 Okay. Okay. So take again, let's, let's complete the degree, complete the degree from RT. So take X in M5, generic. Yeah. And then over this. So I am, I'm identifying X with the Jacobian. Let's put the Jacobian, the Jacobian of X. This is the, the, the, the, the P2 over the, over GX in the, in the exceptional device. Okay. And this is, this is the P2. This is the projectivization. Okay. Let's put the normal GX of M5. Oh, well, let's keep to the Jacobi locus over A5, the projectivization of that fiber. Okay. So let's call R. There is a comment on the, in the chat. I don't know if you saw it. No. Can we put some explicit projections between? Ah, okay. I need to think more about, I, I will look at the end. Okay. Yeah. This is a more precise question. Okay. Yes. Thank you. How they close my chat. Right. So let's call R the inverse image of this exceptional map at the level of exceptional divisors of this P2. So this is in, in the exceptional divisor over the 3-1. Okay. And we want to describe, we want to describe the map from these two, from this R into this P2. So this P2 is actually the, the dual P2 of, of the net of quadrics, quadrics containing the plane quintic. So P, it's a P2 but it's dual to this one. P, it's a plane. It's the plane. Yeah. It's the plane containing the, the curve F, I call it. So the genus, genus curve, genus 6, the plane quintic. Okay. Also wish, I don't know if I need it. This is that is generated by, by the three quadrics, three small quadrics. This is the net of quadrics and this P2 is the dual of that one. Okay. So recall, let me call alpha, put another color because I think it's a new alpha. So alpha is, goes from M5 to RQ bar. So you take X and, and it's the construction I give you above this dual associated the, the singular locus of X and modulo with the evolution. So as pair is with the evolution, minus one. That's right. And actually, I don't know what they put the, this is always a quintic. So you have, you have this situation. You are in P2, two, three, four. This is F, have a line. This is P2. Up this pie, this big pie. So, okay, let's work with the duality. So you know that a point, a point in P2, point in P2, in this P2 that I have here is equivalent as a line in the dual space. Okay. But a line in the dual space. So you see is a net of quadrics is, is, is, is the same as a pencil of quadrics containing the canonical curve. Pencil of quadrics containing the canonical curve in P4. No, we are all in P4. Okay. And vice versa, vice versa. That's my right. It's all very, very, very trivial, but one can get lost. So a line in P2, you have a line in P2. It's equivalent to a point in, in, in the dual and appointed the dual is just a quadric. Actually a singular quadric. A singular quadric Q containing the canonical example. Okay. So let's take, let's see, let's see if I can. Let's take a double covering in RT. Search that the prem city, let's see, isn't Jacobian. So, above this point, you have a P1 bundle. And so you have the P1 about this point. Like this. It's in there. So it's, it's, it's the fiber, a fiber above. Yeah. So remember F is in RT. How did I call it? Tilda is the blow up. Now, you restrict the sectional divisor. So now this is a computation. This is a restriction to this P1. This P1, let's see, into P2. It's actually injective. Because, because this restriction is injected because we just prove it. Is this, okay. This P1 is injecting this P2. Okay. And it's image is a line. It's image. So again, in P2, in P2, so this P2 dual. So again, my P2, I just wrote it here. Aligning P2 corresponds to a singular quadric to, yeah. Such that actually this quadric is going to be an element of the kernel of the co-differential just by duality. Okay. But the co-differential restricted to wherever, to the connormal maps. Yeah. Okay. So now the degree, so this degree, by definition, is the number of points, everything in generality. So over, over my prime moment, over a point, P. So P is here, the P2 generic. And this is the number of curves, C. This is the diagonal curves with a double covering, such that they are in alpha X. Okay. So this is why the alpha X was important. Where is the alpha X? Here. See, you see, alpha X. So it's again, well, one has to understand that this guy parametrizes, parametrizes the G1 force. G1 force. On, on, on, on X. Okay. So, and every time, and then I'm going through the three-way constructions, every time you parameterize a G14, you have a G14 on X, you have a double covering. Okay. So, so then you look at all, all those double coverings, such that P of, of, of, sorry about P, principal, such that the brim of this double covering, as the Jacobian, yeah, that means is, is, is over this point. And the P, it belongs to the image of this line, this line, this line here. Okay. So this is the same as the number of core, I write cores, but actually it's a core with, together with the G14. Sorry. Yeah, is, is the double covering associated to the G14 and alpha X. This is count, not so clear, right? So maybe such that the, I'll give myself a fuse. What is this pi? Oh yeah. So I give it a name actually. So, so alpha X. Yeah, is this pi. It's the plane quintic. And the plane quintic is contained. Okay. So, Mm hmm. So such that the image is the small pie of this curve as a point on the plane quintic is a point contained in the line associated to P. So this is in P2. And the line associated in that is, that is a line on the dual. Okay. So what is this degree? We're almost done. The degree, the degree of PA tilde is the degree of my double covering. So it's two pi times the degree of F, which is a quintic plane quintic. This is the. And that's all. Does it look understandable? Okay. So that's the point. I kept the notations of the nuggets me, but I'm not so happy with the notations C in our fights. So one has, again, one has to understand that. So an element here is a G14 on X. And with the G14 from X to always construct the C tilde C. This is a trigonal construct. Okay. And I refer all these curves. Yes. Any question? Okay. If you don't have questions, I'm going to keep the older. Go