 to the organizers for the introduction and the invitation, and thank you to all of you for coming. Yeah, I'm really excited to give this talk in person. One of the things that I have missed most over giving talks on Zoom is interaction with the audience. So I really, really hope that everyone will feel very free to ask whatever question pops into their mind. I would much rather get a lot of questions and talk about the things that are interesting you, than get to any particular point in my slides. So please, please, please just ask questions anytime. All right, so the topic of my lecture series is Rational Points on Varieties and the Broward-Manninence Obstruction. But this is taking place in the context of the broader theme of the whole PCMI program, which is number theory informed by computation. So for me, that means that in this lecture series, I particularly want to focus on things like, what are the computational questions in this area? How does the computation inform an aid theory? And how does theory inform an aid computation? And I only have three lectures and there's a lot of interesting work going on in this area. So even with this focus, just out of time and it's me giving the talk, this is my perspective on these questions. Okay, so what I hope to do is give a broad overview of things that you can ask and current developments that are happening and then give you lots of pointers that you can go and explore further for whatever topic interests you. Okay, so the starting point of this area is this very simple question. So if we have some variety, some system of polynomial equations defined over the rational numbers, how do we determine if we have, like what the set of rational points are? So you saw related questions in some of the talks earlier. David Harvey's talk and then Drew Sutherland's lectures at the beginning where we were looking at curves over finite fields. You could also, if you're looking for elliptic curves over Q with a particular Galois representation, you're normally end up looking at the equation for some modular curve that you wanna study the rational points on. Maybe you're presented with a very classical problem like the rational box problem which asks if you can find a rational box where the distance between any two pairs of vertices is a rational number. Write that equation down. You get a set of polynomial equations and you wanna know what are the rational points on this. Okay, so a naive approach when you're handed this problem is you could just start looking. You just start plugging in some numbers. You set a bunch of the coordinates to be zero. Then you set a bunch of them to be one. You can make this systematic by searching in some box of bounded height. So you search and search and search. Maybe you find a bunch of points. Maybe you only find five points. Maybe you find no points. But you search and search. And then whenever you get tired of searching, you're led to this problem of how do I know if I searched enough? How do I ensure that I have found all the points? Sometimes you'll have infinitely many and then you'll know, okay, I definitely didn't find all the points yet. But if you know that there's only finitely many, you could hope that you found the points. So you have to know if you found them all. When do you stop? And a very extreme example of this that I wanna focus on is say you're doing this search and you find no points. You search up to height five and then 10 and 20 and a hundred and a thousand and a million and you're still finding no points. At what point do you give up and think, okay, there are no points. How should I show that there are no points? You might have start becoming very pessimistic that you're not going to find any, but we're mathematicians at the end of the day. If you wanna stop and say that there's no points, you have to prove that there's no points. And searching for points will never help you prove that there are no points. You need a completely different method. So I am interested in rational points generally, but in this talk we're focused mainly on this question. Okay, if you suspect that there are no points on your variety, how do you prove that there are no fires, sort of a fundamentally different approach than if you think there are points and you're going to go search for them? Okay, so a general method framework that happens, the set of rational points is often very hard to get a handle on. So what we do is we embed the set of rational points into some other set that is hopefully more understandable and imputable. Okay, and some of you may have looked at these questions before. Some of the first sets that you might take are the set of local points. So you first look if there's real points, right? If my equation is of the form x squared plus y squared plus one equals zero, I know that all squares are non-negative and so I can just give up and go home already. You can do the same thing by looking for piatic fields and you can package all of those local conditions together into the set of idyllic points. First, natural sets S that you might take. Okay, and just fundamental idea of how we're going to approach this is once you have this embedding and if you can compute S and you can show that S is empty, well then you have, this is your proof. You have no rational points. Okay, so the idea is to find some set S that is more understandable or imputable to