilon ijk w1j w2k upstairs. And you also see that there's an obvious symmetry between lines and points. And in general, between planes and points. Planes would be an object in a higher dimensional space. Let's say in three dimensional space would be p3, p4. Points would be things with downstairs indices. One downstairs index of plane would be something with an upstairs index. And so obviously, you can switch points and planes. And everything works fine. I'll just say one last thing before I stop, just so you really see how cool and useful these ideas are. This was, again, as I said, just all meant to familiarize you a little bit more. So this is one that I'll use. I'll do just two more things. One that I'll use in a moment in the next lecture. Let's say that I want to, I have a triangle, z1, z2, whose corners are z1, z2, and z3. And I want to talk about some point on the inside of the triangle. And let's say this point had a mass. I'll call the mass c1. This had mass c2 and mass c3. I'm using c for a funny reason. But anyway, normally I'd write the center of mass here as something like c1, z1, plus c2, z2, plus c3, z3, over c1, plus c2, plus c3. So when we do these, there are these 1 over the sum of the masses. But what is this as a projective statement? As a projective statement, it's that yi is just c1, z1, plus c2, z2, plus c3, z3. And I don't have to divide by c1, plus c2, plus c3, because that's taken care of by the projectivization. If I put that the z's, if I put that each one of these are 1, z1, and so on, then the y that I get would be c1, plus c2, plus c3. And then c1, z1, plus c3, z3, which is projectively the same as 1, and c1, z1, plus c3, z3, over the sum. So you don't have to do these manual ways of making things have the right units and so on by hand. Everything is just taken care of by the projectivization. I'll give you a final example that you can play with just as an exercise. Here we talked about lines and points. What is a conic? Let's say I'm on a plane, and what is a conic? Well, all the points on the conic are something that satisfy a quadratic equation. So all the points on the conic satisfy an equation, like cij, ziz, j is equal to 0. So what specifies a conic is a 2 by 2 symmetric matrix. Sorry, it's a 3 by 3 symmetric matrix c. So a little exercise for you to convince yourself, it's very simple. Let's say you have a conic and you have a point on the conic. Let's say this point x is on the conic. What is the line which is tangent to that point? And now this is starting to become much more of a pain in the ass in the usual way of doing high school geometry. But what is it from this point of view? There's a point on a conic, and from it I have to get a line. So I have to take my data is cij, and then I have this x with a downstairs index j. And out of it, I have to build something with an upstairs index to be a line. Well, there it is. That's the line. That's the line that's tangent to the point on the conic. And that's the point of playing, when you think projectively, all you have to do to do geometry problems is just put the objects together and contract the indices the only way you possibly can, and everything always works. And OK, so that's it for projective geometry 0, 0, 1. Then we'll start with the proper geometry in five minutes. Thanks. OK.