 Oh, let's share the screen first. Okay. Yes. So yesterday, before, actually, I forgot to say something, maybe you noticed, but I didn't make it clear before, before starting with the trigonal construction, I talked about some construction about how do you get from a Jacobian of a very very generic, very generic five genus curve, and how can you get a double covering and the double covering will be over a plain Quintic. Yes. So, using the singularities of the data devices. So this this results this construction actually. So just to just to make it clear. So from yesterday, this is some for some reason I skip that part from yesterday. So what do we have, we have a very generic Jacobian in the Jacobian locus of a J five. And then we do this construction of, of taking them. The, the singular locus of the chapter of the data device of Jacobian, which is the genus. Well, I'm not sure I want to recall everything but that was something like this thing singularities and the singularities. Well depends on where do you see the data device or but this, that is the, the, the plus minus the plus minus. You have always the minus one mapping in any ability and variety. And this is known should be known, known, hyperliptic no trigonal and no vanishing data. So in this context you have this is a smooth. Chinos 11 curve. And this is a plain Quintic. This is what they do describe as the this. Yeah, this will be the discriminant locus Quintic. Yeah, maybe I'm confusing to two things. That's right. Okay. So you, you, you get a print. Just, you call it C tilde C. And then you get the premium C, C tilde C is again the Jacobian. Okay, so that is that is when you know construction need to be the month for. Mm hmm. Oh, I see, I'm seeing all my oh, oh, oh, oh, oh, that's right. I was looking at all the things. That's right. Yes, yes. Did I get the latest, I think so. Okay, actually, yeah. Mm hmm. The corollary of the month for a professional yesterday is that actually the pre map restricted to this Quintics. Mm hmm. So you have P six restricted from to the argue. Remember, these are the, the global covering of a Quintics with them, even, even, even to the ocean point is reject by J. So you is one to one correspondent between this, this is Jacobian say and and the coverings coming to discover is plain Quintics. A little warning here is the thing is not yet. This is just a rejection. And once you have to prove and I'm not going to do it. I still has to prove. That is not ramified there. Yeah. There's no. Right. No, no, no ramified basics is not ramified on. And that that involves again looking at the, at the, at the co differential map if this is so it is some some work to do right, but that is not ramified. So it's not only that it's an objection say okay but maybe maybe on that part will be completely the purpose ramified and in fact is, we are not having a good count on the degree. So once you prove that, then the contribution of this locus is one. Okay. Okay, that's I forgot to say yesterday so let's come back to our other locus the trigonal ones. Okay, for the trigonal ones, we had this trigonal construction, and we had totally to be double the generations. So what happens. Yeah, can leave it to work out the different generations, but actually. So the trigonal construction construction from yesterday again, gives us defines a map. So it goes to define a map and cold towel from I'm going to give some notation from this is always the Nagy Smith notations G14. So that means that you have a car from genus G minus one with one G14. So let me put like this one G14 so tetragonal curve. So you can go to RG. And then you have of course the pre map this is, yeah, you have extended to H minus one. Okay, so you get an allowable. Okay, so that's right I didn't talk maybe I want to leave it for the exercises we spoke yesterday about what happened when the G14 has two points, our fiber with two double points or all the point was together. And when you consider those cases you get things on them on the boundary, but in any case that would be something. Well, there is one case which is singular, what do we don't we don't we were not considered. But then in case you do others will be allowable. So the claim now what they want to show is that you you compose these two maps. This is how is identity. So you get back to the identity. Let me see. Okay, so there's a proposition. This is defined Tao X like above. So, first PG from Tao X is again the Jacobian of X. It's not properly the identity. Forget my claim. And second of all, the, actually, the, the map, the map you have a map. I'm going to call it C from C till up into the Jacobian into Jacobian that define it by. So what you remember what was C tilde in the trigonal and that was that was given by the trigonal construction. The, the were the pairs is above the person said one device or so