Twistor String Theory — Part 2

0:00 I'm coming to the dinner, so we have three of us at six or five in Wallet's Chapel. Okay, so to get to the chapel, we're here at the Mount Institute, this is St Charles, this is where St Charles bifurcates, okay, if you don't know your way around, the simplest way to get there is to go out to the front door, turn right, right again into Keetel Road, all the way to the people road, 90 degrees, down here South Park Road, Park Road, anyway, straight down here, and Wallen is on the corner, this is Broad Street. Broad Street, Blackwells is there, the Sheldon is here, so you can't miss it as they show them. If you are a bit more adventurous, you go out through the side door, through the parking lot, down the road to the back here, and then you place your top. When you get to Warnham, so this is the diagram coming down here, when you get to Warnham, you're going into the fort, around to your left, and go through a passageway as if to to go to the garden, this is the garden, here. So, as if you were interested in horticulture, and off the passage that goes into the garden is an opening to the chapel where there should be a chinking of flowers. And then afterwards we will somehow move around and go round here, and it's complicated. It's down the back. We're in the old library. We're in the old library for dinner. And 7.15, and we've been told we have a warm start-up, so if we want it to be warm, we have to start on time. So we'll start on time. Although this isn't the end of the meeting since it's my last round, so I'll just start by thanking the organizers as well as other speakers before I forget to do so later.

2:30 Ladies and gentlemen, Now, a little bit of redundancy of my starting point that's on previous slide. I remind you that the momentum of a massless particle, which I'll call t, is likewise We made the quadratic equation, p mu, p mu equals zero. And the space of solutions of p mu, p mu equals zero is a cone, which we usually call the lake count, if it were x instead of t. And it's a cone over something that I'll call our quadric. But we'll take our momentum to be complex, so our quadric is a complex quadric. The quadric is what we get if we can write the p's as homogeneous coordinates. So the equation P nu P nu equals 0, with the P's regarding as homogeneous coordinates in CP3, would define a quadric. And that particular quadric, so there are four P's, so there are homogeneous coordinates for CP3, and the quadric equation leaves us two dimensions. So, the projected quadric in four variables, we'll be able to explore the four dimensions, It's actually, the quadric is P1 times P1, or CP1 times CP1. And this idea is usually captured by writing them in a mechanism in terms of spinners. Someone said that P mu is vector-related as a bispinner, by contracting it with sigma mu, which is the tidal part of the rock-down matrices. And what P being white-like says that P factorizes as lambda times lambda-tilda, where lambda and lambda-tilda are homogenous coordinates over two P1s. So there are two P1s because the quadrant is P1 times P1. Lambda and lambda-tilda are one P1, lambda-tilda is the other, and P is lambda lambda-tilda. and if p is given, lambda is the term of electric multiplication by non-zero-complex numbers, by which we should define lambda t. Now, at least for most purposes in this conference, the main reason it's useful to introduce the lambda's is that it enables us to neatly

5:00 incorporate the helicity of a massless particle, such as a gluon, in a scattering process. So, here's a scattering amplitude with N particles, which I've numbered from 1 to N. And if we're doing Yang-Mills theory, and we're doing pre-diagrams, they come with a natural cyclic ordering, which enters in the gauge 3 factor of the amplitude. I'm not going to write it down. I've taken a statement from the fact that the transparency is two-dimensional to exhibit the natural cyclic ordering, which is benevolent, which is like a diagram. Happily, there is an actual cyclic ordering, so the diagram is not the same. And each particle has a lambda and a lambda-tillula. It has a momentum, which is the product of And it has a helicity, which is 1 or minus 1, if there are limits. And when I say that the helicity is neatly incorporated in one of interest lambdas, what that means is that if the I-particle has a helicity H-I, then there's a scaling of lambda and mantillin for the I-particle. We make a separate scaling for each particle because each particle of its own for the i-th particle, the amplitude should scale as Ti to the minus 2Hi. That's a statement that has nothing to do with gauge theory. It would be true for massless particles in any scale. Like in our applications today, we're doing gauge theory, and therefore the Hi are all plus or minus 1. So, if you want to derive this fact, well, what Elicity is by definition is the eigenvalue, or it measures how the state transforms under a rotation around its direction of motion. A massless particle can't be brought to rest by a Lorentz transformation, and any alliance frame is moving, you rotate it around the direction of motion, and under an angle, rotation by angle theta, it transforms like e to the by theta times h, where h is called the helicity. Although I defined it in a particular alliance frame, the helicity is so defined as the direction frame. So, one deduces what I say here, by just thinking about how the lambda is transformed under a rotation around the spatial direction of which can be vector p. And if we set up our orientations correctly, we get this resolved,

7:30 but the opposite orientation, we have a plus line in the next one. So our amplitudes are actually going to be functions of lambda and lambda theta that scale in an appropriate fashion, and for tree diagrams, which means the classical limit of Janus 3, those amplitudes are going to be rational functions of lambda. Remind you, if you were here at my first lecture, that I just explained briefly that I'm not sure what the scattering amplitudes are. They describe the alternative to a transition process between the days and the past and the days and the future. Now, what are the diagonal scattered amplitudes? The simplest ones are actually a zero. If If all external particles have the same multiplicity, for example, plus multiplicity, or if they all but one have the same multiplicity, which I've drawn here with the second case with a lot of pluses and more minus, the amplitude is zero. One thing I should warn you of is that the convention here is that we don't distinguish particles that are coming in in the past of those going out in the future, because in And in relativistic scattering, the same analytic function describes both cases. So what's indicated as a plus is a particle that, if it is coming in from the past, has a plus elasticity, while if it is instead going out to the future, it has one of the cells in it. So that's built into a highly compressed notation. The fact that the all plus or all but one plus amplitude has vanished was originally proved in the 80s using supersymmetry. You actually can prove this using supersymmetry, even if you're interested in non-supersymmetric mainstream, because at tree-level supersymmetry does no effect on the scattering entities anyway. I think you can also prove it, more or less along the lines discussed by Bardeen, by using the fact that the self-flaw and anti-self-flaw equations, f plus equals zero and f minus equal are consistent truncations of the fully angular equations, which I remind you, are second-order differential equations, which you can write in various ways. So, the tree approximation to the scattering amplitude is actually the action for a classical solution that's determined by which initial final part of this you want to consider. So, consider N minus 1 of liloans all have the same elasticity.