use to witness that there's real points. Sometimes the idyllic points for some particular classes of varieties that's enough to tell us everything. So a class of varieties is said to satisfy the local to global principle if for every such variety the existence of idyllic points implies the existence of idyllic points. So quadric satisfy that with this. Also any severity of our variety, so any variety that geometrically is isomorphic to projective space also satisfies the local to global principle. And let me just say, so this is a property that you wanna know for a class of varieties. You wanna know it before you start looking for rational points. So a variety that has a rational point, just like a single one by itself automatically satisfies the local to global just based on maybe some geometric information or the degrees of the equations or the number of variables, the type of thing we're looking. Quadrics and severity of our varieties those are both very, very close to projective space which is geometrically very simple. Unfortunately when we move into more complicated varieties curves of higher genus, higher dimensional varieties. How's it tell me? I said just about this whole thing. So yeah, once we move into more complicated varieties the local to global principle typically fails. So the set of idyllic points is no longer enough to generally tell us when we have rational points or not we need some refined obstruction set. Some set S that better approximates the set of rational points that we can use to detect when there's no rational points. Okay, and so that is really the main focus this particular obstruction set that we'll study that can witness the lack of rational points. So this is called the Brouwermann obstruction and this will maybe come up a little bit later. You can talk about obstruction sets for other properties of rational points. This just not only existence but as I said at the beginning I'm gonna focus on this extreme case can we try to prove that there are no rational points. So I'm focused on the Brouwermann obstruction to the existence of rational points. But that's a mouthful so it will usually just be abbreviated. Okay. So our goal is to define this intermediate obstruction set that tells us potentially more than just the information that's contained in local solubility and this intermediate set is called the Brouwermann inset. And I started this over Q but for the rest of the talk I'm gonna switch to working over some number field K and define it in this generality. Much of what I say also works for global fields of positive characteristic but I just don't wanna keep adding characteristic assumptions at the various places well needed so I'm gonna stick to the case of number fields. But much of what I say holds in greater generality. All right and as I said so the goal is to define this set it's a set that the K rational points embed into and so when this set is empty then we say that there's a Brouwermann obstruction to the set of rational points. So this set is what we call an obstruction set and this is the thing that I wanna study. Any questions up to here? Yes, the Adelic points combine together all of the local points. So you can define this in terms of the Adele ring which comes up, I don't know I guess maybe you see it in a class field theory book but for this talk because I'm always going to be working with smooth projective varieties the set of Adele points, I think I can do this. Yeah, you can just think of this this is the product over all the places including the ones dividing infinity of the set of points over the completions. So it's packaging together all of the local information. Great question, see it'd be really bad if you wouldn't ask that question. It is in general a restricted product but because I'm working with projective varieties there's not the restriction there. So in general you have to, yeah it's a different set but so this is using is because X is projective. Okay, X is always going to be smooth and projective in my talk so I'll just include that there. Great, more questions, yes. Yeah, so when I'm looking for global fields of positive characteristic so like FQ of T much of what I say still goes through but there's some things that I might say that might not hold if like the order of a particular element is not primed to the characteristic or so like some things that I write down you might just have to be careful when you're looking at the characteristic and it's a thing that I'm particularly bad at asking answering questions about during talks. I feel like I always forget to do the various characteristic assumptions so but yeah and I give a lot of references in the notes so you can always look up the more general statements there. And let me just say so I said this is a question about global fields. So for finite fields I mean you could ask the same question and we saw lots of yeah as I said that that was the topic of David Harvey's talk today and Drew Sutherland's lectures so that's also a very interesting question but there's a more straightforward just like effectivity result in terms of an algorithm because you can enumerate all the points in projective space over a finite field so you can buy enumeration, you can check, you can prove that there's no FQ points and you can't do that here so the question is the point of view I'm taking is most of interest when you have already