 I just want to briefly make a connection with the standard treatment of torque geometry. And sort of I assume that you have seen how to, you know, what's a fan and so on. So in torque geometry one usually has M, the lattice of monomials, N. So M is the group of characters, copies of GM into C star. Well, if it's intrinsic, it's just an n-dimensional tool. And N is the dual. And we know how to attach a torque variety, sigma, the collection of cones. And so I want to make the connection what we did the other time. So I assume that you sort of know how to do this. A fan is a set of cones. And if sigma is a cone in an R, then you can form X sub-sigma. You know, it better be a rational, finally generated rational cone. Then X sub-sigma is spec, the group algebra K intersection M. And these, so if sigma, big sigma now, yeah, yeah, so, yeah, you're right. For me, K is always C. Yeah. I know sigma is a certain set of cones. And that need to satisfy some axioms that the intersection of two cones, or the closure of two cones is still a column. There is a bit of an issue whether you work with closed or open cones here, but I work with closed cones, I suppose. Let's not go into that. Then the torque variety X sub-capital sigma is a union of affine torque varieties X sub-little sigma as sigma runs through all the cones of the fan. That's more or less how you make torque varieties from a fan. Now I'm going to tell you how to see this same torque variety as a quotient of a representation of a torus. So the thing to do is the ray map, which I always call rho. And this thing is constructed, so the set, as usual in this situation I tend to call rho sub-i. The image of the i-th basis element. And if I look at the set of r plus as i goes from 1 to n, m, then this set is the collection of one-dimensional cones of the fan. So given a fan, you can always form a ray map. And in what follows, I assume for simplicity, the map is surjective. Not only that, but rho i is primitive. So that's always understood. So this assumption that rho is surjective is not really necessary, but z being a ring of dimension 1, rho i is not surjective. Then one has to do a little homological algebra, which over z is very easy, but still a bit meaningful. So let's not do that. I'm not going to tell you what to do when rho is not surjective. So then if rho is surjective, it's surjective. And I can take its kernel, which I call l, and then I can dualize this sequence. So that's zm star, taking m now. And then the dual is l star. And that's the thing. Unfortunately, I did the mistake. In my earlier lectures, all the l's, all the bold l's should have been l stars. But then because the kernel of rho is usually called l in the literature. You know, maybe you should go back and change all those into l stars. And oh yeah, because this is map D. This here gives sigma. Yeah, yeah, yeah, yeah. You could if you wanted to, yes. Well, you know, if you want. This D here, this is data giving what gives sigma as a quotient. So by what about the stability conditions? Yeah, what about stability conditions? Okay. So let me state now how that works. So notation for a subset i, m, the set 1 to m, I'm going to define sigma sub i generated by the rho i's in i. And this is a cone in n. And I'm going to denote by c sub i, the cone generated by the d i's. i in i. And that's the sub cone of l star r. Okay, that's just notation. And the statement, implicit fans. So the fan is implicit if all its cones, little sigma are simplicial cones. And that corresponds to the stark variety x sub big sigma being in a natural way an orbital. Statement is that the cone sigma sub i is in the fan. If and only if the cone c i complement. So if you like m that minus i, if and only if that cone contains the cone of stability conditions, the ample cone. Of the torque variety. Well, you know, you just call it the ample cone. So, you know, that's it really suppose and moreover, suppose that you're giving the fan the cone of stability conditions. The ample chamber, the, the chamber of stability condition is indeed the intersection over sigma i. Now over those eyes, such that sigma i is in the fan of the corresponding cone i complement. And vice versa, if you're given a stability condition, a chamber. The fan sigma is the collection of all cones sigma sub i complements where the ample cone is contained in c sub i. Okay, so that's it. C, c. Well, you know, just, just, yeah, c for chamber. So we did comment that the, so long as I'm the inside in the open part of the chamber, it does not depend which chi I take inside the chamber. So everything, the unstable locals, the stable locals, the quotient and everything. So somehow, you know, I'm really ambiguous. Sometimes I just mean the chamber. Amp is the cone of stability condition. Amp is the chamber. Yeah. It's a chamber. Yeah. It's a chamber. Even a chamber. You know, it sort of makes sense to call it that because that chamber will be literally the cone of ample line