 You know when I was in graduate school, I lived with a guy called David Carlton who sends his regards to another student of Barry's and You know, I was reminiscing about a night that I think the only night that we had Barry over for dinner and David I like Monzo was a sunspotist as well as being about fish and he had spent a year in Sri Lanka He liked to cook really spicy Indian food So we made this gigantic dish of chicken vindaloo which is one of the spiciest things that's ever eaten and we've Heard so much about Barry's graciousness in every situation even in the situation of maximum extremity and this I think Beyond any mathematical interaction we ever had I think this is what it was maximally display is. I was sitting there on our table Visibly sweating and in great pain and his face getting redder and redder as he politely continued to eat the food that that we had made And The reason I the reason I tell this story besides just to sort of add one more example of various graciousness is to say that That very night Barry wrote me an email titled that wonderful dinner and some math In which he wrote about well what I had been thinking about at the time was you know It was the 90s for those many of us remember the 90s and it was the time of excitement about Fermat on the new developments and the modular method and I like a lot of people was thinking about Generalized for my equations right what happens when you have ternary equation to look like from out with different exponents like a to the p plus b to the q plus c To the r equals zero and the subject of his email that Barry wrote me quite possibly still in a kind of fever to the sweaty state from For this food meeting Was that you can really think of this just as the Fermat equation with p p and b I mean is this value of points on some higher genius curve as Barry pointed out this all the way it looks like an affine surface Is in some sense a curve too because there's an action of gm On the solution on the set of solutions to this equation But it's weighted right let's see if I can write this down correctly so gm acts on this affine surface in three space What I call I guess it sends it out. I could maybe even send that to uh Lambda sends it to lambda the q r a lambda to the pr b lambda to the p q C right and the point is that the quotient of this surface by this action of gm Well the quotient of a surface by a one-dimensional group should be a curve except it's not quite a curve Where there's something funny in the case for all the exponents of the same It's really just a standard story where what I've written down is a gm bundle over an honest curve and we're studying points in that curve but here um Something else is going on because there's some stabilizers in this action. So uh Barry in his email did not use the word stack. You call this a magnified curve But I'm not sure whether it was because we're trying not to intimidate me Or because you yourself preferred Not to use this language. So maybe uh part of the uh Part of the of the lesson is that sort of it's now 20 years later. It's 2018 instead of 1997 and uh We're sort of in a world where people are sort of like more at ease with Calling a stack a stack. So I want to sort of uh say a little bit about Rational points on stacks even though we were afraid to call it by its name At the time. So let me sort of say the motivation uh For why one might want to do this. So here's uh Let me talk about two popular conjectures uh in number theory in the area in an area that's sometimes uh called arithmetic statistics. So Let me start with counting number fields and I'll give a very abbreviated story about an area that a lot of people in this room Have have worked in so Yeah, the question is what does one actually mean by a solution to this equation? So maybe maybe one sort of in order to get a reasonable answer, right? I mean you may want to say I want Integral solutions that are co-prime or something like this, but the reason you find yourself wanting to say that is you're sort of uh It in a mouse the same thing as integral point on this so-called magnified. Okay. So here's a Fundamental question. So um, you can say Okay, number fields are there. Okay, there's a lot. So let's say a little bit more For a long time. So if you're taking notes, don't maybe don't write it down yet. Um, let's say of degree d over q Okay, still a lot, right? There are a lot of quadratic fields. Um, so let's say now and discriminant Bounded by x and now already we're down to a finite set, right? We're discriminant I'm always going to be the norm of the discriminant to the absolute value of the discriminant So now already if I fix the degree and I put an upper bound on the discriminant Um, we're going to have some finite set of number fields by a rather old result of her meets Um, and given a finite set we should of course Ask it its size. It's the only thing that we can't ask. Um, and in particular we can ask how this behaves as a function of x Let me even decorate this a little more. Okay, so first of all um Let's be grown-ups and not restrict ourselves to the field q. Let's just have some arbitrary global field here But if you want to think For at least some of this talk you can pretend it's q. There will be other times when I want you not to do that so let's say k Clare k global fields and now series now so now this discriminant certainly means like the norm of the discriminant down to my down by global field. Um, and now this might not be A number