 So we're very happy to have the last lecture in Bianca Bray's series on Rational Points on Varieties and the Brayur Monument of Structure. Okay, thank you again for the introduction. Is the mic okay in the back? Can you hear? Great. All right, and thank you all of you for attending for the third lecture and I feel like you should give yourselves all a pat on the back, especially the grad students who have been here all three weeks making it through all the courses. Okay, so if this is my first lecture you're attending I'm just going to repeat that the work that I did that's appearing in this talk and also my work preparing the talks was completed on the lands of the Coast Salish, Duwamish, Stela-Guamish, and Soquamish nations and we're currently on the lands of the Isshashoni and the Yut nations. Okay, so in lecture one we started with this fundamental question if you're given a variety how do you determine the set of rational points and we said okay well we could search we just do a naive point search but at some point you want to know that you have found them all so we started considering the extreme case where you're searching and searching and you don't find any and you want a way to prove that there's no rational points. And so over the last two lectures we learned about the Brouwer-Modern obstruction and how you could with enough effort and luck compute the Brouwer group and the Brouwer-Modern obstruction so say we've done that and we determine that the Brouwer set is empty. Okay so where do we go from there? Well I mean depending on how you stumbled across this variety that might be the end of the story for you. You might have wanted to know are there any elliptic curves over Q with this Galois representation and you computed the set of points and that tells you no and then you can plug that into whatever you needed for and continue with your progress. But you might have wanted to know the rational points because you were just interested generally in the arithmetic of this variety X. And well one thing we know from varieties is that they're not just their set of Q rational points particularly when the set is empty. There's more information contained in X than the set of Q rational points and so if you are interested in the arithmetic of X you really want to understand all of the arithmetic of X. So that includes knowing sort of okay ideally in my dream scenario where I can solve every math problem that ever existed I want to know all of the Q bar points and the Galois actions on them and what's happening and what they look like. Okay so what I want to think about today is a question in a different direction which is can we leverage the information that we've learned about X over our ground field and try to extract some information about what's happening over extensions. Okay so we can't if the set of points is empty you won't be able to know well there's a lot you might not be able to know but can we can we leverage the work we've already done because remember not only do we know the set is empty we prove that it's empty by computing arithmetic information about X. We computed the set of idyllic points on X. We computed the Brouwer group. We computed the Brouwer-Mannen set. So in discovering that the set was empty we learned some information already and I want to see whether we can we can leverage that information. Okay and so as I've said through the lectures the set of rational points so over a global field this is a fundamentally difficult problem and so it's very hard for us to study that problem directly so really what I mean is can we use what we know about the Brouwer set over K to study what happens for the Brouwer group over extensions. This gives us like some approximate information on what's happening with the global points. It's not the whole story but it gets us started. Okay so what we saw last lecture is this philosophy that Brouwer classes really want to obstruct rational points that if you're looking at a family of varieties over some moduli space or some parameter space that parameterizes them if you can have non-trivial Brouwer classes then we expect that across that family there should be members in that family where they obstruct. So here we're looking at a slightly different situation. We have a fixed variety. We're not varying over moduli or varying over parameter space. We have our our fixed chosen X that we have under a microscope and are trying to study and then we want to know what's happening as we vary extensions. So as we go up from K to bigger extensions what is happening to the Brouwer classes. So the results that I mentioned I mean you can maybe view them in this context but they were they were really sort of ready-made built for this context where you're varying in moduli rather than varying the field. So now I want to look at this philosophy again and see okay does it still hold when we vary over extensions and if it does in what ways does it hold? Okay yeah so that's that's what I just said. Okay so this is a more recent line of inquiries studying what happens over extensions now that we have sort of are more skilled more experienced about looking over a fixed field. Now we have now where we're starting to look at this question of what happens when you're varying an extension. So the results are newer but we do still have some results that show that this philosophy does still hold. So there's a recent paper of Liang and the way I've stated it it's using a result of Kola Talan-Sansuk and Swinerton-Dyer from 87 and this says that if you have any number field K so you just fix your field so maybe the particular variety that you are looking at this doesn't hold true for but there is a variety a Châtelet service which was featured in the in the exercises from last yesterday's Tuesday's lectures. So there's this Châtelet service over that field and a quadratic extension that