 thanks to the organizers for the invitation to give this talk. I was first in contact with them, I think it was sometime last year and I kind of let it slide for a very long time. So I'm glad when I finally got back to them they were still agreeable to be giving this talk. So three notes before I leave this slide. So this is a drama in five acts, which means there will be brief intermissions between the acts. Those will be some opportunities to ask questions. Second is that the slides are available both, well the link has been posted in chat for my slides. That's just a link to my webpage. There's also a handout version if you prefer not to see the overlays. And third, normally I would start by acknowledging that my campus sits on unceded and such a land of the Kumeyaay Nation. But as Mike pointed out today, I'm in Washington D.C., which turns out if you check nativeland.ca, which you should, it's the ancestral land of the Piscataway Nakhachetank people who Europeans call Anacostian, because that's easier to pronounce. Great. Okay. So act one is where we meet the characters. So the characters in this drama are abelian varieties and they're associated vape polynomials. So A is always going to be an abelian variety over a finite field FQ. And I'm going to refer frequently to the order of an abelian variety, which is a slightly nonstandard terminology. When I say the order of A, I always mean the order of the group of rational points of that. So I'm not forgetting that A is an object of algebraic geometry. I'm just keeping track of how many points it has over its field of definition. And of course, that's a group. But we're not going to be using the group structure very much in this talk. Mostly we're going to be using a vase interpretation of this order as the value of a polynomial. So there's a polynomial P of t, which has integer coefficients, such that this order of P of one, and this polynomial comes from, well, it's okay, it's the characteristic polynomial for Benius acting on the Latic tape module of A for any prime L not dividing Q, any prime L not equal to the characteristic. But what you really need to know for this talk is that it's a polynomial, which is monic of degree two times the dimension of the abelian variety. And as a polynomial viewed over the complex numbers, all of its roots have absolute value Q to the one half. And such things are called Q-v polynomials. And this talk is mostly not about abelian varieties per se, it's mostly about the abelian varieties. So the abelian varieties are part of the motivation, but they recede into the background of the action fairly quickly. They may they will appear briefly again at the end. So I'll just mention that a lot of what I'm doing depends on the fact that there is built-in code in Sage to enumerate v polynomials. This is based on an algorithm that I started working out more than 10 years ago by now, but relatively recently, this code got incorporated into Sage. So you can run experiments like this where you ask Sage for all the v polynomials for a given Q and a given D. D is the degree of the polynomial. So it's two times G. And as long as you don't make these parameters too large, you get the answer back fairly efficiently. So the main thing that makes the abelian varieties not so critical for this talk is the fact that this is very tight correspondence due to undertake between isogenic classes of simple abelian varieties over FQ and irreducible Qv polynomials. So if A is a simple abelian variety over FQ, it's associated polynomial would be a power of some irreducible where the integer can be read off from the polynomial, but most of the time in practice, it's equal to 1. So in particular, if Q is prime, which will be most of the time in this talk, it's more or less always the case that the characteristic polynomial of A will actually be irreducible. So sort of the abelian variety being simple, which is kind of like irreducible, it doesn't factor as a product to isogenic, that corresponds to the polynomial being irreducible in the usual sense. As a reminder, when I say isogenic, I mean, it's a morphism, which is rejected with finite kernel. So we're not going to be talking so much about isomorphism classes of abelian varieties. That's a little bit more subtle. We're going to be talking about isogenic classes because these correspond exactly to vapor polynomials. And the order of an abelian variety will be invariant under isogenic because it only depends on the vapor. So it's natural when we're talking about orders to just work up to isogenic. And yeah, so as I made this point earlier on a previous slide, the upshot of honetate is that problems about existence of abelian varieties of a given order or over a given FQ are really problems about existence of vapor polynomials with certain properties. What that means is that for most of the action, we will not be seeing abelian varieties explicitly, we will be just manipulating polynomials and deducing consequences about abelian varieties from that. So most of the work in this talk is happening at the level of constructing or ruling out the existence of vapor polynomials under certain conditions. So really, this is a number theory talk and not an arithmetic geometry talk except at the end, maybe some curves in their Jacobians will sneak back in. I mentioned on the previous slide that the order of an abelian variety is an isogenic variant, but the isomorphism class is not. So for example, you can have one abelian variety whose group of rational points is cyclic of order two times cyclic of order two being isogenous to another one whose group is cyclic of order four. The calculations that I'm going to be doing since they're only at the level of the vapor