 Good morning, good afternoon, good evening to all of you around the world and thank you for joining, joining us this. And also thanks to the organizers for this wonderful worldwide seminar that has been running now for quite a while. It has, it's wonderful that we all get to gather and today will be gathered here around the world to study elliptic curves. And if there's one thing that keeps us together I think no matter what's happening in the world it's a love of elliptic curves. I was just talking with a student at the end of our spring term about torsion on elliptic curves. And how do you tell when an elliptic curve has torsion, and one of our favorite things to do is to reduce mod p. So if you start with an elliptic curve over q, and you take a prime p that doesn't divide the discriminant of e, maybe take p greater equal to 11 to avoid some small prime phenomenon, then whatever the torsion is of your elliptic curve over q. It necessarily embeds in the FP points of your elliptic curve. And so we were playing around. And I took this elliptic curve. It looks innocuous enough I hope. If you're a LMFDB aficionado you can enter in this label and see all that you would could ever want to know about this elliptic curve. And what do you get if you count points on this elliptic curve modulo a couple of primes well I said I wasn't going to look at seven I guess so we'll start at 11, you start counting you get 181815271530, and so on. Maybe soon, you notice a pattern that all of these point counts are divisible by three. So you might ask yourself does that necessarily mean that three divides the number of points on on the, the torsion points of the elliptic curve. It's certainly allowed by the injective map that we wrote down here. And, well the answer is no. No this implication need not hold. It's too bad otherwise this would be a really nice way to detect what the torsion is if you try for a couple of primes and it looks very often that you have some sort of the visibility statement that you can always be sure that this it's easy to count points over elliptic curve otherwise it torsion might be much harder to detect. So what's going on here can we fix this some way. There's a theorem of cats. That says that if you take an elliptic curve E and you have the property that m divides the number of points over fp for all good p so p doesn't divide the discriminant maybe take bigger than 11. Then, okay well you don't necessarily get E with the property that you're after but you can always find an isogenous elliptic curve. So isogenous over Q with the property that m divides the number of points on e prime the torsion software. So why is this the the right theorem. Well, it's a little unfair to ask the poor elliptic curve to have this property about the its number of points, because the number of points is in its is an isogenous invariant if you look at any isogenous look to curve it also has the same number of points. So within the isogenous class over Q. It's not that it's not obvious that this property of having a torsion of a given order will be preserved. And so it's only fair to really ask on the isogenous class itself, and then the theorem of cats says that you always find at least one elliptic curve within that isogenous class with this property. So for example up above, well here's your elliptic curve that is isogenous it's a little bit bigger I guess, in terms of this coefficients than the one you started with, but it's easy enough to check that this point actually is a point of order three. All right, so maybe you stop there theorems of cats have a tendency of settling important questions, but we didn't stop there. So you might ask, well alright so it's not always the case that your elliptic curve has this property that m divides the global torsion subgroup. So you might ask what were the odds going in that it has this property. So it's kind of a statistical reasoning so somebody hands you a note to curve and says, Well I always get three dividing the torsion subgroup. What is the probability that it actually has a global three torsion point. And that's also the question that john Cullen and asked me. So if you have this property that m divides the number of points mod p for all good p greater equal to 11 what is the probability m divides the number of points on the q tour. John asked me this question over email and innocently enough, maybe you guess it's 5050 or maybe you guess 0% I guess at some point I also guessed 100% and maybe it's a number in between. I like this kind of question it's a, it says whenever you and maybe this is one of the things I'll offer you for for your attendance today. Yes, Andrew the isogenes of degree three that's, you can almost read that off of the proof of cats itself but along the way we will even refine that part of the statement if you move beyond a prime P and look at something like the 16 divide the order we will be able to to bound the degree of the asogyny will always be a power of p but will be even precise be precise about that. All right, so I like this question because it takes something that was, I don't know what theorem that says you can't guarantee the elliptic curve has a property, and then you ask the question all right well can you bound or measure the failure of that theorem to hold so another way that I like this is it's a kind of probabilistic local global principle so if it looks like your elliptic curve has a three torsion point module every prime, what's the probability that it has a three torsion point globally. And that's really local global principle because you could replace not be with the attic on the. That really is the same thing. Okay, so that's your riddle to hopefully you're hooked now on the question of sort of understanding and quantifying the properties of elliptic curves. So, how are we going