field anymore because k of course might be the function field of a curve over a finite field And maybe let me put in one more condition. Uh, even though I've already gotten myself down to a finite set here Uh, I might want to say Contain in the symmetric one on d letters Okay, so this is a slight abuse of notation because I don't mean to necessarily say that these fields are gaula I just mean to say a degree d extension, right? It's gaula closure will have some gaula group Which is a subdued of the symmetric group on d letters and I may wish to refine my count by that Okay, so now I think I've decorated the sentence enough to have this be the Most refined form of the question that I wanted to address so we can call this uh So we can call this number n sub g d is now absorbed into the g where the g is a permutation group on on d letters, um Say again, let's look in the function case. Let me draw a picture of it. I mean I could just say the discriminant is the Discriminant is like when we sort of said that You don't like that Well, let's say it's a divisor on the curve that has a norm Let me draw this picture. Okay, so here's We all agree that's a curve We all agree this is a more complicated curve Okay, so uh, so k it's going to be the function field of this curve l is going to be the function field of this curve and then in this cover there's some ramification way So there's a ramification divisor uh downstairs. I just mean uh, if you like the degree of that divisor, um, or maybe that times log q if we're really going to be uh, Besides we're getting an actual real number But so as in the number field case, it's measuring the ramification In this map and if this map happens to be completely unramified then that would be one or I guess I'm using a blog board It's like it's really easier. Um, am I not supposed to do this and I'm supposed to use these buttons or something? Where are they over here We're certainly not going to have an exact formula for this but we can ask about ask some topics So, uh, the conjecture that is completely supposed to tell the story for this originally due to my also I'll write down Roughly what this says so conjecture sort of a strong form in the weak form. So let me write the strong form. It's that um number fields um Is asymptotic to um some constant g times k And k and maybe I should say that it's the conjecture is not just that there's some constants making this true And the function has this form the conjecture specifies specific values of a and b um, and maybe I'll write down what the a is so um a equals um The one over of g. Okay, so what is the index of an elevator? I don't mean index I mean I don't mean the index like the group it generates in the group I mean index in the sense of every element of g is a permutation and the index Uh, sorry d is what I mean. So for instance What would it be what would a be if um If I had the symmetric one d letters like the Gallo group or the full symmetric which in some sense It's supposed to be the generic case, right? So e g equals sd Element of the symmetric group has the most orbit that's just a transposition right which has n which has d minus 1 orbit And so the index is d minus d minus 1 which is 1 so um So a oh, yeah, sorry if uh if g were let's say the alternating group Then you don't have a transposition right but you do have a three cycle that has n minus 2 orbit so you get one half so If g equals ad then a of g Is one half um and maybe as I've written this I might as well say uh b equals the number of such Uniquity classes That's minus 1 so actually and um in this case there's a unique conjugacy class of transposition So this b this log backer doesn't Arrive so in particular what this predicts is that if I count the number of sd extensions of let's say q found in the discriminant I've just went to less than x the number of those is actually asking the product to x like x to the 1 And uh, yeah, they don't right so this is why I didn't want to say okay So you have to it's really like up to this sort of cyclotomic action depending on how many roots of unity are in k So I already said something that wasn't quite true so e g um um This tells us that uh, the number of sessions with gala group sd Asymptotic to some constant need known to be true now for um Well for d equals 2 this is kind of if you've never thought about this question before then it's kind of a fun exercise To think about why this is true for quadratic extensions for quadratics of two You're essentially those are in bijection with square free integers and so you're just trying to convince yourself that There are about x square free integers less than x which is actually not a complete triviality, but it is true. Um For three, this is a theorem of doubt department how run for the 70s and then for four and four and five Uh This is of course the results of of many people including manjul and his collaborators were sort of invigorated thinking about this question, um Maybe I'll say one more thing to make this plausible, which is that I'll just say it in words Why should this be true? Why shouldn't it be about root x? extensions with gala group ad so Just to make it seem like this isn't totally crazy um Why should there be um root x extensions with gala group ad while having gala group the alternating group that's equivalent to having your discriminant be a square So you're really just saying if I think it's about x extensions with discriminant up to x And then I say oh, but I only want to consider