 And for me, I think this is sort of one of the reasons one one might care about my life curves to begin with So here's this here's the question. So so I'm going to fix I'm going to fix five complex numbers So very general So the question is for how many T Is the hyperliptic curve so the hyperliptic curve of genus 2? So I'll call it CT it's going to be to cut defined by the equation y squared Okay, so it's going to be branched over the six these six points. So x minus eight one Say five D elliptic by which I mean There exists a cover of some elliptic curve of degree D So so he's an elliptic curve, okay? So the first enemy check is why would you expect this answer to be finite? So why finite? Well, okay, so as we know there's a there's a three-dimensional family of genus 2 curves and So how many among these genus 2 curves? How many of them have this property? Well, so such a map So this will have two branch points by Riemann Hurwitz and So if I think about the family of covers So if I think about just the family of sort of all covers of a genus 2 curve mapping to a genus 1 curve degree D So it's a standard fact Well, I can turn this around and say if I specify the base curve and the the two branch points Then I get finitely many covers after well, I get a unique cover after specifying the monodromy So in particular I get finitely many covers If I just specify the base with its two branch points and so we have a two-dimensional family of covers Okay, so Okay, then you have to give some argument to to convince yourself that you know if you have one genus 2 curve You don't have infinitely many maps coming out of it But the point is now I have a two-dimensional family and if I start with a one-parameter family I should expect to see finitely many intersection points So I should have finitely many fibers that live in this family. Okay, so how might you hope to answer this question? Well, so I've defined for you a family. I guess over P1 Inside M2 bar. So over here I have sort of I Have a map like this Which I guess I will call phi and it sends t to this curve Ct And on the other hand I have this locus of de-liptic curves But I'm working M2 bar. So I need to take a closure so this is a closure of De-liptic locus and what you want to do is you want to intersect these classes, right? So you want to take these classes and the chowering of M2 bar and you want to multiply them. So Want to intersect in Chowering or if you want the co-mology ring of M2 bar and by M2 bar. I will always mean the stack All right, so the problem with this as I've stated it is that I've done some something drastic Which I've said which is I've said take some locus and take its closure, right? And this is a And the problem with this as well. I mean there are a number of problems One is that a priori we have no idea which boundary curves are in this closure, right? and Another problem is even if we did you can start with some sort of nice locus and M2 without a bar and take its closure and get Sort of something sort of arbitrarily bad, right? You could introduce sort of horrible singularities. You could you could pick up sort of sort of really bad deformation spaces So I mean so so so the problem is that Is that taking closure is sort of difficult in particular the question arises that sort of what are the limits of Smooth D-lip the curse right so our prayer is not so obvious So let me instead sort of phrase a What turns out to be a more reasonable question? So better question is what if I just remember the whole data of the family of the Cover so instead I can ask what are the limits of branch covers? So this is essentially answered In the paper of Harrison Mumford, which we've already heard about so this is the paper in which they Prove that MG is general type for g-large This is Harrison Mumford 82 so what do they what do they do so they show that there's a proper Okay, so I'm gonna throw some adjectives up so Cohen Macaulay Lee Mumford stack Which I will call Curly ADM So G over H come a D. Okay, so this is just some object Containing a Dense open sub-stack Which I'll call ADM Cirque and So I'll tell you what the dense open sub-stack parameterizes. So these parameterizes these parameterize Or this parameterizes Primitrizing So degree D Simply branched covers a genus G curve to a genus H curve. Okay, so I guess I'll write the genus up here We're excellent XMI are smooth are smooth And let's say connected So I have this open sub-stack of sort of nice objects of sort of honest covers of curves And what Harrison Mumford do is they they compactify it, right? So this is some kind of open stack and then they give some some bigger moduli space That allows these things to degenerate So, okay, so what are these parameterize so? This compactified stack parameterize parameterizes something called admissible covers Okay, so let me tell you what these are. Yeah, simply branched. Oh, yeah, so right So, okay, so at the beginning I said I said pick the points very general so Yeah, pick them general enough so that you only get simply branched covers. Okay, so what are admissible covers so definition? All right, this takes a bit of setup Okay, so I'm going to use the same notation here. So X X Y this will have genus G and H and these will mean the arithmetic genus So this is a going to be a finite Degree D So this degree D means degree D over every component of Y morphism of nodal curves Okay, and I'm going to call the branch