 OK, so let me tell you about the Hore-Baffer construction. The starting point is a final complete intersection, X in F. The data for such a thing, lattice L, a rank R lattice, group homomorphism D from Z to the M to L, and C vectors, L1, LC, and L. Briefly, how you make a total complete intersection out this data, here L group characters of a torus, of a R dimensional torus to C star. So here, TR is the torus spec E of L, the group algebra of L of the complex numbers. And D can be interpreted duly as a bold L as a group homomorphism from TR to C star to the M and then encodes a representation with C star to the M. You can add diagonally on Cm. Yeah, I mean L. So R means L is a rank R lattice. And so then the torque variety is going to be a geometric quotient of Cm mod T. And as I said, I'll tell you later how to take this quotient. For today, just imagine that there's a way to take the quotient. And so Li's give Leiden bundles on F, especially because Li is, again, a character of T. So for example, you can interpret the sections of Li on F. Sorry, I use notation and call Li the corresponding Leiden bundle. Then those are just the polynomials in C x1 xm such that for every P in T, Ga equals Li of G F of A. And then x, vanishing locus of general sections of these lambas. So it is the variety of F1 and the section, the variety C, Fi, gamma, Fli are general. And we take this inside out. There are going to be some conditions on the data that ensure is a final orbital. So by the way, Don, I decided that you're right. And so I'm just going to call these things orbitals and not smooth orbitals. After all, I say manifold and I don't say smooth manifold. Is that the nature, the essential nature of your objection? No, OK. OK, I see. OK, so for me, OK. Well, we're happy for different reasons, but at least we're both happy. So I'm not going to spell these out these conditions because I don't know them, OK? This is an interesting problem in torque geometry, actually. So in fact, I'm just mentioning right here. Startling the orbital case, we don't know how to enumerate final orbital systematically by computer. We don't know enough about the matter to be able to do this. And in fact, we don't even know that in the manifold case, under the appropriate level of generality. We know a lot more about enumeration of complete intersection final manifolds in torque, in torque manifolds. We know a lot more about that problem. But we still don't know how to do it in the appropriate level of generality. Absolutely. It's easier. No, no, no, if it's a manifold, we know some good conditions. Uh-oh, I'm not giving you the conditions, but there are some algorithms and it's kind of clear what to do. Yes, yes, I will give some examples later in the course and the nature of the problem will become clearer, OK? But let me just at least say the final condition. Yeah, perhaps I should say at least, let me just put a little note here. So if I call di d of the ith basis element, OK, I will always at least assume this that if I look at the cone spanned by the di's, so by definition, this is the cone sum r plus di inside l tensor r. I assume that this is a strict cone. In other words, it contains no straight lines. And that ensures that the torque variety f is proper, projected in fact. And so we know by the adjun- so here's the key condition. So we know that by the adjunction formula that kx is the canonical of f plus the sum of the li restricted to x, OK? And so we will want that, you know, one condition is that minus kf minus sum of li. In other words, the minus k of f is the sum of the di's. Maybe I should call these lj's minus the sum of the lj's. I want this to be in the ample cone of f. I will talk about the ample cone of f later when I do torque varieties in a bit more detail. That's also the cone of stability conditions that we'll discuss. And so if that thing is ample of f, then it will follow that x is final, provided that it is an orbital. OK, so in order to do the Horivafa construction, we need a further key assumption. OK? So for every k in the set c, so I'm going to do square brats c for this set of integers 1 up to c, OK? There exists a subset sk of c. And I want sk intersection sl to be empty when k is different from l. And then I wanted my line bundle lk. Well, no, because I want the line bundle lk be the sum over i in sk of di. So indeed, if there are no lines, this discussion includes the case where there are no line bundles. And then in that case, there will be no s's. And we're just discussing the torque varieties of. Yeah, I know. Yeah, so no, because you see lk is just an element of