 completely general effectivity result or a method for computing the Brouwer set. So we can do it in practice, but this is a very, very different situation than David Harvey's lecture series. So he's trying to really optimize the running time and the complexity and getting in deep to like how far you can push these algorithms. This is a fundamentally different situation. We're just trying to see, okay, how do we do this? Can we do it in an example? What, yeah, like can I throw all of my knowledge and ideas and and do it in this case? Rather can I like, as the LMFDB people say, can I push a button and run it for like, I don't even know what orders of magnitude examples? So that's a different situation than I'm going to be talking about. So let me just say, so I said here completely general, so that we do have two effectivity results that I describe more in the notes. They're both by Kreschen Schinkel. So in one, they show that a larger subset, the algebraic Brouwermann subset, which I'll define in a minute, is effectively computable. And in the second one, effectivity for surfaces, they show almost that for surfaces, the Brouwermann and set coming from the end torsion in the Brouwer group is effectively computable for any given end. So we do have some results. But again, these are also, these are really effectivity results, not algorithms with complexity analysis. So it's still fairly different flavor than this morning. Okay, so we don't have a completely general method, but we do have sort of, I don't know, an approach slash procedure that people usually do. And as you'll see, there's parts of it that's like, okay, that's all you're telling me. What do I go from here? And the answer is like, yes, exactly. Where do we go from there? That's that is what is making it so that it's that it's not completely effective. Okay, so before we get into computing the Brouwer set, so the Brouwer set was defined as the set of a deli points are orthogonal to every element in the Brouwer group. So first we want to look at just understanding or computing the Brouwer group. Oh, okay, there we go. So first thing just to note, so I might have said this, but it's not apparent from the definition I gave you, but it is true that the Brouwer group of X is a torsion, abelian group. So we're not trying to figure out very complicated group structure. We're not dealing with trying to understand what normal subgroups are or anything. It's very simple group structure. But as you'll see it still can be hard to compute. Okay, so the way we often begin on trying to compute the Brouwer group is by subdividing it into different filtrations. So Brouwer zero of X typically denotes the image of the pullbacks from the Brouwer group of K to the Brouwer group of X. Okay, so these are in the lecture notes. You don't have to like super get them in your head, but let me just remark. So we have X, X is living over K. So we have this structure morphism from X to spec K. And so then we have a map in the other direction. And so these are the ones, these are the ones that you think that they're, they don't come from X. They come from K, right? They're not telling you anything about the geometry of X. They're, they're constant. They're sort of like there and you have to deal with them, but they're not really telling you what's going on with X. So you should think of these, yeah, as not as important. Okay, and then the second step of the filtration is denoted Brouwer one. And these are called the algebraic Brouwer classes. And these are the ones that are annihilated when you pass to the algebraic closure. So the algebraic Brouwer monon set. These are the ones that are slightly easier to compute. Okay, and one thing that you looked at in your exercises yesterday, you might not have gotten to this particular one because there were a lot, maybe you were interested in other ones. But you can prove that these constant classes, they cannot obstruct any idyllic points. And this should fit with the intuition that they're not seeing anything about X. So why should they be able to tell us anything about the points on X? Right, they really just come from K, they're already there. So you know, and then you can also prove that the Brouwer set depends only on this quotient, that actually you only need representatives for the quotient Brouwer mod Brouwer zero. So really when I talk about computing the Brouwer group from the point of view from the Brouwer monon obstruction, and also just the point of view of what is the part that's related to X, we really only care about the quotient Brouwer mod Brouwer zero. Okay, so I will try to say Brouwer quotient over and over again instead of Brouwer group, but I predict that I will fail often. So you can ask if it's unclear, but you should also just think where everything is about the Brouwer quotient. Okay, questions. Yes, that's correct. Okay, so the question was, if X was a field, then passing to the algebraic closure would just kill everything. So Brouwer would just be equal to Brouwer one, and in fact equal to Brouwer zero, but okay, but the important part was equal to Brouwer one. And so I think the question was two parts. One, guessing from what I've defined that that's not true in general, and two, like, is there something that we know always