points Well, okay, so I'm going to I'm just going to think about for now I'm just going to think about the branch points there in the smooth locus of Y So I'm going to call them Y1 through YB Y smooth, okay, so with smooth branch points Okay, so that's the setup so F is admissible if Okay, they're going to be four conditions Okay, so one is that when I take the cover with the sorry the base with all of its branch points I want I want a stable curve so to I want F to map smooth points to smooth points and nodal points to nodal points Okay, so I'll just write this in the following compact way. So X is smooth if it only if It's images smooth Okay, I want F to be simply branched over to the Y I So so far these are fairly mild conditions Okay, and the crucial one is called the balancing condition so balancing or at least that's what I'm going to call it So for each node of Y. Let's see. So the ramification indices over the two branches Okay, so let me let me explain what this means Okay, so what do I mean here? So So okay, so F is some kind of cover and so if I zoom in on some node of Y So this is supposed to be a picture of a node So I have so okay, so this this condition says that nodes have to map to nodes so I can look at these two nodes and I can say that so each node has two branches through it and well one branch down here has to map to some branch down here and the other one maps to the other one and At this node there's gonna be some ramification index on both sides And I'm just requiring that those two branches have the same ramification index Okay, so another way to say this is well Okay, let me erase this cartoon So if I look at the deformation space if I look at the complete local ring down here I'll get something that looks like this Or maybe I think I called it Z and then the map will go in the other direction Okay, so W and Z are the two branches here and X and Y are the two branches here So the map has to look like something like W goes to X to some power and Z has to go to Y to the same power So the condition is that these two numbers are equal No, I didn't say that. Yeah, so let me give some examples. Okay, so are there any questions about the definition? Okay, so I think I started at 53. Is that right? Okay. I think I started a little later than that, right? Okay. Thank you All right, so let me give some examples so we'll start sort of Reasonable and then I'll try to convince you that these get kind of wild Okay, so one one easy way to make admissible covers is you can start with sort of nice covers of smooth curves and then start gluing together nodes Okay, so the simplest was a simplest example of this I can start with a double cover of P1 so it's like this and then Okay, let's see if I can use And then I can start gluing together some points. So I'm going to just pick two points down here away from the branch points I'll glue these two points together then I'll glue these two points together and then I'll glue these two points together And then I get a picture that looks something like this So I've set this up in such a way that I've made two nodes upstairs that map to the one node down here and The value of E here is just one because I've glued together points where this map is at all and then I still have my two branch points And this is a stable curve, right? This is this has arithmetic genus one and it has two marks points Cool. Okay, let's do something a little wackier So one way to so okay, so I'm going to start Okay, so this is so the base is going to be so this this has a Irreducible base, so let me make one with a reducible base so let me start with a An isogeny of elliptic curves, so this will be just any map from E to E prime and Then on the base, I'm going to glue on P1 with two marks points Okay, so I need to put mark points on here so that the base is stable and Okay, so now I need to add a bunch of components over this P1, right? Okay, so let's I guess this map is degree four. So here's one way you can do it, right? So you can I'm going to attach two P1s that just map isomorphically and Then while I need to have two branch points, so I need I need a two-to-one cover, right? So I can this picture is going to look kind of I'm going to sort of cheat and draw this backwards. So I have two P1s mapping isomorphically I have some kind of okay, so that's going to be my E and then I'm going to attach another P1 But that maps sort of by a two-to-one map down to this P1 So I have I have three copies of P1 and All but one of them map isomorphically to the P1 down here But I have one that has degree two where I get these two branch points okay, and Let me point out this curve this this this the source curve is not stable, right? Because I as a just as a curve it's just an E with three P1s attached to it And all three of them have not enough nodes and Again, I have four nodes up here and my my my E is going to be one This is example one Two and let me do one more example. I'm not going to explain it in detail But I just want to sort of draw a weird-looking picture Okay, so this is going to be a genus one curve These are going to be These are all going to have genus zero Okay, so I'm going I'm first going to map to a P1 the genus zero curves are going to map by Sort of X goes to X to the a All right, so it's going to be totally ramified at two points and unremifed everywhere else So I'm writing in the ramification indices at these points and Then the genus one