 Okay, so early on in the first few lectures, they see all sort of emphasize that our point of view is practical, okay. And for us, maybe we have this ambition, maybe of classifying final manifolds from this viewpoint, okay. And so what we wanna be able to do is we wanna be able to start with the simple side of the picture. So we wanna start with a phano polytope because that certainly seems a lot easier than to go away and try and find phano manifolds. And we've seen in this class that, you know, if you give me a phano polytope, then there are various things I can do to it, like I can mutate it. And that's just the QG deformation from, well, from one type of variety XP to another type of variety, XQ, okay. So that's one of the things we can do. And you know, this seems very successful. It preserves the period sequence and things like that. So yeah, I mean, this seems like a good thing to be doing. And it's all very good. But how on earth do we move from having a polytope to having a phano albifold, okay. So I guess the crucial question is, how do we get P or maybe XP to a phano albifold X? Just as crucial, how do we find the marrow F? And in general, we don't know. But fortunately, there's a large collection of cases where we can certainly do this. And what we do is we do the Hori-Vaffa construction backwards, okay. So one way we can do this, or so sometimes we can do this. I eat, we can find an X and an F via Laurent inversion. And so yesterday I gave you a whole series of examples that we're trying to illustrate this method. So yesterday's examples, I was starting with a polytope P and I was trying to find some way of assigning combinatorial data to P that allows me to sort of back build the Hori-Vaffa construction to recover the phano albifold X and the marrow F, okay. And when it works, it works really well. But making a sort of precise statement about when it works is not something that we really know how to do. So from the practical point of view, this is the best tool that we've got. Frankly, if it doesn't work, then you're in a world of pain, you have to really sort of struggle to do this. And fortunately, the vast majority of phano manifolds can be constructed in this way. So of the 105 phano refolds of Mario Mackay, I think it's at least 78 of them. Oh no, I don't have it written down. So I think at least 78 of them can be built as a tarot complete intersection. We're using techniques like this, okay. And there's the case of albifold delpetzels with a third one one singularities. In that case, we have 26 of these mutation equivalence class of polytopes and most of them on inversion are scaffolding. Like I said, when it doesn't, you've got a whole different problem, but it does seem to be the vast majority of cases. So it's not a sort of technique to be ignored. All right, so let me try and make precise what I think a scaffolding is. Definition. So characterise when, I don't know. Do you have a, but that's really just an obstruction to be in a TG. Yeah, we don't know, basically. Okay, so there is one issue, I'll explain it, why not. So in this case, it turns out there are actually 29 delpetzels with a third one one singularities, which is not equal to 26, okay. And so what's going on here? Well, these three additional cases have H naught empty. Okay, so this means they can't be of class TG. They can't admit a taric degeneration. And the ratio's point is so far, this is really the only obstruction that we know of. No, it's not actually, it's just nobody knows a model for which it works. That's all, yeah. Yeah, I know, I think Dom was asking whether they have H naught zero, which is not the case. Yeah, we just don't know if they have a complete intersection, description, definition. They find this process of Laurent inversion, a way of at least conjecturally moving from the polytope to X and the mirror F, okay. So I'm just, everything's happening in a lattice N and I'm gonna fix a basis for N for all time. So let's just fix a basis, B. Fix a partition of this basis for all time. So it's a partition S1, SC, and U of B. And I want these SIs non-empty. So I'm just gonna define SI to just be one of the SI circles and then I'm just gonna throw in the origin. I'm gonna define the simplex delta I to just be the convex hole of these points. So these delta I, they're basically just a collection of standard simplices. But then a strut, B, is just given by, so I'm gonna take each of these simplices in turn to allow myself to translate the simplex by one of its vertices, okay. I remember that the origin is one of the vertices and then I'm allowed to dilate it by some non-negative amount. And I'm gonna then take the Minkowski sum of this for every vertex and then the Minkowski sum of that for every strut, for every simplex. So as I try and move one of these simplices forever away from the origin, it necessarily becomes bigger and bigger and bigger. And then also, once I've done that, I'm gonna allow myself to translate the whole thing by this collection U of basis elements that we didn't use to construct simplices. So the U, thinking back to yesterday, are the unaliminated variables. And so here we need all the coefficients to be non-negative. So now let's award ourselves a final polytope inside that lattice. So let P be final polytope. Then a scaffolding on P with respect to this basis and this choice of partition is a collection of struts whose