 So, it's a pleasure to be here and well, I'm going to tell you about a journey from twister strings to DBI. It's never a good idea to lecture at 3 AM in the morning, especially if it's on the Blackboard, but let's try. So, any problems with me, spellings and things like that, it's not my English, it's just the time zone, okay? Well, it's really my English, but. So, these lectures will be a journey connecting two unlikely objects. One of them is twister strings and the other one is direct board infill. But what is the aspect that we're going to be connecting while we're going to be talking about the scattering matrices? So, twister strings was first developed as a reformulation of the scattering matrix of Young-Mills. Then it went on to gravity, scalars. Here in scalars, we could even talk about nonlinear sigma models or the Keidel Lagrangian theory. Here we can also talk about board infill and finally, direct board infill. Well, but along the way, sorry, maybe I'll move this a little bit lower here if you don't mind. Okay, so along the way, we will also meet some other objects. One of them is something that I didn't really know very much, say four months ago. And that one is, and it's gonna sit somewhere here. It's called a Galilean theory, okay? Very good. So let me now give a little bit of motivation and historical review. So why twister strings? Well, the whole story started, I think, in 1986 with Park and Taylor. Here I'm going to assume that all you guys are experts on Park and Taylor, right? Because Simone gave lectures on Park and Taylor. Well, sadly, I'm sure the first thing that he did was to factor out Park and Taylor and remove it from everywhere he was doing. So you probably saw Park and Taylor during the first lecture. But then after that, you never saw it again, okay? So I'll try to review it. Then we have Penrose in 1967, introducing Twister Space. And these two things came together with the work of Nair in 1988, who had the idea of connecting these two formulas or these two formulations together. And then, of course, we had the string theory going around. And in 2003, we then came out with the idea of having a twister string, okay? So as I said, this was a formulation for the scattering of particles, or in these lectures, we're going to talk about the scattering of gluons in N equals 4 superjamins with a gauge group UL, okay? Now, how many different formulations did Simone show you for the scattering amplitude of N equals 4 superjamins? Well, I hope that he gave you at least one new formulation that you didn't know about. Of course, you all know about Feynman diagrams. So Feynman diagrams is a way that we all learned in high school, right? If you went to a good high school, you learned Feynman diagrams to compute scattering amplitudes, especially of gluons, right? But you never make it beyond four particles, because it's too difficult. So we had to leave that to Park and Taylor. So Feynman diagrams is one way or one formulation. I hope with Simone you learn some other techniques, say momentum twisters, right? So that's something that you should all be familiar with right now. So I hope Simone also show why we look for alternative formulations, okay? So what do I mean by that? What I mean is that we have a physical system, and we can have our formulation A, and we can also have another formulation B. Or C, B, all the way, of course, the formulation Z, right? So we can have many different formulations for the same physical system. What do I mean by this? Well, I mean that we can have different descriptions. And the different descriptions, well, by now I'm sure you are all very familiar with when we have different descriptions, we usually say that the descriptions are dual to each other. And we usually like to say that dualities allow us to make manifest things in different formulations that are difficult to see in some other formulations, right? So I would like to propose a principle, which is probably very old, but let me just write it explicitly. For every physical property an observable has, there exists a formulation that makes it manifest, okay? So that's my principle. As I said, it's probably very old. But this is something that we have been learning in the study of scattering amplitudes for the past 10 years. That every time you find, at least in the context of S-metrices, every time you find a property, there is always a formulation that makes that property completely manifest. Well, we haven't found any single formulation that makes every property manifest, but there is always one that makes something manifest at the expense of making other things obscure. Well, that's a trademark of dualities. But here, these are very low budget dualities, right? These are not like the fancy dualities that you're learning with Juyi or with some of the other speakers here. These are dualities that can actually be completely proven, because they are very, very simple, okay? So I'm hoping Simone gave you some examples. So in N equals 4 superjamins, you probably learned about dual conformal invariance. Is that a yes? Excellent, yes. And he gave you the formulation in terms of momentum twisters, which made the whole property manifest. But dual conformal invariance was only a piece of something bigger, which is a yang-yang, that N equals 4 superjamins also enjoys, or at least the s-metrics enjoys. And there is a description in terms of grass manions that makes the symmetry manifest, okay? There is yet something else that comes from here, which is called positivity. And there is also a description that makes