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STUDENT: So when you wrote down the forward--
NIMA ARKANI-HAMED: Yes.
STUDENT: So there, you wrote the same down for a single possibility. I was wondering why you couldn't-- you know, there are a bunch of other functions you could write. So this is true. So you use both the--
NIMA ARKANI-HAMED: So the one you wrote down, that's S and that's T.
STUDENT: Ah, I see.
NIMA ARKANI-HAMED: Yeah. So anything which has-- so let's do 1, 3, 2, 4-- ah, let's do 1, 3, 1, 3. This is vanishing little group way. Because of this vanishing little group way, you can write it as a function of the momentum, directly as a function of the momentum. The momenta are things with vanishing little group.
STUDENT: So anything that has vanishing little group.
NIMA ARKANI-HAMED: Yes. anything that has vanishing little group momenta you can write in terms of the planet.
STUDENT: I see.
NIMA ARKANI-HAMED: Exactly. And so they're the standard way.
STUDENT: There are many ways of writing it, it's just changing--
NIMA ARKANI-HAMED: Yes, exactly. So this is what I said. All you have to do-- all I was doing is stripping off some factor which was guaranteed to take the helicities into account.
[INTERPOSING VOICES]
NIMA ARKANI-HAMED: That's right. As was remarked yesterday, you can contract forward momenta also with an epsilon symbol. But look, just for simplicity, let's let this one parrot. But yeah, that's the point.
STUDENT: The other thing I didn't quite understand when they were explaining the states, and then equals 4, and the-- I didn't understand this exponential, whether it was 2--
NIMA ARKANI-HAMED: Oh, OK.
STUDENT: Two Grassmann variables.
[INTERPOSING VOICES]
NIMA ARKANI-HAMED: There's only one Grassmann variable. It's e-- Q is a supercharge.
STUDENT: Q is another name for--
NIMA ARKANI-HAMED: Yeah. Oh, that's right. Let's do this.
[INTERPOSING VOICES]
NIMA ARKANI-HAMED: That's right. So I forget how we're defining it. I forget where I was putting the end--
STUDENT: You notice that's omega there.
NIMA ARKANI-HAMED: Yeah. I'll tell you what the omega is. So yeah. Omega is not a fermion. Omega is actually something which was spinor. Omega is like-- so imagine lambda, tilde, alpha dot is 1, 0. Then omega is probably this 0, 1. It's something who's dot product from lambda tilde is 1. OK?
So it's just there as a stupid normalization pattern. It isn't uniquely defined. Of course, I can shift it by any portion of the lambda tilde. But that shift doesn't affect--
STUDENT: --direction.
NIMA ARKANI-HAMED: Exactly. Just pick side of direction. You see, it's just because-- what we're trying to do-- it would have been easier to say all this if I picked some particular lambda tilde ahead of time and said let's say it's 1, 0. That I could have just written it down, without this exponent spinning into it.
But it's because what Q, often, dot does to the state is give me-- I mean that means it has to go somewhere, right? To the lambda tilde alpha dot lambda. And so that index is picking up with the lambda tilde alpha dot. Right? So that is the I. And so it has to be either the Q something that knows about lambda tilde dot, right?
STUDENT: OK.
NIMA ARKANI-HAMED: And that's why what you got is its [INAUDIBLE].
STUDENT: I just had a question. So you said that--
NIMA ARKANI-HAMED: This is-- by the way-- I mean, this part of the stuff, just as a reference is, I think, very clearly explained in the paper I wrote with Freddy and Jared Kaplan in August, 2000. I should say, the introduction of that paper, we tried very hard to be extremely straightforward. And all the spinor helicity story, the SUSY story, everything, is, I think, very, very clearly laid out there.
I can't say that for everything else that we've written since. But it hasn't been the desire of nothing pedagogical. It's been for the absence of time. Hopefully, we'll write something pedagogical soon. But this part is displaying, I think, I think, quite nicely in that paper.
So if you're following the little details with the SUSY algebra and stuff like that there, just look at it in that paper, and it's very clear.
This on-shell superspace, by the way, it was invented by Nair already in the '80s, but somehow was not used. It was not-- even [INAUDIBLE] and friends, for a long time, were doing the computations and components and summing over all the particles and multiploids. But of course, doing things in a symmetric way is incredibly, incredibly hard.
STUDENT: So can I ask a question?
NIMA ARKANI-HAMED: Yes.
STUDENT: So you said yesterday that the two-point function in this formulation vanishes when the momentum is real, right?
NIMA ARKANI-HAMED: Yes.
STUDENT: But that's a tree-level statement. So I was wondering--
NIMA ARKANI-HAMED: It's an all order statement because of the structure of the three-point function is [INAUDIBLE].
STUDENT: So that's for all the loops?
NIMA ARKANI-HAMED: Infinite loops. In fact, there's no renormalization of the three-point function in any meaningful sense. It is via coupling. It just defines via coupling. Yeah, that-- it's really cool because, as I said before, when you write down Lagrangians, you sort of feel like the three-point function has all the information. But because there's gauge redundancy, it comes along with in the Yang-Mills and the four-point function. And then gravity comes along with an infinite number of higher order operatives, right?
But somehow you see, in directly looking at the invariant amplitudes, that, directly, the three-point function is nailed by proper inference. But then the question is, what higher point functions can you have consistent with that three-point function and consistent with the physical requirements-- factorization, locality, all the rest of it. And unfortunately, [INAUDIBLE] guarantees that that's all going to work out. But if you're starting to try to break away and do it in another layout, it's a really good starting point. It tells you there is something you can't mess with in the structure of the three-point. OK.
So we have a little bit more kinematic stuff to talk about. Before going into a conformal invariance, and twistor space, and things like that, let me actually start even with what we've gone through so far by talking about beginning to compute scattering amplitudes. And let's start at tree level.
And I'm going to describe a BCFW recursion relation. So the BCFW recursion relations are the systematic way of starting from the three-point function and generating everything. So the idea is to imagine that you have-- now, this actually works in any number of dimensions. It's not special four dimensions. So I'll say it in general for a while, but we'll transition to four dimensions in a bit.
The words, which are very familiar from the '60s, is to try to compute the amplitude just knowing its singularities. Now, at tree level, the only singularities the scattering amplitude should have are poles, as we've described. And the poles should be such that the poles are 1 over some sum of the subset of the momenta squared. And the residue of that pole should be the product of two lower-point amplitudes summed over the helicities of the particles that can be in between. That's the residue struc-- the pole structure we know ahead of time that the amplitude has got to have.
The question is, how can we make use of that information to reconstruct the amplitude? And remember, we're complexifying all the momenta. So it's sort of a daunting-- at first order, it sounds like a daunting task. A funny comment is Steve Weinberg likes to say that he was very happy that the S-matrix program failed, because if it succeeded, he'd have to learn all this fancy, higher dimensional, complex analysis.
[LAUGHTER]
Well, we're going to end up doing a lot of fancy, higher dimensional, complex analysis. However, a nice insight is that much of this fanciness isn't needed. And you can explore the big space of complexified momenta one intelligent direction at a time. So you're dealing with one, simple complex variable at a time.
So let's say you have this idea. You want to complexify momenta. And you don't want to do it in the most complicated way. You just want to complexify as few momenta as possible. So what do you do? You have particle one, two, three. And you can take P1 and say you're going to make P1 complex. Let's say you're going to make it complex by adding by shifting it. P1 plus Zq. Let's say q is complex.
Well, of course we can't just make one momentum complex, because momentum's gotta be conserved. So if one momentum is complex, you can't have momentum conservation.
STUDENT: q is complex or Z is complex?
NIMA ARKANI-HAMED: Z is complex, q is complex. Yeah, everything is going to be complex. Let's say Z is complex. The only thing that's not complex is P1. So this is P1 of Z. So you can't conduct it by one momenta. Well, if it were me having this idea, I'd probably just get discouraged at this point and quit. But of course, you can do the next most complicated thing, which is a complexified two momenta.
So let's say you take two momenta. How do you take two momenta and make them complex? Well. So now we're getting this grid. If you have two momenta, we've made them complex, momentum is conserved, because P1 plus P2 is conserved.
However, momentum was conserved, but we also have to guarantee that these momenta are on shell. They continue to be massless for any value at Z. So I have to have P1 squared of Z0 be 2 squared of Z0, which tells me that q squared is now diminished, and q dot P1 has got to equal 0, and P dot P2 has got to equal 0. Those are the coefficients of Z squared, and Z, and P1 squared, and P2 squared.
Now, this is also impossible to do if q is real. If q's a real momentum, this is impossible to do. But in general, it's not impossible to do. For example, if P1 [INAUDIBLE] where P1 looks like that and P2 looks like that, then q can look like 0, 0, 1, i. So that vector satisfies that q squared is 0, and q dot P1 is 0, and q dot P2 is 0. OK.
It's possible to complexify the momenta, keeping them on shell. Now, at this point, if you're going to talk about the general number of dimensions, I should tell you how to complexify polarization vectors, I should tell you how to do the whole thing. And it's a very simple story and it's possible to do, but I won't discuss it here.
Instead, let's specialize immediately to saying what this deformation on the lambdas and the lambda tildes. So on the lambdas and the lambda tildes, this deformation is very simple to describe. It's lambda one goes to lambda one plus Z lambda two. And lambda two tilde goes to lambda two tilde minus Z lambda tilde one.
You'll notice this leaves lambda one, lambda one tilde plus lambda two, lambda two tilde as invariant. The fact that I've described it as a deformation on the lambdas-- I was, of course, guaranteeing that I'm talking about non-shell objects the whole time.
STUDENT: Would it have been 2 lambda two and lambda one?
NIMA ARKANI-HAMED: They're underformed. You just leave them alone. Yes.
STUDENT: But how does the q relate to this preservation?
NIMA ARKANI-HAMED: Oh. So q-- well, actually, nice homework question. So what is q? So what's happening to P1? P1 goes to P1. And P1 is lambda one, lambda one tilde, right? So if P1 goes to P1 plus Z, lambda two, lambda one tilde. Right? So that's q. And you see why it's complex, right? Because lambda tilde is not equal to its lambda star. Yeah. All right?
