 So, what I'm going to talk about is need some motivation, which goes back 30 years exactly. And just before the Knisnikz homologic of paper, actually. So, in those days, I was working with Jean-Louis Gervais at the Ecole Normale, and we were trying to use Polyacos approach to string quantization to really compute four-point functions, at least, in more reasonable dimensions than 26 or 10, using L'Hubert's theory. So in the L'Hubert's theory, which you are supposed to use, oops, there is one at least. So in the L'Hubert's theory, which is supposed to be used for strings, then the central charge, okay, thanks. So the central charge, C, is 1 plus 6q squared, where q is b plus 1 over b in what is now the standard normalization of this stuff. And we were interested in trying to do things in, for C, in those days, less than 1, less than 25 to be in some more physical dimension. Now, which means that b is unimodular, actually, it's not real in those cases. So we are explicitly very far away from the minimal models, where b squared is rational. And I must say that I have kept away from minimal models because I just, impossible to compete with the Russian crowd, working on that subject, even without the excuse of my involvement in the French administration. So what were we trying to do in those days? Well, we're saying, well, okay, we have L'Hubert's theory, and we know that if we want to compute a string amplitude decorating the orbital modes with the L'Hubert modes, then we have to compute some expectation value of conformal operators in L'Hubert's theory. So here is the starting point that we had in those days. You recognize some of the diagrams that Sasha Zamolochikov was drawing this morning, so we have here conformal. So this is the string or L'Hubert four point function that we would like to compute. And then here there are alphas in the standard way. And here, let me call this P2, L'Hubert momenta, P3, P4, something like this. I leave this line floating more or less for the moment. And so we would like to compute this for values of the alphas or the momenta, duville momenta such that we get some reasonable decoration for the orbital modes of the string. Now, so the dimensions of the external of these operators, delta j where j would be 1, 2, 3, 4 in a second. So this is alpha j, q minus alpha j, and that's also q squared over 4. Here there is a sign difference with what Sasha was using this morning. Minus, sorry, minus p squared, pj squared. Okay, so that's pj is alpha j minus q over 2. Okay, so this is what we were trying to do in those days. Now, for the, in principle, useful values of the pj is a first string theory. Then these conformal operators in duville theory don't satisfy any particular differential equation. And so what we said, Jean-Lou and I was to try this thing to add an extra leg here, a fifth leg. And consider then you recognize immediately something that Sasha was writing down this morning. So here I would have alpha 1 and duville momentum p1. And here the emission of the vertex for the emission of one of those null states. So what we are really trying to do in those days was compute this. So you take conformal operators in duville, minus b over 2. So that's the one corresponding to this, taken at point z1, z2, v alpha 2 of z2. So I've already chosen a particular Mibius frame where of the four of the five points where my, where my conformal operators sit, one is at zero, one is at one. And the other, another one at infinity. And v alpha 4 of infinity, and we were trying to compute this. And the idea was that this satisfies the second order differential equation in z1 and z2 because this corresponds to, there is degenerates, there is a null state for that value of the external momentum for this vertex. Okay, now, so because of that, then I said this differential, this differential, actually I must say yes. Listening to Sasha Jamalov-Shikov this morning, when he was writing something very close to this, I was very much afraid that he was going to describe exactly what I'm going to say. So, so fortunately, the structure of these things is rich enough that there is still some place for myself. So, yes, so this, I have to write some relatively tedious formula. It is known that this can be, this four point function can be, satisfies the differential equation for the following quantity. You don't have to really, to really follow exactly all this. I take out some relatively trivial factors, 1 minus z2 to minus 2, alpha 2, alpha 3, 1 minus z1, z2, or b, alpha 3, g of z1, z2. Where g, this function g of z1, z2, satisfies that special second order differential equation, coming from the fact that I have here in our state, which is actually, essentially, the Knies-Niesh-Jamalov-Shikov equation, except that it is written for u-veld. It's the same idea, we