over K the Brouwer elements don't do anything but over this extension they carve out a proper subset. Okay so even if Brouwer classes don't do anything over K when you go up sort of gives them more power that they can't obstruct. Yes oh sorry yes this last one I see you know this part this should be V of AL apologies copy and paste error thanks for catching it okay any other questions yes so Châtelet service is a it's just a particular type of surface on the next very next slide I'll write an equation for it so it was in the X the very very long probably intimidating exercise from lecture two yesterday so it's a certain type of conic bundle so it's a surface that maps to P1 where the fibers are all conics and there's four geometric singular fibers and it's even a little bit more special than just that general case but it's a type of surface where we understand the arithmetic very well so in this paper from 87 Koli-Telen-Sensuke and Swinerton-Dyer proved that the Brouwer Monon is is the only obstruction to the existence of rational points for for these types of services they actually proved the stronger statement that the K rational points are adelically dense in the Brouwer set so since we we understand so much about these services they're often a good test case for our results more questions okay I'm gonna go a little bit farther but I'm gonna pause longer next time so be prepared okay so another result in this direction is very recent just on the archive this summer by Masahiro Nakahara and Sam Rovin so Nakahara appeared at the end of last talk when we were talking about these capturing Brouwer Monon obstructions he's my postdoc he's on the job market in the fall so so this is an example of a Châtelet service so they study these Châtelet services and now remember the last result of Liang you fixed a K and then he found a Châtelet service so here we start with our Châtelet service and I'm gonna I will say what the condition is but sort of for most Châtelet services you always can find an extension where the Brouwer set is strictly smaller so it might not happen over your ground field but typically okay so one of the conditions will imply that the Brouwer group is non-trivial that's that's a necessary thing but as we go up the tower you should expect that that Brouwer classes can cause obstructions so if you're interested this is the full condition so F is going to be the splitting field of this polynomial P of t and if there's a place over that splitting field where either this a is negative oh there we go or if it's a finite place a is a non-square and your Châtelet service has bad reduction then you will always have an extension like this okay okay so just as we saw last time because these Brouwer classes they just like want to obstruct if there's possibly any possible way for them to do so this is again too strong of a question that we're asking so we should ask for a little bit less control about not describing the entire Brouwer set as we go over all the extension that seems like a lot of information but can we get can we ask for something a little bit less and so following the same theme from Tuesday it's sort of a much stronger thing to ask when the Brouwer elements completely eliminate all the idyllic points so instead of asking to describe the Brouwer set over all possible extensions which seems like a daunting task even to write down the answer for let's try to understand the extensions where the Brouwer set is okay either trivial either empty or not empty you know you could ask for either one and we'll we'll consider both okay so this is the the motivating question that I want to consider today any questions for me I promised I would pause longer so I have to follow through yes yeah that will be one of our one of our questions so the Brouwer group can change as you go over an extension so we'll see that as I said this is a newer line of inquiry so we're at earlier stages so most of the results are where the Brouwer Brouwer group doesn't change too much or this Brouwer quotient doesn't change too much as you go over extensions but yeah that's part of part of the thing we have to figure out right so the the question was okay there were two parts the first one that I answered was can the Brouwer group change the Brouwer quotient as you go up an extensions which I said that can happen and then the second part was so you could also ask a different question a slightly weaker question instead of just asking about the idyllic points orthogonal to the entire Brouwer group you could ask just if they're orthogonal to the image of the Brouwer group that you get over K the image of the restriction map and that that is a very good question and indeed a first first step question and that will come up in the talk so great yes no I think it can get okay I don't know if there's an example that's written down that shows that it gets bigger and smaller and bigger and smaller but that theoretically I would expect that to happen so we saw that there are these we have this filtration of the Brouwer group we have Brouwer one my Brouwer zero which is an algebraic Brouwer Brouwer group that's killed over the algebraic closure and so when that is finite that part you definitely expect to get often smaller over extensions you can take the extensions that killed that that whole Brouwer group but this the transcendental part Brouwer my Brouwer one those are really we should think of those elements as those are elements that want to live over the algebraic closure and then sometimes you're lucky and they descend to your ground field so certainly as you go up the tower you expect more and more of those to to appear so yeah it's not monotone great questions yes X here is defined over a little K oh yeah sorry I yes I like decided to switch halfway through from K to Q somewhere and I didn't check check all the things so yeah apologies