to answer a question like this well first we need to formulate it in a more precise way. So we need to some there's infinitely many elliptic curve so to define a probability we have to order them in some way, and then count the ones that have a property that we want. So, let's order them by height. So, the probability that the elliptic curve E over q is uniquely up to isomorphism over q to an elliptic curve of the form, why squared equals xq plus a x plus B, and B, you can maybe start out as rational numbers, but you clear denominators, and you do so in a minimal way, you have over clear denominators if you ever have a prime P with the property that p to the fourth divides a and p to the sixth divides be you can with the substitution of variables and x and y clear that additional power fee and then you've done so, minimally, of course all elliptic curves of this property that they're discriminant is necessarily non zero. So, somehow you could label all elliptic curves uniquely in their isomorphism class by simply a pair of integers capital A and capital B satisfying these properties that's a way of bookkeeping and keeping track of all of the curves. And if you do that it's kind of natural to consider its height to be I don't know the largest of the two terms that appear in the discriminant this is correct by the way in which a and B are weighted in this homogeneous rescaling. And it's natural for for many other reasons. So, we define for an elliptic curve e maybe will this script e is supposed to denote the representatives of the isomorphism classes elliptic curves with these two properties, just to be clear about where, how we're keeping track of them you can think about them as pairs of integers in a similar way. I don't know alright so for a bound x will take the elliptic curves whose height is it most x, and we'll put them into a set so that'll be. I hope it's clear that there's only finally many elliptic curves I bounded a and B in a kind of weighted way so that's really, you can write them all down, you could even take uniformly random one by just uniformly choosing a and B in those ranges and then accepting it only if it satisfies these two conditions so you really can get your handle on elliptic curves, there's lots of other ways of ordering elliptic curves by discriminant conductor other properties, they're harder to get at and this is the right one for the stacky comments that I will make at the end. All right so with this in mind we can define a probability. And, oh gosh this looks kind of weird. We take the elliptic curves with the property that m divides the torsion that's the probability that we're hoping to quantify, and then you on the bottom you ask for those that have this property that m divides the number points Monty for all good P, and you take them just up to high x, and then you take the limit as X goes to infinity, so this is the usual that this is the total space that you're interested in up to hide x. This is the ones that have the property that you're interested in, and then you define this limit and you hope that it exists. What we restrict to is the integer m should be one where you could even have an elliptic curve with torsion subgroup of size m. A major tells you that the list of such m's are just the only ones that are allowed the one up to 10 12 and 16. All right, so our question is now about does this probability exist and if so what is the, what is it. There are two computations, and if you count for example up to 10 to the 12, then there's 7,578 elliptic curves pairs a and b with the property pairs and 3,800 native them. Sorry, with the, you know, we restrict to this set but then you have to ask for this property that they have. And it looks like they have a three torsion point on P for every P. So it comes out to be about 50.3%. So at least for this initial computation you might guess that the odds are even. All right, and the first theorem for today is that this is joint work with john cullinan, making Kenny and the theorem and myself and the theorem is that this probability exists, and it's always positive. P three is a half. We calculated the values. So P two is equal to one P four is about 27%. And we didn't calculate them for all then but we did show that it is effectively computable and necessarily positive. Okay, so. Great. So how do you prove a theorem like this. Well, there's sort of basically two things you need to do. The first one is to reduce the problem to counting elliptic curves with prescribed level structure so level structure will be instead of the probability that we announced up here where you ask for a property of the elliptic curve. What you have to do is to translate not unlike Andrew's question about what's the nature of the isogeny that holds between any prime, but you have to somehow translate this into a question about counting elliptic curve with certain level structure image of Galois. And then I'm going to skip that step for reasons of time I'll invite you to read our paper if you'd like to hear more about it. I'm going to end today on this question of counting them because it's a nice kind of appealing analytic algebraic arithmetic geometry question that I hope you will find also engaging and interesting. All right, I think this is a good moment for a quick question if somebody has learned. I mean, hooked on counting elliptic curves with certain properties and intrigue to hear more. Does anyone have any quick questions. I have one. Can you tell me what yes I can hear you. What's the percentage of electric cars counting according to your height versus according to the discriminant to the conductor. I have not been able to count by discriminant or conductor for the question because of your result. I wonder if that represents. So to speak, the more natural number I mean because, or then according