you have a pair AB that send it to fee. Okay, I have to say what is what I'm going to use this notation fee a plus fee B. So remember, this is the cream of this of this double covering is the other is the other cream up for. So once you identify the Jacobian with the cream. Do you have a, what is the other print? So let me put in a different color. The other print map is, is very similar to the other Jacobi map. So you take is a composition of C tilde, you, you, you take the other Jacobi map into the, into the Jacobian of C tilde, and then you project into the cream with one minus the evolution. Okay, this is the other print map. And the, okay. Actually wanted this more one. And the fee. And let me put because I have notations for canonical ones. So let's call the small fee. It's more. Okay. The small fee is actually the canonical, the Abel Jacobi map for, for the course, because. So let's see. So everything makes sense. So you have. Let's see if it makes sense. So remember the, the, the, that makes sense because if you remember the construction, the C tilde is a subset of the, the symmetric product of it. So this, this AB are points in index. Okay. Well, this is not so. So how do you go with the proof. So this is a typical proof. How do you prove that something is. It's a prim, it's a Jacobian or more generally a prim, a prim, you need to use the universal, we use the universal property of the prims property of primarities. There is an universal property of Jacobian so is one of the very similar. So that means you have every time you have a map from C tilde to C. into an abelian variety that is more general that this is symmetric. Symmetric with respect to the evolution in, in the, so C, C tilde comes with an evolution. So when you are in the global coverage, you have a map, when you have the abel prim map that I define over so it's the composition with, let's see. Abel Jacobian map into the Jacobian when you fix a point C tilde. And then you go to the prim by, so this is our C. Yeah, that I define here, but it depends. And this is one minus. I don't know why I changed Sigma, why I changed Sigma because I call it here Sigma, I don't know, let's call Sigma. The evolution. Okay, then. So every time you have a morphine some C. symmetric, so you can you can put a curve inside the Jacobian, you can extend the map to the, to the, to find this map to the, into the, from the frame C tilde Jacobian, and you close the diagram by a translation. You translate a translation by the minus the point that you use the point the base point so C tilde. Okay, this C tilde that I use for that. That's the universal property of the varieties. So I have such a symmetric map. So symmetric means that the C and the evolution commutes. Minus C. Okay. And this is what, and in our case this is. So this fee has this property just by definition. Observe that observe that. When you say C Yoda from a plus day. So what what is the other of a plus the name of the evolution Yoda. So you see. Let's remember here, you have these eggs. And you have three columns with this the drama right and over over a point you have a B C. Okay, and the tree tree going to construction, I give you the, the, the diagonals, all the possible diagrams. CD. Okay, this is the fiber over. So, if the evolution of a B is, is to take the older two points, complementary ones on the, of the G14 C plus T, and C from plus T. So my definition is fit for C. Plus feet from D. And this is my nose. Minus. Minus fee a minus fee be because, because the sum of all these four is constant. Yes, and the Jacobian of city. So you take fee a plus fee B. This is, this is, well, to be zero. Well, yeah, you can see you can, you can translate it to be seen in more than a nice constant because it's zero actually on the Jacobian of city. Because all the four points lies over, like on the fiber and this fiber depends only on P one and P one in the Jacobian max to zero, this is constant. So, this is zero. So at the level of the Jacobian, the sum of these two points is minus the sum of the other points. Okay, so they have this property of the symmetric. So the universal property of the spring varieties tell us that the city less well defined. Okay, now it suffices to show. And this is, so I want to show that what I want to show. Yes, right. This is isomorphic. So it suffices to show that the homology class of the image of the curve in so this is a homology class into the Jacobian of X, Z. Moment C. Yes. Yes, yes, yes is twice the minimal class. Twice theta G minus two where theta is the class. So it's the class of the data device or in the Jacobia. So it's the class of the principal polarization. Okay. So that is, this is well. This is my civic is criteria. Maybe I want to recall it a little bit later. This is a little generalization of Matusaka criteria once something is something is a