to the boundary so far we haven't touched the boundary. So there is one one locus on the boundary that gives you contributes to the degree boundary components. So RS, RS does want to be my building in covers. To do one in our as no sorry, six. So this is as is for singular, singular covers. So our six allowable such that C is an stable curve here. So actually I'm thinking of those guys. That's, yeah. Okay. And, and this nodal course. This nodal course is a divisor is a divisor over mcs and mcs. So this, this is a divisor. And we have another ones that are the elliptic tails, but it's not going to contribute. Are they are. They're the drawing. So there are allowable covenants. In, in, in our six of the form. So you have, you have a genus by curve x, we have another component as an elliptic curve, you glue it zero for instance, and then you make two copies. And then you leave. With the double covering another elliptic card. This is easy. This is elliptic. This is elliptic in the, this is what we call elliptic days days. And this is a singular course. Well, those are singular as well. Okay. Yes. So first observe well, observe that the, the p6 restricted to RS. Well, RS is already bounded device and RS goes to G five. In a five is also proper proper. So we will ready extended the map to the boundary, and it's subjective. Because every time you have a five genus curve, you can construct such such discovery. So you just take X, you, you take, you make a little knob everywhere, and then you get a cover. So it's very simple. This is subjective. And actually, this has dimension dimension 14. As I said, it's a divisor dimension 12. So you have two dimensional fibers. And when you can identify the fiber with the symmetric product. So for a fixed. Let's see. Again, X and M five. Generic. Thanks. And here now I want to use generic in the sense that has no automorphism. This is very, very, very practical. So generic is whatever you need to be. And now what I needed this is no automorphism. For the following reason. Yeah, because I don't want to, to mix up with the account of the on the on the cover. So is take the universe of this point restricted to the singular ones. So this one you can identify as the symmetry product of the curve itself. So it's just the truth is the show is the choice of, of the points where you will be clue in X to blue. Okay. That's very simple. And So there is a, I'm not going to prove it. I don't know if this is part of the work is that is the following proposition for C. Define it as a snow that course. The kernel of the co differential map. At that point, and this is two dimensional restricted to, well, I don't know what there is. At that point is two dimensional. Yes. Okay, it's to the how do you get the two dimensional so it's two dimensional. And it can be identified with the quadrics containing X. Now, now, containing C analysis is this know that. So, yes. So recall, so recall that call that exists is a complete intersection of three quadrics. Before. Before. So these are smooth quadrics in general position and and the second. Now you take two points. The second line going through this PQ imposes another, another condition. Linear condition, one linear condition on the on the quadrics. So I since, okay, since I wanted to get. So you have X. So I have a three dimensional family of quadrics containing the canonical and made in a mix that I want that they contain also second line. So it's an extra condition where this is the second line that where I'm going to glue to get C. Yeah, this is how do you get from. Yes. You count, let's see how many you have a line, yeah. Okay, let's see. Let's restrict the fiber. Let's restrict the fiber over, over the generic X. So, goes to P to X, JX, yeah, on this side, the fiber. And that is isomorphic. Well, this is the normal bundle, the projectivization of the normal bundle over this Jacobian from five in a five. And again, the dual parametrizes. So the dual P to dual parametrizes the quadris going through the X. So let's call this F. So for simplicity. F. So what is this map. F. When you F of two points to symmetric points here. Plus Q or FQ. The Unix is equal to the pencil of quadrics. Pencil of quadrics containing the secret. P, P, PQ, so you embed the canonical embed and you consider the secret like this. Okay, clear. So every. So X. This is just to make it clear what the neutrality this is. This is net of quadrics, quadrics. So you have, you have a two dimensional family of quadrics going through F. So you have one dimensional family of quadrics that contain a given line. And this one dimensional fiber, this is a pencil, a pencil in big P and the pencil in big P. Means, yes, means a point in the dual. Okay. I think you have to think twice, but this, this, this, this already came. The image of, you see, it's very, very, very, very concrete. What I mean by, by the P till a map and now with this in mind, we can conclude the degree. Let's see the degree of F is the number of course or sequence. Of the canonical curve, contained in the intersection of the two of two quadrics, two quadrics in general position. So two quadris given the two quadris is that is the terminal pencil. Yeah. In general position. So. So one, how many of those we have. So for that you need to let us and maybe, maybe this lemmas is I don't know. What are the things but these two lemmas one can show it. A. The intersection of two quadrics is not really difficult the intersection of two quadrics to quadrics. In general position in P for contain 60 lines before contain 60 lines. Because essentially you can see that as the bed so that's this. So the quadrics in every quadric is three dimensional, and you intersect both of them. What you get is the pencil surface surface of degree four of degree four. So the pencil surface of degree four is this P to blow up in five points. And with this description is easier to count the 16 lines. And the second one. This is one thing. And the second one. The second lemma is that the fix the canonical embedding meets each of these lines of the 16 lines twice. Therefore, so