to the conductor, if you count the number of points, this is more natural right. Yes, I agree that conductor and discriminant are more natural. If I have time at the end I'll make some comments that actually height is the right way to think about this is heights of you count points on a modular curve, according to height, and then height is really the right thing if you want to fit it into a program in general of counting points on variety so I think it would still be interesting. But no, because, you know, number of points is really character by the function. So but conductors much more natural. Right. Yeah, I think it would be really interesting to compare this results and to try to prove something there I expected will be harder for example, even just counting elliptic curse ordered by conductor is already without any question. And I realized, but what's your expectation is that percentage in your theorem would be the same one half. I mean if you just guess of course you said you this difficult I'm not asking that, but just just what would be your guess. I guess that this theorem should be pretty stable that it should exist, and it should still be positive, but I don't think that it will be exactly a half anymore. Yeah, the probabilities should change a bit. So there's a hand raised. Yeah, I was just wondering if cats theorem would already tell you that PM has to be greater than equal to one over eight, because I saw any classes that there except most like a with the curves in the same my so journey class class and it's not clear to me that it does because you don't know the heights of the elliptic curves in that I saw any class. Oh, maybe the one that has a torsion point is gargantuan larger than the other. And as you go farther they become rare and rare we shouldn't expect that. I think your intuition is right that it should suggest a positive probability, but you have to bring something. Okay, thanks. And a question from Ariel Weiss in the chat. I'll just read it at least one piece great and we shouldn't the probability ordering by conductor just be 50% and a five isogeny class where one has five torsion the other will not. This conductor is invariant under the isogeny class. So you would need to know that when you have an elliptic curve with this property that five divides the order but you want to check globally that that is the isogeny class you're talking about is just the pair of two. I think that is the most likely scenario and I think it just follows from like, it just follows from like classic ribbons that you have two distinct characters one of them on the diagonal of the Galaxation one is trivial and one is the stomach. So one isogeny classes that will be trivial and the other isogeny classes that will be so it doesn't. Yeah, that sounds nice I wonder what happens if you take 16 for example if you'll have to investigate a more complicated isogeny class, and it doesn't address the question about discriminants that that's really probably harder. Alright, I better get keep going here. So, now what we're going to do is we're going to set up a general problem about counting elliptic curves with level structure. So in order to do that I need to talk about torsion points on elliptic curve and how the absolute Galax group acts on them so I'll do this hopefully set up notation in efficient way so there's the absolute Galax group of q. This is my algebraic closure of q. It acts on the m torsion of the elliptic curve. It's q bar points as an abelian group it is isomorphic to Z mod mz the Galax group moves them around that gives you a representation depending on em from the Galax group into GL to Z mod mz it acts by change of basis. So, and it's the main example to keep in mind here is if you take the image is contained inside the subgroup I don't know one star zero star. What are you all saying is that the determinant of this representation is necessarily the cyclotonic character so here you even know what the star is. The image this holds if and only if eq has a point of order m. This is up to a choice of basis. So when we write down two by two matrices representing the action in this isomorphism I've chosen a basis of them so I don't know if this thing is like generated by P one and P two, then in this case the point of order and is the point P one, and if I choose a different one I will have conjugated the subgroup. So in order to have this notation that I want to consider the image contained inside this up to conjugation, all right it like this instead of a less than or equal to all right it with the less than and then a tilde to indicate that it is up to a choice of basis. I hope that's okay. I hope it's clear that when gala acts like this on P one, it's really leaving the point P one where it is that's what it means to have to be a fixed vector under the action and by gala theory that says that you have a rational and torsion point. So we want to study things like this property and whatever other images of gala you get from conjugate subgroups from being isogenous to such an elliptic curve. We want to focus on fixing the group G and asking how many elliptic curves have gala image contained inside a fixed subgroup so how do we write down something like that. Well I guess we'll switch from little m to capital N, which is more common in this world so you take a non negative integer n, and you choose a subgroup G and GL to Z mod nz which we only care about up to conjugation. So we want to ask that the determinant of G is fully Z mod nz cross. This is the cyclotom in the character and it's a necessary condition if you have an elliptic curve over Q. Any image of gala g has to have the property that it's determinant of subjective because you don't have non trivial words of unity in Q. And the first thing that we need to do is to count in order to count such