an infinite field and I think I forgot to repeat all of those questions so I'm gonna try to do better next time. Any more? Okay, keep them coming. Okay so I'm gonna define, well work up to the definition of what this Browerman said is but it can be kind of a mouthful to get all at once so I'm gonna work through it in a particular case and just assume some things to get us started. Okay so right now I'm just gonna assume that I have this magic machine, black box, some Oracle in the sky, that given any F rational point on X it produces for me a conic and this probably, if you've never seen this before it probably seems like a very random machine that I want to use. That's totally okay but I'm just gonna pretend that I have this thing and let's see what we can get out of it. Okay so I just have some way like a conic making factory that takes in varieties and just like spits out conics that I wanna work with and if you want you can think of this you can think of this as really giving like just two coordinates if you want because you can, since I'm in characteristic zero I can always diagonalize to my conic and I can scale to make one of the coefficients one so I can always write in this form so my conic making machine I could really just think out of as a machine that's just spitting out these coefficients AX and BX for my conic. Okay so for the set of K rational points which I don't know how to compute I'm trying to understand what that is but if I had it I could plug it into this machine and it would spit out for me a bunch of conics and I can do the same thing for the adelic points so that's really giving me a collection a collection of conics one for every place and then I have a whole set of those for every adelic point that I have there. Okay and I have my inclusion that we already saw that the K points embed into the KB points and if you prefer just you can think of K as just Q everywhere you'll retain a lot of the and all the examples in the exercises are for Q so you can work with that generality. Okay and this is my machine that I've dreamed up so I can assume that it has some nice properties so that I have some way of like some nice commuting properties. So if I have a K point and I view it as an adelic point the box gives me the same set of conics. Okay so now what we're trying to do remember what we want is some intermediate set in here. We want to find some set that sits in between these two things so we want to know well what is something special about the K rational points? What do they have to satisfy that potentially the adelic points don't always have to satisfy? What can I get from them? And you'll see the top of the slide tells us that I'm going to use quadratic reciprocity. So let's try to understand what quadratic reciprocity tells us about these conics and this let me just say this is probably not the way you were taught quadratic reciprocity when you first learned it but I'll explain why I want to present it this way. Okay so one thing that is true and is in one of the exercises to prove over Q that for any conic over, let me just do this over. So if I have a conic over Q so I could look at the number of places where this conic fails to have a local point and it's not too hard to show that this set that this is always and so in the problem session there's an exercise that outlines how to do this and in fact this is something that's true in great generality if you have any smooth variety over any variety over a positive dimensional variety over a number of fields the set of places where you fail to have local points is always finite so that's always true but something even more is true. Quadratic reciprocity tells you that not only is this set finite, it's always even. Okay this is not the way you learned quadratic reciprocity I am like almost 100% positive but I like phrasing it this way one because it's gonna help me define the Brouwer-Mannin set but also because it makes clear like absolutely unreasonable it is that quadratic reciprocity holds like quadratic reciprocity it just has no right to be true and it's true. So let me explain to you why I think it has no right to be true. So say I have this conic which I can think of as just giving this A and B so I don't know what they are I hand them to you. You tell me all of the finite primes just the prime numbers where this conic fails to have a solution. From that, that list of finite numbers where it fails to have a solution I can tell you about the signs of A and B. Now when you work mod P they know absolutely nothing about signs like there's no notion of positive and negative in FP but somehow quadratic reciprocity tells you if you work with all of the primes you look at the solution on all the primes and sufficiently high prime powers it will be able to tell you if A and B are both negative. That shouldn't be true it just shouldn't be true but it is true and that's yeah, so we should use that. It's like this amazing surprising fact that we should be able to use to like to our advantage. Okay, so that's what we're gonna do. So what we know is that if we have a conic over the rational numbers or over our number field the number of places where it fails to have a point is even. So I can look at just now I can take an arbitrary collection of conics one over each completion they may not all come from the same thing but I can just say, okay let me look at the set of