 third lecture around the inverse Galois problem. Thank you. Can you hear me in the back? OK, thank you. So let me recall where we are. So in the first lecture, we talked about the quotient an mod g. And I've explained that when this is rational, you get a positive solution to the inverse Galois problem. But I've also explained that this is not always rational. So then in the second lecture, we discussed the rigidity method, which is another way to apply Hilbert's irreducibility theorem to construct extensions with given Galois group. So this applies in many cases. But it's still a limited scope of application. So for example, it doesn't apply to p-groups. I mean, it applies to many simple groups. But such things as p-groups are generally outside of the scope of the rigidity method. So OK, and what was it about? We were looking for families over p1. So because of the versatility property I mentioned, looking for families over p1 is the same as looking for curves p1 inside of this space. So if this is not rational, maybe we can find a rational curve in it. And if we cannot find a rational curve in it, then maybe we can find rational points directly. So that will be the theme of today's lecture. So find interesting rational points in this question. OK, so actually, we're not going to work with this variety. It's slightly more convenient to work with another one that I'm going to introduce now. But it will make no difference. It's just more convenient. So let's fix a finite group G. So recall here, I chose an embedding of G into a symmetric group. And I let it act on the A and by permuting the coordinates. Now let me do something slightly different. I embed G into SNN of k. So I work over some field k, so number field. So you embed G into SNN of k in some way. And I mean, you can always do this. First, embed G into SN, for example. For some n, embed SN into some SNN of k. And then let's consider G as a subgroup of SNN. So let's consider SNN as an algebraic group of k. So that's an algebraic variety with a group law. And I view G as a subgroup. And I consider the quotient of the group by the subgroup. OK, so if you like, it's the quotient of SNN by the right action of G by multiplication. And why is it better than this one? Well, because the action is free, so this is smooth. Here you have singularities. So it's more convenient to work with this. So here we have SNN goes to SNN mod G. So this is smooth. So this is a g-torsor. And it turns out to be a versatile g-torsor just in the same way that we got a versatile g-torsor with affine space. I emphasize these are varieties of a k. So we're going to work with this. And constructing rational points here is really the same as constructing rational points there. I mean, in technical terms, these varieties are stably barational equivalent. This is not so important. And it's explained in the exercises. So let me recall the relationship between the rationality of this quotient and Grunwald's problem. So recall. So sorry, I will go here. So if SNN mod G is rational, then it satisfies weak approximation. So I recall weak approximation means that if you are given a finite set of places and local points at these places, you can find a rational point, which is as close as you want to these local points at these places. OK, so because it's true for a1, it's easy to deduce it's true for rational varieties. And this, remember, implied the positive solution to Grunwald's problem for g over k for all sets of places S. And just to refresh your memory, so this is asking for the existence of extensions of k with group g, sorry, Galois extensions of k with group g, whose completions at the places of S are prescribed. And just to recall how this works, the local prescribed extensions give you kv points on this quotient by versatility. Then you approximate them. And in this way, you get a rational point, which by IKDAL, you can assume to have a connected fiber. So this gives you a Galois extension of k with the prescribed completions. OK, so it's actually not difficult to see that this implication is in equivalence. OK, so that's the situation. And I've also explained that this fails, for example, for z mod 8z over q. So now, after Bianca's lecture, the natural question is, what is this failure of weak approximation on this variety? Is it a Brahmanian obstruction? Is this a Brahmanian obstruction to weak approximation? And, well, the answer is yes. So to state this, let me introduce some notation. So unfortunately, I cannot use the same notation as Bianca did because her varieties were proper. But these are not proper varieties. And I don't want to introduce smooth compactifications. So let me introduce slightly different notation, but it's the same thing. So x of variety, I will denote by x of k omega, the product of the local points, product of x of kv. So this is exactly what Bianca considered, except she wrote x of ak, the adelic points. It's the same when x is proper. OK, and so then she considered the Brouwer group. And here, the same, it's not the Brouwer group of x, which is going to be interesting, but the Brouwer group of a smooth, projective variety by Russian equivalent to x. So this is the so-called Unramified Brouwer group, denoted Brouwer nr. So this is going to be the Brouwer group of any smooth, proper variety containing x. Any smooth, proper variety containing x as an open subset. So it's an untrouble fact, but it's true that it does not depend on the variety you choose. It really only depends on x. OK, and so then Bianca explained that you have the Bramann in set. So yes? No, no, no. Yes, you can. But I mean, we're