elliptic curves, how do we do it well there's a, an open modular curve I don't care about generalized elliptic curves right now. Well it's doing everything it can to parameterize the elliptic curves you over Q with a property that it's modern gal representation, the image of Galois here is contained up to conjugation in G, the fixed subgroup that I took in GL to Z mod nz. And if you want the precise description it goes like this, you take your group G you intersect it with SL to, and then you pull back under the natural surjective map from SL to Z, you define a group called gamma G, and then the group. So the curve that you want to write down YG is the quotient of the upper half plane by this group. So that's the recipe. This isn't a talk on elliptic curves and their gal representation so that's all I have time to say right now but I hope you'll agree that this is something that is easy or effective to understand. So the curve that we had up above where you take G to be this one zero star star group. Then this gamma G is just our usual friend gamma one event if you're upper triangular modular and that's exactly the condition that we wrote down, and then moreover you want the upper entry to be one. And so this YG that I wrote down is just the open modular familiar curve, why one event. So there's some quite some work in the LMF DB to try to catalog all of the possible modular curve so just if you clicked on Drew's link to get to know the elliptic curve I bet you have this uncontrollable urge. Also to click on classical modular curse to get to know them better. That should be coming to an LMF DB near you. Now there's lots of work and finding rational points on modular curves. And that's, you know, of high genus in order to see exotic things that are happening that's also not the topic of today though we're waving at it as we go by, we want situations where the modular curves have genus zero so that we can have infinitely many points to count. So there's only if it's genius greater equal to two faulting says there's only finitely many and then there's no point in doing asymptotics. If it's genus one. Well, that's sort of, there'll be logarithmically many at most, depending on the rank, and that's also not really the fundamental question that we're asking about we want to look at the cases of Gina zero. As it turns out there's really only finally many up to twist so we have kind of a finite concrete task in front of us. So with that notation, here is the quantity that we would like to investigate. So it depends on my group G which is a subgroup of deal to see mud and Z. I choose my parameter X, and then I say hello elliptic curves. I look at you that have height at most x, and then I say when is the image of your gaol representation contained inside my group G up to conjugation, and I want to count this as a function of X. Alright, so the, the first sort of systematic attempt to understand these according to height was work of heron and Snowden. And we have a more general theorem which I guess I will just show to you. So this will be our first main theorem for the talk today. And it's how it's the main ingredient for our probability theorem so again it's joint work with john and Megan. So we have somewhat strong hypotheses so let me go through those first but the answer will be an asymptotic for this function and G of X enough to count our probabilities. So we asked a very strong condition that this group gamma G is torsion free so it also doesn't have minus one in it. So there's no elements of finite order in the group. So this is gamma G that we took upstairs other than the identity element. There's a condition here about nowhere regular costs but I hope you'll just ignore it is just a silly technical thing that only shows up for gamma one of four and you really should I mean I have to put it there because I want my theorem to be true but it is not relevant for understanding. And then I do I make this less severe constraints that YG has genus zero in the sense that that's what we're interested in any way in counting points on. So this is an integer which is called DG, and it's half the index of gamma G inside SL to Z. This really is an integer because I asked that minus one is not in the group. So gamma G projects into PSL to Z, and that's why it's necessarily always an even index. This is always even so I can divide by two. And then interpret this as some kind of reduced degree of the J map, if you wanted to think about it that way on the set of elliptic curves, which feels like the natural measure for how complicated these elliptic curves are getting. Right. And our theorem is there exists an effectively computable constant CG. It might be zero, it might be the case that your YG is a genus zero curve but it doesn't have any rational points, in which case, there's no problem counting points on that if there's no points at all the answer is zero but then the statement still needs to be true so I, there is a possible quaternion algebra or a conic that might show up that has no rational points. Thank God I got to say quaternion algebras at some point in the stock. All right, and our answer is that this NG of X is asymptotic to X to the one over DG, and the error term is one over to DG so it is like a square root error term which is pretty strong. Given the nature of general asymptotics but that's what we were able to prove as X goes to infinity. Does that make sense? Hope so. And this is enough to prove our probability exists and is non zero. You look at the modular curves that I haven't told you about but which are described in the paper, they, they're finally many you see what these main terms are with the D of G, they all have the same D. And the constants are all non zero. So basically the numerator and the denominator are both the same order of magnitude with