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NIMA ARKANI-HAMED: I want to start today by doing a little sample computation with this Grassmannian integral. And then tell you some ways of thinking about what's going on with it. Actually, I'll tell you some things that are not in the literature, yet, that will hopefully come out in a paper in a little while. Then I hope I will be able to go tell you about [INAUDIBLE].
So let's remember the object we introduced. We introduced something we called L-n, so now I'm doing-- the notation's going to be consistent today. So, L-n, m. Think about m as the number of negative helicity gluons. Negative, minus, m is minus. The number of minus helicity gluons. And this is the integral, e-n times m C alpha a, Over m2 delta m, [INAUDIBLE] m plus 1. And 1m minus 1. Delta 4 slash 4 C alpha awa. Product, [INAUDIBLE] delta m.
So this is in twistor space, and we could go back to momentum space, Back in momentum space, this was a formula that I'll loosely draw, by reminding you that it came from this picture for thinking about momentum conservation with an m plane became the lambda plane, along with the lambda [INAUDIBLE] plane.
And that then motivated us to try to write it as an integral over m minus two planes, in n dimensions, which because of this very special cyclic property of the minors, we managed to do.
And so, what we find, is that back in momentum space, L and m, as a function of lambda, lambda tilde, and eta tilde, [INAUDIBLE] delta 4. So, I could just also write this as the MHV tree amplitude. Times, what we can call r and k of z. Now k, is equal to m minus two, and it's written that way because this k refers to the n to the k MHV amplitude. So it's k equals zero is MHV.
K equals 1 is NMHV, and so on. Where z are the momentum twistor variables. Remember what the relationship was, was that lambda A was lambda A, lambda tilde a was a minus 1 a, ua plus 1 plus cyclic, over a minus 1 a, a plus 1, and eta tilde a, was the same thing. So, this is just reminding you that if I hand you a bunch of momentum twistor variables, you can uniquely reconstruct the function, ultimately, in momentum space, just stripping off the momentum conserving delta function.
STUDENT: [INAUDIBLE]
NIMA ARKANI-HAMED: Yes?
STUDENT: You gave us one of your motivation stories using these pages and the super symmetry to say we don't need to worry about negative or positive looping stance.
NIMA ARKANI-HAMED: Exactly, exactly. What I mean here-- so this is a super amplitude, But what I mean here is that you should think of m as also-- m labels the overall r charge, of this component of the super amplitude. But you can also think of the sector labeled by m as containing at once it's component amplitudes. The component amplitudes that has m negative helicity gluons. It's a whole super object. but if you want, the label, the sector, by the number of negative helicity gluons, but. Is that clear?
STUDENT: Perfectly.
NIMA ARKANI-HAMED: Where RNK is exactly the same formula. So in fact-- So, RNK. 1 over ball GLK. I forgot the 1 over ball GLM here. Product of alpha equals 1k, delta 4 slash 4, e alpha a. So in this way, we see that the object is superconformal invariance, and dual superconformal invariance at the same time.
STUDENT: So it's invariant [INAUDIBLE]?
NIMA ARKANI-HAMED: I could tell you a little bit more about how I see this directly without making any changes to the variable and so on. I could tell you what the Yangian generators are. There's a bunch of linear generators, and there's quadratic generators, and high order generators. I could just tell you what the quadratic generators are, and your could-- I could tell you a check that you can do to see this directly, from either representation. But, it's a little bit boring, so I will skip it.
But as I mentioned yesterday, the mathematical statement is that, all superconformal invariants, are written as an integral over a Grassmannian with some measure delta 4 slash 4, c dot w. That's a quite interesting statement. It's not a completely trivial statement.
It is obvious that things of this form, are superconforming invariance It's not totally obvious that every superconforming invariance is of this form. One of the things that made it easy to run into this formula, is that every superconformal that is easy to write down, turned out to be this formula. But it really turns out that every superconformal invariance is written as an integral or a Grassmannian. In a sense, that explains where the heck this Grassmannian is coming from to begin with. Somehow, closely tied in with superconformal invariants.
STUDENT: Sorry. Superconformal, or super [INAUDIBLE] conformal?
NIMA ARKANI-HAMED: No. Super. Simply superconformal.
STUDENT: So just super?
NIMA ARKANI-HAMED: Yeah. In fact, and the non-trivial part is, the super. Conformal invariance is very easy. you can build it. If I hand you four twistors, you can just build a four bracket out of them, you can build conformal invariance without a problem.
It's the super part which is the problem. The super part is quite non trivial. And it turns out that this is the-- the claim is this is a unique way of making things superconformal invariant. So you can then check. You can then check these quadratic generators in any one representation.
The dual superconformal generators are certainly quadratic operators. Just write it down, it's very simple. But they're certainly quadratic operators. And then you can just ask, what does it take for those quadratic operators to annihilate the object? By annihilate, you mean, when you bring it under this derivative, you turn the integrand into a total derivative. And then you find there's a unique measure which does it, which is that product of the cyclic minors thing. So this is really the object that generates all [INAUDIBLE].
All right. Now let's actually, just so you can see what it looks like, let's do a computation. So let's look at it all NMHV amplitudes. All residues here, they correspond to NHMV. So, as you can see, for practical computations, it's always easier to do it in momentum twistor space. Simply because you have a lower dimensional integrals to do.
So, NMHV is N equals 3-- three negative helicity particles. It's K equals 1. So that's the simplest possible example here. These minors are just 1 by 1 matrices, which are otherwise known as numbers. So, 4k equals 1. We have R and 1 is equal to the 1 over ball gl1. The integral following d1 over d1 of the d e n over dn. The 1 by 1 determinants are just--
STUDENT: Why are you doing k equals zero first?
NIMA ARKANI-HAMED: K equals 0 is just 1. Very good. So, k equals zero, the object is just 1. So, we learned something very interesting, right away. The only Yangian invariant, which is MHV, is the MHV tree amplitude. See, R is 1. And this object, L, is just the MHV tree amplitude.
So we learned instantly that the only Yangian invariant, which is MHV, is the MHV tree amplitude. That immediately explains that fact that I mentioned, earlier on, that when you compute leading singularities at a hundred loops, and you glue objects together, you compute one that's MHV, you find that it's either zero, or it's the MHV tree amplitude. And this is the reason why. The theory is Yangian invarniance-- believing singularities have the Yangian invariance, because they're not spoiled by infrared [INAUDIBLE], and so, it's a fact. The only MHV Yangian invariants are [INAUDIBLE].
STUDENT: [INAUDIBLE] always [INAUDIBLE].
NIMA ARKANI-HAMED: Yes, that's right.
STUDENT: So your [INAUDIBLE] must obey some property which--
NIMA ARKANI-HAMED: All of those properties end up being reflected in those cyclic minors. It's a cyclic minor. It's a cyclic minor that does everything here.
Like I said, we could spend half an hour talking about the Yangian algebra and how it acts. But just summarize the story. There's extremely nice papers by Drummond, and Sokatchev and Korchemsky, who went through this argument showing that these are the only things, and so on. So it's very, very pretty, but the bottom line is that the magic is all in the cyclic minors. It is the cyclic minors that make everything work.
STUDENT: So it goes one cyclic to another cyclic?
NIMA ARKANI-HAMED: Yes, yes.
STUDENT: That's really like [INAUDIBLE].
NIMA ARKANI-HAMED: Yes, exactly. Exactly. That's right. Yeah.
Then we just have now, a single delta 4 slash 4 of d1z1, plus dmzn. That's all. This is going to be extremely simple.
So now let's stop and do the simplest case here, which is n equals 5. You see why n equals 5 is the simplest case. But we know it should be.
We're doing NMHV amplitudes, right? But we know that-- that's three negative helicity and two positive helicity gluons, for five particles. That's secretly just MHV bar, right? So this should be really trivial. Let's just see how this works. I have an integral, so let's just write it down. It's 1 over ball gl1, integral dd1 over d1 up to dd5 over d5, delta 4 slash 4, d1z1 plus d5z5.
This gl1 is something that just rescales all of these by a constant amount. So whatever ball gl1 means that I have to gauge fix. I have to gauge fix that, which I can do by setting and any one of d's to 1.
So gauge fixing really means that I'm not doing one of these integrals. Let's say, I don't know, we set d5 to 1. So I'll gauge fix, I'll use the gl1, and I'll set d5 to 1. So I'm really left with integral dd1 over d1, up to dd4 over d4, delta four slash four. The last guy here is just z5 plus d1z1, plus d4z4.
I just want to make a quick comment here. We're about to do the explicit calculation anyway, which is very simple. But this object is manifestly cyclically invariant in the five guys. In fact, the delta function is even permutation invariant, but the measure this is clearly cyclically invariant. So whenever I gauge fix-- see, here I'm gauge checking the gl1-- I'm breaking the cyclic invariance.
But that's fine. It's going to be cyclically invariant. The GLK is, in general-- The answer is going to end up being cyclically invariant. But no matter what you do when you gauge fix, you're going to make not manifest, in the middle, but the answer will end up being cyclically invariant. All right. This is an extremely simple calculation to do. Let's just look at the bosonic delta functions here. All these variables are bosonic.
So there's the proteonic in with the etas. Let's forget about them. Let's look at the bosonic part. This is very easy to do, because in fact, z5 is a four dimensional vector, right? So I can always expand any four dimensional vector in the basis of four other four dimensional vectors. So that's exactly what that's doing. When I integrate over those d's, there's some unique d that satisfies those delta functions. There's a unique d1 of the d4 that satisfies those delta functions. Let's see how we can figure it out.
So, what are those unique d's? Let's write negative d5 as d1z1. Let's say the special d's, let's call them stars. Those will be 2 star, d2, plus d3 star, d3 plus d4 star, d4. OK? How do I extract the coefficient of d5? We do what we always do. We dot it into its dual vector.
So if I take z5, and I look at 5. And now this is a 4 bracket. So I'll do 5,234. So if I contract it with 234, on the one hand, from here, I get d1 star, 1,234 plus 0 for all the rest of them. So I just solved for d1. d1 is 5,234 over 1,234 with a minus sign.
Similarly, d2 it's always five times whoever's not there. So 134, whoever's conjugate to do 134 over 1,234, and so on.
So I just solved for it. So now let's just finish doing the integral. So, what am I going to get? I get 1 over all the d's. I get 1 over 1,234, 2,345, up to 5,123. The five guys downstairs. Each one of them had a 1,234 downstairs.
So I get-- what am I doing? I get 1 over-- actually, the first one is-- I'll write it as 2,345, 3,451, 4,512, 5,123. There is no 1,234. However, each one of these four guys had their 1,234 downstairs. So there's a 1,234 to the fourth upstairs.
However, when I use this delta function, there's a Jacobian, right? The Jacobian is obviously 1,234. It's just 1 over 1,234. So it's 1 over 1,234. That's coming from the Jacobian.
And then I'm left with the fermionic piece of the delta function, which isn't going anywhere. So there's an eta 5 plus. And then here, I put in d1. So d1 would be 2,345 eta 1 over 1,234, plus the other ones. So that's the answer. And we can clean it up.
This is a very important object in this business, so it's worth it. So we can clean it up by noticing that I can bring the 1 2 3 to 4 to the fourth inside this fermionic double function. This is a delta 4. I bring it inside by multiplying everywhere by 1,234 inside.
So what I get-- so r 51 is equal to 1 over 1, 2,345, 3,451, 4,512, 5,123, times delta 4 of 1,234 eta 5, plus 5,123 eta 4, plus delta plus cyclic. So this is a beautiful object. Beautiful as advertised, cyclically invariant object.
