 Apologies in advance, I appear to have prepared a lecture that hasn't got a single definition and a single theorem in it at all. It's just a string of examples, but there's one lecture left tomorrow, so I'll try and define what I'm doing today tomorrow. So I want to just begin by recalling the Horry-Vaffa construction. So what do we have? We have X contained in a toric variety, F, and this is a complete intersection. And this is given by a bunch of, I want them to be neph vine bundles. And then there's some additional assumptions. So we assume that there's a partition of R, R I guess is the number of rays of F. So we assume that there exists a partition, E, S1, SC, and then U of, is that what you called it? So I don't know if before U was introduced, U though is just a place to put the determinant of your partition as opposed to just this joint subsets. So for obvious reasons, I'm going to call you the uneliminated variables. And then this partition has to satisfy the following, so such that. So the first one is I want E to define a basis on L-dual. So I want J or a genuine E, a basis, L-dual. In the second one, I want each of these line bundles to be written as a positive collection of these DJ in my basis. So I want each I to be in generated by E. Third condition, I just want the SI to actually define the LI. So I want each LI is just equal to the sum of DK for KN. So given the state, we can define the mirror. So for each of these SI, I'm going to pick a distinguished element. This is really no big deal. Basically, this picture is an affine picture. And so this is basically just choosing an origin. So just pick an origin. And I'm just going to write SI interior or circle, however you want me to pronounce this. I'm going to pronounce it interior. So the SI subtract just a little bit of notational shorthand. And we want to write D as a matrix. So it's going to be M minus D, M matrix. And it's just called DJI. And this is written in terms of the basis on to the M, I guess, dual at this point. And the choice of basis we've made on Alster. Then the next step is we just introduce loads of variables. Variables for everything inside. So introduce following collections of variables. So variables on X1, XM, I also want variables, YJ is J's ESI interiors. And now define some relations between them. So define relations by, I want to say, XK is going to be the same thing as all of the 1 plus YJ, or J in SI interior. And that is I. So this is if K is equal to my choice of a distinguished element. So if K is equal to a little. And I'm going to define it to be YK over 1 plus YJ. If K is in SI interior, otherwise no relation coming from this. And also a bunch of products. So I want XK equal to 1. And this is from K equals 1 to M. And this is for all J equals 1 to minus D. So basically this is what I've actually told us before. And introduced the Horovath construction. And the mirror of D of X is just eliminating these variables from the sum of the X's. So by rewriting Q is equal to X1 plus dot dot dot plus XM in terms of these variables that have survived. So let's have a look at what's survived. We have the X's that are in my uniliminated variables. I'm also left with X's corresponding to my distinguished elements. How many of them are there? I've lost track C, of course, core dimension. I'm also left with my Y's when my Y's are in my S interior. So these are the variables that I rewrite W in. And then I just need to subtract off the constant term. So everything today is basically going to be done up to constant term. And subtracts off S's. Yeah, right, exactly. Because they're sort of being sent to the origin, as it were. So they're just sort of corresponding to Wands. And to me, the Horovath construction is still slightly mysterious. All right, so I did do an example of this before. I think it was the cubic surface. But I'm going to do a slightly different example. So I'm going to make DP6. I'm going to start with a type variety P1 cross P1 cross P1. And this has divisor data, weight data, which is given by 1, 1, 1, 1, 1, 1. And six devices, D1, D2, Rheinbender. So I'm going to pick just a single one to be 1, 1, 1. Well, so now the next step is I need to choose basis inside L. The obvious choice of basis I'm going to just pick is D1, D3, D5. So pick a basis. I mean, it's kind of implicit that choice in the way of my matrix. So I have E1, 3, 5. And now I'm going to choose a way of writing L. I'm going to write it as D2, D4, D6. So I have S is going to be 4, 6. And U, well, there's nothing left. So U is just the empty set. And what the heck, I need to choose a distinguished point in here. So I'm going to choose little less is equal to 0. What are the variables? We have about x1 to x6. And then