non zero probabilities, and therefore the probability existence is non zero. That's how we prove our, our theorem. Well, what Heron and Snowden made significant progress on this but they only proved statements of the form X one over DG is less than less than this quantity and GX is less than less than X to the one over DG. So they got the, that isn't even an asymptotic it's just like telling you the estimate on the rate of growth the less and less than is there exists a constant. And so that wouldn't be enough for our results because you would want to take the ratio of these two things on top and bottom, and because you don't have convergence on anything it doesn't actually tell you what the probability is. So, there are a lot of interesting results and counting a little to occurs with certain properties but I guess I'm going to ask us to imagine and hope for whenever we can effectively computable constant out in front of asymptotic and a power saving error term on this power saving error term in the language of probability would tell us about the convergence to the probability you know what are our odds. If you fix a value of X. And so that's that'll be our gold standard for for trying to address those. Those apply to the cases that you are probably most interested in namely torsion groups themselves so let me show you that even though I've dropped the probability statement so this would ask for elliptic curves where the torsion subgroup is isomorphic to a given group T. So it applies in all but the first few cases which you can do separately. And here's the table that you get. I tried to have nice handwriting but as I was writing this out I decided just to copy the table and see could really see some crisp fonts there. And here's where if you have if you just look at all possible elliptic curves, the number of them is X to the five six. That makes sense because a cube is bounded B squared is bounded that gives you X to the one third and one half those totals give you X to the five six and the error term here is a half that comes naturally from that. I guess for the three torsion that we were looking at earlier. This is the main contribution to the term these would be our friendly numerator terms where you have. They grow like X to one third with one fourth. It's kind of interesting to see when these numbers coincide like at five and six torsion occur with the same probability as Z two cross Z four. That's just how the numbers work out in terms of degrees. And anyway this is some kind of quantification of when you look at elliptic curves what kinds of torsion subgroups should you expect with what kind of probability they're all very sparse. So they'll be 0% but at least here you're quantifying the rate of growth as you increase the height. Okay, so I would be remiss if I did not mention generalizations and related and proceeding where so I've already mentioned the Heron Snowden result, broomer and do consider questions like this. Also trying to make the conjectures about ordering elliptic curves by conductor and height that's just about this line here. And more recently, Peter Brown and Phil Najman and Choe and Jung and Phillips, Tristan Phillips had some recent papers on the archive of PhD student in Arizona, are also working on this thinking about elliptic curves over number fields, and I'll have some more references later on, this is a really active interesting area with lots of questions popping out. I'm sorry I don't have time to cover all of them I'll just try to give you my perspective on them and sort of how, how we might aim for a general result. All right. Good so far, everybody's still with me. So I think, oh good a question. This is the effectively computable constant. What a wonderful question. I think you will have a great answer once I get into the steps of the proof so I'm going to use that as a motivation to show you how the proof works, because you really can describe what the constants are. All right so it works in four steps. I put things in red here. If because of well I announced that we should really try to find an asymptotic for ng of x for general G. And that would, you know with the power saying error term and effectively computable constant so that's what we would like to get to our height, our theorem has some hypotheses in it. And so if the hypotheses are sort of hard to remove. I put something in red right there's a technical issue that arises there. And if it's see it works out pretty much the same when we remove those hypothesis I put a blue so blue is supposed to be a friendly color, and red is where we're going to have to do some work in about 10 minutes. So the first thing you do which is kind of standard is to produce a universal curve. Okay, so what does it mean so there's just here I have an example for number one just to show you what it looks like so I don't have to go into great detail. There's more or less an algorithm now, which is implemented in magma and probably also some formulation and sage and you can even do some of them by hand. So the take normal form if you wanted to write down the universal if the curve or all of the curves with a five torsion point what you do, will you change coordinates to make that five torsion point be at 00 and then you take triple the point and compare it to double the point and you ask that those have the same x coordinate, and that there's no parameterization on the coefficients. So what you will find in general or polynomials a of AB and be a baby which come from the homogenizations of some polynomials in one variable, T would be the parameter on your genus zero modular curve. And these would be the values of the Eisenstein series written as rational functions or cleared through homogeneously, in terms of that, what's called a hot module. And the A over B is just comes about from