 I see that some of us survived the weekend. This wants to be very elementary today, and everybody should understand, and please interrupt me if there is a problem. I mean it. I want to do a little introduction to torque geometry, but I want to start from an unusual end, namely from this definition, and the etotic variety is a geometric quotient, whatever is a quotient. Well, let me just say quotient. Cm modulate torus, where the torus acts on Cm, a representation, c star, the m. So there are various references for this, but perhaps the most complete and self-contained is in the book Cox, little. And the shank, the torque varieties. And the main reason for me to quote this book is that the stuff there is in chapter 14 and 15. So this book is about a thousand pages long, and chapters 14 and 15 are around page 800 and so. And you know, like with every book of that length, it's quite a chore to get to get there, let alone reading the stuff. I mean, it really is a chore. It's sort of curious because, I mean, that's the theory of the secondary fan that we're talking about here. And then one gets to the secondary fan in the third jump to usually when you are just already completely exhausted by everything that came before. And then it seems incredibly complicated because it needs to be tied with everything that was done before. But it's really sort of a curious accident. I don't know the history of this at all, because as most working mathematicians, I don't know any history of maths. But still, let me make a statement that is completely unmotivated. It's just my wish. It's an accident of history that torque varieties were discovered from the end of the primary fan and that the secondary fan would be discovered much later. In fact, you could have started from the secondary fan and discovered the primary fan much later, and that's the kind of introduction to torque geometry that I would like to start giving today, just the first few steps. So let's see how it would work if one were to do it from this end, which it hasn't really been done before. So for me, I will think of T as being a coordinated towards C star to the R. And then the representation then corresponds to a matrix D from Zm to Zr. Where here is that M and Zr, I think of them as group characters of C star to the N, C star to the R. And so this thing here is dual to a group homomorphism C star, the R line to the C star. There are some assumptions that you may want to put on D. You don't need to put them, and sometimes you do sometimes you don't. So for example, T acts faithfully and that corresponds to say that the rows of D span a saturated doublethys inside Zm dual. So, yes, yes. Hold on, hold on, hold on. No, no, no, no. That's not what I'm saying. So the dual is torsion-free. Yes. Oh, yeah, yeah. Well, that's a different star. This is a little cross. That's a little stuff. It's the concurrent of the dual that I want to be torsion-free of the dual map. Z is a one-dimensional domain, and so it's not totally straightforward because there is X to 1 turning up. So, you know, don't worry too much about it. The action of the group is faithful, okay? Oh, yeah, yeah, yeah. But, you know, don't worry too much about it. You do because this one is a... Okay, I'll tell you a moment. R times M, and so we're talking about rows of rows of M columns. The columns are these D1 up to Dm. Column vectors. Column vectors. I don't think so. I will do some examples with numbers, and so, you know, in the end, this will be completely clarified. But... Zr is column vectors. These are the column vectors in Zr. Di equals D of the i-th vector. Sometimes I don't choose a basis of this Zr, but I always choose a basis of this Zm. Yeah, yeah, yeah. I'm pretty sure we're cool. Or equivalently, you could say that the hcf of the set of r by r minors of D equals 1. And then, a further assumption in addition... Well, it would be clearly an addition. T, moreover, acts faithfully on old divisors. So there is a lot of abuse in notation. Well, let me just not do it then. Try to... On x, i equal 1. There is a quote of Andre Veil that I thought about for you, Don. He said something like, it is impossible to discuss mathematics without abusing notation. At least a little bit. But maybe he doesn't. So, you know, that corresponds to the fact that then... This assumption here needs to be satisfied for all this matrix D i hat, which is D 1 with the i-th column removed all the way to D m. Satisfy all D i hat, the above. I will not have the time... Yes? I will not have the time to discuss this, what it means and why, you know. But essentially, these two assumptions are equivalent to... To ask these two things is equivalent to want to... To have the same class of objects that are the standard, the tautic varieties. Then if you don't assume these two, then you're looking at something which is slightly more general than tautic varieties in the good old sense. And... Okay, so when these two assumptions are satisfied, we say that D is well-formed or the quotient is well-formed or whatever. And there are some algorithms. If something is not well-formed, if a matrix is not well-formed, you can transform to the one that is. And what exactly is the difference? That when you take the quotient is something that one can discuss and I will not