so here this here you could either think of this as L over over K or you can just set K equal to Q and then everything will be fine thanks for catching up okay so what is the first step into this direction well there's this very nice lemma which in this generality was observed by Wittenberg so in a joint paper with Croix which will come up again we had a couple of consequences of this lemma which we proved in different ways and we sent our paper to Olivier and he said isn't this all a consequence of this very nice more general yeah more general statement so and that is indeed true so if you have any extension and over that extension you take a subgroup of the Broward group the K points over that extension that are orthogonal to this B it always contains a delict points over your ground field that are orthogonal to the co-restriction of those Broward classes okay and so this has two really important consequences one if you take B to be the full Broward group over an extension then you know that if you have non-empty Broward set over your ground field it can never become empty over an extension okay and this should like fit with what we we hope should happen and it does indeed happen but it also tells us because of restriction co-restriction if you look at torsion that has a particular order then any a delict point over your ground field over an extension will be orthogonal to those elements obtained from your ground field of a certain order so I think this already shows like there should be some things that we can say right the functoriality of the Broward group and the way it's defined does give us some things sort of just for free when we know enough information about what's happening over a little K we get information about what's happening over some extensions why does that keep losing this okay okay so let me just explain this lemma because it is yeah very nicely functorial so if you have alpha in the Broward group of K and you have this local point maybe I should have labeled it X V but just for a particular place V so then we can just pull back that map and we can view it as giving us a point over sort of every every place lying over V okay and then it's just sort of purely functorial statement that you'll find in this recent Broward group book of Cola to Lenin score Bogatov that if I pull back along X the co-restriction of alpha that's the same as if I first pulled back along Y and then co-restricted down okay and so sort of this should make you feel like oh yeah surely this can prove the lemma above and indeed you just sort of trace through the pairing and everything works out okay so that's great so as a corollary from this from the first in particular statement that I said if you have non-empty Broward set over your ground field oh sorry what should I oh my goodness this has many typos okay so then the set of things where this is non-empty let's just do this L over K okay you know what I think I'm sure that I copy and pasted this set onto many slides so we're just gonna set K equal to Q that's okay yeah and this is supposed to be okay did something this is supposed to be a not equals so if you have non-empty Broward set over your ground field then every extension has non-empty Broward set so you just like we just completely answer this question it's just all field extension okay so from this clearly the case then we need to do any work in is the case when you have a Broward monon obstruction over your ground field okay and so if you have a Broward monon obstruction over your ground field really breaks into sort of two questions we're asking over which extensions must this Broward monon obstruction persist so when I say that I'm thinking that this means that X of AL Broward still remains empty so that's the Broward monon obstruction persisting and over which extensions can you must the Broward monon obstruction vanish okay so I'm gonna start with the second question when it vanishes and those should be extensions well where we can possibly gain points okay so first let's look at this one of the other parts of the lemma that came up which at the end told us that okay if I have a delik points over my ground field then those are orthogonal to a certain subgroup of the Broward group upstairs they're orthogonal to the a delik points that are the restriction of those elements who have order dividing the degree of the extension okay so if I have a delik points and if this subgroup captures the Broward monon obstruction so meaning that if the Broward set is is empty then I can be already detected here then that immediately tells us that the Broward set upstairs is not empty okay so so what we want to try to do is understand okay are there any cases where we know that this happens this restriction captures the Broward monon obstruction okay and so what I want to do just first at the beginning is just see okay what are some like known results that are already out there that we can apply to this problem and see what's happening okay so this is the the situation that we we know that we're in so here's an example corollary that we get from this statement just assembling sort of basic things that are already known so if you have a conic bundle over projective space so that where the fibers are conics and the generic fiber is smooth and you assume has a delik points well then you don't get it for free for every every extension but if you avoid some fixed finite extension so that's so you avoid this fixed finite extension so then for all even-degree extensions which are linearly disjoint from this fixed one you immediately get that you have non-empty Broward set and for conic bundles like this we it's conjectured that Broward monon is the only obstruction and there are conditional results in this direction so this should give us an idea where we get points yes question I mean that it has a delik points this means x of ak