 Okay, so let's get started for the last lecture. Thank you. There are many excellent review articles on this subject. Here's a collection of notes, actually a book now that you might find useful. So let's just recall where we were at the end of my previous lecture. So the kinematic data for n-particle massless scattering is equivalent to a collection of n-ordered points p3. And one way to think about this is as an n by 4 matrix where you have n columns and 4 rows, and each column consists of the 4 component homogeneous coordinate of some point in p3. Now while you can do this for any massless quantum field theory, something very special happens in particular quantum field theories, in particular n equals 4 super Yang-Mills theory. There is a surprising symmetry called dual conformal invariance, and this amounts to a symmetry that acts on this matrix via left multiplication by, acts via left multiplication by GL4, let me say on the Z matrix. This is a highly non-trivial symmetry. When you work backwards, so what it means is that if you calculate any scattering amplitude as a function of the kinematic data, it has to be invariant under multiplying this n by 4 matrix by any 4 by 4 GL4C matrix. When you translate back to ask what that implies in the other sets of variables that we've considered, it amounts to, remember we defined these dual momenta that I called x, where pI was xi minus xi plus 1. Dual conformal transformations amount to conformal transformations in x space. So I'll just write it here, conformal transformations x space. Notice it's well known that n equals 4 super Yang-Mills theory is an exact conformal field theory, but this is not the ordinary conformal symmetry in position space. Even though these have been given historically the somewhat confusing name x, these are not position space variables. These are more naturally thought of as momentum space variables. In a sense, this dual conformal symmetry is conformal symmetry in momentum space. So what this means is that it's convenient to introduce a notation where we let ABCD be permanent of ZA, ZB, ZC, ZD. So you just take these four component objects, put them together into a 4 by 4 matrix, and calculate the determinant, and that's what this notation will indicate. So under any GL4 transformation, these things will always transform by an overall factor, namely the determinant of that matrix. Now these Zs themselves are only homogeneous coordinates on P3. So these must always appear in combinations like, you know, I'm going to make up something here, ABCD, EFGH over ABGH. Okay, so you can never have one of these things alone in any formula for scattering amplitude because it's not a well-defined variable keeping in mind that these are homogeneous coordinates and you need this to all be invariant under arbitrary projective transformations. So good variables are ratios of this type. So we'll encounter these things throughout the rest of this lecture. And notice that, importantly, if you were to do an arbitrary GL4 transformation on this so that you pull out a factor of the determinant, that will always cancel in these kinds of ratios. Okay, so the main thrust here I just wanted to introduce some notation. Oh, and before I go on, at the end of this lecture we're going to be doing some fun examples where we play games with solving equations involving these kinds of brackets. So it's important to get some kind of geometric intuition for them. So if you go to a patch in projective space, it's essentially R3 plus some pieces of infinity. So you can geometrically think about these points in twister space as if they were points in R3. And then you can use some geometric intuition. So the geometric intuition tells you that A, B, C, D is equal to 0 if and only if the points A, B, C, D lie in a plane. Okay, so if you have any three generic points in R3, generically they will lie in a plane. Now if you add a fourth point it's a non-trivial condition that all four of them lie in a plane and you can think about this as expressing that condition. Yes. It is invertible for generic kinematics. These things might blow up absolutely. So scattering amplitudes are supposed to be complicated analytic functions that have poles and branch cuts. So we do expect scattering amplitudes to have singularities and the study and classification of those singularities is very important. But let's always for these lectures assume we have generic kinematics. Or you can think about it, I mean another way you can think about it if the lines A, B and C, D intersect. Here I'm introducing a new notation. I mean it's implied in that sentence but I'll state it here explicitly. Notation A, B means the line through the points A and B. So you can get a geometric intuition for what these brackets mean and we'll be exploiting that later on in today's lecture. The thing we want to play with are loop amplitudes. So any loop amplitude, if we think about it in terms of Feynman diagrams, schematically any L-loop amplitude can be written as integral where in capital D space-time dimensions over L-loop momenta and then this thing is the sum of all L-loop