 Okay, well, so welcome. Yes, let me introduce you to Dr. Lisa Markant from the Stony Brook University. And she's going to talk about evolutions of a cubic four-fold. Yes, thank you very much. So first of all, I would like to thank all the organizers for organizing a really good conference so far. I'm looking forward to the rest of the week. And again, thank you for inviting me to speak to that. Yes, I want to talk today about cubic four-folds with what's called an anti-symptactic evolution. So before we dive straight in, I just wanted to kind of talk a little bit about why I'm talking about cubic four-folds in a conference about high-scale asphalt. Okay, so cubic four-fold is a final four-fold. So if we look at the Hodge decomposition of the middle co-amulgee, we see that it has Hodge numbers 0, 1, 21, 1, 0. And to anybody who's familiar with K-3 surfaces, this would look very, very familiar. It's very similar to K-3 surfaces. So in fact, there is a very close relation to K-3 surfaces. And the Bo-Villain-Donaghi proved that if you take a smooth cubic four-fold and look at the final variety of lines, then this is in fact a hypercaler manifold and its deformation equivalent to one of K-3 two types. So the Hilbert scheme of two points on a K-3 surface. There are other various constructions that one can do from starting from a cubic four-fold and to get a hypercaler, but I don't really want to talk about them today. So why is this this relation to K-3 surfaces kind of interesting? Well, if we start with our cubic four-folds and we vary it, well, cubic four-folds have a modular space, a 20-dimensional modular space, whereas the modular space for K-3 surfaces, projected K-3 surfaces is a 19-dimensional. So we seem to get more examples by looking at the final variety of the lines on a cubic four-fold. So more examples of these K-3 two-type hypercaler manifolds. So I'm interested in automorphisms, particularly involution. So why is this kind of interesting? Well, from the point of view of studying automorphism, this leads to what's called exotic automorphism. So we might be able to find automorphisms of hypercaler manifolds that are not induced from a K-3 surface. Exactly. And if we're looking at automorphisms, we say that an automorphism is symplectic if it induces a symplectic automorphism on the variety of lines on that associated hypercaler. And so indeed, like following this kind of idea, in 2019, Laverin Zeng classified all possible groups of symplectic automorphisms for cubic four-folds by using the period map and some lattice theoristic methods. Okay. And so of course by studying these symplectic automorphisms, this gives us information about the symplectic automorphisms on the variety of lines. So now I'm not really going to say anything about hypercaler manifolds, and I'm just going to focus on cubic four-folds. And in order to study automorphisms, we're going to need some lattice theoristic methods. So I just wanted to kind of brief reminder of some of the lattice theory that I'm going to be talking about today. And what is a lattice? So an even lattice is a free, finitely generated V-module, a quick with a non-degenerate symmetric bilinear form. So we say the lattice is even if the self-intersection of any element of the lattice is even, and it's odd otherwise. So what's going to be important is this discriminant group of a lattice. So you take the dual lattice and you mod up the lattice. And this is going to be a finite abelian group. And so I'm probably going to mention these ADE laxes. So in this talk today, these ADE laxes denote the standard positive definite root laxes. And you hear is the hyperbolic line, the hyperbolic line. We say that a lattice is unimodular, is this discriminant group I just defined, is the trivial group. And we say that it's two elementary, if this is a two elementary abelian group. So the mod 2z is just on power. Okay, so how can I use lattice theory to study automorphisms of a qubit four-folds? Okay, so let's take notation. So this x, the row is going to be a smooth qubit four-folds, p5. And we're going to focus our attention on the middle column, h4. And so if we consider this with the natural intersection pairing, we see that this is actually an odd unimodular lattice of signature 21-2. So from a point of view of lax theory, odd laxes are not so nice to work with. We prefer to work with even. So if I denote by h squared, the square of the hyperbolic class of x, so x always comes with the hyperbolic class because we're inside of p5. We have a natural polarization. And we just take the square, this will live in h4. And so we can consider everything that is orthogonal with respect to the natural intersection pairing on x to the square of the hyperbolic class. So this is called the primitive property. This is an even lattice. It's a longer unimodular, but it's still even of signature 22. And it carries a polarized hodge structure of K3 type. So we have hodge number 0-1-20-1-0. Okay, so in fact, as a lattice, we can say a bit more, we know exactly what this is is the isomorphic two. So it's isomorphic to two copies of E8 plus two copies of the hyperbolic plane plus a copy of A2. So this A2 here has discriminating group Z mod 3Z. So this whole lattice has discriminating group Z mod 3Z now, rather than being unimodular. Okay, so once we have this lattice, we're going to follow kind of the ideas behind the period map of K3 stuff. That's kind of what's motivating this discussion. And so following kind of this story, we can define the period domain. So we take the complexified lattice and we projectivize it. So we just look at this set here. So this is the period domain