 Thank you. Welcome, everyone. Happy to see you all here. So I'm going to be talking about Broward-Mondon obstructions that require arbitrarily many Broward classes. I'll explain all of these terms that are in the title, but before I get started, I want to say everything is joint work with Jen Berg, Carlo Pogano, Bjorn Prunin, Mikal Stoll, Nicholas Trantofilo, and Bianca Varai. And that's a lot of names to write. So when I refer to anything we worked on together, I'm going to write E-P-P-S-T-V-V. All right. So let me get started. So throughout this talk, I'm going to use little K to denote a global field of characteristic not to. And if you want to be concrete, just think K equals Q and you're really not going to miss anything for this talk. So the motivating problem is the following. Given a nice variety X over my global field K, so nice means smooth, projective, and geometrically integral, decide if X has a K point. So I think we can all agree that in number theory, this is a big thing that we'd like to understand. So let me just do a few examples in the case that our number field is Q. And all of these examples are going to be smooth plane conics. A variety which lives inside projective space of dimension two, the plane, it's smooth and it's cut out by an equation of degree two. So my first example is X squared plus Y squared equals Z squared sitting inside P2 over Q. And if I have this variety X, then it's not so hard to work out that X has a rational point because we could just write it down. So if I take the point one zero one in projective coordinates. This is a rational point on X. All right, a little more interesting, I could take as X minus X squared minus Y squared equals Z squared. And you might have to think a little bit harder about this one. But you'll see that in fact it doesn't have a rational point because it doesn't even have a real point. The real points are empty. Because anything that I make over here is going to be a negative real number, but anything I make here is going to be a positive real number and all of the coordinates are zero but that's not allowed in P2. So I have no real points for this one. So in particular, no rational points. Another similar example, if I take 2x squared plus 3y squared equals Z squared sitting inside P2 over Q. So this one does have real points, but it doesn't have three added points. In fact it doesn't even have Z mod 9 Z points you can check this very explicit. And so that also says that I have no rational points. So this procedure that I sort of outlined right here is the first thing you would try if you're trying to understand this motivating problem if you're trying to decide if you have a point. So this first approach. Another problem is to decide if X has local points for all places. And if you're working with smooth plain conics like these examples I wrote down here, that will work. It will help you, I mean it will answer this motivating problem you'll be able to decide if you have a rational point. And this is the Hasselman-Kowski theorem. If X is a smooth plain comic, then the following are equivalent X has a rational point, this motivating thing we wanted to understand, and the test that would come from this procedure I've outlined above. So for all places V of our field K, if I look at the KV points, those are not empty. And I can package all of this information for all of these places as saying that X has an idyllic point where AK is just the product over all places V of KV. So this is great. It says that there's an effective way to decide if your variety, if your smooth plain conic has a rational point. And one of these directions is easy. So this forward direction, if I have a rational point, then I have an idyllic point. This is easy. And it's just the fact that the rational points embed diagonally in the product of all of the KV points. So if I have a rational point, I can consider it as a KV point for every place V. But the reverse is actually the interesting content in this theorem. Wouldn't it be amazing if the reverse was always true, it would give you this procedure for deciding this motivating problem. And so we're going to formulate that into a definition. So we say that some class of varieties satisfies the HASA principle, which I'll abbreviate HP. If exactly this reverse error works. If I know, if knowing that I have idyllic points implies that I have rational points for all varieties in this class. So the Hasselman-Kowski theorem that I just stated here implies, so let's see, examples of varieties satisfying the HASA principle and non-examples. So the Hasselman-Kowski theorem says smooth plain conics are an example of variety satisfying the HASA principle. More generally, you could take quadric hyper surfaces. So varieties, smooth varieties, smooth quadric hyper surfaces in projected space of some higher dimension that are cut out by a single quadratic equation. And let me just point out maybe something which doesn't fit into this are delpezo surfaces of degree at least five. So some surfaces that again have relatively simple geometry. But once you make the geometry even a little bit more complicated, you start getting classes of varieties that don't satisfy the HASA principle. So if I go from smooth plain conics, which are curves of genus zero to curves of genus at least one. This class of varieties does not satisfy the HASA principle. There exist varieties in this class that have idyllic points but do not have rational points. I can also take my delpezo surfaces of degree at least five. I can make them slightly more complicated geometrically. So