polynomial, they will only see the order and not the isomorphism class. Questions about existence of abelian varieties with a given group structure are also tractable, but they use a different set of ideas in combination with these. So there's a recent preprint of Marcellia and Springer that explains the relationship between these kind of results up to isogyny which prescribe a given group order and results at the level of isomorphism that prescribe a given isomorphism class of the group of rational points. That's the, sorry, that's almost the end of act one. Now it is the end of act one. So let me pause for a moment to see if there are any questions. I don't see any in chat. Seeing none and hearing none. Let me continue. So in act two, we look at orders over F2. So now for the moment, let me not specify F2, but on this slide, let a be an abelian variety of some dimension G over some finite field FQ. Remember what I said about the order being P of one where P is a vapor polynomial. Well, if you just use the information that's implicit in that statement about the roots on the complex roots all being on the circle, absolute value of T is equal to square root of Q that implies a lower bound and an upper bound on this order. So the upper the upper bound is square root of Q plus one power two G. And the lower bound is square root of Q minus one power two G. So the leading order term is Q to the G. The correction is Q to the G minus a half. So if Q is large compared, if G is large, sorry, when Q is large compared to G, there are gaps between these intervals for consecutive values of G. So this number will be close to QG. And if you go to G plus one, that number will be close to Q to the G plus one. If Q is large, then there will be a gap between the intervals that you get. But when Q equals two, there isn't going to be any gap because this number is actually less than one. So the lower bound for Q equals two is meaningless. It tells you nothing because of course the order is at least one because it's a group. It has an identity element. It can't be zero. So when Q equals two, you don't have any intervals that any integers that are obviously excluded from occurring. And so you might ask, well, is it the case that every integer occurs? And this is the content of a theorem of mine with Everett Howe. We prove that, yes, every positive integer is the order of at least one abelian variety over F2. More precisely, we show that in this range, every positive integer going from, say, this about four-thirds to the D minus one to about four-thirds to the D occurs for a polynomial of degree 2D. For some reason, the D here is what I was calling G before. So these intervals, you'll notice, they line up perfectly. So you don't miss anything. You pick up every integer because you've sort of tiled the positive integers with these intervals. We can also ensure that the polynomial is ordinary, which in this case means that the middle coefficient is odd. Geometrically, this means that the resulting abelian variety has two torsions subgroup over F2 bar as large as possible, or rank is D. I see Sarnak has asked, do you use the distribution of roots? I mean, this is a statement about just finding one abelian variety. So we more or less give an explicit formula for a vague polynomial. So this is not a question that involves analyzing the distribution. This is a question about really producing a concrete polynomial for every choice of the order. And I'll show you in a moment how we do that. My question was, can you hear me? Yes. Yeah, thanks. My question was, whether in that example, okay, we'll see how the roots are distributed. Okay. Yeah, maybe when I get to the end of this act, that might be a good time to ask this again. So let me go through briefly how this is proved. Prove is actually only a couple of pages, so I can give you most of it. The key lemma is an older result of de Pippo and how. So this is a sufficient condition for a polynomial hat to be of a polynomial, at least if it has integer coefficients. But this is really a statement about polynomials with real number coefficients. So I take a sequence of real numbers, which are not too large, they satisfy some kind of weight condition. So a1 through an are bounded in some way. And if they're small like this, then the complex roots of this polynomial that I get by using a1 through an as the first coefficients, and then making it reciprocal to finish, the complex roots of this polynomial are pairwise distinct and lie in the circle absolute value of z equals square root of q. So this is essentially sort of a winding number argument. We'll see another one like it later. A very short way to say this, which your impudence suggested to us is that if you look at the function f of z over z to the end on this circle, it's real valued, and it has sign changes at the roots of unity. If you start with if you start with if you start with the ai's all being zero, then you have a polynomial that manifestly has roots on the circle, and the sign changes sort of survive when you perturb this much. There's a question in chat which I'm going to answer on a later slide, so I'll hold on to it for Yeah, so now you have to show that every positive integer actually occurs as f of one for some polynomial like this. Now when we originally were working on this, right, to Pippo and how wrote this limit down in the late 1990s, and they in their paper, they point out that it gives you some intervals. But for a long time, I think everybody including me and including Everett thought that these intervals weren't going to be wide enough to actually cover all integers. And so we thought we were going to have to upgrade this criterion in order to actually prove this. And then we