those Whose discriminant is a perfect square there's square root of x such discriminant So if I believe there's an average of one field per discriminant, maybe this is somewhat reasonable um, but this we don't know even for A4 and actually fun group theory exercise where a4 actually you get two Controversy classes of three cycles and so it's uh supposed to be root x log x But that seems completely out of reach and a4 is a pretty small Solvable group um So um big point about the log factor. Let me just comment that mostly today. I want to talk about the weak form Which is just that asymptotically uh x to the a Is less than this count Is less than x to the a plus something specified epsilon. I'm just going to worry about the exponent of x here Whatever. Oh, yeah, I want x good Uh, you got heck there. Oh this constant. It's a bit subtle So for s for the symmetric group mongeal has a prediction, which I think is a pretty good one Which I believe in is true for d up through five that he makes a prediction for all d um I think in general you can imitate Imitate the heuristic that mongeal used to get that for other groups, but I don't think it's right for all because this is something that um that actually I thought about a bit and then um Signed and wrote a paper about this right for a five. So actually students. I'm going to sign Salcedo wrote a Paper about this in the a five case where there are kind of Expirational factors that occur to be honest. I think I still find them a little bit mysterious and don't quite know what they It's also the case. I mean, I'm going a bit far afield, but um D that it certainly depends on how you count in one of the themes as we go It's going to be that there's different ways in which you can count number fields and that and the behavior of that constant might uh Might depend on them in a very serious way. So for instance for d4 There's an old theorem of on Lee Cohen and Diaz and Diaz and Olivier um And they show the mileage conjecture is true in that case and the cost you're excited for the cost to be some like nice oiler product and like no way it's like nasty It's like some kind of like double deerslay nonsense. It's like a big sum of I mean, I'm sorry for the double deerslay people It's all still here. It's uh, um, no, I mean they compute it, but it's not it's not as beautiful as you might expect so um Is that an okay answer for that question? Okay. Um Kind of see since I'm even afraid to talk about the log factor of this talk I'm definitely like a hundred times as afraid to talk about the constant. So I'm mostly going to talk about the exponent um Okay, so often I get like full talks about that problem, but I'm not going to do that uh today. Let me quickly Uh transfer to another popular conjecture that we study, which is the following. Okay, so um So question two, let's say Over q it's now I'm going to immediately Change this decay, but you can still think it's q um effective variety v um And now of course as we know very well depending on the variety there might be very few Or there might be a lot for the moment I want to think about the case of this variety is very close to being rational where it's a so-called phono variety Whose canonical bundle uh is anti-ample. So with the kind of great trichotomy Uh of curves for instance, this would be like the genus zero case where there's infinitely many points. Um So let's Put it here But then of course I have a question Where the answer is probably going to be infinite. So in order to get a better question Let me say Of height. Okay, so what does height mean? Well, this is a projected variety So these points are sitting inside projected space. Let me take k equals q for a minute and say what does height mean Oh, I did it again you by this I just mean The maximum absolute value of these n plus one integers But now of course remembering that because These coordinates are only well defined at the scaling. I have to scale so that they become co-prime integer so having done that and writing the Rational point of its usual reduced form This is what I mean by height and then it's completely clear in a way that is maybe not quite as clear And they never feel counted case that I mean there are only finitely many points of protective space of height at most x I mean I'm just letting these coordinates range negative x to x And so in particular only finitely many of them are on this variety, whatever it may be And we can ask how many Okay Here too There is a popular conjecture that lots of people believe And it looks like this. This is the conjecture of a tier of the money here. I have to be a little bit careful about something Let's say Open sub variety Since somebody before complained that I didn't say not trivial for a group element Let me also say that I mean it not empty Open sub variety There's an open dense sub variety u of v such that If I count well, I guess I didn't give this a name. So let's call this n sub v Of that's an n sub u of x comma k. So I'm not quite counting all the points on v I'm only counting the points on this on this open sub variety, but that should be it should be okay to throw out Some close sub variety. I feel um tonic to some constant depending on v and k Time must have depending on v to some constant depending on constants are The tier of the money they tell you what they're supposed to be. It's not just to think about the form of the asymptotic It's a precise prediction, which again, I won't Uncommon. Well one thing just I mean I