collections of conics that the set of places where they fail to have local points is even. That is a subset that I can take. And so the idea is that we wanna use this our intermediate obstruction set should be the set of a deli points that gives us this thing in the middle. Okay, this set S which is going to be our Broward-Mannen set maybe slightly generalized. I mean, I think this is not really an intuitive thing like maybe when you get used to thinking about it enough like I've thought about this for many, many years now it's like we're good friends but it feels can feel contrived and it can feel like weird that it works. And I think that's because it's weird that quadratic reciprocity is true. This is not an intuitive thing that like okay we're used to working with it. I feel like it's like a number theory test that you have to love quadratic reciprocity. So it's something that we're friends with but it is surprising. And I think that that surprisingness of this fact makes it hard sometimes to wrap your head around this Broward-Mannen set. And it's not for lack of trying it's just because it's a surprising condition that is imposed, yes. Yeah, so these CXV, they're just a collection of conics over any local field. So the number of places, it's a different conic. I can have a different conic over like Q3 that I can have, oh sorry, the question was what is the difference between this set on the left and the red set in the middle? And the difference is the one on, yeah, you're left. It is, that's just a collection of conics over each completion that may not have anything to do with each other. So I can pick whatever conic I want over Q2 and then a totally different conic over Q3 and a totally different conic over Q5 and so it's possible for them to fail to have local points at an odd number of places or maybe they fail to have local points at an infinite number of places depending on what this black box is, right? I haven't described this squiggly arrow, this is just my imagined black box machine which is doing something. So right now at this stage that machine could do anything, could give you whatever thing you want. Yes, so the question was do I have to ask for each conic in this collection, this black one, to be defined over Q and the answer is no. These are conics defined over the completions. A little higher, okay, great. The question was can I hold the microphone a little higher? Yes, oh yeah, so sorry, when I'm saying that I'm leveraging quite a lot of reciprocity that's really happening over, I guess really that's equivalent to over Q, so for K equals Q and then in three or four slides I will say how I'm thinking of quadratic reciprocity for a number of fields, which is basically just this, but yes, okay, more questions. I love it, so many, keep them coming. Okay, so now I have showed you how to give this intermediate obstruction set assuming I can build this black box, right? This whole thing rested on this assumption that I have this magic machine that takes in points and spits out conics and it's in some nice way so this diagram commutes so that the way of making conics is compatible with the inclusion of the K rational points into the KV points. So now I have to justify that part or show you how we can make a machine. Okay, so conics, as I've said, if you're in characteristic not two, which we are, then you can just think of them as two numbers and you're just looking at an equation like this. So I think in one of the previous lectures you saw something very similar looking to this. Oh, great, yes, yes, that's not clear yet. So right now, yes, thank you. The question was, I said this set S, but how do I get it? Cause I just magically drew it. So really what happens is that I'm saying now I also want to like make my black box machine even better so not only can I run it forwards to kick out these conics, I can reverse it and stick the conics back in and spit out the idyllic points that gave it to me. Yeah, so that is something that I have to justify. Okay, yes, so I think in some of the lectures you saw some objects, maybe they were given by A and B, I forgot to ask this part, but specified by two numbers, Q algebra, whose center is exactly the field and that has no non-trivial two-sided ideals. And this conic is an example of a severity Brouwer variety. I mentioned that earlier in the slide, these are varieties that are geometrically isomorphic to projective space. And it turns out in general, there is a correspondence between severity Brouwer varieties and central simple algebras. And so there is something that sort of classifies both of them. So you can take the set of central simple algebras up to what's called the Brouwer equivalents or you can take the set of severity Brouwer varieties up to isomorphism. And then those two things happen to be the same thing and they're called the Brouwer group of my field. So instead of thinking of this machine as just giving me a conic, I'm going to think of it as giving me an element in the Brouwer group, which is just the isomorphism class of this conic or you could also think of it as giving this quaternion algebra mod Brouwer equivalents. This is just a gadget. The Brouwer group is really awesome. I am a big fan of it. I probably couldn't even give three lectures just on the Brouwer group. So this is basically all I'm gonna say about it. But we're gonna focus on the two torsion