 Thanks very much for the introduction. It's a pleasure to be here. Yeah, so I also wanna thank everybody for accommodating the unusual time when I understand for the seminar. I really appreciate it. So I wanna talk about some joint work with a rural Shankar and we really are motivated and I would say at least I remain motivated by bounding torsion in class groups. That's somehow a problem that I consider quite fundamental and it has many applications and things that I've worked with. So that's what I'll present as our application but really what came out of this work is the sort of realization that it might be more natural to instead of focusing on class groups specifically as objects related to number fields to abstract away and just consider finite Galois modules and consider a sort of larger category of what we call Selma groups of these finite Galois modules what we call and other people call as well and consider the problem sort of divorced from that setting becomes a little more natural. And specifically I'll argue that for these objects it's not actually clear at least to us what the so-called trivial bound should be and that at certain circumstances the trivial bounds for these problems and the being a little bit better than you might have hoped. So that's kind of the big picture set up. So let me now do move on to sort of the actual setup of the talk. So just to kind of get our context in place let's let the K be a number field of degree N over Q and we'll let C okay be the class group of K and everything is measured from the stock more or less in terms of the absolute value of the discriminant of K called a D sub K. And then of course you have the famous class number formula which relates a whole bunch of important numbers to each other above this field K and specifically we'll be focusing on this corollary for our purposes, which is that really if you look at this class number formula my cursor is visible, right? If I move it can I just confirm? Thank you. So if you look at this class number formula from an analytic perspective the main objects of interest are the discriminant power here, this is sort of a big number then you have the regulator in class group over here so we can't say much about the regulator in fact it should in some sense be considered together with the class group but if you drop it because it's sort of at least one you get the famous Broward's Eagle corollary that at least if you fix the degree N of the number field which we'll almost always do in the families we consider then the size of the class group is bounded above by discriminant to the power of one half more or less and this exponent of one half is basically tight you can make field a small regulator and we expect them to occur with some frequency. So that's how the class group behaves and the question that I wanna focus on is well, okay, even though the class group behaves we know it's asymptotics and we know it's always a finite abelian group but if we isolate a particular torsion piece of it so we fix something to your M and we just look at the M torsion elements and the question is how does this group act as the number field K varies? And this question just comes up for various degrees N integers M in various sort of applications to counting number fields, elliptic curves there's a connection to Shimura varieties and Gael already special points there's connections to counting weight one monitor forms and so on. So this sort of is a pretty natural question that shows up in various contexts. The heuristic that's pretty well known to people who work with these kinds of questions is that to first order you would expect this class group to behave like a random finite abelian group and a random finite abelian group should basically be close to a cyclic group. That's sort of not true if you zoom in enough there's interesting questions to be asked but from a first order perspective again you would expect it to behave mostly like having a large cyclic component and then some other stuff that's of much smaller order. So in particular you would expect it to have very small torsion and there's this conjecture which I think was first formulated precisely by Xiaohu Zhang and Bremer Silverman which is if you fix the degree N and you fix this integer M then we had our upper bound of square root of discriminant but in fact you expect the M torsion side of the class group to grow slower than any power of the discriminant. So it should just exist on a totally level on a totally different scale. So this conjecture turns out to be extremely difficult that's far beyond anything we know right now. We have the trivial convexity bound just coming from the fact that the M torsion piece can be bigger than the class group itself and the class group is pretty much of size the square root of discriminant so we can get the K to the one half as an upper bound but it's very hard to do better. So if you formulate the sort of sub convexity problem which is just sort of a name there's no actual L functions going on here but if we call it sub convexity which is just can you beat the exponent of one half for a fixed M and N? This is already a very difficult question and open in the vast majority of cases even though again we expect to be able to take this delta M and to be all the way up to a half. So here's a summary as far as I know I think of basically all the previous