convex hull and the convex hull of the illuminated variables gives us P. No, I'm not, because this is just a point. So it's, yeah, these B's, they're also just points. So it's, yeah, sure, but so this is just a point. So all I'm doing is translating. So it's defined. Yeah, yeah, let me say that again in case some people didn't catch it. So this delta is indeed a simplex, okay? Then so we're just subtracting a point of it. So that's just the translation of this impact. That's no problem. Then we're dilating it. Then we're taking a bunch of Minkowski sums. And then finally, this is just a sum of a collection of points, so we can just translate again. That's defined by that. And then finally a scaffolding is just a collection of struts and the unaliminated variables that fill out P. So very much like we were doing in the examples yesterday. Let me just, using this notation, just do an example. So I'm gonna start with P that looks like this. I'm gonna describe a scaffolding on it. So the origin is sitting here and this whole thing is in N, R. And for those of you who've been keeping track, this is one of those third one, one polytopes. So the first thing is I need to fix a basis. So in this case, I'm gonna fix B to just B, zero one and one one. And now I need to choose my partition. Well, I'm just gonna have S naught to be zero one and I'm gonna have my unaliminated variables to be one one. So delta is just the convex hole of the origin and S naught. So delta is just a little line segment, a vertical line, unit line. And now I'm gonna pick some struts. So I'm gonna define two struts in this language but from the examples yesterday, I think you might know what I'm gonna do. So my first strut in this language is I'm gonna take three lots of delta and then I'm gonna subtract off one one. And then my second strut, I'm gonna take one lot of delta and I'm gonna subtract off zero one. So I've done two different things here. This first one, I've, if we look over that expression, I've taken delta and subtracted off the origin and then dilated by factor of three and then I've subtracted off my unaliminated variable. That can work out to be. The second one is slightly different. I've taken delta and subtracted off one of its vertices, dilated by factor of one and then subtracted off no copies of the unaliminated variables. So let me try and make that clearer. So in this case, my coefficients, my D's and my C's, I have D zero zero equal to three and D zero one is equal to zero. And C one one is equal to one. And in this case, I have D of zero zero is zero. I have D of zero one is one and C of one one. Yeah, there's no I because it's one. Yeah, sorry, just a bit lazy. And so this defines a scaffolding because P is equal to the convex hole of U union T one and T two. And if I just draw a picture of what the net result looks like, just like the pictures we were drawing yesterday, T one is along here, P two is here. And then my unaliminated variable and this circle here is my U delta would be this fine segment. Now the conditions that I've put on how to define the struct and how to define the scaffolding are basically just a bit of a con. It's so that the conditions we have in the Horry of Africa construction will be satisfied. Okay. And so one thing I can do is I can try and reinterpret what the divisor map D looks like in terms of this data. So let me just, just so we don't get into too much of a notational nightmare, let me define a couple of vectors. So for a struct T, we're going to record the coefficients D, A, I, C, B. Remember these are non-negative in some vectors. So in vectors. So D, I will just be the coefficients from the D, I, simplex, D is just going to be the coefficients from the other nominated variables. And I'm also going to define the sum of the DIs. So I'm going to write Li to just be the sum of the DIAs. And again, this is going to be a non-negative integer. Once I've done that associated to a scaffolding, we have a matrix. And it's a N by D plus C plus N. And this is just D. It just starts off by having an identity block. So in the Hori-Raffer construction, this identity block corresponds to the basis of L du. Then I'm going to just stick my vectors D in. So they just live here. And there's a bunch of vectors. And how many of these do we have? We have, what did we say? It's a C of them. And then finally, I'm going to stick in my coefficients coming from the unordinated variable, the trade. Oh no, no, no, these are meant to be the vectors. I've indexed each vector by the index of the strut. So this is meant to be, yeah, this is meant to be the vector corresponding to the first strut. And then this is the vector corresponding to the empty strut. Yeah, I'll just write that down. So we're rows, yeah, indexed. And you know, I can also write down a bunch of column vectors that are going to become my divisors. So I can write L, which is just going to be E's, L, L-i's, oh yeah, that's a really good point. Yeah, it's meant to be the dimension of N the lattice, it's the rank of the lattice. It should be the rank of the lattice. Okay, so there's the N block, then we have, okay, no, it's indeed the size of the lattice, shall I explain? So of course the N is from here, okay. It's because these have all got an extra point added into them, yeah, on the thing, I think at this point helped more than drawing what it is in this case. So let's try and do