 Thank you very much and thank you to the organizers for the invitation to speak. So I'll just say that so Luke mentioned, look you're already laughing I haven't said anything yet. So at the beginning Luke mentioned you know several categories in which people benefited from OFA and I would say I like fall into several of those categories. I'll might kind of tell some stories at the end. So today I want to talk about some questions around Hilbert's 13th problem and their connection to the moduli of the Bieling varieties and this is joint work with Benson Favre and Jesse Wolfson. This is Benson here. This is Benson. You know I make the remark that when I have the graduates students coming in they seem to know more and more and the reason is they start earlier and earlier. This is not Jesse. This is Jesse's son Leo. Okay, so what kind of question do I want to consider? So if you take a variety of an algebraic closed closed field of characters to zero and let's just consider a connected finite etal cover of this variety and we're interested in some sense in the geometric complexity of y over x and what I mean by this and I want to ask myself kind of what is the smallest the smallest dimension such that this cover comes from a pullback of of something of that small dimension so we'll define the essential dimension of the cover through the smallest integer such that there's a dense open u and x and a map u to some u prime such that first of all u prime has dimension e and y comes from the pullback of a covering u prime over e prime some finite etal covering and let me just say here the fact that in this definition you're allowed to take some dense open is absolutely crucial so that's something to keep in mind a lot of these questions become much easier if you just insist that u is equal to x but we don't want to do this here and maybe the reason I mean the reason will maybe become clear a little bit later on then there's a second kind of definition where we also allow composites of such coverings so what do I mean so the second definition resolvent degree uh where I have the same thing I have a covering and then I want to consider the smallest integer such that there's a dense open and a tower of coverings over the dense open such that each of them has essential dimension less than or equal to r so each one of these comes from something of smaller may well possibly smaller dimension r can be r might be the dimension of x and also the at the top I dominate the covering that I started with okay and of course everything is connected everything is well I assume that y over x is connected right otherwise I think from the beginning because of the other question about the empty sense certainly them terminates 12 you don't want some you should better assume that the finite oh you mean there's they should be non-empty if they're finite at all but they could be empty I think a finite is like finite is being subjective but yes, let's uh that's for uh connected and for completeness we have said the connected here means also non-empty yes not like you hope okay already the talk became empty within in a few minutes okay so okay so here we have the definition again of resolve and degree so there are two notions again right one is like for the individual covering whether it comes from something of small dimension and then you have like towers okay and then there's a notion of uh for a group just for a finite group you can define essential dimension and resolve and degree of the group as the soup overall coverings y over x such that the gala closure has group g I'm in characteristic zero my k was characteristic zero okay so later on i'll use some arguments right at the end in mixed characteristic but for now most of the talk k will be in characteristic but the question is in characteristic zero um Okay, so we're up to here um Okay, so we have this definition So in fact it turns out this notion of essential dimension resolve and degree if you have a covering it's the same for the covering or for its normal closure So it's kind of invariant to that so this is for a group g But notice that when you reduce the field definition you don't say that the g What okay when you take the g it doesn't mean that the the g-toss so If you take a g-toss over a variety it doesn't mean the generic lead comes from a g-toss so Yes, you have to prove that though It's true You mean the question are you asking just if you have a covering but now it has some extra structure like it's a g-toss So because suppose you have a g-toss To settle in the gala closure As group g Okay, so your definition is that you can reduce the field definition to a south field or to a south field degree here Okay, but just as a finite talcaval Not as a g-tosser Yes, but but it's the Yes, exactly. Yes. Yes. Yes. Yes. It's the same. Yes. Okay Did everyone Yeah, so first question, I mean which we're going to sweep under the rug is here. I'm just when you have a covering I The question is about reducing it to smaller dimension but as a covering But maybe it has some extra structure. Maybe it's a local system or a g-tosser and maybe you want to reduce The dimension with that structure. In fact, it turns out that it's equivalent to that, but