work that's known on this. So the rotation again is delta M N represents the saving you can get over one half when you deal with M torsion in fields of degree N and you can do a little better if you specify more things like to restrict the Galer group or something like that but I'll just restrict myself to talking about cases where you don't have to do that. So the most one on case is the case of Jimis theory that's due to Gauss. So if you look at two torsion and quadratic fields that's very well understood and there's not very much of that and then if you can bound two torsion you can bound two to the K torsion and so that case is handled. And then beyond that we only have estimates and the estimates we have are as follows. So for three torsion in quadratic fields there's been quite a bit of work on this. So many people have obtained upper bounds. There's work of Pierce, there's work of Helveglad Venkatesh and Ellenberg Venkatesh. All of these three methods are based on sort of different approaches and the best upper bound is due to Ellenberg and Venkatesh is the most recent where they save a full one sixth. And then on the same paper they also managed to deal with the case of three torsion in cubic and quadratic fields and they get some explicit constant that's they don't work out but I worked it out that something like one over 200 I forget exactly what it is but they get some saving over the one half. And then if you look at two torsion in an arbitrary degree field there's a work of Bargava Shankar and Iguchi Thorne myself and Zhao trying to break records but not quite succeeding of we get a saving depending on end of one over two N over the trigger bound. And then if you're willing to assume a GRH this is basically the trick of Ellenberg and Venkatesh is they show how if you assume GRH you can get small split primes those small split primes can't be a very small torsion order cause they're just too small to be principal so you can get a bit of a saving there and in fact the way their paper works it's very beautiful is they show that you can often avoid relying on GRH due to sort of reflection principles where you kind of argue one field or another has small split primes and that's enough to make the argument work. Okay, but these are all the sort of known cases so far at least to my knowledge and the theorem I want to present is the following. So if you assume the refined Bertzmann's and Dyer conjecture for elliptic curves so it's not enough to assume that just the ranks match up we need to assume a certain formula as well then we can give a non-trivial bound for five torsion in quadratic fields to save a 16th and then if you assume Riemann you can in fact do better you can go all the way up to a quarter and if you assume Riemann and get the quarter then it's worth mentioning that this is also true for cubic fields because that beats the current records of course if you don't assume Riemann we just assume refined BSD we can also get Delta 3, 2 equals 1, 16th here as well we just don't mention it because it's not more than what's currently known. Moreover, you can sort of have the same story of refunction fields where you have some ways in some cases of doing better unconditionally but not in general in the refunction fields you can make the same arguments and in fact the results are unconditional in that setting we need to assume neither refined BSD which is not known in that setting nor GRH because GRH is known and I'll say something about that at the end if I have time because the picture becomes a little bit clearer over function fields. Okay, so what I wanna do now is I wanna sort of explain a little bit about our approach and how we think about this problem. So here is sort of the big picture heuristic method that we have it's based on embedding into motives I'm very uncomfortable with motives so I'm going to give this picture with the biggest possible caveat I can and that's blackboard bold. So, but here's the idea so the step one of our method is to forget about the field K. So instead of thinking about entorsion in some number field in the classroom of some of those K we're just gonna remember the finite gamma module and step away from thinking about ideals and then the way we think about it is that we have these two categories. So on the one hand we can look at finite gamma modules A and we want to bound some Selma group that I'll explain which is some generalization of this class group or this entorsion piece of this class group up here. And on the other hand we have motives floating about one motive is like this number field itself but there are others these motives tend to have class groups and these class groups satisfy some class number formulas that let us get at least in principle some kind of handle on these objects. So again, in the case of actual number fields you have an actual class number formula and you can get an actual handle on it with a Broward-Zegel bound. And then you have these two categories and then from time to time you can embed one of the finite gamma modules into some motive as some sort of torsion-like piece. And that gives you an embedding both of these in quotes of the Selma group you want to understand inside the class group that you do understand and that gives you some quote-unquote trivial upper bound just by saying that if this Selma group embeds inside this class group it