somehow discovered initially by doing numerical experiments and then observing what was going on that actually this is enough. And the proof of this is a tweak of a very classic fact from the early history of computer science that every positive integer admits a unique non-adjacent binary representation. So you can write it as a sum of powers of two with signs where you don't use any two consecutive powers of two. So this might be familiar to those of you who are used to implementing elliptic curve arithmetic for crypto purposes, for example. So this is a tweak of that because it's not quite the problem we're solving, right? We essentially have to represent something. We have to put two to the i plus one here instead of two to the i. And then it's not exactly the same proof, but it's very similar. And then once you've done that, then you get a v polynomial that does what you want. So there's the there's the concrete construction. And this, I think, will answer a question that was raised in the chat. What about larger q? So yeah, q equals three. You don't get much mileage out of the v bounds either. But it is at least the case that so for every prime power q, every sufficiently large positive integer is the order of at least one abelian variety of rafq. This is a result of fanbamo, costa, we, punin, and smith. For q equals three, I believe they do get every order. For q equals five, I forget which where I think for q equals seven, you don't get the order two. But up to that, you get all orders. And you can, you can ask for more conditions like ordinary, simple, or principally polarizable, but that will change what sufficiently large means. So for example, for q equals four, you get all orders. But if you insist on ordinary, then you miss one of the small numbers. I forget which one. Yeah. And again, the point is that the v bounds do start to overlap once g is large enough. It's not the case that every interval integer in this interval occurs, but you can make this interval a little bit shorter and get an interval in which you can find everything by a kind of more sophisticated version of what we did with Everett. This is also refinement of results of Uri, Alouille, Lechot, and Kadyitz. The result of Kadyitz is in some sense a motivation for a lot of what I'm talking about. So I want to give a shout out to that. I think that's the end of act two. So this is a good point. Maybe Peter can circle back to his question and I can see if there are any other questions. So where are the roots of this polynomial? Are they running around? As you know, you can make the roots avoid intervals. I mean, Sarah has studied that. So in this polynomial that you constructed, you know where the roots are. We didn't make any efforts to sort of constrain the roots. I mean, in some sense, they're relatively close to the roots of the, you know, they're relatively close to q to the one half times roots of unity because we're deforming that polynomial. So that will give you some condition, but we didn't check very carefully. Yeah, thanks. We're not sort of probing arbitrarily far into the space of a problem. You can't do this with curves. Well, I'll talk about curves at the end a little bit. That's act five. When you have your polynomial constructed in your way, let me say polynomial satisfying all your conditions, are able to give us an example of a billion variety that for that polynomial, we did not attempt to go, we did not attempt to explicitly write down the abelian varieties that come up. A hint how it is looking. Well, it's just what you are doing your opera in five parts, right? Yes. So, yeah, so this I mean zeros on the unit. Yeah, so of course, you in principle, you could invert, you could run through the proof of candidate, right? One of those is to lift a character, you know, sort of do everything in characteristic zero by explicit, you know, analytic complex multiplication, find a model over some number field, reduce, and then descend because that will give you typically something, a basic extension of the thing you actually want. There's also a more algebraic approach by maybe try and or. But neither of those is extremely explicit. So it would be, I have not tried to make explicit examples, except in a few cases that I'll talk about an act five. I mean, something constructed out of electric curves, some kind of products, you know? Yeah. Yeah, I don't know how much that would cover of the whole range. I mean, of course, the problem with taking a product of elliptic curves is that the order will be a product as well. So you won't get all possibilities, right? That way. Thank you. All right. I don't see more questions. So if I don't hear any more, maybe we can move to act three. So in act three, each order happens many, many times. So the motivation for this act is this statement, which is also a refinement of a theorem of Aubrey-Alioula show. This is the version by, well, this, maybe you can deduce this already from Aubrey-Alioula show, but there's a stronger version of this theorem in the paper by Cadets. For every prime power q greater than two, you actually get an exponential lower bound if you insist on simple abelian varieties. So this is another answer to the question about q equals three that came up in chat. So even for q equals three for symbol abelian varieties and q equals four, I should mention, for symbol abelian varieties, you do get an exponential lower bound with one exception. Namely, there are elliptic curves of order one over q three and q four. They fall on the Haase interval. And right, it's due to, right, it's an old theorem of doiring that everything in the Haase interval, at least for prime order shows up, but also for q equals four in this case. So, okay, so for