10:00 That would correspond to a self-tool solution of the annulism actions, or anti-self-tool of the reversal orientations. Now, imagine adding one more blue on that might have either cross or minus ones. So, depending on which one we add, we'll get one of these two annulisms, and we're calculating the variation of the action with respect to that last blue on the other. But since the self-dual equations were a consistent truncation of the solutions, the solution of the pool equations corresponding to the n-1 gluons of the same illicity was actually a complex but self-dual solution to the n-illist equations. And since there was a solution, the action was stationary, and the action being stationary means that it doesn't change when you add one more gluon. So we get zero if we just add one more gluon regardless of its illicity. If we have two more of the norms, we'll be taking the second derivative of the action at a classical solution that won't be there. If I'm speaking too fast, I can't make that argument as precise as I can. But I do believe that that argument can be spelled out a little bit more precisely, and there's a good way to understand why these two entities don't. Of course, there are other ways to understand it. There's the supersymmetric length, and there's also the twister stringer crunch, where, for instance, this one can correspond to a curve of a degree minus quantity, which doesn't exist. So, from this point of view, the first one managing amplitude is what people call the MHV, or maximum helicity violating amplitude, in which the helicities are some termination of two minuses and a lot of pluses or two pluses and a lot of minuses. So, it's called maximally elicity-violating because, well, for example, this one would completely violate elicity. A lot of plus elicity particles come in and disappear, or a lot of plus elicity particles come in from the past and going out into the future there were all minuses. the Ignat-0, this would describe a completely elicity-violating process, and this describes the one that violates electricity as much as possible. Now, there's this nice formula by Park and Taylor. Well, for four particles, it goes back to DeWitt. Park and Taylor computed it for five particles using string theory.

12:30 DeWitt does not say how to compute it. You can look at his paper. He gives the answer. And you're on your own private physics. But Park and Taylor computed it for five particles using string theory. String theory, physical string theory, described Yang-Mills plus a lot more stock, but the more stock isn't important to the bone energies. So they computed a string amplitude, one edge of a, and got this formula for five nodes. And then in principle they would have done that for end nodes, but once they had cases of four and five, they guessed the general formula, which was then proved directly without getting it from string theory. This was an In the early instance, therefore, where strain theory was used, well, at early but not the earliest instance, where strain theory was used to calculate perturbative amplitudes which could be gotten from field theory, but which seems more hard to get from field theory. That was possible because the strain theory in Gena 0, the tree approximation, involves a simple integral on a disk, or the Riemann sphere, rather than the complicated sound-of-finding biographies that the textbooks have given you at the beginning. Similarly, Green and Shorts have computed the first one-loop, four-particle amplitudes in gauge theory about it. Maybe Green-Forts improved for some of those kind of things. And that was again done in spin theory before it was extracted from gauge theory. Anyway, regardless of how you get it, this amplitude is a delta function that says energy and momentum are considered. That's a universal factor in all scattering in the Kelsky space. What's interesting is what it multiplies. What it multiplies, we expect it to be a rational function of lambda and lambda-philo. It turns out to depend only on the lambdas. And the lambdas live in a two-dimensional space. Remember, the lambdas were homogenous coordinates, or Cp1. So they live in a C2, which has a Lorentz invariant which has been denoted by this angle bracket. So from this point of view, the simplest line vanishing amplitude is given by this simple Lorentz invariant function of the Lambas. The fact that it only depends on the Lambas and not the Mendotellus is often described informally by saying it's polymorphic in Lambas. The reason for the terminology is that here today we're taking the Memento to be complex, When you really apply this to physics, the elementa are real in Miklansky's case.

15:00 And then lambda tilde is plus or minus the complex conjugate of lambda. And the amplitude, therefore, was destined to be a function of only lambda and its complex conjugate, but in general, it wasn't going to be a whole and longer function. But because this thing only depends on lambda about the complex conjugate, when written for physical elementa in Miklansky's case, it's actually described by a whole and longer function. Now, Maier suggested an interpretation of this function. It contains a numerator, and he showed that the numerator was a supersymmetric extension of the else function. And it contains a denominator, and he showed that the denominator is a function that appears in the three-dimensional in the current algorithm and in the theory of Katz-Linian. This is what businesses call a current correlation function of Cp1. So, what I did was to... So, Langer suggested that some kind of two-dimensional field query on Cp1 should be used to get the simplest non-trivial gauge theories gathered into it. That image re-entitude, the cocktail line. What I did was to interpret this not as an abstract CP1, but as a CP1 living with CP3 slash 4, the glabial extension of CP3 that we discussed in the first lecture. And then I interpreted holomorphy, which I reminded me it's independent of . As a consequence of the curve having degree 1. I showed that if the CP1 is actually a degree 1 curve, sometimes called a line in CP3-4, then holomorphy would rise naturally. And more of a short, although not truly rigorous, combination, the card with the formula, with my years in combination with the non-vigorous as well as the meaning of it. Now, we can see why the curve should have to be won by pursuing the reasoning of the first lecture. I'm actually, I've been saying various things that were explained by other constructors, but I thought it was a good part. Now I'm about to say some things that were more likely to struggle with them. So we go back to Cp3-4, it has four bosonic homogenous coordinates, I'm calling Z1 up to Z4, and four perionic homogenous coordinates, Psi1 up to Psi4.

17:30 And we recall that it has a symmetry that we call PSU4-4, which is simply the linear transformations. Well, the P means we divide by scales, then we consider linear transformations up to a scalar multiple. The reason being that the scalar will act trivially on C3-3-4. But the S means that we ask that our linear transformations preserve the measure, DZ1 to The reason we want to preserve the measure is that we have to use it in refining the twist or string theory. But, so PSU form form becomes a symmetry of the twist or string theory, but there are additional linear transformations of these coordinates, which are not ones and also the identity, and are not in the PSU form model because they don't refix the measure. For example, if we, there's essentially only one example. If we rescale the size and don't rescale the z's, we can define a transformation, maybe called u, u is the group that generates those things. It doesn't refix the measure. You have to remember that the psi transforms opposite the psi for anti-commuting radicalism. And then you see that the magnetor transforms as lambda to minus four, just because there are four anti-commuting fermionic coordinates. Now, in our first option, we had Christopher's case deal of curly A, that we expanded in powers of the size. We expanded in terminated, because there are only in a part too many size of the anti-commute. For brevity all we consider the bottom and top terms. So the bottom term is no size, we call it a hat, it describes a motor of elicity one, and the top term we call it g hat, it describes elicity minus one. And obviously if you consider a transformation that scales sine by lambda, a hat is invariant, there's no size, but g hat scales on lambda to the minus four, because we have to compensate for the way the size transform. So this is a transformation which is not going to be symmetry

