 with the computation of, with the computation of the, of the degree over a Jacobian, a Jacobian, a generic Jacobian in, in A5. And I want to tell you just briefly about another curve, another, another fiber, sorry, another fiber coming from the modular space of cubic trifles, non-singular cubic trifles. So I defined yesterday, this is the, consider all the cubic trifles in P4, non-singular cubic trifles. This is modular space of dimension 10. So it's not so big inside of A5, but it's pretty interesting. We have a, a theoretical theorem that allows us to embed this in A5 by, by taking a cubic trifle associated to, it looks like an annotation from the other fiber, but is, is, is intermediate Jacobian. So I define intermediate Jacobian with the, from Hodge theory. We take the h12 group complex coefficients over the third homology group of x. So this has also is a lattice. This is a, a C5 vector space. This is a five-dimensional a brilliant variety, which a principal polarization coming from intersection the, the co-product in the homology, in the third homology. This is intermediate Jacobian. So it's, it's not so, for the, for the geometry is not, we are going to, not going to see much about that. We are going to all work with the cubic trifles. Okay, intermediate Jacobian. So we, but we are identifying the, with the, with the, yeah, this is popularized, a video about it. Okay. Okay. How do you construct? How do you construct a covering in, in R6? So you, what you need, take a cubic trifle together with a line, a line inside it. Okay. And consider the projection. So the cubic trifle is inside before. So given this line, you can consider the projection over this line. It's a rational map into P2. Okay. This is a projection from the line projection from L. Okay. So this is not well-defined on the line, but you can always blow up the line x. And this is, this is now well, you get a well-defined map by, I'll call it pi also. And this pi turns out to be, is what we call a conic bundle of P2. This map is a conic bundle. So, but, but that I mean that the inverse image of a point in P, in P2, you take a point, you have two options. The fiber, the fiber, the fibers is one-dimensional fiber. And the fiber is either a conic or you have two options, a conic or two lines intersecting one point. So it's also degree two. So at the generic conic, two lines, the union of two lines, two lines. Let me put this. So conic or two lines. Okay. Of course. So we are going to do the same, the same idea as we did with the quintics. We consider the locus on P2 of the points where, where the conic degenerates. The conic degenerates. And this is the discriminant locus again. So the discriminant locus, locus of this, of this conic bundle is well, all the points in P2 such that this is this inverse, inverse image, a pair of lines, pair of lines. And let's call that C. And this C is again a quintic, a plain quintic. Quintic. Now, do you have a plain quintic in P2? Yeah. I like to draw P2 because the few things I can know how to draw, to draw. Then a quintic. And the quintic, you take the class of a line. Yes. It's a different line now. It intersects five points. And it turns out that that's right. I will tell you something about this, this, this line. The thing is, you can now define in the, okay, let me tell you something more. The funnel in the, in the space of all, all the lines, the funnel surface of lines of a cubic three-fold. Let's put it fx is the set of all lines inside of x. It's a line. So it's a surface, funnel surface of lines. And in this funnel surface of lines, you have in particular the, the fiber of the conic bundle over the discriminant blocks. Okay. So here I'm going to define the covering. So I'm going to define the covering C tilde over C as the set of the two lines, the lines in the funnel surface such that this line intersect. Okay. Now let me put names. I fix a line for the block. Let's call it this L0. Okay. L0. Maybe I can give a drawing here. So you have, you have L0. And from here, you, you, you project your cubic three-fold. This is somewhere there. And how do you project? So imagine that you are already in x tilde. Okay. And you are over, over a point on the discriminant locus. So the fiber of this discriminant locus is a pair of lines, but this pair of lines, since, since, since we are talking about the projection, this pair of lines lives in the, in the same plane generated by well, the pair of lines and this LC. So you have two pairs of lines intersect and L0. So let's call it L prime, L and L prime. Okay. And these two have the same image, the projection that goes to P. You project PL0, you go to P point in the, in the, in the corp C. Okay. Another color. And this is just two lines as the inverse image of this P. So by this procedure, I have, so I just can define C tilde as a set of lines in the final surface such that they intersect my given L0 line. So those are precisely the lines, the lines of, of, yeah, of the degenerated conics and the conic bond. Okay. And this is by construction, this is, well, I call it P is not, not a problem. And it's also a two one. And it comes with