discuss. Okay, so it turns out that the quotient depends on the choice of a tillinearized line bundle. There are two types of men, the Koch people and the Pepsi people and the HCF and the GCD, okay? I'm an HCF kind of. So usually I use this notation here. C is the cone spanned by the column vectors of D. This is a cone in R to the R, okay? And the quotient depends on the choice of an element of C, an integer element of C. That's equivalent to a tillinearized line bundle on Cm. And then the group of sections. So this is supposed to be the group of sections on the quotient, which we don't even know what it is yet, but still we can talk about the group of sections. It's then going to be the polynomials and variables, which behave nicely and equivariately. So for all A in Cm, for all lambda in T of lambda A equal chi of lambda f of A. So chi homogenous polynomials. And out of this thing, you can form this at stable points or rather semi-stable points. Let me just call it u chi. And that's the set of all points A in Cm, such that there exists an n larger than zero and an f in gamma, such that f of A is non-zero. That's the set of semi-stable points in Cm. And sometimes helpful to look at the unstable locus, that chi equal Cm minus u. That's a closed subset. Zariski closed in Cm. And these are the unstable. So let's do the first example. So we take d to be this matrix. 1, 1, 0 minus n. n is a positive. I think of n as being a positive integer. In other words, I want to take the quotient of C4, mod C star squared. And the action of C star squared, it's just a lot of minus selfs. What this matrix encodes, it's encode in action C star squared. We're A0, A1, A2, A3. So the first C star, that's the first row, that maps to lambda A0, lambda A1. A2 stays there and then lambda to the minus n A3. And the second C star, so let me put lambda and mu as coordinates of this C star squared, will send that to A0, A1, mu A2, mu A3. Okay, so that's what that matrix is telling us. We're supposed to take the quotient by that touch. If you've never done it, it's good to do it. As an exercise, to try and see how the set of stable points and unstable points depend on the choice of your chi. So here you plot the columns of D in Z2, or maybe R2, and it sort of looks like this. So these are the columns, D0, D1, D2, and D3. So D0 and D1 are the vector 1, 0, so that's D0 and D1 there. And then D2 is the vector 0, 1, so that's D2. D3 is the vector minus n, 1 somewhere here. So the cone spanned by all these vectors is this cone here. And it turns out the picture itself divides itself into two subcones. So C1 here and C2. So that's C2, that's C1. And for me, C1 and C2 are open cones. Oh, I see. But I did introduce this notation earlier last week. And for me, this hooked notation is the cone spanned by those vectors inside. Do you want a plus? No, no, no, but for you, so we should have the plus here. Okay, no, no, okay, no, but I'm happy to do that. That's cool. Whereas you would, without the plus, it would be the vector space span. That's good, that's good. I wonder for a long time what I should do about this. Now I know. So for me, these C1 and C2 are the open cones. And so there are two cases. Well, several, but I mean, let's just at least look at the open cones. If C is in C1, okay? If chi is in C1, then an unstable look. Z chi is the union of X0 equal X1 equals 0 and X2 equal X3 equals 0. Yeah, in C4. And the stable locus is therefore C2 minus the origin times C2 minus the origin. And the quotient, oh yeah. So this is, this I didn't say, the torque writing question F, the quotient, then it's supposed to be a bona fide, okay, whatever that means, quotient of the stable, say stable locus by the torus T. And so here the quotient F, u chi mod C star squared is covered by four charts. Maybe I call them, well, why not? Not two, u not three, u one, two, u one, three, mod, these guys mod T, okay? Where u ij is the set where X i and X, and X j are both different from zero. This is the quotient which people call the surface Fn, the segre surface Fn. It's nice not seeing your surface. Something very interesting happens if your stability condition is in C2, okay? Then if you look at that, then the unstable locus is going to be X not equal X1, equal X2 equals zero, X3 equals zero. You know the stable locus u chi then is C3 minus the origin star times C minus the origin. So maybe, okay, so maybe I want to call this u one and I want to call this u two, okay? Maybe not. I don't necessarily want to do this comparison. Sorry about this. I'm just trying to describe what I'm trying to... So you see, I'm trying to say here is that when I take the quotient of u chi mod this C star squared, I can actually view this as just the first C3 minus zero mod just one C star. And so if I have a naught, a one, a u chi, then I can use the second action here, the action of the second copy of C star multiplied by mu equal a three to the minus one. And then I can send that to b naught, b one, b two equal a naught, a two divided by a three. Conversely, I can think of this as embedded in here as b naught, b one, b two, one. Then if I act with the first copy of C star, then that gets mapped to lambda b naught, lambda b one, b two stays the same. And then I'm supposed to do lambda to the minus n, which again by the first action