Notice that it's exactly the MHV bar amplitude. Just think about it. Each one of these four brackets downstairs is proportional to-- let's think about what these four brackets are proportional to.
So remember, we said that if I have two lines, w1, w2, w3 w4, that 1, 2, 3, 4 was the x associated with the line 1 2, minus the x associated with the line 3 4, squared. So the x associated with the line 1 2, is the point x2. The x associated with the line 3 4, is x4.
So this is x2 minus x4 squared, which is nothing other than-- now remember that p2 is x3 minus x2. So p2 q-- We do p2 is x3 minus x2, p4 is p3 is x4 minus x3. So p2 minus p3, is exactly x2 minus-- is exactly x4 minus x2, so it's r minus x2 squared, is p2 plus p3 squared.
So, this has a pole at exactly the place where p2 plus p3 squared is 0. With the angle brackets. Angle and the square brackets. So we have the overall MHV factor out in front of the whole thing, if we want to go back to the overall amplitude. But we know that this amplitude is secretly MHV bar. It's exactly the same formula with the angle brackets replaced by square brackets. We have to get the square brackets from somewhere, and this is where we get them.
This is actually very nice important object in general. We'll even see it merge directly. Let me just say it right now. If you hand me any 5 twistors, za, zb, zc, zd, and ze, then this is just exactly the analogue. Just replacing 1 2 3 4 5 by increasing the z. In other words, they don't have to be consecutive. So this would just be abcd, eta e, plus cyclic, over abcd, bcde, delta eabc. So this is called an r invariant. Also ought to refer to it with abcde. Notice that it has a natural antisymmetry property. Nice totally antisymmetry property.
STUDENT: The whole thing is called square bracket.
NIMA ARKANI-HAMED: Yeah. The whole thing is often also abbreviated like that.
So now, let's talk about higher. Let's talk about other residues, for other n. Let's first now look at n equals 6. Look at n equals 6.
Almost the same thing. This MHV case, I should say, is unusually simple, because there's no determinants really, here. But, still, we're going to get lots of intuition for what's going on. All right, now we have delta 4 slash 4 always the same, d1z1 plus d6z6. Now you see that we--
STUDENT: Can you go back to that, k was 0, but n equals 5, reproduce this result?
NIMA ARKANI-HAMED: K equals 0, n equals 5, r is equal to 1. r equals 1.
STUDENT: Yeah, but that's dual this, right? It flipped the plus minus conjugate.
NIMA ARKANI-HAMED: Yes.
STUDENT: So how do 1 equal to this-- how do I say?
NIMA ARKANI-HAMED: Oh, no, of course, if I go back to the full [INAUDIBLE] momentum space, it's the MHV prefactor, multiplied by r. For the actual MHV amplitude, r equals one. For this MHV amplitude, this is r. Now the cyclic is, if you took this result, times the MHV, times the HMV, prefactor, you'll get the MHV bar answer. OK?
And the way it works, is what I was just saying. Is that, these-- you might ask, where the angle brackets downstairs is coming from. Because you know the answer has angle bracket 1 2, angle bracket 3 4, and the other ones, right? And the angle brackets are precisely coming-- you have to do a little more work to convert the whole thing, but they're precisely coming from the fact that these guys have those poles in them. These guys have precisely the square bracket poles [INAUDIBLE].
STUDENT: So really you only need to compute half for a given n.
NIMA ARKANI-HAMED: Yes, you only have to do that. Let me remind you of something that we said on very general grounds. The number of integration variables, was the same as the number of degrees of freedom of the number of k minus 2 planes-- oh m minus 2 planes, and m minus 4 dimensions.
So that's something we've said in general, from our momentum conservation picture, which is just k planes and n minus 4 dimensions, with our new definition of k. And we can see that very clearly here. We have the z's are generic, so the deltas, these are-- here it's 1, in general there's k delta functions. So we're subtracting k from-- So this number is exactly k times m minus k minus 4. Which is exactly k times m minus k minus 4k. Right?
So the dimensionality of the Grassmannian in general is k times m minus k, and I'm subtracting 4k from it, which is what I'm getting from the bosonic delta functions. So we can see, for k equals one, here, and n equals 6, we should have, for any k equals 1, in fact, the number we have is n minus 5.
So that makes sense. With n equals 5, there was no integral to do. Everything was just [INAUDIBLE]. So n equals 6, we have one integral to do. Where does the 1 come from again? The gl1 always allows us to set one of them to 1, and not do the integral, there's four gone from here, so now we have to do the integral around one more.
The point is, we now treat all these integrals as contour integrals. So any integral that I have left is trivial to do as a contour integral. It's just going to set it to 0. I'm doing integral dd3 over d3 set. But I'm just doing it around the contour that encloses the pole and d3 equals 0.
I should also say something else. All of these integrals look perhaps funny, because we have delta functions in them, as well as things that look holomorphic. These delta functions really need to be thought of as poles. Everything here is holomorphic. Absolutely everything is holomorphic.
The delta of z, it's really 1 over z, where you're giving a prescription, that you must enclose pole on z equals 0. Everything is holomorphic. These are not-- this is not a real delta function. It's a pole. So really what we have is a whole punch of poles, but I am forcing myself to enclose some of them. All right, so it's very easy to figure out what to do. All we have to do, is put--
I have six possible contours I can choose. Or any layer combinations of them. The contours are, I put, d1 to 0, d2 to 0, d3 to 0, d4 to 0, d5 to 0, d6 to 0. If I set any one of them to 0, then I have five parameters, and I just compute the one that's left. Then everything is soaked up by the [INAUDIBLE].
So, the most general residue that we can have for six points is then, one of those basic invariance, where abcde, or any distinct set of five of the six particles. So there's six residues that we could have. In general, you just pick-- You pick any five, and then you set all the rest of them to 0. So you can label the residue by all the guys that you're setting to 0, that's the sort of official way of labelling the residue, if you're doing complex [INAUDIBLE].
Or you can label them by the guys who are not setting to 0. It doesn't make any difference. It's giving you exactly the same variables. So let's look at it. Let's actually examine the residues for six particles.
There's a residue that you should call residue 1. Which is to remind you that you're setting minor 1 to 0. In this case, minor 1 is just d1. This would otherwise be called 2 3 4 5 6. Or 2 could otherwise be known as 3 4 5 6 1. And so on. 6, otherwise known as 1 2 3 4 5.
STUDENT: [INAUDIBLE]
NIMA ARKANI-HAMED: No, no. This is all supersymmetric. This is all completely supersymmetric. So they're all super amplitudes. If I said another way, had we first just solved for the delta functions, then we'd find there's one tree variable left. And then we'd have a contour going over that one variable. This is one case where it's not multidimensional complex analysis, just normal complex analysis.
In that one remaining variable, we can just call that remaining variable tau, there are six poles. And those poles correspond to whether my [INAUDIBLE]. If I If I put the four bosonic delta functions to 0, all of the d's become a function of one remaining-- one free variable, tau. So I have d1 tau, d2 tau, 3, 4, 5, they're all linear in tau. So we have just linear, simple poles. Six simple poles.
Now, let me just tell you some facts that you see what you inspect these things. I mean, they're a very simple expression, incredibly simple expressions of momentum twistor space. You can take them back to ordinary momentum space and stare at them. And what you find, is that 1, 3, and 5, are the three terms that appear in the BCFW formula for the tree amplitude.
Remember I wrote that expression on this board here? Pretty expression with three terms, right? Amongst other things, it was a little surprising they were all just cyclically related to each other. When you compute using BCFW it's not obvious that they're cyclically related to each other, but they ended up being cyclically related to each other.
Here's it's completely obvious, because it's just equal to one, you cycle it over 3, you cycle it you cycle over 2, you cycle it over 2. And remember, we wrote that second form, that parity conjugate form of BCFW, those end up being these three guys. 2, 4, and 6.
STUDENT: So the first form is the--?
NIMA ARKANI-HAMED: The first form-- these are just the big three terms that occur in the BCFW formula, and these are three terms that occur in the parity conjugate of the BCFW formula. Yeah?
STUDENT: The formula we work out is plus, minus, plus minus, plus, minus.
NIMA ARKANI-HAMED: Yes. Though that was for a particular component. This is the super amplitude. If you take that component of the superamplitude, this is what you find. You can also-- I didn't write it, but you could have written the whole superamplitude on the board. Yes?
STUDENT: There is no integral [INAUDIBLE], right?
NIMA ARKANI-HAMED: No, but these are also not real. Nothing is real here. Everything is complex. These are complex, the delta quantums or poles, everything is complex.
STUDENT: But it's a contour integral. You are supposed to integrate a line [INAUDIBLE]?
NIMA ARKANI-HAMED: No, no. Whenever you have a complex integral, the thing to compute with it is always the residues. The residues are always little circles. so what I'm doing-- If you want to do it in the steps that you just said, we can first satisfy the delta functions. Satisfying the delta functions-- there you could say, either do the integrals, and set them equal, imagine that they're really delta functions. They're really poles.
So I'm telling you, just do the little circle integrals there, it doesn't make any difference. I'm left with one variable, and that one variable in that plane, I have six poles. And they contribute to residues by little circles around each one of the six [INAUDIBLE].
STUDENT: You knew each residue.
NIMA ARKANI-HAMED: I knew each residue. Exactly.
STUDENT: But how do you know which ones to--?
NIMA ARKANI-HAMED: How do I know which ones correspond to these-- oh, beautiful! I'm just telling you, I'm just making a dictionary. Have you noticed, hey, this is cool. These things coming from this crazy Grassmannian integral are the BCFW terms. These things are the other terms. Why am I putting together in this-- I have not told you that. I've not told you that.
That becomes the very cool question, right? Because I know, any one of them, isn't something physical. But somehow, there's something interesting that your-- there's got to be something interesting about the ones we're choosing, that end up corresponding to the physical amplitude. So I'm going to tell you what that is in a second. Yes.
But before telling you that, simply observing that these are-- Simply observing these are BCFW, and those are the parity conjugate of BCFW, immediately proves that 6 term identity. That six term identity is just a Cauchy's theorem because the sum of these residues is equal to the sum of the other residues. So that bitch of a six term identity is just Cauchy's theorem.
So we see that this object has generated these interesting terms for us, but also tells us the relations that they satisfy. OK. Before going on, I'm telling you what's actually special-- oh, actually, in the particular case of the six particle amplitude, the very particular case of the six particle amplitude, this is the unique thing that you could have chosen. And let me explain why.
You might have thought, If you want the answer to be cyclically invariant. You might have thought, for it to be cyclically invariant, you should just enclose all of them. But then you get zero. If we add up all the residues, you get zero. By Cauchy's theorem, the sum of all of the resides is zero.
It turns out it is very easy to check, that once you go to these variables, cyclic invariants is a slightly different statement. That when you clock over one, you pick up a minus sign, if n is even, and not if n is odd. So you actually have to pick up a minus sign. You have to pick up a minus sign, since n is six here, you've got to pick up a minus sign when you clock over a one. So, there's a unique contour, which is then manifest in cyclically invariant. That contour is circle the odd ones, minus circle the even ones. So that's the only contour which is cyclically invariant.
And then, by Cauchy's Theorem, you could write it as either 2 times circling one of them, or two times circling the other one. So in this case, the answer is actually unique. All right. Now, let me tell you what all the terms-- so, if you use BCFW,
STUDENT: So just, in principal, for any kind of amplitude, is there an understanding about how to choose the points to get which [INAUDIBLE]?
NIMA ARKANI-HAMED: It depends on what you mean by an understanding. It is absolutely known how to identify all the BCFW terms. So if you knew about BCFW ahead of time, you know exactly which guys you'll be adding [INAUDIBLE], OK?