that. So the example continued. So in this case we're gonna have D is equal to, so there's two struts, so N is two, so it's gonna be a little identity block, and then I need to write down the first vector, which is free zero, and then for the second strut it's zero one, and then I write down the vector coming from the U, so that's one for the first strut and zero for the second strut, and I have in this case just one L, and for the first strut it's this sum, so it's free, and for the second strut it's this sum, so it's one. So from this collection we can also construct a Laurent polynomial. So to a strut T, associate a Laurent polynomial, and it's just gonna be given taking, oh it's gone now, hasn't it? Taking the definition of a strut, and turning all the additions into multiplications and our subtractions into the divisions and so on. So what we're gonna have is the product over all of the struts, sorry, the product over all of the sympathies, and then the product over the vertices of those sympathies, that's the first double sum, and then at multiplication becomes exponentiation, and subtracting off that point just becomes dividing by a monomial, and then the simplex delta i, we're just gonna reinterpret it as a sum of monomials, and then we need to subtract off the points of U, so that's the same thing as multiplying by the product of one over x to the Cb. That's really just a literal translation of the polytopal description into a strut of Laurent polynomial. And so by construction we have that it's Newton polytope is just equal to T. You're gonna have to stop. So finally to a scaffolding T1 to Tn, we have, which is just gonna be the sum of these FT's. And then if you remember, I have to add on the unaliminated variables, so I just do, and by construction, because this is a scaffolding for P, we have that the Newton polytope of F is equal to P. And the way I've set this up, if the hori of affer construction is gonna work for this divisor data in this collection of line bundles, then this should be the mirror that the hori of affer method works. Okay, so I'm gonna have to stop on that ratio takeover, leaving a bit in the verge. Okay, so can you hear me? Let me sort of try to wrap this up somehow. I can clear this board. And let me remind myself, what are we doing? So we started out with a polygon polytope P, yeah, polytope polytope. And then we, I'll told you what's a scaffolding of P and then associated, okay, so maybe I'm gonna summarize this as one, sorry, associated to a scaffolding, we have one N times D plus C plus N matrix D plus some column vectors, L, I's. And what that gives us is the A, D, and these L, I's, this is the data, complete intersection, X and F, some dark variety, giving a scaffolding, we have a complete intersection data. And then Al has also told you how to make a lo-ranpoin. Again, that would be, if you like, the same sentence here, associated to a scaffolding in T1TN, we have one complete intersection data, and two lo-ranpoin, and two are mirror to each other. They are so under the Horewaf construction. Let me sort of point out, if you start with a polytope and the scaffolding on it, there will be some assumptions, some further assumption on that scaffolding that would ensure, for instance, that X is a quasi smooth and well-formed orbital, and well-formed orbital, and we don't really know what those assumptions are. We do this on examples, in most cases, when we do it, X is well-formed quasi smooth orbital, but sometimes it isn't. There's another thing about this construction. Okay, so in some sense, this is a kind of conclusion, okay, except that we'll further wrap it up later, in a moment, but then let me say another feature of this construction. There is further this construction. Let me say it in words, and then I will make it a bit more precise. Actually, from the polytope P, you can form the final variety, the historic final variety, XP, and XP is one of these complete intersections in F, is a member of those complete intersections, of the family of complete intersections, and we can write down its equation, and let me tell you how to write it down. Suppose that I do the scaffolding, and then I get this data here, then XP is one of these. It's one of these, yes. No, no, it's one of these. There, family, the one given by the scaffolding. And so, we'll introduce variables corresponding to the columns of the matrix D, as we always do. And so, this is an N by, I would like to say, not D plus C plus N, but N plus D plus C matrix, okay? So, I'm gonna call the first N columns Y1 up to YN, okay? Then these guys, there are D of these, and so, I'm gonna call these ones YN plus one, all the way to YN plus D, and then these guys N plus D plus one, all the way to YN plus D plus C, okay? So, we introduce variables Y1 up to YN plus D plus C, corresponding to the columns of D. Yeah, it is the size of the vector C. It is the size of the vector C. And how many S's do you have? Yes. Exactly. For each one of those, I have one column at the end, because, sorry, that's correct. In fact, okay, okay, right, right. You are right. So, what do you wanna call it K? What's D for you? What's D for you? So, you know, but the S's were a partition of, you know, you really should have asked Al about this, because, you know, instead of asking me now, I mean, you know. Okay, so in fact, actually there are a few more variables then, because the