20:00 of Yang-Hill's theory or of crystal string theory, but is useful for determining what kind of contributions will give a given scattering. So suppose we want to get an amplitude with q of the neurons have negative velocity, while the rest have positive velocity. As I told you, the answer is zero if Q is zero or one. And when Q is two, we will get the Park-Taylor whole market deficit, and the higher Q will get more complicated, in general. But anyway, for any Q, how would the amplitude scale with this non-symmetry? Well, the negative velocity and the amplitude gluons are represented by G. So if we need two of them and G scales as lambda to the minus four, then Q back as G scales as lambda to the minus four of G. So to get an amplitude with Q negative velocity of gluons, we'll need to find a process in the Twister-Springs that transforms, like so. Now, the strategy for doing so is what I very briefly stated in the first lecture. So, we let C inside CP3-4 be a kernel of genus GM3-3. For the moment, it will be a connected kernel. Now, let M be the molecular space of kernels Ideally, we would find a whole measure of du on n, and then the contribution of n to a scattering amplitude is obtained by expanding the integral over n du. We'll have to figure out what that means. Perhaps it's going to be a contour integral sometimes. What we integrate, we take the A, remember was a 0-1 form on CP3 slash 4. We just restrict it to the curve C, and then we make the D bar operator, D bar sub A. Well, D bar sub A restricted to C, or the D bar operator on C. We take its determinant. Then we integrate that over M. Well, if it was a function of M, we could then expand the powers of A. And when we expanded the powers of A off to a certain order, we were supposed to get the scattering amplitude for the appropriate number of gluons, depending on how far we take that expansion. Now, in genus 0, this is actually going to carry out successfully, though not

22:30 completely grimacing, as Fradlin explained. GINA0 is important because GINA0 describes the classical limit. As is always the case in the BIM theory, GINA0 is the starting approximation, and the quantum corrections come from curves and hyogenes. Well, the first step is to define the measure. In fact, we're in luck because on the space of GINA0 curves on CP3-4, there's a natural north of the measure. To describe the modulate space of curves of geno zero and degree d, actually I'm repeating things to rather than 10, but we take an abstract CP1 with homogenous coordinates u and v, and we describe the geno zero curve in CP3 slash 4 by a map of the abstract CP1 into CP3 slash 4, which is given by expressing the coordinates of CP3 slash 4 as functions of u and v. To make the degree d, the functions and alpha should be homogenous polynomials of degree D. Well, we have to understand alpha is big anticommuting because psi is anticommuting, so the coefficients of polynomial alpha are anticommuting variables. The measure, if you find this notation clear, D is 0, for integrated over P and alpha, it just would be P and alpha. If that isn't completely clear, it can be more explicit. We take the basis F gamma of the space of degree D on alpha. So there are D plus one of them. The choice of basis doesn't matter, we'll disappear another. We write, we expand P in terms of this basis and we expand alpha in terms of the basis. There are a lot of coefficients, which are of the little alpha. And the integration measure is literally an integral dp in the alpha, the overall components of little p and little alpha. It's independent of the axis of faces because under a linear transformation, the measure for bosons and fagnols transforms oppositely. So this measure is actually independent of the unnatural choice of faces. And moreover, So, it's invariant under the action of GL2 acting on U and V.

25:00 U and V with homogeneous coordinates in, on the abstract CP1. So a linear transformation of U and V would change the bases at gamma, and that would be a special phase of what I just told you, that this measure is completely independent of the choice of bases. So, since it's invariant under GL2, we can divide, If we took the space of coefficients, that would be c d plus 1 slash d plus 1. Well, that's 4, because there are 4 p's and 4 alters. The bond aspects is this thing divided by gl2. I wrote a measure upstairs. It's gl2 invariant, so it descends to a measure of closure by gl2. And that's how we get our natural hormone measure on the modulus. So it's a natural measure, but now we need to know how it transforms onto this operation that's not a symmetry. This operation didn't even preserve the basic measure in CP3 slash 4, which is like the degree zero case of what we just written, and it certainly won't preserve the higher ones. If we rescale Psi without rescaling the signal, nothing happens to the bosons polynomials, but Alpha scales like range to the D, because it was a... it's got a whole gamut, isn't it? The whole idea of definition of Alpha was that Psi was Alpha over U of D. So alpha transforms like sine. It's linear in sine. So each component of alpha transforms like lambda. So that means that d alpha transforms like lambda inverse. And alpha had four d plus one components. Four times d plus one. So the measure, finally, transforms like lambda to the minus four times d plus one. Dividing by gl2 doesn't change that. That's scaling the valley of both upstairs, on C something slash something, and also downstairs on the monolith space. So we got a homo-orbit measure of the monolith space of degree D curves in Cp3 slash 4. That transforms like this. Now, we can therefore see what must be the degree

27:30 of a genus-zero curve that can contribute to a scattering process with q gluons of negative elicit. We decided before that we had to scale lambda to minus 4 here. So the important relation is that q is equal to q. So if the number of negative elicit gluons is q, the degree d has to be such that q is d plus 1. And so we see a new explanation of the vanishing of the all-plus amplitude. If q is 0, which would be all-plus, the degree would be minus 1, and we'd be out of luck. There aren't any curves of degree minus 1, so the all-plus amplitude vanishes. And moving on to the case where q is 1, then the degree is 0. That's highly degenerative, means our curve is collapsed to a point at tp3 slash 4. to study the degenerate case a little bit more. The precise statement is that those amplitudes vanish when the overall plus velocity of lumens is more than 2. We do get the plus plus minus from this degenerate collapse curve, but we don't get anything more generic from that degenerate case. I won't explain that more quickly. But when Q gets to be 2 or more, well, I remind you, 2 is the NHV case. We get honest numbers of positive degree. Curves really exist as monoscopes in Cp3-4. And, for example, for q equals 2, the degree is 1, so we would get straight lines in Cp3-4. And Boydman, LeGon, Spradlin, and Volovich extracted the tree amplitudes of any q from this formula. And I regard that as the best evidence that twist or string theory really has something to do with the annual theory. It's the best result that's come from twist or string theory. It can't be interpreted as a result you could formulate in Rinkowski's class. It's a very beautiful formula. It hasn't been so far as useful connotationally as some of the other formulas, but conceptually it's the most beautiful formula in twist or string theory. And it's