q is at least seven, this is immediate from the bay bound. For smaller q, this is a kind of classic strategy in algebraic number theory. This goes back to Castles, this kind of approach to identifying algebraic integers of small trace and norm. This kind of question was also studied in detail by Raphael Robinson. And so, indeed, when you run into the case q equals two, you hit a sort of threshold for this problem. This kind of threshold is what Robinson studied very carefully. And you actually hit the threshold for q equals two and something very different happens, which I'll mention on the next slide. So, this is all kind of related to things like the sure as eagle Smith trace problem. I wanted to give a shout out to Alex Smith. He's been working on this. He has a not yet public, I think, preprint making some progress on that question. But that will not directly intervene in this story because That's on the archive now, Kiran. Oh, it's great. It's on the archive. Congratulations, Alex. It's a really nice result. Yes. But I'm not talking about that because I'm focusing on q equals two here. And in q equals two, as I mentioned, the situation is kind of different. So, there's a good reason why you can't improve the way bound even for simple ability varieties for two equals two. And it's because this old theorem of modern, this combines a theorem of modern Paul with a theorem of Robinson. If you take a simple ability and variety over fq of order one, then first of all, this can only exist when q is at most for zero. The zero ability and variety is not simple. So that doesn't that's not a counter example. Right? Just like one isn't prime. If q is three or four, then the only thing you get is the elliptic curve that I alluded to earlier. But for q equals two, you get infinitely many simple ability and varieties of order one. And you can describe their Frobenius eigenvalues rather explicitly. They are things you get by taking a root of unity and then solving this quadratic equation. Most of the time, this gives you a quadratic extension. There are a couple of exceptional cases, n equals seven and n equals 30, where a root of unity of order n gives you a split quadratic equation. And so, in those cases, you get two different isogenic classes. But most of the time, you get a unique isogenic class corresponding to each root of unity. And this more or less the proof is that if you look at alpha minus one over alpha bar minus one, right, the order of the ability and variety is essentially the norm of alpha minus one. And so, since that's one, alpha minus one is a unit in the ring of algebraic integers, as is alpha bar minus one. So the ratio is an algebraic integer whose conjugate all have absolute value one. So, Kronecker's theorem tells you it's a root of unity. So that implies this equation. And then the contribution of Robinson is to use the strategy of Castles to show that you only get a splitting when n equals seven or 30. So you compute the discriminance of this equation and ask whether it's a square. This is a problem about writing something as a sum number as a sum of very few roots of unity. So it's kind of classic algebraic number theory to solve that problem. Yeah. So there are these infinite family of ability and varieties of order one. And so the main result of this act is this theorem that says that every positive integer, not just one, but every positive integer occurs as the order of infinitely many simple abelian varieties over F2. Now, for m equals one, we have an explicit recipe for doing this. For m greater than one, it's, we don't give, we give a somewhat explicit recipe in that I can, we construct infinite sequences of vay polynomials realizing a particular order. So the sequence of vay polynomials, you know, giving you a given order is explicit. It's a kind of tweak of the polynomials that come out of the modern Paul construction. The catch is they don't a priori come to you as irreducible polynomials and we're looking for symbol abelian varieties in this case. So you have to massage things in order to actually be able to prove that they're simple. And essentially what you do is you rig everything up so that you can prove irreducibility over q2 either directly using the Eisenstein criterion, or which is due to Schoenemann originally, or, you know, a sort of generalized either side criterion where you say, I cook things up so that the Newton polygon, the two added Newton polygon has only a single slope with no interior vertices. So maybe it has rise three and run co-prime to three. So that's not the vanilla Schoenemann Eisenstein criterion, but it does still imply irreducibility over q2 and hence over q. Now, we don't quite do this for the original polynomial. We show that there is a factorization with a cofactor of bounded degree. And then the rest I claim is what I want most of the time, because these polynomials will satisfy some second order linear occurrences of Chebyshev type. And you can show that any factor that occurs more than once actually gives you order one. So pulling those out doesn't affect the order. So that's the strategy. Now, let me show you how to write down these polynomials. So I'm going to start with this, you know, the number theorists normalization of the Chebyshev polynomial, which is the one that makes it monic. So Tn evaluated 2 cosine theta equals 2 cosine n theta, right? It's also the number theorists normalization because these are supposed to be algebraic integers when you plug in a rational multiple of pi. So I'm going to sort of tweak this polynomial in a particular way. This factor x plus x inverse minus four, I don't think I included a graph in here. So I'll just say this Laurent polynomial has the property, well, maybe