Feynman diagrams. Now again we hope to never actually have to add up Feynman diagrams and compute them but we know that they exist and we know that you could do this computation for any amplitude you're interested in and if you did do that computation you would add up all your Feynman diagrams and you would get something and this is the thing we're interested in. This is what we call the integrand or let me write the L-loop integrand and importantly it is a rational function and we know that just because you can choose your Feynman rules to be manifestly rational functions. You can have various numerator factors that depend on momenta that typically happens in gauge theory and then you have propagators in the denominator. So it is a rational function of the external kinematic data that would be P1 through Pn and the loop momenta L1 through LL. We're integrating over these loop momenta from minus infinity to infinity in all components but in today's lecture I mean there's a huge body of mathematics about carrying out that integral and we're not going to have time to discuss any of that. What we're going to discuss today are the properties of this thing as a rational function and how to determine it easily, absolutely not and in general it doesn't converge. Alright so I want to do one example in order to motivate how fantastic this momentum-twister notation is. So often we like to compute integrands not by summing Feynman diagrams but by guessing. Here I'll explain computing integrands by educated guessing and a more sophisticated and technically correct word would be called generalized unitarity. This is a method that has been used for decades especially by leading practitioners of the art like Bern-Dixon and Kosovo and we're just going to do an extremely simple example. So let's consider a one-loop four-particle amplitude. So here this gray blob what it means is the sum of all one-loop Feynman diagrams. Now if you're in a complicated theory like n equals 4 super Yang-Mills theory that can be a rather non-trivial sum. There will be several dozen Feynman diagrams contributing to it and each one will be quite a mess. Let's guess this turns out to be true but let's imagine that this sum of Feynman diagrams if you would literally add them all up what if they somehow cancel and collapse into a single scalar integral. So let me finish drawing this and then explain what I mean. It means a constant to be determined. I'll need to define an L here. Okay this picture stands for this Feynman integral. So if you literally translate this scalar Feynman integral as an integral over the loop momentum that's what it means. Okay so let me take a step back and explain the motivation what's going on here. The poor man's version of the generalized unitarity method is simply to say well gosh I really don't want to compute all these Feynman diagrams. Let me try to guess what they could possibly add up to. So what I'm going to do is I'm going to try to solve for this coefficient C and if I get a solution that's good. That doesn't prove that I'm correct but maybe I'm on the right track. The basic idea is let me be a little more sophisticated. We know this thing is a rational function so what I can do is I can make some educated guesses about some relatively simple rational functions that might appear. I could make a list of them. I could take a linear combination of them and try to determine all the coefficients. In this particular example it's kind of trivial because we'll see that in fact you only need one single rational function. In a more complicated example that we'll do in a moment you would have some more complicated amplitude here and you would make a guess. Oh well there are six obvious rational functions I could think of that it might involve. So let me try to guess that it's a rational function of a linear combination of those six and then try to determine the coefficients somehow. And if you find a self-consistent solution you're on a good track. If you find there's no solution then your basis is not big enough or what you thought was a basis was not actually a basis. So the basic idea here is we're going to probe these rational functions by looking at their singularities. So we'll probe this equation and compute C by looking the singularities of both sides as rational functions. So specifically we'll integrate both sides over L. But instead of the physical contour if you were actually wanted to calculate the value of the scattering amplitude you would integrate either side over loop momenta ranging from minus infinity to plus infinity with a particular I epsilon prescription that's absolutely crucial for getting the physics right but I'm not going to do that at all. I'm going to integrate both sides over a completely different contour in the complex plane. I'm going to integrate over a closed contour in the complex, well actually it's in C4 because I'm going to treat each of the four components of L as an independent complex variable and then instead of integrating over the physical contour which is the one that goes mostly along the real axis in all of them I'll integrate