bundles on that orb. It makes sense. Yeah. So this is the chamber. Yes. And so, you know, given sigma, then, you know, this is a chamber. And no matter who I take inside that chamber, the quotient will come out as x sigma. That's sort of the statement here. Yes. Yes. But then I would have to assume that chi is not on any wall. So, you know, I'm giving a simplified version here. For simplicial funds. And, you know, those historic varieties correspond to taking chi in the interior of a chamber. Then I can recover all other funds by allowing chi to be on a wall. But then the statements become a bit more complicated and there are some checks and balances. And, you know, I don't really want to go there besides, you know, well, anyway, that's the thing. So let me make an example. I'm going to illustrate all of this. Okay. And so I start from the Ray map fan. Okay, let's look at this fan. Sigma is this fan here. So let me put in N. I'm just going to draw it. Here N is Z2. A little cross at the origin. I'm going to draw the, and I mentioned two for surfaces. The Ray map completely determines the fan. And so, you know, this is your fan. It's the face fan of this pentagon. Let me call those. So here the Ray map, I can write like this. Oh, I have five. Yeah, so I'm going to call these guys row one, row two, row three, row four, and row five. And so the Ray map sends Z5 to N, which is Z2 here. And so, you know, row one is the vector one, zero. Row two is the vector zero, one. Row three is the vector minus one, zero. Row four minus one, minus one. And row five is the vector one, minus one. So that's the matrix for row. And if you work it out, D, the action, you have to take the kernel of row, and you can write it like this. And so let me sort of copy it. I did it home. Okay, so that's the matrix D. Okay, so the columns of D are again D1, D2, D4, and D5. I want to draw, this is a rank three action. I want to draw the secondary fan for this action. These are vectors in R3, D1 of D5. And I'm taking a two-dimensional slice and a generic cone, and I'm drawing a two-dimensional slice of that cone on the plane of the blackboard. D3, D4, and D5 are the vectors one, zero, zero. Zero, one, zero, and zero, zero, one. They are the three coordinate vectors in R3. And I'm going to take my slice to be the plane that contains those three vectors. That's the plane of the blackboard. And then those three vectors then appear on this plane like this. That's D3, D4, and D5. Okay, then look at the vector D2. D2 is the vector D4 plus D5. So it lives somewhere in the middle of the line between D4 and D5. So let me put it here. That's your vector D2. And then I see the vector D1, which is D3 plus D4 minus D5. So it's sort of here. So now what are the walls of the secondary fan? Are just all the two-dimensional cones generated by pairs of these vectors. And so let me draw them all. There is this wall, there is this wall, there is this wall, there is that wall. And then from D2, there is this wall, there is this wall, there is that wall. And from D3, there is this wall, this wall, there is this wall, that wall. And then from D4, I think I have them all. And so you see now there are some chambers and these are the walls. And I'm telling you that the chamber that I'm interested in is this one. And so let's illustrate the statement there. So what's the fan? The fan is the collection of the following cones. You can see them there. Sigma one, two. Sigma. I'm going to make a list of the cones of maximal dimension of the fan sigma. And so that's sigma one, two, sigma two, three. Sigma three, four. Sigma four, five. And sigma five, one. Okay, so these are all the cones of maximal dimension of the fan. And then the anti cones are the cone C345, one, four, five. C one, two, five. C one, two, three. And C two, three, four. Okay? And so what I'm saying is that this chamber here, the one that I've shaded, is the intersection of all of these cones. Precisely. So you see the cone 345 is this big cone here. And it does contain my little chamber. The cone 145, 145 also contains it. Similarly, the cone 125, 125. And you know what I'm saying is that that chamber is the intersection of all those cones. You may wonder here now that something is wrong in what I've told you. Because given the ray map, at least here in two dimensions, there is a unique fan that has that ray map. Okay? And so why do we have all these chambers here? There should be just one chamber. Why do we have all these other chambers? What do they do? So the situation with these other chambers is, let me tell you what happens. The situation here is very similar to the example we had yesterday, where we had these two by four matrix that gave, then there were two chambers. In the secondary fan, one gave the quotient, the segregate surface Fn, and the other gave the quotient P11n. And if you remember, the quotient P11n, the action was redundant somehow. Of those two C stars, I could have killed one. And then, you