I'll come back to why this Laurent polynomial is a special one in a moment. So this basically gives you the construct, the polynomials that come from the modern Paul construct. That's in some sense why I'm doing this, but I can say more. And then I kind of rig up some weird linear combination of these things and then divide out a big factor of x minus one. So this is some recipe that you can follow. And this is partially rigged up by, I did some experiments and it worked, but there's a better reason why it worked, which is on the next slide, which is that there's something kind of like the De Pippo-Hau lemma that applies here that says the following. If I take a sequence of real numbers, ending in one, such that if I use these as the coefficient of a polynomial, I get a polynomial whose complex roots have absolute values at most square root of two. So again, this is a condition that says that these numbers are not too large. If I use this, then if I make a coefficient, if I use these as the coefficient of a linear combination of the polynomials g and k from this slide, which again are kind of modified tweaked from Chebyshev polynomials in some way. Because I have a moving index n here, I get an infinite sequence of polynomials and all of these polynomials give you roots, which are real pairwise distinct. And in this interval, three minus two root two, three plus two root two. And again, this is some kind of winding number calculation that proves this. And the point is the following. So on one hand, this is an interval whose end points multiplied one, right? Because nine minus eight. On the other hand, if I translate, then I get minus two root two plus two root two, which is where you end up if you take a complex number of absolute value root two and add it to its complex conjugate. So this polynomial shifted will then give me the what I call the trace polynomial associated to a two-v polynomial with the right order. Whose order is whatever this thing is, the trick is that g and i, did I say this? Oh, I don't think I said this. g and i has value two to the i. So whatever binary representation, whatever this binary representation of sorts fits out, that number is represented as the order of infinitely many vapor elements. Yeah, so now, right, so the polynomial, the mysterious Laurent polynomial x plus x inverse minus four is the one that carries this interval to plus two root two down to minus two root two at the value one, and then back up to plus two root two. So it's rigged up to have to sort of shift the usual graph of a Chebyshev polynomial over to this interval that's relevant for constructing Bay numbers. Okay, so that's the end of this act. So if there are questions about this, I can take those here. Okay, I don't see anything. So let's talk about act four. So act four is another look at Abelian varieties of order one. So in this act, I want to distinguish between simple and geometrically simple. An Abelian variety is geometrically simple. If it stays simple when you base extend it to any finite extension of the base field or equivalently to an algebraic closure. Now again, this is really a condition on Bay polynomials, at least if the characteristic polynomial of A is irreducible, which more or less always is in my situation. So this condition says that no two of the the Frobenius eigenvalues so no two roots of the Bay polynomial have a ratio which is a non trivial root of unity. So this is a for any given polynomial this is a relatively easy thing to check computationally. If you want to check this for infinite families, this is also tractable as we'll see. For any given m greater than one, I expect there are infinitely many geometrically simple Abelian varieties of order m, but I have no idea how to prove this. So I'm not going to focus on that question because that's my approach on the previous in the previous act doesn't seem to give this. At least I don't know how to how to control that. So instead, let me look at m equals one where we don't have any choice of the Bay polynomial. We just have a list. We know what all of them are. And we just look at that construction. We see what happens. So here's what we do. So the main result in this act is a theorem with torrent and L a warady who was a summer research intern with me. So they are Berkeley undergrad who's applying to grad schools right now. So you should admit them. Great. So this theorem says the following. If I pick a positive integer in that gives me one or possibly two in a few cases, I saw any classes of simple Abelian varieties of order one over F2. And I ask, well, what happens when I decompose that over F2 bar? Well, it's always going to be isogenous to some power of a simple Abelian variety over F2 bar. And so the real content in this question is what is the power? And it turns out that you do get geometrically simple examples infinitely often when n is a power of two, you do get something geometrically simple. Most of the rest of the time you get F equals two. The exceptional cases give you exceptional answers. But yeah, so most of the time you get two factors and there's a good reason for this. And this but this reason doesn't apply to powers of two. And in that case, you usually only get one factor. So here's an interesting corollary of this, which I don't know how to prove directly. It's just a corollary of this classification. You can also compute the Newton polygons of these things without too much trouble. And you show that A is ordinary if and only if n is not a power of two. And so what that means is that there was no simple Abelian variety over F2, which has order one, which is ordinary, and which is geometrically simple. So those conditions together run into each other. And it's a