Now, if you want to know, is there some deep, in principle, understanding of why you're asking that question? There is even an answer for that, at the moment, for all tree amplitudes. And I'm about to tell you what it is. But the generalization of that, all loops, we're just starting to understand.
STUDENT: So the basic challenge is to get some kind of theory.
NIMA ARKANI-HAMED: That's right, exactly, but you're starting to see what the basic point is. There's all these residues. This is a menu of all the objects that will ever occur in scattering amplitudes. That's a way of thinking about it. Them, and various operations you do on those, is a menu of everything that is ever going appear. They are all non-local, they have all the symmetries of the theory, they are the building blocks. They're the building blocks that are the invariance.
Now there's some interesting rules about how they are put together. And the rules are clearly going to be some rule about an interesting contour. And you see, because it's a contour, it's going to come in many varieties, right? You can rotate it this way, you can push it that way. So we have to learn how to characterize these homology classes. How do you characterize the sort of invariance that the contours represented a lot of? But all of these questions about the emergence of space time, and unitarity, and things like that, get turned into homology questions, topology questions, about the nature of this contour in the Grassmannian.
STUDENT: Just to be clear, if we know the dictionary of choosing those contours and all that for all these r and k objects, then we're done?
NIMA ARKANI-HAMED: No. We know it for all tree amplitudes. Yes, we know it for all tree amplitudes. You see, this is grown into a relatively large story, and I want to tell you part of it. There's actually-- there's starting to be actually, at the moment, there seems to be two very closely related, but two moral ways of understanding, without ever saying the word field theory or BCFW or anything like that, why you choose a particular contour.
One of them is associated with this idea of giving the integral over the Grassmannian, a particle-like interpretation. So giving the integral a particle like interpretation, is-- let me just say the words for this one, because I want to tell you about another one. And this other one's the one that isn't in print yet, but which is easier to explain. It's associated with some very pretty pictures, and may end up being the deepest way of talking about it.
But let me just very quickly say some qualitative words about this particle interpretation. The strangest thing about the Grassmannian integral, is that the integral for n particles, the space you're integrating depends on n. It's gkn, and it changes discontinuously as you go from one n to another. That normally doesn't happen in physics. we normally have n particles, they live in some space. So you don't have to change everything all of a sudden, you add one more particle.
Now you can ask, is there some way that we can make the transition from one particle to another particle to another particle more smooth. And you can, if you choose the contour correctly. The correct contour has the feature-- and this principle is the principle that ends up uniquely determining all tree amplitudes. The contour has the feature that to specify, you have to specify-- in general, remember, to specify a general contour, you've got to specify a whole bunch of polynomials equal to zero.
So the rule is that you have to build it such that-- you have to choose the contour such that, having specified the polynomials that are zero, for the problem up to n variables-- up to n particles, if you an n+1 particle, you don't change anything about what you did before, you just add a few more polynomials. That's technically what we mean by giving it a particle interpretation.
STUDENT: So in a sense then we can deform the polynomials depending on--
NIMA ARKANI-HAMED: No, so you had a contour before, and you're just adding e-- you're just adding information to the contour. You're not spoiling what you did already. So far the integral is now in the world of [INAUDIBLE] space. But you just choose the contour to allow you to think of it as adding particles one at a time.
I don't want to clarify this any more, because then I'll spend the whole time talking about it. So I'm just making statements, you can look at-- and I'll tell you where to see it in the literature. But this line of thought-- this line of thought, first of all, uniquely specifies the tree contour. it's quite remarkable.
It basically becomes completely nailed, what you do, once you start doing it. First of all, completely nails the tree contour. Secondly, having expressed the contour in this particular way, you notice that there was a very canonical way you can deform the integrand, that converts what we did here, into Witten's twistor string theory.
So there's a very interesting connection between this picture, and twistor string theory. Now, twistor string theory doesn't make the Yangian manifest. This makes the Yangian manifest. So it seems to be the kind of UV theory, if you like, that there's something that's making all the symmetries manifest. But this particle like interpretation makes a very close connection to the twistor string. And in fact, had you just thought about integrating over Grassmannians, you didn't think about cyclic minors.
Had you just thought about integrating over Grassmannians with this idea of doing the integral over the Grassmannian with a particle-like interpretation, you would have started from a different starting point, that would have directly led you to twistor string theory. If you never thought about topological a models, you would have just directly written down the formula for Witten's twistor string theory, attempting to do an integral of the Grassmannian with a particle interpretation.
But it would land you on the second form, that's then related to that this one, by the deformation that I was talking about. So this is a very rich subject. But the basic lesson, is that attempting to get a particle like interpretation is one principle that nails the contour. I'm going to tell another one, in a second. So often only there are some geometric object here, and there's going to be a number of invariant ways of describing what it is.
But I want to emphasize, that there is absolutely no doubt, at tree level, and loop level, and all loop level, all the objects are in here. And BCFW tells you how to compute them in tree level, and I will tell you how to compute them at loop level by the generalization of BCFW in a way that makes the Yangian invariance matter less.
STUDENT: So say I start with a five particle order [INAUDIBLE], which means I remove a particle--
NIMA ARKANI-HAMED: Removing a particle is very simple [INAUDIBLE].
STUDENT: I just project the contour that I have onto [INAUDIBLE] space by that logic.
NIMA ARKANI-HAMED: That part is trivial. What's not trivial is going up. So let me just tell you what-- let me tell you what the elementary amplitude ends up being for any n. For any n, I'll tell it to you in two languages. First, which minors are you setting to 0. Second, which minors are you not setting to 0.
So, here, with 6 particles, we've found with the BCFW form, was that you can add 1, plus 3, plus 5. Or, there was some other form of PBCFW, which was 2, plus 4, plus 6. I'm not worrying about minus signs, and things like that. We're all up to minus signs. OK. For n equals 7-- so in other words, you put minor 1, 3, 5, to 0. 2, 4, and 6 to 0.
For n equals 7, there's two variables, we have to set two minors to 0. Always n minus 5, is the number of variables. So the minor that I set to zero, are 1, 2 plus 1, 4 plus 1, 6 plus 3, 4 plus 3, 6 plus 5, 6. Or, in the other form, it's the other way around. 2, 3 plus 2, 5 plus 2, 7 4, 5 4, 7 6, 7.
And this was equal to this, by residue theory. Right? This ends up being equal to that, by the global residue theorem. In general for any n, here you can write it as any sequence of odd, even, odd, even, up to n-5, or it's strictly increasing-- some overall these things strictly increasing.
Or the other way around, even, odd, even, odd, even, odd, n minus 5 sum strictly increasing. I know you're struck by how combinatorial all of this. These are amplitudes.
These are amplitudes of three negative helicity gluons and the sea of positive helicity gluons. And they're given by this incredibly combinatorial data. And it's also clear that we don't have any given unique representation of them. Right? We can write them this way, we can write them that way, and that's because they're a contour integral. So there's many representatives of the same object.
I should have mentioned, this was one of the qualitative things BCFW that I said was so striking. It didn't look like BCFW was coming from a Lagrangian. Because remember, on all the diagrams that we had, particle one and two were always separated on different sides of the diagram. That never happens in a Lagrangian.
And associated with that, there was never any one unique form. You could write it in many different ways. So, when you're first thinking about it, you wonder, what the heck kind of theory is not going to give you a unique answer for an object, but will give you all sorts of different answers for it? The answer is a contour integral. A generalized contour integral doesn't give you some unique form. It's a representative of some class.
All right. Now let me tell you-- I'm just labeling all these residues. Let me label them for you in the complementary way, where I just tell you all the objects that don't make an appearance.
So that way, I can write it as, for example, the sum of 1 i i plus 1, j j plus 1. Sum over i and j. This doesn't look cyclically invariant.
There's a 1 sitting there, and that's fine. Every BCFW representative is not going to be manifestly [INAUDIBLE]. It turns out that residue theorems tell you, that nothing depends on that 1. It could've been 2, or 3, and so on, and you get exactly the same answer.
So that's great. We know all the [INAUDIBLE] amplitudes. It's a very pretty formula. I want you to notice something else about it. As we've remarked before, what objects are growing the denominator here? So there's a delta 4 upstairs, that I won't bother writing down. There's beta 1, 5 plus 1, jj plus 1.
Let's look at the poles. The poles are like 1 i i plus 1 j. Then there's i i plus 1 jj plus 1. And there's i plus 1 j j plus 1 1. Then there's j j plus 1 1 i. And then there's j plus 1 1 i i plus 1. OK?
Of all of these poles, that one is physical. All the rest of them are spurious. You see, that one's physical. Why is that one physical? Because it corresponds to x i plus 1, minus xj plus 1, squared, which is nothing other than the sum of all the consecutive momenta starting from i going up to jj plus 1. OK?
So that one's a physical pole. All of these poles are physical poles. But the rest of these are spurious. As we've seen, BCFW is riddled with spurious poles.
But these residue theorems that tell you, that I can replace 1 by 2, or 3, or that 1 could be anything, once again I'm telling you, the spurious poles can't be there. Because in the different representations, the spurious poles differ, the physical poles are all the same. Very good.
Now, let's say I'm trying to find the more invariant way of characterizing what this contours. You see, I can say, here was a given representative. And then the residue theorem says that it's equal to another one. Or I can try to just more invariantly characterize what is the-- what's the class of all of these things that are equal to each other. How do I talk about it? OK. This is a homology problem.
So I'm going to say it a little bit abstractly first. I'm going to say something a little bit abstractly first. I'm going to motivate it, by noticing that I can interpret the residue theorems that these objects satisfy, in a very simple way. I can notice that these give bracket objects, satisfy a natural six term identity. This six term identities is what having a natural condition of the boundary of something equals zero. It'll be like a formal boundary statement.
As a formal boundary statement, we then take boundaries of boundaries of zero, we're going to play around with it for a few minutes, and we'll find that the invariant characterization of some of residues, is not the sum itself, but its boundary.
So we'll play with that. We'll see what the boundary is. It will be very, very beautiful. It will be formal. It'll just be motivated by, how do I figure out, if you give me this set of residues-- this sum of residues, someone else gives me that sum of residues, how do I know if they're really equal to each other?
We'll just set up a little calculus for solving that problem. After we do that, I'll show you why that calculus exists. It exists because these objects are actually the volume of some region, with some boundary. They're actually, not just formally-- they're actually the boundary of some polytope. And this is one particular way of chopping up that polytope.
You can chop it up in any way you want. You can triangulate it any way you want. And you'll get different answers. In fact, you will find the different ways of triangulating this polytope give you the BCFW form of the amplitude, they give you the CSW form of the amplitude. All the different forms of the amplitude that anyone ever thought about turn out to be different ways of triangulating the same polytope.
Furthermore, you actually find new ways of trying that people hadn't written down before-- actually, they had. I think I mentioned this. These old variance [INAUDIBLE] recursion volitions, for computing a scaling amplitude based on Feynman diagrams long ago, came up with some nice forms of the six particle amplitude, which are exactly what comes out of this yet new way of triangulating the polytope. But which then generalizes to all atoms. So you can write down-- so there's even a way of writing down the answer so there's no spurious poles, everything is physical poles.
So that's really the basic lesson here. The reason why this is looking like a contour-- So, it looks like a contour integral. That tells you that everything you're writing down, any specific representative, is some representative of the class.
And now the mathematical problem begins, how do you identify what these classes are? What we have right now, and I don't know how much more we'll get by the time we put out the paper. I hope we just get everything. But what we have right now, is a complete understanding of that problem for NMHV tree amplitudes and MHV1 loop amplitudes.