STUDENT: So just to be clear, q is some object that we have to solve for in some sense before we do anything with Z--
NIMA ARKANI-HAMED: Yes, yes, definitely. We have to find a q. We've got to pick a q that makes this possible. I just told you how to do it canonically.
STUDENT: Yeah. And then we get this new variable, Z.
NIMA ARKANI-HAMED: Now we've promoted our amplitude from what it was before, which was the function of momenta, to a function of momenta and Z. And now what we're going to do is keep the momenta fixed and just explore what it does in the Z.
STUDENT: q is gonna be fixed so--
NIMA ARKANI-HAMED: q is fixed, q is fixed. Yup. q is fixed. In fact, just forget about q now. There is no q now just directly. I've only introduced the Z here, right? OK? q is some fixed thing which satisfies these properties. I'm writing it this way just to emphasize that this part has nothing to do with spinor helicity variables. You can do this in any number of dimensions, et cetera, OK? The important point-- it is possible to complexify momenta, but you really have to make them honestly complex. OK.
So now what we have is I will tell you in a second tell you what to do supersymmetrically. For the moment, imagine we're doing this non-supersymmetrically. The amplitude we had before now goes to a function of Z. And now that's the question. If we understand what kind of singularity Z has, m has in the Z complex plane. Well, remember, all the singularities the actual amplitude has are poles where you have a sum of a sum subset of the momenta on the left and some subset on the right. So for every way-- this doesn't even have to do with planarity. So this is incredibly general. Everything here is very, very general.
But the poles are associated with just some way of breaking the external particles into two sets. And we'll call them left and right. And the pole is that what you might call 1 over P left square, where this is the sum of all the momenta on the left. Those are the poles that the amplitude has. And again, the residues are the product of the lower point amplitudes, all right? OK.
Now what we're doing is we're just deforming P1 and P2. So if L contains both particles one and two, there is not going to be any Z here, because the Z cancels out on that sum. So the only times we'll have a pole is when one is on one side and two is on the other side.
So here, one is on one side here. And again, if the theory isn't planar, then there isn't even any border there necessarily. If it is planar, I will put it so it looks like what it does when it's planar. But in general, it doesn't matter. So we get poles where one is on one side and one is on the other side. Where is that pole? That pole occurred at some value Z star that depends on exactly who's in l.
Where? Well, where what? Where P1 plus Zq, lambda one, lambda one-- well, let me write it this way. P1 plus Z. Whatever P left was before is getting deformed to being plus Z times lambda one, lambda two tilde. So it's when this squared is equal to 0. That's where the pole is. So I can just figure out the Z star that makes that happen. And that Z star L is just negative P left squared over one P left two, which is just P left dot q.
STUDENT: And it's 2x1.
NIMA ARKANI-HAMED: So for every way of separating the external particles into two sets with one on one side, and one on the other side, and two on the other side, there's a pole in the complex Z plane. And furthermore, the residue of that pole, the residue of that pole is just the product of lower point amplitudes, on-shell amplitudes. This is the important point here. Lower point on-shell amplitudes.
STUDENT: So why you have many possibilities?
NIMA ARKANI-HAMED: Because there's many ways of splitting the external particles into two sets, OK? So for every way of splitting the external particles into two sets, where particle one is on one side and, yeah, PL is different from place to place. OK. OK. So is that clear?
STUDENT: Yeah.
NIMA ARKANI-HAMED: Great. So this is wonderful. We appear to understand all the poles of this guy in the complex plane. Yeah. By the way, this is one that we have to do some deformation to keep things on shell. You might say, don't Feynman diagrams-- why don't we just use Feynman diagrams? Feynman diagrams have singularities. That's true.
But the problem is that we want to be able to interpret the residues of the poles in terms of lower point amplitudes. So that's why you have to keep everything on shell throughout the whole process. It doesn't do any good to randomly complexify things if you're going to compute things in terms of some off-shell quantity, that you don't know what it is anyway. OK. Yeah.
STUDENT: Angle bracket square bracket notation there.
NIMA ARKANI-HAMED: Here? If I write angle bracket, some vector, square bracket, just contract them the only way you can. So it'd be A, alpha, alpha dot lambda two tilde, lambda one alpha. You can sometimes write it like that, because a times 2 is something, effectively, you will-- which is an angle bracket. OK. OK.
So wonderful. So notice, not once have I said the word "Yang-Mills theory, bi-cube theory," anything. This is just true of any, any theory at all. We know the residues of all these poles. So when we know the residues of all the poles, how do we determine what the function is? Well, you just remember, we're interested in the function at Z equals 0. So you remember your elementary complex analysis.
So you write that as a PZ over Zm of Z where the contour encloses the pole at Z equals 0. So m of Z over Z now is a pole of Z equals 0. But now I can form the contour away from the origin. And so this is equal to the sum over all the other poles of 1 over Z star L times the residue at Z star L. The residue of m at Z star L. And we just decided we knew what the residue of m at Z star L was. It was just the product of the lower point amplitudes. So that seems to give us a way to compute higher-point, on-shell amplitudes just using this information of lower point, on-shell amplitudes. So that's spectacular.
What's the catch? There's only one catch, which is there's one poll that I didn't talk about. It's the pole of infinity. It's what the amplitude does as Z goes through infinity. If m doesn't go to 0, as that goes to infinity. Let's even say m goes to a constant as Z goes to infinity. In this integral, there's also a pole at infinity. So it's this plus the pole of infinity. All right? And yeah.
STUDENT: Sorry. Does Z go to infinity?
NIMA ARKANI-HAMED: At pole, Z goes to infinity.
STUDENT: And it doesn't matter maybe like a [INAUDIBLE] around?
NIMA ARKANI-HAMED: Yeah. It's just the usual-- if you want to compute the pole of infinity as usual, you say, Z equals 1 over w, and you compute it as the pole around 0 in the W plane, OK? So just the standard notion of pole of infinity in the complex analysis. Really, it's a Riemann sphere. And so you're looking at the north pole rather than the side. OK.
So I'm not writing this very precisely. But it's clear this is computed using lower point objects. I'll write it in the second. But this, we have no idea what it is. So this looks wonderful until you actually start thinking about this pole of infinity. For example, let's take a scalar theory-- 5-cube theory, 5 to the 4 theory.
You might think, oh, this is going to be great, m will just vanish in infinity, obviously, and so everything is fine. The problem is that, in fact, not only does m not vanish into infinity-- sorry. Not only does it not obviously vanish into infinity, naively, it blows up infinity like hell. OK? And even in the best theories-- and what you would think of as the simplest theories like 5 cubed, 5 to the 4 theory-- let's take 5 to the 4 theory, OK? 5 to the 4 theory is a good example. The amplitude in 5 to the 4 theory goes to a constant at infinity, even for the four point function. Even for higher point functions, there's always some diagram where it goes to a positive infinity. So in order to evaluate what F is at infinity, you still need some Feynman diagrams. That's why it's not very useful.
STUDENT: You have it propagate on, let's say, 1 over z.
NIMA ARKANI-HAMED: Yes, but there's always some diagram. For example, if this is 1 and that's 2, look at a diagram like this. In such a diagram, z doesn't make an appearance anywhere in these [INAUDIBLE]. Because once again, the sum of two momenta--
STUDENT: Oh, you always separate them.
NIMA ARKANI-HAMED: No, no, no. No, that's what I'm saying. The poles that we know are the ones where 1 and 2 are on one side and the other. But I'm saying for the behavior of infinity, there's always diagrams like that. There's no z's anywhere there's a constant term.
OK, it's in fact worse than that. Because if you take Yang-Mills, there's derivative couplings. There's momenta at the vertices. And naively, M actually blows up at infinity. It goes z or z squared or z cubed depending on which polarizations you're talking about.
And if you talk about gravity, then there's z squareds at all the vertices. And so not only does it naively blow up, individual Feynman diagrams naively blow up, but they blow up worse and worse and worse with more and more particles. So it goes like z to the n as the number of particles becomes large or erratic. So that's the saddest. Scalar theories, it goes to a constant, it doesn't work. Gauge theories, individual Feynman diagrams blow up like at least z. And gravity, they blow up like hell. So it looks like a great idea. But it looks like it's dead upon arrival because of the pole at infinity.
The wonderful surprise is that this conclusion is true for scalar theories. Yep, they go to a constant. But if you sum up all the Feynman diagrams, the full actual amplitude vanishes at infinity for Yang-Mills. It vanishes at least as fast as 1 over z in fact. And not only does it not blow up in gravity, but it vanishes even faster in gravity. It vanishes at like 1 over z squared.
So that's yet another-- by the way, gravity is gain squared again. So if you're here at infinity, it goes like the square root behavior of Yang-Mills. So this vanishing at infinity is a very efficient way of capturing the fact that there's tons of cancellations when you compute Feynman diagrams.
STUDENT: So they prove it for some specific n, or they prove it for general n?
NIMA ARKANI-HAMED: So let me tell you the actual. I'll just tell you the correct statement.
STUDENT: It leads to a Feynman diagram, right?
NIMA ARKANI-HAMED: No, ultimately, the nice physical proof is not using Feynman diagrams. The nice physical proof explores this region by realizing that as z goes to infinity, you can think about particles one and two as being one hard particle blasting through a soft background.
And then you use some background field arguments, and you notice that at infinite momentum, there's some enhanced symmetries, which aren't there when you have particles of spin, but are there when you have particles with spin. So there's something that we call the spin Lorentz invariant symmetry. There's one copy of the Lorentz group for spin one. There's two copies of the Lorentz group for spin two.
There's a big, big symmetry that emerges in this large z limit. And that symmetry is what's killing all the large z behavior. This is the long story. It's not a long story, but a story I don't want to get into. But it's not a Feynman diagram. It's hopeless with Feynman diagram. It's another argument. It's a physical sort of background field argument.
STUDENT: But they check if you use Feynman diagram before.
NIMA ARKANI-HAMED: Well, yes, so there was also an argument, a direct argument, using Feynman diagrams, which was a lot more complicated. But it's true-- and not using Feynman diagrams, using tricks and recursion relations of other sorts and things like that.