little surprising you can prove this given that there are infinitely many simple Abelian varieties of order one. But it's a consequence of the fact that we have an explicit classification. So we can do this. Okay, so how do we do this? Well, this is a type of problem that may be familiar to some people in this audience. I have, you know, say I take two for many as eigenvalues. For this, I don't have to assume they occur in the same simple Abelian variety. So this actually gives you some information that I'll use later when you do this with different Abelian varieties. It gives you information about, you know, homomorphisms over extensions of F2. So I take two for many as eigenvalues, both of which have this form, which is say they both are roots of quadratic equations involving roots of unity 801 and 802. And I also want to assume that their ratio is a root of unity. So that's a third eta three that I put in here. So now I have three equations in five variables. What do I do? I eliminate alpha one and alpha two. That gives me one polynomial in three variables. And I'd like to find solutions of that equation in roots of unity. And there's a sort of standard techniques for doing this. We use the method of Conway Jones, which involves writing down all the additive relations among short sums of root of unity. You can classify those. I think at the moment, the record is that those are classified up to 24 summands. There's a preprint by Christie and Klepp, I think, that goes out to 24. We actually use some non-trivial subset of that result. We also use some sage math code that I wrote for this earlier paper, paper from last year with Kolpokalpon and Rubinstein, where we classified tetrahedral with rational dihedral angles. That was actually a much more serious instance of this problem that was a polynomial in six variables because it involves six variables corresponding to six dihedral angles of a tetrahedron. So I had to write a whole bunch of code to solve intermediate problems involving three variables. So that code off the shelf handles this problem relatively easily. So that did a lot of the heavy lifting for us. Then we have to interpret what the answers are. You do get some parametric solutions. So one of the parametric solutions is the reason why you get a two here and not a one. So there's a systematic collision that happens between two of these roots of unity. It's a fun exercise to find it or you can just look in our pre-print. Then there are sporadic cases that you use to explain the other entries at the table. So that's the end of Act 4. So let me pause once more to see if there are questions about this. I have a question to Act 3 actually. Oh yeah, we can go back to you said that all numbers are represented well under the conditions. Yeah, so right, so I'm sorry. complexity, right? And do you have any idea how not necessarily uniformly but although of magnitude other certain values special that appear very rarely or some other I mean just lower at the bound. I did not look into this question about the distribution but much easier and but still you know whether there's certain special values in this. Yeah, we I didn't look I didn't try to make any effort to look at lower estimates. So in Act 2 the the work of fanbomal etc. does maybe give you some information like this. For this problem right it's it's I found it very difficult to to do anything quantitative because I mean right I'm I mean if you look at the space of the polynomials I'm in some very small corner and you know the number of points in this corner is a much lower you know is a is a smaller power than the main term. So I have to I have to use an optic where I really sort of look from this point of view and make it more quantitative and I haven't entirely figured out how to do that. So it would it will take more ideas I think to really get quantitative estimates about you know lower bounds for how many how many times you can represent a given integer. I mean I had a question I am really not familiar with the subject but if you I mean you don't you don't discuss that the examples of this are being the right is having having the very point again you would have to go through haunted tape to find you know it's up to the isogenic classes but the question is is and if you don't have these examples how do you how do you know that that somewhat equal you know in size the multiplicities I mean your construction in your mind behind the back you are not giving it's just only shows that these numbers do are represented indeed. Yeah but statistics would probably require interference into the construction you know. Yeah I mean I think that's a question for future investigation I think is to try to either by constructing example on the baleen variety side or more likely doing a better analysis on the vape polynomial side to get more quantitative information. Okay that's okay that's all thank you. All right if there are no other questions about parts one through acts one through four let me talk a little bit about act five so act five is of course where I close up some loose ends but also tease a sequel so so here's where we actually get we'll start to see some abelian varieties back in the picture and so in particular I want to start talking about Jacobians of curves we talked a little bit earlier about to what extent you can do things with Jacobians of curves so this theorem of various authors says that well there aren't very many curves over finite fields at least of positive genus if you exclude p1 so there are not very many curves of positive genus over finite fields for which the class number of function field is equal to one which is to say the order of the Jacobian variety is equal to one so there are one