 Okay, yes. Before going further, I want to do a little computation that I will need later. You see, I told you about this orthogonality of complementarity of the Jacobian downstairs with the print, but also they are orthogonal with respect to the polarization. For instance, at the level of the tangent planes, the tangent spaces, and the zero, one is orthogonal to the other, because the tangent space to the Jacobian C is orthogonal to the tangent to the print with respect to the polarization to this valignia form. So by construction, let me put this as a remark. By construction, the tangent at zero of C tilde is the space of holomorphic differentials of C tilde, because it's a quotient. Jacobian is a quotient of this space. Actually, it should be dua. So the tangent space, and you have also an action of the involution, yota, into this space. So that give you two split as a representation in the invariant part, as I said, plus and plus one, and then the involution acts with minus one, minus. So the plus one is isomorphic to the tangent space to the Jacobian downstairs. So C omega C. So I should be dua also. And the other is this would be the tangent to the print. But more precisely, if you apply the projection formula, so let's see. This is the C tilde, omega C tilde. That is isomorphic to the H0. You push forward the curve. So you have C downstairs. Push forward of this line bundle. Okay. But actually, you can write this bundle. You can also tensor the trivial one and nothing changes. But on the other hand, for etal cobrings, for ramified type cobrings, you know, this is also Riemann-Hurwitz. You know that the pullback of omega C is isomorphic of C tilde. This is because it's etal. You have ramifications. So that's actually this is C tilde. And then here you apply the projection formula that tells you that this line bundle is equals omega C tensor push forward of the trivial bundle. But the push forward of the trivial bundle is given by this line bundle that we construct. Omega C tensor. C tensor it. So that's just moment. This is plus direct sum. And well, this is distributed and H0 of the direct sum of line bundles is it also is the sum of the H0. So you get H0 of C. We can see direct sum of C tensor tensor. Yes. Okay. So you have the two parts of the composition. And this is precisely this is composition I was talking about. So this part is precisely the dungeon to be dual. It has to be dual because yes. Yeah, it's omega C tensor. Omega C. Thank you. Yes. Okay. Okay. Omega C. So this is this is what I let's just to put it the historical context is this what the Supreme, the Mr. Supreme, Friedrich Supreme, he was an analyst, an analyst. He didn't know anything about coverings or something. He studied what he studied. It was this print differential. So later month for color, the print differential. So the print differentials are the sections of this line one. And this is what how Friedrich Supreme was studying at the time as an analyst. This is I'm talking about 19th century. Okay. Now, this was a little remark. So you know also the tension to the print. So let me define the print map. So the print map, because this is classical print map, it will be more print maps later, is a PG from RG, the terrible coverings into a G minus one. I never said that I mentioned, right? Okay. I will make a remark. So you take a class of a double clovering or equivalent a curve plus a two torsion point and you associated the print with the principal polarization. Okay. So to observe, so maybe I can put it here. So in this case, the dimension of the print. And by hood with formula, this is two G minus one minus G. So it's G minus one. So this is why you get G minus one. Okay. So that's very nice. We have a global wave to from going down again a curve plus an extra data that the two torsion points and you get a principal polarize a billion variety. Let's come back to my table. Where's my table? This table here. And this table. Now, with the print map, you can, you get from MG to AG minus one. So this is dimension of AG. So I go from one up. So the print map, for instance, go from here. This is P four from here to here. This is P five. I'm from here to here is P six. Okay. So so just comparing dimensions, we expect that now the the general pre the principally polarize a billion variety of dimensions. Well, at most, at most five are primes of some double coverings, the double coverings, just because of the dimension. Of course, you need some theorem to do that. So let me write it, summarize the theorem. So this is, this is a big theory. So so the print map is dominant is dominant if genius is a mode six. And this was already known by Virtinger. Also, German mathematicians from the end of the 19th century, beginning of the 20th century. And yeah, that that was already studied. And then PG is known is generically finite. Yeah, not only finite, generically injective. If the genius is at least seven. So just look at the dimension is where you have room to put RG inside of the G minus four. And in those cases, as soon as the dimensions allow it, you get generically injective. And which is more striking is that the pre map is never injective. So it will be always some law side inside of a G where the pre map is not is never injective. That was first noticed by Boville. But then is more generally stated