kills the elements of the Brouwer group? And okay, so let me answer the first one. Yeah, so in general, Brouwer is not equal to Brouwer one, but in several types of varieties it is. So if you have something that's geometrically rational, so geometrically birational to projective space, then Brouwer is Brouwer one also by Sen's theorem for any curve. That's not an obvious thing, it's just like a thing that I'm throwing out there, if you want to look at it more, that for curves there is this theorem that implies that Brouwer is equal to Brouwer one. So that helps in those situations, we're in a better situation for competing the Brouwer group. If you had a variety where Brouwer was not equal to Brouwer one, and the Brouwer quotient Brouwer mod Brouwer zero was infinite, then say Brouwer mod Brouwer one was infinite, I don't know if you can, okay, well, if you base change to the algebraic closure of the function field, it will kill everything, but maybe you wanted a smaller extension than that, and I don't know a smaller extension that is guaranteed to always work off the top of my head. More questions. I really want to get the questions going again, so if you ask a question, you are doing me a favor. That is a great question, so, okay, so the question was, if you base change to a field extension, can Brouwer one become like as small as possible? So yeah, so Brouwer one always contains Brouwer zero, so if you base change to a number, like a larger number field, it won't be trivial, but it, so when Brouwer one mod Brouwer zero is finite, there is always a finite extension that trivializes Brouwer, wait, no, am I saying this correctly? Okay, under an assumption which I will mention later, which implies that Brouwer one mod Brouwer zero will be finite, then there is always a finite extension of your base field that trivializes that Brouwer one. Yeah, but lots of times Brouwer one mod Brouwer zero can be enormous and infinite, and so for like, for abelian varieties and for curves, there is no finite extension that will ever make Brouwer one as small as possible. Great, thank you, anyone else? Yes, yes, each piece in Brouwer one still dies in a finite extension, that's right. Okay, the question was does each piece of Brouwer one die in a finite extension? An answer is yes. Okay, I'll pause again shortly so you can keep thinking. Okay, so really we're interested in computing Brouwer mod Brouwer zero, and just by definition, we have this nice exact sequence where we have Brouwer one mod Brouwer zero injecting and then we surject onto Brouwer mod Brouwer x. So normally we think of dividing those two pieces, and then as I explain in slightly more, but not too much detail in the notes, from a spectral sequence, you get some exact sequences or isomorphisms for these quotients that relate it to the Galois cohomology of pick x bar. So, particularly if pick x bar is finitely generated and torsion free, then this Galois cohomology group is fairly computable. Assuming you have generators for pick x bar, then you can compute this h1. Okay, this part is much harder to compute, and that's just the way it is. So you could work on making, yeah, understanding this exact sequence better from a computational point of view. Right now, I think this hasn't been leveraged so much for computational purposes, only for theoretical ones. Okay, so that's the approach, and as I said, you may feel very underwhelmed by this approach, because it's far from an algorithm, you're still left with computing these cohomology groups, you're left with computing pick x bar. Okay, we have, yeah, some effectivity results for different pieces, but that's why you have. Okay, but sometimes you get lucky, and you can sort of guess an element in the Brouwer group and verify that it is actually an element in the Brouwer group. And I outline one of these approaches in the exercises. So sometimes you get lucky, or maybe you're in a situation where someone else has already worked out these cohomology. Now you have the elements of the Brouwer group, how do we go from there and compute the Brouwer set? So assuming we've computed this quotient, how do we compute this Brouwer set? And so let me just put this up here to have it. So this is the set of idyllic points, such that the sum of the invariance of alpha, okay, how did I do note it? I think pv upper star of alpha is zero for all alpha. Okay, and as someone asked yesterday, there were a lot of questions yesterday about this map. So does the pullback depend on alpha? Does it depend on x? How do we compute it? What do we have to know about it? And I said, yes, you have to know all those things. It's hard to compute in general. So this is what we have to figure out. And just, I think I introduced this notation. I often want to think of this as a map from the idyllic points to the Brouwer group that depends on alpha. So I'm going to call this map f alpha from x of kv to q mod z. I guess really this should be like f alpha sub v, but the v will hopefully be clear from context. Okay, so we know some things that are helpful. So we know this evaluation map is always locally constant. So if you want, you can always throw away as the risky closed subset of your x and just compute the evaluation there. And then you know you won't have this