a natural involution, of course, Utah. And this is, this is the associated, you have a plane, Quintic. Okay. Quintic. And every double cover, a double covering comes from, comes from two torsion points. So this is related to a two torsion point on C. And it turns out that this two torsion points is an odd torsion point, odd in the sense that when you take the sections of the line bundle, it tensor with this L that I define here. So, yeah, well, okay, I say this is the line, but this is more precisely is, I should say is, so you call it that fee is the pullback of the class of a line. Okay. May I ask, so is this discriminant Quintic in general as well? Yes. Good question. It's in general smooth. Yes. That's true. That just has to be shown. Yes. You can, you can expect the, yes, general smooth. Thank you. Yes. I'm wondering about when this can degenerate. You don't have double lines as fibers. No, no, I don't have double lines as fibers. Precisely, because I wanted the anecdote over here. This is it. Yeah, they don't have ramification points. So in general, yeah, let me write it here. Smooth in general. And what I want to say is that it's all these, in the sense that this is always with the parity one. So the number of sections of this tensor, this tensor is, is even, is always odd, not odd number. Okay. That is, that is what they mean by the way. So that, that, that takes us in the, in the other component of the Quintile, of the coverings of the plain Quintics is of different nature. So, and moreover, well, okay, this is something I'm not going to prove. But this is my proposition to move in, I think, is that the prime of this constructed double covering is a somorphic to the intermediate jacob. This is a very beautiful. And, okay. So what they wanted to, I wanted to, to study now the fibers of over this, this, this, over this intermediate jacobian. Observe first that by construction, so to construct this double covering, what I need is a pair, is this pair, X and a line. So if X is fixed, if the intermediate jacobian is fixed, or the cubic threefold is fixed, you have such a double covering by the choice of lines. So in this sense, you can identify the fiber over any fixed intermediate jacobian with the surface of finalize. So it's two-dimensional. Just choose a line, choose the line inside of X where you want to project. Okay. So let's, let's look at them at the map. So you have, let's see, R6. Angela. Yes. It's obvious that if you change the line, then the, that the covering will change. Because you say isomorphic, so it's. You can identify. No, I agree with you. It's not adobius, but it's true. It's not, you get non-isomorphic coverings. Yes. Yes, yes. No, I am making a lot of shortcuts on this, but it's, yes. Yes, this is, maybe, maybe the construction will be more clear. A5. So this is P6. And now we are going to blow up on Rc. So Rc remember is the double coverings over being fixed with all theta, all theta torsion point, all theta torsion point. This is the old ones. And we blow up P6 over, over these locals. This is exceptional divisors. And the image, by this construction, the image is in, in the locals of intermediate Jacobians that I identified with the cubic leafels. Okay. So I blow up here. And this is still exceptional divisor. Okay. So as I said, this is called dimension five. This is small. This is called dimension three. Okay. So you have two-dimensional fibers. So here is a P2 bundle. And this is P1 bundle. Oh, P4, sorry. P4. That's right. And then there is, so I just giving you the result, but it's very, it's very, it's not so difficult to do. It's purely geometric. The description of of this map P6 at the, at the level of exceptional divisors. So what is the point here? So a point here. So what is this P2 bundle? Also, you can identify C eta L. Yeah. So you have a point C eta L in RC. It's still, and it's not here. This C is for, for, for, for cubic leafels, huh? Not for Cs. It should be like this. Okay. So this, this point is over the point C eta. And L, L is, is, what is it? C, C is a plane quintic. Eta gives, eta is the two torsion point that gives you the double covering connected, different from zero. And this is plane quintic. Quintic. So the L, the L is a line is, is the class of a line in the plane quintic. So you can write it as L in the dual line in P2. Every line in P2. And every line in P2 defined is degree, degree five, degree five line bundle with a, so I mean, then define the line with the line bundle. It's not such a great idea. But such that this is called. Okay. This is the one side. On the other side, a point here, what is a point here? Okay, is, you can identify with a cubic trifle together with hyperplane H. So let me write it here. X H in the section of the device or over, over the cubic trifles is X. So X, X is cubic threefold. And H is any hyperplane in P4. So which P4? The P4 where, this P4, where the X is, is in bed. Hyperplane. Okay. So these P4s are to each other. Okay. So what is the map? Yes. So the P, the P inverse tilde on the sectional device. So let me write E for the sectional devices of a point in C tilde. That should not be