 Yes, the answer is correct. So let me, so someone answered in the chat. I don't know how to repeat. Yes, two times, that's right. You compute, what do you do? You compute the degree of the ramification and you, the branch locus and you, and you subtract for the automorphism group of P1. So, let me see. So the degree of, that's what they say, 16, that's correct. And then moduli count is 16 minus three. In any case, it's bigger than 12. So you also have a five-year-old dimension, one dimension. Okay, so we are now in business. Let's see, let's see how, yes, yes, please. The thing is, the first thing, okay, for this, when you write this exact sequence with a normalization, I guess you have a Jacobian of normalization. So I assume that if a normalization is disconnected, you just, you take the product of too, too many smooth components, yes? So that's just the first thing. Okay, and the second thing, when you write those for loci, yes? I guess you need to show that the image, that the restricted prem map is still dominant on those things, yes? Because they are not generic, yes? In a sense, in the whole... Yes. Yes. So this is not trivial, I think. No, no, no, no. So first of all, the list itself is not trivial. Actually, the list itself is bigger, but there are some cases we are not considered because they go to a smaller loci, so the smaller dimension, they doesn't cover, is less than 12. And as you said, the restriction has to be, has to be proper as well. So it has to be, yes, so, yes. So how do you compute, so here's the question, how do you compute the degree? So what does it mean the degree over a fiber, which is half the whole city dimension of the fiber? So you have to blow up. So I think it's worth to spend time on this lemma because I think it has, yes, let's answer that question. Okay, so how to compute the locality? Okay, let's do a little bit. How to compute a locality? So let's consider a proper map. First, you start with the proper map and don't mean it. Between same dimensional varieties, same dimension, oh, yeah, n-dimensional varieties, n-dimensional varieties, x, x, y. So in particular, f is generically finite. So take an irreducible, irreducible two-variety of co-dimension k and then take the inverse of this variety. So this breaks out in, finally, many components, many components, the set E of different co-dimensions, l, i, n, x, okay? So you can compute the local degree of, local degree, the i of f along every component set i, yes, by localizing x at the set E, then you compute the local degree this map. So it's such a way, so let's say the local degree of z along set E, so the degree of the map, the degree of f is the sum of all these values. Okay, so I have to do it. Okay, let's analyze, take one of these components, and then look at the diagram after blowing up. So you blow up set, blow x along z. Let's call it is x tilde over x, set. So it has a sectional divisor called set tilde, that maps to z, sectional divisor, and then goes, you have f goes to y, you have the image here, and you blow up on this side, v of y, this is the blow up, and let's call it w tilde, the sectional divisor here. So you can, you have a map between the blobs, f tilde. Okay, now, you have an induced map, so that can use a map. Actually, you have induced map f tilde between the sectional divisors, and this is the degree you want to compute. Yeah, actually, let's call it in the Nagas notation. Yeah, this is f tilde, just take z tilde. So with what is this set tilde? So from the definition of blow up, set tilde is the projectivization of the normal bundle, normal bundle of, I don't know what this said over x. Yeah, this is the definition. So you have a tangent space to x, so in each, yeah, tangent space. So the composite tangent space to z plus a normal bundle, z over x. Okay, good. So take a z, a small z in z, w the image, and look at the differential of z at the point. So this is a map between the tangent spaces at the point, and then these maps here inside, you have a tangent set of x set goes to the tangent of w, small module, okay? So these set induces, so you have induces, you have a map between the fibers, so actually a map between the normal boxes at the point, small set, with the normal bundles, what, okay? So here is the lemma, this lemma follows from the universal property of blown ups, that this map f tilde, so you want to extend this map in such a way that is now regular, f tilde is regular at the generic point in z tilde, if and only if this map at the level of the normal bundles is not identically zero, zero, a generic z in z, okay? Ah, well, that's actually didn't, I didn't really need this, but now