feel I I must Confess one thing these containers have in common is that neither of them is true They both have counter examples that have been found over the years um, but um But the the counter examples have to do with the delicacy of getting this power of log Right, so there are examples where we know that the power of log is not quite right Let's cut we're going to concentrate for today. I'm just sort of weak forms of these conjectures that I think are quite Likely to be okay. So similarly with the tier of money because they we're just going to worry about the exponent of x and not worry so much about what the uh what the log factor is um Why do these two conductors Have such similar forms and there's a few reasons it could be right? I mean like one is that There's just not that many functions in the world And maybe if you're studying the asymptotic for something like, you know, what's it going to be except the power of x and the power of log? That's a somewhat unsatisfying answer, which could be correct for all we know um It could also be That both of these problems we might discover are sort of somehow governed by Uh, some kind of outfunction that has a continuation and we're sort of going to use some taberian theorem And then read off this power of x and the power of log from maybe like, uh um the uh The location of some pole of some L function and it's and it's order. Um That's probably not really true though. I mean, I think that these it's true for very special examples You know varieties that are the homogeneous basis You can really I mean the cases where this conjecture has been proven are often Brides with very large group actions where you really can draw the connection Between the rational points and the variety You can make a sort of zeta function out of them that actually has good analytic properties and then you find You actually are, uh finding the right most pole and the uh And the order of the degree of the pole of some L function and that's what's giving you your power of x and power of log When in general, I think these problems are sort of too floppy To be described in that way So the story I want to tell today and this is maybe I'll say that everything I'm going to say What so everything I want to say today, uh Is joint with David Zurich-Brown And with Matt Satriano, let me write and David Um is to try to express a sense in which these two conjectures are really Uh species of one more general conjecture that can be interpolated between these two Um and the challenge you face when trying to do that is that they seem to be about different things, right? What is about counting? They're both about counting, but one is about counting field extensions Uh, and what is about counting solutions to some equation rational points and some variety? So I think somebody mentioned earlier that sort of one of the Aspects of working with barriers. He sort of trains you to do number theory like a topologist So What does that mean in this case? It means that it's if you are a topologist It's not at all strange to think of the set of g covers of something as maps of That space to a certain thing and what that certain thing is the classifying space of g whatever that may be So let's just say um So how can these how can these two problems? two problems uh be unified um And the answer is by means Of the observation a g cover of field k Is a point on bg Now but now what is bg? I mean what kind of thing is it? Um Well, it's not a scheme anymore. It's an algebraic staff So, um Let me sort of draw the topologist's picture. I'll just sort of take one board to do this so So what is bg? I mean, I think the picture from topology that we're all very familiar with um, or or can be made familiar with It one board is that um, so okay everything from here on it's like not algebraic gentlemen real It's just a good picture to draw to give you the idea if you haven't thought about classifying spaces before um For instance the classifying space um of z It's just a circle How do maps so how does this work? I mean I seem to be saying that if I have um some space The thing about a circle is the circle Certainly has a nice z cover namely it's covered by r so if I have anything Any map to the circle? Give me a z cover of s just by pullback the fact that this cover is actually contractible makes it be the case that actually Z covers of s Are just a bijection with homotopy classes of maps from s to s one that's what it means to be a classifying space It means I can take this project if for some reason I wanted to sort of study Z covers of s I could do so by studying S points on s one if you like maps from s to s one So what makes this go is that s one is a contractible space modular z so in sort of homotopy land we would just say It's a point mod g mod z and so for us, what is this bg gonna be? um, so now I'll sort of back to being upright geometry So for us bg is nothing but a point Modulo the action of the finite bg acting triviality so I mean there are and But of course this quotient It seems a little funny if you're not used to it, but I can uh and use along the way geometry We just say well the quotient by trivial action is just the thing itself, but as an algebraic stack That's certainly not true. It's kind of its own thing and indeed The sort of fundamental picture here is that if I have any scheme s and an apps to be g bg has an atoll g cover By just a point and I can pull this back to get a g cover of s in particular if s is spec k Then I can pull