uh it turns out that not only over f3 and f4 there are one isogeny class each but there's actually one isomorphism class each and so all other cases are going to be over f2 so uh in the in the 1970s lights of modon and queen found five examples in general one two and three uh and then proved uh that this list was complete and this proof in air quotes stood until the mid 2010s when stirpe noticed that they were missing a case stirpe found an example in genus four so that's why seven has to be crossed out here this theorem was published with the word with the value seven but then it had to be corrected to the value eight about 40 years later so mercury stirpe gave a corrected proof and Shen and she gave an independent corrected proof so now it really is eight but this is an interesting cautionary tale oh you should uh trust but verify when you use old results uh but uh anyway so here's uh here's a relative version of this problem so uh so in the in the rest of this and the rest of these slides i want to talk about the relative version of this so the relative problem is if you have an extension a finite extension between two function fields when can it be the case that the relative class number is equal to one i.e the the two function fields have the same class number so the when you make an extension of function fields the class number um of the bottom function field is always a divisor of the the class number of the top function field the ratio is is the order of the generalized prim variety if you like and so you can ask well when is that ratio equal to one well one version of this question is when you take a curve and you take a base field extension so that's in the in the language of function fields that's like taking a function field and then you know taking a constant extension taking the compositing with it with a constant extension of fq and so um you can prove uh that there are relatively few cases except um i should have said g is not equal to zero on this slide so i'll say that out loud um because if g equals zero of course there's all you always get one on both sides but if g is bigger than zero and the order of the jacobian does not change going from fq to fq to the d then you only have um a specific list of cases for these values of q d and g and i have an explicit table there's a whole bunch of values a whole bunch of cases over f2 and a few cases over f3 and f4 again genus one in both of those cases but over f2 there is an example of genus three which i believe was found by lights of modern and queen and back in the 70s so the way you analyze this is to say well i mentioned this that there's an abelian variety a that measures the ratio between these two things in this case it's it's you take j of c you take its base extension then you take the v restriction of that and you compare that to j of c um the co-kernel of the map between the two is an abelian viad of order one so when d is bigger than two this thing has firmanius eigenvalues that involve d throats of unity uh in some way and so you have to have pairs of eigenvalues whose ratio is the d throats of unity and so act four comes in especially the part of act four where i allowed you to compare different simplest simple classes so that that that's secretly why i needed to do that in act four that takes care of all the cases where d is bigger than two now when d equals two that doesn't help because when d equals two a is just a quadratic twist of j of c so within a there's no anomaly it's j of c has one set of eigenvalues a has another set of eigenvalues i can't sort of do any internal comparison of eigenvalues to observe anything anomalous but what i can do is use the fact that occur because there's a curve and a curve over fq squared has at least as many points if a curve is over fq then it has at least as many points over q squared as it does over fq and this puts a restriction on a that turns out to be enough to to to reduce this to a finite computation so it's quite critical that we're talking about curves here this is where we use the fact that it's actually a curve uh the Jacobian of a curve and not just something generic abelian driving okay so this is this is a of the relative function field the class number one problem for extensions of function field that only differ by a constant field extension the opposite extreme is if you actually have an extension which comes from a a morphism of curves with the same base field this is this case is much harder but um i can tell you what happens when q is bigger than two this should also be in preparation i should point out um um so if i had so geometrically i have a morphism of curves of degree greater than one um and um i guess i'm also assuming that uh i also have to rule out the case where both curves are genus zero and birth curves are genus one um so ignore those exceptional cases um so if if c and c prime have the same class number and q is not equal to two then there is a complete list of possibilities so you can find examples of you know genus three to genus five over q equals three i think this might have also been an example in light somata and queen and over q equals four the best you can do is genus two goes to genus four um so i believe over i didn't write this on the slide but for genus two to genus four you might be wondering whether that's a degree two map or a degree three map i think both possibilities can occur so you can have a ramified map of degree two or an un-ram or in a tall map of degree three and there's a complete list of possibilities um so the way you you approach this is again there's an abelian variety that whose order measures this ratio it's the it's the prime variety for this covering and again you use information about