writing the fraction T, in terms of numerator denominator and then writing them homogeneously so this is what you would get for why 105 and the theorem is that every elliptic curve with a five torsion point comes about via some T equals a over B. Okay, so you can write down exactly those. There's a little bit extra structure here namely that this is elliptic curve equipped with the five torsion point. So that is handled in our point for. All right, so once you have an elliptic curve. What do you do when you do something which is called apply the principle of lip shifts which is also dab and poor slimmer. So a few words on that but what now what you need to do is to take this A of AB and B of AB and imagine plugging in integers a and B asking for these quantities to be founded. So now you have a bunch of, I don't know we'll call it lattice points in a region where you need to count them with some explicit inequalities. And that's a more or less standard thing in analytic number theory show you that in just a second. You then you have to sieve afterwards you've done some over counting. There were the conditions that p to the fourth and p to the sixth to not divide a and B respectively. So you have to treat those conditions and I call this. First you groom them. And then there's something else that occurs which is you have to avoid something called the minimality defect at certain finally many primes. And finally account for automorphisms which is over counting if you just want to count the elliptic curves not elliptic curves equipped with something. And maybe that that's asking for multiple values of these polynomials that come about from the same value of T. That's a relatively easy step but also itself interesting. So that's the basic strategy I hope that makes sense to show you what the constant looks like. It comes from the area of the region and this sitting which will introduce a zeta factor so I'm going to show you what that looks like an example. So here is the principle of lift shift so in general, if you have a bounded region defined by nice inequalities in an art to the end and you want to count the number of lattice points here we'll just have two dimensional region so I'll do a slightly easier one. So how would you estimate the number of points in a region like this, well you would estimate it as being approximately the area of the region, whatever this region is. That makes sense because if you put a little square over each part of the region that should do a pretty good job of estimating the number of lattice points there. What could go wrong what would the error term be related to. Well it's sort of near the boundary. If those squares don't exactly line up with the region, you should be expecting an error term which is as complicated as the boundary of the region. And so the principle of lipstick which is also called down ports lemma says that the error term exactly the length of the boundary. For which you could take the maximum of the projections on to the coordinate axes that that would be another way of thinking about it. So if we write this out so if I have a region are and I want to count the number of lattice points in the region that should be the area of the region. Together with the length of the boundary. Okay, so how would you count elliptic curves. Well we have some kind of homogeneously expanding regions with A's and B's so if you do you want to count just elliptic curves without any extra conditions, you would be counting something like pairs AB, where for a cubed and 27 b squared are less than or equal to x. So if you count something like that will I have an AB plane over here. And what I want to do is to make the region that encapsulate those lattice points and then the height here will be something like a goes up to x over 27 to the half, and the B goes up to four to the one third. And you have to double those I guess. And so the area of the region is just going to be four times these quantities, one third, one over 27. That's the x to the five six that's the area when you multiply these two quantities together, and the area of the length, which is the error term is the x to the one half. So hopefully I've just given you now the explanation for what the exponents are, both in the main term and the error term they come from the description of the area of the region and it's homogeneously expanding description. That's the first answer to Jacob to your question, what is the constant represented represents the the area of the region where you take you know R of one. So you take some fixed region and then it homogeneously expands. We're not done yet because we have to save. So in the sitting step, we've over counted. How many do we need to remove. Well we need to remove any a pair and be where a prime P to the fourth divides a and a prime P to the sixth divides be that sort of one out of every P to the minus 10 so we need to multiply by one minus P to the minus 10. So for all primes P. We also have to remove the a and B where for a cube plus 27 b squared is equal to zero but that term truly is negligible. And what you get from this I hope you guys recognize the Zeta factor here it's Zeta of 10. So the number of lifted curves, not with any without any torsion subgroup is given by this quantity you arrange your constants, you have a term down here Zeta 10 x to the five six plus big O of x to the one over one half. So I hope I answer your question Jacob, all of our constants should look like this there's an area term, and then a term that comes from sitting together with in this case no extra weirdness with minimality defects at finally many primes, but if you ask for like m torsion subgroups at prime dividing M there's also kind of local factors that show up there. And in general what you'll have to do is to take polynomials like