30:00 It's truly twisorial because it's very hard to imagine describing a degree d curve in twister space as any kind of entity in Mikalski's space. So this formula really shows the diagonal scattering at three-level in Mikalski's space. Well, first of all, it can be understood in numerous ways, some of which are already known and some have been discovered in twister space, but this is the way of describing it, but really is twisorial. So, whatever else it is, three-level diagonal scattering has something to do with twister Now, while that's definitely the most beautiful formula, and the best evidence that twist or strain theory has some substance, there are other formulas that came out, and Peter Srirczak told us about one of them. Actually, Ruth Grito wrote another, but I won't go on that one. So, here I'll just say that in my original paper, I considered a problem. So, there was a degree D curve. Okay, and it had a certain transformation under that scaling of the curve length. But I started worrying about the following D could be D1 plus D2, for some D1 and D2. And then I could make a disconnected curve of degree D. So, here, we attach all the curve. Here, we manage to attach some here and some here. We're actually going to get zero generically if we leave it at that, because momentum won't be conserved on either side. But, we want this twister space field curly A, and we can connect the two currents by Yeah. So this one's scaled as lambda to the minus 4d plus 4. This one scales as lambda to the minus 4d1 plus 4. And this one is lambda to the minus 4d2 plus 4. But to get anything, we need to connect them by a propagator. And we pick only one propagator because we're amplitudes, two propagators would make a degenerate case, and a wood amplitude. And if you work through what we had last time, you see that the propagator transforms as an amplitudes before.

32:30 So then I noticed that this would, that the same amplitudes that received a contribution from here, could receive a contribution from here. And actually, in my original paper, I thought the the correct amplitudes would be gotten by summing them up. But, Wojtmann's experiment of knowledge showed that the connected geno-zero curves alone give the complete answer. So, that suggested that other types of configurations also give the complete answer. You might do a configuration of two, what we could go all the way to a lot of degree-one curves connected by propagators. D of them, if the degree is D, and each one, each one scales like lambda to the minus eight, the propagator is lambda to the four, and you'll see it all adds up to lambda to the minus 4 e plus 4. So, um, the chances were checking on, um, analyzed this completely disconnected case. Okay, since the connected one had been shown to give the complete answer, that suggested that maybe the other ones were too. It clearly was wrong to just add them up. If the connected one gave the complete answer, maybe the complete And we analyzed the complete disconnected diagrams, which I started yesterday, and that not only gave the complete tree diagrams, but it's the best evidence that this description is useful, because it gave very simple formulas. The one with the three, the connected curve, here this will occur as a degree d, gives conceptually the most beautiful formulas and the most convincing evidence that the twist restraint tree means something. They're harder to write. Here, you get formulas which are less inherently twistorial, but if you have the right inspiration, you might have guessed them in the council space, rather than motivating them from twist restraint theory. But they're very explicit, and for some purposes even useful. I'll go back to the disconnected

35:00 diagrams a little bit later. We've got a little self-requestions for a moment. Does the fact that you're in genus zero make the computation of these determinants of So that's one of the reasons why you're doing it. Oh, yes. Well, there's an even worse reason. But one of the many things that's nice about you, in general you can calculate everything as possible, obviously. But completely different. You can make everything as possible. In the connective case, there are these polynomial equations that you can solve for as possible. I should say there's one more thing, which is that there's a paper that attempts to show directly that the connected and disconnected by-grounds are equivalent. There's also a paper that was briefly alluded to this morning, I think the Dixon side, that considered cases that were neither completely connected or completely disconnected. So it's believed, perhaps not completely understood, that each configuration actually goes a different way. One thing I should tell you is that no derivation is completely earnest. The present make that pretty clear for the connected line. There's this elegant formula that actually works, that is motivating for the twister string here, but the derivation isn't completely clean, and the same is true for the disconnected So a variety of formulas that are actually correct, some of them have occurred to be correct, but in each case the derivation could stand some clarification. Any other questions? Well, I'm going to devote most of the rest of the lecture, most of the rest of the lecture, will be devoted to surveying what I consider unsatisfactory. I'm going to try to summarize it in four topics, at least. I'll summarize what I consider unsatisfactory in four topics, which I labeled from 0 to 3 rather than from 1 to 4, because the zeroth one was so basic. The zeroth one I really just told you, but in general, the derivations lack the priceless

37:30 and precision of standard string theory. And standard string theory, or preterminal string theory, they are completely clear-cut and well-savowed rules, and everything is exactly right. But here, all the derivations are a little bit murky. I hope that's because we're not because the theory doesn't exist, although I can't offer any money back on this. Well, topic number one in our list from zero to three is that there's this nice definition of the measure of G is zero, but it's hard to believe that an analogous formula exists in our case. So, describe the variable higher genesis in CP3-4. One simple thing to do is to start with an abstract curve C and aligns bundle L. And to describe the math fine from C to CP3-4, L should have positive degrees so it has no sections. And then we write Z and side as sections of homework, sections of L.A. or C. So the both in H.R. of C. and L. Except algebraic genres who discuss anti-food violence, Boyd Camp is Operation P that reverses the pattern. I wasn't sure if I wanted it or not. So L was a terminatic section of L and C. So we described the map from C to C3-3-4 by choosing these sections. Now, as in Geno 0, there's a natural measure of EZ-3 if we keep C and L fixed. But there's no way to integrate those with choice of C and L. Or at least, no way to In the B model, I can't see a cure for this. In the Berkowitz model, you might get a SATS-hack measure from the C equals 28 matter system, but yet, although that hasn't been demonstrated in the work that. Now, I haven't had any thoughts about this that I consider good. So, I therefore wasn't sure I should tell you anything. But I decided that if I told you my own ideas, you'd see I'm scraping the bottom of the barrel. And that would at least encourage you to believe that you wouldn't do less.