Jacobian. Well, and in this case, what I want to prove that is a pre the Jacobian is the plain variety is equivalent. Okay, so we are going to choose the following the generation. Okay, use the generation. This is, so one can compute the class by hand and show that this has the right, the right class. But one way of, you need to look at this is by the generation. So we are going to degenerate X, put X in a family of such generic, generic curves, and we are going to degenerate to the union of a course of genius. One less was putting names. Yeah, G minus one. So the difference is that this, this has, okay. So this is a family of tetragonal curve. So you have a family of tetragonal curve 60. Each one comes with a divisor DT G14 or a linear system. And at this, in the center of fiber is X, no one with the P one such that X, such that the points, three of them are here in X, no one and the other is the green point here. Okay. So this is something you have to assume. Assume that the limit divisor, divisor class of this DT is the T zero plus not this quality zero is the class of a G13 plus the point, the green point P zero. For some, for some P zero in any zero. In such a way that so every FD is a four to one, but the X, so when you restricted F zero restricted to X zero is a trigonal one. Okay, and then the other the other brand, the other, the other sheet is the P one. Okay, so I apply the trigonal construction to this, this the generation. So to the center fibers so the trigonal construction keeps you the seat tilde is, is, is going to be precisely our verting and coverage. So you take zero copies of X, X zero copies. And then you blew like we did yesterday. So like this. So that goes to X zero. Modulo P one, two. Okay. And this. So this, this one is is to do one. This comes with the trigonal covering. And the normalization of C is X zero. So normalization of C is X zero. Okay, so business. That's right. So what is the class of the class of class of what we are going to compute the class of C, C does not change the generation generation. So you did the trigonal construction for every element in the family. And this is what I'm showing you is not it's just the central five for the others. You have him always this city, see, you can do it for, for everything does not change. In the generation family, because essentially, yeah, this is this is a discrete element. This is a discrete, this is a discrete group. So they all have the same class. And in the limit. In the limit. In the limit. C of C tilde. Oh, sorry, this should be C tilde C of C tilde. And how does he look. Well, is the union of fix of the copies of these guys. So this is why it's very simple proof you do by generation because you see immediately that you have, you have two copies of the of the embedding of x zeros. So, let's put that. So it's zero prime. That's precisely C tilde. And so the class, the class, this class is the same artist class. And this is twice the class of the x under the Jacobi embedding. And then you know that, okay, at the class, the class of the curve inside inside inside of the Jacobian. This is again, well known fact so much to soccer, for instance, this is twice the minimum price. Yeah, so x. Yeah, have to clear the chinus of x is G minus one. So this is one one less. You can look at much to talk as good day. And that's the proof. So this is the generation is not an expensive argument. So you just you have to apply this criterium that we know for for Jacobian so the premise twice what happens in the Jacobian. Okay. Yeah, so we have now, let's RT, the RTG. So double coverings allowable double coverings. Well, no, let's, let's say smooth ones. First of all, in RG, such that C is trigonal and then consider the closure inside of RG. Okay. So we have this map tau that goes from pairs of, of curves with the gene one four into RG. Now, the inverse construction so we have also a way of going from here to here by the inverse construction of the trigonal of the trigonal and trigonal construction. We can conclude as the image of tau is the is is the full RT. Because, so every, every, every, every double cover allowable double covering. Alpha Quinti and I'm sorry alpha trigonal curve and it is is is the Jacobian or something. Mm hmm. So. So as a corollary for okay for. So here's the observation this isn't this is general but the main observation is, when you have genus five cor is always the dragon remark. X in F5 is always to travel. And in fact, the general, the general one will have the generic. And we will use this calculated the generic X has exactly five you won't force. This is by a pretty letter to you. Okay. Yes. At most that's right on our because hyperliptic is not the dragon in. Okay. They will have always the diagonal map, maybe not. Yes. Depends how you put the dragon has always a g one four. Okay. So that's g one four. As