these, these a and B, and you'll have a different kind of more complicated region, but still you'll have the same kind of general strategy for for pre. Okay, great. So good a moment to pause for just a minute to make sure that I'm still reaching you so now I've basically told you how we prove our, our main result. I'm going to review the steps here. There was more or less standard computational thing we just write down all the universal curves we really do need to get those in the LMF DB you guys that's that's really essential when we when we do that part. The principle ellipse it tells us about a homogenous expanding region we end up counting lattice points. And that's relatively standard so the hard part will be boy what happens when you don't have a universal curve there's some issues there. And then there's really problems in sitting, especially with this minimality defect when we start to peel away our hypothesis so now I'd like to try to imagine what a general result would look like so can we get an asymptotic with power saving for the functions and g of x for an arbitrary g of Gina zero with so what were the hypotheses that I need to delete the hypotheses are okay well I guess I already in advance deleted the no irregular costs but okay I guess there. That's not that's that remains a technicality that isn't interesting. So it's really this torsion free that we have to worry about so. The reason why modular curse where the gamma g is torsion free is great is because they really are a fine modular space that really are universal polynomials there's no. Stacky weirdness that that happens but as soon as you remove this hypothesis for example if minus one is in the group. Then any elliptic curve with the image contained in G will also all of its quadratic twists will also have the same the same property. So you really have to count in a different way. So on the way to a general result maybe we should keep should keep some cases in mind. So, I may be the most the next most natural thing you might want to count our elliptic curves with the cyclic anisogyny defined over Q. So the difference here is the point itself it's X and Y coordinates are not necessarily defined over Q anymore. So if you ask for a subgroup of order and which is stable under the Galois action. So that changes your upper triangular matrix with the one here. Now you just ask that that element maps within that cyclic subgroup. The associated open modular curve is now called why not event instead of why one event, and I hope you'll agree that if this program is to succeed these are emblematic of the kinds of with the curves we should be counting. If you have an elliptic curve with a cyclic anisogyny so to will every quadratic twist so minus one is in the group and you really are not in the nice case that we already dealt with you genuinely run into new and interesting obstacles. Alright, so I have two more theorems to show you. So here's the next theorem which represents the desiderado that I announced in sort of the first interesting case where you look at three isogene so if you have a if you have a two isogene you just have a point in a non trivial point that's basically not harder than two points themselves. So if you ask for three torsion points I decorated this and called it g zero of three for the group. So this is counting elliptic curves that have a rational three isogene. And this is joined with Maggie Pizzo, Carl Pomerance with and Carl Pomerance, and the statement of the theorem really does look quite different than the one that we had earlier. So it starts out with an x to the one half term which is kind of crazy. This corresponds just to the curves with a is equal to zero. Okay that's equivalent to saying that the j is equal to zero. See there's all curves of the form y squared equals x u plus B for some value of B. B squared is supposed to be less than or equal to x, and that's what you get already x to the one half as many elliptic curves. And this constant is no harder than the one we just wrote down, looking at elements in a square. Okay, and this is really weird that sort of one elliptic curve and its twists already account for. So if you ran up to a lifted curve on the street with the three isogene with all of them that you run into are going to have j equals zero for whatever reason. And then after that, I don't know they're effectively computable constant C and now there's a C prime, depending on this, the group G zero three that have well now there's a log term, and then a term just below it, and still we have a power saving error term so each of these terms will have some meaning when we go down below, I would say this x to the one third is that that one third is the same as the third that we looked at when we did three torsion points. And the reason for the log is because of this quadratic twist it's like a Dirichlet divisor problem if you have a familiar with that aspect of analytic number theory introduced sort of divisors up to a given point, and that's why you have both an extra one third with a log x. I don't know that I've ever proven a theorem with like, not just a secondary term but a tertiary term and a power saving error term after that but that's what you get. When you write a paper with Carl Pomerance, you really get top quality results. Now Carl and I and Maggie we did not prove this theorem, using the principle of lip shits, because we didn't find a way to make it work. Instead we just consider the problem on its face. You want the three division polynomial which is a quarter polynomial to have a root, and we just got to town sort of proving what we could. I wanted this to give us some momentum some some tailwind to launch the general project of the NGs. And this was the first non you know non trivial case where you could actually do something. Since this