here. The inverse point. What is that? Okay. Let's look at, looking at our construction is, well, is, is on the, on this covering into, in order to give the double, double covering, we need a line where to project. So it's going to be a line. It's going to be the set of lines in the final surface. Yeah. With the property that this line belongs to the cubic trifle intersected with the hyperplane. That is the answer. So maybe it has to do with your question, Pavel, because so every line will give you a different, a different way of projecting. That's right. But I'm not considering all the lines, but only the lines in the intersection. And if, if you, you choose the X and H generally enough, this intersection is a smooth cubic surface for X and H general. Okay. So respect to have nice, nice intersection. And as it's a very well known fact that the every, every smooth cubic surface contain exactly 70, 70, 27 lines. Okay. So let me write it down because this is the whole point of this discussion. Every smooth cubic surface contains 27 lines. So the, the, the interesting part of this, looking at this fiber is that the 27, the lead already computed appears naturally. It has, it has some, some interpretation that this, and that was what Donaghi thought that this should be. It has to do with this, with this, with this cubic surface. Now, yes. So let's, let's, let's put it more concretely. So, so the cubic, so now this is not only about set 27 only, it's not only a number. It has to do the, the competent incidence corresponds to help with an incidence. For instance, we know that every line intersects in all the lines. Yes. So is, is this structure of the lines reflected in our fiber? That was the question. And at the time they, they, they knew the answer, that the degree is 27. They didn't know yet. So the, what was the, the relation between the, the incidence of the lines and, and, and, and the points on the, on a, on a fiber. Yeah. How. And this is when, when Donaghi come up with this tetragonal construction. It appears a little bit later. Okay. So I'm going to give you a recipe of how to, how to go from one tetragonal, double covering of a tetragonal and you get two orders. Okay. And this is, this is very, very similar to the trigonal construction. And maybe a probably Donaghi got the idea from there. You have two to one ethyl, double covering. And assume that this, this, this curve is tetragonal. Okay. I don't know if that could Okay. This is the drug. This is very general. Okay. Any, any genus, any genes for any genes. Okay. You can always. And now what do you do before this is so P1. So you have a pencil of G1s, G1 force. So the G14 is parametrized by P1. So for every point in P1, you have four points on the curve. So you can embed it, because before you can embed the G14 in the symmetric product of C four times. And you can define this associated P4 map from CT level in the way that you have a device or on the fiber here. Before I call it, let's call it P1 plus P2 plus P3, P4. And over this point, you have the choice for every, every point you have two choices. You can choose to, I don't know. Pj, Ij, K, L, the others. Yeah. And every index can be one or two. So you have to, to, to play image of each point. So in general, you have two to the, two to the four times two. So this is a to one map. So let me put it clear. The inverse image of a point, P, you have two. So if P tilde one, P tilde two. Maybe the notation is not so sure. Anyways, and then you consider, you restricted that to the P1. And you call it X tilde. It's embedded here. So you take the restriction over P1. And it turns out that this is also curved, but this curve is disconnected, has two, two pieces. Let's call it to, let's call it C, C0 tilde union C1 tilde, which is the joint. They are really nothing to do. One nice to see each component is by defining a linear and equivalent relation on, on the set of, of the points of X tilde. So we will say, does D, you have two divisors, D, D prime in X tilde. So since in, since point here. So device of degree four and C tilde, we say that two, two of these divisors are equivalent, some, some, some, some relation, equivalent relation. If it only if they put, first they push down to the same divisor, push down to the same divisor in C4. So are in this fiber. Yes. And, and they share an even number of points. So this is a way of, of, of, of describing this linear, this linear equivalence of C, C, C tilde. And with this relation, so this, this relation comes from the monodrome. Okay. This is, you have to look at the monodrome around the branch, around the branch locus. What happened around the branch locus of, of F and how did it, it, it lead to the, to the monodrome into P4. Because you have to do careful. But this is the result. And, and it comes with a natural evolution, because this guy has an evolution. Let me see. I call it his sigma. Yes. And the sigma extends to the, to the device source of degree four. So you have an evolution here, just to change the two of them. And, and given this equivalent relation, in respect to this equivalent relation. Yeah. So each of this component