But anyway, the statement is true. Now non-supersymmetrically, it turns out to be true only if particles one and two will have fixed kinds of helicities. So if particle one is helicity plus and particle two is helicity minus, it works. If it's plus and plus, minus and minus, it works. But minus/plus, the amplitudes do blow up. They go like z cubed for gauge theory and z to the sixth for gravity.
STUDENT: Why are the two interchangeable?
NIMA ARKANI-HAMED: No, they're not interchangeable. Because I decided to deform the lambda for 1 and the lambda tilde for 2. So that's where the parity ranking was coming in. So I should've emphasized this. This deformation is manifestly parity, you guys. Because I'm deforming the lambda for particle 1 and the lambda tilde for particle 2.
STUDENT: So when exactly does it not work again?
NIMA ARKANI-HAMED: It doesn't work when it's minus/plus helicities. But you shouldn't worry too much about this. Because this turns out to be a massive red herring. So really it just works all the time. But it works all the time when you think about it supersymmetrically. But I'm just telling you the facts.
So the fact that it worked already for those helicities was enough to make it work for all amplitudes. Because you could always choose the appropriate helicities to go all the way down. Notice that what this allows you to do is compute a higher point function in terms of the product at lower point amplitudes.
Then you can keep going, keep going, keep going all the way down until you get to what? What can't you split anymore? The three point function. So it allows you to build everything up, eventually knocking it down to three point functions.
So this is the way of building up the whole amplitude from [INAUDIBLE]. So you know the three point functions. You put in the desired factorization properties. Out spits out the [INAUDIBLE]. Now let me just tell you there is a super generalization of this, which nicely eliminates all of this funny thing about which helicities you're talking about and so on.
So actually something very important I should have mentioned yesterday before getting into that is one of the lovely things about this superspace formalism is that let's say we're doing Yang-Mills or gravity, doesn't matter. Now we're labeling, because we're using these Grassmann coherent states, all of the external particles in the same way.
So using these super variables allows us to expose symmetries that the answer has, again, which are obvious that it should have. But any particular helicity component breaks it. For example, the superamplitude for Yang-Mills is cyclically invariant. In the planar limit when we were just looking at these single trace operators, remember there's a natural cyclic order into the external particles.
So the superamplitude is really cyclically invariant. It has a cyclic symmetry. Of course, if you picked any particular helicity component, you'll never see that. Because it's plus 1, minus 2, plus 3. That thing is not going to be invariant when you cycle i goes to i plus 1. But the superamplitude will be. So that's quite lovely.
For gravity, there's a whole big permutation invariance. I mentioned that one of the reasons there hasn't been much progress in gravity is that these simple kinematic things about the amplitude haven't been fully understood yet for gravity. One of the big problems that no one has presented-- even the simplest gravity amplitudes in a way that makes a permutation invariance manifest. It's an incredibly basic thing to want to do, and it hasn't been done, even for the simplest MHV amplitudes.
I think that's the central stumbling block to gravity. And when someone figures that out, the gravity story will begin to explode. There's already a huge number of interests in there. But I really think that's one of the main conceptual stumbling blocks right now.
But anyway, so it's wonderful that these cyclic symmetries are made totally manifest using-- that the cyclic symmetry's permutation invariance is made manifest. Now you're labelling all the external particles in uniform.
But it also suggests that when we did this shift on lambda 1-- by the way, conventionally always use eta tilde. It's a convention that we chose a long time ago, and it ends up being better for many, many reasons. So I'll use the eta tilde representation all the time. But when we deform lambda 1 to lambda 1 plus z lambda 2, and lambda 2 tilde to lambda 2 tilde minus z lambda 1 tilde, why did we do both of these deformations? Because we wanted to keep momentum conserved.
But the theory is supersymmetric. It would also be nice to keep the supermomentum, the super partner of momentum, conserved. So remember there's a delta function. There's a delta 4 plus sum of lambda lambda tilde. But there's also a delta 8 of the sum of lambda eta tilde. So this is clearly not a supersymmetric operation. For it to be supersymmetric, I should also deform the eta tilde in order to keep that thing invariant.
So since I'm sending-- so there's a lambda 1 plus z lambda 2 eta tilde 1 plus eta tilde 2 plus lambda 2 eta tilde 2 plus dot dot. So to keep this invariant, the coefficient of lambda 2 is now eta tilde 2 plus z eta tilde 1. So I should also shift eta tilde 2 goes to eta tilde 2 minus z eta tilde 1. So if I do that, that will keep this supermomentum there.
So this is the supersymmetric BCFW shift. It's much more natural supersymmetrically. Of course in components this does something interesting. For example, for exactly the plus/minus amplitudes, it's not relevant. This eta shift isn't showing up. But for any other amplitude, it's shifting not just the gluons, but the scalars and the ganginos and all that stuff in some correlated way.
But it turns out that with this shift, all of the superamplitude manages as z goes to infinity, like 1 over z, or 1 over z squared for gravity, 1 over z3 angle is 1 over z squared for gravity. So the whole superamplitude manages to go the infinity-- no need to talk about any helicity components or anything like that. So that's very nice. It's a very uniform way of talking about it.
STUDENT: So is the fall off at infinity always governed by the spin, and doesn't-- supersymmetry doesn't allow any-- you just stated it doesn't allow you extra cancellations or--
NIMA ARKANI-HAMED: Yeah, in fact, it's actually really cool what supersymmetry is doing for you here. Normally early on, especially in computing loop amplitude, people use SUSY, roughly speaking, because gluons were the hard part. So we're using n equals 4 to take the gluons into account. Then they were subtracting scalars and fermions because the scalars and fermions were the easy part.
From this point of view, it's really the other way around. The gluons are the awesome part. The gluons are doing the beautiful thing. They want to vanish at infinity. What you do is you use-- what the SUSY is doing is allowing the scalars. Remember I told you phi to the fourth sucks, right? You know this theory of phi to the fourth interactions.
So how are we getting around that? We're getting around that because there's this very funny correlated shift involving both scalars and-- I mean, if you want to say, the components, it looks miraculous. But it's the SUSY that's allowing the scalars to inherit the good behavior of the gauge bosons. It's totally backwards. It's really the high spin particles that are awesome. And SUSY is helping all the other spins inherit the good behavior of the high spin particles rather than the other way around.
STUDENT: In this case, do you just do it by scaling?
NIMA ARKANI-HAMED: Sorry, in which case?
STUDENT: The supersymmetric case?
NIMA ARKANI-HAMED: Oh, what you actually do is the following. Remember yesterday I mentioned that we can always use SUSY to set two of the etas to be anything we want. So what you do is use the SUSY to set the etas such that this turns into the plus/minus problem. So the plus/minus problem, you know it vanishes at infinity. And then you transform back to this background field argument, and it proves it for everything. So even the proof is literally using SUSY to inherit the good behavior of the gauge bosons for everybody else.
OK, so this now means that for super Yang-Mills and for supergravity, in principle, the problem of tree amplitudes is systematically solved. So you can build them all up starting from the three-point function. OK, let me show you what the answer ends up looking like for Yang-Mills. I'm not writing the full super glory of it.
It's very nice to go through one BCFW exercise. And I could do that. But doing that would take around 20 minutes just to do it really step by step by step. But it's really simple. If anyone wants to actually do something in this business, just computing that six point amplitude using BCFW is-- if you do that, then you know how to do anything. And it's very, very simple actually.
STUDENT: How many particles would you like-- without running into any problems.
NIMA ARKANI-HAMED: There's never any problem. It depends what you mean-- by hand?
STUDENT: Yeah, by hand or by computer. I don't know.
NIMA ARKANI-HAMED: Well, by computer we do up to 40 particles. It's just how long you're willing to wait for Mathematica to spit things out. No, I mean, I can't emphasize enough. It's nothing. It's so recursive. And now we're not even yet using the best variables. When I talk about these momentum twistor variables in 20 minutes, then it's just a dream. You fall out of bed, and you compute the eight particle one.
[LAUGHTER]
That formula, the momentum twistor analog of it is so much simpler that it doesn't even look as complicated as that. But anyway, I first want to just present it. Because it's highlighting some of the important physical points. Yeah, so this is literally a 10 minute calculation. So it looks scary, but it's literally a 10 minute calculation-- maybe 20 if you're doing it for the first time. Maybe an hour if you're doing it for the very first time. Maybe three hours if you're really doing it now.
OK, but this is what it ends up looking like. So I'm writing down the component amplitude for a six particle scattering where the helicities are plus/minus, plus/minus, plus/minus. It's next to maximally helicity violating. I should have said that. Because the amplitudes just is part of the lingo. The amplitudes with all plus vanishes. All plus and one minus vanishes. The first ones that don't vanish are the two minuses and all the rest of them plus. Those are called MHV, Maximally Helicity Violating. I don't know the historical reason for it.
Then if you have three of them, it's called next to maximally helicity violating, and so on. So oh, I could have said that supersymmetrically too. I'm sorry. Let me go back one last time to this super-amplitude. This distinction between MHV, NMHV, NNMHV, and so on is very simple supersymmetrically. Supersymmetrically, the point is that whatever this is is a function of the eta tildes. And we can sum over sectors that have different net numbers of these eta tilde fermions.
So this is sum over k of what we might call M, n, and k, where M and k contains 4k eta tildes. The eta tildes have four indices on them. They've all got to be contracted up with each other. Because everything is SU(4) the invariant. So every piece of the amplitude is at least 4k eta tilde, has a multiple of 4 number of eta tildes in it. But we can separate things off from the things that have 4 eta tildes, 8 eta tildes, 12 eta tildes, and so on. So it's like some overall R charge. It's really some overall fermion number carried by the eta tildes.
The sector with k equals 0 is the sector with all-- that superamplitude contains, in its components, the one with all plus helicity gluons. So k equals 0 vanishes. k equals 1 vanishes. k equals 2 is MHV. k equals 3 is NMHV, and so on.
In fact, let me change notation here a little bit. And this is a better notation. It's not the one we used originally. But let's call this variable m. Because it reminds you that m is also the number of minus helicity particles. So m equals 0 is all plus. m equals 1 is all plus and one minus. m equals 2 is what we call MHV. And there's another variable, k, which is m minus 2. And we could talk about N to the k MHV amplitudes.