40:00 So, one thought that's crossed my mind is that maybe the target space of the B model shouldn't be all of CP3-4, but only the plus part, that's only the part where the intermission form is positive. Or equivalently, remember there's this 2-2 intermission form. But it was in Stephen Hubbard's lecture on Muget. So, we can take only the cross part, but only the minus part. Here's what string theorists would call a flop-like transition between them. Only we usually do flops in positive signature calorie geometry. This is a flop in indefinite calorie geometry, so... Is that also a clobiality? It's a clobiality because it's just a piece of the other one. So it still has the same holomorphic measure, just as shown. Well, first of all, one question is, are curves in C plus 3 slash 4 a reasonable enough amount of length of it? Deals with that question may well be known for a long time. And is there any chance that we have a holomorphic measure? I doubt it, but at least, as I told you, I explained my rationale. It's explained as a matter of people's questions. I also sometimes wonder if it's misleading to consider an abstract curve C and a map to CP3 slash 4. Because C can, you see, CP3 has bozoni dimension 3. So when you vary one parameter, a curve will intersect itself. And maybe we want to measure without a chance of pulling out something else to look at. And if that was true, it might be important to try to describe this measure in terms of curves in TP3-4 rather than an abstract curve together than that. But I have no good idea, and I consider both I really only explain them, well, there's no chance that somebody will say something useful about it, and make one of those ideas from the last one, actually. But otherwise, maybe I could encourage you to do that. No other suggestion would be after yours. What did I do when I said this? I had no question about it. Well, so, topic number two is conformal supergravity. So, in the B model, conformal supergravity appears here, well, it appears here rather like the N-Mills wheels here. The way Penrose

42:30 would have told you it should, it appears by deformation of the complex structure of regions of CP3-4. And in the B-model, last time we were really analyzing what string pairs would call the open strings on the B-model, and those gave gang-mills rules. But a similar analysis of closed string nodes of the B-model would have shown that the basic node is an intangible deformation of the complex function together with a conjugate variable. The conjugate called G, which will describe the opposite of this one. So in the B model, what happens in the B model is what happens in the standard string theory. In standard string theory, there are the heterotic strings, which are only closed strings. But if you type 1 strings, which are older, you get gate chills from the open strings and rapidly from the closed strings. A variance of that is what happened with the B model. The open strings may gain in those fields, which I described in some detail. analysis, which are a little bit of time to explain, and what is on today, give deformations of the complex structure of CB2-4, and therefore they give the formal synchrogravity of space to it. In the Vorkovitz model, the gravity as well as the H-bills, a couple of instances, and Vorkovitz explains this a little bit in his lecture, and it was worked out most Now, it's familiar that closed strings constrict the idea of gravity, getting gravity from open strings is bizarre. It's unusual to see conformal supergravity. And that's because physical string theory generates sensible theories. After all, that's because people outside of that. And regrettably, conformal supergravity isn't believed to be some super theory. If it's got higher derivatives, the action is Weill squared. The field equations are fourth order. And we don't have unitary and positive energy and all those nice things. So it can appear here in contrast to physical spin theory, because we started with this

45:00 wasn't guaranteed to have a physically-sensible intermination in spacement. And it would appear that these closed-stream conformal supergravity modes prevented from having one. On the other hand, conformal gravity is kind of interesting although it's hard to imagine as physical intermination. String pairs have never had the chance to study it in string pairs because we usually generate physically-sensible things. Also, we don't usually have conformal That's another reason that physical strength theory can't usually generate conformance of the language. It might mean something, but I don't know what. Whether it means something that is worth letting or not, I think that the United Secure's version of physical strength theory doesn't give conformance of the language. The reason that the 90s has to do with the motivation I gave in the future of the production. Whatever else we want to do, one of the things we want to do, is to find a reasonable, strong curvature cousin of the usual, say, ABS-COT correspondence. And that should be a theory that doesn't have gravity. So, I would like to get rid of the gravity. If we're going to study it, one of the first questions we should ask is that, well, there's actually a little bit of a literature on this, I think quite slightly, one-fourth maybe. In conformal supergravity, there are one-loop anomalies. And it would appear that only certain gauge groups are allowed. One would like to see this. In Dworkowitz's approach, only certain gauge groups would appear to be allowed because of the matter system having seen this 28. But the restriction there, at first sight, doesn't appear to agree with restriction from the anomalies. So that's perhaps the most basic question to answer. Why would I have exploring the conformal subrography? But in addition to exploring it, one wants to know if you can get rid of it, and I've had only one real idea about getting rid of it, which hasn't worked. Let me remind you that although physical strength theory generates gravity, you can sometimes use physical strength theory to discover the behavior of non-gravenational systems of standard particle forces. So lots of results about dynamics of gauge theory, clock and phenomena, and all kinds of things, have come from string theory, even though seemingly string theory is changed

47:30 out of it. That's been achieved because there might be a limit of physical string theory, where gravity decouples from certain processes. That's usually done by an arrangement, so the gauge fields live on a brain while the gravity propagates in a higher dimension. And if the transverse dimension is infinite, if the extra dimension is infinite, you can effectively decouple gravity, and study a subsystem that can be treated just as a gauge theory, and it's actually been one of the most helpful ways to study quantum gauge theory in the last decade. So, the most naive version of this would be considered a V model, not of CP3-4, but of CP3-4 times C, for example. With a gain that would be localized at point So then we'd hope to get Yang-Mills on the brain and get better for gravity while they get propagated on the block. Technically, I haven't found a word from the back that actually works. But, hope springs at all. As I expressed, whether or not it's important There should be a theory related to the small g squared unlimited of the correspondence between quantum gravity and quantum infinity, the India's COT correspondence, the Mars-N correspondence. And, while there's certainly no guarantee, it would be nice if this small g squared limit is a twist or strain theory of some kind. Hopefully, we could have caused of the words we were talking about. Finally, this is Randall. We were listing things that did work from 0 up to 3. So this is the last one. Finally, I sometimes think that the best one is a point between arguments rather than something. So, in Geo0, this integral over the moduli space has motivated the formula to work. But there's a small sight in it. To motivate the right formula, well, after all, we have a holomorphic pressure which is the right thing so that