delpezo surfaces of degrees two, three, and four, again, are classes of varieties that do not satisfy the HASA principle. And okay, well, once you make the geometry even more complicated examples about so K three surfaces and request surfaces. All right, I could go on and on. But what I want to point out here is that if you want to show that some class of varieties does not satisfy the HASA principle. If you want to do any one of these things that what you need to do is you need to exhibit. You can exhibit some example of X in this class, such that X of AK is non empty, but X of K is empty, which means in particular that you need some way of proving that the rational points are empty without using that the local points are empty. So most examples in the literature that do this, that do this exact thing, use the Broward-Mannin abstraction so let me very, very briefly recall that. So most, most examples failing the HASA principle, have a Broward-Mannin abstraction. Most examples, maybe I should say in the literature, have a Broward-Mannin abstraction. All right. So what does this mean? Well, if I have a class in the Broward group of my variety X, which for the moment you don't have to know that much about what this means. But the Broward group is a functor. So by functoriality, I can pull this class back under any maps of another variety into my variety X. In particular, this map that will come up a lot that comes from having points. So by functoriality, if I have a point over any field, while this point is the same data as a map from spec K into my variety X. I'll call that map little X indexed by the point. And so because this Broward group is a functor, I can take my Broward class alpha in the Broward group of X. And I can pull it back under this map and get something now in the Broward group of my field K. So by applying this idea of functoriality, you see that you get a map from the rational points on your variety X. So each of these are maps from spec little K into X. So I can pull back Broward classes and get Broward classes for the field K. So I can just pull back map evaluation along alpha. And what do I do, I just send a rational point to the Broward class, which is pulling back alpha along this map. So that was on the global point side, but I can also do that on the local point side. And take these Adelaide points. So these are just the product of the KV points for all V. And I can do the same idea I can evaluate my Broward class alpha. And now I'll land in the product over V of the Broward group of KV. Well, the rational points sit inside the Adelaide points. And the Broward group of K diagonally maps to all the Broward groups of KV. And so, just from functoriality, you see immediately that the rational points on X. They have to be constrained. They have to sit inside this set of Adelaide points. And when I take these points, and I pull back my class alpha, and I consider this sitting inside the product of all the Broward groups that this lives inside the image of the Broward group of K mapping to the product over all places of the Broward group of KV. Again, this just comes from functoriality. But if you know some class field theory, you're probably getting kind of excited because we do actually know some more. So from class field theory, this is an injection. And the co kernel is Q mod C. And this map here is the sum of the invariance. So the Broward group of KV it always injects into Q mod Z. If V is a non-archimedean place, that's an isomorphism. If it's C, well, then the Broward group is trivial. If it's R, then it's Z mod 2 Z. But anyhow, we always have this map. And if you take the sum of all the invariance, then the kernel is exactly the Broward group of K. So what this says is that I can phrase this a little bit differently. I can phrase this as all of the local points, such that when I take the sum over all places of the invariant at that place of the pullback of my class, that this is zero. And that motivates the definition of this set or the notation for this set. We're going to call this the set of Adelaide points that are orthogonal to the class alpha. Because we have this pairing where we take the sum of the invariance and I have to get zero. And again, this just came from functoriality, but the magic is that this really captures some important relationships between the different points over different places that the rational points must satisfy if they're diagonally embedded inside the Adelaide points. So this actually does give you some information. All right, so a little more notation. So, given a subgroup B of the Broward group of X, define the set X, aka B to just be the intersection over all Broward classes in my subgroup B of the set I've just defined of the classes orthogonal to the to alpha. So we would say this, these are the Broward, sorry. These are the Adelaide points, orthogonal to all alpha in our subgroup B. So, in particular, I can take B to be all of the Broward group, and then I'll get things that are orthogonal to every Broward class. And what you just see up here I've sort of written it but let me just write it again the rational points live inside. Well, they we've already seen they live diagonally inside the Adelaide points. But in fact they live inside the Adelaide points that are orthogonal to the Broward group for every Broward class in the Broward group. So, if this set here is non empty. So I have local points. But this intermediate set which contains the rational points is empty. Then that implies that the rational points are empty. So this is one way that I can