use the fact that there is another state to obtain a differential equation for this. And then, so I should say, z1 and z2 are the parameters, the integration, whatever the, I should say, the cross ratios corresponding to these legs in this dual diagram representation. And what we are after in principle is, on this g of z1, z2, is to take the limit z1 equal to 0 to factor out this extra piece, and obtain, ultimately, the function, the four point function is a function of z2 only, a conformal block as a function of z2. Now, when z2 is 0, actually, then we are sitting precisely on one of the poles here, and we know that we have a hyper geometric equation as a function of z1. But, and then we can try to solve in powers of z2, and we can solve recursively, which we did in those days. But this is not very helpful, because what we really want is z1 equal to 0 as a complete function, and a complete function of z2. So ultimately, well, let me actually write down the differential equation for this g. So it's the following form, z1, 1 minus z1, 1 minus z1, z2, dz1 squared g plus, then there are some coefficients, bz1 plus cz1z2 plus dz1 squared z2 times dz1g. Then there is a z2 derivative, d times z1 squared. This is d, yes, this is a second order, sorry. The square is up there, yes, sorry. bz2, 1 minus z2, dz2, g plus e plus fz2 plus hz1z2, g equals 0. Now, and a, b, c, d, these a, b, c, d, e, f, h are functions of the external deltas or the external momenta and of the internal momentum here, p. And delta here, this delta is the delta of the internal momentum. This is really essentially what Sasha was already introducing this morning. Delta is just alpha, q minus alpha is also q squared over 4 minus p squared. And so when you actually work this thing out, which is not very complicated, then what you find is that this z2, the derivative with respect to z2 when it's there, which is the classical limit, if you wish, when b goes to 0, then this is the Heine equation, as is well known. And the accessory parameter of the Heine equation is related to the internal momentum. OK, now, so what can we say about this? Well, when z2 goes to 0, yes, one thing that we want to do, I have to say that the boundary condition is that g of z1, z2 is of the form g0 of z1 plus z2, g1 of z1 plus et cetera, when z1 goes to 0, with all these gis regular at z1 equals 0. So when z2 is 0, then this is the hypergeometric equation, which, again, was used by Sasha and Alyosha Zamolychikov in 1994 to compute the structure functions of the conformal operators in UVIL theory. So but we really want it for z2, for completely z2, different from 0. Now, OK, now, so the idea is then, suppose that we have one solution of, we happen to lay our hands on one particular solution of this equation for values of the conformal charge and the internal momentum here. And then, what else can we find other solutions like this? See, the aim at the end would have been, of course, to compute a four-point function interesting for a string theory. But that is not the case, at least not yet. But at least obtain some explicit formulas where we could have some explicit formula for a function of these external alphas and of the internal momentum in which we could check explicitly the conformal bootstrap by integrating over p. And because in 1994, Sasha and Alyosha showed the conformal bootstrap by numerical integration over p, and we would like, perhaps, it would be amusing to find cases where we could do it analytically. And so this is actually the case. OK, so then the basic formula for, yes, one thing I want to say first is the following, that. So I say, we are really interested in the case where g is regular when g1 goes to 0. But however, just like this equation has many properties which are quite close, quite similar to those of the hypergeometric equation. Especially, you can find that, let me maybe write down a formula for the hypergeometric equation, which some people will probably recognize, a, b, c, z. The standard hypergeometric equation satisfies the following identity. So that you see here that there is a relationship thanks to this extra power of 1 minus z between the behavior of the hypergeometric at z equals 0 and at z equals 1. So here, something similar occurs. And it's interesting. And you see that when you do this change, then the coefficients of the hypergeometric function are changed a little bit in this way. And here, this is the same. Actually, you can, what? 1 minus z, sorry. What? No, no, no, I'm sorry. It is still z. Well, look. I think this is still true. But this is a little bit, I should say. Maybe I should write it this way. Because this is regular. No, no, this is exactly true, I think. Yeah? I think this is correct. OK, so this, you can do the same. You suppose, forget about the boundary condition