if tilda is regular, I wanna actually need more, I don't know exactly, for every z tilde in the fiber, as of the blob in the fiber over this small set, if only if, actually it's what I need, this map is injected on the normal bundles, on the normal space, oops, oops, oops, oops, on the normal space to z, so here, where, here, this one, yeah, I wanted to be injective, so this map is injective, that's equivalent to say that actually the f tilde is regular in this case, and in this case, in this case, you can f tilde restricted to the fiber over the point set is the projectivizer of the linear map, so you projectivize the linear map f tilde, so the computation of the degree of the f tilde goes through the compactificate, through the, to the degree of this f tilde, so let's assume, so here's an important lemma, it's not so difficult to show, but I'm just gonna give you, assume that this map is injective, so this is an extra on the normal bundle x over the point set at each set in set, and that's gonna be very, very important, I was playing y, you have to, you have to guarantee this, this map is injective on every point of set, then the local, the local degree of f along this component set equals the degree on the sectional devices of this at the level of special devices, which are of the same dimension, now I have equipped both our devices over the same dimension, so here's the warning, why we need that warning, if, so you need this, this condition, if you have a point where it's not injective, that means that it's not regular, so if it's not injective at set, then you have, then, the map is not regular, because it was if only, it's not regular in any neighborhood, on a neighborhood, on a neighborhood of, of, of, of the fiber, c tilde, so c tilde is over c, it's a point in the fiber, so what happens at, so it's not regular at this point of the fiber, so in order to define it, you have to blow up again over this point, yes? So you have to blow up, yeah, for some, in some smaller dimension there, you have to blow up again, and if you blow up again, you get more components, so you have a super IT, you have super IT, another super IT, another super IT mapping to, to w tilde, yeah? So in that case, so c tilde is not the only, yeah, it's only one of several components, of several components mapping to this, so, so yeah, so what happens is that possible, so it might happen, might not, but the possible, the degree of the f tilde, when you compute the degree, it might be smaller of the actual degree you are looking for, also the degree on the fiber, or in a neighborhood, neighborhood of the point set, around this fiber, which is, that was, it was the local degree, the degree of this component. So yeah, you are computing something, something small, and you are missing some other components. So this is why, this is why, in the computation, one has to check this condition, that for every c, z, this map is tilde set, it is actually injected, and the good, that is very general, but the good news is that in our example, this is all computable, and it's going to be essentially, as we mentioned the other day, in terms of multiplication of sections, in terms of quadrics containing the canonical embedding, and so on and so forth. It's very, very specific. Okay, let's me, okay, let's, let's start with the plain quintics. Quintics. I didn't say what the data characteristic is, right? Well, I don't know if I need it, but let's, let's, let's start with the curve in M6, the smooth curve, or plain quintic, plain, plain quintic, Quintic. So you have a map into P2, given by G25, yeah. So actually this is given by a completely linear series to have L. So it's going to be both data characteristic. Let's see. So there is a natural, a natural line bundle, data characteristic, L, given by the hyperplanetized. So, so it's essentially take the, I'll give a name. So the pullback of the hyperplane class, phi pullback of OP2. Okay. So data characteristic means that the U square, the line model is can, is isomorphic to the canonical model. Let's see. So in particular has a, this guy has degree five. Okay. So here's the picture. So usually there are more pictures, but so you have the embedding. And then as a plain. Yeah. And then plain Quintic, you intersect with a line, the class of a line. Let's see. Yes. So you, you, let's just give name to the, this is the Donagis notation in Donagis Smith, but then in the other paper, change of notation, but I will keep the notation of Donagis. So it's later, you want to read that paper carefully. So. Yes. Let's put it like this. See. All the double corbrings. In our six, such that C is a plain Quintic. So maybe