get a failure of the Haase principle. I could have the Adelaide points be not empty but this Broward's Broward set the classes orthogonal to the Broward group are empty as the rational points are forced to be empty. And so, if this is the case, in this case, say that X has a Broward monon obstruction to the Haase principle. So the thing that I'm going to be interested in for this talk is what Broward classes you actually need to give a Broward monon obstruction. So if I'm in the setting where this is not empty, and this is empty. Can I can I say something more effective about what Broward classes I actually need. So this is a question you might care about in practice, because for this motivating problem, you want to understand, you want to actually decide if varieties have rational points. And, for instance, if you have a class of varieties there are quite a few classes of varieties that have relatively simple geometry, where conjecturally the Broward monon obstruction is the only obstruction to the Haase principle. So this is the only thing you have to check if you want to understand if your variety has rational points. So you care about how many classes or what you can say about these classes. So let me make one definition about that. So say that a subgroup B captures the Broward monon obstruction, if it's enough to just take this subgroup to get empty rational points. If I just consider this subgroup, I don't need to consider things orthogonal to the entire Broward group, I just work with this subgroup. So the question for this talk is basically what can we say about the size in an appropriate sense of this subgroup B. And I think there are lots of ways to interpret this that are very interesting. For example, you could ask, what can you say about the orders of the elements in B. Questions of this flavor have been studied before by Scorobagatoff and Zara and Varai and Kreitz and Volak and Nakahara and they have very interesting results about how the necessary orders reflect something about the geometry of your class of varieties. But something different which is not really present in the literature is about the number of generators that are necessary in your subgroup B. Actually, if you look through the literature at examples of Broward monon obstructions, so that has a principle, basically they are almost all given by a single Broward class. There's just one Broward class and if you look at the Adelaide points orthogonal to that Broward class, then that set is already empty. So natural question is, well, is that indicative of what goes on, or can you in fact, necessitate many Broward classes in order to get an obstruction. And, okay, that's our main theorem that actually that is necessary. So here's this theorem. Again, this is joint work with Jennifer Berg, Carlo Pagano, Bjorn Poonen, Michele Stoll, Nicholas Triantofilo, Bianca Varai. So for all integers n greater than or equal to one, and k, a global field of characteristic, not two. There exists a nice variety over this global field, such that X has a Broward monon obstruction to the Hasse principle. Well, I'm first going to say the Broward, the Broward set of the classes orthogonal to all of the Broward classes are empty, but for all subgroups. I guess I should have mentioned something up here. Sorry, say subgroup captures this. Here's the first observation. Maybe I can state this after the question. Compactness of X implies that we can always take b to be finite. So this is actually a really, I'm always going to only need a finite number of generators. But what can I say about the size, the number. Okay, so now let me go back to stating the main theorem. So a nice variety. The entire, if I consider the entire Broward group, then there are no Adele points that are orthogonal to the entire Broward group, but for all subgroups be contained in the Broward group of X generated by strictly less than n elements. If I look at the Adele points only orthogonal to this subgroup, then this is non empty. So what this says is that there's no upper limit for the number of Broward classes necessary to give an obstruction. Broward monon obstruction, it can really require arbitrarily many Broward classes, a finite number, but arbitrarily men. All right, are there any questions at this point just about the statement of the theorem, before I go on to say something about how you prove it. All right. So, here's the theorem. Then you can unmute and ask away if you have a question. Sorry, this was a mistake. Sorry, I didn't mean. Okay, no problem. So, alright, so I think the rest of the time I'm going to explain some of the ideas that go into the proof of this. And it's very constructive. So we're actually going to construct for you. So I'm going to talk about some of these, which means that I'm actually proving a stronger theorem where I'm restricting what the class of this variety is. So in particular, it's going to be some class of varieties that that themselves cannot have Broward monon obstructions with a bounded number of classes. So, but to height and suspense, I'm not going to tell you the class of varieties for a little bit. And first I just want to say something about how we think about this property here that there are no points that are orthogonal to every Broward class, but for all subgroups generated by most n elements. There are a delicate points orthogonal to just those classes. So how do we think about this. So the first step in the proof is to reinterpret. A Broward monon obstruction in terms of the Broward monon pairing. And that will allow us to see this. This will