by the, if we have time in the course, I would like to talk about tetragonal construction of Donag. Let's put the problem. So this is Donag is tetragonal construction. So over over the currency where the C is tetragonal, the print map is not going to be injective. And you have so such low size in any genius. Of course, higher is the genius, this low size is more smaller construction. And I should put some names because I think the first one to show the generic conductivity is a work of many years, many mathematicians. I think the first one to prove it was Friedman-Schmidt for some, it's not one of these genus, I think genus at least 16. And then for genus at least eight kind of did it. And then, and those are by the generation. It was very much by the generation. And then Veltas have another proof. And the last proof I know is the bar has another proof. And those are more in a more geometric one, as in the Torelli and the Torelli map. So those are the names. You have many papers related to this. That's right. Here's the open question that I think Pavel already mentioned, still open. What is exactly the non-injectivical locals of the print map? Describe completely the non-injectivity locals. So although the theorem is a little bit now classical from the 80s, we don't know the answer to this question. Okay. Okay, let's see. So how do you prove that something is a generic injective, for instance? And then we'll be interested in this question. You have to look at the codifference of the print map. And so to show that something is a generic injective, you need to show that the differential of the print map is injective and generically finite or equivalent of the codifference is subjective. So let me write it down. To show that PG is, first of all, generically finite. One has to show that the differential of the print map is injective or equivalent the codifference, so the dual of PG that goes from the cotangent space of AG-1 into the cotangent space of RG is subjective at the generic point. And this is better to look at the codifference because we have a nice description of these cotangent spaces. So let's see. We use the following identifications. So let's put an element in the image. P is the image of C eta. Mm-hmm. So something in the print. So this is in AG-1. We have the identifications, identifications. So one size, you see the tangent, the cotangent at the point C eta of RG. And I'm using the fact that RG is finite to 1 to mg by this forgetful map. So at the level of cotangent, you put yourself in one point, you can identify the tangent plane to the tangent plane of the image. It's just generically, it's just finite. So this is the cotangent at the point C of mg. And this is also well known is the space of quadratic differentials on C. Okay. I am skipping a lot of theory here behind. I just want to apply to my map. On the other hand, the cotangent space to P of AG-1 is the symmetric product of the tangent space to P, the dual of the tangent space to P that I just compute is the symmetric product of A0 of C of the print differentials. Mm-hmm. In such a way that the co-differential map, when you use these identifications at this generic point, at the generic point, it's just, well, the most natural thing to do. You have the symmetric product of these two. So you have two sections of of the print differentials. And you go to H0 of C omega C, well, just multiply. It's given by the multiplication of sections. So in principle, these multiplications of sections, you go to, you multiply them, and you will end up in this omega C square tensor at the square. So the multiplication of two sections here is a section of this line bundle. But then you identify these two spaces because it square is three. Okay. So this is the co-differential map. And the question is when this map is subjective, when it's a subjective. And it turns out it's going to be subjective as soon as the dimension allowed. But as soon as the dimension of the left side is bigger than the other. And that's how you get the theorem. So let's just write it down. Lema. And it's not that difficult. The co-differential is subjective for genus of these six. Okay. Good. Now, from now on, okay, this works for energy. Okay. You have a full theory. And for the course, I want to concentrate in the beautiful case. Look at this. And where's my, sorry, this case. When both dimensions, this dimension is the same as this dimension. Okay. It's the only case. And for double course, what it happens, we know in this case, the map is generically finite because it's also dominant. So the same dimension that means it's generically finite. And we want to compute the degree. And this little question of what is the degree of this map leads us to a worth of geometry. Really opens the door to all possible classical geometry. And this is what I'm interested in to show you in this course. So let's concentrate to the case g6 from r6 to a5. Okay. So let me tell you right away the theorem. This is proven in the paper of Donaki Smith, which is also on the website. p6, r6 to a5. Well, I know it's generically finite of the degree 27. Okay. And I'm not going to get to, yeah, maybe everybody knows why this 10 is 27 means, but I'm not going to tell you right now. I'm just going to go as the authors did it. Okay. Let's do it step by step. So what is the