that's a little lemma Which took me a little bit longer to prove Okay All right So these definitions are due to a Resolven degree are due to Broward, Shamar, Anold and For an essential dimension are due to Bula Reichstein But in fact, they formalize a circle of questions going back to Hamilton, Hilbert and others Actually, as as you'll see like this subject is a very strange subject because you know, usually you have like one definition And the second definition using the first one comes after that in time Or maybe you have like a conjecture and then the theorem gets proved after the conjecture in this subject Everything is kind of in the opposite order. You'll see so for example here Sorry, I've got here the definition of resolvent degree was formalized by Shamar, Anold and Broward And the definition of essential dimension was given later by Bula and Reichstein I mean they gave it in a kind of more general context Okay So let's look at an example This was the classical one of the classical examples that was of interest to Hilbert So let's look at the space of monic separable polynomials of a degree n So it's given by affine space in n minus one coordinates just given by the coefficients minus the discriminant locus And this has a covering where I look at the solution of the polynomial, right? So you have f as a polynomial and then z as a solution to the polynomial. So this gives a covering And well the essential dimension of this covering is the Of course this is a this covering its normal closure has gala group sn And it's kind of the generic sn covering So its essential dimension is the essential dimension of the group same for resolvent degree And notice here this is going to be kind of one of the rules of the game The resolvent degree of sn in fact turns out to be the same as the resolvent degree of an Because this is just an index two subgroup And as we'll see in a second When I have a cover of degree two, I can always kind of reduce it to Something one dimensional. So I mean that's kind of easy, but I'll mention in a second So what is known about the resolvent degree? So for n equals one It's due to the Babylonians that this is equal to one This is the paper Okay, so Sorry Can I enlarge it? Yeah, this is I mean this is a um I of course, I can't literally if I try to do it now. It'll be a disaster, but uh I'm not sure that the ideas that you use for that In the formulation of the notion like you have to speak about algebraically close Yeah But I already mentioned earlier of that in the subject everything is lacking reverse order So it's true that the Babylonians didn't know what a finite detail cover was Nevertheless, they had proved this theorem As you can see because the paper is right here Right, so the point is that um When uh n equals oh, uh, sorry did I mean n equals one? I mean n equals two Sorry n equals one is not so interesting. Sorry, that's a typo n equals two n equals two Yes No, no, no monic Monic. Yes. No, no monic, but you can delete you can you can get rid of one We're not doing that. Yes, but we're not doing that. Yeah, it should be n because Oh, sorry. Sorry. Sorry. Okay. I knew there would be some typos here. Yeah. Okay. Anyway Um, that's fine. Okay. So this should be an n. Anyway, so this should be n equals two That's terrible. Okay. So uh, but let so the point is you have a quadratic formula, right? So if you have a covering of degree two Uh, then you know that this is basically given by extracting a square root Okay, and which means and then this function f Gives you a map to gm And so then this comes just from this application by by degree two That's the result of the Babylonians Okay And then for n equals three and four you have a similar thing um By kind of there was this classical problem of solutions of polynomials of some degree Where there could be solved in radicals. So you have to target a cardano and Ferrari Solved this in degree three and four notice there you in the formula you have to kind of compose Functions right because you have to take a sequence of radicals Um Okay And then for n equals five there's a result of bring from 1786 that it's still equal to one So the point there is after you extract Some roots for example a square root of the discriminant. I think there are also some other things You can bring the polynomial into fifth degree polynomial into this form So you you kind of only have one parameter Uh, and then well Hilbert was interested in this kind of question He conjectured that for n greater or equal to six the Resolvent degree was greater or equal to two and in particular for small values He had computed and he made some conjectures for small values So if n equals six is two n equals seven is three This is the one that has to do with if you look in Literally the 13th problem of Hilbert and the icm address. It has to do with n equals seven Um And then uh, it's four if n equals eight and nine and one thing is he had already shown an upper bound in these small cases he'd checked that uh That the resolvent degree was at most this value in each of these cases In particular in the case of nine the fact that nine is the same as eight As a slightly tricky argument You mean there is an argument showing that