over a closed contour around the locus where L squared equals L minus K1 squared equals L minus K1 minus K2 squared equals L plus K4 squared equals 0. In other words basically I want to compute a residue. It's a little fancier than a residue because it's four complex dimensional. If I had two rational functions of a single complex variable Z and I told you I'm going to integrate both sides over a little circle that goes around some point Z naught in the complex plane that's a fancy way of saying I'm going to compute the residue on both sides and compare them. Here I'm just doing essentially the same thing but in a higher complex dimension. Each of these is a ration... So the sum of all Feynman diagrams is some rational function of L and this thing is a rational function of L and what I'm going to do is I'm going to compute the residue on both sides of this equation around the locus where these four are zero because I need four conditions to completely localize my loop integration variables to a point. So that's the logic here. Yes, the locus will be... So let's determine what that point is. So these four equations... Actually it's two points. They're quadratic equations so you get two solutions. These four equations have two solutions and they are given by L1. That means the first solution and L2, the parity conjugate. Okay, let's check that these are solutions. Okay, first of all they have to satisfy L squared equals zero. They have to be null. Both of these are clearly null because I've manifestly written them in a form that's an outer product of two component objects. So thinking about this as a two by two matrix in AA dot indices, yes, the determinant of this is zero. Check. Okay, so we've satisfied the first equation. Now let's look at the second equation, L minus K1. Well, and let's just look at this one for a moment. If we do this, that's obviously also null. So let's do L1 minus K1. This is 1, 2 over 4, 2. I'm going to go ahead and factor this out. It's lambda 1, 1, 2 over 4, 2, lambda 4 tilde minus lambda 1 tilde. That's also null because, again, it factors into an outer product. There's nothing to check. This one is also obviously null. And the same holds for L2, obviously. L minus K4, sorry, L plus K4 is obviously also null for both L1 and L2. So the only non-trivial thing to check is the last one. Let's check L1, L, and I'll do the first solution. The solutions work essentially the same way. So this is 1, 2 over 4, 2, lambda 1, lambda 4 tilde minus lambda 1, lambda 1 tilde minus lambda 2, lambda 2 tilde. This is equal to lambda 1, 1, 2, lambda 4 tilde minus 4, 2, lambda 1 tilde minus 4, 2, lambda 2, lambda 2 tilde. I'm combining this all over the common denominator 4, 2. Now this piece here, you can simplify it with what's called the Scouton identity. Scouton identity says that AB lambda tilde C plus BC lambda tilde A plus CA lambda tilde B equals 0. This identity is true simply because the lambda tildes are two component objects. So if you have any three of them, they can't all be linearly independent. There has to be some linear relation between them. So if I've got lambda tilde A, lambda tilde B, and lambda tilde C, they can't all be linearly independent. There's a linear relation between them and this is the linear relation between them. So you can simplify that with the Scouton identity and this simply becomes 1, 4, lambda 1 minus 4, 2, lambda 2. Everything factors now over 4, 2. And this is again null because here it's written in factored form. Are there any questions about this? So I've hopefully convinced you that this and this are the two solutions of these four equations. They're complex conjugate solutions or parity conjugate. It's the same thing. So remember that in general we think of our lambdas and lambda tildes to be independent complex variables. So this is some point in loop momentum space that's in general nowhere near the physical integration contour. It's way off somewhere else and this is at the complex conjugate point. But what we're going to do is compute the residue of that thing around each of these points. Separately. So first let's compute the residue of the right-hand side. This is a fairly simple calculation. It's 1 over, it's a multi-dimensional residue, so it's 1 over the determinant. If I define f1 to be l squared, f2 to be l minus k1 squared, f3 to be l minus k1 minus k2 squared and f4 to be l plus k4 squared, then this residue is 1 over the determinant of the partial derivative of fi with respect to dl mu. This is a 4 by 4 matrix. I runs from 1 to 4. This runs over the four components of the loop momentum, 0, 1, 2, 3. So you just compute that matrix of partial derivatives as a 4 by 4 matrix and then you evaluate it at the pole. It's a very simple exercise and I'll just quote for you the answer. It's 1 over k1 plus k2 squared k3 plus k4 squared. So the residue on the right-hand side is going to be literally c times this. What about the residue on the left-hand side? We did a similar calculation in my lecture earlier this morning. It's the same idea that goes into the residue calculation of the tree-level BCFW