that manifest, and is called by their creators, the amplitude hydrum, okay? So what are we gonna be talking about here? All that sounds very fancy and very exciting, but we're gonna go back to Earth, okay? So I'm not sure we're gonna make it all the way to the amplitude hydrum, so if that's what you wanted, you can talk to me and we can discuss about the amplitude hydrum. But let's go back to Earth for a second, and consider something very boring, as we said that we learned in graduate school this time, which is U.N. yang-mills and the scattering of gluons. So for the time being, let's concentrate on three levels, so we're doing perturbation theory. So I'm also assuming that you all learned about color decomposition with Simone. So what is color decomposition? We know that each gluon carries three pieces of data. One is the momentum, the other one is the polarization vector, and the third one, the color index, right? So we can consider the scattering of N gluons, so these are all gluons, and each gluon is described by a momentum vector, a polarization vector, and a color index. So the color index can take any of N square values, and as you can imagine, if I try to write this down, it's going to be a complete mess, right? So I hope Simone explained that you can take the structure constants, write them down as traces, or as a difference of traces, and then start to merge them as if they were little pieces of a big Lego puzzle, and you start to merge them until you end up with single traces. And at least at three level, you can write any scattering amplitude, or you can reorganize your calculations as a sum over permutations of N points, modulo cyclic permutations. So this should start to look familiar, hopefully, to one of the formulas that Simone wrote. Trace of, so here, my T's are N by N matrices, so these are the generators of UN in the fundamental representation, say. So for me, they are N by N matrices, and we can decompose any scattering amplitude in terms of the color structure and a piece that is purely kinematic. So this part only depends on the polarization, on the momentum vector, and the polarization vector of each particle. The color structure is completely gone. Is there any advantage of doing this? Yes, the advantage is that this object can be computed using Feynman diagrams that can be drawn on a plane with a particular ordering for the external particles. And what's the ordering you're going to use if you're computing this particular one? Well, the ordering given by the permutation of the labels that you chose. So this is at three level. And at three level, yes, everything is plain. So the answer is yes. So most likely, the object you concentrated on, most of the time with Simone was this object, which is the canonical ordering, let's say, the canonical partial amplitude, which is something that he probably wrote as the part Taylor formula times a function of momentum twisters. So but today, we're going to discuss a property that these objects satisfy that I'm sure Simone didn't tell you about, or I hope he didn't tell you about. And the reason he didn't tell you about is that this formulation makes it very, very obscure, OK? So we're going to discuss, for the time being, what is going to be called a mysterious property. So what's a mysterious property? Well, it's not a property of just one of these. It's a property that relates different partial amplitudes with different color orderings. So note that given that the trace is cyclic, the number of different partial amplitudes is not n factorial, as you have thought, because the number of permutations is n factorial. But note that here, we're dividing by zn. We're dividing by cyclic permutations of the number of independent partial amplitudes. Once again, these are called partial amplitudes, is what it seems to be, n minus 1 factorial, OK? So the mysterious properties that we're going to study today are related or are relations among these different partial amplitudes. So here is mysterious property number one. You take an amplitude, one of the partial amplitudes, say this canonical one, and you choose one particle, say the last one. And you add to this the amplitude where everything else is in the same location, except that n, the particle n, just hop one space to your left, OK? Now you add the next one, and so on. The particle n keeps hopping until it gets next to particle one. Now, note that this combination is a crazy combination, right? So these particles all carry different helicities, different momenta. So it's like adding apples and oranges, OK? So we're adding apples, oranges, pineapples, and a lot of stuff. And then in the end, you get 0, OK? So that's a little surprising, of course, just told like that. Now, is this completely insane, or is this something that we can actually sit down and prove? So let's start with MHB amplitudes. So MHB is also something very familiar by now, right? Maximally Helicity Violating Amplitudes. Those are the amplitudes that Park and Taylor computed. So here, once again, I'm going to be using this notation. Each null momenta can be written as the product of two spinors. I'm going to use a notation AB for the contraction of the spinors and this notation for this one. But of course, you know that I don't need this one today, or at least for the time being, OK? Now, if you don't like this, I know some people don't like it because for some reason it looks strange. You can also think, and here is your first exercise, show that you can also parametrize the momentum vector of any particle as follows. So this is a