NIMA ARKANI-HAMED: So MHV is m equals 2, and NMHV is n equal to 3 and so on. So that's going to be very important. But remember that the amplitudes with m equals 0, 1, n minus 1 and n all vanish.
So here, we're talking about the NMHV amplitude then, 3 minuses. And they're sort of maximally symmetrically arranged here. I chose this one, because this amplitude as a component amplitude, of course, doesn't have the full cyclic symmetry, but it does have a cycle over by two symmetry obviously.
So it's one of the shortest ones to write down. And this is the answer. That's the answer if you compute using BCFW.
STUDENTS: Is that a constant that comes from treatment measure?
NIMA ARKANI-HAMED: That's right. So I'm not going to write down g's, overall age couplings almost anywhere in these lectures. So they're all going to be there trivially.
So s here is just the operation that clocks you over by one. So S squared is the flopping over by 2, S to the fourth, flopping over by 3. So you see that it's manifestly the sum of three terms, but it's really one term plus it's 2 cyclic over plus its fourth cyclic over friend. And this is the one term.
I've highlighted-- I've written it as a product of two factors. There's this factor and that one. This factor has all physical poles. All the poles that appear there are things that you're allowed to have. So p4 plus p5 plus p6 squared, that's one of the allowed sorts of ground poles that we have. 1, 2, if I multiply it by angle bracket 1, 2 downstairs and upstairs, that's really p1 plus p2 squared. So that's a lot. p1 plus p2 squared plus p3 squared plus p4 plus p5 squared, these are all nice physical poles.
These are crazy objects. That's p5 plus p4 dotted into lambda tilde 3 and lambda 6. If you attempted to rationalize this, divide it just like you did here, and turn the p1 plus p2 squared. If we get that here, I forgot exactly what it would be, but it's an object of the form p squared, q squared minus r squared, t squared. It is not a local pole.
So these are non-local poles. These are local poles. So there's a sum of three terms. Remember, these were the [INAUDIBLE] diagrams tens of thousands of terms, this is the answer you get from BCFW, just the sum of three terms.
But as I mentioned before, the three terms each have both low poles and non-local poles. Now, the non-local poles must somehow cancel in the sum of these three terms. And it's really an amazing exercise to try to check the algebraically If someone even just handed you that answer, and said check for the non-local poles cancel, it's a hellish calculation. But of course, they do. I mean, we can put it on the computer and they do it. But if you're trying to see it analytically somehow, it's lots and lots of playing around with identities, and it takes a long, long, long time.
Now, actually, there is another form of BCFW, which leave all the parity conjugate form of BCFW. What's really crucial to use is the fact that this came from a super amplitude. Remember, I could have labeled the states with eta or eta tilde. And had I done it with eta instead of eta tilde, I would have another deformation, another BCFW deformation, and a different formula. So there's actually two equivalent forms, which are related to each other if you like by parity. Because going from eta to eta tilde is like parity.
So the BCFW form isn't manifestly parity. So this is its parity conjugate form. And it's parity conjugate form comes out looking very similar, but as the sum of three other terms. The numerator is different. It also has local poles and non-local poles. So this is three terms, this is three terms, and they're supposed to be equal to each other. And the fact this came from field theory guarantees that these terms are equal to each other. So there's some six term identity that's satisfied by these objects. 3 equals 3.
You can ask, can you verify that six term identify if someone hands it to you. Hellishly difficult, OK? Again, you can screw around for a long, long time playing around with identities and you'll eventually find it, but you would never expect that it's true.
Well, there is a proof, of course, which is reversing our derivation of BCFW and going all the way back to quantum fields. So that proves that these identities are true. But you somehow get the feeling that there must be-- these are interesting looking objects, there must be some internal logic here that doesn't much care about finding diagrams and all that stuff, that says why the six term identity is true.
Now, I want to point out that if we somehow knew why six term identity was true, we would immediately also prove that there's no non-local poles, and that the object is local. The reason is, if you look carefully, the non-local poles that appear in this form are different than the non-local polls over here in that form. See here, we have like 6, 5 plus 4 3. Here we get 1, 6 plus 5, 4.
So here it's like odd guys, here it's even guys. When I cycle by two, I only get even guys in the denominator here. I only get the odd guys in the denominator here. They're not the same.
So the non-local poles are not the same on the two sides. The local poles are the same on the two sides. While you say that, I mean, this is p4 plus p5 plus p6.
That's odd, that's even, that's odd. It's true. But because of momentum contradiction, p5 plus p6 plus p1 is equal to p2 plus p3 plus p4.
So the local poles are exactly the same on the left and the right. But the non-local poles are different. And therefore, if you establish this six term identity, then you're also, along the way, proven that this object is local. All the non-local polls have got to cancel, because they're absent in the other form.
STUDENTS: So it reads as even versus cycling?
NIMA ARKANI-HAMED: Huh?
STUDENTS: Just even versus odd cycling.
NIMA ARKANI-HAMED: That's right. Notice that they're not simply changing, making even, odd. The numerators are totally different. It's not just that. But the poles are just cycling even versus odd.
So as I mentioned before, this is very nice, because now these deep highfalutin questions about localities, and space time and stuff like that have turned into a very sharp question. What's the mechanism that's guaranteeing that these poles cancel? Where do these identities come from? Why are there such funny objects out there?
By the way, this is the simplest case. For the seven-particle amplitude, there's six terms in one form, six terms in the other form. There's a 12 term identities that it's got to be true. For the eight particle amplitude we could have NMHV or n squared of hv, depending on which one is the 20 term or the 40 term identity that's probably true. So there are some interesting suite of algebraic objects that have satisfied strained identities. And those identities are what allow you to put them together in a way that local physics comes out.
So we're going to return to all of this later and say what the object is that's generating all these things, and where these identities come form. So any questions about that?
STUDENTS: Just one thing. Could you just explain why the in the second one doesn't even seem to have the same number of contractions as the numerator in the first case? I mean, the structure of everything is analogous. They're the same number of-- now I'm happy.
[LAUGHING]
NIMA ARKANI-HAMED: This I can't explain, because it was wrong. So now-- I can erase this. Let's now proceed to talking about computing loops.
And the immediate puzzle is that, as it's very well known, because of infrared divergences, loop amplitudes are not well defined. They have been for [INAUDIBLE]. So then you start wondering. You have you regulated. Should you use dimreg?
One of the wonderful things in this business is there's no more dimreg. Dimreg is totally gone. Thank goodness. Life was so much nicer and easier without it. There was a very physical regularization branch equals 4 which is moving slightly out along the [INAUDIBLE] branch of the theory. It's all the particles of mass.
And people have always said you use physical regularizations, but it was never clear that it was practical. Turns out it's extremely practical, extremely simple and practical. And at least for n equals 4, it's not for QCD yet.
Yeah. But for n equals 4, we're never going to leave four dimensions. Everything's going to be beautifully four dimensional.
There will be infrared regulators that have masses in them. They're very transparent interpretation. We'll come to that later. We'll come to it later.
But it seems like the amplitude itself is not the sharply well defined object, loop level. And this has always been the source of confusion. If you want to imagine that there's a dual theory that computes scattering amplitudes, because it's funny for a dual theory to compute and fill the fine things. It's nice to have something sharply well defined to be computed.
Fortunately, it turns out there is such an object that can be computed in a planar gauge theory at least. Hopefully more generally, but in a planar gauge theory, there is something that can be sharply computed. And that something is the loop integrand is a well-defined object. You see, tree amplitudes were, in the end, as we saw explicitly, a rational function of the external momenta, right?
The loop integrand is also a rational function of the external momenta. It's the process of carrying up a loop integral, which gives you the divergences. But the integrand is going to be a nice, beautiful rational function. And we can hope to get it from somewhere, and it somehow. So let me explain why.
STUDENTS: What is the point with the loop?
NIMA ARKANI-HAMED: Let me explain why. So first of all, most naively, you might think there is a loop integrand in any theory. What's your problem? So just take this one to the nearest Feynman diagram. Just add them all up together.
This is the d4l. This is the d4l. So just put a d4l on the outside of everybody, and voila, that's the problem. The problem is that in individual filing diagrams, you're perfectly free to shift out by any constant you like. There's no natural origin of loop momentum space.
Who do you call l? Would I call that l? Would I call that l? In that diagram, do I call that l? Do I call that l?
I'm perfectly free in an individual filing diagram to shift any way I like. So it should be well defined, putting them all into the same side. Of course, for any particular choice, I can put them all in the same side. But it's clearly very arbitrary, because I could have chosen to shift one of them and put them all on the same side. The integral will be the same. integrand will not be.
And we're really just directly hit you in the face, it wouldn't gauge invariant. Very, very simple. Just something from the diagrams, nothing fancy. Just imagining [INAUDIBLE], and it simply isn't gauge invariant.
But the very important point is in a planar theory, in a planar theory, there's the well-defined bordering on the external particles. Makes it well defined to talk about what the loop momentum is.
STUDENTS: So it can be? [INAUDIBLE]?
NIMA ARKANI-HAMED: No, the integrand will not be gauge invariant. The integral will be. The integrand will not. That's just clear. Just take any one diagram You know it's very important. You sum all the diagrams for the things to cancel. Any one of them, shift the momentum. Now you've changed the integrand. What we're seeing is something now slightly surprising. So now, you should be a little bit surprised that you can talk about the integrand.
STUDENTS: Because completely undefined.
NIMA ARKANI-HAMED: Right. Because it seems totally that there's no natural origins of the loop momentum space.
STUDENTS: Yeah. The diagram is after the integrand, so is the physics.
NIMA ARKANI-HAMED: Yep.
STUDENTS: But even if I have that role, can't I still reshift loop diagrams in the planar loop too?
NIMA ARKANI-HAMED: As you'll see, you cannot. As you'll see, when it's planar, there's a canonical way of combining everyone together under one integral side. So this is an extremely basic point, but which I think was not fully appreciated. And it was major help of planarity in simplifying the physics.
Everyone know that the planar limit physics becomes simpler somehow. But I don't think that it will ever really exploited in perturbation theory. This is its real use in perturbation.