paper, there's been lots of further work trying to address these questions in varying kinds of level of generality, the field K could be an arbitrary number field or even global field. So Brandon Brogas and some of your sank are have some results of the analogous to the Heron Stoden results with an asymptotic with the less than less than Tristan has also been working really hard to generalize in certain cases. Carl and Ed Schaefer did some results with four isogenes and more than one for isogenes so again there's lots of there's probably other ongoing work as we speak to try to address these these kinds of questions now that we've seen that it is possible. And I have one final theorem to announce one that I'm very happy to share with you today it's sort of emblematic of what I think will be a general result. So this is joint work with my PhD student Grant Mulnar on who we on the job market next year. And we prove this is the count for seven isogenes. So there was it stood out as a case that was not that could not be addressed using the general framework, all of the issue was in sitting and that step three that I announced. I managed to get the civ to work I'll have more comments on that below but anyway the result is similar. There's no weird term out in front here. There's just the X to the one six log acts and X to the one six. The constants are effectively computable and in fact we have them to the first few decimal digits, and we get a nice power saving error term and maybe with a bit more time. We can improve on this 745 but that definitely is of the shape that would be desired for all, all of the possible cases. So, we do expect that this will allow us to address our strategy to address basically all of the remaining genus zero curves, using again the principle of data approach to sieve. And here is the new idea. We count elliptic curves, not up to isomorphism over Q, but in this case where it's well defined on the twist class, it's naturally to find them sort of minimally within that twist class. So the green, the lowercase green here is sort of up to isomorphism over the algebraic number so that is up to twist. So if you avoid the usual candidates j equals zero and 7028. Instead of asking for the condition above where you have a four and a six, you ask, not up to isomorphism over cubit up to over the algebraic closure. Now you ask for no prime p such that p square divides a p cube divides a. And then up to the twist by minus one, there's exactly two elliptic curves that are minimal in their twist class and then we're going to count those. I hope that seems natural if I can count those, then I can further twist to count all of the, all of them up to isomorphism over Q. And this may be a separately interesting question if you have a problem as well defined on the twist class maybe you want to choose a minimal representative in that twist class, just like we chose a minimal representative of the elliptic curve, and then count them this way. So, how do you formalize that will our technique is to form the Dirichlet series where you have the coefficients given by your elliptic curve, where they're twist height, which is what I just defined is of a given in the Dirichlet series and codes the asymptotic if you want to count the such up to X. And then if you want to put in the twists there's a Landau's version of the Tauberian theorem this also goes by the phrase melon perone heuristics. So all you have to do is to add in the twists, which is to multiply by certain zeta factors, of course you wonder if you have a power saving term, you understand the pole of this L function with the given with a bit of room, and then you understand the analytic properties of this ratio of zeta functions and that's, that's all you need in order to finish the asymptotic to understand the steps. So that's all the time I have to say about that theorem. I promise some stacky comments, but I'm only going to say this in in 30 seconds so there's it's sort of my main motivation which is recent work of Jordan Ellenberg matzatriano and David sir Brown they define a notion of height for rational points on stacks. And they formulate a conjecture of what they call a bad man type but but dear of many Mala type. And this is for the number of points I hide out up to x, this should be a general gadget which allows us that should make predictions of a wide generality and arithmetic statistics. So within this framework, we have a stacky curve, the modular curve. And our energy of x is really counting points on this y of g. So the exponent that occurs in our theorems really matches the Fujita invariant that they define in generality. So you could see all of our work on the energy of x as being proofs of the predictions that they make of this general conjecture about stacky points so that's why I'm less interested in discriminants and the conductors and more interested in in heights. All right, well, the final takeaway. If you have an elliptic over the rationales and it looks like it as the three torsion point when you reduce mod p, the other 5050 that does. And in general, I've shown you a couple of theorems, which we hope to have the most general one at least over q. If you have modular curves with Gina zero there's an asymptotic with power saving error term for the number of rational points by height. So if you if you found this an appealing subject. I hope that you will join us for a math research community, explicit computations with stacks you're really counting points on a stack. What you're doing so really with your bare hands, trying to understand some explicit arithmetic functions. So we're organizing an MRC, Andrew cobin sumia sank our Libby Taylor, David sir Brown and we would love to have you. If you would like to join us in solving some of these problems. Thank you very much for your attention.