50:00 you can integrate it, in the case of the Baudrillard example, on a middle-dimensional a real cycle. In the super case, you can integrate over the fermions together with the middle dimensional real cycle. But to really do this as an integral, you need a cycle. And the most obvious middle dimensional cycle is to consider the real curves. So we take a real structure on CP3-4, we take the points in hand that are real, which is a real structure. Now, CP3-4 has a couple of infinite real structures. But here, one is used the most naive real structure where CP3-4 is the complexification of RP3-4. So, the real curves are just the where all our polynomials were real. So what I call the de-and-alpha, which is homogenous polynomials, and the homogenous point is u in the view of the de-and-alpha of everything that's other is real. Now, we do our integral in the real axis. Then we almost got the right answer. And Spratman explains exactly what we have to do to get the right answer to. But I'll just write it down in two dimensions. Really, it happens in many dimensions. But we're doing the integral of A-V version of what she says. We're integrating dx dy from minus infinity to infinity, and what we're integrating is the x-dimensional of ix times the polynomial f of y. Well, we all know how to do that integral. The x-integral is 2-5 times the delta function of f of y, and the y-integral gives us a solution to the equation f of y equals 0, and each one is weighted by 1 over the absolute value of df, dy. So that's that integral. I might tell you that I presented this iteration of the real axis as an odd axis, and that's what it is from pointing to the v-line. But in the Berkowitz model, it appears that that's what we should do, because the variables were all around. Now, there are some barriers to work with this ball, whether or not, and it would be on the answer. Now, as Pratt has explained, what you actually get here is a multidimensional version of what we just did. But, in any event, what you get by doing the integral this way

52:30 is almost what you want, but not quite. To get the right answer, you have to remove the and you have to use the sum of roots of one over f prime without the absolute value. And one over where you have to sum over all roots real or not. So there is a region of momentum space where the roots are all real, but there's a region where they're not all real. Whether they're real or not, you have to sum over all roots. Even when the roots are all real, the f prime do not cause a definite. So we're cheating slightly when we remove the absolute value. The twister string theory, as we have it now, would appear to generate the absolute value, at least with the method of performing a calculation that's motivated by claiming that the integral is an integral in real axis. But when I say that a twisterical formula that gives the amputees has been stuttered, it's been stuttered but not properly derived, because to get the have to remove the outside line. Now, what kind of interval gives us some over complex regions of the polynomial, where you want the way to intrude by one over f point? Actually, in one complex dimension, there will be a lot of ways to do it. But our problem is really in n-complex dimensions, where this problem has an analog that I'll mention in a moment. So, I'm going to skip the various ways to do it in one dimension. I'm just working on the way to do it for generalization and dimensions. What was discussed by me with the student Chris Beasley, where we were studying not these instantons in the B model of CQ3-4, but voltage instantons in the physical heterogeneity. So, I don't really want to explain it to you, but you can... The roots might be complex. So, a contour, we can't really do it as contours, but in one complex dimension, you could imagine a contour that surrounds all the roots. But in n-complex dimensions, a contour that surrounds the roots would have to be a dimension 2n-1, a big sphere around all the rings. So, you're not going to be able to do this by integrating over a middle dimensional cycle. Instead, we're going to integrate dy dy bar. But integrating dy dy bar is too much, so we have to add some fermions that cancel part of it. I call them sine time. What we integrate is Gaussian, the exponential of ys and yf. Then we have

55:00 to add a term involving the theorems. And it doesn't look as holomorphic, but it's got a little topological symmetry that's been written in side effects and low length of the other. And if you use that topological symmetry, you can first of all prove it's holomorphic. And secondly, you can evaluate it, and it shows that it gives this sign. Now, this generalizes the N-complex dimensions. Well, we don't have to be on CM because we a complex model. But for example, for Z here, with coordinates y1 up to ym, which is called a y vector. Then we have an equations f1 up to f1, which are called f vectors. And we also have a polynomial g of y in general. And you can write down a topological integral that computes the sum of all roots of the polynomial g evaluated at the alpha group, So, if you've heard Scrabble's actually, if you look at the table, you'll see the depth so on this is the right point. More generally, you can replace CN with a complex handball, N, and you can let V be a bundle of rank N, F is a section, so the equation f equals zero reduces locally to n equations that all the components of that should vanish. And now, something that, in the book of Griffiths and Harris, you can see described as a residue theorem in Albany's journey, but physicists might want to review about it in all things in terms of the integral, says that the sum of all roots of g evaluated in a loop divided divided by the determinant of EF is zero. That's what compact means. So, if it isn't compact, the sum over the roots might be non-zero. But then if you have a sensible way to compactify it, the sum over finite roots will be minus the sum at infinity. So, my philosophy, which I have to do, I haven't properly carried out, is that, basically, the planet groups correspond to the connected diagrams, and the roots of infinity correspond

57:30 to the disconnected diagrams. And the residue theorem saying that the sum over all roots is zero, should be a good way to interpret the relationship that was, for some example, between the connected and disconnected graphs. So I told you my philosophy, but, okay. It was presented in the context of explaining what was wrong. So what was wrong is that the present twist or string theory generates the absolute value with the closest thing we have to the derivation of the formula. So the trick is to modify twist or string theory the end of the duration, hand-worthy duration, to get rid of the opposite value. And the clue is that you're going to get some kind of integral over the moduli space that gives the formula without the opposite value, and I told you about what kind of integral gives that kind of value. So I would conjecture that we need to modify twist-restrictly to lead to an integral in the real sense, not just dm, but dm, over the moduli, with that Okay. I've completed what I wanted to say about what doesn't work. I don't have any idea if I did a ton of things. I'm not sure what I said. But I wanted to have a little fun again. At least I think it's fun. I can't promise you. But I want to apply in correspondence means phase 29 prescriptive, something that every physicist knows, which is that the Lagrangians morphine yards over the field quotients. And we'll see what we'll learn that way. So that's point B. And point B is we're going to compare check and de-bar homology, not in proving the equivalent, but in comparing what insights they need. As Roger Penrose explained in his lecture, Czech phonology will often lead to the simplest phonics. But there's a conjugate virtue of B-bar phonology, which is that it's sometimes more systematic and would give you some insights that you might not have otherwise, although maybe you're an item. So here we go. Now we're going to consider a massless scalar field in Rinkowski's base. We could use other fields, but I think I can illustrate this point here by just considering the scale.