on g and then say, well, suppose I have a solution of this and then try another solution, g1, whatever, which would be z1 to some power times g. And try to find a power here of z1 such that g1 satisfies the same equation but with different values of the coefficients a, b, c, et cetera. And see what this means in terms of the external dimensions which are here. Well, what you find is that indeed there is this possibility that there is a power here such that g1 satisfies the same type of equation. And when you do that, you find that this amounts to change p into minus p, p1 into minus p1. So this is a very simple interpretation in terms of the external momentum, p1. And then you can find another g1, which would be like 1 minus z1 times g, again to some power. And this turns out to change p2 into minus p2. And then there is another one with 1 minus z1, z2 to some power again. And this changes p3 into minus p3. And there is no corresponding formula for p4 because actually in the equation only delta4 appears. So that p4 appears only squared. OK, so if I am in some kind of four-dimensional space for all these momenta, p1, p2, p3, p4, then I see that I have symmetries somehow by reflection with respect to one of the coordinate hyperplanes. And I can use this if I have a solution, if I know something in one case for this differential equation in order to find new solutions. And OK, so that's one part of what I was going to say. What I want to say, oops, it's a bit dark. OK, next. Well, next is actually coming from one of Bittman's formula. Suppose that I have, again, suppose that I have a g of z1, z2, which satisfies the equation. Now let me try this, this integral transformation. Let me call this one u. du over z1 minus u with some power x, x to be determined. And let me call this g sub integral, i of z1, z2. Equals this. Then I say, well, can I find an x such that if g satisfies an equation of that type, g i satisfies the same type of equation. I don't leave things here. Either it's a contour integral or it's a contour which ends at points where on the right-hand side there is no integer power or negative integer power. So that I can throw away integration by parts terms. So I put this in the equation. And then actually, when you do this in the hypergeometric equation, you find that for any x, you get a new hypergeometric equation with different coefficients, of course. But in the case of the Hohens equation, I may call it so, even including the z2 piece, the dz2 piece, then you find that x has to have one of two values only, which is amusing. And these two values have actually a very simple expression in terms of the momenta p. x is 1 minus b squared over 2 minus b times p1 plus p2 plus p3 plus or minus p4. And so plus or minus p4, because obviously, because again, as I said, it's only p4 squared, which appears in the coefficients of the equation. So that's one thing. So x has to satisfy this. And gi, so of z1, of z1, z2, then will depend on pj prime, pj equals 1, 2, 3, 4, which are transformed from the p's. Just like in the hypergeometric equation, when I do the same thing, then the new coefficients depend also not only on the original coefficients of the hypergeometric, but also on x. So p prime j. And what's amusing is that. So the vector p1, say, p prime 1, p prime 2, p prime 3, p prime 4 is equal to some matrix, which is minus 1 half, 1 half, 1 half, 1 half, 1 half, minus 1 half, 1 half, 1 half, minus 1 half, times p1, p2, p3, p4. Ah, sorry. Missing and very important thing. Plus b over 4, times 1, 1, 1, 1. Is this an SOA trilogy transformation? Pardon? This is SOA trilogy transformation. Well, that's not my interpretation immediately. But what this is, m, this matrix m, is the reflection with respect to the vector, to this vector. A perpendicular, in the hyperplane, perpendicular to this vector, m is reflection, reflection with respect to the hyperplane, perpendicular to 1, 1, 1, 1. So I start with an initial value of the p's, then reflection with respect to the hyperplane, and then translation perpendicular, well, perpendicular to the hyperplane, parallel to this vector. P minus p. Pardon? Well, that's a version of the H is different. No, no, it's different. Minus. Right, that's a version. Well, let's say not, well, p into minus p. But here, this is not really p into minus p. It's a little bit different. I'm going to add that to q minus r. No. No, I don't think so. See, here, I'm not changing p into my p's into minus p's. p into minus p would be a matrix of the type minus 1, 1, 1, 1, for example. This would be changed p1 into minus p1. Reflection to the one of the coordinate hyperplanes. Here, it's reflection with respect to another hyperplane. So this is what we find. So you can repeat this provided that, well, you don't hit cases where you couldn't