curve maps to the genus zero curve by some map That's totally ramified at two points and simply ramified at two other points Okay, so this isn't quite enough right because of a bunch of nodes here and no nodes down here But I can do the following I can say well look at the two points Look at the two sort of highly ramified branch points and glue them together and This will be a curve arithmetic genus one and this is now an admissible cover where at this node I have five modes mapping to it and I've made this ramification index as large as possible and you can imagine I mean you can make this polygon have any size So these things even though this just has arithmetic genus two and this just has arithmetic genus one Which are the numbers we're dealing with in this in this talk? The admissible covers can get quite complicated great any questions So the upshot of all this so I have a forgetful map, so I have a map Where I can just start with admissible cover Okay, so let me specialize the case of genus two and genus one and I have a forgetful map that only remembers the source Right, so I can just remember The source and what I want to say is well, okay, so take some cover and just remember x But this is not quite good enough because as we already noted Often you get non-stable curves, so you can't just take x but you have to take x and sort of stabilize it right, so you have to contract non non non-stable components right, so for instance in these cases, I think you get a Sort of Irreducible no, but no to curve genus two and say same over here Because you here you start with an elliptic curve, but you sort of glue two points together after you kill all the rational tails And here it's the same and so why do we like this map? Well the image, so let's call this map pi d so the image of Pi d is exactly So exactly compactifies Okay, maybe exactly is not useful so compactifies The d elliptic locus Right because I'm taking the stack of all Covers from a genus two curve to a genus one curve suitably compactified and then I'm only remembering the genus two curve Okay, and so why is this better than the formulation? I I I stated the beginning, so there are two reasons so upshot So one reason is that well, we actually have a modular interpretation of this now, so have a modular Interpretation yes of the image It's a divisor. Yeah, so I mean this is the computation that we did at the beginning Yeah, yeah, it's a divisor So we actually have a modular interpretation maybe I also should have said let me say generically finite because that's So we have a modular interpretation so I can tell you in principle Or I mean not just I can tell you what the image of this map is in terms of what the points are and Not only that we're doing intersection theory So in order to compute intersection multiplicities, we need to know something about deformations and in fact Harrison Mumford give Explicit explicit descriptions for its deformation spaces or of Information spaces and I said before I threw in the word co-mikali and because we're doing intersection three That's the property we really want. All right, so in the Three four minutes I have I can now state the theorem So for those of you who like formulas, this is for you. Okay, so theorem so the class of I d in the chowering Is okay, so these sigma d's is this is the sum of divisors function And these deltas are the are the boundary classes. We've seen before so this is the class of the closure of the the locus of irreducible genus two curves and this is the locus of Curse where you have a Union of two genus one curves that meet at one node and My class of this map while I told you was generically finite, so I mean the class of the proper push forward Okay, so and I'm using this okay. I'm gonna write this is this is gonna be sort of a cop out, but I'm gonna write Maybe okay, maybe I just won't say this but I'll say it more. It's So using this right So if you want to answer the original question now you can do something sort of much more reasonable than before So now you can just make a fiber square So this was fee this was pie and you can See and this will be some finite. This will be some finite stack scheme scheme And now you can really just multiply these two classes right And you can get a formula once you integrate for this intersection Okay, and I won't write it down But let me maybe say in the last 60 seconds how you would prove such a formula So and this is really made possible by sort of Mumford's Description of the chowering of of M2 bar So, I mean so what do we know about it? Well, we know that it has dimension two in degrees one and two so a one and a two With a frank two So what do you need to do? Well, you need to take this locus and you need to intersect it Well, okay, so how do you pin down a class in here you intersect it with? You intersect it with two classes over here that are linearly independent and you invert some matrix And so how do you do this well? You can make a list of admissible covers of this type and it it's rather complicated We know they're deformation spaces and once you make this list this sort of tells you it sort of pops out Which classes in here are sort of the correct things to try to intersect with? and then using this rather delicate analysis of Deformation spaces you can compute these intersections you can compute their Multiplicities and then you can recover this formula Okay, I think a lot of time so thank you