you said, and yeah, could be older that they are not formally the dragon up. That's right. So that that is that is why you see this. Yeah this this construction in general but if you're lower genius. So you have to have more more jobs maps but in higher genius in higher genius not every, it will be a smaller locals of those will have an engine one folks. Okay, now. Let's compute the degree. Okay, so let's concentrate again and g one six g one six in the case g one six. If I write it here is going to be. I'm sorry. Yes, because it's the end of the page. So I have to write them a little bit lower. So, let's look at the diagram again. We blow up our, let's blow up this RT, the closure of, of double covers of trigonals here. This is what I mentioned to. Remember, our six has dimension 15. So this is dimension 13, but we computed yesterday properly. And then this is the blob over RT for six. Are. And inside here you have the sectional device so that maps to RT. This is a P one bundle. This is one side. In the other side, these are the maps over the Jacobi and Jacobi and locals a five. J five. This one is of God I mentioned three. We have much less. So you have a one dimensional fiber here to blow up on these locals. Let's call it a five. And inside the exceptional device. So here the optional device or is now a P two one bundle. And you have the final. So in principle is only rational. This p p six p six tilde. And we are interested in computing in computing the degree at the level of exceptional device. So now this was the first device was of course as divisors they have the same dimension. And this is the real is well the final, of course, one has to check first that the map is the map is injective in every point. There was the condition yesterday. So let's see, should I check that. I think I don't know if it is included. I don't think so, but so recorded record as a co differential. Well, this is the right to recall. Let me compute the co differential of the premap is at the point, a particular point. This is the symmetric product of a zero of the c omega c data. So it's a multiplication of sections into a zero c omega c square. Okay. So what we're doing for this particular points. So in this case, the kernel of the co differential, you see, it has a kernel because now this is called I mentioned three. Let me see. Okay, is the quadrics kernel want to be all the quadrics in p four. That contains the canonical, the pin canonical core containing C. So I want to, I hope this, I mean, don't miss out with other see this see I'm going to use this grow the big C is the canonical canonical print print canonical from from C into the projectivization of C omega C. So in general, so it will always have one dimension less than the canonical. This is PG minus one. This is called print canonical. Okay, I think I define it, and I will call the big, the other big thing is going to go from X. And this generic actually is going to be an embedding. This is actually an embedding. That's goes to make the X. But this also PG minus one. Five, in my case. And sorry. What I'm saying. This is PG minus one. This is four, everything is four, four, four. This is the canonical. Okay. Now. Yes. So I'm going to give a very similar construction as the other plain quintics. So X is very general. So a non hyperliptic. Just to call X is X in M five. Not hyperliptic. No trigonal. And no trigonal and no vanishing data. Data. So vanishing data null is when you have a data characteristic with positive kappa. The characteristic with sections. But the data characteristic is even data characteristic. So the, the, the square is the canonical. This is, this is kappa. Sorry. Okay. This is a vanishing data, but we don't have those ones. So since we don't have this vanishing data, we don't have similarities on the singular. So the other singularities. Of the data X. So now I can, I can write what they were doing right now. You can consider this, then, then. This is what they were saying. This is a smooth curve. Of genius. Genius 11. That can be computed. It comes with the evolution induced by. By the, yes. Oh, this is not for me. And I'm going to call this is just as to stick to the. I'm going to call it alpha X. I wanted to think about alpha X as, as all them. Is the family is the family is the curves occur that parameterize all the G1 force. In, in, in. In on the curve. Or more precisely pick force on X. Yes. With, we have sections of L. This is a singularity has to be at least two. Actually will be two. Let's see. So. Yeah. And then you have the quotient. It's called. To. To distinguish from the other. Yeah. This, this you have a map. Okay. That's right. This is a little map. Two to one. And this is then this is a genius score. And actually it's a plain Quintic. Okay. So, so alpha. Alpha X parameterize all the G14 and every G14 on the X. It