relations, hand side. We have the residue 1, 2, 3, 4 equals... Okay. Now this gray blob means all possible one-loop Feynman diagrams. There are many of them. Clearly, the only Feynman diagrams that are going to contribute to that residue are the ones that have four propagators of the form... Where was my... I was pointing up here. Only Feynman diagrams with this topology, meaning only Feynman diagrams that have this propagator and that propagator and that propagator and that propagator. Only those can possibly contribute to the residue. They just throw all the others away. They don't contribute. And so here you literally get... Again, anytime you have something non-singular, you can pull it right out of your residue. So when I draw this picture, this picture means a product of four tree amplitudes. So when we did BCFW this morning and we calculated the residue, we got a product of two tree amplitudes, one on the left and one on the right. Here we're doing a more complicated, higher-dimensional residue calculation, but the idea is the same. The residue of this is going to give this product of four tree amplitudes times the residue of the remaining propagators here, but that we've already computed as one over ST. Let me quickly assemble everything here. Let's specify a definite helicity configuration. For definiteness, let us look at one loop, one minus two minus three plus four plus. You could make any choice you want, but then we have two possible helicity assignments. I also mentioned this earlier today. Here I should really have a sum over helicities of the internal leg, because if you look at all Feynman diagrams that have this topology of propagators, there are several possible contributions. You could have positive helicity there, negative helicity there, or the reverse. You have to allow all possible helicity configurations. Here, if we've got one minus two minus three plus four plus, and we look at the possible helicity assignments, now remember the three-particle amplitude is special. There are two possible non-vanishing three-particle amplitudes, plus plus minus and plus minus minus. There are only going to be two possible things that we can draw here. Over here in the left-hand column, let's pick minus minus plus for the... Let me just draw it in here. Here we can do minus plus. Let's pick the case where this is negative helicity, and that's positive helicity. Over here we'll do the opposite. Over here we'll do plus minus. These are the two possible diagrams, and now you can fill in all the rest of the helicities because we know the only non-vanishing amplitudes. If this is minus minus, the only way to get a non-zero answer there is a plus. If you have an outgoing plus here, that turns into an incoming minus. Here you have minus minus, so the only possibility is plus minus. Here you have plus plus, so the only possibility is minus plus. Here you can play the same game. Here it's a little bit more complicated because there's some ambiguity. Here I could put plus or minus. In general, you'd have to sum over all possibilities, but I happen to know the only one that gives a non-zero answer, so I'll just quickly... The residue of left-hand side equal to this diagram plus this diagram, evaluated at L1 or L2. Sorry, times 1 over ST, but that's going to cancel out of both sides of the equation. That's right. Yes, thank you. That's a better way of saying it. I have the 1 over ST on both sides of this equation, so now I've determined C. Let's keep in mind it's all of this evaluated L equals L1 or L equals L2. Here's the important point. There are two separate residues. We can do the calculation separately. No one tells us to add them. At the pole L equals L1, and now I need to remind you what that was, at the pole L equals L1 equals this, let's look at... We have one corner that looks like this, 1 minus minus plus, where this is L and this is L minus K1. Now let's remember our formula for the tree-level three-point amplitude. It's the two negative helicity gluons in the numerator cubed, and then it's L minus K1, L, L1. That was our formula for the three-point amplitude with two negative helicity gluons. But this is now zero. It's zero because of this factor in the numerator that the inner product, you see, because the undotted index, this angle bracket denotes the... I remind you 1 comma L means epsilon AB lambda 1 with LA, keeping the other index untouched, if you will. That's very poor notation. But the point is L1 is already equal to lambda 1, so you get a zero. Similarly, on L equals L2, the second diagram vanishes. The two non-vanishing diagrams are the same. So at L equals L1, the second diagram evaluates 1, 2 cubed over 2, 3, 3, 4, 4, 1. So I won't do that exercise on the board for lack of time, but if you take this three-point amplitude, times this three-point amplitude, times this three-point amplitude, times this three-point amplitude, and evaluate it at the solution L equals L1, you'll see that it collapses to this. At L equals L2, the first diagram also evaluates to exactly the same thing. All we can conclude now is that C equals 1, 2 cubed over 2, 3, 3, 1, 