in character 6-0 anyway. If you're in characteristic p, you can do other things, but you have to be careful with the p-primary torsion subgroup, where p is the characteristic. So basically, if you look only at what's prime to p, you can do, as we say. OK, so recall. We have these inclusions, the rational points, contained in the Bramann in set, x of k omega, exponents un-ramified, contained in x of k omega. OK, and so recall, this is closed. So the closure of the rational points in the set of collections of local points has to be contained in there. And so if this inclusion on the right is strict, then weak approximation cannot hold. This is a Bramann in obstruction to weak approximation. And so the question here is, take x equal SNN mod g, is this inclusion strict or is it not? So the answer is yes. And more generally, here's the theorem. And please don't hesitate to ask any question you might have. And let me know, because I can't see you. So here's theorem 0. So this is due, I think, to Voskresensky and Sonsuq. So you take x equal SNN mod g. So we are working over a field, a number field, over k, a number field. And g is a finite, a billion group. OK, then the rational points are dense in the Bramann in set. So this is a situation where the Bramann in obstruction controls everything. You can do weak approximation of a collection of local points as soon as they satisfy the Bramann in conditions. So this doesn't directly tell you that you have a Bramann in obstruction, but because you know that weak approximation fails and because you have this density, the only possibility is that there is a Bramann in obstruction. OK, so more generally, we hope that such a density statement is true in much greater generality. So Bianca alluded to the case of geometrically rational surfaces. So let me state the general conjecture, conjecture 1. So due to Colliot-Delen, so this should work for so-called rationally connected varieties. So k is a number field for all the lecture. I will not repeat it. So x is smooth rationally connected, which I will abbreviate by RC, variety of k. Then the same density statement should hold. It's dense. So in other words, the Bramann in obstruction tells you about the existence and weak approximation of rational points. So OK, what is rationally connected? So it's a purely geometric notion. So you can extend the scalars to the complex numbers if you like. So x rationally connected. So let's say you embed k into c. So this means that if you take two general points, you can connect them by a rational curve over c. Two general c points can be connected by a rational curve over c, a rational curve, the image of a p1, some open in p1. So for example, if x is geometrically rational, that's true because if it's geometrically rational, an open will be an open in p1. And in p1, you can use lines to connect two points. So that's very easy. But this notion is much better behaved than rationality. And so for example, also our SLN mod g is rationally connected because it's unirrational. So unirrational means it's dominated by a rational variety. And SLN itself is rational. And again, if you have two complex points, you can leave them. And because it's rational, use lines in the projective space to connect them. OK, so this conjecture, as Bianca mentioned, is very much open even for surfaces. But still, there are some significant positive results. So just to give an example, let me say that this is known to be true. So that's a theorem of Borovoy. True if x is a homogeneous space of a linear group, linear algebraic group with connected stabilizers. So what does this mean? So you have a linear algebraic group which is acting on your variety. And homogeneous space means that the action is transitive on the k bar points. And OK, but you require that the stabilizers be connected. So for example, SLN mod g is a homogeneous space of SLN. SLN acts on the left. But the stabilizers, they are isomorphic to g, which is a finite group. It's not connected at all. So it doesn't apply to this. OK, but this is still quite a large class of varieties. So let's come back to our motivation. So this conjecture is directly related to Grunwald's problem because of the same situation as in the first lecture. So namely, there's this theorem of Ikadal, which is telling us that the Hilbert's irreducibility theorem is true not only over P1 or rational varieties, but also over rationally connected varieties that satisfy the conjecture. So let me state it like this. Take a connected g torso. Then if you look, sorry, an x is a smooth rationally connected variety satisfying conjecture one. Satisfying conjecture one. Then not only can you find rational points approximating any points in the Brahmanian set, but you can even find one such that the fiber of pi is connected. So if you take the set of rational points such that the fiber of pi is connected, then even this is still dense in the Brahmanian set. Sorry, k omega brah nrm phi x. OK, so what does this tell us? Well, it tells us that if you want to solve Grunwald's problem, what you really have to do is to prove this conjecture for SNN mod g and then understand the Brahmanian abstraction. So consequence, if you have conjecture one for SNN mod g and you can compute the Brahmanian set, then you get a complete answer to Grunwald's problem for g over k. You get a complete answer to Grunwald's problem. Again, because your local prescribed extension give you local points here. And then the only question you have is can you complete this collection of local points at the other places in such a way that the resulting family is in the