then I can act with mu equal lambda to the n now. Sorry? Yes, yes. The lambda b zero, thank you. Then that's equivalent to lambda b zero, lambda b one, lambda to the n, b two, and then one. So you see this quotient is then the same as this quotient where now C star acts with weights one, one, n. And so here f, the quotient is none other than p one, one, n. So if you're in the chamber C one, the quotient is the surface f, n. And if you're in the chamber C two, the surface p one, one, n. I did say for me n was a positive integer. So you don't need to assume anything completely generality though. I usually assume that C is a strict cone, sort of yeah. So usually you want C to be a strict cone. In other words, it contains no straight line. I told you about what happens when chi is in the open chamber C one and C two. Now there are some boundary cases, these three lower dimensional cones. And there are various discussions to be had there. And let's just ignore that. So anyway, that's just an example. If I were to develop the theory of torque varieties from this definition as a quotient, then the proper treatment will have to start from a discussion of stability conditions and the discussion of different quotients that you can take if you vary the stability condition. And that would start by doing this bit of theory here. So wall and chamber decomposition of the cone C. So this wall and chamber decomposition is the thing that people call the secondary fan. And so the walls, the cones, those cones, the I1, the IK, have co-dimension one, R to the R. So in fact here, therefore, we could take K to be equal to R minus one. In this example, the walls are going to be just these three lines here. Who dimension one cones generated by your vectors? And the chambers are the connected components of the complement of the walls in C, in the cone C. Because the co-dimension one cones will be generated by R minus one vectors. It's possible, but then that would be the same. So anyway, sometimes try to be too clear when it's confusing. Yeah, indeed, absolutely. And then I'd have to prove a number of things just from scratch right there at the beginning. I'd have to say that I can make an ideal. I, chi. And that's the ideal generated by monomials. X, I1, X, I, R. Oh yeah, I have to choose a chamber, see? A chi in that chamber. And then I would associate the ideal generated by monomials like that. Where chi is in the cone spanned by the corresponding vectors. This is an ideal in the ring of polynomials. I'd have to define Z chi with a variety of this ideal. It's just a union of coordinate sets. And then I would take the quotient. I would have to show that there is a reasonable bona fide quotient of Cn minus Z chi mod T. I'd have to prove that these three objects only depend on the chamber and not on the choice of chi in it. I would want to argue that f chi is an orbifold. And that it is an orbifold and it's covered by affine charts. And these affine charts are parametrized by simplicies that contain the chamber. And for each one of those simplicies, I have a corresponding open chart. U I1 IR mod T. Where that chart U I1 of IR is precise to the chart where the corresponding coordinates X I1 is non-zero. X IR is non-zero. And I'd be insisting that though I'm dividing that by an r-dimensional torus, which is a connected Lie group, nonetheless that thing is an orbifold. And so I'd be saying that that chart U I1 IR mod T is in fact V I1 IR mod mu. Where V is the set where all the corresponding coordinates I1 up to IR are equal to one. And mu is a finite subgroup of the torus. In fact, f chi is projective. If C is strict and f is projective, the chamber C is canonically identified with the ample cone of f. So if I were to give a proper treatment of torus varieties from this point of view, I'd have to prove all these things right there at the beginning from scratch. Whereas if you read the book, Cox Little and Schenck, they prove those things by using everything that they've done in the 14 preceding chapters. So this is meant to illustrate that this is just, you know, one of these things is basically like a glorified projective space, yeah? You know, projective space Cn plus one minus zero. Zero is this thing here. Well, there is only one subgroup condition there because modulo C star. It's covered by open charts, parameterized by these xi. Different from zero, modulo C star. And that's the same as xi equal one, modulo nothing. In that case, this mu is nothing. And that as you're used to working projective space by taking local charts and looking what happens, you can do it on one of these things. It's just a bit more laborious, but not in principle any different. So I want to give another example. I show you these charts, some genuine or before the example now. But I want to conflate that together with an example of a complete intersection, a torque variety. So let me then tell you just a bit further than this and tell you a bit more about complete intersections in torque variety. So, you know, fix a torque variety, namely an action together with stability condition now. And now let's talk a little bit about complete intersections. So here you choose in addition a few more characters, chi one up to chi C in z r. And then you take functions f i in gamma chi i and then the complete