the answer for eight is equal to the answer for nine Well, yes, but With the eight with the answer. No, no, no, I don't know if Yeah, that I don't know if that's true. Maybe um I mean he he just got an upper bound so And the and the royal bound is broken No, this is a conjecture conjecture Okay, it's part of the conjecture. So the statement is still this is still sorry. The whole thing is a conjecture. Yeah Isn't I mean you can see that from the types at latex typesetting see it's like all conjecture Okay And here's a remarkable fact. There are actually no examples of covers which you where you it's proved It's proved that the the resolvent degree is bigger than one So for essential dimension, that's not the case there are you can do something for essential dimension There are non trivial upper bounds for a resolvent degree So this is going to be one of those theorems where everything happens backwards. Okay, so there's a theorem which you do Hamilton That if you look at the resolvent degree of this cover for the solutions of the polynomials Then well, of course, this is at most n Okay In fact, you can do better than that But what they what he showed is that if you take any positive integer r Then there's some h of r such that Oops such that n for n bigger than h of r you have this Okay, sorry about the typos Unfortunate you can't kind of go up there with a torque and correct it. Okay So basically if you so if you fix r and n is big then you kind of can improve the bound Okay, but in Hamilton this this h of r grows incredibly quickly And then this result of Hamilton was forgotten. It was conjected by Segre in 1947 And it was reproved by Brouwer in 1975 And this is Hamilton and Brouwer But again in the days of Hamilton, I believe that things were The notions were the more primitive That's right, and I'm going to explain what the primitive notions are so I kind of deliberately I mean I introduced it this way maybe for my benefit because I always get confused with the The primitive notions I'm going to try to Sorry, what was that? Well, I I will in a second, but before I confuse you I will you know, I'm going to tell you how I think about it. So Well, also how I'll say Brouwer thought about it. Okay So here's the historical interlude. This is the question. How did they think about it? Okay So they thought about this invariant in terms of whether if you have say a solution of a general polynomial of a degree n It could be expressed in terms Of a composite of functions of r variables where r is less than n Okay, so let's try to translate this into that notion Okay, so if you have a cover, let's first try to think about the the notion of a central dimension in that kind of setup So if you have a cut a finite cover and let's assume we've already kind of replaced x by this open Okay, so let's consider first a section Of this cover now this cover we thought of as connected So of course, there's no literal section But you can think of this section as either kind of being a multi-valued function or maybe it's defined locally okay And let's suppose now that this y to x comes from some y prime mapping to x prime with the dimension of x prime less than the dimension of x Okay, then you can choose. I mean, so y is the fiber product of y prime and x of x prime So you can actually choose a section here and then this guy comes from here And then f is a function It's really determined by f prime So if you think about what's going on here, it means that f is a function and this number of variables if you think of Some local coordinates here So if you also think about this question, let's say in terms of real or complex analysis, just take functions In the classical example, yes, whether it is a composite just of analytic That's right. Yes, and this is something which doesn't require to have this algebraic That's right. Yes. I'll get to that too There I mean, there are many variants of this question Um, depending on the kind of functions used Okay, uh, so there's a function in this number of variables That's how you think of f Okay, so there's our diagram and now let's think about the situation of a composite So if you have a tower like this and each one of these has essential dimension equal to r say Then you choose sections like this And then if you think about what's going on then this composite Section is a composite of these functions these sections and each one of them has our variables so So this is a this kind of section is given by a function in our variables. Okay And so what Hilbert actually asked about in in the thirteenth problem Is he discusses? I mean Hilbert. I should say Hilbert has um He wrote about this question in other places So the this thing about I think algebraic solutions is what he was interested in But what he wrote in the thirteenth problem? He asked whether This was equal to three and then particularly asked whether you could just show this in In kind of a naive way namely whether if you have we have a function in Sorry, so you have this general equation of degree seven Okay, and you have the you have a solution And whether it could be written as a composite of function in Functions the less than equal to two variables even locally even at the level of