is non-empty great question yeah so that that question was what is locally soluble really I should write everywhere locally soluble but it didn't fit on the line yes sorry what do I have to join to a small k2 okay so really what's happening here is if you're away from this finite extension I'm trying to think whether I should really put p1 here if you're away from this finite extension then the restriction map generates the Broward quotient over the extension so really what is happening is the question is whether we get this condition and if what's happening in this case is that if you have L over k linearly disjoint then this restriction actually generates the Broward quotient and I am entirely confident of that if the n is 1 and I'm like having trouble thinking through the whole argument when n is higher dimensional but that's what goes into this argument okay so let me just first show you like a few more results in this direction and then we'll come back to whether we should think of this as a surprising statement or not okay but I'm just going to show you that it has very similar forms okay now instead of a conic bundle let me take a locally soluble cubic surface then again I have this finite extension that I have to avoid and then for all extensions that are linearly disjoint for this and where three divides the degree of the extension then I get that the Broward set is non-empty and this is just exactly applying the Swinerton Dyer result that I mentioned on Tuesday that the retortion of the Broward group captures the Broward monon obstruction and again if you're disjoint from this fixed finite extension this restriction will generate the retortion of the Broward quotient over over the extension yes oh great yes the existence of big k is constructed so in this case k you can take k to be you can take it to be smaller than this but at least you can take k to be the field of definition of the 27 lines on the cubic surface so like it's not like Q adjoint squared 2 but you have to do some work to get it but it's completely constructive yes okay and then another one in this direction if we have a locally soluble quartic del petso surface then again from well I guess this isn't just from the classification of Broward groups we know the Broward quotient is completely too torsion and so again away from this fixed finite extension if L over k is linearly disjoint then you get this non-empty for every even degree extension you get this non-empty Broward set okay so for certain types of varieties so this often works if the Picard group is torsion free and if the Broward group is completely algebraic and either the m torsion is known to generate the Broward quotient or you have some classification result that shows that it captures but like there's many things known in the literature that you could assemble to prove a statement like this where you fill in the different types of things okay so what I want to spend a little bit of time on is just talking about whether whether this is surprising or not and it it sort of depends on this the strength of what you got for the m so I'm going to do two extreme cases I'm going to do the conic bundle case and then the the quartic del petso services case okay so for this conic bundle we're saying something about what happens with even degree extensions so it's not so surprising that we should get a lot of points over even degree extensions right because I have a bunch of conics and conics have a bunch of points over even degree extensions but what kind of extensions do we get and how much control do we have over them so if you take your point on projective space then the fiber is a conic and you can use this well basically we can completely describe if you have a given conic over which even degree extensions you get points you just have to have a certain condition at finitely many places so using using the information we know about conics then you can show that if you have adelic points over your ground field you can find points over every even degree extension that approximates a finite set of local conditions okay so for any finite set of places you can give me whatever even degree extension you want and then I can find I can find a global point over over that extension okay this is not obvious from what I've just said you need to do a little bit more work but it's not it's not too too intense so on the one hand this second argument is better because we get actual global points and the first argument we only get a non-empty Brower set but the second argument even degree extensions approximating any finite set of local conditions is a smaller set a possibly smaller set of fields than the ones we get in the top in the top it really says you should get every even degree extension that's linearly disjoint from this fixed one okay so it's not too surprising but it's a little bit more than you can get through other methods okay now let's look at the Cortic del petro service case so this one is a little bit different so Cortic del petro services are intersections of two quadrics in p4 so it's very easy to get degree four points but it's not very obvious that you should get degree two points and in fact if you work over arbitrary fields there's well you can show there's no general construction of quadratic points because you can write down del petro surfaces Cortic del petro surfaces over like fields of higher transcendence degree that have no points over a quadratic extension so there's no like easy geometric way to get these quadratic points so already this is telling us something a little bit more than we might expect however if you have a delic points then you can use those a delic points to view this del petro surface as a double cover as birational to a double cover of p2 and then that gives us a way to make quadratic points right I just take a rational point on the base and I take