time component, and the space component, or the space part of the four dimensional vector is the following. Here I'm using a pair of complex numbers, or actually I'm using a single complex numbers and its complex conjugates to make up real numbers. This is a real number, real number, a real number. And know that I describe my vector with only three degrees of freedom, because z is a complex number and w, a is a real number. So your exercise is to show that this satisfies us, that this is zero, and moreover that these strange objects, our spinors here, are nothing but. So now we are ready to write down the part Taylor amplitude once again. Let me denote it as mHp. So I've been talking about polarization vectors all the time, but I haven't told you anything about helicity. But you all know that the information about helicity is contained in the polarization vectors. So I could start by saying that I'm computing an amplitude where particles 1, 2 are positive helicity, all the way to particle i, which is negative. Let me write it like this. i minus 1 has positive helicity, particle i negative, i plus 1 positive, all the way to, say, j minus 1, that has positive, j minus, j plus 1 positive, all the way to n, that has positive helicity. And this very, very long expression, the answer is actually shorter than the expression that I wrote down there, is just this combination, given in terms of these variables. So this is something that you're all familiar with. Now, do we have any chance of proving the mysterious identity number one? Sorry? Yeah, it's sensing our imagination. Very good. Yes, it's implicit. I didn't write it down, but so we can have a calligraphic ball phase amplitude, which is this amplitude times moment of conservation. Thanks, that's very important. So let's try to prove the mysterious identity number one, using this formula. Well, the first observation is that the i, j to the 42 is doing nothing. Every single amplitude in this identity, here, the gluons are being flipped. The ordering is changing. But they move with their own helicity. So particle i, wherever it moves, is still a particle with negative helicity. And particle j still has negative helicity. And no matter where they are, the formula always has i, j to the fourth. So in the proof, I can first of all factor out the i, j to the fourth. That factors completely out of this formula. Now, the next thing that I want to do is to factor out the bracket 1, 2 in the denominator. Can I do that? Well, you would say, wait a second. No, you cannot do it. Almost every term has the factor 1, 2 in the denominator. Literally, almost every term, except one term. Except this guy. This is the only one that doesn't have it. But don't be so picky. And just factor it out. We'll take care of that term later. How about the term 2, 3? Well, almost everybody has it, except for one guy. But again, don't be picky. Just keep going. And then we're going all the way to n minus 2, n minus 1. Now we open our parentheses, and we start to see what are the things left out. So from the first term, what are we missing? Well, we're missing n minus 1, n. I think that's the only thing we're missing from the first term, right? Good. Now look at this formula. Every time you see something like this, you say, oh, it's so tempting to complete it and make it into a nice object. How would it look nicer? Well, it would look nicer. You see, you guys were lying. I was missing this guy, too. So every time you see something that goes around, and it wants to finish going around, it's begging you to be completed. So let me also complete this here by multiplying by n minus 1, 1. So every term here will also have that in the numerator, because none of them have it in the numerator. So this is what I have to do to this object to get the first one. How about the next one? Well, the next one is going to be n minus 1, 2 divided by, let me think. The next one is going to be 1, 2, 1n, n2. How about the next one? The next one is going to be 2, 3, 2n, n3, and so on. All the way to, so I started with n minus 1. Go through 1 up to n minus 2, n minus 1, n minus 2, n, n minus 1. So that was left after I factored out basically the part Taylor amplitude when I remove particle n from the computation. So if you take this and you multiply it by this, you should get the first answer, and so on. So now we're left with proving that that thing is 0. Well, how do we do that? One possibility, which is exercise number two, is to use a telescopic. So the telescopic proof what it does is that you add up the first two terms, and then you get something that looks like a term that you would have if you didn't have particle 1 in the equation. So these two guys, you can show that they can merge into a formula that has n minus 1, 2 over n minus 1, n, n2. And you can keep going, and using a telescopic argument, you get that this is 0. So that wasn't that difficult. Another way of proving it, which is the one that I prefer, actually, is to use this formula for particle n and think about my function as a rational function of a complex variable zn. So this is a rational function of the complex variable zn. Now zn only appears in the denominator. So it's a rational function that vanishes at infinity, because it has two powers in the denominator, zn, in every term. The other thing is that, well, if you compute the residue of that rational function at any of the possible poles, what are the possible poles? Well, the possible poles are where zn approaches any of the other z's, z1 or z2 and so on. You can easily see that when zn approaches