really, really bad fractal or ill-described pattern. Everything that the preimage of every value is going to be an open set in the idyllic topology, open and closed set. And we know a little bit more. Also, if you know that your algebra, your Brouwer class is killed by an unrammified extension of your local field, then you know that it's always zero. And that is very, very useful. And particularly, you can see this for elements in Brouwer 1. So you know that they'll be killed. Everything in Brouwer 1 is going to be killed by some finite extension. That extension will be unrammified away from a finite set of primes. And boom, immediately, you know, for all of those places, this class is zero. So that is a huge, a huge bonus that helps us. The downside is that outside of this, so, in quotes, this is, unless there is a reason why not, typically these evaluation maps have the largest order that they possibly can have. So you don't, you don't usually just get lucky and like it's constant everywhere. Okay, so some of you who maybe have read a little bit about Brouwer 1 and obstructions previously, or maybe you're doing research in this area, or you decided that you were going to do the last big problem on the problem set exercises yesterday, you might be thinking like, hey, but we just did this problem and it was constant everywhere. It was always constant. It just like, I worked out and I only had to check at one point and it was, it was super easy. That's, that's because we made it that way. Okay, the examples that you see in the literature are examples that are rigged to be easier to handle. So in the literature, they are often constant all the time. And from looking at that, you could be like, oh yeah, no problem. I'm just going to check this one point and see what's happening. But that is not, not what happens in general, unfortunately. So just don't, don't be led astray. Okay, so what I would like you to take from what I just said, I like threw a lot of these definitions I use, some exact sequences, all these different properties. What I want you to take away from that is that it is doable, but it can be hard. So yes. Yes. Through the local topology, through the viatic topology on XKV. Sorry. Yes. Oh yes, thank you. The question was, what does a locally constant mean for the viatic points and is it the same as it being continuous? And so locally, I mean locally for the viatic topology. So if you haven't seen this before, each X, you can cover by affine covering. So you can think of the different pieces that's sitting in KV to the N, which inherits the product topology and the topology from KV. And so that gives you this viatic topology on X of KV. And it's like the only thing that you could think of doing if you look at, at some points. And because Q mod Z is discrete and actually for each alpha, it's going to map to a finite group, that's the same as saying this map is continuous. Okay. So what I want to do is, what I want to do for the rest of the talk is talk a little bit about this theme that I mentioned on the first day, like how do these theoretical and computational pieces interplay? And can we, can we try to bypass this somehow? Okay, so what I want to just think about right now is, yeah, so why is this, why is computing the Broward-Mannan obstruction difficult, sort of on a on a heuristic level? Okay, so the way that I think about this is the, the no free lunch principle. So, yeah, I looked this up on Wikipedia. Actually, the history comes from, I guess, like, pubs or something giving drinks for free at lunchtime. No, lunch for free and then people would buy drinks. Yeah, and you just have to pay somewhere. Okay, so if it says it's free, there's something secretly hiding. And what is going on here, like, you cannot dodge, you can't, you can't, you can't bypass the work. When you're trying to compute the Broward-Mannan obstruction, you can't ignore the elements. So that would be getting something for free that you can, like, just throw away some elements to worry about. These Broward classes, sometimes it feels like their, like, purpose in life is to obstruct Adelaide points. They just, like, if they can, that's what they're going to do. So, what I said is obviously not a theorem, but let me show you some theorems that illustrate this. So the, the, the bigger principle is unless there is a reason otherwise, you should expect x of ak, the set of Adelaide points orthogonal to alpha to be strictly smaller than the set of Adelaide points. You should expect it to carve out some subset. So we already saw some reasons otherwise, right? So elements in Broward-0, elements that come from Broward group of k, they don't, they don't obstruct Adelaide points. But that's because they're not about x, right? That's why. Okay, so the theorem, the, like, legend for the theorem is the things in pink are illustrating the reason otherwise. And then the, the theorem is showing that why we should expect this. Okay, so the first theorem dates back to Hari from 94. He says, if you have some family of varieties that has a non-trivial Broward group on the generic fiber. Okay, so it's like for many things in the, in the family, you expect the Broward group to be non-trivial, but the Broward group globally is trivial. Then for infinitely many points