this degree to cover this is very similar to the ideas that we saw coming up in Wittenberg's lectures I don't understand why this is happening okay hold these in okay so and this is over an odd degree extension then you get this to be this double cover but anybody can somebody just check that that plug is all the way and maybe something got loosened okay it's possible we're gonna be switching to a board talk okay maybe I should just start doing that and someone could potentially help me get get something to show up okay let me see where I was okay so I think I was saying so for a quartic del petro surface if it's everywhere locally soluble you do have some construction of quadratic points but you don't have much control over what points you're getting so that the already the example corollary is is giving us something much more than we know in other cases okay but still the existence of a delic points like over your ground field is helping us a lot in these situations so now I want to turn to this question what if you have no a delic points over your ground field this is the question okay so if so to show that you have a non-empty browser set you need to construct points right this is like the fundamental difficulty in showing that we have rational points is like we often can't just conjure them out of thin air you have to like do something with them and especially to show that I have non-empty browser set I need to exhibit some a delic point over L thank you that has that is orthogonal to the Brower group and if you have no a delic points over your ground field you just have less points to start with that you're given for free to try to like modify and work with okay and also when you're doing this computing the Brower group is still hard okay so I have sort of like brushed this under the rug but I didn't suddenly have an epiphany since Tuesday that makes me like a much much better at computing these Brower groups and Brower quotients so when I'm treating these questions I'm just gonna work in the in the cases where we where we understand the Brower group better so where our lack of general method is not hampering us we can just use one of the many papers that I exhibited okay so okay so I'm just gonna show some results in this direction these are harder to prove so I'm not gonna give a sketch of the proof but just to show you what we can do okay so Sam rovin who was my PhD student he showed that if you have a conic bundle then you can show regardless of whether it has a delic points over your ground field or not you can show that for all quadratic extensions the Brower set that you get from the Brower group of K so that was from Zev's question earlier just that part that's non-empty as soon as you have a delic points over L okay so there let the Brower elements from K cannot cause any obstructions over these extensions okay and if you have a particular type of conic bundle the special one known as a Châtelet service then we can actually do even better it's not just quadratic extensions it's all extensions of even degree so the Brower group from K cannot obstruct cannot obstruct a delic points and based on what I said so for Châtelet services the Brower group is purely algebraic and so wait something okay sorry so navigator this is so weird I'm having the same thing oh here we go okay it's coming so if you want to upgrade this to studying the full Brower set then what you have to do is avoid this fixed finite extension but then you get that actually the whole Brower set can never obstruct a delic points okay so this is much much stronger than the results that I was saying before oh is there a question yes no you don't have equality and we can't hope for and in fact rovin and Nakahar and rovin prove that you often don't have equality so the the Brower set can be strictly smaller and it often will be strictly smaller than the set of a delic points but it can't be completely empty so yeah so sorry when I'm saying it cannot obstruct the a delic points I'm saying it cannot completely obstruct them it can't obstruct the existence of rational points thank you that's a good clarifying question okay so and now wait what is happening let's just skip there we go okay so now let's look at the Cordic del petro surfaces case so remember this is when we have no known construction of having any quadratic points so we shouldn't expect something as strong rovin was proving over all of these quadratic extensions all of these even degree extensions we get we get points here even just getting some quadratic point would be good so first enjoy and work with Crites we prove that locally you always get quadratic points so over any local field there's always a quadratic extension where you obtain points and we can do it almost always over case so there's one case that we're can't do this is even more restrictive than we need but I didn't want to write the exact assumption but we do show for many types of these families there is a quadratic extension where you have non-empty Brower set okay and so we should should have global points over that extension okay so one thing that I think this these results lead to which I would really like to know the answer so one is are there always quadratic extensions so here we've done extensive computations in the case that we can't cover to try to like example searching to try to find one where you don't get these quadratic points we have yet to find one so I think maybe you do always get quadratic points over global fields but we haven't proved that yet and then one thing that's unclear to me is what is special about del petso surfaces because we really we should expect only quartic points and this is just wild speculation on this question just to like throw people out there I would guess there are families of varieties where the Brower quotient is completely too torsion and but you still can't find