In other words, in these two cases, we really understand what the geometric object is that you're cutting up, in different ways. We fully, fully understand. And it gives rise to new forms of the amplitude that had not been written down before.
That's the problem that actually, we're most obsessed with right now. Trying to generalize picture to all the amplitudes, and that would be the final, final thing that we're after. The final, final, most invariant representation of what this object is. I've said it a hundred times already, for the actual, just for knowing the answer, we have the BCFW formula for all loop amplitudes. That makes the Yangian manifest. So the answer is there. But interpreting it as invariantly as possible, has not been completed yet. Let me tell you how it works in this case. First, let me do it-- first let me do it-- First let me do it using these formal boundary operations.
So let's say someone hands you this abcde. And let's say, that you take this, and you deform za plus za plus some little z --abcde-- zf. OK? So this turns out to be momentum twistor space, the analogue of BCFW deformation. Actually not, the BCFW deformation, but-- well, like BCFW deformation, but never mind.
The question is, if you do this, then you're going to find, in the z plane, that this object now becomes a function of z, and you find that in the z plane, it has six poles. And the six poles-- well, it's obvious because there are five poles down here. So we'll have five polls down there, but it also has a pole in infinity, because z goes to infinity, za becomes equal to zf, projectively.
Remember these are projected variables. So as z goes to infinity, it just tells you that za equals zf. So this just becomes-- you replace a by f. It's extremely easy to see, either from direct calculation or just from the Grassmannian form, the other residues just correspond to replacing the other guys by f.
So it's like afcde, abfde, and so on. So just Cauchy's theorem here, tells you, a six term identity that all these guys satisfy. Which is just abcde, plus bcdef, plus cdefa, plus, the final one, eabcd, is equal to zero. It's a six term identity. This just comes from Cauchy's Theorem, are under that deformation.
Which is also a global residue theorem, in this simple case-- in the simple cases, of all these things are the same. Again, if you didn't know where it came from, this is a very non-trivial identity between rational functions. But, it's a very nice statement. It's very nice.
But it's also a problem. These objects satisfy these relations between each other. So if I hand you two different combinations of these invariants, you don't know. It could be they're actually equal to each other, by linear combinations of those identities. That's the question again. How do we decide-- is there an easy way of deciding, whether two linear combinations of these residues are really the same, up to these residue theorems. Is the question being asked clear? OK. So these are all the identities that these objects satisfy.
STUDENT: So essentially the challenge now is to identify the unique things, the unique objects?
NIMA ARKANI-HAMED: Well, before identifying any unique objects, we're trying to figure out, if I hand you a linear combination of 17 residues, someone else hands you a linear combination of 43 residues, right? They might actually be exactly equal to each other. They might differ from each other by just adding many, many, many copies of that residue theorem or combinations of those residue theorems.
So that's a question, how you decide? It's like a gate symmetry. Right? It's like someone hands you two different a mu's. What we're trying to do is figure out how to compute the f so that you know they're equal to each other. Right? It's the analog of that. In fact we're about to do exactly that by using a d in order to do it. But the d is going to be a boundary operation.
OK. So the way to think about this is to think about that residue theorem is actually, formally, the statement that the boundary of a six object, a, b, c, d, e, f, is equal to zero. Now this is a formal statement. What I mean by that boundary is when you normally have a boundary operator for a simplex, it's the usual thing. You delete one. It's the anti-symmetric sum of b, c, d, e, minus a, c, d, e, plus da, da, da. So if someone hands you a six simplex and asks you to take it's boundary, that's what you do.
This doesn't have any numerical meaning, right? I'm just using it as a mnemonic. When I take it's boundary I get a five object, right? And the five object that is get is exactly all the five objects that occur in the sum. So just as a formal statemen-- yes.
STUDENT: Does it have to be-- because usually, this [INAUDIBLE] is 0. So you should have minus signs around.
NIMA ARKANI-HAMED: Remember, these things are anti-symmetric. I've flipped them to make sure that that's exactly correct. OK.
STUDENT: So d squared--
NIMA ARKANI-HAMED: D squared is equal to zero. And d squared equals zero is going to be very important.
STUDENT: How did you come up with zero?
NIMA ARKANI-HAMED: Huh?
STUDENT: I'm sorry. I don't quite understand the last term there. It looks like a cycle of the first one?
NIMA ARKANI-HAMED: Oh, that's because that should be an f.
STUDENT: OK. Thank you.
NIMA ARKANI-HAMED: Sorry about that.
STUDENT: For the observation of this abcdef--
NIMA ARKANI-HAMED: This abcdef is just a formal straight of six objects that you put around some parenthesis. And then you just remove them one at a time and when there's five it has a numerical value. It's that amplitude. It's just a mnemonic device for writing things down.
And for the moment this is all formal, and what we're doing is exploring residue relations in the Grassmannian. in this case, as I said, we're going to see that this formal thing turns into an actual volume and it'll all make sense. And the six objects make sense, and the five-- everything makes sense.
But since we don't know the generalization of that picture for any n yet, it may well be that it's a little bit of a happy accident. But for the simple case of NMHV and MHV, it has a bit of a dramatic interpretation. And in general it's this more abstract homological statement. Anyway, at the moment we don't have a perfect geometric polytope picture. But this abstract picture could be developing nicely. So I'm trying to give it to you in both ways.
STUDENT: When you say homology, which one is the space which is--
NIMA ARKANI-HAMED: I don't know even know what homology. Actually means. So let me just-- let's just do not. I mean, I do know what it means, but I don't want to spend time on the things that don't have substance, given that I have very little of it.
It's homology. Well, let me say a little more specifically.
[LAUGHTER]
The way to think about contra integrals is that contra integrals are associated with homology possible. That's because whatever your function is it has some set of zeros. And what you do is you imagine the complex plane minus the set of zeros. Punctured by the set of zeros. And now you have a bunch of contours that may or may not be smoothly deformable into each other. OK? So those are the homology classes that were talking about. But it's nothing of other than words. It's exactly the same as find things that are equal up to residue theorems.
And it would be helpful if these words allowed us to open some math book and say, aha here's a homology class. But there is no such thing. The mathematicians have not touched this stuff so we have to do it ourselves.
All right. So, this is what the residue theorem is. Now, this is an extremely powerful precisely because d squared is zero. Because let's say you have two expressions, expression 1 and expression 2. And they differ by a bunch of residue theorems.
How can you decide whether they're equal to each other? Just take there boundaries. Because the boundary of a sum of residue theorems is the boundary of the boundary of something is equal to zero. So the boundary is the thing which is invariant. N itself is not, you could have any representative, but it's boundary is something invariant.
Now what do I mean by the boundary of a five object? Well, it's the usual thing once again. Again, this has a numerical value. By it's boundary at the moment I just mean list a, b, c, d, e plus b, c, d, e, plus c, b, d, a, plus d, e, a, b. So just list those four numbers in brackets like that. It doesn't have any numerical meaning necessarily. It's just what it is.
But notice something very important about this. Each one of the number that occur in the boundary here is in a one to one correspondence with one of the poles in this is expression. Right? Oh, I didn't write down the last one. Plus e, a, d, c. The poles of this guy were exactly a, b, c, d, b, c, d, e, da, da, da. So if I hand you the boundary of some five object it's also a list of the poles that it has. The boundary is also a list of the poles that it has.
OK, very beautiful. Very nice. So we know the answer for the BCFW representation of the tree amplitude. So just for fun, let's take it's boundary. So the formula was the sum over inj of 1 i i plus 1 j j plus 1. OK? So let's take the boundary of this guy. The boundary of this guy is the sum over inj. Well, there's five terms. So there's 1 i i plus 1 j, plus i i plus 1 j j plus 1, plus i plus 1 j j plus 1 1, plus j j plus 1 1 i plus j plus 1 plus i i plus 1.
Now I know you noticed that there are terms here which collect in pairs. You see there's a 1 i i plus 1 j. And if I bring it over here there's a 1 i i plus 1 j plus 1 with a minus sign. I can push that j plus 1 through the three of these guys so there's 1 i i plus 1 of j minus 1 i plus 1 j plus 1. That telescopes to zero. That sum telescopes to zero.
So these are going to cancel against each other when I just add the sums. These are going to cancel against each other. And what I'm left with is nothing other than the sum over inj, all inj i i plus 1 j j plus 1. That's the boundary. Namely, it tells us all the physical poles. It tells us all the poles of the object, which is all the physical poles that are on it with equal weight.
That's what the amplitude is. The amplitude, invariably characterized, is give by the boundary. And it's this dead cyclically invariant-- it's the only thing it could possibly have been when you start thinking about it this way. The sum over all physical boundaries with equal weight.
And you can have many representatives of that. In fact, we've just proven that if I place one by two by anything I'm going to get the same answer. So If i replace one with two with seven, that's showing the equivalence of all the BCFW forms.
It turns out if you replace one by a random twistor, it doesn't have to be any one of these strong particles. That gives you another representation of the answer which is exactly the CSW formula for the same amplitude. So different realizations of the same boundary are giving you different representatives of the amplitude.
Now you can ask another question. You can ask could the answer have been something else and still only have physical poles? Can i take linear combinations of things that look like i i plus q j j plus 1, which are not all of them equal to each other. Can I just take 1 2 5 6 plus 17 18 27 28, not take all of them.
Well, you could drag such a thing down but it has to satisfy a constraint. It has to be the boundary of something, right? So the way you can check whether it's the boundary of something is to take the boundary of it. Take one more boundary. You take the boundary of it, you now get a three symplex. And that's a very simple exercise to show that the only linear combination i i plus 1 j j plus 1 with vanishing boundary is that one.
So we're not putting cyclic invariance in [INAUDIBLE]. Basically you just ask the following question. You demand that this object, this is invariant, the thing that invariantly characterizes the contour, the thing that invariantly characterizes the model [INAUDIBLE], the thing that invariable characterizes the sum of the residues. You ask that it only depend on the external particles through the points x associated with twistors. That's it.
That's the demand that it should depend on pairs-- i i plus 1, j j plus 1. That's the one place that you're inputting a tiny bit of space time. Your saying things should depend on the corresponding points in space time. You saying nothing else. You say nothing else about the structure of the theory. You put that information in and this is the unique answer. So, for example, we learned that there's no other linear combination of these residues that are free of non local poles. Unique combination which is free of non local poles is the amplitude.
So I hope you see that this is a very satisfying picture for what the object is fundamentally. It doesn't fundamentally satisfy recursion volition. It sits there. It's something by itself. Once you see what it is, and once you triangulate it in a particularly way-- I haven't told you that's the polytope-- here, once you realize it in a particular way then there is visible interpretations-- BCFW, CSW, and so on.
All right, so now I want to tell you the polytope way of thinking about this. And in order for me to be able to draw things, and also because it makes it much more riveting, let me just do it for an analog of this problem first, and then at the end I'll tell you what the answer is.
[INAUDIBLE] different [INAUDIBLE] of it, which are BCFW or CSW. But of them the more efficient one still is BCFW. But inefficient is just counting how many terms are there? OK, it's the most efficient one. We seldom learn a completely new way of calculating.
From the picture I'm about to tell you, there's actually a completely new way of calculating as well. Yet another formula for the amplitudes.
SPEAKER 5: More efficient?
NIMA ARKANI-HAMED: Well it depends what you mean. It has more terms, but there's no non local poles. So it's manifestly local. So it depends what you care about. It doesn't have that many more terms. For example, for these [INAUDIBLE] tree amplitudes it has a factor of four or more terms. That's [INAUDIBLE] than BCFW. But when you're talking about 20 versus 80, it can all be done by the computer anyway.
But the important point here is the conceptual. There's some way of representing it with no non local poles at all. Which doesn't seem to come from any Lagrangian.