know, the quotient in that chamber was actually a lower rank torque. This is exactly what happens here. All these chambers, these outer chambers are what we call hollow chambers. And there is one of the actions of these three C stars becomes redundant. And those are torque varieties of smaller rank. So they're given by ray fans with fewer rays. And so let me not go into that. You can explore only if you want. But that's the only important chamber that's really doing something. Okay, so I stop here, and I let Al... I was fearing. Don't worry about the rest. Okay, so maybe you remember in the distant past, we introduced the idea of mutation. And this was, you took a polytope, and you imposed the grading on your lattice, and you sliced the polytope up, and then you sort of applied socialist principles to it. You take from those that have, and you give to those that have. So I just want to start by doing another example of mutation. Okay, so I'm going to take the following. Take the cone over the hexagon in three dimensions. Here's my hexagon. And then below the hexagon, I'm going to put the origin. And then below the origin, I'm going to put the next vertex. So let me just label these. And here is the origin. And down here, minus x. And then this should just join up. So what I want to do is I want to investigate a couple of mutations of this. So I'm going to take this top facet of the polytope, and I'm going to try and mutate it in two different ways. So the hexagon has two different Minkowski factorizations. So the first one, I'll just redraw the hexagon. I'm afraid this is going to be taught with lots of drawings and pictures, but that's okay. So I can decompose it as a pair of triangles, or I can decompose it as three line segments. So this is going to give us two different mutations that we can do. So we have two possible mutations. And so the first one, I'm going to just use this triangle decomposition. So if you remember, we describe a mutation by giving ourselves a primitive vector in the dual lattice. So the gradient, which in this case is going to be 00 minus 1. And then we award ourselves our factor that we're going to choose. And our factor is going to be at height zero with respect to the gradient. And slightly softly, I'm just going to write it as a two dimensional shape with the origin of that. But this is really meant to be living at height zero with respect to this gradient. Here, I remove a copy of that factor, and then down here, I stick in a copy of that factor. So of course, when I remove a copy of this factor, I'm just left with this. So at the top, on that top facet, I'm going to have my triangle left over. I'll label these straight away. And then the origin is around here. And then down at the bottom, I've now added in my factor. So it's going to look something like this. And taking the convex hole, it just joins up like what's happening at the back. With all of these things, it's so much better if you draw them yourself and copy someone's flatboard drawing. But here's the origin. If you stare at this, you'll see, actually, there's a change of basis that makes this a lot more understandable. To be honest, I don't know what the change of basis is, but I did stare at this and see this. So we're going to get a square at height zero and then just E3 and minus E3. You're going to end up with this origin thing there. Just going to be E3, minus E3, E1, E2, minus E2, minus E1. And remember, we're working in N. So if you want to turn this into a tarot variety, you need to take the spanning fan. And if we do that, well, we have these rays like this and maybe recognize this. This is just P1 cross P1 cross P1. I'm going to just name this for reasons of my own, Q3. And what we have here is XQ3 is just P1 cross P1. So we should do the second one now. So for the second one, I'm going to do a mutation that removes one of these line segments from upstairs and moves it downstairs. And so when I remove a line segment from the hexagon, I'm left with a square. So we see what we get. So it's going to be exactly the same grading, of course. And this time, well, this is even more sloppy than me drawing the triangle, but she's in a line segment. What happens at the top? I could say we end up with a square. Let me just label these. And then the origin is maybe here. And then this is the point at the bottom. And we've added in this line segment that's going to come to here somewhere. And then when we take the convex hull of this, it's going to look a bit like this. This one's even more weird than that one. And just stare at it for a bit and hope for your spot that it's not actually as bad as it might seem. He says that. Yeah, so here, here, here and here, although it's a bit hard to see in this picture, we've got a square just like we've got this square here. And so if you get this right, what you end up with is this. Let me try and draw it again. I'm going to give myself the square. And then the origin here. So this is a plane