this then this is an atoll g cover of k Which is just a g extension of k so to sum up Say again the covers are not necessarily connected So yes, and that's I that actually Will become important probably not in what I'm going to say in this hour, but like it does become important Um, right, so I mean that g cover not in the sense of a connection g cover So you're right. It's me more general the way I should call it against g atoll algebra Is really like what it is on the algebraic side That's secretly what I mean. Yeah so so to sum up If k is my global field the k points of this stack Are equal to the set of Maybe I'll even write what you told me to say mad g eta algebras um Over k but really sort of you're not far off if you just think of this as like gawad g extensions Okay, and now this is great because now it seems like we've said aha perfect like indeed These two counting problems are species of the same problem in both cases I'm trying to just count rational points on something which maybe is not quite a variety But it's like the next best thing an algebraic stack um But now we encounter a more serious problem Which is that if we count we have to count in a certain order we need to get a finite number Which means we need to bound the height and now when we say this We have to understand what do we mean by counting points of bg of bounded height? And that's really the meat of what I want to explain today like how one can make sense of this because I think in the end It was very psychologically confusing, but I think we I think it now makes sense and I'll try to convince you guys of of the same so, um So what you might call? Batira of money for bg Oh, so what do I oh, I have to pull this one down and then the other one up. Is that right over here? We should do Batira money points on bg now we face the fundamental problem, but What does this mean and of course sort of somehow in the spirit of our discussions yesterday about You know, what are we allowed to say? Maybe really the question is What should this mean? I mean there is the definition. We have to make one Um, I mean I'll emphasize the right what we're so used to doing is to say like well Okay, we just like use like in protective space, but I mean Bg is not a closed sub scheme of protective space because if it were it would be a scheme And it's not so in this more general context um We don't have the ability to use our usual definition. Yeah, Matt. I mean that's a digital atom of our definition, right? so, um, so I secretly of course I have to pretend I don't know what I want to get But secretly I do know what I want to get But fortunately I get that so If you have to define it in a way that makes it look like you could have come up with this definition without being motivated by getting the discriminant Which hopefully I'll succeed in doing Okay, um So, um to describe this definition Let me actually start now by because it's a little bit easier to describe if I think about the case um Where k is just let's say the function field of a curve about in particular rational function field, okay, um So some variety I don't really start calling my varieties x. So sorry for using that and using it as the final height before But that's what I'm going to do. Um Well corresponds To a map from I mean this I can think of this as a field or I can think of it as a function field of a p1 And so a kt point on x So x remember this is in x of it corresponds to a map From p1 to x now we come to an important point before I told you that when I was talking about the height of a variety I had a projective variety One other way of saying that is that I chose an ample line bundle on my variety in this case It was uh, that's just the o of one from projective space. So in this abstract setting I had better give myself that's here And now in this context what the height is is like really simple namely also called x And then this line bundle whatever it is back to my p1 now I have a line bundle on p1 a lot because p1 has a degree and that's the height So it's a very nice geometric thing um picture I'm going to draw on this talk is going to be in this uh function field case, but if you really want it to be over q, uh You should feel free to kind of quietly Mumble the word metrized every time I say line bundle or right before and then Mumble the word are a Caleb like right before I'm going to say degree and then you'll be saying something that's Roughly correct. Um, but it's easier to sort of Say what's going on in this geometric case Okay, so this seems like a pretty simple definition that we should just try to extend to the case where instead of x I have some stock eaves. So let's try to do it and we'll just see everything go to hell immediately. Okay So so let's definition or let's try To a stack and let's say I mean Let's say this is uh Let's say this is a proper smooth the lemum for stack. Um, we can do far for those things I don't want to say so I want to say the lemum for it. Okay Okay, be a point on this stack Over the field k of t so I'm going to be a little bit more careful on what I right now So in other words, I have a map from spec k of t um To this stack this thing sad is that already we're kind of dead at step zero Because what I did what's so used to doing it that I didn't even comment that I was doing it is um I sort of wrote spec k of t and was like, okay So you have a rational point over the function fields. So you have a map from p1 to x And what I was using there is the value of