on the base, the Broward set is going to be strictly smaller than the set of Adelaide points. There is going to be an obstruction. Okay, and so the, the idea, the reason otherwise would be that the Broward element actually extends to be a, an element of the Broward group of the whole total space, not just your, your single fiber. But if that doesn't happen, then you should expect the Broward classes to carve it out. And this tells you, so this tells you that failures of weak approximation, so that's just when the Broward set is strictly smaller. It tells you that you just, you sort of know for free they exist in families. You often don't have to write them down. If you know that there is a family where you have sort of generically non-trivial Broward group, then you're going to get failures of weak approximation. Okay, yes. Oh, that's V sub eta. Sorry, that's the generic fiber. So this is a1. So that's, there we go. So a1 is speck of the polynomial ring. So I could just view, this is a fiber thing over the function field, over some bigger field. So now, rational function field in a single variable. So this is my eta. And this V sub eta goes there. So you could think of this, say you have, your V could be a family of elliptic curves over the J line. And say you have a Broward class, say I wrote down a quaternion algebra that was like five comma function divided by J minus 1728. Okay, so that would give me a well-defined function, probably a Broward class. But when J is 1728, there's, that function vanishes in the denominator that won't extend. So that would be an example of a Broward class that you have generically but doesn't extend over the whole family. So it's like, you have some, and the reason why you want it to not extend over the whole family, well you need some condition because you need to rule out things that just come from Broward K, right? There's no condition here. I don't have any quotients here. So I have to, I have to somehow throw out these elements because we already know that they don't do anything. But if my algebra just came from the ground field, it would just extend over the whole, the whole total space. The whole family of elliptic curves over the J line. Okay, here, and then there. Go ahead. Yeah, so, okay, so the question was, what do I mean by failure of weak approximation? So the, the short answer is I'm going to say it and then also you can feel free to ignore it because this is just a sort of like bonus thing around. So I said at the very beginning, I'm going to focus on the Broward monon obstruction to the existence of rational points. But you can ask about other properties instead of just asking about the existence of rational points. You could ask about, like, well, how many are there and can I approximate any idyllic point that exists? So that would be weak approximation if the closure of the rational points in the idyllic topology was equal to the set of idyllic points. Okay, so WA is weak approximation. So if this is the closure in the idyllic topology equals the set of idyllic points. But this is one thing you'll do in your exercises today. Well, if you want to do this exercise, the Broward set is closed in the idyllic topology. So in particular, it contains the closure of the k-rational points. So it can not only obstruct the existence of rational points, but it can obstruct the, it can obstruct weak approximation. It can obstruct this density. Yeah, great question. Yes. Yeah, okay, so he writes something, yeah, that's right. So by trivial, I mean that it's equal to the, that it's equal to Broward or not. So what you need, so really the condition is you need an element of the Broward group. And by non-trivial, I mean not just Broward or not. You need an element of the Broward group that is ramified on some fibers. But I didn't want to define ramification. So I tried to translate to the theorem statement. Yeah, but so by trivial and non-trivial, I mean equal to Broward or not. I'm not equal to Broward or not. Sometimes. So this is also a question. So I just started the motivation by asking about, okay, if you want to prove that there's no rational points, you need a way to witness there's no rational points. And if the Broward set is empty, then that proves to you that you have no K rational points on your variety. You can also ask the converse question. Say I compute this Broward set, and I know that it's non-empty. Do I then get for free that there is a K rational point? So that's normally called the Broward-Mannan obstruction is the only obstruction to the Haase principle. So that, that is a different, a different direction. And yeah, in some sense, that's a, that's a harder question because it's giving you the existence of a rational point, which is a fundamentally difficult question. And then you can ask the same thing about density. So it is conjectured that Broward-Mannan is the only obstruction in several classes of varieties. So Kola, Tallinn, and Sonsuk have conjectured for all rationally connected varieties. Broward-Mannan obstruction is the only obstruction. What is known falls far short of that, but Olivier Wittenberg has the best results to date on when, with Yonatan Harpas have the best results on when we know that is the only obstruction. I was gonna say something