and the two torsion in the Brouwer groups, aquaternion algebras and conics. And so that's the correspondence we're gonna work through. So it's, so the Brouwer group includes generalizations of conics, but it also includes conics. So this machine that I can think of, okay, I have this thing that splits out a conic and now I'm gonna put another machine next to it that takes the conic and spritz out its Brouwer class. Now I'm gonna think of like the two things together. Okay, and then how do I replace this condition from quadratic reciprocity that the number of places that the conic fails to have local points is even? Well that comes from what's called the fundamental exact sequence of global class field theory, which hopefully sounds very important. It is a very important fact, it encapsulates a lot. It's the bottom row that I have here and it just relates the Brouwer group over a number field that embeds into the direct sum of the Brouwer group with the completions and we understand exactly how it's cut out. It's that the sum of these invariance is, it goes to zero. Okay, and if you look at this, the two torsion of this sequence over Q is exactly quadratic reciprocity. It is equivalent to quadratic reciprocity. So if you want, you can just, if this is the first time you're seeing this exact sequence, you can just think of this, this is like quadratic reciprocity bonus, like quadratic reciprocity on steroids. Okay, and so I guess this is how I think of quadratic reciprocity over number fields. I just think of the two torsion of this exact sequence, which is maybe cheating, but it's my talk. So that's how I think of it. So now this condition that's in the middle, that the set of places is even, I just change that to the sum of these invariance is zero. Okay, because the two torsion, the two torsion of Q mod Z is one half Z. And so if I add up, I get zero. I get an integer. So that's zero and Q mod Z. Okay, so this is a fancier looking, more generalizable way of saying what we had before. Okay, so now I've, but I, we still have this machine now, this, these two machines that fit together. And so I have to tell you how we, how we piece those together. Yes. The in, this is called the invariant map. So this is the fundamental exact sequence of global class field theory. There are also like theorems in local class field theory about the, that tell you about the Brouwer group of KV. So this is written up more in the notes, but if V is a finite place, then the Brouwer group of KV is isomorphic to Q mod Z. So this, this MV is an isomorphism for all finite V. And then for Archimedean V, it's more boring. The Brouwer group of the reals is just one half Z mod Z, otherwise known as Z mod two, and the Brouwer group of C is trivial. So then those maps are less interesting. But so it's a, for finite V in sub V is an isomorphism, but it's a particular choice of isomorphism that fits together in this global way. So for us, we just need to know that it fits into this exact sequence. So it carves out what Brouwer K is. And then in the exercises, you'll work through what this means for the two torsion. So this is sort of a translation of the things that I said about comics, just into things, into the Brouwer group terminology instead. Yes? Yes, that's coming next. I still haven't explained this. Red assume is still, I have not justified it at all. Right now it's like in my mind, I dreamed up this thing. Actually, so I didn't dream up this thing. This is coming. Manin made this obstruction. This is not my thing. But in the way I'm presenting it right now, this way of getting these Brouwer classes is not justified at all. Purely in my imagination. We have to bring it. We have to figure out how to do that part. Oh my God, I forgot again. Okay. So the question was, yeah, what is this black box? That wasn't the exact question. But more questions? Yes, and please just shout out if I miss. The lights are sort of bright, so sometimes I miss the hands. Okay, so the way we do this is it turns out this Brouwer group that I like hand waved a bow and told you to refer to my notes is quite a general construction. And you can define it for basically any algebra, geometric object that you want. You can take the Brouwer group of. So I talked about Brouwer groups of fields previously, but you can also define the notion of a Brouwer group of a variety, Brouwer group of a scheme. And so, oh, so if we have a Brouwer group of a scheme and the definition is comological, so that means that it's completely functorial. So if I have an element in the Brouwer group of X, which is just some thing, and we'll talk about what those things are, and then I have an F rational point, which I can think of as a map. The functoriality of the Brouwer group just says, so my machine is functoriality. That's the way to think about it. My black box is I have a comological definition of the Brouwer group of X, and that is functorial. So it produces for me this black box, which takes an F rational point and gives me an element of the Brouwer group of F. Okay, so now instead of doing these squiggly black box in my imagination arrows, I'm gonna use the functoriality of the Brouwer group. So this squiggly arrow correspondence, now I get my little X, and I just map it to X upper star of alpha. And I mean, really you can just think of this as like radical notation for a black box, if you want. This is just