comes with an evolution. And by definition, those are italic. So this is free, free of fixed points. So from each one, you can define italic over covering. I call it P0. And then each of them maps to P1, 4 to 1. So very simple recipe. Very important. So you have, if you start with a triple like this, a double covering and let's call it here, F0 and F1, you, the tetragonal construction. So the only, the only thing that you require is that this guy is tetra, is the tetragonal. Yeah. You can always construct two older one ramified double coverings with that tetragonal map. This is very, very nice. And this is so called, is so called triality. And I think it's very interesting to know. So why, why this, okay, why this thing works? Why this is triality? So the reality is, is an ocean in group theory or, or yeah, algebra, the algebras. So you have a description of all the little algebras depending on the thinking diagram and so on. So when you look, when you look at the monotomy of this tower. So the tower corresponds to a monotomy corresponds. So what is the monotomy? So it's a representation or, or a map from the fundamental group of P1 minus the branch locus. So how do you describe this, this, this tower? So you have to give a map into which group, the group that you are using for the monotomy is the bi-group of D4. So I will come back with the D4 again, but so D4 has a thinking diagram like this. It's generated by four roots. So it's rank four to the thinking diagram. And it's associated to some Lie algebra. And so the automorphism of this Lie algebra corresponds or the outer automorphism of this Lie algebra corresponds to the automorphism of the thinking diagram. And this thinking diagram has an action of, of S3. Oops, this is not this. S3 acts on this thinking diagram. Yes, by, by moving, by moving this, this root. This is the, the reality that it was told. This is someone non-concept in, and also in representation theory. So, so that means also that every, every representation of the group W, the bi-group D4 comes in package of three. You have always these three, maybe because you have this outer automorphism that moves one, one representation into the other. So this, this fact of going to, when you come in package of three, you see it in, in this three, threeality of tetragonal double covariance. This is like the reason, the deep reason behind of the construction, this group theory. Okay. And this is very, so I just learned recently that WD4, it has this property. Essentially, it's the only, only group with has automorphism, outer automorphism, isomorphic to S3. They are very few who has no trivial automorphism, outer automorphisms. Yes, let me write it down. This is outer automorphisms. And, and so you cannot expect to have this for other groups, all the type of construction. So in this, for the same reason. Yeah. Okay. Yeah. Okay. So this is the tetragonal construction. That's right. So a small remark, you have time to, to elaborate, but just to tell you that the trigonal construction is at the generation of this one, of the tetragonal. You can see it like this. Okay. Now, let's not with, with this tetragonal in mind, let's do it geometrically. Now let's do it geometrically in the case of the cubic drift. So you have, let's see, start with a cubic three-fold in P4, smooth, three-fold, and align. I don't know, I want to, yeah, I fix the line. So I can construct the, the canonical, the conic, conic bundle, PL, into, into P2. You have the, the restriction to the, to the, to the discriminant locus is a, it's a double covering. Okay. And now choose, choose a plane, a plane in P4 that cuts the meeting, meeting the line where you're projecting and towards. Okay. So a intersection X is three lines. So this is where the, you can, the tetragonal cost is, you can see it geometrically because, so the first, the first, the first double covering, it, it was, you, you, you got it by, by, by projection with the line L. But when you project with the other two lines, you get other different, two different coverings. Okay. So you get three different, three different double coverings. So you get three quintics, quintics, C, C prime, C D prime, and then each one comes with a double cover. Okay. And this is what the, the comment of Pavel that you will have to show that these are not isomorphic, non-isomorphic as double coverings. Now observe. Okay. Where is the tetragonal map? Observe that when you project, okay, you have the line and this line is contained in a plane A and you have, so this, this, these three lines is the intersection with A. And when you project, when you project with PL from this line, okay, it's not good. You end up in P2. But the projection of these two lines is a point. Yeah, because they, they, they all live in the same plane. So it's a point here. And once you fix a point in the quintic, you are in a quintic. So you can project from this point. Yes. And when you project from this point, you get a G14. So I project from here and essentially you take all the lines going to this point like this. So in general, she's just as a quintic, the lines intersect while you don't see