applying this to them as the degree of F. Okay, so you have. Yeah, drawing is not my strongest point, but yeah, essentially, you have one of these lines. Lines in the in the intersection of two quadrics. Yes, you have 16 of them. And in these lines meet. Fx. So fix is somewhere here. The canonical everything is in before meets into points. Those are the PM P a cool right. So for it for each line, you have a course of fix you have a second. Okay. That's right. I think that this lemma is this is easy to check. This is easy. Okay, let me see have time. Yeah, have something like 10 minutes. Yes. Yes, thank you. Okay. So, um, look, um, nothing come just I'm just going to write it down. I'm not going to say more about it just to tell you that what is the issue with the with the lipstick tails. The lipstick tails don't contribute to them. The lipstick days. No. Mark. This this locus of the lipstick days. Does doesn't contribute to the degree. Because and this is how the don't like it works out this is. Well, this is actually it's kind of half of the papers just to do to do that just to make sure that there is no more contribution. So we looked at a new compactification. They call it our prime. This is a new compactification. Of R6 such that the, the, the, the elliptic locus is blow down to a point or I'm not sure this is a point but this is blow down blow down. This is a smaller locus such that P6. This is diagram commutes a five. This commutes. And yes. So when you went to those in contribute, because you can you can find a better a better compactification where the diagrams factorized. And you don't see anymore this this device is blow down. That's what I'm saying. And it requires also, also requires a new compactification of M6. Yeah, so just to give you an idea that there are a lot of the technical details that I don't have here is no time for that. The time I have, I want to talk another special fiber, which is somehow nicer. Another special fiber. So you remember, I had two components of, of the plain quintix double corings and plain quintix, the even ones, and the odd ones. And the odd ones, they're going to, they map to, they are going to map to see, let's call it see the moduli space. Of non-singular cubic three-fold. Also, you have, you have a, I'll call it also x, I hope they are not. We have x now and before, but this is dimension three is cubic. It's defined by, by a polynomial degree three. And you have a map, I'm not going to give more details, but you have a map from this space into a five, which is actually an injection. By taking x, you can define another Jacobian, but this is very special Jacobian. And this is also a Jacobian defined through Hodge theory. So analogous to the Jacobian of a curve, you can, you can define Jacobian, intermediate Jacobians, this is called intermediate Jacobians of other varieties, or their hypersurface, for instance, intermediate Jacobian. And it's defined well, we have by a Hodge theory, you have a filtration of the Hodge structure in Comology. So you have the one two Comology except, and you made a quotient of the three Comology group of x. So in this Comology, the integer Comology of x, you can, this is the intermediate one, you can define a bilinear form given by the cop product, and it can be extended to the full, to this group. In any case. Maybe the numerator is in complex Comology. It's in complex, yes, yes, yes. Yes. Everything is complex anyway. Complex. Ah, you mean here. Yes. Sorry. Yes. What they want to say. Okay, so. That's right. So you have also a special loci, via this, this is to have some sort of to really am up here so you can identify the, the cubic trifles with the intermediate Jacobians in C, so I'm just going to identify them. And this is moduli, the dimension count is 10, if I'm not mistaken. So, moduli count is 10. And I'm just going to give you the description, just to do. It's a little bit like a motivator in the next, the next, the talk of tomorrow about the diagonal construction and structure. You blow up, you have to blow up again. R6, A5, over R6, A5. But now you blow this, this, okay, the, an RC G5. The basics of the, the, the plain quintix, the plain quintix, the old plain quintix, go to, go to the, to see to the non-singular cubic trifles, intermediate Jacobians or non-singular. This, this is a long story, but then I can tell you how it works. That works through, so every, so in other words, every, every intermediate Jacobian of this form is a prime variety of some double covers over a quintix. And this is performed by a conic bundle, conic bundle structure of, of this cubic trifle. Okay, if I have time, maybe tomorrow we'll say a little bit more. Should I say now? Where is my talk of tomorrow? Yes. So, yes, let me tell you what is, how it works. So you take a point just to prepare for tomorrow. Take a point RC, so a double covering. Delta L in RC. Yes. Oh, I wanted to give from, yes. Actually, it was already in the, on the blow up. So let me tell you right away what is, what is the map at the level of blocks. So this is five. Okay. So, so here you have, you have a simple double covering data. And by choosing L, L is a line. Angela, if you can. Yes, I have to show that. Yes, it's not going to happen this today. Okay, I will, I will stay here. Yes. No, no, it's fine. It's fine. I, I wanted to say something in two minutes, but there's nothing I can say it's going to be either quick, quick or understood. So, yeah, okay, just tell you week week, the plan for tomorrow. I, as you can see, is there's a change of plan. A plan of tomorrow is I want to tell you about this other fiber. I want to tell you about the general structure, how does it work more general in the, in the context, and I will go for the second part into a wild generality of all the possible bring final pre maps, and I want to finish with some open questions for all of you. Yeah, but so the, the, the, the computation type is finished now and then now we are going to go more in more generality. Okay. Thank you very much. Thank you. Any question. I think this is a good time to stop recording.