has to be at most dead big. And then the game of finding sort of the correct trivial bound one game you can play is you can look for the best M the best motive for a given A for a given finite gamma module. So in other words if you look at the entorsion of the class group of K you can of course re-embed that back into the class group of K itself. But the other thing you can do is you can go splunking for some other motives that you can embed it in and maybe they have better trivial upper bounds and then you win or at least you get some better saving. Okay, so that's the approach that we're going to take. And so let me tell you a little bit about these objects that we're dealing with. And then we'll just- There was a question by Igor Barlinsky about your results. So the question was, you need GRH or BSD and GRH? Maybe Igor if you can- Oh, we definitely need GRH as well as BSD. That was a question. Okay, thank you. So you need both, yeah. Yeah, yeah, no worries. We absolutely need both, yes, yes, yes. The main, the way we get anything is by using BSD and then we can do a little bit better with GRH. This will become completely clear once I describe the proof but yes, we definitely did both. Thank you. Yeah, thanks for the question. Okay, so let me describe these Selma groups that we're going to be talking about. So I should mention that the algebraic number theorists have very precise Selma groups that they deal with and they work very hard to get everything just right at all the primes including the very difficult ones. For our analytic purposes, we sort of don't really care what happens at the difficult primes. So it's a much easier concept because we only care about sort of rough sizes of things. What we do is the following. We look at our homology groups. So if you have a Galo module A, you can look at the global homology. This is some infinite set. And then you can restrict for every prime V to the local homology over the completion of V. And then what we do is we do our best job at asking for this representation to be un-ramified. So in other words, for every V, we have an embedding of the homology of the finite Galo group, the Galo group of the finite field acting on the inertial invariance. This maps in and we just ask for, will the Selma group be all global homology elements which at every place lie in the image of this group? So for almost all V, the inertial group acts trivially. So this is just H1GFV mapping into H1GQV. And then at the nullified places, here what we're doing is we're sort of thinking in some sense the largest possible reasonable thing. You can also just say, well, I don't actually care. Let me say the nullified primes that I want the image here to be trivial. And that will make everything I'm saying work just as well. Just because at this level of magnitude, there aren't enough ramified primes to make a difference. So just to emphasize again, there's nothing very subtle happening here. There's no careful choices being made. This is a very loose object we're dealing with. So here are some basic properties of these finite Selma groups to keep in mind as we work with them. So given a finite group A, we need some way to measure complexity. So for another field, we had discriminants. And what we do here is we're gonna say that L be the splitting field of A. So it's gonna be such that this Galois group is the kernel of the action of the Galois group on A. And then we're gonna take a discriminant of that field and call that D sub A. It's again a very rough kind of choice. And then to make things kind of nice for analytic purposes, we're gonna write these squiggly, greater, less than and equal signs to mean things that are true only up to factors which grow sort of sub-pollinomally in this D sub A. So we're going to ignore small factors systematically through these conventions. I want to point out that I'm being a little precise here of course, because whenever I write little O of one, it means something goes to zero as you vary in some family. So I have to tell you what family I'm varying in and I don't really plan to do that. But in almost all applications, what you can think about is you're just moving in families where the size of the finite abelian group A that you're dealing with is bounded. So we're just not changing the size of A, but we're changing the action of the Galois group on it. And this incorporates sort of everything that I'm going to be dealing with in all applications. It incorporates this general machinery of like, if you have number fields of degree N, you're basically looking at different SN quotients of the Galois group and then doing summer presentation theory with those. But the size of the underlying finite abelian group is not going to change. Okay, but if that doesn't really, if you're not worried about that, then just continue not to be worried and we'll both be happier. So here are some basic facts that are true about these summer groups that make working with them kind of nice. So first of all, in short exact sequences, they're not quite multiplicative. If they were, you could prove, essentially Jiang's conjecture, but they're not. But they're not so far off. So you have this, first of all, sub multiplicativity. If you have an extension B of C by A, the Selma group of the middle guy is at most the product of the Selma groups of the outer two guys. It's not true that it's at