do a reasonable, if I may say so, choice of integration boundaries or contours or whatever. And if you repeat this, then you find that you can go on a lattice with spacing b over 4 or b or b over 4, more or less. And so, for example, if you start from an arbitrary p1, p2, p3, p4, and then apply the following string of transformation. First, i, this integral transformation that I wrote here, then reflection with respect to some of the coordinate hyperplanes. And then, so it's successively r4, r3, r2. So this means changing p into minus p, p, yes, which is p4 into minus p4, p3 into minus p3, p2 into minus p2. Then, again, i, then r4, r3, again, this is the same set of transformations. Again, i, and then finally, r1. Then what you find is that this, for arbitrary values of the p's, then this is changing to p1 plus b, p2, p3, p4. So this means that starting from an arbitrary, from some set of p's, then I can navigate in the lattice. This is a fine value group of SOA. Pardon? What you generated is a fine value of SOA. A fine value, OK. SOA. So p goes to minus p, an exchange of. An exchange of. The finite value group for finite SOA, g4, and this is the. OK. So obviously, so that means that you can go on a lattice with, here you could obviously have p1 plus n1b, p2 plus n2b, p3 plus n3b, et cetera, and a fine shift. And then 1 and 2 and 3 and 4 can be negative or positive, as you wish, because you can invert this animal. OK, so that's one thing. That's one of the things that you can do. This is not the only one. I mentioned perhaps something even more interesting at the end. OK, now the problem is, what are we going to apply these set of transformations to? What will be the starting point for this p1, p2, p3, p4? And OK, maybe I can continue here. Now this then, the starting point, I think the best, whoops. For the starting point, let me use what appears in the paper that we wrote four years ago with Volodya Fateyev, Alyosha Lilvinov, myself, and Enrico Onofri. And now, which uses some manipulation of the starting point, which is the Hohens equation, plus that dZ2 term. Now as you know, the Hohens equation can be changed into elliptic variables. And they include elliptic functions. And I particularly love elliptic functions, because more than 40 years ago when I was in Princeton, John Schwartz remembers. With Joel Sherk, we had a lot of fun playing with elliptic functions, which appear in loop diagrams of the string. So I have a love affair with elliptic functions. And I said, well, let's try using a little bit more elliptic functions to see what happens to this equation when we go to elliptic variables. Now, so make the following changes. And this is really quite remarkable. In my opinion, what happens when you go to elliptic functions? So define nu k is the complete elliptic integral. I'm going to say in a minute what it is. du over root of u1 minus u1 minus uZ2, complete elliptic integral here, where k is 0 to pi over 2. 1 minus k squared sine squared t to the minus 1 half dt. And z2 is, which is k squared, this is the elliptic modulus k, theta 2. That's where the theta functions come in, of 0 and tau over theta 3, 4 of 0 and tau. So you make this strange change of variables, OK? And then, furthermore, define a g. The original g of z1, z2, call it g, make this extra change of function of then of nu and tau times this capital G of nu and tau. And then you find that this capital G satisfies a relatively simple equation. This is actually known to, you see, the fact that there is this dZ2 is going to turn this extra derivative with respect to Z2 is going to come out rather nicely. You find that G satisfies the following equation, d squared g d nu squared. So this is what remains of the dZ1 squared and the dZ1, more or less, if you wish. And then there is the dZ2 piece. And Z2 is a function of tau, of the elliptic tau, plus 4i pi b squared dG dZ2 d tau, sorry, dG d tau. So this looks like a time-dependent Schrodinger equation. And then there is 16 k squared. So k squared is a function of tau. Then e plus fZ2 plus h minus b squared q central charge. e, the other elliptic integral. And let me write this in a special case, in 4 and 4 plus 1 over 4 dN squared. And now I should say that this is the case, alpha 1 equals alpha 2 equals alpha 3 equals minus q over 4 minus 1 over 4b. I have chosen this particular values of the external dimensions. You'll see why in a second. And alpha 4 is, I choose it to be alpha 4 minus q over 2. Let me say it, minus 1 over 4b minus n4 over 2b. And so e, f, and h, as again, as I said, are only functions of the alphas. So this is a function of Z2 only. And I have here, this dN is the elliptic function, the ratio of the third elliptic functions. dN, well, the standard thing. Yes? Oh, yeah, yeah, sorry, sorry, sorry, sorry. There is G here. So