z1, you get a pole here, but you also get a pole here. And I just told you that these two terms combine to something that doesn't have that pole. So the residue is 0. So you've got a rational function that has no poles and it vanishes at infinity. There is only one such thing. So that's a proof. Now I have to say that this looks pretty simple. I mean, if it's actually something that you can do on the blackboard, it's something pretty simple. So I would say that Park and Taylor, or that formula, makes these properties manifest, this particular property. Now, how about more complicated amplitudes? Let me give you a quick example. Take this amplitude as your starting point and try to prove this identity. How many terms? Five terms, because particle 6 has to move five times to complete the circle. Now, just to make my life a little simpler, let me introduce a notation. This means that. And what I'm going to do here is write what this looks like in its most compact form. So this is the formula for this term. And this is actually the simplest of all the possible terms that we have. So take this formula, change the indices. You are going to get this one. This one, you can also get it like that. The next one, where 6 moves here, is even more complicated. But I encourage you, if you want, as an exercise, to have a look at the expressions for these amplitudes in this paper and try to prove this identity. Right now, I can tell you, well, good luck. If you try, probably you won't come back or you won't talk to me for a long time. So this property seems to be very mysterious indeed, from the standard point of view. It seems to be very manifest at the level of Park Taylor for MHV amplitudes, but it seems to be a completely miraculous identity for this. Well, I can tell you what this identity means now. Probably some of you know, so now I'll tell you the identity of this mysterious identity. So it's the following. Remember that having studied UN, imagine that we only have gluons, it's pure young meals, no matter, no fermions, just pure gluons. And this is our gauge group. And remember, we wrote down our amplitude as the sum of traces. Now, what happens if I want to compute the amplitude for scattering m minus 1 SUM gluons and producing a photon, producing a U1 gauge boson? Why should that amplitude be? Well, that amplitude better be 0 because the U1 completely decouples from the SUM part. Well, what's the generator that I have to put in here if I'm considering that particle n is a photon? By that I mean that it is in the U1. It means that the generator that I have to put in my formula is the identity. And remember that my formula didn't care about the color structure of the particles. These things were computed once and for all. That was the beauty of the whole calculation. These things are completely independent of the color structure. So you compute them and they are written in a stone. So they are there. And now the different color structures only come in here. But now we're saying that when this happens, if particle n is a photon, the amplitude has to vanish because the photon decouples. So how is that possible? Well, now we look at all these terms, all the traces. And we plug the identity every time we see particle n. Now imagine that you have trace of T a1 all the way to T a n minus 1 and T a n. Well, this trace goes to trace of T a1 all the way to T a n minus 1 because this is identity. Now imagine that you have the trace for particle n is somewhere in the middle. But otherwise, it's the same trace that I was considering at the beginning. Well, guess what? This guy will disappear because it's the identity. And I'm going to get exactly this trace. So I'm concentrating on a particular order for all the other particles. And I'm only looking at the terms that differ by the location of the trace of the generator of the nth particle. That particle can be in n minus 1 possible locations. And all of them will give me the same trace structure. So all the partial amplitudes will be multiplying the same trace structure. And this object is nothing but the coefficient of the trace with the canonical ordering. So these partial amplitudes better add up to zero because the traces are independent. And if I want to get zero, then this has to vanish. So once again, this is a property that is, of course, obvious from the Lagrangian. But if we don't know anything about Lagrangians, it's a completely mysterious property. In fact, there are more mysterious properties that these objects satisfy. So let me give you the actual name, which I can actually, I could never get right the spelling. These are called the Claes-Kuif relations. And we're written down in 89. These are relations that can be derived exactly in the same way. But now you start to go crazy with the gauge groups. You can say, well, instead of u n, imagine that n is the sum of two numbers. Imagine that your u n can be broken into two groups, into u m 1 and times u m 2. So again, the two groups of gauge bosons cannot talk to each other. So computing amplitudes with different color structures should also vanish. So those will translate into identities. Now, Claes and Kuif sat down and put all those identities together into a single object. And they realize that they allow you to do something very interesting. Imagine that you have an amplitude where particle 1 is here, and particle n is sitting somewhere in the middle, and they are separated by some groups of labels, alpha and beta. So these are just some groups of labels. So KK realized that you can use all these identities