least as big as their product. However, it is at least as big as each of them individually. So it can't get smaller by extending something. I also want to point out that in some sense, all of this exists in a counterfactual hypothetical world that probably we're not living in. So in reality, all of these summer groups probably just do grow sub-pollinomally in D sub A. And so everything is quickly equal to one. So the only reason that I'm bothering to write this down is because we can't prove any of that. So up to what we can prove, we do have to worry about the fact that these guys might be big, even though in reality, they are all quite small. Probably unproven. And then you have this, the other thing we're going to use is this Puittoté duality translated to our language. So if you look at the Cartier dual of a finite Gala module, then its Selma group is the size of the Selma group of the original Gala module. In fact, there's some sort of almost duality between them, but we're not really gonna care about that. We're not gonna care about their size. Okay, so now that we have that set up, let me give some examples of the kind of thing that I'm talking about. So this is our first example of a motive. It'll be the case of an algebraic torus. And I'll work with it in a very sort of ad hoc manner. So let's let T be our algebraic torus, some torus of dimension D. And then we get an integral Gala representation by looking at the co-character group of this torus over Q bar. As long as the torus is not split, you can get some non-trivial action, which you can study. And again, to fix complexities, we're gonna look at the arc and conductor of this representation called that F sub T. That'll be the analog of our discriminant in the number field setting. And then once the class group of this motive is just the usual class group of a torus. In the idelic sense, you look at the sort of idelic points of this torus and you model by the rational points on the one hand and the maximum compact on the other. Okay, so then you have a Brouwer-Zegel-type formula, which is due to many people, the analog of the class number formula is due to shear. And then myself, Umo and Jaffa, have worked out this analog of Brouwer-Zegel. The main input really is how to deal with the Tamagawa number. That's historically the hardest part of the formula that shows up. And that was dealt with by Ono, which is why I'm including his name. And again, we write sort of squiggly signs, squiggly inequalities to mean equal up to the factors of, in this case, the arc and conductor. So then you have the following fact. If you take an isogeny between tori, we'll call that phi, then you can look at the induced map on co-character groups and the co-currinal of those maps will be some finite a billion group M sub phi. And now we can relate all these objects together because it turns out that if you look at the induced map on class groups, then in fact, these two tori have roughly the same size of class groups. That's known unconditionally. That's basically due to what I've written up here, the analog of the Brouwer-Zegel formula. And so the kernel and co-currinal are about the same size. And in fact, it turns out the size of the kernel and co-currinal can be measured in terms of the Selmer group of this M sub phi. So if you want to, if you have a torus, you have a class group and you want to study certain torsion pieces of it or more generally sort of Galois variant torsion pieces. What you do is look at the corresponding isogeny and you study that map between class groups and that will recover for you the Selmer group of this finite a billion piece. Okay, so that's very general. So let's do some examples to see the kind of thing we're talking about because this philosophy I mentioned at the beginning of embedding finite Galois Selmer groups sort of in the class groups of motives. It's already, it has a non-trivial sort of shadows of it just by studying Tori. So as an example, let's give applications to torsions of small degree fields. So let's start with three torsion of cubic fields. So take K to be a cubic field. Let's make it S3 for simplicity and let's take L to be its quadratic resolvent. Then just to align things with the previous language we're going to let T sub K be the corresponding torus the restriction of GM from K to Q. And then if you look at the corresponding representation row then if you reduce that mod N that's just the usual representation row sub KN. So, and then you can recover the torsion pieces of K by looking at the Selmer groups of these finite a billion modules. So in this case, if you look at, we're going to focus on three. So if you look at row K3 which represents the sort of finite Galois module I've written up here. Its Selmer group is essentially the same size essentially isomorphic to the three torsion of your class group. And now we have an exact sequence of Galois modules. It turns out if you work it out where you can look at the three torsion corresponding to the field L you can embed that inside the Galois module corresponding to the field K and the Co-Colonel is just the trivial Galois module F sub three. And so things with trivial Galois actions aren't interesting the Selmer groups are all roughly one because they all come from Q, Q has no class group. And so what you learn in a very rough analytic way is you get this transfer principle for