plan? To prove that is that first, we have to extend pg to some partial compactifications. So pxg extends to a proper map. So to compute the degree, you need to have a proper map, essentially. And if to get the proper map, you have to allow more more double coordinates on the boundary. And this is the part which is more difficult, but in some sense, beautiful, when you allow things on the bound. So when you have that, you extend the pre map to you have to extend. So this is also on work, extend the pre map to the boundary. So the boundary is what we will call a lowerable double covenants, double covenants. And essentially, that will be now covenants between stable curves, but not any stable curve. I will say which ones. And we have to study the generalized. So you have also a data divisor. So how to study the generalized data divisor, data divisor, and then you compute also the local degree over a long different low size on a special file or a particular file. Oh, yes, I have to tell you they do all the construction over one file. Okay, good. Okay, now is where the course starts to be a little bit more technical complicated. I will try to give you the main ideas. This is the plan. I think at first I have to tell you what the stable curve is. So stable curves, they were invented to compactify the modulus based on g. It was not compact. So we say that a curve C is stable. It could be smooth. Is it smooth or is it smooth? Or it is connected. I don't want this connected one connected. And the only singularities are nodes, are ordinary double points. And the automorphism group is finite. And the automorphism group of C is finite. Mm-hmm. Yes. So for instance, you have for all rational components, they must have three mark points to make because you don't mark the points, you have infinitely many automorphisms of this rational curve. So either have a node or meet another in other components and three points. Mm-hmm. That's right. So this leads us to some modulus space, a compactification of mg. This is the modulus space of stable curves. Curves of g and c. And let me do the following assumption. So this is how we will start it. But I think I want to do this little computation because, yeah, I mean, to do something, prove something, let's do something. You have a C tilde stable curve. Yes. With an involution, with yota, involution. And assume the following condition. This is this is the fixed points of C on the disembolution are exactly the singularities, the singularities, the singular points. Mm-hmm. And at a singular point, two branches, the two branches, so you have a singular point, the two branches are not exchanged. That was the first conditions to work with. So under this condition, you have the following situation that the quotient by the yota has only ordinary double points, a singularities. Mm-hmm. So that means I call C, C tilde yota. So I get again, actually, I'm going to get again a stable curve in g bar. Mm-hmm. And the proof is a local computation of the nodes. I don't know. Let me see if I can do it or I leave it for the exercises. Mm-hmm. Yes. Let's do it. So take a singular point S in C tilde, singular, which is fixed by the involution yota. Okay. So the completion ring, the completion ring at this point, locally, is isomorphic or can, actually, can one identify to the formal power series in two variables, call it UB, modulo, the ideal U times V. So essentially, you have a singularity like this. And in one branch is to have the U, another branch is V. Mm-hmm. U equals 0, V equals 0. Okay. This is how it looks locally. And, okay, you have two possibilities. If yota exchanged the branches, that means that one can choose, one can choose this U and V. So how does it act the yota in these parameters? In U is V, and in V is U. So I think the drawing is, so you will just exchange yota. Well, this goes to this. Okay. That's it's clear. But then the subring of invariance is on the, so invariance on the yota of this, this completion ring is, is the formal power series. So C is the sum U plus V. This is the subring, which means, so, which means that, so the invariance on the E is, is that gives you the local, the local description under the quotient. So this is what you want to happen when you go to the quotient. So essentially, if this happens, the quotient is smooth. Okay. This is a smooth ring. This is a regular ring, which is regular. Okay. If it changes, it changes the branches. Office of S. So the, you can choose these local parameters as minus U. Ah, doesn't it change? Sorry. Sorry, sorry. It does not exchange the branches. So it acts in this way. And then the subring of invariance, the invariance on the yota of this completion ring is, is given by the power series. Before. Yes. Because you are, you are identifying yota with the, yeah, this is for those ones that it doesn't, the minus doesn't, it doesn't matter. Modulo, uh, no, that's all. And then, but this, this subring can be identified to the ring also for my power series in two variables, but now modulo is the product. And, and, and that means that we have a node on the coach. So let's do that. So I, I draw a picture like this, in this case. So this is the fixed point. Yota doesn't exchange the branches. So you get a node downstairs. You get a node. Okay. So what I want is that, okay, this is my little computation. I