that I mentioned to write this as a linear combination of partial amplitudes where 1 and n are adjacent. So you can fix the order of two of the particles. For any partial amplitude, you can always write it down as a linear combination of amplitudes with this particular ordering, where 1 and n are adjacent, or at the beginning and at the end. But remember, these guys are cyclic, so that means that they are really adjacent, with the other particles scrambled in some way. And the way is a very particular way. And they even gave us a sign. So the number of elements in beta determines the sign. And these are called order preserving permutations. Well, beta transpose is that set transpose. And what do I mean by order preserving permutations? These are all possible permutations of the labels in alpha and beta transpose, so that when you see them, the alpha set, even after the permutation, keeps its corresponding, is relative ordering. And the beta transpose also keeps its relative ordering. So let me give you an example. So imagine that you want to reorder this, so that 1 and 6 are adjacent, or 1 at the beginning and at the end. So you will start and compute this term, plus some other terms. Note that I took 4 and 5, and I reversed the order, and I put 5 and 4. And then I started to scramble the alpha and the beta sets in such a way that they preserved the ordering. So why is this interesting? Well, at the beginning we started saying that the number of independent partial amplitudes was m minus 1 factorial, because they were cyclic. But now we see that we can find linear combinations so that we can always move one label to the n, keeping one fixed. That means that the number of independent partial amplitudes has gone from m minus 1 factorial down to m minus 2 factorial. Is that the n? Are there any more relations? Well, that's what everybody thought. And you also thought that I was going to forget, perhaps. But here is another exercise. Prove that this is true using MHV amplitudes, once again for the part Taylor amplitudes. You probably don't believe me if I told you, but it's actually lots of fun to see how it happens. How this permutation has to be transposed is quite interesting. So I encourage you to try. But once again, it's something that is a 10-minute exercise, maybe even five. But I won't do it because we don't have that much time. And this is something that I'll still propose, something that Park and Taylor make manifest. So the last property is something called the Bern, Carrasco, Johansson, or BCJ. And for this one, we had to wait until 2008. So what they realized was that m minus 2 factorial was not the end of the story. There is actually more. You can take, you can use the KK relations to bring any two particles adjacent to each other. Say one again. But BCJ wanted to be different, so they wanted to bring together one and two. So they brought together one and two. And they claimed that you can even bring together adjacent to each other a third particle. Now you know why they did that because it's natural to think that the third particle is number three. So they claim they can take any amplitude of this form and they can write it as a linear combination of amplitudes where one, two, and three are adjacent to each other. Once again, this is a sum over all permutations. This time is order preserving permutations of alpha and beta, no transpose, times some functions that only depend on Mandel-Stamm variables and the permutation. You would say, well, this is trivial. I mean, if you're allowed to multiply by any function there, you can basically do anything you want. But don't be so quick. Because I told you, these are functions that only depend on the Mandel-Stamm variables. They know nothing about helicity. They have no idea which helicity gluon 1 has or gluon 3 has. So that's where the non-triviality of the BCJA identity comes. Of course, there is something more non-trivial about the identity, which is the actual identity. These functions are not that easy to write. I could do it. But you can look at equation 423 if this was fast. Now, this tells you that the number of independent partial amplitudes goes down from m minus 2 factorial to m minus 3 factorial. And this is a number we're going to see again and again appearing for apparently no reason. So m minus 3 factorial seems to be something interesting. So the last comment that I want to make about these kind of properties is the following. And this one we will actually prove is that the BCJA identity follows from a basic set of identities. And these ones, maybe a little bit to be pretentious, they call them fundamental BCJA identities. And the identity tells you the following. Here I'm assuming everywhere that SAB is a standard Mandelstand variable. You sum the momentum of the two particles and you square them. So standard Mandelstand variables. The claim is that if you look at this equation in this paper, this equation follows by taking linear combinations of identities of this form. This looks much simpler to prove. I assure you than this. So this is the one that we will actually try. Well, given that we only have five minutes, but in fact, this would take only five minutes. But I don't think I want to use my five minutes doing the proof. We'll do the proof tomorrow. Trust me for a second. And if you want, you can try ahead of tomorrow and try this for MHB amplitudes. So if you trust me that you can actually do this in five minutes, I think you would agree with me that the conclusion of all this exercise is that Park Taylor is excellent for identities