three torsion in cubic fields which was originally due to girth and then has been refined in various ways. But you learn that the three torsion of a cubic field is roughly the same size as the three torsion of its quadratic result. And so this is already a case where sort of the trivial bound, upper bound of decay to the one half might be improved by instead of plugging in DL to the power of one. Okay, so you can set up basically the same kind of thing in other context. So here's another kind of general one you can do if you let K be an S4 or an A4 quartic field doesn't really matter and you look at its cubic resolvent then again, if you're interested in two torsion what you wanna do is isolate these finite Galois modules with Selma groups, recover the two torsion in the class groups and then you get these two different Galois modules Rho K2 and Rho 3 2. Again, you can write it down. Sorry, they're very explicit modules. They're four dimensional and three dimensional respectively with some Galois action. Now they don't quite fit in the same exact sequence as last time but if you take out all the trivial stuff if you sort of take out all the trivial modules which don't affect sizes then it turns out they both contain the same two-dimensional irreducible representation. And so you learn that the two torsion of a quartic field is roughly the same size as a two torsion of its cubic resolve. Which by the way is the reason that all the way back here or something similar to this is the reason why the bound of the vectors get here for quartic fields is the same as for a cubic fields. It's a shadow of this kind of transfer principle that you can access in this language. Okay, so before moving on, I want to mention that we've been working very, very sort of vaguely or sort of dropping terms that are sub-planomial in the discriminant which is great in a theoretical context because if you're trying to beat upper bounds you're in sort of counterfactual land where everything is maybe very big and scary. In reality, these things are quite small. So you want precise formulas. You want to keep track of everything. And everything I've mentioned, you can work more carefully. You can keep track of ramified primes and try to see what's going on. So here's an example of a theorem due to Grahe and Geirth. If you have a cubic field with quadratic resolvent then as long as you input some sort of niceness condition at primes that their join is unramified over the quadratic resolvent you have this precise comparison theorem between the two torsions. And there's this conjecture of Lebermeyer which is still open that says if you have an A4 quadratic field and you consider it's cubic resolvent then their two torsions in fact differ by at most two. So something extremely concrete and small. So it's not known but my student, Jack Liss who graduated a couple of years ago proved the same, the correct quality at the correct order of one but his 10 off from what the right bound should be. Okay, so now I want to explain how to do this kind of setup in elliptic curves which is what leads to the theorem that I stated at the beginning. So suppose you have an elliptic curve E now I'm gonna write it in minimal law stress form like this Y squared is X cubed plus AX plus B. So for the curves it's a little tricky because it's not exactly clear how to measure their complexity. One typical way is to use the height. So the height of an elliptic curve is the maximum of A cubed and B squared. And we'll stick with that for now but also there's another measurement by the discriminant which can be kind of annoying in certain cases because they can behave quite differently. This won't show up for us. I'll explain why in a bit. But let's take a slip to curve E. There is a question from Joel Rosenberg. Joel, would you unmute please and ask away? Yes, my question was about this thing for quartic fields. You didn't put any conditions on the ramification unlike the cubic with a quadratic resolvent. Do you mean this statement here? No, going forward. You had a statement for LK, yes, this thing. That's right. Well, that's hidden in the fact that it's an A4 quartic field. That's a very good point. But it turns out that I don't remember exactly how this works out of my head but there's a statement for an S4 quartic field still and then you have to worry about ramification conditions. But if you insist it's A4, it turns out they're all automatically satisfied. Don't right now have a good reason as to why that's true in my head. But all I can say is that like in the cubic field setting, if you make an A3 cubic field, so a sickly cubic field, then it's quadratic resolvent ends up just being Q plus Q. That sort of drops down to a very degenerate case. So likewise here, there's some sort of generation that's occurring and somehow that's removing all the better ramification that you might in principle worry about. I know it's not a great answer. I don't have a better one. I'm sorry. That's an excellent answer. Thank you. All right, so yeah, I'll get back to elliptic curve. Let's let R be its rank and we'll let omega E be its minimal period. And then we have this refined BSD conjecture, which we will think of as our class number formula in this case. So there's a bunch of terms here. You can ask about which ones are scary or not. So obviously knowing what R is, is a little bit scary. The torsion