this is amusing, because here what we find is when b0 here, when I don't include the tau derivative, then this is for integer d, for integer n4, this is a traibich-verdier potential, a particular one. Lame is the lame equation, yes. Lame equation, and this particular case of a traibich-verdier potential, and we'll see a generalization of that. So amusing by itself. And so what's even more amusing is that we can also, so when there is no tau derivative, then the solution of this is known for arbitrary values of the internal momentum p, or of the supplementary parameter of the initial drawing equation. So this is known. But what's amazing is that there is a solution. So you see, this is very strange, because this is like a Schrodinger equation in this elliptic potential, but time-dependent. But the time is related to the elliptic potential itself. So this is quite strange. And it turns out that this has an explicit, in this case, these particular values of the external dimensions. This has an explicit solution with an explicit integral representation, g of nu and tau, which is exponential 2i Bp Bp pi nu times an integral d nu 1, rho of nu 1 and tau, theta 4 of nu plus nu 1 over theta 4 of nu. And where rho of nu 1 and tau is some relatively strange looking animal, but which is found by explicit calculation, exponential 2i pi B over P nu 1 times quite a few other terms, which depends only on tau, which I don't really want to make explicit. This is just a various theta functions again, et cetera, et cetera, but which depend only on tau. Now here, this is nu 1, not very well written. So we have an explicit representation of the solution in terms of elliptic functions and for all values of the internal momentum. And amusing remark is that when B goes to 0, which is the classical limit, B times P finite, then the solution, you can do steepest descent on this. And then you find that nu 1 is fixed to be 2i pi B times P plus theta prime 1 of nu 1 over theta 1 of nu 1 equals 0. Because you see here you have a 1 over B square and here you have a 1 over B. So when you do steepest descent, you find that this is the solution. And the solution is exactly the solution of the initial La Me equation, which was found by Hermit 150 years ago or something. OK, so we have here an extension, including this tau derivative and an explicit integral representation. So what else? So in that paper with Volodya and Alyosha Livinov and Enrico Oronofri, then we have other cases where we could find, indeed, explicit integral representations. So are there explicit solutions? Or I should have said yes. So sorry, this is in the case n 4 equal to 1. So still, again, so if I take these alpha 1 minus q alpha 1 minus q over 2 is equal to minus 1 over 4B, in general, minus n 1 over 2B. And same for alpha 2, alpha 3, and alpha 4. So this equation here was in the case n 1 equals n 2 equals n 3 equals 0 and 4 equals 1. Then you see there is an integral representation with one variable, new one, one integration variable, which I should have called it new 4, actually. And we have explicit integral representations when these n 4 arbitrary belonging to n, say, an arbitrary integer. And then you can also have two of them equal to 0 and then n 3 equals n 4 in n. And finally, n 1 equals n 2 equals n 3 equals n 4 in n in arbitrary integers. And actually, you get these other cases from the first one by using duplication formulas for elliptic functions. So these are relatively trivial extensions of this case. Now, the amusing thing also is that we have that. So these are the usual, these are some of the trebuchet potentials, where you can find an extension with the tau derivative here. Now, what's amazing is that we also found more such values of the n's, where these could be solved with an explicit integral representation. And so you see, what does this mean? Minus n 1 over 2b. And what we actually find is that you can also add minus mb and 1b like this. So this is a large potential, if I may say so. And this is a small potential, 1b is small. But we have an integral representation for, say, going back to n 4. n 4 and n 4, n 4 and n 4 belonging to n. And here, similarly, we could have m 1 equals m 2 equals 0. And m 3 equals m 4 belonging to n. And here, all of them also belonging to n. So we have explicit integral representations for a relatively large number of values of these n's, which is, of course, these cases where m is non-zero disappear in the classical limit. And we're not immediately interested in them in the classical limit. And we have all these. And then we can, to all these particular cases, we can apply the integral transformation that I mentioned in the first part of the talk and get even more explicit representations of these conformal blocks when z 1 goes to 0, and, ultimately, when z 1 goes