be some combinatorial condition. Okay, so let's get started. So what do I mean by the Broward monon pairing. I mean, we already observed that we have this map, we have this pairing between Adele points. So product, product over all places of the local place and the Broward group of X to Q mod Z, where I take a tuple of local points and a Broward class alpha, and I map this to the sum over all places of the local variant of that place of the pullback of alpha along that Adele point. And the rational points have to be sitting inside the Adele points as things which pair with every alpha to give me zero under this pairing. For any finite subgroup be sitting inside the Broward group of X. Let me write the hat for just the dual the harm from B to Q mod Z. So then, equivalently, I can think about this Broward monon pairing as a map. Okay, I'm 5B from the KV points to the hat, where I send a local point at V to. Well now I have to give you a map from B to Q mod Z. So for every alpha in V, this is the map that sends alpha to just the invariant at V of the pullback under my local point of the class alpha. All right, and I'm going to write SV for the image of 5B. And notice this target, right, this doesn't depend on V. That means I can sum over all places V and I can get a map phi from the Adele points to be hat. And I'm going to write S for the sum of all these images SV. And I'm going to use this map to reinterpret a Broward monon obstruction. So B captures a Broward monon obstruction. Well, by definition, this is the same thing as saying that if I look at the points orthogonal to be that this is empty. What does this mean? This means there does not exist an Adele point that's orthogonal to every class alpha in B, which in terms of this map here, says that there's no point here that when I map it to be hat, it gives me the trivial map from B to Q mod Z. So I can just raise that as saying, this is the same thing as saying that zero, the trivial map is not in the image of V, which I was calling S. There's no Adele point that trivially pairs with every single Broward class alpha in B. So that was one part that I had to understand. But the theorem said that this Broward monon obstruction was not captured by any proper subgroups that were generated by some number of elements. So I raise that combinatorially. So, so for all the prime contained in B, or maybe let me say no B prime contained in B but not equal to be captures the Broward monon obstruction. This is the same thing. You can work through this again and you can see that what's omitted by the image of Phi does not contain a non trivial subgroup. So this is the combinatorial condition that we have to understand if we want to understand if we want to prove that Broward monon obstructions can require arbitrarily many Broward classes. So we have to understand the image of this pairing show that it omits zero, but it does not omit a non trivial subgroup. And, and further for us, for us for the X's that we're going to consider. This B is always just going to be Z mod 2 Z to the N. So it means that the hat well okay it's also isomorphic to Z mod 2 Z to be and, and this condition is the same thing as saying that S is exactly the hat so Z mod 2 Z to the end hat minus zero, because, well it's just as an elementary if you omit any other element than you admit a non trivial subgroup. So, so this problem that we're trying to understand is really going to be about having very large image maximal image of the Broward monon pairing. Up to the fact that I have an obstruction so I omit zero. So I wanted to be as big as possible, given that. And so this S, let me just say this in words this S is the sum of all of the local places of all the local SVs. And so if I want this S to omit only zero, I want that for these local SVs which are only zero for all but finally many. I want these non zero SVs to sum up to only to everything except for zero. There's one kind of obvious way to do that. It's to prescribe. So one way to do this one combinatorial solution is to prescribe SV not to be everything except for zero Z mod 2 Z to the end hat minus zero. For some place V not, and that SW is simply zero for all W not equal to be not. That's kind of the most. That's the most basic way to have this happen is that it happens all at one place, but there are other combinatorial solutions so we proven an appendix to our paper that there is some constraint on the combinatorial solutions to this problem. There are other places where the set SV can have size, at least two so be sort of non trivial is bounded by the F two dimension of V. So that's N minus one. So there's some bound on this and that's a sharp doubt. But in any event, if you're only caring about the main theorem that you can require arbitrarily many generators, then it suffices to use this combinatorial solution. Given the setup. You're probably motivated that the way that we're going to concretely solve this problem of exhibiting varieties that require arbitrarily many Broward classes to have an obstruction is that we want to prescribe what these images of the Broward modern pairing SV are. So this is our more precise theorem that implies theorem one. Theorem two. Again, it's joint work. So let K be a global field of characteristic. Not to you get to choose any N and R which are greater than or equal to zero. And any sets S one through SR, which are subsets of Z mod to Z to the end hat. So this is like prescribing. Well, you might imagine this is prescribing what the images of these Broward modern pairings are at finally many places. That's what we're