else. Oh, but we also know that it's not the only obstruction all the time. So there are examples of Skorbagatov and then of Poonin and variations after that that show that it's not the only obstruction. Yeah. And just as you can tell from my very long answer, I find those questions very interesting, but I just have to choose something for three lectures. So that's why I'm not gonna talk so much about that unless you all ask me more questions, which would be great. Okay, more questions. Okay. I think in the interest of time I'm just gonna blow through these next questions, but this is another result of Martin Bright more recently that sort of looks at these things called residues and shows when they give you that the evaluation map has a large image and then Bright and Newton and Rachel Newton is here. She's giving the research program talk later today. They show that you can push this even farther and even show at places of good reduction. We should expect to get these failures and Pagano wrote down an explicit example that demonstrated this. So when you're computing the Brower set, you just you just cannot bypass computing the evaluation map for these alphas except in these other, you know, unless there is a reason otherwise. So if it's killed by an ramified extension, you know it's gonna be zero. Otherwise, there are a bunch of cases where we know the evaluation is gonna be surjective and that's just what we've proved so far. Yes, question. That's right. Yeah, so the question was here I'm just saying a single alpha should carve out something smaller but maybe we get lucky and like a bunch of the alphas are redundant. So if we compute one of them, we don't have to do the extra work for the others. And I think that exact question hasn't been looked at but I would guess that if you look at some of these reasons, some of these theorems that I put up by Hari and Bright and Brighton Newton, that if you looked in there, their proof, you could get it to tell you that like, okay, if I took this collection of elements, then the evaluation map for the like going to the product of Q mod Zs from the product of the alphas would have a large image as well. I'm not totally certain but I would guess that there would be some conditions that trigger that. Yeah. Yeah, that's a great question. Okay, so this is why it's hard. It's just like there's all these elements. They're likely to do something. You have to compute them. I mean, what do we do? But so I think it's not so talked about secret that one of mathematicians greatest strengths is laziness. So if we want to compute the entire Brower set as a subset of the italic points, we have to do all this work. But that's not what we asked for at the very beginning. What we asked for at the very beginning is, can we prove that the set of rational points is empty? So what if instead of asking to describe the whole Brower set, we instead just ask this decision problem, is the Brower set non-empty or not? I don't care what it is. I don't care what points are in there. I just want to know is it empty or not. So why is that a better problem? And probably I can't completely say it's 100% definitively simpler, but why are some reasons why it might be simpler? Well, let's go back to some of the topological facts that we talked about. So if X is projective, then the set of italic points on X. So here I'm taking the italic topology, which is the product topology on the theatic topologies on each factor. So under that topology, this is compact. And then that tells you, so any compact set, if you have an empty intersection, it can be realized as a finite intersection. Finite empty intersection. So that tells us if the Brower set is empty, which is the intersection of all these X of a k of alphas, then there's some finite set of Brower classes, which will exhibit this is empty. Okay, and so there are some cases where this is definitely already better. So if you have an elliptic curve with finitely many points, assuming the chef of each group is finite, then the Brower set is the closure of the set of rational points. And since you have a finite set of points, the closure is just the finite set of points. And to cut down the set of italic points to a finite set, a finite non-empty set, you have to use infinitely many elements. You can't just use finitely many. So already we know that we're in a better position here if we're just interested in this problem. Okay, so you could ask, like, can we hope to understand what this B is, or some properties about this B, ahead of time? And then restrict our search from the beginning to make the computational problem easier. Okay, this isn't going to give us a, well, okay, it's not an idea for a different algorithm. It's just if we understood the answer to this theoretical question, then it would reduce the computational problem that we need to solve. And that would be a win. Okay, so I'm going to introduce two definitions. And so we'll say that some subset, so for the definition, it doesn't have to be a subgroup, but maybe you might prefer that you just take a subgroup. But we'll say that it captures the Brower monon obstruction. If the Brower set is empty, means that the elements in this B already detect it. So I'm not saying anything about the elements outside of B. In