it from my drawing. But then, then you get the projection from P, gives you from P, keeps you a G14 on C. Okay, that's the G14 that associated to L. So when, when, by doing this procedure to get here, also, sorry, they all come with a four to one. And this is the tetragonal construction, geometrically. So the, the, again, the, the tetragonal map is given by the image of the other two lines. So they are related. And, and the three objects are the trigonally related is what they said. So, and, and these three objects, C tilde, C, F, C0, C0, F0. This is F0, F1, F. Oh, I changed my notation. Sorry. Let's do it all with primes are tetragonally related. So I can go from one to the other by the tetragonal construction. So this is F, F, F prime, FB prime. Okay. But moreover, these three maps can be realized simultaneously by, by the following construction, construction. So you take the A is given, you have this A here, a P2. Then consider once, once you fix a plane in P4, you define a pencil of hyperplanes, hyperplanes, let's call it C, equality of T alpha. So they are parameterized by P1, the hyperplanes in P4. So they are all P3s that containing, containing, containing A. So this is a pencil. And when you move the pencil, you can, you have also a pencil of cubic surface by intersecting with, with X. Yeah. So it's parameterized by this is cubic three four, cubic three four, cubic surface, cubic surface. Yes. And in the genetic, the genetic, the genetic, in the genetic element of the pencil, this cubic surface is going to be smooth. So don't worry. So, but you see, and every cubic surface comes with 27 lines, but you are moving this 27 lines along this pencil. And when you move the 27 lines along this pencil, you get, you get the lines in the different C-tildes. And that's how you can see that, that the tetragonal construction is very much related to the incidence of these lines, how we want one from the other. Okay. Let's say, so, so here is, here is the theorem. This is Donaghi. First book with Donaghi. I have time. That the PRIM, now when you take the PRIM of tetragonally related double coverings, they have all the same PRIM. And this is how we're interested that we are, we are finding the, the explicitly the 27 elements on the, on the file. The proof, well, the proof goes in, we can prove it by the generation as well. Let me just sketch. So what do we need to start with is a map. We have a map. You have, for instance, oh no, I see zero. Change my notation again. Well, okay, for instance, you can go from the tilde here. You can embed that in, in one of the others in four points of the order. And I will tell you how. This is, I think, the, the interesting part. You can always have a map from C tilde prime from one into four divisible. And from here, that's right. And then you apply the universal property of the PRIM. You can extend it to to the PRIM. And, okay, not, not here, but then you go to the big zero, C tilde. This goes from C four. That's a late P1. Sorry. Okay. What is interesting is this map. So you have, this map can be described as geometrically. So I'm using this, this description. So a line, a line in, a line, let's call it M, M in, in one of these, I put it while, ah, sorry. In one of the, this cubic surface that, that meets meeting the line L prime and is not, and not in A, also meets four of the other A lines of the A lines that they are in, in the cubic surface, but not in A. So A contains this three, but I'm considering the old meeting L and each, each in a co-planar one in each, each, however the time we are over time, one in each, one in each four co-planar plates. Let me, let me, let me draw that. So, okay, what they want to say that, that give you the map, you take the map M and then you send it, I don't know if it's an injection, but if you don't need an injection, you need a map in CT lab four for the other curve. You just to go into the four lines, four lines meeting L. Okay. Let me see if I did correct. So you have, I tried to draw it, but I'm not, so you have M for any M line in, in the cubic surface that, that meets L prime. So it's a, it's a line here. This is, that, this is the definition sort of to say that this is there. And they are exactly all their four lines meeting M and the, but they are not meeting A, A is somewhere else. So, and this, these all of them, they meet L. So I am sending, I am sending this M to the four lines meeting L. So four lines meeting L, they are, they are here, these those lines. Okay. And they are all co-planar, but this is, this is independent of L prime, L prime somewhere else. They don't intersect somewhere else. L, L, LB prime. Okay. That, that is, that is a construction coming from the geometry of the lines, these incidence lines of the, on the, on the cubic surface. So I want to stop here for the break and this is essentially what we have done for the triangle now. But they just want to say, how do you obtain a map from one curve into the, the divisors of the order in order to construct. Yes, in order to construct the, the isomorphism. There is any question? Is there any question? Okay. So let's do the five minutes break.