with the origin. Suspended above it is going to be E3. And now there's going to be a facet down here with the other point here. So let me just join all this up. I'm going to end up with just label these, I guess, the same way. So E1, E2, minus E1, minus E2, zero, E3. Then this point here is going to be minus E1, minus E2, minus. This is the square face that was at the top. And then I guess this is the line segment that we moved downstairs. So you're going to ask me what is this topic variety. And I don't really know if I'm going to be honest. But you can write down the weights without too much difficulty. So in this case, D is, well, let me label my vetsies in Samada. Maybe I'll label them. Add it, does it? Row one, row two, row three, row four, row five, row six. So here we're going to have D1, D2, D3, D4, and D6. And what are the relations out of between these? We've got row one and row three is a linear relation. We've got row four and row two. Then finally we've got row six, row five, and then row three and row four. Row six, row five, row three, and row four. So this is, for what it's worth, this is the white matrix. But it's topic variety. So let me just draw the fan in. So the fan looks like that, this, this, this. So there's two different mutations that we can do out of this polytope. And if you remember, the polytopes were really, in our minds, just being Newton polytopes for Laurent polynomials. Now, I guess the important observation is, up at that top facet, it's impossible for our Laurent polynomial to factorize in two different ways. It can do one or the other, but it can't do both. I guess the main point, f is around, and such that Newton polytop of f is p, can follow at most one of these mutations. It can do both of them. So let's just write down two Laurent polynomials that will do these two different mutations. So we'll just write a to be a generic Laurent polynomial. I'm going to put coefficient one at every single one of these vertices, and then I'm just going to put a variable a in the middle of that top. So we just have z of x plus y plus y over x plus 1 over x plus 1 over y. Plus x over y plus a plus 1 over z. Then this is the one that admits the triangle mutation. So just a little bit of tidying up. It's z over x, y, and then it's x plus x, y plus y times by one, well, the factor, 1 plus x plus y plus 1 over z. And if we go away and, it's getting a bit long, if we go away and we calculate its period sequence, just successive powers of the constant term, remember, then we get something that begins 1 plus 6t squared plus 90t to the 4 plus dot dot dot. Remember, a ratio sort of made this analogy at the beginning where we're kind of developing the yellow pages of final manifolds, okay? And we've got little bits of our telephone directory sorted out, and you can go away, you can look this person up in our yellow pages, and you'll find that this is a mirror for p1 cross p1 cross p1. And that's to be expected completely, because of course when we did the mutation, we ended up with p1 cross p1 cross p1. So you'd hope that that's what we'd find. So what about the second case? The case where we mutate the line segment and leave ourselves with a little square at the top, well, it's just as easy. So we have f2 in this case is equal to z over x, y, 1 plus x, 1 plus y, x plus y. Okay, so this allows the second mutation. And here you can go away and you can calculate the period sequence. Well, we don't expect the same thing, because we kind of don't expect this to also be a mirror for p1 cross p1. And indeed, we don't get the same thing. So we get 1 plus 4t squared plus 60p to the 4 plus dot dot dot. Okay, and again you can go away and you can look up this telephone number in the yellow pages, and you can find out what final manifold this is a mirror for. And I'm not going to tell you just yet, because it's kind of part of the point of what we want to tell you is how you can figure this out from the polytope yourself. Okay, but anyway, this also turns out for a final manifold, cross p1. So yes, what's going on here is the torrid variety xp has two deformation components. And particular choices you make of the Minkowski factors, basically pick which of those components you're allowed to live on. I suppose not really mutation factors, it's the whole mutation data. So mutation data, picks which factor, which component. But it's mutation data, you know, it's very subtle. I don't think we really understand what we're talking about anyway at the moment, but it does seem to be that you have to award yourself more than just the polytope. You have to award yourself also this additional way that you're going to allow your polytope to be mutated. So very quickly, just to sort of finish off, if you remember in telegeometry, there's two lattices that play a big role. There's N, where the family lives, and then there's M. The dual lattice were minus K, that's where the divisor stuff lives. So mutation has got a rather beautiful