this case, I'm just saying it can be, the essence, the obstruction can be captured by the subset. Okay, and then I'll say something stronger, that it, this set completely captures if whenever you have some obstructing set, any subset that obstructs the full existence of rational points, you can detect it from the elements in that subset that already belong to your B. Okay, so completely captured is like the golden standard, but captured already reduces the computational problem. Okay, and as I said, this is really capturing the Brower monon obstruction to the existence of rational points. You can't ask, well it doesn't make as much sense to ask for a capturing set to weak approximation, because that's what I, that's a, you need to understand more things. Okay, so ideally we would be able to find a B that captures or completely captures and that this subset is more easily computable. So, but at the first sense we find a B and maybe if it's, okay, B equaling Brower X is not fun, that definitely captures and completely captures, but if we can find some proper, possibly proper subgroup, then, then we're getting something better. Okay, so this terminology was introduced, I introduced it fairly recently with my collaborator, Brendan Kreutz, but you can look back in the literature and there were results in this direction, so if I rephrase them in terms of this language, also it follows from what, from what Monon did when he originally introduced the Brower Monon obstruction, that if you have a degree D genus 1 curve, so genus 1 curves can be realized or embedded several different ways, but if you have one given to you as a degree D curve, then the D, the D primary torsion completely captures it. You don't need to look at any Brower classes with order co-prime to D and Swinerton-Dyer proved that if you have a cubic surface, then the three torsion of the Brower quotient completely captures. So the, for a cubic surface in this paper, he classified the possibilities, so it can either be two torsion or three torsion and he showed that whenever the Brower group is two torsion, then there's no, actually he showed that it satisfies the local to global principle. And then Kohle-Telenen-Punin proved that if you have a cubic or quartic del peto surface, then there is always a single element that captures the obstruction. So you actually only, you don't know which one, but there's always only one. Okay. Okay, and so let me just say like one thing that's slightly difficult about thinking about this or what you might want to prove is you want something that you can describe ahead of time, that is sort of intrinsically described. But there's usually not that many, because we have a torsion-Abelian group, there's often large automorphisms of these Brower groups, so there's not like really a way to pick out a single element. So for instance, the Brower, in a cubic surface case, the Brower quotient is one of these five possibilities. Okay, and so what we can say is that the three torsion completely captures it and there's a three torsion element that captures it on its own, but you can't hope to say anything more about which element it is because there's sort of like before you start computing, there's no way to distinguish what the different elements are. So we couldn't even like hope to say something stronger, because how do you tell apart the different elements in Zmod 3 squared? Okay, I want to leave more time for questions, so maybe I'll just skip some of the things about like how you might directly compute the n-torsion and we'll just go to some of the more recent results that show what we have. So in this paper with Brendan Kreuz, we prove that if you have a twist of an Abelian variety, so this generalizes the genus one curve statement that I had before, and then instead of the degree of the curve, you put its period that that subgroup completely captures it. So this should just be taken as an illustration of what we know. Why is it? Okay, and then we know for kumar varieties, which are closely related to Abelian varieties, they're desingularizations of a quotient of a twist of an Abelian variety. But anyway, it's something that's related to an Abelian variety and it's related by a two primary cover. So in that case, then Skor Bogatov showed that the two primary subgroup completely captures it. So that's very strong. And then Masahiro Nakahara proved, and all of these, I put publication dates for really like all of these papers came out right around the same time, that if you take a vibration of twists of Abelian varieties, so you could take a family of genus one curves, or you could take a higher dimensional family, then the period of the generic fiber that torsion completely captures the Brouwermann obstruction. So this tells you already from the beginning what primary order you have to look in. And particularly in the case of when you have all these Abelian varieties flying around, Abelian varieties have torsion in the Brouwer group of every order, and the N primary torsion is always infinite. So restricting the torsion there, which orders you're looking at, that's really a strong statement. Okay. Oh, and one more. So this is a subgroup. Something is going on with my remote. But this is a subgroup I didn't mention. It's usually denoted by a