I'm going to fast, but this, remember that this Etta is, is the two torsion point associated to the local code. Okay. This one is isomorphic. You see, you can, you can give an isomorphism from, from here into, you tensor by, by L. So you have a naturally, you have always, since you have a plain Quintic, this L, this degree five, this Etta characteristic. And you can tensor C, hit the tensor with L. And now the difference is this, this line bundle now lives in, in peak five, in G minus one. So you can, you can check the parity of the, of the space of sections. So essentially what I'm saying here, I'm going to, I'm going to, I'm going to, essentially what I'm saying here is that this equivalent to, to, to, to the pairs like this, such that C is plain Quivic. So I, is the couple of a plain Quivic plus a data characteristic, plain Quintic. And we say that, yeah, I see a zero of, you have two types of such data characteristics is one of us. And then we use this, it is, we say that data is called, well, then we said, it, sorry, it is called. And with this congruent to zero, it is, even, but it depends on the choice of, of, of embedding on the plain Quintic. Yeah. So then you can stick with it. This is the so called the spin course. But you have plus a minus. And this is another story. So this, this distinction between, between all then even gives you at a composition of this space. It has to, to connect the components, the old part. And I will use Q for, for, for what? For Quintic, Q for Quintic. And with C, that's a story. This is notation for, from the Nagismith, but it has a reason. So this is the even. And this is the off. The even. Well, as we, we know under, under the data map, they want to map to Jacobians. And the others. We, I hope to see later, maybe in the last lecture, they are going to cubic truffles. So cubic truffles, they are another nice example of. So the intermediate, I mean, I mean the intermediate Jacobian of a cubic, cubic truffles define it as a, in terms of a whole story, gives you naturally and principal price, a million variety of dimension five. This is another nice locals that they expect. I hope to say a couple of words about that. And one can compute also the degree there. Okay. So where am I? So, so let me tell you some, some, so it looks a little bit aside, but we have the following result. Give me more fun. Oh, maybe it's not so relevant, but yeah. You have a double covering in R6. Oh, an RQ prime. So double cover and a Quintic, such that the Bream. So we say that the Bream is in the Bream. Sorry, I'm changing notation all the time, but you take the Bream of this double covering, is in the, and wrote in major loss. I look us in one. I will tell you what it is. If only if the associated data is even. It's another way of characterizing. Yeah. So that means, that means if you're covering actually was seen in the, in the RQ without, without alkystroph. And what is N1? This is, this is the Android, Android remain major. We say that the principle polarize a billion variety. A theta lives in the NK. This is, and the O.T. Mayor. No, sorry. This mayor is mayor. So probably. Sooner. If the dimension of the singular locus. So the singular locus of the data divisor. Yes. At least K. So he is. In general, you take a general, a generic, a billion principle polarize a billion variety. The data divisor has no singularities. Okay. The general one, but you are interested to understand the, the geometry of the model space of a G, you start to look for a special law size inside there. And one natural way of distinguishing special law size is to looking at the, at the places where the data divisor has singularity. And then you have all sorts of. So the first case is when the data divisor has at least one singularity. And it turns out that. And. The prems of quintics. They have all. Actually the Jacobians, they will have also singularity. So where it comes from. Okay. So more and give you more precisely. So. So if you have the Jacobian. Of X. Is the brain. Of discovery. Okay. With the generic. So, okay. Maybe here I need to say more precisely this is non-hyperliptic. And North. She is going to be pretty one. But the start mix is not. Okay. So. You can look at the singular locus. So. Of, of, of the, of the data divisor. Maybe, yeah, sorry for the notation, but okay, singular locus. So. By the Riemann singularity theorem. The singular locus of the data device. It can be identified with the line models of degree. Four. Onyx. Which have more section as expected. Namely, at least two. And it turns out. Well, that is one dimensional. Actually is a curve. This, this gives you a curve. So. This is. And these