l. Yeah, yeah, yeah, sorry, sorry, sorry, sorry, you're right. This is no, no, you're absolutely right. Sorry, sorry, sorry, sorry, sorry. For each k, these are subsets of m. Thank you very much. Very good. There is a tragedy in my written work done that there are misprints everywhere. And I just don't know how to fix this problem. You know, this is a strange condition and we'll discuss it a little bit. It is often believed that we know how to do mirror symmetry for completeness structures in torque variety. Nothing is further from the truth. We need this key assumption, all right? To write the mirror, let's choose a basis of l. Then your d is now a matrix, d, the ji. And you take y to be given by these equations. I take product, the dji, and this is from j equal one to r. So I have r equations here, equal one, sorry. And I take sum i in sk xi equal one. And so this is k in c, so that's c more equations. And I take this thing inside c to the m, in fact, c star to the m. And I take the function w from y to c to be defined by the sum of over i naught in the union of the sk of xi. And this is the Horivath mirror. Yes, the ds is not a basis, it's not a basis, it's just a set of vectors, 0, 1, yes. I will give you an example today of something that does not satisfy the extension, I'll ask some questions to people about it. So you will come back to this. Yes, so let me make here some comments maybe. So what would be, well I told you that this is the mirror, but what does that mean? So what's a mirror theorem, what should the mirror theorem say? So the mirror theorem should say the g hat of x equal the pi w, the period of this thing, y with the function w, let me put it like this, of t. Okay, under some conditions, and we don't know this, we know this in many cases, in some cases, for example, the case where we know it is when x is a manifold, sorry, f is a manifold, the torque right itself is smooth, and when all the line bundles are neff, well then in this case we do it's a theorem due to give and talk, but not in the desired generality. In fact, even when f is a manifold, it's not natural to assume that the l i's are neff, but that's maybe a wrinkle, more importantly, when f is an orbifold, when x is an orbifold, then really especially x is an orbifold. Okay, and so let me just say the reason that we don't know much in the case when, in fact, we know extremely little in what x is an orbifold is that we understand very little, there's many open problems in the Gromov-Witzen theory of orbifolds, so we don't know how to handle this g hat function. There are some conjectures and blah, blah. But we know very little, Andrea there is an expert on this problem. So let me, now let's discuss an example. So let's consider cubic surface in P3, so P3 is an example of the torque variety and then x3 is a torque completeness section. Given tau, it teaches us to do the following. So first we take the i function of x and this is just this power series here, some 3D factorial. Okay, maybe before I even look at this, so what's the data for this torque completeness section? Well, d will be a map from z3 to z, z being my l. Now, and it's just the element 3 of z, okay, the line battle of degree 3. And d is the matrix 1, 1, sorry, z4, we're in P3, 1, 1. Then using this data, given tau teaches us to do this, the i function is this thing here. So let me, so 3, because we're looking at a cubic surface. Not if we're looking at the quartic, the clubial, and there we take 4. So later in the course towards the end, if I have time, I may teach you some of these things. For now, I'm just saying where do these numbers come from? Well, the 3D is this 3 here, d to the 4 is because I have 4 ones. And so note that this thing here is 1 plus 6t plus ot squared, okay? And then given tau tells us that the g function, g x of t is e to the minus 6t times i. Sorry, let me just write instead of 6, I write c, some constant c. And then c equals 6 is uniquely determined such that to ensure that then g of t is 1 plus t squared plus, sorry, 1 plus bigger of t squared. So you know what the g function is. And then oribaf gives us, tells us, you know, follow the instructions written there. Yes, of course, because the graph written in variance are deformation in the, they don't, they are deformation in variance. I was supposed to take, well, you know, the variables for p3, I call them x0, x1, x2, and x3, and so I have x0, x1, x2, x3 equal 1. And then x1 plus x2 plus x3 equal 1. So here I take s to be the set 1, 2, and 3. And so these last three ones. And so that three there is the sum of these three ones. And w is the function x0 from y to c. So I want to briefly