surreal like B, it means the locally constant algebra. So these are Brouwer classes that when you base change to KV, they land in Brouwer 0. So they become locally trivial. So for these actually, once you know them, then computing the evaluation maps is easy because you only need to evaluate at a single point. So this is really great. And so Croix showed that if you have a twist of an Abelian variety, then the locally constant subgroup completely captures. And so then actually you can take the intersection of the first result and the second result and it tells you the locally constant algebras of order, a power of the period they completely capture. Okay. So that's the good, the bad. So we would like this. I mean, yeah, we saw all these results were like, wow, maybe this holds like all the time. Like what? And so I feel like there's something missing from my slide. Oh, yeah, here. So in joint work with Croix and Philippi Volak, we've tried to look at this question for curves of hyogenes. And basically, like the paper is just a series of counter examples that show like many things that like any possible statement that you might hope for generalizing from the previous slide doesn't work for curves. So I mean, maybe there's a different way to do it or like that these these subgroups that capture should depend on more than the genus and the degree or in some subtle way, but we could not figure out how to do it. So I think it's it's it's hard. Particularly when you go to general type varieties, it could be hard to get a handle on on these things. Okay. And yeah, I'll just I'll just put up one one slide. So I have a bunch of open questions. I forget from David Harvey's things. Maybe these are like single lightning bolts that are open, but he hasn't thought very hard about them. Am I remembering correctly? So that's what these these are for. So I'll post the slides on the website. We won't get to look at all of them. But I think there's a few other cases that seem tractable that you could ask to what to what extent this behavior extends. Yeah, so I will just go to my summary slide. And end there. Okay. Okay, thank you. Are there more questions for Bianca? Yes, it is Brouwerman. I don't Oh, yeah, sorry. The question was if you have this one of the first examples, counter examples to the local to global principle is this curve three x cubed plus four y cubed plus five z cubed equals zero, do I say correctly? And whether that is a Brouwerman and obstruction. And the answer is that it is a Brouwerman and obstruction. But I do not off the top my head. Remember what the Brouwer class is or even what reference explains it? Oh, Monon's paper. Okay, that was one of the three on the bottom. Great. Yeah. So but does it does Monon's paper write down the Brouwer class or just show that it can be realized as a Brouwer Monon obstruction from theoretically? Okay, right, right. So for as I said for torsors under elliptic curves, we know that Brouwer Monon is the only obstruction assuming finiteness of Tate Chevrolet rich groups. So and then Bjorn is telling me that in Monon's paper, he checks that. Well, okay, this example is a Brouwer Monon obstruction. But I think it's not. I don't know if there's had someone has written down like this collection of Brouwer classes gives you the obstruction. Yeah, you're welcome. Yes, question. So the question was, is there an algorithm that's conjecture, but we don't know that it terminates, but it's conjecture that it terminates for computing Brouwer groups or Brouwer Monon obstructions. I don't think there is. I mean to my it's possible that maybe like something alluding to this is in the effectivity papers that I mentioned by Krush and Schinkel, but I have never actually tried to turn those into an algorithm. They're like very intimidating to me, but I think Jen Berg did you do this for your like sort of made it algorithmic in an example. I think it's pretty far from a general algorithm that is expected to terminate, but we don't know. Yeah, I think it's a different situation. But so someone else can correct me if I'm wrong, but that is if there is an algorithm that is conjecture that conjecture to terminate, it's at least not very easily implementable because I don't know what it is. Is there another hand? Oh, yes, here. So the question was, if I take a Brouwer class that comes from an Azumai algebra, does it help in any way to do any concrete computations? So for me, when I have computed it, the best way, like the way that I typically do is I realize, try to realize the Brouwer class as a central simple algebra over the function field. I guess if you knew that you had it already as something that like, that honestly lives over all of x, so that might be a benefit for an Azumai algebra. Then you can really sort of evaluate, you just take the fiber at any local point to do the evaluation, whereas if you're given a representative over the Brouwer, as a central simple algebra over the function field, there are necessarily going to be, there's necessarily like a Zariski closed subset where you can't evaluate. But since it's locally constant, that doesn't cause a problem, but I think that would be the advantage for using a more global object. Alright, if there are no further questions, let's thank Bianca again.