going to do. So given this data, there exists a nice variety X over K. And an injection. So this is going to be Z mod to Z to the end into the Broward group of apps. So this is going to be our subgroup B and places. The one through VR. So these are the places at which I'm going to realize these sets as the image of the Broward modern pairing. Well, exactly what you want the image of Phi VJ. So this is what I was calling S, sorry, SVJ. This is my set SJ that I got to arbitrarily decide for all J. Two, for all other places, the image. W is SW is simply zero. We're all W not equal to be J. And finally, there's no there's nothing else in the Broward group that could give you an obstruction. This, if I take this injection from Z mod to Z to the end into the Broward group of X, and I compose with the projection, where I quotient by the image of the Broward group of the ground field sitting inside the Broward group of X, which I'm going to call Broward X bar because it will come up again, then this composition. This is an isomorphism and these things in the Broward group of K they can't give you obstructions to the Broward to the to the Haaser principle. So what this third point is saying is that there's no other classes that could give you an obstruction. All right, so in my remaining time, my remaining 10 minutes, I think I want to say a little bit about how you prove this theorem but first let me make it more precise here's the. I said I was for dramatic suspense I wasn't telling you the class of varieties but you might guess based on all the setup what this class of varieties is so this is going to be a contact bundle over P one that split by a constant quadratic extension. So, so more precise theorem you can realize any images of the Broward modern pairing that you want by such comic bundles. Okay, so let me say something. First about these kind of bundles and then I'll say something about what goes into the proof of this more precise theorem. So the first contact bundles split by this constant quadratic extension. So I am adjoining the square root of some element a from my field K that is not a square already. So it's not already a split contact bundle. So, how do I give this so this is going to be attached to a choice of a polynomial F, which is going to be square free and for for nice for nice reasons where I'll mention why I'm always going to assume that this is even degree. So then, this kind of bundle given by the choice of this non square class a and this F, it's nothing other than smooth compactification of y squared minus a z squared equals F of you. All right, so let's stare at this equation a little bit and notice that there's a map from x to P one with coordinate you that just sends y z and you to just you. Here's P one I'm fixing you if I fix you that when I take this compactification, the fiber over any point. It looks like the conic which is y squared minus a z squared equals whatever that F of you was times w squared in the compactification so generically. But there's finally many points in the base where F of you is zero. And if F of you is zero, then you see that I get the singular conic, which is why squared minus a z squared, not a smooth plane conic. And if you had a square root of a then you could then you could take a difference of squares, and this would be the union of just two lines. But I specifically assumed that we don't have a square root of a in the ground field. So these geometrically look like the union of two lines, but they're, they're not, they're not defined over the ground field. So I was to pass to this constant quadratic extension where I joined a square root of a, then all of these do become split. All right, and this even degree assumption. This is exactly saying that the fiber over infinity is smooth. It's not one of these. Not one of these singular conics. But why is this a nice class of varieties I mean this, these kind of bundles flip by these constant extensions. These are some of the first varieties where Broward monodobstructions were exhibited. And one reason that makes them so nice is that we very concretely understand what the Broward group is for this variety. So, so if my f of you factors as f not of you times f one of you dot dot dot all the way through fn of you. So if this is a factorization, where each of these f eyes are even degree. And this extension that splits my kind of bundle so k alpha is not contained in the splitting field. This is explicitly irreducible of even degree, not contained in the splitting field. Then under this assumption, the quaternion algebras a comma fi of you, which a priori only live in the Broward group of the function field. There are in fact pullbacks from the Broward group of the function field of P one. These in fact live in the Broward group of my variety xaf. I have a source of Broward classes that you can see right here I've written down for you n plus one Broward classes they're all two torsion so this is looking kind of good for what we wanted. Furthermore, we can say that these classes a f not through a fn, they generate the Broward quotient. And by, by constant as they, and there's only one relation, which is that they all sum to zero. So if I take a f not plus dot dot dot plus a fn this is the same thing as the Broward class, where I take their product which was f, which is, you notice the class of the model itself. And so this is, is zero. And that's the only relation. So in particular, this implies that the Broward group of xaf the Broward quotient is isomorphic to Z mod to Z to the amp. So we have a variety, which has lots of