comes from a, from a family of planes of planes. On a singular quartic, quartic. That contains the canonical curve. Okay. So it's how we construct the other. Yeah, playing, playing Quintics. But now I'm looking at them. Yes. Family of two planes. So you, you, you go through these two planes. So how does it look you have a family. You have a singular quartic. This is a singular quartic. You see this is in, in, in before. So this singular quartic is three dimensional. It has, it's actually, it's a cone. That's a cone over a smooth. Two dimensional quartic. Sure. Yeah, sorry. Quadric story. Yeah. So I, I, I am excited to make in the drawing because it's not going to be quite accurate. But. Mm hmm. But this quadric. This quadric contains actually two, two families. Mm hmm. So actually you have. Mm hmm. Yeah, this is, this comes from, from, from, from the evolution. Right. So you have this. And you have K minus you are four. This is the evolution. And so do you have to actually took this quadric. Singular Quadric S. S contains actually two families. So you have two families of planes. And each, each family of planes caught out. Cutting out. And you want for colleagues. Yeah. So they always come by pairs. The, the, the one force. So as a conclusion, okay, this, this, you can identify this alpha X. As the fiber. Of, of. X at the. Six into five. Has one dimensional fiber. Somehow we know. Over a generic. Mm hmm. Because, because precisely because one. Because the generic X. The generic. The generic X. The generic Jacobian. Mm hmm. Has the. Has one dimensional family of singular quadrics containing. One dimensional. Family. Of singular. Containing. Okay. And this is, you can identify that. So now you can identify the quadrics. You see, I am, I am proving this proposition. The quadrice containing before. Is precisely. Parameterize also the fiber. The fiber over. Over a fixed generic. Okay. Now the computation. Now understand better the geometry. The local. Degree. Of the six. On 30. Is 10. Mm hmm. So let me see. I wrote something. Yes, I think it's going to be. More or less clear. Okay. Remember, remember. I don't know if you remember this. My notation that we have to show that. So we have. That this map. That they said. So that is the, the restriction. The restriction of. Of the P. Tilda to the. To the fiber. Mm hmm. Is injective. In the normal space. At said. If only if. Is to check the conor. On conormal. On conormal spaces. So let me put it here. What is this. So the project. The station of this guy. Is precisely. On P. On the sectionals over the point. On the fiber over there. And this map here. From here. To here. So it's called it. P and the sectional. P T. On the section. Yeah. So it goes from P1 bundle to P2 bundle. Okay. Yeah. But the spaces are. Are same dimension. Okay. Same dimension. Mm hmm. So it's subjective on conor. And the conormal on the conormal looks like this is the six. In corner. Cause from. I don't know if I'm consistent with my notation, but this is conormal. Into. The normal bundle. R T. R six. If this is subjective. So. Mm hmm. At one point. You do think one over over one fiber. One fiber. One fiber. This is of rank. Three. Well, it's a bundle is a rank two. But this is the most we already computed them. So we already computed C. You see, we computed that the kernel of the full thing. And in the kernel is. Is one dimensional. It's almost one dimensional. So it's going to be subjected. This is subjective. So. You see, this is just, just a part. From, from the full. The. Got tangent space. Yeah. So this is just. We have the full potential space in five. Or G five. Sorry. G five. And the content of the normal. And here is. And this is the. The cotangent at the, at the full a five. So you have the tangent to our team. Yeah. This, this normal here is conormal. Of my is. Maybe it's better to put the. Start here. Okay. And this is, this is the time. The cotangent to our six. Okay. And here. This is the differential. The P six. Yeah. And he has a most. So this is subjective. Because. So this year. This is subjective. I am over time. Because. The, the dimension. Of the kernel. The kernel of. The kernel of. The kernel of. The kernel is almost one. So we come here is one. So if you restrict to two, to one, one, some of the. Of the, the bundle. It cannot, the kernel cannot grow. Can just the. The dimension of the kernel. It's almost one. The kernel of, of, of. Restricted to conormal. Is. At most. So if it is ranked three and ranked two and the company's almost one. So we can really, and at least in this case, explicitly check the condition that the, our condition that tells you that the degree, we are computing the local degree problem. So we make a break, of course, a break and just, and then I make the, I finish the computation to give you the 10, okay? Thank you. Thank you. Okay, come back.