the criterion of properties That if I have a proper scheme and I have a map from the function field of a curve to that scheme Then it extends to a map of the whole curve And the sad fact of life is the value of the criterion of properties in that form does not hold for stacks And you can sort of see the to know that can't happen. You might really like it I mean, let's say for instance That this x or b of z mod 2z in some sense like all of the A lot of what's going on here. You can tell just for this very simple example. They're classifying back a group of order two Okay, so if I have a map from spec of k of d to that Then that's a quadratic extension, right? I mean, this this would be a z mod 2g cover So this corresponds to a quadratic extension of k of t Well, if I can extend that to a map from p1, let's sort of pretend for a moment. I can do this This would be an eight tiles z mod 2z cover of p1 But p1 doesn't have these all covers or rather To that point it only has the disconnected one So it certainly doesn't have the one who's generic point is the quadratic extension you want The quadratic extension you want is somehow the function field of some hyperlip to curve which corresponds to A double cover of u1 that is not a tile that is branched. So You might think at this point already like what the hell am I going to do? Okay, so it turns out And this is one of the things That's in the paper that You can't quite extend this map to a map from p1 to this that ekes, but What you can't do is extend it to a map from Something i'm going to call script c, which we call in the paper the tuning stack Where c is A stacky curve so By that I just mean it's an algebraic stack which is generically the same as p1 So it's like a one-dimensional nice smooth thing, but But it has some points on it That are stacking so you used to draw pictures like this very or you would talk about these magnified curves The picture you should imagine is in a case like this. Let's say I was Studying the quadratic extension of k of t Which corresponded to some I don't know some paper looked a curve of genus g So there's like supposed to be two g plus two branch points But what you're going to send this to is a map from some kind of curve, which It looks a lot like p1 do they make an eye But actually there's two g plus two points where this little thing i'm drawing here is actually only half of a point And that's the way these stacky curves look. I guess I shouldn't have drawn an odd number of them, huh? Maybe I should Put on one more okay to make that realistic and then you know this funny curve with a bunch of half points This is indeed covered by a genus two hyperliptic curve And that is an atoll cover because what looked like a ramification point is actually an atoll cover of a half Point that's kind of like the yoga of stacky curves okay, so this seems promising right we have We've we've gotten over the first barrier, which is that we actually have A curve mapping to our stack which we can pull back to okay. Oh, and of course we also better have a line mungle We're going to compute height. We've got to have a line mungle on our thing so Ready we've got our line mungle this i'm still going to call x. Let me call this extension x bar I'm going to give it a new name now that there's actually some content to it and so it seems like Completely obvious what you should do Well, namely at this point you may say so idea to be the degree On this stack c Of the pullback, you just got to believe me that you know on a stack curve There's still a notion of a divisor and there's still a notion of degree of a divisor Everything is good So for instance, we could do that for this case here and see what we get and Some of this definition is so simple that this clearly has all the properties You want height to have and you can sort of Try it in this example and everything about this definition is great except that it's identically zero So it has all the conditions that you want, but it's just not the right thing so So problem with this definition When e g equals b of z mod 2z always the test case you want to have in mind This does not give the discriminant to the extension it gives zero What is that it's sort of for a very simple reason um Just let me casually write down that line bundle on that stack because If you either know what a line bundle on the stack is there if you were like intimidated away from asking like What the hell is a line bundle on the stack? Okay, I'll tell you at least in the case of classifying stack of a finite group So a line bundle on b g is a character of g A one-dimensional representation Okay, so I didn't tell you which line bundle I want to use Let's pause it that I don't want to use the trivial line bundle And if g is z mod 2z, there's only one other One dimensional representation right the non-trivial one so For b Of z mod 2z We may take l to b But then here's the issue The issue is that that representation tensioned oneself is the trivial representation if you like the Picard group of bg is just The ambillianization of g which is a torsion group So but l plus l Equal zero so what that means is that whatever this number is um, if I double it I get zero. Okay, so here we face a difficult choice You could declare that you're going to try to make some kind of definition of height with torsion coefficients. I refuse to do that That just feels wrong height should be a real number, but then You have this problem that you sort of that this is Apparently is