All right. So let's see why that is. Actually, it's going to be very useful to talk about all of these things in using these dual variables. So remember when things were planar, I can define momentum.
Ta could be xa plus 1 minus xa. As it turns out, as is pretty clear, it's also very easy to write loop momentum, loop integrals, using these variables, as integrals in this x space. Let's go through a simple example just to illustrate the point, and you'll get it right away.
So here is the standard, simple loop diagram, where, again, we arbitrarily say that's l, l plus p1, l plus p2, l plus p2 plus p3, l minus p1 let's say. We read it like that, right? So one of the annoying things, there's always these strings of sums of p's everywhere, for example. So we write this as d4l, 1 for l squared, l plus p2 squared, l plus p2 plus p3 squared, all minus p1 squared.
But there's a totally different way of thinking about it, where, imagine what would effectively amount to being the dual of this diagram. So instead of the loop momentum, imagine just talking about some internal point x in this x space. Imagine that this is a point x1, this is a point x2, that's a point x3, that's the point x4.
You see, this is literally imagining that these are the sides of that polygon. I'm just drawing that polygon. I'm just drawing that polygon. And I'm using the fact that, look, x2 minus x1 is just the momentum that's flowing through there, right?
Let's say I have two lines here. Let's say it was 1, 2, 3, 4, 5. Well, I would say that here I put an x1, here I put an x3. Whoever comes next, they're already putting an x4. Here, I put an x5.
And now you see why I do that. Because x3 minus x1 in the polygon. Here's 1, 2, 3. X3 minus x1 is just p1 plus p2. So when you draw these dual diagrams, you never see these big strings of sums of p's, which very, very nice.
And now, for example, if I want to just match this answer, let me just say that x minus x2 is l. But I would write instead here, I'd write d4x over x minus x2 squared. Now, what is l plus p2? That's just x minus x3 squared, x1 minus x4 squared, x minus x1 squared.
So you see, without the dual variables, there's always arbitrariness. Who do you call l? I mean, I could have put l anywhere else there, and I'd get a different looking integral.
With the dual variables, it always looks like d4x over x minus x1 squared, x minus x2 squared, x minus x3 squared, x minus x4 squared. So the dual variables give you, for any given planar diagram, a unique way of writing it.
Now, of course there is arbitrariness that's shifting the origin of the loop momentum. That arbitrariness is nothing other than translations in x fits. I could, of course, translate everybody. If I translated x to everyone else, then I'm not changing anything, right?
So if I have two diagrams, I could still, if I was perverse, take one of them and take the other one and shift everyone. However, I can stop myself from doing that by fixing-- by removing the translational degrading simply by fixing all the external particles to one set of x's once and for all. So if I just declare that the external particles are the fixed set of x's, then I'm done. There is no more translational freedom, and every single planar diagram is uniquely associated with a given rational function.
STUDENT: Why you cannot wait for non-planar?
NIMA ARKANI-HAMED: Well, for-- so here I was using planarity just to show how nice and simple it is. But I could say it back here. I could say at least at one loop back here by saying that-- there is a canonical way of saying that l is the loop momentum that flows from one towards two, because one and two are always cyclically ordered, so two is always next to one. And you could always say, I'm going to make l be the guy that flows between one and two. So that gives me a canonical choice of who l is from diagram to diagram.
But if I don't have planarity, then I can't even do that, right? I mean, there is no such thing as-- in some diagrams, the three is next to one, four is next to one. There isn't even any such thing as the momentum approaching one, one towards two, or-- so there is no canonical way of doing it. There might be something nice you can do in a theory that has permutation symmetry, so there might be something very nice you can do in gravity. But no one has come up with it.
STUDENT: And if you go for two loops--
NIMA ARKANI-HAMED: Yeah, two loops, it works exactly the same way. So at two loops, there's just one more thing you have to think about. At two loops, For example, in diagrams that look like that, now I have two momenta-- two internal x's, x and y, let's say-- x1, x2, x3, x4. So I would write x minus x1 squared, x minus x2 squared, x minus x3 squared, y minus x2 squared, y minus x3 squared, y minus x4 squared.
STUDENT: [INAUDIBLE]
NIMA ARKANI-HAMED: But now I have a problem that-- why did I choose to call that one x and that one y and not the other way around? If I do it the other way around, I get a different rational function. So the only invariant object you can talk about is fully symmetrized. So that's a role. That's what the integrand is. So you draw the planar diagrams and then you fully symmetrize them in all the loop integration points. But that is now a well-defined rational function of internal loop momentum variables and the external momentum. Just like the tree amplitudes are functions, rational functions are just the external ones.
--the scattering amplitude. You integrate it to get the scattering amplitude. But there is a bigger object, this nice big rational function. It's a rational function. It doesn't have any branch cuts, log divergences, infrared convergences, nothing. It's a beautiful rational function. Depending on what you do to it, you will get different answers. But you should expect that this rational function has all the symmetries that the theory has, because it's not being broken by anything.
The BCFW terms are rational, and as it turns out, and we'll understand why, they are Yangian integrand. They're conformal and dual-conformal integrand. And as we'll by the end of the lectures, now that we understand how to compute this integrand to all loop orders, so what we now have is the analogue of these recursion relations to all loop orders in the integrand. So we have, in principle, a complete definition of the planar theory in terms of super one [INAUDIBLE] variance. Everything is determined. No Lagrangian no gate symmetry. It's manifestly Yang invariant. So we'll see it's manifestly Yang here. But, that's a very nice thing, that we're dealing with an object which is perfectly well defined.
Then we've got to deal with the issue that the actual amplitude is choosing a particular contour of integration which forces the introduction of infrared divergences. We'll see that actually those infrared divergences are not only not some kind of annoyance, but they're in fact the entire story of what the scattering amplitudes turn out to be. So you'll understand that by the end. It's not like amplitudes exist, and they're also infrared divergent-- too bad. It's the only reason they're nontrivial is because of infrared divergences It gives us a very big clue about how to actually do the integral.
So the integrals are even less complicated than they seem. They're not just four-dimensional integrals, eight-dimensional integrals. They're much, much simpler integrals, which are dominated by places where the infrared divergences happen. The infrared divergences are like an anomaly for this Yangian symmetry, and the statement is that the entire amplitude is actually completely controlled by this anomaly. That's a hope, which we haven't made-- it's starting to become sharp. But that's not something I'm gonna talk about in these lectures, how to do the integrals.
But just backing up, the integrand is a nice, well-defined function of loop momentum and the external variables that should exhibit, and we'll see it does exhibit all the symmetries.
STUDENT: Nima?
NIMA ARKANI-HAMED: Yes.
STUDENT: For regular, we say the infrared [INAUDIBLE].
NIMA ARKANI-HAMED: Yes, that's right.
Oh, we'll see all of that. We'll see all of it. Well, I don't know if we'll see it, but it's--
OK, one final set of words to talk about is there is his nice big integrand that depends on the external momentum and the loop momentum, and we just decided that if we do it on-- if we integrate it-- usually [INAUDIBLE], it gives us infrared divergences. So we should maybe look at the integrand itself.
But the integrand itself is more complicated than just the tree amplitude, because it depends on more variables, the loop integration points. So you can ask a natural question. Let's say I just want a set of functions that are coming from the loop integrand but don't depend on the loop. In other words, can I do the integral on some other contour that isn't infrared divergent? If you can, then that's some other data associated with this integrand. It's like knowing its poles-- knowing the poles and the residues of the poles of the integrand, now thought of as a multidimensional complex integral.
Those objects-- so this integrand is well defined. But we can do the contour integral. We normally think of the integral as a real integral-- l goes from minus infinity to infinity for l0, l1, l2, and l3. But you can think of it instead as a complex contour integral. And if you do, then instead of doing the integral from minus infinity to infinity, you can instead do the analogue of the residue integral-- the integral that encircles the poles of the object.
The integrals that encircle poles of this object are called the leading singularities of the scattering amplitudes. There are also data associated with loop amplitudes now that have no infrared divergences, are perfectly well defined. Everything is fine because they're coming from the integrand. In fact, they're just the poles of the integrand. Yeah.
STUDENT: So when you say poles, you mean poles in the internal momentum variables?
NIMA ARKANI-HAMED: Yes, poles in the internal momentum variables. So just so you have an idea of what it looks like, let's say you have a d4l integral, have one loop.
Let me first make some general comments about-- this is going to be-- for five minutes, we're going to switch and talk about the multidimensional complex analysis. So let's say you have many variables, z1 up to zn, and you have some function g of z, and a bunch of polynomials downstairs. I don't care how many polynomials you have downstairs, but however many you have, let's just write them as a product of n factors, so p1 of z up to pn of z.
The point is that in order to define-- now we want to figure out how to define a residue for this guy. So in order to define a residue for this guy, we have to find a point where the denominator blows up. So there's n variables, so we should put n polynomials to 0. So the residue is defined from the point z star, or p1 z star up to pn of z star is equal to 0.
How should we define the residue? Well, go to the neighborhood of where p1 up to pn vanished. In that neighborhood, just switch variables from z1 to the p's themselves. That's a good idea, because in that neighborhood, just say that p1 of z is equal to u1 plus dot, dot, dot in the neighborhood of the z star, right?
So this thing-- so if write dz1 up over the dzn, so I want to figure what to mean by this integral, then now it's becoming in that neighborhood, I would write it as du1 over u1 to dun over un, because all of these are now u's in that neighborhood. But there's a Jacobian, which is 1 over the determinant of dpi by dzj, evaluated at z star. And the numerator is just g at z star. So I'm just writing the integral, and I've just gone through the neighborhood where the z vanished and written it like this.
Now, you would have to be pretty hard-hearted not to say that the residue of that is equal to 1, or 2pi i, which will be one in these lectures. So I'll observe the 2pi i and what I mean by these d's, d slash.
[LAUGHTER]
So this residue is 1, and so we have a nice definition, then. We have a nice definition that the residue of this function, of g of z over p1 of z and the pn of z at z star is just g of z star over the determinant of d di by dzj at z star. So this is a simple generalization of the standard Cauchy residues for one-dimensional integrals.