But I think this geometric picture will make things a lot clearer. So we have been playing here with these five index objects. Right? So let me just present a little toy version of everything we did with three index objects that is easier to think about.
Also, this was a supersymmetric. And I'm just going to do something here which is totally bizarre. What I'm about to do is going to correspond, in the case of NMHV amplitudes, looking at one of the particular components. The so-called split helicity amplitudes, which are when the three minuses are right next to each other. Something simple happens in that case. But you'll see the analogy immediately. OK? So just let allow me to do it here.
So I wanted to find an object, abc now, not abcde, but just abc. Which is little abc squared. So these are now variables in CP2, not CP3, in CP2. Little abc lived in CP2. OK? And I'm going to introduce an auxiliary, a fourth guy whose going to be a spectator throughout all of this, but--
So I could label this abc sub a to remember that it came from a, but since it's always got the a there we can just leave it like that. You can see why this might look like one of the component amplitudes of something like that. If I do a component of those objects I would have to pick out one of the components of eta and then there would be some fourth powers upstairs. So here there's not fourth powers, there's squares upstairs. And it's got the same type of denominator structure with three denominators rather than five.
Now, just take these as objects. Just some nice operator objects. And now it turns out if you do exactly the same exercise, if you deform z a equals z a plus z z b, then you discover of these objects satisfy four term identity. So Cauchy's theorem discovered that the poles in z have the same character, the same kind of object. And they satisfy four term identity abc plus bcd plus cda plus dab is equal to zero. OK?
Now some of you I told you ahead of time that there was an amplitude, quote unquote, written in some BCFW form which is write the sum over i of 1 i i plus 1. And then you discover using these identities that it's the same if you put 2 there.
And then you would go to exactly what we want through here and you'd say aha, this is really the boundary of abcde. This is really the statement that the boundary of abcd is equal to zero. We once again find that the boundary of the sum over i of 1 i i plus 1 is just the sum over i of i i plus 1. So only physical poles.
In fact, in this case it's particularly obvious. Only sum of i i plus 1 which doesn't have a boundary is just this one. Because the i i plus 1, you can just think of them as being the edges of a polygon. So obviously, the only linear combination of these guys with vanishing boundary is the sum over all of them. So in this case, that's particularly trivial.
OK. And so then, you would discover just what these algebraic manipulations, that all that matters is the boundary, so that, in fact, the general representative of this formula would be that m is the sum over-- you can put anything here you want. Star i i plus 1. That star is like taking this polygon and just triangulating it with any old point. Star in the middle.
In other words, this is guaranteeing-- so just think of this polygon of the moment as it's just a formal way of writing down the sum of that three-term guy, that three-term guy, that three-term guy, and so on, because you know that that is the boundary. Then you know that all that matters is that the boundary of this formal sum of triangles is the correct one.
So in this way, we derive, if you like-- we've derived lots of interesting rational function identities. So this thing has many representations. If it's a totally general guy, some auxiliary twistor that's like a CSW-like representation. If it's one of the external guys, it's the BCFW-like representation, and so on.
So that's the literal analog of the steps that we just did a moment ago. But now, I want to tell you how to think about this as literally the area of a polygon. That's because it is literally the area of a polygon.
To see this, let's ask the question-- if I hand you-- so this is the za. So now this is in CP2 So I have a point za, I have a point zb, and a point zc.
Now, let's forget about that in general. Just forget about that in general. And suppose, in CP2 I give you-- so these have lower. Let's say they have upper indices. Forget about this for the moment. Let's imagine, in CP2 you have three points with lower indices. So these are three points in CP2.
Of course, they have a point. I can dualize it, using the epsilon symbol to get something with two upstairs indices. So a point in CP2 is the same as a line in CP2 and vise versa. But I'm choosing to label these things so that there are points in CP2. Let's say I have three points in CP2, so that they define a triangle.
Now, if I take a fourth point it CP2, a, then I can choose to project this triangle out, down along a, literally into some triangle on a plain. Literally, some triangle on a plane.
So what do I mean, literally? What I mean, literally, is image you choose a to be 1 0 0. So just, we can always rotate a so that it's 1 0 0, and say that w1 is 1 some little vector here. Little x1, little 2 vector. w2 is 1 little x2. w3 is 1 little x3.
Then, given this data, just in ordinary, two-dimensional space now, I have a triangle whose vertices are x1, x2, and x3. And now you can ask, what is the area of that triangle? And the area of that triangle is just x2 minus x1, cross x3 minus x1, which is x1 plus x2 plus cyclic.
It's 1 plus x2, plus x2 plus x3, plus x3 plus x3 cross x1. Well, so this is just-- I'm just making it look as much like junior high school as possible, so we could just write it directly, because we know everything we need to know. Projectively. There's a unique invariant I could make out of these guys. Completely unique invariant I could make. What is it? It's just--
So za here, by the way, has an upstairs index. So I'm saying it has an upstairs index. So let's say l. So what I'm doing is just writing down w1, w2, w3, over a dot w1. This dot is because one is upstairs and downstairs. So that's a l w1 l a a dot w2 a dot w3.
So this is an object which is vanishing. It's a function on CP2. It has vanishing weight under w1, w2, and w3. It's a nice function of of CP2. Course, it has some weight over a. That's fine. We need something to give us a notion of area, what the units of area are. And that's just from the overall scaling under a is. Nothing, more or less.
So this expression-- notice that I take this expression, and I choose this representation, that expression just turns into, this upstairs just becomes a determinant of 1 1 1 x1 x2 x3 over 1, because all these things are 1, and that's nothing other than x1 cross x2 plus cyclic. So that's exactly this area of this triangle.
There's a more abstract way of writing this, which we won't use here. But the more abstract way of writing this is as a contour integral over CP2. We're not going to use it.
So very simple. So this thing is an area. That thing is an area. OK. Now we can imagine, then, how would we write the area if, instead of having been given the vertices of the triangle, we were given the edges of the triangle?
So let's say, now, we have this triangle. And let's see how we would write it, if we were given the edges. Well, so the edges you can think of as another vector in CP2, but with an upstairs index. So the added 1 2 would correspond to z 1 2. So this is 1 2 3. z 1 2 upstairs i is just epsilon i j k w1 i, w1 j, w1 k.
Conversely, w1 is the intersection of the line 3 1 with 1 2, and I can think of w the other way around. So I can think of 2 as an epsilon with a z z. So now, let's say I've given you the edges. And let's say that the edges-- So instead of giving you the vertices, I've given you the edges. So I've given you this line. Let's call it the a. That line, let's call it little b, that line, let's call it little c. In other words, this is the za, zb, and zc.
So what's the area of that triangle, which is now just that area? Well, it's the same formula. This a dot w1 is now an a, ab, abc, aca. And this thing upstairs becomes abc squared, because each one occurs twice.
So we may not have been used to it in school, figuring out what the area of a triangle was, given its edges, rather than given its vertices. But if we did learn it in junior high, we would discover that abc is exactly this, is exactly the area of the triangle whose edges are a, b, and c.
Is that clear? Because this would be, the w1 would be sum ab, bc, ca. They all have to be antisymmetrically dotted into each other. So the only way you can do it over a, ab, abc, aca. So this is nothing other than abc squared over everything.
So now, let's go back. Let's say someone handed you the-- Well, so now I think it's obvious, but let's go through it anyway. So it's obvious what the object we're computing is. The invariant object we're computing is nothing other than the area of the polygon. So let's actually just do it from four particle [INAUDIBLE].
So we have-- remember-- our data that we started with were the external edges, so 1, 2, 3, 4. So it's the area of a polygon whose edges are 1, 2, 3, and 4. So those are the edges.
The vertices would then be, if you like-- that would be 4 1, 1 2, 2 3, 3 4. Now, the point is that when we give objects written in terms of abc when we get-- for example, let's see.
So there was this BCFW-like formula, which would say that it's 1 2 3 plus 1 3 4. So that would be the BCFW-like formula for this case. Remember, in general, it will be 1 i i plus 1, sum over i. So for four particles, it will be that.
Let's see how that is actually computing to the area inside here. To see that, we have to see who is 1 2 3. 1 2 3 is the area of the triangle whose edges are 1, 3, and 3. So the vertices of that triangle are 1 2, 2 3, and 3 1.
So the first term is 1 2, 2 3. And now, where is 3 1 in this picture? Let me draw a little [INAUDIBLE]. So here's a line 1 2 3 4. So here's the point 4 1, here's the point 1 2, there's a point 2 3, there's a point 3 4, so there's also a point 3 1. That's the intersection of line 3 with line 1.
There's also a point-- let's see-- What am I doing?
STUDENT: It's correct. [INAUDIBLE]
NIMA ARKANI-HAMED: Yes, yes. That's right. No, but I'm just trying to find-- oh, that's right. Everything's fine. Thank you. Yes, yes, exactly. Oh, no. Right. So what I wanted to do is get--
Great. So, let's also put in the point 4 2, though. That's right. Exactly. So here is the point 4 2 exactly. OK. So the first formula is 1 2 3. So 1 2 3 is this big triangle. And let's actually put an orientation around it, because they're oriented.
And then we have 1 3 4. So that's the triangle 1 3, 1 3 3 4 4 1. So we have to start from 1 3, 2 3 4, 2 4 1. So it's this area with the opposite orientation. So that cancels in here. And I'm left with the area in there, which is correct.
So this BCFW representation is computing this thing by subtracting that from that. The other representation, using point 2, for example, would give me 2 3 4 plus 2 4 1, and that's this picture. That's that triangle minus that triangle. So we see there's many representations of the same object.
We could add a fourth point. We could add a totally other line, and include things relative to that one, and that would be the analog of putting any old twistor in here and summing over all the four terms. OK? That's like the CSW representation.
But we see that these are all different representations of what's just, essentially, the [INAUDIBLE]. Notice again, all of these representations have spurious poles, the spurious poles being everything which [INAUDIBLE] a 1 i, everything that isn't a i i plus 1.
But it's obvious that none of them were there, because the basic object is fully specified by giving you the boundaries [INAUDIBLE].
But the final point is that there is yet another way of computing this, which is, compared to which, this looks, what we're doing now, looks idiotic. Just that. Just add those two triangles to each other. Don't try and do it on the outside. Try and do it on the inside.
Now, what is that formula? That will tell you that if the area of, you know, 4 1. But, you see, the point is that the reason we don't get that in our usual forms is that these triangles are not conveniently labeled by the external lines.
They're really conveniently labeled by the points. So there's 4 1, 1 2, and 2 3. That's one of them. Plus the area of 2 3, 3 4, 4 1. But we can work out exactly what these things are.
So this is yet another representation of the object, which is manifestly only as physical poles. Every single thing that appears here, the only things we'll have downstairs are a i i plus 1.
So this is a toy example. So this is our basic object. And we can try and relate it in one space. So roughly speaking, it's like trying anything in the W and the Z space.
So in the W space, we can try and relate it in a variety of ways and we get BCFW, CSW and so on. We can try and relate it in the Z space. In the W space, you can also try and relate exactly the same object. And, in fact, there's just a new formula.
And so here, if it isn't obvious, of course, this picture is making completely clear what these boundary operations were doing. There really is a boundary. That really is a physical boundary, and it's the boundary of this. It's a boundary-- that formal boundary is giving the boundary of some polytope. And the amplitude is actually the volume of that polytope.
So that's the case in that toy example. And I'm sure you can believe that if we just tack on two more dimensions, all we're doing in the NMHV case is exactly the same thing, except we have higher dimensional tetrahedron, and we're triangulating them to some lower dimension of smaller tetrahedron. Those are BCFW CSW representations as well as a new representation that didn't exist before but which gives you the answer directly in a local form.