uneliminated variables are even beyond that, it seems to me. The size of U, and I have to append it at the end of everything, and I'll just do all this stuff, in other words. No, wait a minute, no, no, I don't know, because I have to do something now that has to make sense, and you know, if, why didn't you ask Al about this? No, no, there is, I think there is a K here, or a U maybe, okay, right. There is an identity, there is an identity. You're paying just at the wrong time. Sorry, I mean, so at the beginning, there is these S I knots, okay. Okay, so there are C of them. They form a partition of, together with you, I see. I see, I see, I see, I see, I see. I mean, I want, well, sorry, I mean, let me just put like this, at the beginning, so what was the thing that we had at the beginning, and what was the dimension of N? The dimension of N was D, okay. And this is somehow the thing that I don't terribly like. So I want this to be K plus E, and then, indeed, absolutely. And then I want to be something like this, so I want to say that K plus C equal H, so to speak, okay. And then I want to call these guys all the way to Y, N plus H. And then here, H plus one, all the way to H plus U, okay. Maybe now you agree with me. Anyway, so this is all some large academic, because I'm just going to write down two monomials now, binomials, C, C, I'm going to write, ha ha, C binomials. And so I'm going to write like this, hopefully, let's see. Or I, in Y and C, I'm going to write binomials like this, product J in S, I, Y, N plus J. And then minus product K equal one to N, one to N of YK to the, and my claim, the claim here is that these binomials define XP in F as a complete intersection of the correct limons. Let me try to wrap it up, wrap it all up. So remember at the beginning, we spoke, I told you in fact, about this mirror symmetry program. And so, you know, mirror symmetry, as I said, is about torque degeneration. We always start a final part of P, and then we want to understand enumerate, classify whatever, study, understand, we call this families M, of deformations. Of XP to a final orbital. And then to each of these families, or each of these families, attach the random polynomial F in, you know, C, Cp is a polytope in NR, and we want F in C bracket N with mute F equal P, okay? Okay? And there is a certain mutation in variance to these things. And so we want to be able to say things like that, for instance, for two different polytopes, P and Q, then the final XP and the final XQ deforming the same family, M, we want to be able to say that that's if and only if the corresponding mirrors F and G differ by a chain of mutations, those algebraic mutations of Laurent polynomials to the family of deformations of XP to an only. Okay, so they are attached to the polytop and the family. They're attached to the polytop and the family. And, you know, this implies it's stronger than saying that P and Q differ by chain of mutations, okay? So this is what we're working on, and we're working on some of the theory of this, okay? But we're especially interested in the practice of this. What did I say at the beginning, these lectures that we would do? Well, I promised that it would give you some practical things that works sometimes. And that's what we've done. So here's what we actually did. So start from P, okay? We defined this notion of scaffolding. You know, there is a complicated, like always with combinatorics, you know, it's sort of hard to get your head through a formal definition. But in fact, these things are reasonable, okay? You can draw pictures of them and, you know, that makes sense. And out of the scaffolding, a scaffolding gives one a family of total complete intersections. And XP is a member of that family. And so in good cases, the general member of that family is a well-formed orbifold. So this sort of answers part of question one. You know, I don't understand the whole family, but here's a way to give, to write at least one family down. And indeed, it gives this mirror Laurent Poirot. So we can see that at least this goes some way in doing what we want to do. That's it, really. We didn't really talk much about three. We don't understand too well where we're working on it, but it's not so, so we can, we can, this is more like a question, okay? So in practice, but it doesn't work very well, okay? I'll just put it like this. But, so given a polytope P, you can look at various mutations of P. And typically in that mutation class, very few polytopes will admit scaffoldings. That sort of answers your question. No, no, let me just work out whether I'm finished with everything, or whether I want to say something more. Well, I suppose that, you know, maybe I was somewhat defense. I mean, in practice, even though this is limited, not all deformations are into a complete discussion. This is a fact. And we know several examples of that in some sense that's the most interesting case. Nonetheless, if you are interested in doing the yellow pages of all possible final LG pairs, this works. And, you know, it fills a large, you know, a large chunk of those yellow pages. It's not a theorem. It's a feeling. We, you know, in the, in the corners that we've been looking at, we were able to fill almost all the table using this method. And then there is a few extra cases where you have to work much harder. And perhaps at this point, I can't stop. I've stopped. That's it.