term here is quite small. We know how to bind torsion even before measures work. This wasn't the problem. Now it's very much not a problem. It's absolutely bounded. These local terms are also something you can ignore analytically. So essentially you get the class number, which is sort of the Tamagawa number here. That's the way we're gonna think of it. You still get a regulator, which you can't ignore. And then you have this minimal period. This is what I'm writing down over here. And then you have the minimal period, which you can't ignore because it's actually pretty big. In fact, it's of size, sorry, this is a typo. This should be HE to the minus one-twelfth. The next slide is correct, but the size of the minimal period is the height of the elliptic curve the power of minus one-twelfth, more or less. So if you have this optimistic conjecture where you assume refined BSD, sort of this formula is true in the first place, you assume GRH, so that you don't have to worry too much about the L value here. And you assume bounds on ranks, so that you don't have to worry about the R too much. Then the thing that you expect to be true if you assume all that stuff is that the sort of class number analog here, the Tehsheff-Ravage group times the regulator is of size, the height of the declared to the power of one over 12. So this conjecturally, you have your class number formula and it's as precise as you might hope for. Now, I wanna sort of point out that elliptic curves usually have Selma groups already and Selma groups. So we now have two things being called Selma group, the sort of cell sub N of E, or we could look at the Selma group of the N-torsion-Gavel module E of N and they're defined in slightly different ways, though they're not too different, they're both some sort of local restrictions on the same global homology group. So it turns out that in fact, these two are almost the same. And this is because at all good primes and all primes where E has good reduction, it's not immediately given definition, but in fact, you're requiring the same kind of local condition at all of those primes. And there's just not enough of the primes where E has bad reduction for us to worry. So we can use these two interchangeably. Sorry, one more question from Arman Brumer. Arman, would you please unmute to ask away? Yeah, when you were saying GRH, I thought for only the L functions of Dirichlet series or something, you are actually using GRH for the L function of elliptic curves. Correct. Okay. Thank you. You just made a decent order bound on the L value or do you really need zeros to actually get one? Oh, no, no, no, no. Just lend the loss of convexity stuff. We don't care about zeros. Just does the L value for sure. Other questions? No, I think it's okay. Thank you. Okay, so let me move on now and explain precisely how we get this bound of the five torsion using these kind of methods. So here's the only part, very frustratingly, the only part of the proof which requires five, three would also work. Assume you have an elliptic curve or Q which has totally split five torsion. But totally split now because the way pairing exists, you can't hope for Z mod five plus Z mod five. The next best thing is Z mod five plus mu five. These elliptic curves exist for various reasons. One is you can write them down. Two is the modular curve with full five torsion structure is rational. So it's got many rational points. And in fact, these elliptic curves don't exist if you replace five with any larger prime. So frustratingly, this method will not work for any larger prime. This exact method that we're using, the general philosophy might, we're not sure. But start with that. And we're gonna consider the family of quadratic twists E sub D. And if you look at their five torsion, you just twist everything by the quadratic character corresponding to keyword join root D. So write it in this way. And because it's split, this is again where we use five. The summer group is just the direct sum of these two summer groups. This is actually an equality because this is an equality. So we don't even need squiggly equals here. This is on the face of it equal. And then here is the sort of trick which makes everything work. It's very simple. It's just that chi D five is a Carti dual to its own twist by one because it's a quadratic character. And therefore each of these two summer groups is of size the five torsion in Q adjoint in the class group of Q adjoint root D. And so you learn that if you get a bound for the summer group of the five torsion of this elliptic curve, you in fact get a bound for the square of the class group, the five torsion peaks that you're interested in. And that's very good. And that will lead to us getting an acceptable bound at the end. So, okay, we've sort of successfully removed ourselves from dealing with the field Q adjoint. We're now interested only in this elliptic curve. And so what's left is to just relate the summer group to the pitch of ravage group and plug in the trivial upper bound and see what you get. So let's go through a little bit of analytic details just to see how it all works out. So first of all, the summer group is very close to the five torsion of the pitch of a ravage group. You're off by the actual rational points of the elliptic curve. So you're fine as long as you can bound the rank, right? You get the pitch of ravage group