Brahmanian set? If you can do it, then you have a positive answer. And otherwise, you don't. So in a sense, theorem 0 in this way is a reformulation of the Grunwald Wang theorem that tells the complete answer to Grunwald's problem for a billion groups. It's a much cleaner statement. But on the other hand, you still have to compute the Brahmanian set. And I mean, that's where the part which is not so clean goes in the computation of this set in this case. But it can be done in this way. You can recover the statement of the Grunwald Wang theorem. So in practice, if you have another group and you know the conjecture, well, you still have to compute this Brahmanian set. And as Bianca explained, this can be hard. So well, the first thing you would do is to compute the Brouwer group, the Unramified Brouwer Group here. And well, in general, this is hard. For this variety, SNN mod G, at least we have formula that we can try to apply. So there's the part of the Brouwer group that survives over K bar, for which there's a formula due to Bogomolov. And there's the algebraic part of the Brouwer group, the part which dies over K bar, for which there's another formula due to Arari. So in practice, usually you can use them to compute the Unramified Brouwer group in this case. Still, this doesn't tell you what the Brahmanian set is. So this formula only gives you the structure of the group to compute the Brahmanian sets. You need representatives, and that's not so easy. But still, in some cases, it gives you the answer. For example, it can be that you find that the Brouwer group here is just reduced to the constant classes. So then there is no Brahmanian abstraction at all. And then you have a complete positive answer to Grunwald's problem without excluding any place. So this happens. So there's an example like this in the exercises, where you apply the two formula and you find this conclusion. So in the lecture, the only thing I will say is there's one general fact that gives you some information about Grunwald's problem starting from this. It's a theorem due to Luchini-Arteche theorem, which says if you have conjecture 1 for SNN mod G, then you have a positive solution to Grunwald's problem for G as soon as you avoid the places that divide the order of G. Positive solution to Grunwald's problem for G, for any S that does not contain places dividing the order of G. So this is the hope that I formulated at the beginning of the first lecture. But we hope that this is always true. So this is the explanation why. So the proof of this theorem is really a study of the evaluation of the classes in the Brouwer group of such a space at the local points. And well, he proves that nothing happens at the places that don't divide the order of G. OK. So the theorem that I wanted to state in this lecture is a positive result for conjecture 1. Sorry, so this was theorem 3. This was theorem 2. And this is theorem 4, which I proved with Jonathan Harpass, which is that conjecture 1 holds for SNN mod G. If G is nilpotent, or even supersolvable, if G is, say, supersolvable. So let me remind you what this is. So supersolvable is in between nilpotent and solvable. It means there is a filtration, G equals G1, G2, and so on, Gn, such that each GI is normal in G, and the quotients are cyclic, normal in G, and the GI mod GI plus 1 are cyclic. So this includes nilpotent groups, so all P groups, for example. And so as I've explained, this gives a positive answer to Grunwald's problem, at least if you exclude these places, the places that divide the order of G. And in theory, if you are able to compute the Brahman in abstraction, you can get more precise answers. So in a sense, this is the best approximation you can get to the Hilbert-Knotup method. I mean, they wanted to construct rational points on the An mod G or SNN mod G by requiring that this be rational. OK, it's not rational, but you can still construct enough rational points to make the method work, in a sense. So OK, is there any question about this? OK, if there's none, so now what I will try to do is to explain how you would go about proving such things, what type of tools you can use. And that's what I will do in the remainder of this talk. So OK, so the main tool, one of the main tools is so-called Descent theory, so I will try to explain this. So for this, I need to introduce the notion of torsos in a slightly more general situation. So let X be a smooth variety, and G an algebraic group. I'm not assuming that it's a finite group anymore, an algebraic group of a K. So it could be a finite group, but it need not be. OK, and K is a field of characteristic 0. So a G torsor over X, the definition is essentially the same as I gave for finite groups. So it is a smooth, surjective map pi from Y to X. Y is some variety of a G. Smooth is telling you that the fibers are smooth. And an action of G on Y, such that this is a covariant, so such that this is a covariant, and such that G acts simply transitively on the fibers at the level of K bar points. Pi is G equivalent. So again, here the action of G on X is trivial. I really mean that it's invariant if you're pre-composed by the action of G. And G of K bar acts simply transitively on the fibers of Y of K bar goes to X of K bar. OK, so when you have this, in particular, X is the question of Y by G. So the notion of torsor like this for algebraic groups is very convenient, for example, for stating the Hilbert 90 theorem. So example, so Hilbert 90. So one reformulation of the Hilbert 90 theorem is telling you that GLN torsors over a field over K are isomorphic to GLN, are trivial. GLN torsors over the