So this gives you, I think, a feeling for what the general fantasy is now, right? So I think it's completely clear that this is what the amplitude is for NMHV. It's this abstract object. It's specified by specifying this homology class. And different ways of slicing it up make it obvious that it has a space-time interpretation, make it obvious that it factorizes properly, et cetera. But it really doesn't care so much about that.
So let me close by telling you what the story is at all loop order.
STUDENT: Still for NMHV?
NIMA ARKANI-HAMED: No, for everything. Well, now what I'm telling you is when I the story, I'm now telling you how to compute the all-loop order integrand recursively. Going to give you a recursive formula for the all-loop integrand, which is a generalization of BCFW and which furthermore is ultimately Yangian symmetry [INAUDIBLE]. So I'm going to show you how to start from these Grassmannian objects and get the whole loop integrand out of it. Go ahead.
STUDENT: Cyclic means planar.
NIMA ARKANI-HAMED: Huh? When did I say cyclic? No, everything is planar here. Everything has been planar. Of course, the tree level, everything is planar. But what I'm talking about now is the all-loop planar integrand. So the all-loop planar, the all-loop planar integrand. Right.
So what this is is not so-- so it's a recursive definition. And it's really just doing for loops what you do for BCFW. It's doing for loops what BCFW does for trees. OK? So why isn't it the whole story? Well, if you just want the answer, it's the whole story. If you want the answer for the integrand, it's just the crank. You turn it, you get the integrand.
In particular, there's no more-- for this problem, there's no more need to make an [INAUDIBLE] of integrals, check these cuts, check those cuts, see if it works. You just write down the answer. You don't check anything. You just write it down. And furthermore, as I said, it makes the connection for the Yangian symmetry to the Grassmannian [INAUDIBLE]
OK. In order to motivate it, I'm going to motivate it by-- in principle, we could talk about just this part of the story, only this part, if you just wanted to get the integrand and nothing else. You didn't care about understanding of symmetries. You didn't care about anything. If you just wanted to get the integrand, you can do it without talking about the Grassmannian. It would be harder even, technically. But logically you could do it without talking about the Grassmannian.
So let me first describe it like that, and then you'll see what the Grassmannian adds. So what we're after is we have an integrand, which is a rational function of loop momenta as well as external particles. We can imagine writing the integrand in momentum twistor space.
So let me describe how this works. So let's say a loop integrand is an integral d4x over some function of x, right, as well as the external particles. I can perfectly well think about integral d4x, even those integral d4x, or completely equivalently I could think of it as an integral over lines in momentum twistor space, because x is a line in momentum twistor space.
So that means I can think of it as an integral d4za, d4zb. So AB is going to give me the line. So it's a line AB. Except this is too much, right? This is a four-dimensional integral. This looks like an eight-dimensional integral, but that's because I really have to make sure that it's only function of the line. So I have to model out by the GL2 transformation that would move these guys around the line. So I mark it out by GL2 and now I have a function of A, B and everything else.
This turns out to be, even though it's trivial to go back and forth from X to AB, it's just much nice to think about them in the loop integrals in momentum twistor space. It makes the dual conformal properties very easy. It allows you to write these down. Anyway, it's just much, much nicer.
But either way, you have a rational function of all these sort of variables. But now, I've written [INAUDIBLE]. It's natural to think about all the other variables in momentum twistor space.
Now, it's a rational function. It's a rational function of a bunch of Z's. So what's your strategy for computing it? Just perform one of the Z's and perform one of the Z's, see what happens. In fact, even before doing that, let's go back up here when it was a function of lambdas and lambda tildes. Let me, before you jump here, let's stop here with function of lambdas and lambda tildes. Just do the BCFW deformation again.
So I'm just doing a one-loop amplitude. I'm just doing a BCFW deformation on particles 1 and 2. Now, if you do that, now you have this rational function, f of x lambda lambda tilde you're trying to determine, and it has two kinds of poles.
One kind of pole is the kind that we talked about before. The kind that we talked about before where 1 is on one side, 2 is on the other side. And it's a pole where the sum of a bunch of momenta, external momenta squared to 0. And what that corresponds to in this picture is something that would have a lower-point tree amplitude on one side-- not amplitude-- lower-point tree integrand on one side and a one-loop integrand on the other side and vice versa.
So this is through the whole rational function. And so it's like a BCFW-type term where you would have something linking something with loop order L1. If you did it at any loop order, you would have terms that have loop order L1 and loop order L2 on this side, where L1 plus L2 equals L. Total loop or, if you're interested, you have 1 and 2 here. Those are the unsurprising sort of obvious poles that you have.
But there is another sort of pole that you can have, which would involve the loop momentum X, that as you're deforming these particles, it's possible to put some of the internal momenta running around in the loop. It's possible to put some of them on show.
Now, if I'm deforming particle 1 and 2, and they're adjacent to each other, then if I'm cutting a line that's joining them, then what I'm getting is a lower point amplitude where this is now tree amplitude, the whole thing is now a tree amplitude, but I have two extra particles. The line that I cut has now turned into two extra particles. And the two extra particles have momenta Q and negative Q.
So the residue of the pole, the residue of the pole, is the lower-point amplitude with two extra particles but in a particular limit. And the limit where the two particles at equal and opposite momenta. This is called the single cut. So a single cut of the amplitude. Single cut because it's putting just one propagator on show.
Now, these single cuts have been funny objects for a long time. People talked about them a lot. They're interesting, but they always have the weird property. They're not quite-- in general, they're not quite well defined, because these lower-point amplitudes have colinear singularities.
The really precise statement is I'm lying a little bit when I say the single cut is always well defined. That's a well-defined rational function. It has a well-defined pole, so the single cut is well defined. But the single cut is not exactly a lower-points tree amplitude with two extra particles.
The reason is, the lower-point tree amplitude with two extra particles would include diagrams that look like that, but if that were to come from a single cut, that would mean that you would drop a loop that looks like that that you'd never drop for very good reasons. So you don't have that. And also in the tree amplitude, you do have this. And this does develop a divergence as these guys become equal and opposite to each other.
So in general, the single cut is not related directly to the forward limit of a tree amplitude. The single cut is some other interesting object, some other age invariant interesting object, but we don't know exactly how to compute it using on shell data yet.
In fact, this is one of the main open problems. If someone figures out how to do this, then combined with everything else, we can do planar QCD to all loop orders. This is the only obstacle is to figure out how to understand single cuts. If you can do single cuts, non-supersymmetric planar QCD is done at the level of the integrand.
But let me go on. Something special happens in N equals 4 super Yang-Mills which is something Simone [INAUDIBLE] worked out very, very beautifully in the context of thinking about something called find the tree theorem. But he figured out that for N equals 4, something nice happens. For N equals 4, you can include these diagrams in the tree amplitude, but when you sum over all the polarizations here, there are more 0's upstairs then the 0 there, and this contribution is actually gone.
That means the forward limits are well defined. And the forward limits do compute the single cuts. So for N equals 4, the forward limits are under control, and they do compute the single cuts, and they're related to lower point and they're just-- the lower-point tree amplitudes and the forward limit, some go over the whole multiplet becomes well defined.
He also noted that this is true even for N equals 2 and certainly for N equals 1 for massive theories. This is not some Yangian fancy thing. It's just any amount of supersymmetry seems to be good enough for this. This means, by the way, amongst other things, that this formula isn't all ready. No one has computed one of these amplitudes but already lets you calculate any N equals 1 amplitude recursively. Any N equals 1 amplitude.
STUDENT: So what we're about to do right now, that would [INAUDIBLE].
NIMA ARKANI-HAMED: Yeah. So what I'm about to write down for N equals 4 will also work for N equals 1. Planar N equals 1 at one loop, probably at all loops. The real challenge is that in pure QCD, we just have to figure out this one more thing to make it fly.
OK. But that's it. That's the other kind of pole. So the other kine of pole is like a forward limit with two extra particles over one lower particle amplitude. It's good that we have that sort of term, because just these terms would allow you to recursively show that you can relate higher-point loop amplitudes down to the three-point loop amplitude that vanishes.
So you need to have a homogeneous term. You need to have a source term in the recursion relation, and the source term in the recursion relation is two extra particles added-- two extra particles added and L minus 1 loops.
So to compute the one-loop amplitudes, you need to understand-- for six particles, you need to understand the lower point one-loop amplitudes and the eight-particle tree amplitude. But with that data, you can compute the one-loop amplitudes and so on.
OK. So we have a recursive relation for-- now, this is true with one loop. If you define the general integrand, as I did before, my symmetrizing over all the possible integration points for the general all-loop integrand, then the combinatorics is just trivial, and you can see that the all-loop amplitude is actually defined recursively in this way.
So there's a formula for the all-loop amplitude-- sorry, integrand. The all-loop integrand is given by a formula that looks like that. So you have the sort of BCFW-type terms with low-point amplitudes jammed together plus this term, which is the one lower-point amplitude with two extra particles in the forward limit.
Not once here have I said the word Yangian, Grassmannian, nothing. So the really important point here is that if it's planar, so if the integrand exists, that's the biggest point of all, and that the N equals 4-- if N equals 4, the forward limits are well defined and compute the single cut so that this inhomogeneous term is under control.
STUDENT: So this has only a poles in Z?
NIMA ARKANI-HAMED: Only poles in Z, exactly. Right. You see, there are many nice things about dealing with the integrand. People for a long time tried to use factorization properties of the loop amplitude to help in computing them. But famously, the full amplitudes don't have very nice factorization properties because of [INAUDIBLE]. However, this is at the level of the integrand. Everything is perfect at the level of the integrand. None of those issues, the fact that the integrand, so you can just determine the integrand.
OK, now why does Grassmannian Yangian was-- why are they helpful? Well, first of all, every one of these-- both of these terms, all of these things, turn out to have a very pretty interpretation in terms of canonical operations that you can take.
In fact, you can hand me a Yangian invariant, I try to make other Yangian invariants. I can fuse two of them together, I can add particles, I can subtract particles, and there are three canonical operations that you can do on them. You find that every one of these terms is one those canonical operations.
In particular, the origin of loop amplitudes is there's even an origin for-- there's even an interpretation for what the loop amplitude is.
You don't put in by hand. You see that there is something funny here. If you think about it in twistor space, the loop momentum points, the loops are associated with these lines in momentum twistor space. We're integrating over these lines. Everything is just points. So where is it coming from?
[INAUDIBLE] coming from the fact there's phenomenal operations for removing particles, essentially integrating over them. And if you remove them one at a time, it does nothing interesting. It just takes soft limits of the amplitude. It takes tree amplitudes down to trees, down to trees.
But if you remove one and then you remove another, but you remove them by doing integral in the following way. If here's a and b and I'm integrating over both of them, first integrate a and b over all points, over a and b lying anywhere along the line joining them, and then do integral over all the lines that turns that gives you loop integrals. So a loop integrals really arise from hiding pairs of particles, but hiding them in a entangled way. The contour of integration you use to integrate them over, is this interesting-- it's basically the only other contour you can possibly choose.
There is nothing else you can do. If you have three particles, you can try to integrate over all of them on being on a plane, but that's just [INAUDIBLE] again. So the only interesting thing you could possibly get in CP3 is just such sets of integrals over lines, and then terminate to loop integrals.
It's very striking. It's a new physical picture for what loops are. You imagine you have a huge number of particles, and then you're removing some of them. You're just hiding them. But you can hide them in these canonical ways.
Which, if you hide them in the simplest way, you just get nothing. You just get trees. You hide them and you dial a knob. And you hide them in other way you can hide them, and you get loops.