curves come with a natural. Involution. Yota. Which is just take L. K minus L. It's an evolution. And it's going to be an evolution. Of, of. Free. Without fixed points. Because. Well, because the curve is generic. So it has no financial. But. Okay. To have. Mm hmm. Okay. Now. The generic. Also tells you that. For instance, The canonical map is an embedding. So it's not hyperliptic. Goes into before. Okay. Canonical embedding. And as you know, this is a classical. Geometry. For a generic. Genus five curves. In the canonical embedding, you can describe as a complete intersection of three. Quadrics in general position. Small quarks. Okay. Mm hmm. So, actually any, any, any G14. On, on X is cut out. By one parameter family. Of planes. So there are planes. Swiping. Swiping out the curve. Swiping out. Okay. So you have, you have a family. A family of planes inside. Inside the aquatic. And when you caught. Yeah, you caught the. Yeah. Your, your, your curve. It defines four points on the curve. This is a G14 analogy forms comes in. And the quadrics are actually singular quadrics. So they have. Could be ranked. Three. Or four. Before. Of course. So quadrics. So the three dimensional quadrics. Containings. Okay. So, okay. So you can always define. You can parametrize all the quadrics containing your curve. Your canonical curve. By a plane. Let's call it P. Generated by this. Two one. Two. Small quadrics. So it's a P two. Mm hmm. And. Mm hmm. Mm hmm. Three dimensional quadrics. Containings. Okay. So you can always define. You can parametrize all the quadrics containing your curve. And so P comes with a discriminant log size locals. The discriminant locals of P. Is. Of P. Is, is the set of all the quadrics or all the points. Representing a quadric which is single. Let me correct. So. So. So let me be more concrete. So every element here. Right. You can describe it as. So for instance. Zero. Zero. Just. Make concrete. Two. And the lambda zeros are. Parameters is lambda. The points on P. Two. Two. Two. Two. And the lambda zeros are. Parameters is lambda. The points on P. Two. The points on P. Two. Okay. Mm hmm. Okay. So. Mm hmm. So this is the discriminant locals of. Of this net of quadrics. Is. It's a Quintic. It's a Quintic we're looking for. Quintic. Because it's given by. The banishing of the discriminant. Of all. Of all these. Yeah. The. The banishing of. Yeah. The determinant where the entries are, are the partial. The partial derivatives of this. So it's linear forms inside. So there are two planes. So for each. For each lambda here. So let's call it that C. For each. For each lambda and C. So we have a singular quadric. The singular quadrics. It comes with two, two, two families of planes. They are two. Two planes. Of cool lambda. Two families. Cutting the. The G14. All right. G14. Okay. So this way you can construct. C tilde. To see. Two to one. Covering. So the C tilde parameterizes. So you can. Yeah. You can associate it also to. All the G14. Coming. Coming from the singular quadric. When you cut out the. The families of the singular. Quadrics. Yes. Two. No, two planes. There are two systems of planes. Systems. Two. So there are two systems of two planes. Two systems. Of two planes. Mm-hmm. Mm-hmm. Okay. Right. So this is the construction. So, uh, from the singular locus. You recover. You recover. C tilde. Plus the data. Plus the data of the G14. Yeah. Mm-hmm. Okay. How much time they have. Um. Is any question. Here. It's a lot of. It's a lot of. Okay. Okay. Right. So this is the construction. So from the singular locus. You recover. You recover the city. Plus the data. Plus the data of. Of the G14. Yeah. Okay. It's a lot of. Around 10 minutes. Okay. Right. Okay. So. When you say plane quintics. Okay. Let's let's talk about the trigonal curse. So. Mm-hmm. So this is going to be trigonal. Trigonal curse. So I'm going to tell you the residual construction. Construction. Mm-hmm. And it works in arbitrary G. Yeah. So I start with the curve. X. With G minus one. Such that the extra. Okay. So it's going to be applied for a genus five. When X is of genus five. Because all, all the, all the genus. The general X. We'll have a tetragonal. We'll have a, actually we'll have a finite number of tetragonal. Maps. So G one force. So it's the first case you can apply it. In the generic generic way. And to, to every extra diagonal curve. You can associate the double, double covering. So let's say. Let's. It exists a double covering. Such that the pre-modus double covering is a size. So. To the top of the two. So I hate to see. To see. To every extra diagonal car. You cannot see the double, double covering. So let's say when it. It