for cubic fourfold. And if we take the global monotony group, so this is a subgroup of esometry of the lattice that preserves this domain and also acts trivially on this group, this on this discriminating group. So this group acts properly discontinuously on this domain. And so we can take the quotient and this is actually a quasi-projective variety. So this is called the global period domain for cubic fourfold. Okay. On the other hand, we have the modular space, the classical modular space of smooth cubic fourfold. So we can construct this using GIT. So we're just taking the smooth cubic fourfold here. And this is a quasi-projective 20 dimensional variety. So this is also 20 dimensional. And how are they related? So Wosan first showed that the global terrarium is true for cubic fourfold. So the period map that takes a cubic fourfold and associates to its hodge structure. And what the lack of hazard, laser and ojenga, worked on showing what the image was of this period map. So this period map, cubic fourfold, is an isomorphism from the modular space of smooth cubic fourfold onto its image inside of this period domain. And what's its image? Well, it's everything apart from this union of two irreducible devices. I don't want to talk about these devices too much, but one can think of this C2 as parameterizing singular cubic fourfold. And this C6 is the generation of cubic fourfold. So what does this give us? This gives us a way to associate to a smooth cubic fourfold its hodge structure. Okay, so in fact, slightly stronger is true. So I think this is more recently, maybe 2019, Zheng, this is due to Zheng. He proved the following strong global terrarium. So we have, if we let X1 and X2 be two smooth cubic fourfold. And we assume that we have an isomorphism of polarized hodge structures between middle co-homology. Then we can say that this is actually induced by an isomorphism of the cubic fourfold themselves. So in particular, we have this isomorphism of groups. So on the left hand side, we have the automorphism group of a smooth cubic fourfold. And on the right hand side, we have hodge structure isomorphisms on H4 that fix the polarization. And this is exactly going to give me an automorphism of my primitive co-homology. And so what we can do is we want to study automorphism of X, we can just study instead automorphism of the primitive black, which is essentially a lattice theoretic question. Okay, so that's what we're going to do today. So instead of studying automorphism on X, we're going to study the automorphism on the primitive co-homology. So L here throughout is going to denote the primitive co-homology. So this was an even marker. So whatever, I'm going to focus a little bit more on involutions now rather than automorphism, we're going to zoom in on on to on to involution. So if we have an involution of a lattice, then we can take it determines two eigenspaces that is L plus here is what's called the invariant sublassus. In the primitive co-homology that is fixed by my involution. And this L negative is the co-invariant. You just take the orthogonal to L plus with respect to the intersection pairing. Yeah. Okay, so if we take the direct sum, this is a sublassus of the primitive co-homology. And if we take this kind of quotient, we see that we actually get a two elementary group. This is kind of a bit technical, but this is what's going to allow us to study these. The automorphism. A bit. Lastic theoretically. So here I want to be a bit more precise. So if we look at middle co-homology and we with the Z coefficients and we intersect with the H2 two. This is what's called the algebraic sublassus of X. So classes inside of here are going to be classes of hodge classes. And so they're going to be surfaces that are contained in my qubit profile. On the other hand, I can take the orthogonal lattice to this, but it's inside of the co-homology and this is going to be the transcendental lattice of my qubit profile. And so I say that an involution of my qubit full-fold is plastic, it acts trivially on H3 one. So in other words, this is equivalent to saying that this transcendental lattice, the pathogenetic level, the algebraic classes is contained in my invariant lattice. On the other hand, I'm interested in anti-complexic involution. So anti-complexic, the transcendental lattice will be contained in the co-invariant sublassus of my qubit. And so this just allows us, along with this technical kind of fit about this embedding group, this just gives us a way to compute the rank and the signature of these lattices. So if we have an anti-complexic involution, so here the transcendental lattice will be contained in the co-invariant lattice, the L-negative. Then we can show that our negative is actually a two-elementary lattice, and it's indefinite, so signature R-2-2, whereas L-positive, L-plus, the invariant lattice is a positive definite rank two, rank, sorry, whatever the rank is, and this is discriminatory. But then my important part is this positive definite. Okay, so when I say I want to study an involution on the lattice, what am I really saying? I really want to identify these two lattice, I want to identify the invariant subclasses. And so if I have an anti-complexic involution, this L-plus is going to be contained inside the primitive algebraic classes. So if we take this lattice I define here, the algebraic lattice, and intersect with the primitive homology, L-plus is going to be contained inside of that. And so for a general cubic full-fold with an involution, this L-plus is going to be equal to the primitive algebraic homology. Okay, so now let's talk about the geometry of the lattice. So