quadratic points but maybe maybe not so I think this hasn't really been looked at because the examples that we know of where the Brower quotient is completely too torsion are conic bundles where you definitely always get quadratic points and quartic del pets of surfaces where it's unclear so I think it's just we haven't looked further okay any questions so far yes yeah so we can actually prove something stronger than this we can prove that if there is not a point over the unique unramified quadratic extension then you get a point over every single ramified quadratic extension and experimentally it seems that you you can fail to have a quadratic point failed to have point over at most one quadratic extension but I don't know how to prove how to prove that in general well we don't yeah more questions okay so so that's over which extensions must the Brower mountain obstruction and vanish and just in the last a few minutes I just want to say a few words about which extensions the Brower mountain obstruction must persist which again is this question that if you know that you have empty Brower set over your ground field then you still have empty Brower set over your extension now thinking about this intuitively this should be a much more difficult property to hold right normally the way we make points is by taking extensions so as we take extensions we expect there to be global points over those extensions so then that would give us non-empty Brower set so it should be harder to guarantee that the Brower mountain obstruction persists and probably it's only reasonable to expect persistence of the Brower mountain obstruction for geometrically simple varieties for varieties that are well-behaved so the Brower mountain obstruction persisting should be telling us something about about the relationship between the property of having rational points and having zero cycles of degree one and for general varieties those concepts should be far apart but sometimes sometimes they're more closely related okay so what can we say in this case also what is one very geometrically simple variety well that's quadrics and for Brower groups of quadrics they're just trivial so the Brower mountain obstruction doesn't even tell you anything but Springer's theorem tells us that quadrics have points over your ground field if and only if they have them over an odd degree extension so then for free from this you get okay well you get that you have a local point if and only if you have a local point over some odd degree extension and so from that you can prove that you have a delik points over K if and only if you have a delik points over some odd degree extension and since the Brower set is non-trivial so this is like a check mark example okay well revealed the punchline but okay so quadrics are good for Cordic del petso services we get the same property about that they have a point if and only if they have it over an odd degree extension this is a you have to add something ingredient into Springer's theorem but together with a theorem due to Amar and Broomer we know that this statement is true we have a rational point if and only if you have a point over some odd degree extension and so then you can ask does that imply persistence of the Brower mountain obstruction and in this case it doesn't directly follow from Springer and Amar Broomer because the Brower group can be non-trivial but this does follow from work of Kola-Telen and Coray and the classification of the Brower group by Swinerton-Dyer okay so that there you get it still persists so okay what's another geometrically simple variety so we could look at cuba cubic surfaces so in this case we there's a conjecture of a similar property from Springer's theorem but it is now a conjecture no longer a theorem so this is the conjecture of castles and Swinerton Dyer that a cubic surface should have a point if and only if it has a point over some extension of degree coprime to three oh yeah I guess okay I said it as a existence but then okay I think what I wrote is correct so then you can ask the same thing about the Brower set so does the the Brower mountain obstruction persist over extensions of degree coprime to three and over local fields Coray proved this in his thesis that the castles and Swinerton-Dyer conjecture does hold for local field so we immediately from this we know that you have a delict points over K if and only if you have a delict points over some extension of degree coprime to three and then in joint work with Carlos Rivera we proved that the Brower mountain obstruction does persist for this extension so there are some cases where you can prove the Brower mountain obstruction persists I think it's it's less likely that it holds in general but it's still nice to know these cases okay so I'll just end with sort of summarizing what I think of as the main themes that I was trying to get across so the arithmetic of X when we want to understand the arithmetic of X it's much more than just understanding the points over the ground field really we'd like to know something what that's happening as you go over extensions and then so far what I've shown but as I've said this is a newer line of inquiry they may be suggest that as you go up in extensions it's easy for the Brower set to be strictly smaller than the set of a delict points but it's harder for it to be cut all the way down to the empty set okay this is like a very soft soft speculation of what is possibly happening but so far I mean as I've said all the things that I've shown you all the results that I've shown you they're only for geometrically rational surfaces so this is a very when you think of the all possible varieties this is just giving you a very small slice so we don't really know what's happening so what I just leave you with is we should explore this more and I hope some of you will take this up okay thank you very much