zero so the question is what to do about this and now I think I can Basically tell you what the actual Definition is of height with respect to this line bundle. Um So yeah, so So twice this degree is is the degree of the pullback of 2l Which is zero and that's uh, and that's a good so instead I might have first seen a little bit at Hawkin that I'm going to try to convince you that it's right so instead Define I got to put one more so here I have this stack to curve into the generic point, but also A stack to curve has what's called the course moduli map. So everybody who studies with the curves knows this very well Right, there's sort of x of one the moduli stack of elliptic curves But there's also the j line, which is an authentic p1 that that moduli space Maps to and we all the time try to describe elliptic curve by its j invariant We know there's like all kinds of technical nastiness that comes from the fact that that's not really the moduli space It's the course moduli space and we just kind of grid our tubes and deal with it. That's what we're going to do now Okay, ready? So this thing maps to some p1 by a A course moduli moduli map pi and the definition of height Is that we take l We pull it back by x bar star to get a line bundle on this stacky curve c and then we push forward To get a line bundle on an authentic p1 and then we take that And of course, I should say that how I started with the function field of some other curve that other curve would be Here so this definition is not restricted to p1. I'm just like the easiest case to describe it. So Let me say a few things about So let's let's draw a picture of it. So here's the it's it's David taught me how to do this very fine Yeah So what you're thinking I think I'm going to try to go into mark's mind You're going to say like wait a minute if I have like some degree r map like the push forward with line Bundle is going to turn to like a right guard vector bundle. This map is degree one I mean, it's still a degree one. It's a degree one map. That's not a nice morphism If you like it's a birational map because it's it's a nice it's a nice morphism the generic point So it does in fact take line bundles to line bundles the kind of thing happens, right? So if I um So if I have a Line bundle on here, let's say it's like I can think of it as represented by a divisor and that divisor is going to have It's going to look just like we're used to at the non-stacky points It's going to have some pluses and minuses of points But then at these half points, it's allowed to have a one-half It's allowed to be a half in the jerk if it wants to and Roughly speaking what happens is um pushing forward a divisor into its course modular space p one sends the sum of ni pi to the sum of Floor and I actually does something to the degree exactly So what is it? It's exactly it does exactly the right thing in the here You want to get I mean you want to get some number that keeps track of how much verification there is Um this pullback line model actually you get something that has degree zero here When you push forward you end up losing one half for each of these two g plus two points So you lose a g plus one So you basically are keeping track of that No, it's not and that is in some sense our difficulties with this We're in some sense very psychological because exactly what Evan has pointed out It's really disgusting, right? I mean we are used to the very height machine And we're used to almost as like a physical law of life that the height with respect to two The height with respect to l1 plus l2 is the height with respect to l1 plus the height with respect to l2 um It's very hard to let go of that But if you don't you're completely screwed because the argument I just gave shows you that there is no non zero height For a torsion line my mother respects that so either you just have to define things identically zero Or you have to give up additivity. We made the latter choice I think it's the right choice, okay So so absolutely push forward is not like a linear thing to do. It's not a homomorphism um so So warning warning is having quite the doubt. Um It is no longer the case what we're so used to That the height with respect to l1 plus the height with respect to l2 is the same thing as the height with respect To l1 plus l2, but I think that's unavoidable if you want to get non trivial theory Well, I mean I think that I mean I think if I have a torsion thing I mean that can make problems go away, but I think it powers up But as you say that it makes the problems go away by setting everything equal to zero Okay, so I'll say this what I'm going to say in a second is that So first of all one issue is that we don't want to do this just for line bundles We want to do it for vector bundles and there it's a little harder to see what to do I mean there's sort of tricks that help you for line bundles like taking a lot of high powers, but it's not so clear So maybe I'll say that and then see if I mean so Of a point with a height and in your sense, just give a give a numerical example Okay, so, um Yes, let's uh Okay, let's see if I let um, I'm buying some time Okay So let's let's try to do this example that I just wrote down over there. Okay, let's try to take um Let's try to take uh So, um The quadratic extension k of t a joint root. Um, I don't know x to the six minus one Um, right. So this is a good quadratic extension of k of t corresponding to a genus two curve Um, and now fortunately I already draw on the