let's talk about the geometry of what's going on. So if I have a cubic full-fold, this is just defined by a cubic equation and five variables. And in fact, if I have an automorphism of such an X, I can actually lift it to an automorphism of the ambient P5. If we lift it to an automorphism of P5, we can apply linear changes of coordinates and we can diagonalize our involution. So it becomes very easy to write down what the possible involution of a cubic full-fold are. And so this was done in 2011 by Gonzales Aguilera and Leondo. I think they also did it for cubic three-fold and not just the involution. So what can we say? So if we take an involution fee of my cubic full-fold after a linear coordinate change, we have one of the following possibilities. So this notation here is just the diagonal matrix. So you have a bunch of ones and then a bunch of minus one, and it is acting on the coordinates X0 through to X5. And so from this kind of description, we see that we have two possible anti-symplotic involutions and one symplotic involution. So I'm mostly going to be focusing on this third anti-symplotic involution today, but I'll mention this one too. I'm not sure how much time I had left, but I'm not sure when I started. I think you're around at least 10 minutes, right? Okay, cool. Okay, so if we want to try and identify some invariant primitive classes, I should start by looking for some invariant surface classes that are contained in eggs. So the goal is to try and find some invariant surfaces that are contained in my cubic full-fold. So this description of the involutions enables us to write down more precisely an equation of X with these involutions, and it becomes very obvious that this X contains a lot of invariant planes. I don't know why it's not letting you change slides. Okay, so what can we do? So if we let X be a cubic full-fold with an anti-symplotic involution, but I'm going to focus on P3, but the same method works for the other anti-symplotic involution. Then we can immediately show that X contains at least 19 invariant distinct planes. It contains a lot of planes. So firstly, this P3, how is this acting? So it fixes the first coordinate, so X0 to X3, and then just act by minus one on the last three coordinates. And so if we write down the equation of X, we see that the X3, the X4 and the X5, they can only appear when we don't have any like cubes and we can't have them appearing by themselves in order for the equation of X to be invariant. And so these LIs and this G, these are all equations in just the first variable, so X0, X1, X2. And so, immediately we'll see that such a cubic is going to contain this plane. So in fact, this plane is fixed by the involution. And so once we get a single plane, we can consider linear projection from that plane. And so what does that give us? So if we blow up the plane, this gives us a quadricep vibration over P2. So how should we think of this projection? Well, this base P2 is parametrizing P3 slices of X that contains this plane P. And so if I'm intersecting X with a P3, I should get something of degree 3. And so the fiber here is going to be the residual quadriceps that you get when you intersect. And the singular fibers are going to be parametrized by a plane sextic curve in P2. And now what's good about this, the involution is that from this equation of X, you can immediately lay down the equation for P. And in this case, it splits as the union of two irreducible cubic curves instead of an irreducible sextic. And so if you look at a smooth point on this plane sextic, well, the fiber is just going to be a quadric cone. Whereas if we take one of the intersection points of the smooth of the cubic, this is going to have worse singularities. And because X is a smooth cubic, you can actually show that the fiber has to be a union of two planes. So we have nine intersection points, so we have nine pairs of planes. Plus the one we projected from. Okay. And so these, these planes are what's going to enable us to identify our invariant sublacris. Okay, so great. So we mentioned this C1, the other anti-simplastic involution. So this was already studied by Lazar, they use a slightly different methods to my method. But what did they show. So here they classify the invariant sublacris of the primitive cohomology. And in fact, they, they say a little bit more. So they say a cubic four fold with this involution, this is equivalent to having what's called an Eckhardt point. And so, they can show that this X contains a cone over a cubic surface. And if you take the lines on the cubic surface and take the cone over them, you get a lot of by 27 planes. So what did they show. So possible they contain these 27 planes, and then they, they, they show that this invariant sublacris to the algebraic primitive classes is isomorphic to an E6 to an assigned by exactly the differences of these planes. And I also have a nice description of the transcendental common g of X. So this will be the Cohen variant office of the classes which aren't invariant. And this is isomorphic to three copies of D4 plus two copies of new. Yeah, so what, what did I show. So we did the same thing for three. So, first of all, we have a similar statement we contain at least 18 invariant planes and they all intersect. The six planes. And again, if we look at the differences between the planes, they generate the primitive caramology along with an extra class which isn't immediately obvious. And so we don't have quite nice description but we have that this lattice is isomorphic to the unique positive definite lattice of right hand which is obtained as a index to overlap. And we also can describe the