points, over K. And are isomorphic to GLN. And SLN torsors over K, that's a consequence. Are isomorphic to SLN. I mean, you can just take it for granted, it's reformulation. So next thing I need to talk about descent is the notion of twist. So definition, so if you take a G torsor over K like this, and sorry, a G torsor over X and a G torsor over K, then it makes sense to twist pi by P. And this means the following, so the twist, which we denote by Y sub P, but sub on the left. Well, that's just you take the product of Y by P, and you mod out by the diagonal action of G. And this still maps to X. I mean, you have the first projection here. And because X is Y mod G, you still get the map to X. Right, so now I have everything I need to state the conjecture of descent in this context. So it doesn't look like it's very much, but it's really the same as descent in the context of elliptic course. It's a generalization. So conjecture 5, descent. So suppose you have a smooth, rationally connected variety over a number field, X, over a number field K. And suppose you have a G, some linear group, some over K, linear algebraic group over K. And suppose you have a G torsor over X, a G torsor. So the conjecture, well, I should maybe call it a hope. If conjecture 1, the rational points are dense in the Brahman insets, holds for Y, and for all of its twists, then it holds for X, for all the twists of Y. So it means for all G torsors over K, P like this. No, so this is a slightly subtle point that I was hoping to avoid. If G is a billion, then yes, they are still G torsors. Otherwise, they are torsors under an inner form of G. But here I'm not using this fact anyway. I'm just looking at them as varieties. OK, so if conjecture 1 holds for all of the twists of Y, then it holds for X. So this is a way to descend conjecture 1. I mean, it can happen, and that's what happens in many situations, that you have your over ITX, and you are able to produce some G torsor over X. Whose geometry is simpler, and whose arithmetic is simpler, and maybe you can actually solve conjecture 1 for Y and for all of its twists, and then you're done. So it's not clear at all a priori why such a thing should be useful. But in practice, it turns out to be. OK, so here's a theorem about descent, which is due to Cooley-Othelan, Sonsuc, Arpas, and myself. Is that this descent procedure, the conjecture 5, holds is true if G is a torus. It's true if G is a torus. So let me recall what the torus is. So it's a commutative algebraic group. So torus means it's a commutative algebraic group, which geometrically over k bar is isomorphic to a product of GM, GM cross GM, cross GM, cross GM. Or if you like, it's the group of homomorphisms from a lattice L to GM. OK, L is Z to the power n, but maybe there's some Galois action on it. OK, so with this, let me show you how using this, you can prove theorem 0. So sketch of proof of theorem 0. So recall theorem 0 says that conjecture 1, density of rational points in the Brahman inset, is true for SLN mod G when G is finite a billion. OK, so this is just a sketch. So first, there's a fact, which is not obvious, but it's not difficult. And it's an exercise in the list of exercises. So let's just take this for granted. Fact that you can find an exact sequence of commutative groups of a k. There exists an exact sequence. 1 g t q 1, where t is a torus over k, and q is also a torus of a special kind. So where q t are a tori over k. And sorry, I have to go there. And q is a so-called quasi-trivial torus. Sorry, and q is quasi-trivial. So what does that mean? Well, it means that if you write q geometrically as home of a lattice to gm, there is a basis of the lattice which is stable under the action of Galois. So that means really q over k bar is home of some permutation module to gm. So for example, gm is quasi-trivial. The product of gm is quasi-trivial. The restrictions are quasi-trivial. OK, let me not insist on this. I'll just tell you why it's useful just for two things, quasi-triviality. Well, first, it implies that q is rational. It's just like gm is rational. And also, q torsors over k over a field are isomorphic to q over a field are isomorphic to q. And this is again Hilbert's theorem 90 in yet another guise. So all of this, if you like, is an exercise. If we take it for granted, it's rather formal. And let me explain how using this, you can prove these statements for SNN mod g. I'm sorry? Yeah, it's the two of them. I mean, you need Shapiro to get to gm, and then you apply Hilbert 90 for gm. Yeah, sorry. Thank you. So OK, so here's what you do. You have SNN mod g, and you consider SNN. So we've embedded our billion group into this torus t. So that takes a product with t. And let me take the quotient by g, by the diagonal action. OK, so I claim, well, the first projection induces a map here. And now, if you think about it, of course, t acts on this by multiplication on this factor on the left, right? You mod it on the right. But there's still an action on the left. And it's obvious that it's going to act simply transitively on the fibers. So this is a torsor. This is a t-torso. OK, so here's our t-torso. And we're going to apply descent to it. So theorem 6 tells us we can do it if you want to prove conjecture 1. For this variety, it's enough to prove it for this one and all of its twists. So apply descent to this. OK, but then why should it be easier for the top space, the top variety? OK, so to understand this better, let's look at the second projection. SLN cross t goes to t. So I mod out by g. This goes to t mod g, which is q. But now, for the very same reason, this