numbers of points so specifically the upstairs curve has to have a non-negative number of fq points between uh the structure of a which is limited by the theorem that i stated earlier about abelian variety's order one in this case since q is greater than two a is just isogenous to a power of a certain elliptic curve so once you know the order of c you know the order of c prime and vice versa so c prime having at least zero points implies that c has quite a few points and uh then you compare that to uh sort of classical by now um lower bounds on the number of points these are the ones that come from the the so-called linear programming method um of ihara and especially ustelae and what which you find in uh the notes of sarah on uh rational curves rational points on curves that were finite fields from his harvard lectures in the 1980s which are now published uh by the smf so you use a lower bound on number of points uh on a curve for its genus and that runs sorry you use an upper bound on that you compare it to the lower bound that comes from the constraints of this problem and that tension creates um a finite list of possibilities which you then investigate so so that sort of solves the geometric version of the relative class number one problem for function fields over f three or larger what about f two um well you can do something similar over f two but you get a worse answer you get g less than or equal to seven g prime less than or equal to 13 this is sharp by the way uh there actually is an example of a degree of a of a genus seven curve with a double cover an unramified double cover therefore a genus 13 um where j of c and j of the double cover have the same class number i found exactly one example so far i'm starting to think it's unique but i haven't managed to prove that yet um here it's a it's a similar argument in that you start with a prime variety that has order one but now there are lots of possibilities because there are infinitely many simple ones and i have to allow combinations of the simple ones too so this seems like a mess uh the linear programming method uh over f two is not enough um because there's an example with there's there's an abelian variety of order one with trace zero but the linear programming method conveniently gives you more than advertised it actually gives you a lower bound an upper bound not just on the number of points on a curve over f two but it gives you some correction terms that involve the number of points over f four over f eight and so on with smaller weights and we actually use uh those error terms to our advantage to to to tighten up the argument that i sketched on the previous slide over f two and that actually ends up being enough to give you a bound um that doesn't quite give you this bound you then have to um i think it gives you g equals nine and then you have to do some searching over vapor polynomials to say that uh you have to spend like two hours searching vapor polynomials to show that you can't get g equals nine or g equals eight so uh this is something you can do um and now you can ask well okay i have a bound can i find the complete list well uh not yet but i can report one more partial result to where it's finding the complete list which is that um i can prove that all of these all of these cases where you get the um you know where where if g is bigger than one so i'm not messing with genus zero curves or genus one curves because you know i can those cases i can kind of make lots of i can you know insert isogenes and things so if i have hyperbolic curves uh and i have a not an isomorphism of course again to avoid trivialities and i have this and both curves in the cover both the curve and its cover have the same class number then i actually have to have a cyclic galois covering uh it doesn't have to be at all so galois here is in the sense of allowing ramification but it is a sick the extension of function fields has cyclic galois so this involves looking at all possible pairs of vapor polynomials for c and c prime so you have to do some more searching of the of the sort that i described on the very first slide and then of course you can ignore the cases where d equals two because quadratic covers are automatically galois so you just worry about covers where d is greater than two um there are about 70 of them uh then you you kind of go through this is a kind of argument that's frequently used to uh refine the upper bounds and the number of points on a curve over a finite field you say well i have some some vapor polynomials i look at how close points of c split in the covering and then because i have a non-trivial galois closure there are other curves that appear as quotients of the galois closure and so i sort of transfer information to get points counts about those curves and then they have isogenic factors in their Jacobian so i get some more vapor polynomials and i kind of mash all that together and uh to complete the classification uh based on this you need to find all the cyclic covers i'll just mention there actually is a cover of degree seven that occurs so there there are some interesting examples this would be straightforward in bagma just computing class filter function fields except that i need a i need a table of curves of genus up to seven over f2 and this does not currently exist there's a table up to genus four in the lm fdb and there's work on genus five that's ongoing genus six and seven should be doable because mukai has these beautiful descriptions of canonical curves in terms of homogeneous certain homogeneous spaces but there's a computational problem that is not yet done to to finish this classification and i think except for my references that's the last slide so i will end there