demonstrate that the period for this thing is indeed what you want it to be. And so that the thing works in this case. And so I prefer to reparameterize this thing as a torus. And the function w is a low-run polynomial. y as a torus, c star squared, and w a low-run polynomial. I do this, I write, well, see, I solved for x0. And so now I eliminate x0, so y is just x1 plus x2 plus x3. Yeah, so this y here lives in c star to the 4, as written. And then now I look at some sort of 3 equal 1 inside c star cube. And then on y, the function is 1 over x1, x2, x3. I'm going to reparameterize y by a torus, c star squared, with coordinate x and y by declaring that x1 equal 1 over 1 plus x plus y, x2 equal x over 1 plus x plus y and x3 equal y over 1 plus x plus y. And then I work with y equals c star squared and w equal 1 over x1, x2, x3, which then comes out as 1 plus x plus y cubed over x times y. Yes, it's a subset. Why is a subset? Absolutely, yes. Oh, that 1 plus x plus y is, yeah. Neither is technically a subset of the other, I suppose. These are all in C cross by definition. Yes, yes. So let me sort of picture this La Ramp polynomial. These are dots in Z2. Z2 is the space of monomials in x and y. And I put a cross here where I put the constant monomial. And so this is the Newton polygon of w. And now I'm going to write the coefficients of each of the dots. And there's a 6 there in the middle. And here you can see that the period of w defined as yesterday, integral 1 over 1 minus tw, integrating along omega, equals indeed the i function. And both satisfy this differential equation here. d squared minus 3d plus 1, 3d plus 2 equals 0. No, it is absolutely right. Sorry, right, right, right, right, right. I hat where I put an additional de-factorial here. And I think that does correspond to the g is w minus 6. Let me call it w prime. All you have to do is to shift it by 6. W prime for w minus 6 gives pi w prime of t equal g hat of t. And I don't want to write the differential equation for that because it isn't as nice. But it's no longer a hypergeometric differential equation. There is something here I don't know. It's a curious thing, OK? But in some sense, a shift by a constant is not a very important, it's not a very fundamental obstacle. And so what I'm saying is that the Horivafa procedure does not give me the perfect thing that I want. Yes, right, right. So this process here is, in fact, one could say a bit more, is the mirror map. And there is a mirror map both on this side here and on that side there somehow. And it's somehow the same mirror map. I don't know what that means. Let's forget for the moment Horivafa. I just wanted to give you a construction of mirror symmetry that we will elaborate on during the course. And so I want to finish this bit of contextualization here. So I want to tell you what we are about, what we are trying to do, we being myself and my collaborators, Al, and so on. I can't hear you. Doing the cubic. I'm not doing the quartet, I'm doing the cubic surface. But we're going to try to tell you, try to tell you. So we understand that mirror symmetry is about torque degenerations, OK? So that's something to take home from Gross-Zebert and lots of other people. So let me just give you a couple of definitions just to allow me to discuss this. A final polytope is a lattice polytope in some z to the n. And such that, we are going to ask that the origin of z to the n is strictly in the interior of p, sorry. It's a polytope in rn, OK? And being a lattice polytope means that the vertices of p are in zn. And two, I want that the vertices primitive in zn. To give the final polytope is the same thing as to give a toric fan of variety and not smooth. And so a final polytope p gives me a toric fan of variety, which I'm going to note by x sub p. And this is the toric variety with fan, a spanning fan. And so don't worry too much about this. If you don't know about toric varieties, this is something I'm going to tell you more about. Let's just think about the data set here. It's a lattice polytope. So in fact, these are n. I always think of it in the toric geometry setting as n tensor r. n is n. So then I'm going to say that a final orbital, x, is class Tg. And so Tg means toric generalization. There exists a qg degeneration, x sub p. So here qg stands for q Gorenstein degeneration. And that's a technical condition that means that the degeneration is nice. And I'm not sure I'm ever going to talk about it in this course. But there's a nice degeneration to a toric variety. And so the idea is that this is the correct class of things for which it makes sense to look for a mirror. And