expression in the dual lattice M, which should satisfy Don because he doesn't have those remainders in it anymore. Let me show you. So let's let P be a polytope, okay? And the dual polytope, just in case you've never seen it before, is just defined to be all those points in the dual lattice that evaluate to at least minus one on P. So this is the dual. And in telegeometry, this is the minus K polytope, and so we have that the volume of P-dual is just equal to the degree. The number of points in successive dilations of P-dual is just H0 of minus Mk. If I mutate P, is there some way of seeing how I can change P-dual without just having to mutate P, then dualize what I get? And the answer is yes. So we can see how mutation acts on M. So I'm not going to prove it, but the key observation is that if I take any lattice point U and M, then there exists some integer M, such that one over MU is a support in hyperplane of P. So once you've made that observation, what you can do is you can sort of look, so here's my P maybe, here's my one over MU, so I can look locally at P at that point and I can just figure out how P changes under mutation and this lets me see how this hyperplane changes under mutation. Okay, so we can see this changes, we mutate. Okay, that's true. Okay, so if you want, I can write exactly. So I can just write HU1 over M, by which I mean the set of all those points in N, such that UN is equal to one over M. Is Y equal to minus one, sorry? No, it's not even that. So it's the set of all those points, A and N, such that U of A is equal to one over M. I'm just taking one over M, you know, okay. All right, let me say this one. And let me say this, such that U is in the member of the boundary of MP dual. So what we have is mutation defines a lattice automorphism phi, basically given by piecewise linear function. So it's H times the min of UA for A in the vertices. So come back to the point that, don't raise last time, you know in the definition of N side mutation there's a lot to decide about it, okay. It's not really a map, it's just a process. You take your polytope and you size it up and you do some stuff and then you put it back together, okay. And it involves these little remainder terms that are there basically to make sure that you can do this. And it doesn't really matter what remainder you choose so you'd think it would be nice to remove them from the definition and all that sort of stuff. And here we have this really nice description. It's only involved in my factor A and my H. There's none of this remainder junk. It is just a piecewise linear map, you know. And you say why didn't I just use this? And I suppose for some purposes that's really useful. Of course we're thinking about Laurent polynomials so it is more natural for us to think of what's going on in M. And there's another issue too. I don't know if anyone actually knows the answer to this. Of course I can just take any P-dual. I can take any H and any A. I can apply this to it and I can end up with a new object, right. And maybe throw away the cases where it's not convex but you know I can end up with a new convex object. And when I dualize back is it still a lattice polytope? Does it still stand any chance of being the Newton polytope of anything? And I don't know how you can tell that just from the data that you have in M without doing the whole dualization process and so on. So all the hard work is done in the N side definition basically to ensure that when we apply this we still end up with a convex lattice polytope in N. So this acts linearly in each of the chambers given by the inner normal defined by A. It's nice. It's explicit. It's really concrete. For example, we're not making this a true example. Let's just suppose A is equal to some triangle. And this is inside H. Then we want to draw what this chamber decomposition looks like in M. So I'm going to rather cheekily draw my triangle in M where it doesn't belong because it does lie in N. And I'm going to figure out what it's in the normal. So of course the normal vectors would go like that, that and that so I just need to take the negation of them. So I'm going to end up with a decomposition that looks something like this. And again maybe I think it's perfectly fair for some people to get a bit upset that I said that this was a fan and these aren't strongly convex cones. They all have at least one, a one-dimensional vector space in them given by. So here at the middle, here, I thought it was a different colour, is the linear space spanned by H. And then within each of these chambers, let's just call this one sigma A, this chamber here, well it corresponds to this vertex A. So the chambers of sigma A are just in objective correspondence with A in the vertices of A. And in each one, I don't need to do this, it just looks like this. And so this is a shear transformation. I just mean it moves in the direction of the vector H. One over once will stop and next time I'll give some examples.