exists a double covering. Such that the BREAM of this double covering is isomorphic to the, to the Chaco. And this is a, this is a bijection. You can go up. Okay. So let me let me tell you all the city. So pairs of points but unordered points is why put some on the symmetric product of X such that they are in a fiber. So P1, P2, P3, there exist all the two points, so such that this is in the G14. So what does mean the G14 is that you have P1. And that's right. So C tilde is going to be a map 6 to 1 to P1 because you have six ways of taking pairs of points among four unordered ones. And so it's taking some little drawing. So you have four points. Yeah, P1, P2, P3, P4, and they're all mapped by the G14. This is a G14. So what you're choosing, you're just choosing pairs, one, two, three, four of the diagonal. So but it comes also with the natural involution, namely the P1, P2 points you map to the order two. So it comes with an involution that I think I call it Sigma. It's not your anymore. Sigma natural involution. And the quotient of this we call it C. So this is 2 to 1. This is without fixed points. Sigma is free of fixed points. So I said that, and this is 3 to 1. So you get naturally a trigonal curve C. And when you, in also the branch locus of the diagonal curve, the branch locus of the trigonal is the same. Yeah, this branch locus are in the same points on P1. So when you do the computations, you get that, okay, first, as I said, this is trigonal. This is tal. And H has, which is H, I call it H is this. The branch locus has, as I call it F, same branch locus as F, which is equal to 2G plus 4, I think. In any case, when you make the computations, the genus of C is one more than the genus of X. This is the diagonal construction, has an inverse construction. Yes, so now you consider a double covering C in the trigonal locus. So they are CT, let's see. In R6, such that C is trigonal. And how do you go? So C is trigonal. So now it's a C who has a map 3 to 1 to P1. So I'd like to give this diagram. So you have a G13. So this G13, of course, is isomorphic to P1. In G13, you can embed it in the third symmetry product of C. Yes, just take three points over one on the fiber. And the double covering pi here induces a double, no, it's not double, but you have the induces a covering from CT to 3. Which degree? Well, from over each device or degree 3 here. For every point you have two choices of the covering. So the degree is 2 to the 3. So this is a degree 8 to 1. Okay, and then you take the restriction of this map to P1. And just call it X tilde, which is embedded here. So this is just P3 restricted to, this is the restriction to do this to P1. And it comes also with an evolution, sigma, which means because induced by the evolution here. So C tilde has an evolution so you can exchange the points, so you exchange the points over each point of the device. So that gives you also is fixed point free. And you get the quotient, X, 2 to 1, so it's 2 to 1, then you start with the 8, so from here you get degree 4. So you get an tetragonal curve. X. Okay, so maybe I don't have, let's see. Where am I? My notes. Yes. Yes, maybe I should stop because there are now, now we are interested was it was a question. So using this trigonal construction, I'm going to show that there is only one exactly only was not so difficult now to show this is only one, only one double covering over three one of course and not to the jackpot. And some point we need, we can study also the degenerations of the trigonal construction. So which generations can happen. For instance what happened when, well, in principle, the tetragonal covering is generic so it's simply ramified, but you can have points where double ramified, triple ramified, and the double covering looks different. And in some cases though this double covering is going to be in the boundary, it's going to be disconnected and some cases you know, so what has to analyze all these cases. So what I'm going to do tomorrow, and I think is for the theory is very important is to give you a proof of the fact that I mentioned that the preem of this double covering is isomorphic to the jackpot. This is the proof that I can give it. And it's a very nice argument by the generation actually. Yes. Do you have any questions. While we wait for questions there is something I would like to say for those who are still with us, and we're not giving a talk. It would be since tomorrow we have a free morning we should all be a bit less tired. We, our aim is to meet that 130 on gather so that those who are not giving a talk can introduce themselves. There is a podium where you can speak to everybody. And you just