so mirror symmetry starts by choosing a polytope p. And then it talks about rates. The mirror symmetry works for deformations. If you like more precisely, perhaps, deformation families, x p, Laurent polynomials, certain Laurent polynomials, nu to f equal p. And so here are some of the questions that we want to address in the longer term for the longer term. So for example, given p, understand deformation x p to orbitals. And question two, given p1 and p2, two polytopes, when x p1 and x p2 have deformations in common? These are hard questions in algebraic geometry about which we don't know a lot. And then ideally, at the end of a long time, we'll have some result that says something like this. There is a one-to-one correspondence between QG families of class TG. In other words, the whole deformation family of Fano orbitals of class TG. And this is going to be in one-to-one correspondence with certain Fano polytopes, those for which x p admit a deformation to an orbifold. And then to identify those p that have deformations in common. And so let me just call this mutation. And then that, in fact, let me do another one-to-one correspondence between a certain Laurent polynomials, F with new to F equal p, also up to mutation. Well, I don't need to call the new to F equal p. Certain Laurent polynomials are not mutation. So this we hope to do in the very long term. We don't know that. So let me come to that in a moment, OK? In this course, we will tell you a combinatorial recipe. So this is in the paper called the Laurent inversion. Owl, casp chick. I don't know who all the authors are. There is Tom Coates. There is also Thomas Prince. And maybe some other people, I'm not sure. Owl. That's it. OK, very good. So this is on the archive. It's a combinatorial recipe. You start from a polytope. And then there will be some combinatorial games to be played on this polytope. And when they work, there is not going to always be possible to do it, they will produce both a deformation of xp as a complete intersection, as a toric complete intersection. And also, we'll give some Laurent polynomials that are the mirror, the Landau-Gissel mirror, off. The serious thing to do is to do this program inside the gross zebras machine, which we're working on. But I'm not talking about that. Instead, I will just tell you this particular combinatorial recipe that in practice works in a lot of cases, sometimes in the majority of cases. And it allows to make long lists of final GPS. So that's what we're going to try to tell you. To conclude, I just want to give you an example. So consider a toric complete intersection, x66 and p, weight a projected space. This is really simple, 22333. This is a very nice orbital. It's a surface. It has four singularities, four quotient singularities that form a third 1, 1. It's final. Minus k of x is 0, 1. And it has degree k squared equal a quarter, a third. 6 times 6, yeah, a third. Very nice orbital. The i function of x, if you follow your nose, you follow given tau, you will write sum 6d factorial squared over 2d factorial squared, and then 3d factorial t to the d. OK, so this happens to be 1 plus 600t plus o t squared. So if you want the g, then you're supposed to take e to the minus 600t times i. Then anyway, it's a harmless operation. And this indeed has to do the orbit fold Gram-of-Witton theory of x, though in a slightly subtle way and you want to ask an explanation, you can ask Andrea, you can tell you about that. And this does not satisfy the condition of Horibafa. So we cannot make a Horibafa mirror of this thing. And perhaps I should try to wrap this up. We cannot make a Horibafa mirror of this. We know that the mirror exists. g hat, perhaps you should also say one thing, it does not have a toric degeneration. It's a theorem. It's obvious because minus k is empty. g hat, i hat, if you like, are of geometric origin because they satisfy a hypergeometric differential equation. And we know by Nick Katz that all hypergeometric equations are of geometric origin. And therefore, a mirror exists somewhere, even in the low level sense, even though I cannot write down one explicitly for you. Yes, but as we learned from the example of the cubic, that type of shift just means that if I find one for i hat, I just shift it by a constant. Indeed, it will not be Laurent Poirot. I agree. I agree. Well, I don't know because if I know. Yeah, so perhaps it's a function. I certainly can subtract a constant. Perhaps I'm not 100%. Fine, fine, fine. I take the point, it's OK. I want to, I don't know. This is a question to everybody to come and discuss this matter with me if you want. Stop here.