4, 4, 1 is consistent. So let me take a step back, and here we're trying to compare two different rational functions. We found that each rational function, I'm sorry, this rational function obviously has precisely two poles at L1 and L2. Then we computed the residue, and on the residue of those two poles, it takes the same value, except for a crucial minus sign, which I'm really suppressing the minus sign here, but it's important. Then I computed the residue of the left-hand side on those two poles, and I got the same value. Had I gotten different values here, then this formula, there's no way it would have worked. There's no way that you can have one rational function that has different residues on two poles be equal to a rational function that has the same residues on two poles. So the conclusion is, unfortunately I still have this on the board, that this is equal to 1, 2 cubed over 2, 3, 3, 4, 4, 1 times this integral, so let me put a time sign, plus possibly terms with zero residue, L equals L1 and L equals L2. I'm comparing two rational functions, and I've only checked two poles. The left-hand side can have plenty of other poles, because who knows what the Feynman diagrams look like. Now here I'm going to invoke a little bit of magic that I'm actually not going to prove it all. These terms vanish in N equals 4 super Yang-Mills theory, but can be non-vanishing in other gauge theories, either with less supersymmetry or with less supersymmetry with various matter contents, pure QCD, whatever. Okay, so now that I've done this kind of calculation the hard way, I definitely have improved that this is correct, but I've given you hopefully some indication as to how these kinds of calculations go and how we approach this kind of problem. Yeah, so the point, oh, but I'm talking about a one-loop amplitude. Yeah, sure. So the statement is that the sum of Feynman diagrams is a complicated rational function, but you can move pieces around and regroup terms, and it's possible to express it just as this single one term. This was originally derived, this formula was actually originally derived by Green and Schwartz decades ago by taking an alpha prime to zero limit of a one-loop string amplitude. But yeah, from the point of view of Feynman diagrams, it's not at all obvious that everything cancels, you know, you can have all kinds of crazy triangles in them and all kinds of stuff, but the supersymmetry, one way to think about it is supersymmetry allows all that to cancel leaving just this. But if you have a non-super symmetric theory or even a less supersymmetric theory, those cancellations will be less exact. Ah, yeah, so you correctly noted that this thing here appearing is exactly the tree-level amplitude. Yeah, that turns out to be a general feature for MHV amplitudes to all loop order. That you could be at 27 loops or whatever, any, and then you wouldn't have a, you wouldn't, if you were at 27 loops, you would not have a residue in C4, you would have a residue in C4 times 27, which is 112, and you calculate the residue and you would always get the overall MHV amplitude times some pure scalar function. So, now that I've done this calculation the hard way, since I only have 10 minutes or so left, that's plenty of time to do it the easy way. And I'll even do a more complicated example. I'll do a five-point example using momentum twister geometry. And it's here that we're going to appeal to intersections of lines and planes, etc. First of all, how to write loop integrals in momentum space. Well, let's look at that example again. Let's observe, so in that example I had defined this leg to be L. A more symmetric way of writing that integral is to switch to these variables where you associate the integration variable not to one edge of your loop diagram but to the face, okay? And when you do that, let's see, let me be consistent here, X2. Remember that momentum P2 is in my notation X2 minus X3. So notationally this is indicated in this diagram. This is the nice thing about planar diagram that every edge is sandwiched between exactly two adjacent regions. So the momentum flowing along this edge is the difference between X2 and X3. That also works for the loop momenta. This propagator here, this propagator is 1 over L minus K1 minus K2 squared or the same thing as 1 over X1 minus X3 squared. How did I get that so quickly? Because this line is the boundary between regions marked by X and X3. So maybe except for an overall factor of 2 pi, this integral here is exactly the same as the integral over there. So now when we transform to twister space or momentum twister space, we need to know what happens to this integral over D4X. And the answer is that you have to introduce two auxiliary points, momentum twister space. So what this means is that integral over points space time translates into integral over lines in P3. That was our twister correspondence that we talked about earlier. How do you specify a line in P3? Well, you can specify two points, B, A, and Z, B, and specify that that means the line going between A and B. Now that's redundant because if you give me any two points A