think this is the most complicated computation I'm going to do today, but I just want to explain the importance of these changing branches or not extending branches. So this condition allows you to go nodes, goes to nodes. Okay. And this is, this is what they want. Okay. Okay. I have to do some choices now. Okay. Maybe, maybe, maybe I could jump some, I want to give you the definition of allowable coverings. One is my initial moment. But yet, I have to work on the, yes, yes, yes. Okay. Okay. Let's, let's do, I think it's all interesting anyway. So independent of the, of the purpose. So assume, assume that we are in this condition. Assume the condition asterisk. So the quotient has only the ordinary double points. Only nodes. Let's put it shortly. And you have a, you have a commutative diagram like this. So you have to, you have to go through the normalization of your curves. So this is Pi. That's called, let's call F. F is the normalization of the normalization map from C. And so you have a map, let's call Pi prime on the level of the normalization. So as a little lemma on the conditions, the push forward, it behaves a little bit like a, like a metal case, because the push forward of C, even if it's not, this is a stable curve, it's still the, the canonical divisor on C tilde. I'm cheating a little bit because we have to work out, still work it out. You see, you see that the, the issue here. I am now on the boundary. My curves are stable. I have nodes. Okay. Nothing else, nodes. And we have to make it, get an idea, or clear the idea of what does it mean to have line bundles of these tables. And so what does it mean, the shifts? I will give you this idea. First, let me tell you something that the arithmetic, because of this lemma, the arithmetic genus, now we have to talk about the arithmetic genus. I still have this property that this is twice the arithmetic genus of C minus one. Okay. It still holds. So you, you have this relation. Now I have to tell you what are the Cartier divisors just to get an idea of what they're on C tilde. And I will give some examples as well. Okay. Very simple one. So don't worry if it looks technical. Let's go K tilde. So that will be the ring of functions on C tilde, K, the ring of functions on C. Yes. So if C tilde has several components, those will be the product of the, of the, of the field of functions of the component. And then I have to define the group of Cartier divisors on C is, yes. So you can think of a shift of, of, of, of ring of functions of, of, of, of, yeah. So when, when the curve, when the curve is, is connected, you have a field of functions associated to the curve. And here, since I have several components, I cannot talk about what I feel, but they can think about the ring. So just take the product of these fields on each component. Yes, you can think of, oh yes. Yes. Yes. I think he will. Yes. More or less. Okay. Okay. So, so the group, maybe I'm more familiar with Cartier divisors. So the group of the character divisors of this curve. And so is the C tilde. So you have a part coming from the, as we see the divisors, normal divisors. So on the smooth parts, so the formal sums of points where X is on the regular part. Plus something coming from the nodes from the singularities. Okay. And then, okay. And they have to, so do you have these, these, these ring of functions. So you have at the, is the, is the, is the fiber, modulo, O, S. Okay. I mean, it's all local. Yeah. You see, C tilde and the point S. Okay. So, so let me see what happens. So you have the normalization. The normalization. Assume that you have these two points. These two points under the normalization map get identified in one node in C tilde. So this is C1, C2. And let's call new one and new two the valuations of these points. So the valuation of, at the point is, is, is, is the function that tells you you have a pole or a, a zero of a function there? Yes. So locally of C1 and C2, respectively. Okay. So, and so you have the following, you have the following exact sequence. This is the non-zero elements in C goes to X, C tilde, O, S, C tilde. So what do you do in S? So this is only at the point S. You take the two valuations and then you map it to Z times Z, O. So our function here, you take the valuation of the function. And these are integers. So for instance, if, if the valuation in F, the, of F at the point S is zero, that means that you have no zeros and no poles. You have negative, you have a pole of this, of this, with this multiplicity, I mean, with this, with this order. And the, and the, and the quotient is, can be identified with C tilde, the kernel, because you have F, F in the kernel of these valuations. If only if the valuations at the point are zero, both of them. So as I said, no zeros, no poles of F at, at both points at S1 and S2. So you can associate it to this complex number name. So in S1, F, oops, oops, sorry. No, I mean F in S1, F in S2. This, this is what, so in conclusion, in conclusion, you should not be scared of this, local description, because even a nice way of describing, after you choose, choosing parameterization, T1, T2, local, yeah, local parameters, parameters around the points, there's one in S2. So these points are over the, over, over my node. So you can identify is C tilde, this is isomorphic to C tilde, okay. So a