say your name where you are what you're working on, so that we have a little bit of the knowledge of each other we would have if we were at an in person conference. Angela, please. I just had this announcement to make anyone has questions. I've had one, one question. Oh, yeah, so but I think that you already somehow answered it, because you said that in your tetragonal construction, yes, that trigonal construction that the sigma is free of fixed points yes. I think that in order to, to, to have the sigma free of fixed points you have this tetragonal this g14 to be as general as possible yes because I can choose probably x the g14 such that you have fixed points yes p1 plus p2 plus p1 plus p2 somehow. Yes. Since you assume that x is generic in G in a five in J five yes in the Jacobian in M five. You will, you will assume you will get it somehow for free that the g14 is as general as possible. You can still have funny cases where you have more ramifications but yeah you can still have them. Yes. But, again, they are not going to be general inside of the, of all g14. And, and when you made a modeling count. And so in the end is, we can just reload out because they are going to map to something smaller. But funny things can happen for instance you can get singularities on the city that's not something that we don't like so much, or, yeah, but we can talk to more of you want more so thank you. A larger picture of what happened on the, you can happen. Yeah, for for the. Cannot have for the form to be the feed there is. That's right. Yeah, some of the answer. Yes, it's what I mean. So she in the case for him because you have a fiber to two p plus to two, or do you still have those are precisely the cases where the Associated Corp is singular. Yeah. Let's see how much singular and yeah. Yes, Angela, and I also that there are some problems if the, the foregone series is not simple so the 41 covers stock factorizes as two to one to two. I think that the trigonal construction does not. Because. It's also commented in the chat. I don't know if you saw it is something is called. No, no, this is was no you mean something else you mean. That's right so. But I think in that case you will get something hyperliptic. Yeah, they will do will have a you factorize this x to some double covering. Well, you will get some special curve in any way. So you have the factorization them is a exist is special in some way. I think that you don't get a curve you get a rule surface because there is also another approach, but maybe we can. That's interesting yeah okay. Uh huh. So if I understand correctly, if this you mean if this map for two one actually factorizes to something else. Yes. I don't know. Yes, yes, through a hyperliptic curve that has to be hyperliptic. And this is well depends on the genius political bio elliptic. But if you get an elliptic one, why hyperliptic yes, my hyperliptic right you have an if that's right. And so my understanding is if you can factorize you have a special course in some sense. So it's not that general. You have an extra out. Yeah. You may have extras automorphism there. Yes. Yes, so. Yeah, yeah, but yeah, that's nice. Okay, this is another way to consider the trigonal curves, trigonal construction on this. You get something different. Yeah. Yeah, true. Okay. Another comment. If the tetragonal map factor is then it necessarily has to pick up two in the fiber. I think. Yes, but then you need it through in every fiber. Yes, so this is again the game. And then, and your answer is that you take the generic X and generic X is not of this. No problem with that. But, you know, since you said that the trigonal construction works, you know, more generally. No, that's then you have a problem of sometimes if you have problems sometimes. Yes, but I think it is still war war to look at the generations, maybe for all their problems for all the things, but it's too war to look at the generation of the construction. For the degree will not going to be important but you get the interesting things. And this is when you allow ramification on points and yeah, you can you can play with that. Yes, but depends depends what you are looking for. Don't dismiss the specific case is very important. Yes. Okay. No more questions. Okay. Continue tomorrow and the plan is just finished the computation with more more geometrical arguments. I'm ready to give you the answer. And then I will talk a little bit about the other specialist side cubic triples and Yeah, and then why you have this 27 more. Yeah, let's discuss about this 27 more precisely. Okay. Thank you. Thank you. Thank you very much. And as you all in gather very soon. Bye bye.