and B, I can slide either one of them along the line and it still describes the same line. So in other words, if you take Z A and Z B and you multiply by any two by two matrix, you'll get new points Z A and Z prime that describe the same line for any GL2 matrix. So we need to mod out by that symmetry and that's what's indicated by this volume factor. Okay, anyway, so that's how you transform the measure into momentum twister space. I dwelled perhaps too long on that because I actually won't use the measure. I'm just interested in discussing rational functions. All right, so this is literally my last half page. Let's consider a five point example. And I guess I mean gray blobs here. Let's calculate. Let's look for the locations of singularities where these four propagators are on shell. So the idea is exactly identical to the previous example we did except now I have a five point amplitude. And instead of using L, L minus K1, L minus K1 minus K2 and L plus K4 plus K5, we want to do this everything in momentum twister space. Momentum twister space, the propagators are very simple. AB12 equals zero. This is the propagator AB12. How do I know that? Because this face here is the face 12. The X point corresponding to this face is the line in twister space between points one and two. This face is AB. That's my loop integration variable. So this line, this propagator goes on shell precisely when this line in twister space and this line in twister space intersect. So once again, putting this propagator on shell is saying that you demand this line and that line to intersect. And I do the others. So here are my four propagators and I need to find all possible solutions. I'm really in my last minute now because I'm just going to write down the two solutions and you'll see how simple it is. When I did this in momentum space, it was actually a little bit of an exercise to check that I was indeed satisfying all of my conditions. Here it's really simple. These equations have two solutions. Suppose here are my five points in twister space. Let's draw the four lines one, two, three, four and five, one. And let me do two copies of this because I'm going to claim there are two solutions. There's some inherent in precision in the blackboard. Okay, now your goal is to... So the problem of finding a way to put all four propagators on shell transforms into momentum twister space as the problem of finding lines that intersect these four given lines. Well here's an easy solution. The line one, three clearly intersects those four other lines. So that's one of my solutions. There's a second somewhat more subtle solution which looks as follows. If you let the blue here be the plane one, two, five and you let the yellow be the plane two, three, four and I'm running out of colors. But if you let this be the intersection of those two planes then clearly way over here, I'm sorry I ran out of space, but inevitably there's going to be, well there can be two different points where this intersection line intersects all those. So the second solution is the line five, one, two intersect two, three, four. This notation means the intersection of that plane with that plane. This was an example meant to indicate that solving these kinds of on-shell problems, these geometry problems where you need to put a collection of propagators on shell, becomes completely trivial or relatively trivial. I mean there are more complicated examples in momentum twister space. And so you can use these kinds of methods. So let me just write the final formula. Product of four tree amplitudes vanishes on one, on the second solution, but not the first. We saw a similar phenomenon in our study of the four point case. So this suggests that an appropriate rational function to use in constructing the amplitude is, and here's my very, very last formula. You've got the propagators downstairs and we need something upstairs in the numerator that will cancel one of those residues so it doesn't appear. So we can explicitly put a factor like this in the denominator. That's a factor that vanishes when the line AB intersects that line. And then that's almost correct. There's a little bit more here that you can only guess because you needed to have the right scaling, two, four, five, one. So this is required to give correct projective scaling. And then we have this integral d4zA d4zB over vol gl2. So here with almost no work, we've computed the integrand for a five particle one loop amplitude. And I haven't specified the helicities. There's one particular helicity assignment for which this is the correct answer and there's another helicity assignment for which it's the conjugate answer. But anyway, just to conclude, I hope I've given you some flavor of how these momentum twister variables really help to expose a lot of the geometry of n equals four super Yang-Mills theory. Most of the material in my first three lectures has been fairly generic to massless gauge theories. But these tools that have been pushed really, really far are quite special to n equals four which has the most mathematical structure. Okay, so thank you for your attention.