But there's a very important dimension. There's a very important difference between higher-dimensional and one-dimensional complex integrals. In one dimension, there's a useful sense in which you can think about the integral encircling, surrounding the pole. In higher dimensions, this is not a sphere encircling the pole. It's really the product of n circles. That's exactly what we did. d1 over u1, d1 over dn is the product of n circles.
In no sense is it enclosing the pole. You can say that it's encircling the pole, but it's not enclosing the pole. And related to that is an extremely important feature of this residue. It's not a number. It's antisymmetric in p1 up to pn, because the determinant is antisymmetric in p1 [INAUDIBLE].
So it's a naturally alternating object. The residue depends on the order in which you present p1 up to pn. So if someone just handed you the function, the function all by itself, there isn't a well-defined residue associated with it. But a function together with an ordering of how you present the denominator, there is a well-defined function associated with it.
So this-- it's perhaps a little bit surprising. You would have thought that the theory of these higher-dimensional residues would have been worked out in the 1800s something. But it really only started being properly understood by [INAUDIBLE] and his friends starting in the 1950s. Yeah.
STUDENT: Why do we need a z star set, because that all these [INAUDIBLE] go to 0?
NIMA ARKANI-HAMED: Because we're trying to determine-- and we want to find a special point for n variables. You want a present zn. So I'm trying to determine n nz, so I have to give n equations to determine nz. It's just equation and unknown counting.
STUDENT: OK.
NIMA ARKANI-HAMED: So I have two variables. Then to find a point, I have to set two polynomials to 0. OK, so that's the definition of a residue. Yeah.
STUDENT: You said it wasn't determined just by functions, but is it true that you could get it just by doing an integral? If that's telling you what the function is--
NIMA ARKANI-HAMED: Well, by doing this integral, but at this integral, you now have to tell me an ordering and an orientation for how you do it.
STUDENT: I see. That's the--
NIMA ARKANI-HAMED: That's the extra information you need, exactly. That's right. Yes.
STUDENT: That's right. Because originally, z is just [INAUDIBLE]?
NIMA ARKANI-HAMED: No, no, no, z is many variables. I'm just abbreviating z to be z1 up to zx.
STUDENT: Oh, right.
NIMA ARKANI-HAMED: Sorry. I should have said that. Now, just to complete our crash course, there is a higher-dimensional analogue of Cauchy's theorem that if the function vanishes at infinity, the sum of all the residues is 0. That higher-dimensional analogue is known as the global residue theorem. And it's basically the same statement. And if you sum over all the residues, of res of f of the z star, that's equal to 0. If you sum over all the z stars, residue of every z star is equal to 0.
STUDENT: Under the assumption that--
NIMA ARKANI-HAMED: That the function vanishes at infinity, appropriately. So g of z doesn't blow up to infinity. It will always be true for our examples, so I'm not having all the time to work that out. You just count. It's true.
But again, there is an important distinction between one and higher dimensions, which is that if I hand you some given function like that, there are several residue theorems associated with it, not one. Let me give you a simple example. Let's say we have a function of two variables, f of z1 and z2, just some g of z1 and z2, but there are polynomial factors downstairs-- p1 up to p7. Let me write that as q1 up to q7.
Now, I get a residue. How many residues do I get? To get a residue, I have to write this as the product of two factors. So there's 7 choose 2 equals 21 different residues that I could be talking about. So this function has 21 different residues in the complex z1 and z2 plane.
But now, for every single way of breaking the denominator up into the product of two pieces, I get a different residue theorem. They're related to each other, but I get several residue theorems and not just one. So these higher-dimensional residue theorems are much more powerful than their one-dimensional-- than the one-dimensional.
And as you remember from your complex analysis courses, Cauchy's theorem was a very good way of proving some cool identities. And it will turn out that all of these fancy identities are just the global residue theorem, because as we'll see, all of the terms that appear in BCFW and appear in the loop amplitudes and everything are actually residues of some contour integral over a given space.
So that was the crash course on complex analysis. So let's return to just talk about the leading singularities. So if we're talking about a one-loop integral, and then to define a residue, I would have to find a place where there's four poles downstairs. Because there's four l's-- l0, l1, l2, l3-- I have to define four poles downstairs.
Furthermore, I know that all our poles that define the diagrams-- again, they're all just propagators. So what are the possibles poles of this integral? Well, what they correspond to is every way of dividing the external particles into four sets.
Let me just draw the picture. So for every way of diving the external particles into four sets, there is some set of Feynman diagrams-- of all the Feynman diagrams there are, there's some set of Feynman diagrams that look like this-- have this four set on one side, and then some lines connecting to this set on the other side, that set on the other side, that set on the other side. And I will have associated with putting back propagator on shell, that propagator on shell, that one on shell, and that one on shell. So that's called a quadruple cup. You can't cut any more than that because there's only four of four variables.
But the moire formal way of thinking about it is this is nothing more than a residue of the loop integrand. It's one of these complex resolutes of the loop integrand. Now, you can actually compute the residue just from tree data. What is the residue?
It's the product of that. Because these are now on shell, whatever that is, the sum of all those Feynman diagrams is just a tree diagram here. So it's the product of that tree amplitude, that tree amplitude, that tree amplitude, that tree amplitude. Some overall helicity is going around the loop.
And evaluated-- at defining value of the internal momentum that puts the internal lines on track. So that's what a leading singularity is. And in fact, you can define it without even knowing what the full loop integrand. How do you do it?
You just glue together. You draw what looks like a loop diagram. So that's the one loop [INAUDIBLE]. Here's the two loop [INAUDIBLE].
I need to have eight poles now. I have two loops and then eight variables. So need to draw something that looks like a planar loop diagram. But now I cut all of these lines. So there's eight physical lines now. And I can cut all of them. That's why I've got to make it look like a pentagon box to get eight lines.
But this too loop leading singularity is just the product of tree, tree, tree, tree, tree, tree, tree. But it knows. It has a memory of the fact that it's coming from the loop because I'm evaluating that these internal lines, of the value of the internal lines that puts all of these things on shell.
And of course, the particular value of those momenta we'll definitely know because of this topology of the two loop diagram. So yes?
STUDENT: Why does the location of all these poles turn out to be on shell?
NIMA ARKANI-HAMED: Oh, no, because remember, we're trying to define the residue. So to find a residue, we're finding the poles of these functions, the poles of this one over p squares in the Feynman propagators. So when we go in the neighborhood of those poles, it's precisely putting all these internal lines on check.
So what the leading singularities are, they are numbers, series of rational functions associated with the theory. And I'm just telling you how to compute them even if you didn't know the actual integrator. You just take three amplitudes. You glue them together as if you were drawing loops, just as if you were drawing loops. But then instead of the internal propagators being one over p squared, they were delta p squared. That's another way of thinking of it.
STUDENT: I lost you at this point. When you have to draw the diagram, it looked like it was four internal properties?
NIMA ARKANI-HAMED: You have to draw a loop diagram of four kinds of L internal propagators where L is the number of loops.
STUDENT: It goes for one loop?
NIMA ARKANI-HAMED: Yes. For one loop, you just draw something that looks like a box.
STUDENT: If the box diagram is specific, if you draw in diagrams, not--
NIMA ARKANI-HAMED: No that's right. No. That's right. Exactly. There's no such thing. There's nothing to five. You just draw blobs. You draw blobs that look like they would be a Feynman-- they're not Feynman diagrams. You just draw things that look like planar diagrams with four L loops.
And you have to distribute the external particles around the outside however you can. The vertices are not Feynman-- they're not Lagrangian vertices. They're full amplitudes. They're full lower point tree-- they're full tree amplitudes.
STUDENT: [INAUDIBLE].
NIMA ARKANI-HAMED: Yeah. So there are no big similarities associated with those. These leading singu-- you see, a general amplitude will have many other poles and many other singularities. This is the sort of most singular piece that any amplitude must have. It must have a leading singularity.
Generically, we'll have some leading singularities. There are always sub-leading the singularities as well. What turned out to be special about n equals 4 is the leading singularities determine everything.
STUDENT: OK.
NIMA ARKANI-HAMED: But I don't hear too much into the details of this because this is not-- it's not as important now as it was before, as it was six months ago. But I just want to stress that there's another nice set of rational functions associated, which is closely related to the integrand.
It's like the poles of the integrand. The poles of the integrand are also well defined. Of course, what integrand is well defined. So if you have that, that's the best of all. But even the poles of the integrand a set of data associated with the loop amplitude.
It turns out that all of the generalized unitarity methods that people have been using for computing loop amplitudes are basically using these leading singularities to actually calculate the amplitudes. I won't go through how that works because we don't need to do it that way anymore. We just write down the answer.
But I wanted to mention something. I wanted to mention something, which is that something that people have noticed, but not very explicitly. But it kept being surprising over and over again when people did calculations with these things. It's that these leading singularities as well-- well, let me say one thing quickly. The BCFW terms are special cases of particular source of leading singularities. I won't go through the reasoning for that.
But one loop, a particular loop, cost of one loop leading singularities turn out to compute the BCFW terms. So BCFW terms are a special case of leading singularities. That's not totally obvious. But it's true.
But people had noticed that the leading singularities satisfy strange sorts of relationships. Let me give you an example. If from the rule that I told you, you could draw arbitrarily complicated-- 100-loop leading singularities, you just draw a very complicated diagram, and you do-- I don't know-- everywhere, and it spits out some rational function. Now, if you do this for MHV amplitudes, one loop, two loop, three loop, you start noticing something strange, that all the rational functions that you get from this process are either 0 or 1 times the MHV tree amplitude, or negative 1 times the MHV tree amplitude.
You're never going to get two times the MHV tree amplitude. You just get just the MHV tree amplitude. It's strange. You think you've generated a whole, bizarre, huge class of objects. But no. For the MHV, you just get the MHV tree amplitude all the time.
For the six-particle MHV amplitude, you can do the calculation, and in the one loop you find new objects. At two loops, you find the same objects appear again. You run out of them. So even though the definition I gave you from field theory gives you-- and here's a nice set of objects. They really are.
But the field theory definition of these objects does not make obvious they satisfy certain really interesting relationships between each other. Another set of relationships is that these objects also as functions, as rational functions, satisfy lots of linear relations between each other. We saw that already. That six term identity is a special case of a linear relation that's satisfied between leading singularities.