is at least as big as the summer group up to five to the power of the rank of the elliptic curve. So now you have to figure out how to bound the rank. And to bound the rank, we use this clever trick which is due to a paper of, I think it was in the paper of Silverman that we saw it and I don't remember, I apologize. But it's a sort of standard trick in the subject where because you're looking at quadratic twists, you actually understand the two-summer very well. So the two-summer group of quadratic twists is bounded by number of ramified primes. But then once you bound the two-summer group, that already is enough to bound the rank. And so then you can turn that around into a bound of the rank, which you can plug back into here. And that tells you that the pitch of ravage group is essentially the same, it's five torsion is essentially the same size as the five torsion of the summer group. Whereas in principle, if you weren't dealing with the family of quadratic twists, you would legitimately have to worry about the rank. Now, the regulator you can show also is not too bad. Again, in principle, you have to worry about it being very small. In this case, you don't because the rank is not very big. And to control the central value, you could either assume Riemann and be done with it, or you plug in the best-known subconvexity estimates plus like a tiny amount of work and you can save an eighth off of the trivial bound, which in this case is d to the one-half, let's do the hard-coach right now. You calculate the height, which because again, you're in a quadratic family, this tension between the height and the discriminant being different doesn't occur. It's whatever it is, d to the six. And so finally, you plug all that in, you work out what you get and you get at the square of the class group is bounded by d to the power of all told seven eighths. So once you square root this, you get what you want. And again, I wanna emphasize that we only win because we have the square factor here. If for example, our elliptic curve only had a five torsion point, but the five torsion wasn't totally split. So you only had an extension, you wouldn't be able to put a two here. You would only have a one and then you would be even assuming Riemann, you would only recover the trivial bound, but you get from the number field case. Okay, before moving to my next slide, which is there, I want to emphasize that we're very much in the dark as to why this works. So we had the idea to try it and then being diligent members of the community. We went ahead and tried it and we got numbers that worked out and so we wrote a paper about it. But April or I, before starting this computation, it wasn't clear to us, it wouldn't get a bound of like d to the 20. We have no heuristic telling us what to expect, even assuming all the right things at the end, we just kind of hammer away and get what we get. So if somebody figures out what to expect with this kind of method, that would be interesting in my opinion. All right, so to end, I have, I still have 10 minutes, right? Yeah, to end, I want to explain how this all looks in the function field case because in the function field case, everything is somehow a lot more concrete. Like in other fields, you can talk about other motives and you can look at these blockada conjectures and they're the echo variant, the McGowan number conjecture and you can try to make sense of what is, but it's kind of complicated and I found it hard. But in the function field case, everything is very straightforward and it seems like at least if you can find motives where things behave as you expect, it potentially proves that conjecture function with case in computer outdacity. Let's do a plausible approach. Send me a plausible approach to it. So let me talk about what things look like there and I'll focus again on the case of bounding torsion in class group static fields. So here's a setup that I want to work in. Let's let K be a finite field and D be some divisor of even degree on P1 and let's let KIDL be the local system on P1 minus D which corresponds to hyper elliptic cover. There's a bit of a choice there that I'm ignoring but let's not worry about that. And then essentially what you want to do if you want to bounce sort of the analog of L torsion in the class group over the function field case is you want to bound this arithmetic homology group where you look at this FL local system over P1 minus this divisor and you want to bound this H1. Okay, so here's what our method of embedding into different sort of motives looks like in the function field case. So more generally than power minus D than P1 minus D let's let U be any open set and let's let L be a ZL local system on U now and we're going to assume later on that L really comes from geometry. So L will be some sort of co-amology of some variety over U, some relative co-amology group. So L can be very interesting but let's assume that if you reduce L, local system L, mod little L, that is trivial. And in fact, in general, you probably want to assume not quite that it's trivial but that it's a sum of twist or something but for simplicity, let's assume that it's trivial. Okay, and now once again, we're going to let L sub capital D be the twist of L by the quadratic character corresponding to P1 minus D. So now bar here means we're looking at co-amology over little k bar, so geometric