we have y4. And we have y6. And we have relations. We have x2. So 1 over 1 plus y4. We have x4, y4 over 1 plus y4. And we have x6, y6 over 1 plus y6. And we just look at these rows of the weight matrix to get products equal to 1. So we see from the first row that x1, x2 is 1. The second row, x3, x4 is 1. And then the final row, x5. So we have W is just x1 plus x6. And so this is equal to, and just going to do this a little bit slowly, because it's an intermediate step that's actually useful to have. So x1 is just 1 over x2. x3 is just 1 over x4. x, which one? 5 is just 1 over x6. And then this is going to be x2. Just rewrite this in terms of y. I'm going to get pedantic and write that over 1. And then by construction, this is equal to 1. Let me just label this with a star. But anyway, this gives me writing x equal to y4, y equal to y6, we recover our Laurent polynomial, f, x over y, et cetera, et cetera. I just subtracted off the constant term, which in this case is 4. If we just set p equal to the new polytope of this, it gets something that looks like this is the origin. Taking the spanning fan, I recognize this. Looking at this star, I mean, I can forget the constant term here, because it's not going to matter. Like I say today, we're going to be very loose with our constant terms. But I've got these three distinct expressions. So let me just call them A, B, and C. It's not 2, it's 4. And nothing stops me just drawing their new polytopes, too. And I'm going to draw their new polytopes on the new polytop of p. So we have A is this one. Remember, x is y4, y is y6. So we have A is this one, and B is this one, and C, and B is this one. So I mean, you can see that these three new polytopes, if they're risen in a natural way through the construction, because they arose in the mirror, and they also look a lot like they've got a lot to do with the resulting polytop. You know, they fill it out and give it its shape. So we're going to give them some names. And this terminology, perhaps not the best, and it might not be stable. But we're going to call those three things three struts. And we're going to call the collection of those three things a scaffolding, because it sort of holds p up. So we'll call them C struts. Rather, I mean the new polytopes of A, B, and C. And a collection of struts is a scaffolding. We could happily do all of this without these names, but I'm going to call them this, whether I define it or not. So I might as well get it out of the way. All right. So this is our gain. Our gain is to invert this process. I want to start with a polytop, and I want to see if there's a way of putting a scaffolding on it, so that I end up with other ampolynomial. And I want to do this in such a way that it gives me the correct amount of ampolynomial. It'll give me a toric complete intersection description of my variety. Now, because we're starting with p, and we really don't know much else about p, it really is just a sort of combinatorial gain. And I've started writing my lecture a little earlier. I might have got to the point where I could have given you a sort of definition that made sense. But for now, we're just going to do a series of examples, and at the end of this, pretend that we can extract what the correct definition is. Maybe we don't even know what the correct definition is. So an example, I want to do the cubic surface backwards. So I'm going to start with p. This is in N, and I want to somehow see that this is the cubic surface. I'm going to give it a strut or a scaffolding. Now, there's a very obvious choice of scaffolding on this, which is just the triangle itself. So a single strut, that works good enough. So I'm not going to draw that again. So we have a single strut. So how can I turn this into something like a Laurent polynomial? Well, this is, if I just choose x and y to be that, this is just 1 plus x plus y to the 3, and then translate it down to there, so x over y. OK? And now I need to sort of invest this process. So I'm going to define x2 to be 1 over 1 plus x plus y. And I'm going to define x3 to be x over 1 plus x plus y. And I'm going to define x4 to be y over 1 plus x plus y. And I'm going to be slightly slapped. That's about my constant term. And I'm going to get this is equal to 1 over x2, x3, x4. And now I'm just going to simply add on x2, x3, and x4 because I'm trying to recover w through this process. So I'm still going to use an equal sign. But I'll try and at least make a token effort to note that it's not really equal. And so this is going to tell me that I should set x1 to just be this product. And this is going to give me the intersection data. Well, I've got d1, d2, d3, and d4. And then I've got my single line bundle. And it's going to be 1, 