Broward classes. So we hope that this is a variety where we can use these Broward classes to prescribe arbitrary images of the Broward modern pair. So, given that setup, I'd now like to just briefly sketch the steps in the proof of theorem to so steps in the proof of theorem to. So, step one is so theorem to says we can arbitrarily prescribe the image of the Broward modern parent at finitely many places. So I want to do this, that all of these finitely many places I'm trying to prescribe, they're all actually just everything in Z mod to Z to the end. I'm not, because I'm taking the door. So, this is very different than what you would want. If you wanted to write down a Broward modern obstruction, right. In that case, if you want a Broward modern instruction that's, that's captured by by. Well, just if you want a Broward modern instruction zero can't be in the image. And I'm saying that the image is everything. I'm thinking about the problem in this way, where I think about arbitrarily prescribing these images of the Broward modern pairing. It leads you to also consider these surjective cases and it turns out that's, that's going to be enough by a by a trick that I hopefully will say something about to get everything. So first I do this special case. So let me just say verbally because I don't have a lot of time. So I'm going to choose F not through F and to multiply together to get F. I'm going to choose them all to be monic and irreducible and separable of even degree. But other than that, you have a lot of flexibility. I'm going to choose my places where where I'm going to get these so these are going to be for I equals one. To R. I'm going to choose my places to be. Let me just explain this in the case cables. So I'm going to choose my places to be primes that are sufficiently large. So P one through PR are going to be sufficiently large. And this a that goes into the theorem is just going to be the product of all of these P one through PR. So once I have this setup, then this is enough to guarantee that this Brouwermann and pairing map alpha and Phi P is surjective for all P, which are these finitely many places. And the reason for this is is is kind of nice. So here's this Brouwermann and pairing. It goes from, well it goes from the QP points on X, which are the same as the ZP points on X. And if I want to so surjectivity, and it's enough to look at the points that just live over a one. And this goes to infinity for right now. And this goes to the map out, which explicitly is home from this group. Well it's the, it's the Z much to Z to the end generated by these Brouwer classes so AF not through AFN with the one relation that the sum of all the points is zero. So this is really a Z much to Z to the end. It goes to Q mod Z but all these classes are two torsion so it goes to one half Z mod Z. And what do I do? Well I send some tuple YZU to, okay well I have to evaluate my class AFI on this tuple YZU. And you see that the Y and Z they totally don't matter here. All that matters is the U. And so where does this actually go? This sends a class AFI, well evaluation I just plug in U into FI and I consider if this is a trivial Brouwer class or not. And if FI is not zero modulo, FI of U is not zero modulo P, then this is the map that just sends this to the Legendre symbol of FI of U mod P on P. So all that I care about here are the square classes of these FIs. So the special case is implied by the following lemma. So if P is sufficiently large, depending only on N and the degrees of these FIs, I can arbitrarily set the square classes of the FIs. So of F1 through FN and FP star mod FP star squared. So I can arbitrarily set these square classes with the only relation that of course they all have to multiply to zero because I have this relation in the Brouwer group. And the proof of this lemma is just the Langvay estimates for counting points on varieties over finite fields because setting square classes can be interpreted as a set of equations where I want FI of U to be equal to something, some X times X squared, etc., and the intersection of all of those defines for me a variety over FP and I can count its FP points. So if P is large, the Langvay estimates will imply that I have a point there. All right, I'm mostly out of time so let me just briefly say where what step two is. So in step two, so we've constructed this thing, Xaf, mapping to P1, where for finitely many places, the evaluation is surjective. So if I look at the, if I look at, say, the ZP points here, this breaks up as some collection of discs where, which are given by the elements, congruence with the elements that gives me these arbitrarily many square classes. But so these discs give me everything in Z mod 2Z to the N hat when I evaluate over those discs, but I would just like to prescribe my S to contain just some of the discs. And so what we do is we show that you can always take a pullback. You can always take a pullback by some map G. So now this is another one of these conic bundles, but I have my new polynomial is F composed with G. And what's special about this is if I look at any one of these ZP points, it will map me entirely into the set where I have the image that I want for us. So, of course, there's some details to show here that the Broward group that you can always choose this G so that the Broward group here is generated by just the classes a comma F composed with G. But these are all finitely many local conditions and we can satisfy those. And so taking pullback, once you have surjectivity, you can get something arbitrary. All right. Thank you.