co-amology groups as opposed to arithmetic co-amology group. And if you take co-amology groups of the ZL local system, you can do that before or after reducing mod L and a nice thing now, of course, is you have Frobenius. And so the left-hand side here is just the co-kernel of one minus Frobenius acting on the ZL co-amology group, tends to do with F sub L. On the other hand, the reduction of your local system mod little L by assumption is quite trivial. It's just this chi DL to the power of little r. And so what you conclude now is that the size of this co-amology group, which is ultimately what you're after, is bounded by this right-hand side, which can be measured in terms of eigenvalues of Frobenius acting on your ZL local system, plus the number of one eigenvalues. You have to be a little bit careful because you have one minus Frobenius acting on this group. In general, you expect this to act on q-o-comology to be invertible, right? Except for you might have some pieces where Frobenius actually acts trivially. And so you have to sort of manually deal with them. But what I want to point out is that unlike in the number-filled case, in the number-filled case, you can sort of pray for BSD to be proven, and then you can pray even harder for the refined BSD conjecture to be proven. In the function-filled case, you can get at the analytic part, at least, of the refined BSD conjecture, without actually needing to prove BSD. Because what's BSD in this context? BSD, which is basically the tate conjecture in the function-filled context, is telling you something about how many eigenvalues of one that Frobenius has when acting on various cosmology groups. That's in its most sort of general form, the kind of statement that it is. Now, if you look at the setup, then yeah, we have these one eigenvalues, but as long as we can give some kind of trivial-ish bound for them, which you can in every setting that we've encountered, you can basically ignore this as an error term and get right at the midi part of what you want, which is this right here. This is some sort of global class group, which is measured in terms of product of one minus the non-trivial eigenvalues, the non-identity ones. So somehow you have access to this formula directly without needing to worry about the tate conjecture. Right, so that's what I'm pointing out over here. So examples of this local system that concretely you want to look at include, well, if you let this local system be actually trivial, so one way to ensure its reductionist trivial is to make it trivial. So if you make it be z sub little l, that amounts to just doing the thing I mentioned at the very beginning, where you embed the torsion of the class group into the class group itself and use that upper bound. On the other hand, if you look at the universal elliptic curve equipped with five torsion structure with full five torsion structure, and you look at the relative h1 and you take that to be little l and you apply this whole argument to five torsion, that amounts to what we explain in number three case in the first part of the talk. You just applying this reasoning to this little l. So what do you need to make this work? Well, you have to understand the power of l that divides this product. So in general, you have to worry a little bit about denominators, because things might not be algebraic integers. And so to get a little bit more technical about it, you have to use Newton above hard statements to sort of control the slopes of Frobenius, i.e the denominators occurring in these alphas. But otherwise this product is something global, you can read it off from sort of global geometric information, you'll get some integer, you can bound this integer in terms of its actual size and get some inequality. Okay, so final remarks over here for primes P above five, what can you hope to do? So like I mentioned, there's no elliptic curve satisfying this. So the exact method is not gonna work. What would work is if you assume BSD and subconvexity or Riemann, maybe you can find the billion variety over Q with full level P structure. So Z mod PZ plus mu P to the power of G or something. That seems unlikely, but if you find even one such guy, you can make this argument work. Yeah, more generally, if you have a motive, then the, to actually make this class number formula heuristic work, the closest thing we've found is you have to assume block cut-off and the equivalent number conjectures. Even then it's not quite clear how things behave analytically and it's not quite clear how to find these embeddings. Even though in the function field case, it really is very clear. And so concretely, what you're looking for is the following kind of thing. You're looking for smooth projective varieties such that if you look at their co-homologies, whoops, if you look at their co-homology groups, then as Galois representation, at least if you reduce mod subprime, you just get something trivial plus some power in the trivial plus some power of its Cartier dual. And we haven't managed really to find any of these guys besides these universal families of elliptic curves with five torsion and three torsion or sort of mild variations thereof. But if you find at least one such guy, then you can automatically run this argument and get some result. So it feels very tantalizing. Okay, that's all I had. So thank you very much for your attention.