1, 1, 1. Then how do I figure out what else is just the sum? So I just sum those to give me. So I've just inverted the process. So we have, OK, maybe to make it really concrete, I have e is equal to 1 is equal to 2, 3, 4. And no one eliminated variables. I'm going to just put a little dotted line here. And this is just going to be a notational device. What we're going to do is, when we choose our e, we're choosing a basis for L dual. And so I'm always just going to rig it up so that my basis is on this side of the dotted line. And so I'm always just going to know I can write an identity matrix here, whose size is just equal to the rank of the variety. Just going to do it as a little placeholder, OK? All right. But also knowing that, I can read off the binomial that describes, well, actually, no, too soon, too soon. So we have L is just equal to d2 plus d3 plus d4, OK? And we have f, let's say this is p3. And so I know what the rays of f are. So I could just take the kernel of this. So I have the fan of f has rays, obviously, up to a choice of basis. But it's got rays, row one, row two, row three, and row four. And I'm just going to do e1, e2, e3, and minus e1, minus e2, minus e3. And these are in nr, which is three dimension. And if you remember from the adjunction formula, what we need to look at is minus kf, minus l. So this is just equal to, so always because it's a total variety, minus k, you just sum the di's and take the negation of it. So it's minus the sum of the di's. And then minus l, what was l? I've got d3, d3, d4, OK? So I'm just left with minus d1. And I can turn this into a polytope in the dual lattice, m. That's where you can draw devalues. So we have, so I'm just going to call it q of minus d1. So this is just going to be defined. Just think this is the definition for all cases because it is. You just need to choose all those. Yeah? That's the sum of the di's. Yeah? Minus. Yeah, good. I'm going to take, I'm going to take, you is here. OK, I'm going to take u of rho1 greater than or equal to minus 1 because I've got exactly 1 d1. And then I'm going to take u of rhoj, it's going to be greater than or equal to 0. And this is for j equals 2, 3, and 4. If you work it out, this is a polytope with the following vertices. L was defined by using p2d3d4. So that defines what the binomial is that gives us x in p3. So x is given by the binomial. I'm going to have to choose some variables, aren't I? I can't use x's. I can't use y's. Call them. I can't use capital X's. OK, so I'm just going to assign some variables to p3. And then the binomial is going to be given by z1 to the 3 minus z2, z3, z4. What I can do is I can quotient n by the corresponding vector 1, 1, 1. I get q dashed is going to be equal to the convex half of q1, 1, 2, and not not. This is now inside m prime. So I can at least draw a picture without too much fuss. Looks like this. And remember, this is in m. So I need to recover the fan xp. I need to take the normal fan, and the normal fan is going to look something like up to a sine issue, which probably I should have just done all of these as best they're equal to. So up to a sine issue, the normal fan recovers spanning fan. All right, let's do another example. Go into this in much less detail. Let's just start with what I happen to know is dp6. I want to construct, I guess most of you know, there's two ways of describing dp6 as a complete intersection. And I'm going to construct the other way. So I'm going to start with this in nr. And I'm going to give it a scaffolding. And I'm going to use two structs. I was trying to do the scaffolding in yellow. So the scaffolding in this case, I think when I draw it you'll see I'm just going to do this. And I'm going to call this one a. So notice that square is just equal to the Minkowski sum of a line plus a line. And this information tells me that what I'm going to get is cut out by two line bundles rather than one line bundle. And so we're going to have struct a is just going to give me 1 plus x, 1 plus y over x. And then struct b is going to be over y. And then up to constant term, I'm going to write this as 1 over, so x1, x2. I'm going to write this as 1 over x4, x5 plus 1 over x3. And then plus x4 plus x3 plus x4 plus x5 plus x6. And then what have I done? Well, I've done the obvious thing. So x3 is 1 over 1 plus x. And x4 is x over 1 plus x. And together, this is going to define one of my line bundles. And then x5 is 1 over 1 plus y. x6 is y over 1 plus y. And this is going to define a second. And then the relation that gives me x1 and x2, I just read off the denominators. And so we have x1, x4, x5 is equal to 1, and 2, x3. So we've covered a complete intersection data. It's down to six devices. I've got two devices. And