It makes it manifest. This Grassmannian Yangian way of thinking about it makes manifest that all these terms are Yangian invariant. And it also gives you a very effective way for actually calculating them.
The way I've told you, especially here, when you think about forward limits, It's a tricky limit to tick. You've got to actually take a limit and it looks like, well, it's just a little bit delicate in actually doing the calculation.
Whereas the way it comes out of this formula is it's done as a contour integral. And then when you have a contour integral, nothing is ever blowing up. It's a very canonical algebraic operation. There's no integrals anywhere here. These are all simple, canonical, algebraic operations.
So I haven't described it in detail, obviously. But that's the basic philosophy. So given that the integrand is a rational functions, we can determine it if we know its poles. We do know its poles. These poles are universal. These are simple. These poles are single cuts.
In n equals 4, we know the single cuts are related to the forward limit of tree amplitudes, which is a nice fact about n equals 4. And furthermore, both of these firms have beautiful interpretations in terms of simple operations on Yangian invariance, fusing them and removing particles. So they're manifestly Yangian invariant.
So in this way, we see that this Yangian symmetry is there as a property of the full integrand of n equals 4 super Yang-Mills. It also gives us very efficient way of actually computing the integrand. And this way of thinking about them in momentum twistor space also lets you express the answer of the way that exposes lots of simplicities that were not seen before.
For example, we computed all [INAUDIBLE] MHV amplitudes. This was done in the literature in a beautiful paper by Christian Gregio, about six months ago maybe, but using the standard methods of generalized [INAUDIBLE] and so on. And it's a big computation, the answer is eight pages of diagrams. That box of this coefficient, that boxes is this coefficient that box is this coefficient, and we find that it's a single diagram.
STUDENT: How did he take it?
NIMA ARKANI-HAMED: Well, it's just true, so there's nothing to take. No I think everyone's very happy that there's [INAUDIBLE] official going on.
Now, I'm not explaining what these diagrams mean, but there's really like a single term. There's some funny diagram that looks like that, which is a very specific integral in momentum twistor space. This one diagram is all the terms.
And now, it gives you a good idea for how we could do it. We computed the amplitude using BCFW. We've got some answers. We could just leave it like that. It's fine to leave it like that.
But then what we do is we manifestly put it in a local form for a variety of reasons. We just want to see what looks like manifestly locally. What the momentum twistors let you do is put it in local form in a very nice way, which was not available to people without momentum twistors. It was not easy to do without momentum twistors.
You could, in principle, have done all of this in the ordinary space. But it was just not an obvious thing to do. Momentum twistors made it an obvious thing to do. So there's a new basis, a new nice basis of integrals, that you can use to express the answer.
In this new, nice basis you easily see patterns. So we compute them using BCFW. There's the answer. We have the answer.
But then just to stare at the data, we present it in this local form. In terms of this nice, new basis of integrals. And in this basis of integrals, you immediately see there's this pattern. All of them are exactly the same integral.
And then you check your guess. Because you have the answer so easily, you check the guess by computing it using BCFW go up to 30 points at two loops. And then just checking that this answer works. And it does. And it's possible to compute it up to 30 points in two loops because the recursion is trivial. Not by hand. There are 80,000 terms in two loops in BCFW. But that's no problem for Mathmatica.
And by the way, there's a very powerful check on the whole formalism, which is that there's 80,000 terms, all of them with spurious poles. They all cancel perfectly in the sum. There is not one minus sign that's up to us. Everything is just completely prescribed.
And of course it works. I mean, of course it works. We derived it. It works. But it works in a quite non-trivial way.
So you can see patterns. We computed the integrand for 6 and 7 point NMHV amplitudes. No one had ever computed an NMHV two-loop integrands before.
So I want to emphasize it's really, truly nothing now. It's just a question of how long you want, to how often you press shift enter, OK? But you can wait a little bit. You can wait a couple hours and get the nine point NMHV two-loop integrand if you like. You just don't care that much anymore once you that you can get it.
But there's probably lots and lots of patterns hiding in there. Now the fact that these local forms turned out to be so incredibly simple, the BCFW form is incredibly simple. The local forms are incredibly simple. So it seems like the simplicity of these local forms is actually reflecting this underlying polytope structure.
There's clearly something like-- in other words when I say that we're not quite done, we've given a representation that makes all the symmetries manifest. But it's like giving the representation of that square, in terms of the difference of those two triangles. It's like you've given that without saying it's the area of a square. So we have the analog of that for all loop amplitudes. We do not yet know what the analog of that underlying geometric algebraic object is. That it's really, really a computer.
STUDENT: So all loop, not just NMHV, but--
NIMA ARKANI-HAMED: No everything, all loop, all amplitudes. Yes. All loops, all amplitudes.
STUDENT: Is it sensitive to how the z momentum flows?
NIMA ARKANI-HAMED: No, no.
STUDENT: Different flows gives you different formulas?
NIMA ARKANI-HAMED: Oh yes, yes. It does. It is sensitive on who you deform.
STUDENT: Right.
NIMA ARKANI-HAMED: It is sensitive on who you deform.
STUDENT: [INAUDIBLE].
NIMA ARKANI-HAMED: Yes
STUDENT: But then how that momentum flows through the loop?
NIMA ARKANI-HAMED: Oh no. You don't have any choice. You see, because you the integrand, see that's one of the points. Because the integrand is well-defined, it's just a function, it's just a single, rational function.
STUDENT: How many loops and external [INAUDIBLE] does it start to take a reachable amount of time on?
NIMA ARKANI-HAMED: I mean, we haven't even tried to optimize it yet. It's running on a laptop. We went up to three loops, for the paper we did three loops, five particles. We can probably do three loops.
I really don't want to say, because we haven't even tried. There's completely, totally, bone-headed things we were doing just because we wanted to get it out. Which, if you just fixed them, will speed it by a factor of 100.
I think, for anything you care about, like three loops, 20 points will happen. Four loops, 15 points will-- I don't know. I don't really know. But I think you can basically assume that it's no longer the data. It's not data limited anymore. If you want to see some pattern, if you want to actually see it, the integrand can be obtained, and can be presented in the local form. You can look at in the local form and see if it's telling you something.
I think that's the piece on the integrand side which is really left to understand, is what is the underlying geometric object that this is cutting up, because it clearly exists. We figured it out, by the way, for a one loop MHV. It's literally these two dimensional CP2, pulling in all integrals.
Quite literally, it's the square of those, the product of two of them. It's like taking two of these areas and jamming them together. Remember there was an a there? Actually you can just add a b to everything. Which is just a specator.
You can start in CP3, just ad a b to everybody. Project that along b. Now you have a and the rest of them. Then you can project down along a, you get these triangle areas. That gives you the functions of a and b. And it's literally one function of a b, the other function of b a, you multiply them together, and that's the [INAUDIBLE] amplitude. So it's like the square of a polytope.
And the final thing I want to say is that this nice spaces of objects that we write the integrals in terms of, there seems to be many reasons to suspect that the integrals themselves are very simple in the spaces.
And there is an amazing paper by [INAUDIBLE], where they show that the answer for the two-loop MHV amplitude at six points, which had been computed in the literature, and had been written in terms of 20 pages of complicated four polyogarithms, could be simplified to sit on three lines in terms of no complicated polylogarithms at all, just four logs, [INAUDIBLE].
And this turns out to be related to some theory for polylogarithyms. There is a mathematical theory that controls polylogarthims developed by Sasha [INAUDIBLE], which is related to something called the theory of motives, and very, very fancy sounding things, but sitting in the middle of all this kind of algebraic geometry. And it's related to Grassmannians, and it related to-- I mean, everything seems to be in some big soup of this sort of object. Screwing around with planes, lines, intersections, and some extra data, can generate as we've seen an enormous amount of rich, rich structure.
Now you've all suggested that I mention some very specific open problems, so let me just mention a few specific ones, and then I'll stop. I really need to stop actually. So let me just list-- Yes?
STUDENT: So what has the amplitude for [INAUDIBLE]? Is there a simple way to sum it over all of them together? [INAUDIBLE].
NIMA ARKANI-HAMED: Yes, absolutely. That's a really interesting question. There's was this object called the integrand. Which exists and makes sense, right?
Now, what could it mean at strong coupling? So that's funny. There's an object which is not the answer, not the whole answer. Some auxiliary object, that when you integrate it you get the answer, right? Which exists at any coupling, right?
So what could it possibly be at strong coupling? I don't know the answer. But I know one other object that exists at strong coupling, that isn't the answer, that you integrate to get the answer. It's the [INAUDIBLE] action.
So if there's any justice in the world, somehow the integrand at strong coupling summing everything up somehow turns into the [INAUDIBLE]. Now, that sounds like the words that people said since the '70s. That you have these fishnet diagrams and that large [INAUDIBLE], they condense and turn into a [INAUDIBLE].
Except before, it was a qualitative picture. Now we have the integrand. Now it has a chance to become more quantitative because we have a function sitting there right? And it could be that some property of that function of strong coupling, you can see this condensation picture makes sense.
STUDENT: So if you do four point it's very do the best you can, you can compare with [INAUDIBLE] as a result?
NIMA ARKANI-HAMED: Of course, that's what I'm saying.
STUDENT: You didn't compare?
NIMA ARKANI-HAMED: No, no, we did not, no. At the moment we have the integrand, we have the integrand.
[INTERPOSING VOICES]
STUDENT: [INAUDIBLE]? Ten?
NIMA ARKANI-HAMED: You can get anything you want.
STUDENT: But in principle, in practice?
NIMA ARKANI-HAMED: In practice it's already been computed to five loops. Son it's up to four loops, and it's being done for five.
STUDENT: So you can compare already, right?
NIMA ARKANI-HAMED: No. You see, four points is a little boring. Four and five points are a little boring. In a sense the whole answer is known for four and five points. I didn't have time to do it properly. This is the so called [INAUDIBLE]. And it's because at four and five points, there's no cross ratios that you can build.
So it is completely determined by some infrared anomalies. It's completely determined by things like [INAUDIBLE]. So that done. And that just exactly matches [INAUDIBLE].
But that's almost, it's a little trivial. It's really remarkable. It's a little trivial because it's totally, completely determined by an anomalous symmetry.
At six points and higher is when the comparisons are starting to be made. So [INAUDIBLE] have done some incredible work over the past six months. Computing all the [INAUDIBLE] loop observables, so that's the MHV amplitudes. They've computed all those amplitudes at strong coupling for any n.
STUDENT: For any n?
NIMA ARKANI-HAMED: For any n. And it's related to these y systems and all sorts of incredible integral structures, and TBA equations, all sorts of crazy stuff. And then they realized that there's certain quantities that they can compute at any coupling involving taking various co-linear limits. So if you take various co-linear limits, you can compute some things at any coupling.
So they've even started this process of taking this integral stuff and making some predictions compared to the known two-loop results, and they actually agree. So there's a lifeline coming in from the other end as well.
Now for a while, there was an obstruction that people didn't know what the general supersymmetric version of the Wilson Loop was. In other words, how to generalize the Wilson loop to include any helicity information.
Now in the past month, that problem has been solved. So Mason and Skinner, and Simon [INAUDIBLE], independently-- although Simon's paper just came out yesterday. Mason and Skinner's came out three weeks ago, but Simon more or less had the same idea about the same time-- figured out what the supersymmetric Wilson Loop is.
So that's very nice. Now there's a complete symmetry. We knew it had to exist. We knew it was there. We knew the integrand was computing something. It's called the scattering aptitude in space time it's called this supersymmetric Wilson Loop in the dual space time.