is an SLN torsor. You still have an action of SLN on this, on the left. And it's clear that SLN is going to act simply transitively on the fibers here. SLN torsor. OK, but now here's the geometry. It's very simple. We have this base, which is q. And q is a quasi-trivial torus. In particular, it's rational. And here, we have this SLN torsor. And let's look at the generic fiber. It's an SLN torsor over the function field of q. And by Hilbert 90, this is SLN itself. So there are rational points in it, which tells us there are rational sections here. Rational sections. So all in all, this top space is by rational equivalent to q times SLN. So it's rational. SLN cross t mod g is rational. And so it satisfies weak approximation. And conjecture one, for obvious reasons, satisfies conjecture one. OK, so you still need to look at the twist, not just this. But if you look at the twist, take a torsor, say p over spec k, a torsor. If you twist this by p, you will get the same with p instead of t here. So maybe I just replace it. And then everything else is the same. You still get a projection to p mod g. So p mod g is not q, a theory. It's just a q torsor. But q torsors are q. So in fact, you get q. And then exactly the same argument applies. So that's how you do it. So you see, it's a typical example where we don't have like this a natural torsor to apply descent. But it turns out that you can construct one in some way, then apply descent to reduce yourself to another variety whose geometry is simpler. So OK, so here's the actual theorem that is behind the theorem I stated on super solvable groups. So theorem seven, which we proved with unit on our pass. So in fact, it's that the descent method is true, not just when g is a torus, but also when it's a finite super solvable group. Conjecture five is true if g is a finite super solvable group. So this is what we call a super solvable descent. So I'm sorry, I just erased the statement of conjecture five. So let me just repeat it. You start with a torsor over a rationally connected variety, a g torsor where g is a finite super solvable group. If rational points are dense in the Brahman insets for the top space and all of its twists, then the same holds for x. And it's very easy to see that theorem seven implies the conjecture for SLN mod g. So this implies this was theorem four, I think. You just apply it to this torsor, SNN goes to SNN mod g. So this is a g torsor, but the top space is rational. So obviously it satisfies the conjecture. And the twists, well, they are exactly as here. If you twist, you still have the action of SNN on the other side. And the twists are SNN torsors, so they are SNN. So the twists are isomorphic to SNN. So they are rational. So I will not say much about the proof of this. I just want to say that the basic idea starts exactly like this. So in fact, now g is a super solvable group, so we are not going to embed it into a torus. But OK, we have this filtration with cyclic quotients. You take the first quotient, and you embed it into, and you apply this fact to it. So you embed it into a torus with quasi-trivial quotients. And then you do this construction, exactly this, except that, so is there a color? No? So OK, so I'm just going to rewrite here, except that here, of course, because you're looking at the quotient, so now this is going to be, sorry. So we had this g equals g, sorry, g is 0. I forget, g1. I forget what I wrote, g2, and so on. So we embed g mod g2, which is cyclic into a torus. And so here, we look at the silent mod g2. So, OK, cross t. And we mod out by the action of g mod g1, or g, if you like. So, and then you have the same diagram, and you apply descent for torsos and dottora on the left. The only thing is, then, the analysis of the arithmetic of the top space is more complicated. It's not just rational. But you still get an action of SLN on the left, but it's not anymore a torsor under SLN. It becomes a homogeneous space, and the stabilizers are g2. So at least you've gained something. The stabilizers are smaller than what you had at the beginning. It's not g. Now it's g2. But that's how the proof starts. Then you have to study the arithmetic of this vibration and work inductively. So I don't want to talk more about this. Instead, I just want to point out that super survival descent is actually quite more general than just this fact. So you can apply it to other variants of the inverse problem. Let me just give one example. So theorem 7 has other applications. So for example, it's somewhat of an anecdotal statement, but it's a challenge to prove it. So here's the theorem. So g, a finite super survival group, survival group. And you fix k and number field and some elements of k, alpha 1, alpha n in k, k star. Then there exists a Galois field extension capital K of k, Galois with group g, such that the alpha i's are norms from k. So this is something that had been proved first for a billion groups by Fry, Laugren, and Rachel Newton. They proved it by actually doing, by counting the number of a billion extensions that do this. And they found a positive number. But to do it for a super survival group, well, I mean, first you have to find a way to construct Galois extensions with given super survival Galois group. So the only known ways are Shafferovich's proof. And this one, as far as I know, only this one can give you such a statement. So how you deduce this from super survival lessons, well, that's one of the exercises. So I encourage you to go to the exercise sessions. OK, so we have a bit of time for questions if you like. And I'll stop here. Thank you for your attention.