1:00:00 And it wouldn't add much to be more general. So Penrose told us that a Maslow's scanner is a field . Now, in the Pythola, I thought the chat would put in the deep archipelago, H1 of Cp3 with values in O of minus 2. Now, having this correspondence between the solutions, we now want to ask the following question. Should Green's function apply corresponding to what? Here, by the way, we aren't necessary, but we don't really want an answer that involves an explosive formula, because we could be doing this in a field of an arbitrary nonlinear rabbit hole, in other words, conformally self-doinstein we could have turned on to Yang Mill's age one. So, the general context of this question, even though I'm not writing it, won't invite an explicit answer. It means a theoretical answer sometimes. And the question has been, it was answered 25 years ago. You can read about it in the T.S. book, John, G. of Yang Mill's fields, among other sources. I think that you might plan to give us to read how it was answered 25 years ago. But back in those days, the answers seemed frustrating to be abstract, at least to those of us who are immersed in the proper realms of algebraic geometry. So, how would we answer this question as physicists? Well, first of all, we'd read breadth and breadth by total by a D-block in other words, a zero-one point that's homogeneous of degree minus two, we're closer to the way we usually do physics than what we call it a check goes on. So let's do that. So by total, there's a zero-one point that's a homogeneous of degree minus two. So by D bar, it has this linear gauge of X, and it's applying to an element H1 of CP3, a region in CP3 that dies in O minus two. The reason it's a region, by the way, is that this equation should really be the conformal invariant scale equation with the scalar curvature. It doesn't have any global solutions in Rukowski's space, including the conformal classification of Rukowski's space. And the Penrose transformed maps after that, that if you literally take out of Cp3, there aren't any of these either. So, we've solved this equation in some region, which might be the finite part in the classical space.

1:02:30 But anyway, it's not the whole conformal chromatic region. And then, duly, we've solved this equation in some region in Ct. which might be the most irrelevant on this small piece. But, we know in physics that it's much better to start from the Lagrangian than to start from the field of atoms. The Einstein action was a better starting place than the Einstein equations, and so on and so forth. So for the scalar, instead of just winding down the equation of box phi equal to zero, we divide it from a Lagrangian into the integral of phi box phi. And in Twister's face, instead of just saying that we have a D-bar phonology class, we write down a Lagrangian, which we use our mother, and a Lagrangian is phi tilde wedge v bar phi tilde. You see, phi tilde was a 0, 1 form. So phi tilde's v bar phi tilde is a 0, 3 form. And therefore, to integrate it, we need a 3, 0 form of negative value to 0, 4. It has to be holomorphic so as not to spoil the gauge invariants. When you consider the gauge invariants, you've got other invariants. When you're integrated by parts and try to prove the independence, you need d bar omega to be solved. So, um, so omega has to be a homomorphic entity, so if you just try writing here a homomorphic volume for it, Here's a homomorphic 3 form, which I've written in this, I didn't know if you were out of that. While I wrote a Z times three VZs, I contracted it with an epsilon to make an SL4 invariant, which we wanted to achieve conformal invariants, since conformal transformations in spacetime correspond to SL4 transformations in crystal spaces. Well, it turned out to be homogeneous at degree 4. And therefore, for a Lagrangian to make sense, phi-zillita has to be homogene as a degree minus two. Of course, that's a standard result, but it's kind of fun to get it by trying to write a Lagrangian. You see, among all of India, Maslow's linear wave equations, the The scalar is the only one where you can write a Lagrangian with only one field.

1:05:00 The other Lagrangians contain two equal opposite molisters. So, it's because we have just one field, phi-tola, that phi-tola has to have degree minus two. If we were willing to use two fields, we would say phi-tola and phi-tola, Then 5s of them would have any homogeneity, let's say k, and the other one would have homogeneity minus 4 minus k. And then in space time, they would have multiplicity h and minus h for some value of h that, in fact, would turn out to be k. Anyway, this is certainly the result. You know if you're studying this history at all. But it's amusing that trying to write a Lagrangian tells you that the scaler must be a degree understood. But I wanted to show you something which is, I think, self-reviewed, which I'm telling you about because we need it to make sense of the disconnected guidance, and which I know frustrated me 25 years ago because it was hard to digest what was being said by the Twister exists about the property of it. Now, having written Borgangia, many physicists would say that there are five correspondences to Phi Tilda under some transformed plane from Pausse's basement, Twister's basement, that the analybe of the proprietor of Phi is the proprietor of Phi Tilda. So, the proprietor of Phi is one over the Laplacium, or one over with the Gaussian plus the scalar temperature, to make it conformally invariant. The other propagator is 1 over D-bar, but sets it to the space unit. You have to do 1 over D-bar in the space, the transverse to the age transmissions. And in fact, this will help. Now you can read what the algebraic genres are. Look at Mathias Wook, for example. And you can explain it in purely holomorphic entities. And it's true, as Roger said, that that will often lead to the simplest formulas. But if you want to motivate it as a physicist, it's good to understand that it's 1 over the e-bore, because that's the inverse of the KEDGOT, and having learned that much, then you can try to interpret it as a holomorphic entity, which will be useful for some calculations. So this gives us the propagator that's been used, regrettably and driving the M-H-V tree diagrams, in other words, and driving the prescription for the

1:07:30 annual scattering from disconnected curves in the twister space, as for a championship. So, if the degree is 2, the disconnected diagram looks like that. So there are two CP1s, each of which here has degree 1, so it's a line in the twister space, although I'm drawing it as a sphere. They're connected by one of the U-bar. And now I've drawn some points on each side. The points are meant to be points where blue ones are uncertain. Remember, we felt this determinant of the U-bar theory, or restricted to C, and we're supposed to expand that in perturbation theory. And if you think the good wave functions, and I don't but I think it's probably, it's not perhaps in the woods, it's probably one of those things. This expansion involves some distinguished spots that have some correlators involved before the field offers to those spots. So we get this kind of picture which we've been evaluated, a little bit non-reversely, and we get this nice MHB tree formula for the angles at the least, which has actually been visible as we've heard from some of those points. So this is what we get here, but we're going to have one last part of the fun, which is we're going to remove the neurons and take the zero point from them. So now we've got the two CP1s, and they're simply connected by 1 over Dbomb. So if we do this, well, each This CP1 corresponds to a point in the Kelsky space, in the penrose transfer. So we should be calculating, literally, the propagator of a scalar field in the Kelsky space. By the D-bar operator connecting two CP1s that don't have any of the marked points that we're here used to calculate scouting. When we remove those dots, we get to what Tina called H1 within here, and it's actually a cousin of our penrose. So, well, in combination with those fives. So, in Michalski space, we want to calculate the property of the five, the inverse of one portion between states which are delta functions of x or y. In going to twister space, x corresponds to a cp1 that I call e sub x, and y corresponds to the CP1 drive called e sub 1. And e sub x and e sub 1 have homogeneous coordinates,