And in fact, physical things, like unitarity, the infrared, consistency of infrared divergences, and all that stuff, when you compute these amplitudes using generalized unitarily, those physical statements turn into directly or in one-to-one correspondents with these funny linear relationships that are satisfied between these leading singularities.
So there's a suite of functions out there that leading singularities are the amplitudes, which their field three definition does not make obvious. Also it's a remarkable properties that they have. That's exactly what-- that's really what we can suspect that was a duel theory.
Even though everything is weakly coupled, there was no obstruction to defining what these are in field theory, but the objects have bizarre properties. They are not many obvious when you give their find the diagram origin. So what we're going to talk about first is a theory. This part is not completely understood. It's the theory that gives you the dual theory for all leading singularities.
STUDENT: OK. I have a question. So here, everything that we've been doing, all the integrals, these are already in dual space?
NIMA ARKANI-HAMED: Well, yeah. So going back and forth, l is just x minus somebody. That's right. I should have written it in x. Yes. Yes. OK. OK. So yeah. So we'll talk about the-- so we'll first talk about something that computes all the leading singularities and tells you from where all these relationships comes from and everything. And then from there, it will be a quick step to using this machinery to compute the whole integrand.
But as I mentioned before, we still don't have the-- I think still the most invariant, deepest way of talking about the integrand. But at least we have the answers. So we can start it. OK.
Oh, so the leading singularity should also have all the symmetries of the theory, by the way. So leading singularities should be conformally variant, dual conformally variant. OK. So then we finally-- and we can have a discussion by talking about the last set of things that's going to be important, which is twistor space.
So this is actually-- there's a lot that you can say about twistor space. There's very little that you actually need to know. So I'll tell you exactly what you need to know. So the quality which was Penrose's motivation, which is incredibly modern and holographic, is that instead of talking about points in space time, you want to talk about light rays in space time.
His motivation was precisely that gravity makes space time well-defined, all that stuff. So you should instead talk about things out at the boundaries, things that go to infinity, light rays go to infinity. He said all these things in 1964. But that was the essential motivation.
So let me tell you the most basic [INAUDIBLE] about this twistorial correspondence. So let's say you want to talk about a light ray in space time. Everything is going to be complexified now. So if for a second you'd think about the real null ray and Minkowski space, that's a five-dimensional space, five-dimensional space. Because the null ray can point in some direction.
That's two directions. And then you have to give-- there's three-dimensional vector origin from any origin. So that's a five real dimensional space. The realization, Penrose's realization was that it has a very natural complexification for something that's only six real dimensional, three complex dimensional.
The most naive complexification for the set of null lines will be ten real dimensional five complex dimensional. But there's a much more natural one, which is six real dimensional, three complex dimensional.
So let me just tell you what it is. It's extremely simple. It says that to define the null ray, the null ray or all the points in space time that satisfy the equation, mu tilde alpha dot minus x alpha alpha dot, lambda, tilde-- oops. You can fill the alpha minus x alpha alpha dot, lambda tilde alpha dot is equal to 0.
So there are two parameters here, mu and lambda tilde, which are free. So that's the data. If you give me a mu and you've given me a lambda tilde, given a mu and a lambda tilde, I then look for all the points in space time that satisfy this equation, all the x's that satisfy that equation.
By dimension here, by the way, whenever you see mu, it will have alpha indices rather than alpha dot indices. And so even though it has a tilde, it has an alpha index because it's a mu. OK. So let's quickly see that this, in fact, defines a null. It, in fact, defines a-- it does what I say.
So let's say we have two points, x1 and x2. This is clearly a line. It's obviously a line. Let's say you have two points, x1 and x2. So let's look at-- so x1 satisfies this equation. X2 satisfies this equation. So x1 minus x2 immediately satisfies. But x1 minus x2 lambda tilde alpha dot is equal to 0. But that means that x1 minus x2 is a 2 by 2 matrix has a vanishing eigenvector, which means that it's determining vanishes, which means that x1 minus x2 squared is equal to 0.
So that means that x1 and x2 are now separated. So any two points on this slide are now separated from each other. So this is a complex line. Mu tilde and lambda tilde are complex. Everything is complex It's a complex line.
So we've seen that there's a correspondence between a-- there's a correspondence between a complex line-- I want to keep saying complex-- between a line and space time, a null line and space time. And there's a flight, mu tilde lambda tilde in something called twistor space. So just these four variables sticking together, you can call them-- so this is mu tilde alpha lambda tilde alpha dot. Then we can call this z or w-- sorry-- w. And we get an index i. So i runs from 1 to 4.
STUDENT: So you do it only regular Minkowski space?
NIMA ARKANI-HAMED: Right here on the Minkowski space.
STUDENT: And with twistor space, it's also the metrics of Minkowski?
NIMA ARKANI-HAMED: No. It's not going to have any metric at all. That's one of the beautiful things. So there's no metric, nothing. Physics in Minkowski space is all just projected geometry. There's no metric, nothing.
STUDENT: Have we complexified anything yet, or no?
NIMA ARKANI-HAMED: Mu tilde and lambda tilde are complex, these points extra complex. Everything is complex. There is a real slice. So Minkowski space corresponds to a particular section in this twistor space. But understanding that very well right now isn't going to be important.
STUDENT: So this is a generalization of twistor space--
NIMA ARKANI-HAMED: No, no, no. It's just taking-- it's giving you a simple set of complex null lines in Minkowski space. If I specify mu tilde and a lambda tilde, which are general complex numbers, it gives me a line in Minkowski space, complex line in Minkowski space.
If you want to visualize it more easily in terms of real variable, if we're in 2 comma 2 signature, than everything would be real. Mu tilde and lambda tilde could be real. And what x is would really be a sheet. It's really two real dimensional. There's a two real dimensional sheet in two-two symmetry, which is null. And all the points are now separated from each other on that sheet. But we call it a complex line, so complex null 1 in Minkowski space. Yes?
STUDENT: So this eight real dimensional right now or is--
NIMA ARKANI-HAMED: No, no, no. Oh, yeah. Oh, that's right. It's eight real dimensional. It's about to become six because you'll notice, of course, there's an obvious redundancy here, which is this equation, I'm describing exactly the same null line if I multiply this by any complex number.
So mu tilde lambda tilde and any complex number times mu tilde lambda tilde should be identified with each other. So mu tilde, so w and tw, or net should be identified or the same. So really, w is not nc4. It's ncb3. It's just-- so complex points and four dimensions [INAUDIBLE].
STUDENT: So you're pulling this off with a twistor.
NIMA ARKANI-HAMED: I'm sorry?
STUDENT: You're pulling this off by a twistor. But it looks like a direct space.
NIMA ARKANI-HAMED: Yeah. Yeah. It looks like a direct spin. Yeah. Yeah. The Brits often call [INAUDIBLE] twistors too. So but no. Of course, they're not components of some-- yeah. These are all the [INAUDIBLE] variables.
STUDENT: Sorry, a a brief organizational question. So the next talk is at 12:15. And the question is do you want to get some food before now?
NIMA ARKANI-HAMED: No, I don't, actually.
STUDENT: So we can go for another 20 minutes here and--
NIMA ARKANI-HAMED: Is that OK?
[LAUGHTER]
NIMA ARKANI-HAMED: That's all I need, actually. I need 20 minutes.
[LAUGHTER]
STUDENT: [INAUDIBLE]
NIMA ARKANI-HAMED: So-- no, no. I have a personal cutoff, which is I have to finish writing my next talk. So, uh--
[LAUGHTER]
STUDENT: --for the next time. Yes?
NIMA ARKANI-HAMED: Yeah, it's the cutoff for the next time.
STUDENT: The next talk is 12:15.
NIMA ARKANI-HAMED: This one can only go to 12:14 as far as--
[LAUGHTER]
STUDENT: So this one hour.
NIMA ARKANI-HAMED: Oh, OK.
STUDENT: The next talk is probably--
NIMA ARKANI-HAMED: A talk.
STUDENT: I mean, the relative [INAUDIBLE].
STUDENT: I think if Nima keeps talking.
[LAUGHTER]
NIMA ARKANI-HAMED: OK. Right. OK, so, there's this amazing correspondence, then. So I have-- so I'll keep writing you this "mu, tilde, lambda, tilde," but remember this is modular over all the scale. So this is twistor space. OK?
So a line in space time is a point in twistor space. Now, another important piece of the correspondence is the other way around. A line in twistor space corresponds to a point in space-time. So let's see why that is.
Let's say I have two twistors, now-- two twistors. So, mu tilde 1, lambda tilde 1, mu tilde 2, lambda tilde 2. Each one of these is associated with some null line. OK? And we can-- null complex lines.
And we can wonder what the interesection of those two null complex lines is. OK? So we can see, is there an x which is on both the line associated with twistor 1 and the line associated with twistor 2. So we're trying to solve the equations mu tilde 1 minus x lambda tilde 1 equals 0 and mu tilde 2 minus x-- same x. Lambda tilde 2 is equal to 0.
Is there any solution to these equations? The answer is, of course there is. We just multiply this one by lambda tilde 1, that one by-- uh-- that one by lambda tilde 2, that one by lambda tilde 1.
Subtract from each other, and you get that x x alpha alpha dot is mu tilde 1 alpha lambda tilde 2 alpha dot minus mu tilde 2 alpha of lambda tilde 1 alpha dot over 1 2-- lambda tilde 1 lambda is over 2. OK?
So, given two twistors, given two twistors, there is a unique point in space-time associated with two twistors. OK? But notice that, if you hand me any solution of these two equations, if I do anything two-by-two linear transformation on these equations, I'm getting the same solution. I'm just-- it's exactly the same equations I'd be solving.
So that means that this point in space-time, if I do any two-by-two linear transformation that mixes 1 and 2, I'm getting exactly the same point. So, if you think of these as vectors in C4, then, if I have-- then I have two vectors in C4. I can do any two-by-two linear transformation on them that I like, and it's ending up at the same point in space-time.