picture. It's over there, but I have to tell you, uh, what happened Uh With this line or it's maybe I'll maybe I'll draw it again. So it's right by here. So Here your c is going to be One two three four five six half points to b mod of z mod 2z and over this I'm going to consider the line model which corresponds to the non-trivial representation. Um, and when I pull back, what are we going to get? Oh, um Whatever we want this variable to be what did I call it before I thought I called them t Oh, yeah, this they're both t. Yeah, okay so When I pull this line model back, um, I'm going to get something I know has degree zero but you can also check that it's sort of it's uh The plus that it has in each of these branch points It actually is not integer. It's a half integer I mean when you look at linearly equivalent dividers on a stack, it's just like P1 usually is you can sort of move integral amounts of point from one place to another But if you're one half you can maybe make a three half by taking one from someplace else or negative one half But you can't ever put it into z So in fact what this divides what's going to look like is it's going to be linearly equivalent to A half each of these points and then you know somewhere over here A minus three or maybe a three different point doesn't matter which One and then when we push forward um to P1 You floor all these numbers And you get yes, thank you for making me do this very these house all become zero and you get negative three Okay, so this is why after we'd written the entire paper. We realized we had to go back and be like Oh, yeah, a better thing would be to say it's like negative of what we said it was you get positive numbers Actually, if you define it like this, it's you get negative numbers rather than positive as you might Be cosmetically more appealing No, because you sort of dualized the lawnmungle do this and then take negative of the degree that you get Yes, we also went we went through this exact same Series of changing the signs until like David would answer in all cases. So yes, that's what you have to All right, I'm going to write it right. It's uh, where is it? It's it's negative There we go. That's the actual definition in the case of a scheme. It agrees those negative signs Cancel each other out Very near the end of my time. So in two minutes, I'll give my pitch for vector bundles, which is that We're so used to sort of saying height is a property of a lawnmungle on a stack But if I had not done these zima to zi and instead Done, I will say the symmetric group on d letters Well, what lawnmungle is an interesting one with which To compute the height and the problem is there's not that many interesting lawnmungles on the symmetric group on d letters Because the symmetric by d letters does not have a very interesting one dimensional representation theory All it has is the sign character and the sign character is certainly not telling you everything you need to know about Uh About an sd extension. It's only telling you it's quadratic subfield. So in fact the So to get this right to compute with respect to On sd in which case the height of x is defined to be um Negative is defining the exact same way. It's negative degree of pi lower star x bar upper star But now this thing is like some rank R vector bundle on the base Well, I have told you I'm saying this is the definition of height with respect to a vector bundle and now I'll tell you which representation I want So all this comment that we could have done this for a scheme But it would have been a silly thing to do because this is just really the degree of the top wedge product of v And I could have just done that to v in the first place and used out But because of the non linear idea of pi star I can't do that here. I actually get a different answer than if I'd start it then the wedge product can't cross this So this is like actually now a new thing um Okay, so to conclude I'll just say that d or some subgroup of sd And v is B of a finite group is just a representation of that group And you just take the permutation wrap then You can compute that the height with respect to v of x is the discriminant Uh the corresponding puts out there's lots of representations in this metric group many more than What if just d by 2z and actually one thing that's cool is there it's sort of been Understood by member fields counting people that there are lots of different invariants that are likely discriminant With respect to which you can count number fields and from this point of view This is just a matter of choosing different vector bundles on b of sd And I think in some sense this is like a guide to sort of which Counts are interesting ones. So I'm at the end of my time. So I'll just say um This is the definition when you go through different situations in which you want to study Rational points on a stack like for instance the modulite stack of lifted curves or something like that It seems that in every case this definition with an appropriate vector bundle In almost all these definitions people already have some sense of what they mean by the size of a rational point On that stack And it seems to me that by large we sort of recover them all like somehow like people are already right about what they met By the size or complexity of a point on these stacks, but this uh Gives you sort of a way of thinking of these all in a relatively clean Uniform definition, which I think is the right one. Okay, so I'll stop there. Thank you guys