Those are two different names for the same object. The fact that the Wilson Loop is equal to the amplitude is proven. The proof occurred in yesterday's paper by Simon. And the proof proceeds simply as following. You take this Wilson Loop that you defined and you show that satisfies this recursion volition. It satisfies the [INAUDIBLE] properties, and so it's done.
This is the whole theory. This is the theory. Now you can do anything you want with it. You can compare things at the level of the integrand. And it's way easier then comparing them at the level of the answer, dialogue, and infrared [INAUDIBLE]. Just show they're equal at the level of the integrand.
There's two stories going on. One of them is still one more step toward very deep understanding of the integrand. That's finding this question of what's this invariant object that we're talking about in different ways to get at the integrand. So that's the question that most interests me right now.
STUDENT: That's the geometric interpretation.
NIMA ARKANI-HAMED: That's this geometric interpretation. Algebraic geometric interpretation, whatever you want to call it. That's one question.
The other question, which is more immediately related to getting the answer, is how you do the integrals, right? And, as I said, these are not random integrals. This basis of integrals tuns out to have extremely special properties. It's very, very likely integrals are really simple.
And it's very likely that Mr. [INAUDIBLE] will be able to stare at the integrand and tell us what color log goes with it. OK? So that's something that needs to be developed. But if that starts getting developed-- you see, it's now the answer which is going to start making contact with Juan and his friends. Because they're not seeing the integrand, they're seeing the answer. They're doing the answer.
Now it's very interesting that there is this object, the integrand. Which is more than the answer, and which has so many nice property. It's this integrand that has all these symmetries that matter to is. It's this integrand that does all these things. So it is a nice question what the integrand might mean in strong coupling.
And that's what started this discussion by-- I don't know how, and it's a vague set of words right now, but it could be that it's the integrand that turns into the [INAUDIBLE]. In some sense, it's the integrand that turns [INAUDIBLE] action. And if this is true, this would be a way to seeing how the string comes out of the CFP. So this would be the beginning of a proof of ADSCF2.
STUDENT: So in a case that has strong coupling, right?
NIMA ARKANI-HAMED: Yeah.
STUDENT: So in principle, you can write down all of the integrands and essentially you can write down a form which can be equal to the answer they have.
NIMA ARKANI-HAMED: Yes, the integrand. But, you see, they don't have an integrand. They have the answer. So if we learn to do the integrals, then we can start comparing.
STUDENT: If you know the answer, that should help you, right?
NIMA ARKANI-HAMED: Yeah, maybe. What's clear is that especially this local warming is the correct presentation of this object as loop integrals. It's a very canonical representation of these objects as loop integrals. So hopefully, there's some nice way of doing the integrand.
Something else I can mention just briefly is that another big advantage of writing things in this Yangian invariant way is that it makes it completely clear that the integrals are not complicated. The reason is that had you been doing-- let me just give you a--
The reason that these objects are formally Yangian invariant, in the sense that the integrand transforms into a total derivative under the integral. But the contour can be dangerous and can force you into [INAUDIBLE] divergent regions. And so it's just choice of contour which is screwing you up. So it's clear somehow that the integral is completely dominated by the regions where there are infrared divergences.
I can say this much more specifically. Let's say we're talking about MHV loop amplitudes. The MHV loop amplitudes can only be one if they were Yangian invariant. That's just what we saw. The only Yangian invariants are one.
So these integrals that we get for MHV, including that wonderful bunch of objects, are all MHV loop amplitudes. There objects which, if you do them on any well-defined contour, you'll get 1. They can't be complicated. If you compute them on any contour which doesn't give you infinity, you get 1 or 0. So what that means is that all of its dependants, all of the fact that it's not trivial, come from the infrared divergences.
STUDENT: From the tree level, right?
NIMA ARKANI-HAMED: No. Of course, all divergences are related to trees But more specifically all the loop amplitudes have divergences. But the infrared divergences are exactly localized along the edges of the Wilson Loop.
So it's telling you that these integrals are not big complicated integrals. They would've been one or zero had you not had these infrared divergences. So somehow they're fully localizing the neighborhood of where these infrared divergences happen. So that should give a very good clue [INAUDIBLE] Wilson loops.
[INTERPOSING VOICES]
This picture of the polygon [INAUDIBLE], I realize, that if you computed a Wilson loop with this configuration that that was computing the amplitude. So that was the observation that goes back to them. And what Mason and Skinner and Simon have done is show what the generals supersymmetric form of that is, after three years, four years where this was an open problem.
So as I said, at a certain part of this progress has been relatively discontinuous in the last five, six months, right? So something's, that all the boarder are known, the name of this object is known, how to define it in momentum space, how to define it in the dual space, amplitudes, Wilson loops. Everything which is well defined, if you like, is really under control. This generalized polytope picture that exists is, I suspect, going to be the best way of thinking about what this theory is. And will probably even tell us the best way of doing the intervals. But that's it. I really think that is all I have to say.
So a number of specific open questions are, for example, any amount of reduced SUSY. 1 to n equals 2 1 0.
[LAUGHTER]
STUDENT: Can I ask, what's the definition of open at this point? That means that they have a paper coming out soon?
NIMA ARKANI-HAMED: No. This is open, at the moment none of the 10 people on the world who are working on this are working on it.
STUDENT: I just wanted a definition.
NIMA ARKANI-HAMED: Right. I'm specifically saying goals because the rest of them are either going to be-- I would say, as far as the rest of the problems, either they'll be solved in the next month or two or there hard and then t it will take a little while. So I've told you all the ones directly related to this stuff that I think are open. Do the integrals, find the polytope picture. And that's something that basically everyone whose been working on this for six months is working on now. But there are equally, perhaps even more important problems, that just for the sake of momentum are not being worked on right now. And I'm telling you what they are.
So doing any amount of reduced supersymmetry in any way. Like, no one has even taken these things and orbofolded them just to see what it starts looking like. Just a random a simple example. We know the amplitudes are going to be the same. If you orbofold The n equals 4 down to n equals two or n equals 1, or even n equals zero.
But, for example, what kind of Grassmannian formula do you write? We don't have etas anymore. So how is the GLK working out? How is it all-- is it nice? What does it look like? It will be nice to see what some non supersymmetric formula looks like which manages to have a GLK, manages to have a Grassmannian interpretation somehow. Because it might give us a clue for how to think about non supersymmetric things in general at this pint. Just as a random example.
An extremely interesting problem, which I would really like to work on, and I might because I keep getting worried that it's more important than anything else we're doing, is to figure out how to deal with these single cuts for non supersymmetric theories. If you come up with a formula that allows you to compute single cuts for non supersymmetric theory, we've solved QCD. Planar QCD at the level of the integrand. In particular, the generalized unitary program for planer theories is just replaced. You just don't have to do it ever again. It's replaced for n equals 4. You might not care. But it might actually be useful one day. It could really, potentially, be useful one day.
There's this really central idea, extremely simple central idea, that the integrand is well-defined and the theory's planer. It seems crazy that we shouldn't be able to just determine it.
So [INAUDIBLE], what you have to do is come up with some way of just defining what the single cut is. Take tree amplitudes minus this class of diagrams. And then the resulting objects would still be gauge invariant in the [INAUDIBLE].
But how do you compute that? Is there an efficient way of computing that recursively? Is there some way of doing it? If you come up with a way of doing it we're done with QCD at one move. We're already done that one loop. But I think we could perhaps extend it very easily into all of [INAUDIBLE]. So those are two questions about through to a lower supersymmetry.
Another question is figuring out how the effects of masses start being included. So in n equals 4 you can go out little along the coulomb branch of the theory. And you get masses for all the particles.
And at very, very high energies, you expect, of course, the scattering amplitudes return to the amplitudes of the usual theory. But there must be some simple, nice formulas, universal formulas, for the leading [INAUDIBLE] correction site. I mean, they're totally obvious things. They're not obvious things that are second order. They're not SP versions of something that has been done for SU. They're just things that haven't been done. But I think they are important.
Of course, we can say the word gravity and start an entirely new discussion. But that's something where it's not totally obvious how to start. But let's see.
So obviously, anything non-planar, even in Yang-Mills. And here there was actually something a little bit interesting, which is that at least a low loop order, I don't know if it's true at all loop order.
But at low loop order, it's true that if you know the coefficient of the single trace operators to all orders of 1 over n then you can determine everything. You can determine everybody else. In other words, the coefficients of the double trace, the multitrace operators, are related by simple identities to the meaning trace operators.
That's kind of interesting, because if you restrict yourself single trace operators there's still a natural cyclic ordering for the external particle. It still makes sense to work in momentum twistor space. And it still makes sense to see exactly what is going on. So I think getting any handle on what the 1/n corrections on the [INAUDIBLE] operators are would be an interesting question. Oh, a whole other question--
STUDENT: [INAUDIBLE] would say one loop that would trace-- that you won't have the cyclic order, right? [INAUDIBLE]. Once you have the cyclic view, you [INAUDIBLE].
NIMA ARKANI-HAMED: No, but this is what I'm saying. There are these [INAUDIBLE] that relate. You want the [INAUDIBLE] to relate the various things to each other. So I'm saying it's not the same diagrams, but their identities that allow you to determine the coefficient of the double trace and higher operators from the single trace one. If you know the single trace to a higher loop order
STUDENT: So [INAUDIBLE] apply to loop [INAUDIBLE]
NIMA ARKANI-HAMED: Yes. You want [INAUDIBLE] applied, and it's cousins, applied to the loop diagrams.
There is an entire set of questions relating all of these things to twistor string theory. So here there is there is a set of papers, just a few papers back in December. There's an absolutely fascinating connection there, which I have no idea what it means, why it exists. What the role in life of the twistor string is in this picture at all.
But it's clearly playing some role because the way it's related to this Grassmannian picture is extremely interesting. As I said there's this particle interpretation picture that I didn't really tell you in any detail. But this particle interpretation picture tells you how to build the tree contours. It tells you ahead of time what to do.
And the form's not a contra deformation. It's an integrand deformation. Deforms it to be related to the twistor string. So understanding any of these things in twistor space. You take any of these loop integrals, any of these things. Just seeing what they look like in twistor space is a very nice open problem.
I really strongly suspected that understanding them in twistor space will ultimately be the very, very deepest understanding. And we've been relying on all these other things as a crutch because we can recognize momentum space results easily. But this whole momentum twistor space is crucially tied to planarity. And ultimately we know that if this whole thing only works in the planer limit that would also be very disappointing. Right? But there's lots of hints of something remarkable happening away from a planarity as well, and understanding them would be great.
And of course there's many, many other things going on that have nothing to do with this, apparently, at the moment, related to these BCJ ways of writing the amplitudes. I'll just mention one big thing, which is that it's funny that BCJ suggests new ways of writing the amplitude which are completely local. And we're also finding new ways of writing the amplitudes that are completely local
STUDENT: Same?
NIMA ARKANI-HAMED: No, it's [INAUDIBLE] They don't look the same at all. But that's, I think, a very surprising thing. It's one thing to find these new ways, non-local ways, of writing the answer. It's very surprising to find completely local ways of writing the answer. If they're are nice and completely local ways of writing the answer, why isn't there a Lagrangian that can do so?
STUDENT: [INAUDIBLE]
NIMA ARKANI-HAMED: Definitely not. No, we don't see anything like that. That whole story is so orthogonal. They have colorization [INAUDIBLE] right? It's very closely related to Feynman diagrams, and it's very closely related to-- yeah. There's probably some nice Lagrangian that makes BCJ manifest. Or maybe it's the [INAUDIBLE] string that's making it manifest. But it doesn't seem, yet, related to any of these. I think that's all I can think of off the top of my head right now. Thanks a lot.
[APPLAUSE]
The last 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.