by construction, d1 and d2 give me the basis for L-dual. And then I just read these off the denominators. So I've got 0, 1, 1, 0, 1, 0, 1. And then L1 is going to be those two. And L2, so again here. Sorry, I should have pointed out this is up to constant time. Let's try and answer a question using this construction that's been hanging over us from a few lectures ago. So if you remember, we started with the cone over the hexagon. And there were two distinct mutations. And one of them gave us p1 cross p1 cross p1. And the other one gave us, we don't know. So we'll use this struct method to find out what the other one is. So let me draw it carefully. So it was a square here. And then these that gave us a square face here, if you remember, looked. Just to label these, this was the origin. This was z. And then I think I had 1, 0, 0, 1, 0, 0 minus 1, 0. This will be 1, 0, 0, minus 1, minus 1. This was in. So I need to put a strut on it. A scaffolding on it. And actually, you'll find that there's only one way you can do this. You'll think that you see one way and you'll discover that you can't really do it. There is, in fact, only one way to do it. And it's not what I just drew. And it's given by taking, let's call this one a, or this one b. And then this time, I'm going to use the unaluminated variables. So I've not been using those before. So I'm going to make this one and this one unaluminated variables. So we've scaffolded with two struts to unaluminated cyclincl. And I'm going to not choose the obvious basis for n. So for a basis, I'm going to use the following. I'm going to use, well, 1, 0, 0, so far, OK. 0, 1, 0. And then 0, 1, 1. And when I write down my Laurent polynomials, this one's going to be x. This one's going to be y. And this one's going to be z. Now let me try and write down my scaffolding. So I have, OK, so a, going to give me 1 plus z. Divided by xz, 1 plus z. Divided by y. Then I have my two unaluminated variables, which are just x and y. So the same trick as always. So now I'm just going to write this as, let me think. So it's just going to be one ion bundle. So I'm going to write one ion bundle rank 2. Over 1 over x3. And then x3 up to constant n. x3 is going to be 1 over 1 plus z. x4 is just going to be z over 1 plus z. And together, it's going to define l. And then x5 is just going to be x. 6 is just going to be y. And we have x1, x4, x5, x2, x3, x6 is equal to 1. So we get the divisor data. A, B, 1, 2. Here it's going to be an identity. What do we get? 0, 1, 1, 0, 1, 0, 0, 1. And then l, this time, only comes from d3 and d4. So I'm going to get 1, 1. 1, 1 divisor in 2 plus p2. So this answers the question that we were wondering before. How can we figure out what that variety was? So there's just time I'm going to do a final example. And then next lecture at the very beginning, I'm just going to try and give a formal definition for struts and then see if, or at least, you'll have anything you want to say. So final example. So do you remember we had those third 1, 1 surfaces? And the question was, OK, very good. We've got polytopes. But what are our various varieties? How are we going to understand this? And we just do the same thing. So I'm going to take one of the examples, this one on the list, which just looks like this. And I think it had the origin here. And my scaffolding that I'm going to use, again, there's only one choice I can make. I mean, it's completely uniquely defined. I'm going to have to choose this strut here. And I have to choose this as an unaluminated variable. So scaffolding, one strut, one unaluminated variable. So the basis I'm going to use is, I'm just going to use this and this is my basis directions. So I'm going to use 1, 1 and 0, 1. And this one, I'll just associate with x. This one, I'll just associate with y. And I'm going to get 1 plus y to the 4 over xy plus. So this here is, and this here is my unaluminated variable. And then just doing the same trick. So it's going to be just rank 1. So this is going to be 1 over. So it's going to be 3x4x5x2 plus x3 plus x4 plus x. Wait, wait, wait. Seriously, let's get the number. x2 to the 3 plus x3 plus x4, x2, x3, x4, good. So up to constant term. And then this is where x2 is just 1 over 1 plus y. x3 is y over 1 plus y. And together, this defines my divisor L, and then x4 is equal to x. And we have the relation x1, x2 cubed, x3, x4. So finally, just write out the divisor data. So d1 is going to be my basis dc4. And it's going to be 1, 1, 1. And then L is just coming from dc. So it's going to be 4. So what I have is x4 and 1, 1, 1. So that is what that variety deforms to. OK, so I'm definitely going to stop there. Like I said, it consisted of no definitions, no theorems. Any questions?