So that means that this is really associated with a planning in C4. And I give the two vectors, it specifies a plane. Any two-by-two linear transformation is specifying the same plane, so it's really a plane in C4. Which corresponds to a point in space-time.
That's actually the most invariant way of saying it. This two-dimensional plane in C4 is-- since we're going to be talking about Grassmannians a lot later-- in general, a k plane n dimensions is-- the space of k planes in n dimensions is known as the Grassmannian Gk,n. And this is a very simple example. G2,4 is the same as Minkowski space. So this is the correspondence.
By the way, all these things-- so, Penrose emphasized all their importance and talked about them a lot, and so on. But all these issues go back to the 1800s. This is known as the "Klein correspondence," for example. And [INAUDIBLE] did it in the 1890s, this realization that G2,4 is the same as four dimensional space.
Anyway. OK. But, because we like to think about these twistors individually as living in CP3, we say that it's a line in CP3. So a line in CP3 corresponds to a point in space-time. So a line in twistor space is a point in space-time.
So if I-- any point on this corresponds to some point x-- some point x. Now, one of the beautiful things about twistor space is that the action of the full conformal symmetry becomes incredibly simple on these variables. Let's remember that conformal symmetry has the sort of trivial part, which is just rescaling everything. And then it has the exciting part, which is the inversions. And, when we normally talk about the way inversions act, in ordinary Minkowski space, then the inversion-- then the-- well.
So, inversions are x mu goes to x mu over squared. OK? So that's a big nonlinear transformation. And you can do an inversion, and then do a little translation, and then do an inversion back, and that gives you an infinitesimal. That gives you an infinitesimal generator of the special conformal transformations.
But the point is that the special conformal translations are not associated with linear generators. They're like "d squared dx squared." They're not "ddx." So just the usual space-time way of talking about Lagrangians and-- well--
Normal Minkowski space really makes an artificial distinction between the Poincare group and the full conformal transformations. The special conformal transformations act in a more complicated way. OK? And that's because inversions are hard.
Now, an obvious fact among inversions is that, if you have-- so, if you have two points-- and let's say you're just in Euclidean space. If you have two points, the distance between the points is not invariant under inversions-- obviously. So the distance squared goes to distance squared divided by r1 squared r2 squared. So it's not invariant.
But if two points are null-separated, the null separation is conformally invariant. OK? So that's interesting. Conformal transformations leave null-separated points null-separated.
So that makes you think, well, they must do something nice in twistor space. In fact, they do something beautiful in twistor space. Let's just go back to this very basic definition. OK?
So let's say you have mu minus x alpha dot lambda tilde alpha dot is equal to 0. Now, so, all the points x over x squared will also be some null line. Right? Because of this fact that, if two points are null-separated, then their inversions are also null-separated, then, if all the points that satisfy-- this guy's a null 1, there's a difference null line, which is the null line you get by applying an inversion on it. It's not the same null line, but it's some other null line. OK? OK.
But what null line does that-- what mu tilde and lambda tilde does that null line correspond to? It's really really! Take this equation and divide it by x equation.
Then we get lambda tilde alpha dot minus x of alpha dot over x squared mu tilde alpha is equal to 0. Just dividing the equation by x squared. So you see that the null line associated with x over x squared has, instead of mu tilde and lambda tilde, it's given by lambda tilde and mu tilde!
So inversions are nothing other than swapping mu tilde and lambda tilde. Inversions are nothing other than the special, simple linear transformation that switches lambda tilde and mu tilde. And, in fact, the full conformal group, which is SL4, is just the four-by-four linear transformations that act on W. OK? So the conformal group is just linear transformations acting on these variables.
Now, when Penrose did all of these things, it was very related to position space. But one of the great things that [INAUDIBLE] did in his paper is allow people to much more easily make contact with standard momentum space amplitudes, and twistors, by realizing that things are very simple in 2,2 signature. So now I want to tell you how to go to twistor space-- um-- what amplitudes look like in twistor space if someone hands the to you in momentum space.
And, for simplicity, let's say someone hands them to you in 2,2 signature. So we have a function of lambda and lambda tilde. OK? We'll put the SUSY in in a second. Then--
And what I'm about to do, by the way, is yet another way you could have discovered twistor space. OK? So, if Penrose had not existed and these things not happened, this is something that you might have been motivated to do, in order to make the action of the little group as simple as possible. So the action of the little group has this ugly feature-- that it makes you want to think of lambda and lambda tilde as projective variables.
But you can't. Why? Because you're rescaling the models that lead to each other. There's really three degrees of freedom, here, but there's no nice, canonical way of seeing that there's three-- exactly three. Right? I can-- OK?
It's because lambda and lamba tilde are rescaling oppositely. Wouldn't it be nice it we could write the amplitudes in a way where the action of the little group is just some totally homogeneous scaling of everything? Now, what's the very simple way of doing that? Just Fourier transform with respect to the variables.
So we can define an M of mu tilde and lambda tilde to just be the Fourier transform with respect to lambda e to the i lambda mu tilde of M of lambda and lambda tilde. If these variables are real-- which there are-- in 2,2 signature I'm allowed to do this. And this then takes you from a function of lambda lambda tilde to a function of W. OK?
Furthermore, now, the little-group action is exactly W goes to T times W. Right? So, great. So W's actually in CP3. W's a projected variable. Exactly as it should be.
And a final lovely thing, when we put SUSY in-- let me just put an eta tilde here, and an eta tilde here-- you still have Fourier transform with respect to lambda-- is you find-- and this is an extremely simple exercise, so I'll just quote the result. You'll find that M becomes a function of a super-W, if you like, where the super-W's components are the W and the eta tilde.
And you'll find a, for all the particles-- I'm sorry. I should have said this [INAUDIBLE]. You'll now find that the little-group property-- and now I'm talking about Yang-Mills-- you'll find that M of tW is just M of W. There's no weight. There's no minus 1, there's nothing. The amplitude is just a function of weight 0 on CP3/4. OK?
So you would have wanted to do this even without any deep motivation-- just from little group, you'd be motivated to the same, and you'd find, to your delight, that the whole super conformal group acts incredibly simply. It just acts as four-by-four linear, super linear transformations on this W. OK? To boot. But just what the properties of the function are-- they're just a function of weight 0 on CP3 and then both 4 Fermionic variables is fine.
So you couldn't ask for a simpler thing for amplitudes-- a simpler characterization of what they are, just kinematically. There's nothing deep going on, here. It's just a basis choice. Just a basis choice. But sometimes a basis choice where you're giving the amplitudes as much opportunity as they need to look simple. Yeah.
STUDENT: So where do you find your first W before you went to super symmetry? So are you going to determine that in the system like you just described? And, if so, then what is that null line that corresponds to that?
NIMA ARKANI-HAMED: Sorry-- what does the null line-- the correspondence do--
STUDENT: I mean, you started off with these two variables describing a null--
NIMA ARKANI-HAMED: What this is doing-- so-- there-- so-- these mu tildes, lambda tildes, are also associated with null lines and space-time, and stuff like that. It's just that we're not used to thinking about scattering amplitudes in position space. We're used to thinking about them in momentum space. OK? So this is giving you a very effective way of going from momentum space to the amplitudes and is exactly the same twistor space.
You should really think of those-- you just think about W as also being associated with a null line in space-time. I mean, it is. If we find singularities in twistor spa-- you know, we also interpret those things as singularities in the space-time. But we got there starting from momentum space-- just much more convenient. We don't have to Fourier-transform back to momentum space and figure out how to go-- that was the big, just, technical obstacle to making progress on scattering amplitudes, before.
But, thinking about it in 2,2, signature, we can just do this straight Fourier transform that's very straightforward. You see, Penrose hated 2,2, signature. And this act of going back and forth like this is normally called the "Penrose transform" and involves contour integrals and check of homology and all sorts of crap. But, in 2,2 signature, any idiot can do it. It's Fourier transformation. OK?
[INTERPOSING VOICES]
NIMA ARKANI-HAMED: No. So we actually won't be spending a heck of a lot of time in ordinary twistor space. I'm telling you these things because it's going to allow certain things to become obvious. It'll make it obvious that certain things are super-conformal-invariant, for example. OK? But we won't be spending much time doing computations there.
STUDENT: All this is in 2,2 signatures?
NIMA ARKANI-HAMED: This is an easy to say in 2,2, and it's true in any signature. But I'm telling you how to say it in 2,2. OK? Let me just give you a quick-- but just a quick-- very quick example of how lucky things are. Let's take our favorite three-point function.
Now, the whole function has also this delta function of momentum conservation and also the delta function for-- also the delta function for a super [INAUDIBLE]. So that's what it is. It's pretty simple and has those delta functions, too.
Now, let's say we Fourier transfer on this guy to twistor space. And I just want to just write down the answer. It turns out, in this case the most convenient way of writing it is to take one of the particles and make it a W and the other two particles and make them Z's. So the duals to the W's. So, use the eta representation for two of them and the eta tilde representation for the other guys. And Fourier-transform in respect to lambda tildes and not lambdas.
It's not very important, but I just want you to see how nice the answer ends up looking. It's sine W1 dot Z3, sine W2 dot Z3. So it's 1 and minus 1. OK? As nice as it is here, I think you'll agree that that's even nicer.
And, in fact, what led very directly to this Grassmannian formula was playing around with the amplitudes in this space, seeing what the generalization was here, figuring out the diagrammatic rules in twistor space, writing the BCFW terms, realizing that these were the same as in the diagrams Hodges was drawing for many years, and then seeing that the simplicity of the structure is now ultimately seen is actually reflecting this underlying Grassmannian structure, in particular gauge.
So we're not going to pursue that line much further, but just to say, it's not all garbage. You can actually do some of these Fourier transforms, and you see something really wonderful happen. OK?
STUDENT: So maybe this is a good point to stop.
NIMA ARKANI-HAMED: OK, great. So, tomorrow I guess I have to start with momentum twistor space, and then we'll get to talking about Grassmannian. And then Friday we'll do the [INAUDIBLE].
The third in a 5-part series of technical lectures on scattering amplitudes given by Prof. Arkani-Hamed in conjunction with his Messenger lectures on fundamental physics at Cornell University. The focus is application to N=4 supersymmetric Yang-Mills Theory.