intersection is going to be the variety of those functions f one equals c equals zero. That's a subset of cm. So I am supposed to also remove the unstable locus z chi and then I divide all this thing by t and that's a subset of f. And that's what the complete intersection is. And so there are some definitions that me I'll just put them here even though I won't have a huge amount of time to I'm going to say that x is quasi smooth. Either the variety of these guys is all contained in z or in fact is this variety minus z chi is smooth. And you know, this this lives in cm minus z chi of the expected co-dimension of co-dimension. And then in addition, I'm going to say that x is well-formed for every stratum s in f, non-trivial stabilizer. If that stratum is contained in x, then its co-dimension in x is at least two. Okay, sorry about this part of these exercises to at least give this definition. Yeah, yeah, of course. And we had these two conditions at the, you know, maybe I shouldn't let me not even try to say that. You definitely most of the time want to work with well-formed complete intersections because otherwise certain things will go wrong. So I'm going to now discuss the first problem, a problem in the example sheet and I'll sort of set it up and so here's an example. So consider this data for a torque complete intersection. So first I write the matrix of the action. So this is my matrix D. So, you know, D1, D2 all the way to D6. And then I put a vertical bar and next to it I put two characters, two for 0, 1. Okay, so these are the two line bundles. Okay, kai 1 and kai 2. And so the tasks are, what's the, the exercise consists in fix stability condition to ensure that X is a final object. So by the way, so this is C6 mod C star squared. So it's a torque fourfold and inside, you know, I have two line bundles. So they're supposed to be a surface. Then find all singularities of X and make sure that X is a well-formed orbital and maybe compute kx squared, the degree of the canonical class. Okay, so let's try and do all these things. You know, you plot your vectors D1 up to D6. So D1 is here. D2 is the vector 1, 1. So that's right there. D3 is the vector 1, 2. That's up here. D4 and D5 are the vectors 0, 1. And D6 is up here. So you see there are 1, 2 and 3 chambers. Those are the, those are the, the walls and there are 3 chambers. So the anti-canonical class of X, so let's go, let's get down to the first task. I want to fix the stability condition such that X is final. I want to choose one, one of these chambers. The anti-canonical class of X is, you know, minus kf minus l1 minus l2, restricting to X. And minus kf is the sum of all those Ds. So 3 minus 2, that's 1. And then here have 5 plus 3 is 8 minus 5, that's 3. Okay? So the vector 1, 3 is somewhere here. And so if I want, remember that the chamber then is going to be your ample cone. And so if you want that to be ample, you better choose this chamber here as your stability conditions. Okay, so that's your chamber, c equal amp. So that's task 1. So this thing has charts. Uij mod t is in the set 1, 2, 3. And j is in the set 4, 5, 6. So we have 9 charts. And in this game, you really have to check them all. Sort of let me demonstrate. So some of them are not as interested in some other charts. So I want to look at the chart now. I want to look at U16. So mod t. So U16 means x1 different from 0, x6 different from 0. Okay? So you just sort of look at this action here. And in order to see this, so you put both of those non-zero. And then you see I can use the first group action to set the first coordinates to 1. And then I can use the second group action to set up the sixth coordinates to 1. But there will be an ambiguity there of a mu3. I had to take this cube root and there is no unique way to do that. So in fact, this is the same as x1 equal x6 equals 1. Modulo and mu3. Okay? So in fact, this chart is the thing that I write like this. I write a third. So I put x1 equal x6 equals 1. So on this chart, I have coordinates x2, x3, x4, x5. And then how does that mu3 act? Well, you see that action does preserve the fact that this guy equal 1. And so essentially that mu3 acts. Think a moment about this with weights 1, 2, 1, 1. So here I write 1, 2, 1, 1. So my notation for these things is that I write 1 over r, a1, an to mean the quotient of cn by mu r. Where mu r, the group of rth roots of unity acts with weights a1 up to an. So you see in this chart that your torque variety is a genuine orbital. It's not a smooth thing. And you have to look at nine other charts. And I don't have the time to look at many of these charts. But let me just say this is what happens in this particular chart. So I want to study a little bit the complete intersection now. F1. So F1, remember that's the section of the first line bundle, chi1, l1. And F1 contains, let me write some monomials that F1 contains, that F1 can use. So in chi1 I have to be in 2, 4. So for example, x1 to the fourth times x4 squared. That's in the correct weight space. This guy to the fourth is in 4, 0. And x4 square is in 0, 2. So together in 2, 4. Let me make a list of some monomials that F1 can use. So not contains. Can use. May use. The following monomials. May use. x1, x4, x1 squared, sorry, x1 squared, x4 to the fourth. x1 squared, x1, x5 to