relating partial amplitudes. Now, we know these identities are supposed to be true for any amplitudes, not only for MHB amplitudes. So if any object satisfies these identities, you could think about Park Taylor as being almost a defining object that satisfies these identities. If all amplitudes, regardless of their helicities, also satisfy these identities, perhaps there is a formulation that uses at its core something like Park Taylor and trivializes all these identities. Or at least it makes them manifest or as manifest as they were for Park and Taylor. So and here is where Nair comes, as we said in 1988. And he says that, well, Park and Taylor, this formula that we have been writing down all this time, looks like a correlator of fermions on a Cp1. Well, we had a complex number, right? I told you, write this vector as something that had a real scaling times, something that only depends on complex numbers, Za, and its complex conjugate. So there seemed to be a sphere somewhere. But which sphere is this? So Nair also realized that the sphere was somehow the celestial sphere. But the celestial sphere was something that Penrose used as the building block for building twisted space. So Nair envisioned that this Cp1 actually lives inside twisted space, which is Cp3. And then he went on and said, OK, this is it. Everything, all scattering amplitudes in four dimensions, can be written as correlators inside Cp3, inside a sphere in Cp3. So he tried for the next one, NMHB amplitudes, the next two MHB amplitudes. But he couldn't make it work. I have to say he came very, very, very close. Actually, he came extremely close to developing something that is known as the MHB expansion. But then Whithing in 2003 said, well, maybe this Cp1, where these correlators are being computed, is not a Cp1 in twisted space. It's actually its own Cp1. So maybe that Cp1, where we are inserting these operators, is the world's sheet of something. And twisted space happens to be the target space. And then you can go to space time using the Penrose transform. Now, how many different ways are there of embedding Cp1 inside Cp3? Well, they are classified by a number called the degree of the map. So you can have degree 1. Degree 1 means that you take your Cp1 and you basically embed it identically inside your Cp3. And that happens to be 9 years Cp1. As you increase the degree, you get different objects. And Whithing conjectures that they happen to compute the different levels of how far away you get from MHB is related to the higher and higher degree. So I'm going to end the lecture by writing down the formula, not in twisted space, but in space time. And tomorrow, we're going to look at it. So this is a formula that was written down by Whithing, but also very studied carefully by Royburn, Spradling, and Volovish in 2004. And it's now known as the Whithing RSV formula. So let me write it down. And this is a formula for amplitudes in n equals 4 superjam miles. Of course, you are experts. So the formula starts exactly with a per-tailor formula, but written in the variables that I show you where each of the landas has a scale and a complex number. So I'm going to start by writing what would be the per-tailor formula if you were using those variables. Well, but these variables should agree with landa only when the degree is 1. So in general, we should be integrating over these variables, n of them and n of them. Well, we have to find a way of mapping our CP1 into twisted space, or in this case, it's going to be space time. So I'm going to define the embedding of the CP1 as follows. So these are all spinners. They contain two components. So a map of degree d would be this polynomial map. So I have to integrate over the components of the map. And I have to impose that the punctures in the original sphere map to their corresponding values projectively in space time. And the last piece of the formula is this supersymmetric piece. But I've defined this supersymmetric object to be, and these are the four components of supersymmetry. That is something that I forgot. But as you know from the space, I didn't, which is this formula has a redundancy. And even though this is all holomorphic, I'm pretending that I'm dividing by the volume, whatever that is, of gl2. So what this formula is, don't worry about it. But what's important is that look at the formula for a second, actually in our minus two minutes that we have. Look at this formula for a second. And know that everything is permutation invariant, completely permutation invariant, except the only information about the ordering of the particles, where does it enter? Here. So all the information about which amplitude where it's studying, or which ordering, enters in this factor, which is identical to the Park-Taylor factor. Which sector we are studying? So once again, d equals 1 is mhb. d equals 2, degree 2 is next to mhb. Now you can see the pattern. So all the proofs we did for mhb amplitudes, or that you're going to do, will carry over to this formula. So we have succeeded in writing something that contains Park-Taylor at its core. And everything else is permutation invariant. And hence, all the properties that we have discussed become completely manifest in this new formulation. So this is the formulation we're going to continue discussing tomorrow. And if you try the fundamental BCJA identity for mhb, try to think how it happens to be true for this object as well. Know that this time there is something else you have to take into account. So tomorrow we're going to discuss that particular issue. All right.