transcendental caramology of X here. Okay. So I don't really want to talk about the method of involve this I want to kind of focus on. So whenever we have a family of cubic four-fold, kind of a next question is to ask is about whether this cubic four-fold could potentially be rational. And one conjecture is that if the transcendental caramology can be embedded into the K3 lattice, or can be realized as the transcendental lattice of the K3 surface, then your cubic four-fold is potentially rational. So I think the conjecture is that it would be rational. And indeed, you can see that this transcendental lattice, this can be, this can be embedded into the K3 lattice. So there is those existing K3 surface with this transcendental lattice. So, so what can we, what can we say in terms of rationality. Maybe I won't say this. Okay, so this is the bullet point I wanted to say. So, first of all, we know that the cubic four-fold is going to be rational if it contains two distinct planes. We can use these two distinct planes which do not intersect. We can use these planes to construct a construct a map. In our case, these cubic four-folds with this intersection, they do contain lots of planes, but unfortunately they all intersect. And one way to see this is if we look at this vibration and if we had a distinct plane, this would give us a rational section of this vibration, which would mean it would have to intersect the fiber oddly. In our description of the, of our lattices and A plus the shoulder, this just can't be the case. So we just cannot contain two non-intersecting planes. So all our planes intersect. However, it is, it is actually rational. So it's kind of proven a little bit of a roundabout way. So what do I want to say, so if we look at these. So whenever we have a surface class with inside of my cubic four-fold X, which isn't homologous for computing to section, we can take the sublattice, which is just the intersection, which is just generated by the square of the high plane section and that surface. And this will have some determinant D. And so if we let CD denote the cubic four-folds with such a sublattice inside of that algebraic algebraic classes, then it is going to denote this is going to define a irreducible divisor in a modular space of cubic four-folds. This is due to happen if and only if these conditions on this determinant. And so the description of the algebraic lapis enables me to identify which devices, which half the devices that this family of cubic four-folds are contained in. And in fact, they're actually containing every single possible non-empty divisor. And in particular, this implies that they're rational. So lots of these divisors, for example, C 14 is the closure of the fashion. And it's known that every element inside of this divisor is rational. So we kind of get the rationality kind of free. So it's interesting to notice that I think it's been shown that the intersection of all of these half the devices is at least 13 dimensional. And so, if you remember these X's with this involution three, there was a 10 dimensional family. So it was a 10 dimensional sub family, I guess. And previously, I think it was known that the instruction contains only the format cubic, which is, which indeed has one of these involutions. But I think I'm probably out of time. So I guess I will help. Thank you very much. Is there any questions? Thank you, Lisa, for the talk. Is there any question? Maybe I have one question. Sure. What about the map giving the rationality in that case? I mean, is the one that gives the unirrationality or is another one? So that's a good question. So, because X is contained in the C14 locus, and it does not contain two distinct planes, one can show that it actually contains a quartic scroll, a rational nor quartic scroll. And I think this is enough to give the map unrationality. I think it's going to be birational to the symmetric products of this surface. But I haven't, it's hard to see it geometrically from the equations and stuff. I would like to try and find this surface inside of X more explicitly, but I haven't been able to. Okay. Thank you. Any other questions? Sorry, Lisa, so at a certain point you said that this discriminant locus of the quadric bound is the union of two cubic and then over the singular points of the discriminant locus, there are a couple of planes. But is this for every? In general, when you do projection from a plane, this sex-stick curve here will be like an irreducible sex-stick curve if your X is general. But the fact that it has this involution and forces this sex-stick to split into these reducible. I think this curve is always smooth if there doesn't exist a plane that intersects the plane that you're projecting from. But is it possible to realize all these cubic fourfolds as quadric bundles over P2 or not? No, so this is, in general, when you contain a plane, you need to contain a plane. Okay, thank you. Are there any other questions? Yes, I have a question. Yes, please. Who is talking? Sorry. So do we know the fixed locus of this involution? Do we know anything about the fixed locus? Yes, so the fixed locus is going to be this plane peak that I have here. And then it's also going to be a cubic curve. And the curve is going to be given by this equation G. So this bit G here is this is a degree three polynomial in X0, X1 and X2. And so if you take the plane X3 equals X4, X5 all equals to zero and intersect with this cubic, you'll just get this curve and that's also fixed. But that's the fixed locus. Okay, thanks. Thank you. Other questions? If not, thank you Lisa again for the beautiful talk. Thanks. Yes, we will have a break until 11.