point here, so yes, you have a function here. So a term of a function, it can be written as this parameterization, this parameter into, so you have two different points. And then you send it to U, V, M, N, those M and N are integers. Okay, so, so conclusion, over our node, you can describe the cartel divisors as, as this triple. And we are assuming, and I can tell you exactly how does he act, your time this node, by the following formula, Z, M, N, at the point S is minus one, M plus N, Z, M, N. So it doesn't act, it doesn't change the order of the, the M and N that the branches doesn't change, it's just to change the minus one on the, plus or minus one on, on Z, depending on M and N. Okay, so you have, you have as usual some sort of diagram. So you have the cartel divisors to each cartel divisors, you can associate the aligned bundles on Z tilde and you can go to, with the norm map. Here you take the push forward of the cartel divisors. So on the part where smooth is clear, when the part is not smooth, you can use this formula. This is, this is also, this is a norm map also for, for extensions. Okay. So let me tell you exactly what it is. So on the, on the smooth ones, it's just to push forward and over the nodes is minus one plus plus N. You take the square of this and that's all at the, at the node downstairs. Okay, let me have a little time when I should finish now, right? Yeah, yep. Okay. Yeah, I had to be more. Okay, well, just to conclude, my plan is to just to give you an idea of what happens in the boundary. Look, I don't want to do all the details, but at least you know, I will show you next time some examples of allowable covers. So what would be an allowable cover would be something such that the kernel of the norm map is an abelian variety. That's what they want. Yeah. And I will tell you exactly which allowable coverings are, I mean, how they construct. They will give you some simple examples and I need to know them in order to compute this local degree. But it's like an excuse, this computational local degree is an excuse to have a look at the boundary and understand all these these constructions on the node. This is fascinating, but it's technical as well. So, yeah. Okay, questions. Can I ask a question? Yes. So in the middle of the presentation, you mentioned a lemma, this is a cotangent map of PG. Yes. It's a surjectile for genus at least six. Yes. Is that a surjectile or generically surjectile? Generically, as a surjectile for the generic point is what they need. Oh, okay, surjectile at a generic point. Yes. Okay, that's my question. Yes. And then that would mean that is generically finite. In other words, sometimes for example, genus six, it can have some positive dimensional fiber. Exactly. Yes. And we will see them then in the computation. Yes. So, because this computation's degree, I will do it, I will tell you now, I will do the computation of the Jacobian of a genus five curve. This is a principally polarizable in variety. And I will look at the fiber of this generic Jacobian. And the fiber is not finite. This is the beauty of the thing that they computed the local degree over a fiber that is actually not finite. But then they do the blob. And I find it beautiful that you understand many things. I mean, something that looks abstract, what is the blob? It can give you exactly what does it mean, the normal bundle, what is all in geometric terms. So there is a full description of what is the blob, what is the local degree at this blob, and it can make the count, precise count just by looking at the geometry of those objects. So that's the beauty. So, yeah, I mean, this hard work that is rewarding. Yes. Okay. Thank you. Yeah. So, for the angel, so in this P6 map, is in this case the contracted locus completely understood? No. No, the contracted locus. You mean? So where the map is not finite? For that case. That's a good question. I don't think so. There are many loci that is very well understood. The cubic trifles, for instance. But I don't think it's known that, okay, that is, I don't think so. No, no, no. I will have a look at that, but I, no. Okay. Yeah. And for higher genus, are there any conjectures about the injectivity locus? No, I don't know. So, Herbert Lange and Isadi Elham, you can ask Elham. They have a nice paper on the, they show many examples that they don't know, they are not coming from the tetragonal construction for any genus and I think any clifor index. So they are examples of, you know, no, yeah, in any, it's a very, very general type of examples in any genus where it's not generically finite. So that, that, that contradicts original conjectures, originally people conjectured that all the non-injectivity locus come from the tetragonal construction, but it's not the case. This is not, not true. This is not true. So we don't know, I don't know, I haven't hear any particular conjecture. It's really mysterious. Yeah. I mean, if you can put your hands on low genus, it would be nice. Yes. But no. Okay. Thanks. More questions? Okay. So see you tomorrow then. Or see you at what, for, for, for 40, for the picture? For 50. For 50. Okay. For the photo. Remember, everyone, we have the group photo today. Okay. For 50. Okay. And now we say goodbye to Angela and thank you. And we move over for half an hour on gather. And then we will be back here.