like, it is something that I'm allowed to do, and I have it, and it gives me this property. Okay, and then I do the same thing on the Adelix point side, and it turns out that the map actually lands into the direct sum instead of the direct product. That's not a big deal, but it's a thing that you. But now you can sort of pattern match. Why does this work out so nicely? Well, we have Brouwer group of K here, and we have the direct sum of the Brouwer group of KVs there. So I should rearrange this picture and put my exact sequence instead there. And now from this map you see, well, if I have a K rational point, and I map it all the way to Q mod Z, instead of going across the top and diagonal, I can go down and over, and because the bottom is exact, that means that the K rational points have to go to zero. So there is this intermediate set, which is exactly the inverse image of the points that go to zero. That sits here. It's gonna let people look at this diagram. So this defines everything, but it's sort of a lot to take in. So that's why I went through the examples starting with the conic and going through, but I'll just leave this and take questions people have. Yes. Sorry, does the image of the rational points under? Yes, it's very dependent on alpha. Yes, and that is gonna be a big theme coming up later. That is a great question. So the image of the rational points, this map, phi sub alpha, it really depends on alpha. This black box thing that I put, I mean, that's like sort of a joke that the pullback is a black box, but it's also not so much a joke because it's hard to predict what thing you're going to get. You really need to know a lot about the element alpha and what you're putting in and the points you're putting in and how it works. Other questions? Yes? Yes, that is something you have to prove. So, okay, the short answer is because alpha extends to an element in the Broward group of a model of X over some ring of S integers for some S, and so then it goes there. I understood that answer was not supposed to make sense to everybody, just to be clear. Yes, yes, yes, yes, phi alpha depends on alpha and the inverse image depends on alpha. Yes, so that's the next thing. So this set, X, A, K of alpha, so often this is thought of as a pairing between the Broward group and the set of Adelic points going to Q mod Z. So we usually refer to this set as the set of Adelic points orthogonal to alpha, like using linear algebra terminology, and then as many people are pointing out, we can do this for all alpha. So actually the Broward-Mannin set is all of these. You take the intersection over all the alpha and the K rational points still live inside of there. And so the Broward set is the set of Adelic points orthogonal to the whole Broward group. Okay, all right, let me try to think what I wanna aim to get to. Any more questions on this definition? Yes, the Broward group of the Adel ring is the direct sum of the Broward group of Kv. That's another way that you can prove that it maps instead of using what I was doing, which I said, okay, since it's projective, the Adelic points of X are the direct product of X of Kv. And so then you map to the direct product and then you prove that it lands into the direct sum. You can instead do it functorially if you want and just say, okay, the set of Adelic points is a map from spec of the Adel ring and then we pull back under that and Broward group of the Adel ring is the direct sum of the Broward group of Kv's. That is another approach. More questions? Yes, that's right. Yeah, the top line is a map of sets because I'm just looking at the K points of X or the Adelic points on X, so I'm forgetting all of the other geometric structure on it. So phi alpha is just a map of sets, which is why I pedantically wrote phi alpha inverse instead of kernel of phi alpha because it's not technically, well it's not a map of group. When I was a grad student I always wrote just the kernel of phi alpha. I just pointed out that I should really write phi alpha inverse more correctly. But everything lands, if you work with two torsion elements, so if you start with a two torsion Broward X, which would give you this map where you go to a conic, then everything lands in the two torsion. But I could, and in all of the examples, I'm just gonna pretend the only elements in the Broward group are two torsion. But there are higher order elements of the Broward group of X and so then that you land into higher, like one over nz monz for some larger n. These are all really great questions. Maybe I will, okay let's just remind us where we started. What did we wanna define this for? So we have this Broward set and we've proved that it's an intermediate set. I mean by definition it's a subset of the Adelai points and by global class field theory it contains all the k rational points. And our idea when we first started, the motivation for this was to find a set that we can use to witness that X has no q rational points. So the set we've got now that it contains X of n. q or X of k depending on what field you're working on over. But we want to actually be able to understand it or compute it. So that is like the million dollar question. Can we actually do anything with this definition? Enough to be able to witness it. And so I think just given the time, I'll stop here, leave you on the edge of your seat and we'll, to be continued Tuesday tomorrow.