the fourth. x1 squared, x4, x6, and so on. Incidentally, chi1 is the vector 2, 4. Sorry. This line bundle is on the boundary of the cone. It's right here. That's where l1 is. It's enough. It's enough bundle. Okay. But notice here that. Okay. So the first thing to note is that no pure monomial. I'm sorry. I'm going to have to overrun for about five minutes. Hopefully that's all right. There is no pure mon... So here, yeah, perhaps I should have said, I want to study x intersection u16. Okay. I studied the complete intersection x in this open set here. v16 mod mu3. What happens in this open set? So I do not have any pure monomial. F may not use any pure monomial in x1 and x6. And you see, to be in this chart, you have to put x1 equal x6 equal 1. So what that means is that somehow when I look at this chart, there is no constant term. And so the center point of this chart, the origin of this chart, let me call it little x16, must lie in the variety of f1. There is just no way out. So even though it's an f-line boundary, it has this base point in its base locus. And so, you know, Bertini, non-Bertini, we have to verify that the thing is quasi-smooth and well-formed. And so, but fortunately, we can use, may use a monomial of the form x1 to the a, x6 to the b times the linear term, xj. Some other j. For example, x1 squared, x4, x6. And this implies that the thing is, well, at least the variety of f1 is quasi-smooth. You know, it's f1 equals 0, constant term 0, but the Taylor expansion contains a linear term, so it's a non-zero gradient. And so, you know, by some version of inverse function theorem, this thing is quasi-smooth. Here I'm looking at the complete intersection, though, of f1 and f2. And so, we have to make a similar analysis for the other, the other line bundle, the line bundle L2. And, you know, that's a pretty strict requirement there. So, f2, these are just, have just two sections. So, this may only use two monomials. May only use the monomials x4 and x5, but that's okay. And so, we have a similar analysis to this. And so, when we push this to the end, then the center of that particular chart, the point x16, has to lie on the complete intersection x. But you see, f1 can have linear term x4, and f2 can have linear term x5. And so, using the inverse function theorem analytically, locally at that point, I can solve for x4 and x5. And so, then x6 equals 0, I solve for x4 and x5, I am left with x2 and x3. And so, in fact, x itself is quasi-smooth, well-formed, and has a singularity of third, one, two, at that point. That just finishes the discussion of this particular chart here, u16. So, you have to study all other eight charts. You really have to do it. And the conclusion is going to be that x has two singular points, one of these four, a third, one, two, and the other one is a third, one, one. And so, at the moment, we do not automatize this dusk, we have to do it more or less by hand, okay? You don't do this by computer. There isn't a difficulty in principle, but this is looking at all the charts and making doing this analysis is a lot of work. Okay. How to compute kx squared? Okay, so, instead of doing the computation, let me sort of... Okay, so this, you know, once you've done all the nine charts, this will verify that all the singularities are computer all the singularities and that's a well-formed orbit. To compute kx squared, let me lay out some principles. The Chao group of f of the Tartic Variety f, at least over the Rationals. Well, again, you know, you have to take this seriously. If you want to develop the whole theory of Tartic Geometry from this point of view, then you have to do everything including homology, homology, Chao group, and everything without talking about fans. And so the Chao group is generated by the divisors with a bunch of relations and the relations in questions are just di1, diR equals zero, i1 times, you know, times. These are c dots. diR equals zero when your chamber is contained in the cone di1, diR plus. That's your Chao group. It still doesn't tell you how to compute degrees. Am I saying this right? Yes, I think I am. But then you can also write other things. So if c is contained in this cone, di1, diR plus, okay? But these guys are in, you have to think of them as living in, yeah, you're right. So in other words, they are in dead R. So let me say this one is a bit more important. The degree of the product of the dj, j different from iK is one over mu, okay? Where mu is that finite group such that this already corresponds to some chart, v i1, iR mod mu, and mu is the order of that group there. And that's because those divisors will only meet in this chart, at the center of it, and then that's going to be the degree of that intersection. So this is not in terms of the chart group, but also the degree function. And if you use this, you have to practice a bit, but maybe in the problem class we'll tell you how to do it if you're interested, if you're not then. So you compute that kx squared equals 10 over 3 for this surface. It's perfectly beautiful, good, nice, orbit fold, well formed, and it has that degree. So you know, maybe next time, I don't know whether I'll do next time, but I will also tell you how to connect this treatment of tautic varieties with the standard treatment. Sorry for keeping you so long.