1:10:00 so I'll call u and d, and u prime and d prime. And if you want to add a whole market measure, the best we can do is to make a one form that's homogeneous with degree 2. So we've got u' to u' and u' to u' and u' to u' and u' to u' and u' to u' and then we take, so we've got these two CP1s, but u and v are coordinates on one of them, u' and v' together. So you and me define a point in yes, which is inside of the CP3. Likewise U' and V' we find a point in the other one, which is inside the other CP3. So U, V, and U' and V' are points of CP3 among all of these. Then we take the inverse of D bar operator between those two points, and we integrate. And that gives the scalar probability. And as a check, let's see if you think it through, that 1 over d-bar is a 0,1 form of each value that's a degree minus 2, because phi tilde has those properties. And we multiply by the whole amount of measure, that's a 1,0 form of degree 2, we land on our feet, we get a 1,1 form of each value, and we integrate. And that's right. So we've got the scalar property here. And for extra credit, you can try to determine how it transforms under conformal transformations, but I want to talk to you right now. Thank you. I think it's puzzled me about the solution of the wave equation in the cost of space. You'd have to change the sign when you go across the field, and there are loads of solutions. Well, if there are, there would be loads of elements of H1 with dynas and O of minus 2. Yeah, but this is the real Necustin state. Because the elementary states will do that, yeah. It was a slip while I was talking, but you know, it was complexified. Okay, well, I agree with you. Anything that would give us, in the Tristory Transform, all the Tristory states, to make what I said correct.

1:12:30 We are trying to get a measure on the space of all curves, I'm just wondering if some Some of the difficulties might have to do with the fact that usually when you move a curve, of course, you have the moduli of the induced complex structure. It might be a good idea to try as a baby case to just see what happens if you just look at complete intersection curves and see if you can get a measure on those. Because there you wouldn't be trying to, you know. I can't say anything quite right. I thought about it a little bit. Okay. I'll say something differently. But I'm going to ask you something. You would need, okay, our curves are meant to be a bosonic connection. So they're defined by two bosonic at the fourth on the end of And to get to have a measurement, we need the degree of the bosonic relations to equal the degree of some of the degrees of the family relations. So, some cases work. For example, the degrees 3-1 of the bosons and all degrees 1 of the families worked. Some cases, by the same measure, wouldn't work. But the one I've just said is a fun case, because it corresponds to the plain people. I didn't consider it, although we actually heard it in a fixed resection, I guess, about new results, about entities that should come from the plane to the plane to the plane. So, it really would improve this picture a lot. It's nice to have formulas for the that are related to the breathing curves of all data, even though the genus is in. But we've enriched the picture a lot, it starts to see genus learning curves, who lies in the scattered processes. The closest we've heard is that we've heard about that triangle, which had a lot of neurons past year. A triangle is in a generation of a tiny cubicle.

1:15:00 And although it's not completely understood in this business, a lot of times when the amplitude is structured for a certain point of hope, The answer looks like it lives on a regeneration of it, but it's not a serious fashion, it doesn't have a proper expression. So when you see an amplitude on a triangle, that leads you to what they saw here. And some of it actually could have been starting from . I don't know, we've thought of looking at Kaley forms. You can certainly find curves of a given order in complex tree space. I mean, then you'll have some scale, but they won't be just an absolute measure. What is the equation? Oh, this is the equation on a line, in three dimensions. The equation on a line in a unique... It's really the child form, it's usually called the child form. It was really invented by David, though. It was invented by David, so it's called the child form, yes. Well... It's more usually called the child form, you said, don't you? Well, you'll have to tell me about... Yeah, well, it's... It's a polynomial equation on the line coordinates. And it has degree n where the curves of order n or degree n. But it's subject to a messy set of polynomial equations. It's not a generic... Well, if they were a complete intersection at least... Yes, that's right. But this is a general case. No. The twisted february isn't a complete intersection. But it's still a good case. The reason it's still a good twist is there. In that case, the space of twisted february can be primarily by just linear coordinates. So even if the curve wasn't a complete intersection, I imagine it's so much greater. Well, it's interesting. So, um, you said, if you said, you said, if the code function, you said, you still have a single polynomial, that's what I mean, for the, uh, for the K-D form, the child form, so it's, it's, it, maybe it's, it defines a hypersurface and a cross-monitor. That's right, yes. Because cross-monitor has the property. Are you claiming the modular space of codes in C-P-3, in general, is a hypersurface and a resonator? There's a way of assigning a single equation of Grasmanian to each curve, but the problem

1:17:30 is that it's not the case that every hyperservice in the Grasmanian comes from a curve. So when you perturb the Chow equation, it's not necessarily something that comes from a curve. That's the tricky point. Yeah, that's a mess. There's another class of Weiss curves that you could look at, and I wonder if it breaks the formal measurements in an interesting way. I don't know if it's helpful. But if you take a curve in P1 times P1, and then you lift it to its perpendicular cotangent model just by taking this sort of canonical lift by taking just the derivative of the Those exactly correspond to curves in P3, which are tangent to a contact structure. And that happens to be exactly the thing that we get from taking, say, imposing the round metric on the force here. Now, I don't know. You're trying to get rid of the conformal invariance of the theory. I don't know if restricting these kinds of Lejongrenian curves might lead to something or not, but perhaps it's what we're trying. So you think the contact structure, when you say contact structure, it's defined by a long form of P3, 2. Yes, that's right. So a curve, which is tangent, the tangent vector is always... They're always killed by that one form. And then curves, you've got roughly... They come from curves, and those are given by... Right, but it's still a trick. You have to introduce the bosonic variables to the right way to see if you can make this work. Well, one thing about what you're saying because that, I think, that contact structure actually makes performance, although we don't want to do that with most of the animals there. But if there's some way to get ice and gravity, it should be very important. Yes, that's the way to get ice and gravity. Okay, you're welcome. That's a good thing. Well, if I was to think about just a way into that, I'd try to tamper my instrument I was warned that it wasn't allowed to be a context-catcher. And then, if I found the way to do that,

1:20:00 it would be time for your business. Otherwise, it costs me this. Let's take a moment. Thank you.