to 0, we get a conformal block for the Nouvelle theory, a four-point function. We can get even more such explicit integral representations. We can't yet go everywhere because we are restricted on this lattice. So the conjecture that we had in the paper was that actually we could get more integral representations involving, well, away from these particular cases where two of the n's or four of them are equal, we can go in principle. We hope to get everywhere for all values of n and all values of m, which is not the case yet. Even though all values of m, we can already reach more or less thanks to the integral representation I was mentioning. And let me finish. I'm not yet almost there. Let me finish with one amusing remark, apparently, is that this integral representation that I wrote down can be used to reach values of the n's, which are, for all values, for example, apparently, apparently, I haven't checked all the details. But that reflection formula with respect to that hyperplane, perpendicular to the 1111 axis, can be used to reach values of the n's here, which are all of them half integral, where there is not supposed to be known a solution of even the case where tau doesn't appear, except that there is a strange remark in a paper by Gaston d'Arboud dating back to 1882, who says that he has found a solution indeed of the, well, not exactly the Lamy equation, but the case where there are all n's and they're all half integer. And so maybe that's what he had found in those days, but he hasn't published the details. And apparently, it's not in the literature yet. So maybe like Fermat, maybe like Fermat is somewhere in a footnote of one of his notes. So what? This seems to be not. What? Conjection was correct. Yes, but it's not clear that his proof was correct. Not correct? Well, we'll have to discuss that in private. Thank you very much for your attention. Questions? Do you have an algebraic interpretation of the equation you wrote? Sorry? Do you have an algebraic interpretation in terms of the answer? Of this equation in terms of your answer? It looks like a sugar rock construction because of tau. It looks like the Bernhard equation. Yes. Well, certainly this is the same as the original one. No, because this one is on the torus. Yes, this is on the torus now. This is on the torus, yes. Sugar rock construction on the torus. So L0 is a... It's a change of variables which looks like they are going from you will tau as L tau as the mean. Yeah, but then you move from the new vectors at level 2 to the formula for L0. Well, initially I was... I'm on a sphere, I'm on the sphere here, and here I'm on a torus. There are some function of the sphere which are related to one-point functions. Yes, yes, that's what it is in this case, actually. This is the one-point function of the torus. Yes? It's a transform. It's a transform, yes. So this 5-point kaffron block on the sphere is called the 4-point kaffron block of vessel 2. On the sphere. But then it's probably for special... It can probably be related to one-point functions. But then, if it's fully of the one-point functions of the torus, you should have an explanation why you have an integral representation. You should be able to write all possible integral representations. So one-point function of some vertex operator on the torus, so you can compute that. Well, it depends on the vertex operator. Then it will tell you when you have an answer. When you can do it and when you cannot do it, yes. Well, probably Verodia would... If he doesn't speak with his neighbor, he would answer. But, Sasha? I have a very technical question. Probably I was confused with the first part of you. You demonstrated us this transformation which ended up with the shifting of p1 by B. So my understanding is that you can shift every one of this piece by positive... Or negative. Or negative because you can invert the transformation. I mean, each... Well, because each of these operators are reflections in this four-dimensional space and which have an inverse. This is really... The transformation which you have shown us was the... I don't know how we invert it. It applied twice. It was a reflection, so... If you applied twice, it should be the identity. Right. You should apply it to the other side. You could get... Oh, you are... You are separately just changing p to minus p by this elementary operation. So that's why you say letters. So really, this shift of p generates not the... The integer belongs to N, but the integer belongs to Z. Yes, that's good. Yeah, all right. That's N. N means... What? N means non-negative numbers. N is non-negative. Well, N is to 1 minus N. Here, N into 1 minus N is just... So my understanding is that this additional piece adds another integer which can be now positive or negative. Yes. Yes, yes, yes. Other questions?