continuous functions. So locally means I mean We're over p seven which is let's think of it as a complex variety But you can you just look at a little region Which is a little cube and you ask us so there you have a solution And you ask yourself. Can you even write it there in that way? But uh, it's also zero zero one is more like In rm. No, right. So I mean well, it's complex manifold. So it's a real manifold m is Yeah, m has a particular value that's right. That's true No, but if the functions are to be analytic or real Well In his thing, I mean that he discusses actually different kinds of functions Well, continuous ones are one of them one one class function. He works as in a real domain or Or in the complex domain. It's not so I mean, I think at one point he's asking this question where he's just asking about continuous functions But that would work with the real polynomials or complex polynomials Well, that I don't quite understand. I mean you have like you have this setup, right? So If I restrict to a cube in here, I can ask whether I have the solution function I can write as a composite of continuous functions or not for example, yeah, but when you use real variables, you have twice as many variables so it's not um And how to formulate the question with the world variables when you work when you change complex variables, yeah So no, so right. So I think you have to think of it as as as 14 real variables. I guess, right You won't You want to know it's not possible to write as a constant function It's most two variables with in terms of complex variables not in terms of real variables Or maybe it's it just can do their continuous functions that he's asking about So I don't Yeah No, no, he's really asked about I mean he asked it also for analytic functions, but let me just Okay Yes, he asked he asked to show that it can't can't be written that way Okay Yeah, so maybe here. It's here. It's ambiguous if I if I don't specify the kind of functions But he's let's say he's asking about continuous functions here Okay um So this is what an old and colmar grove Proved could be done. So they prove that if you have any continuous function on a cube It's a composite of continuous functions of one variable In fact, this is more precise The statement yeah And now you might be a little confused because if you're thinking about it classically like this In terms of these coverings then of course if you restrict the covering to a cube, it will be a trivial covering So what is left to do? Well, the point is that if you're thinking about solutions of your polynomial You have really not just a covering but also an embedding into the app online. So it's really this structure That kind of you have to reduce So all so if you risk so if you restrict to a cube then kind of This covering structure disappears and you just have this function left and of course in the other formulation You don't you don't care about this. So it turns out in the algebraic setting Which I formulate the problem this kind of choice of primitive root doesn't doesn't change the problem either So in some sense, this is kind of orthogonal to To the other question The question is if you can reduce the covering to covering of things of dimension at most If you want to think about the multi-valued solution functions, then Essentially, we'll have to follow this but then we'll have to apply some algebraic morphism. It means polynomials And so we have to use polynomial and this polynomial is Like you have addition right functions of two Yes, that's right lose the one. No, no, that's right So so so there's a question here, which is when you're when you're uh, sorry when you're um Adding and multiplying that's strictly speaking a function of two variables. Okay, but if but for us we kind of that's allowed for free Okay, so it's true. In fact, it's if you want to get into the history you might wonder Why did he'll consider seven here given that he didn't even know the answer for six And one reason could be that he wanted just a clean formulation like this And since you always have to have at least two variables This was kind of the first case where you have a non-trivial bound if you're allowed to if you allow the stupid two variables where you have additional multiplication right Okay Happy all right Okay, so that's the end of the historical interlude So here are some examples from geometry From classical examples So let h3 3 be the modular space of cubic surfaces and b3 Modular the action of pgl4 So this is a four-dimensional space and then we're going to consider the kind of so if you have a a cubic surface And p3 It's a kind of classical fact that if you look at lions and p3 which lie on the surface There are 27 of them. So this is meant to be a picture And uh, so you can consider the such such lions, right? So this is a 27 sheeted cover Okay, so every point here you have the 27 lines And it's group is the vial group of v6 So, I mean it has to be a subgroup of s27 and it's the vial group of v6 And so when I say it has the group w of v6 what I mean is the the the normal closure has the scalar group And the conjecture is that the resolvent degree of this covering is three And there's a theorem of Burkhart and Klein Which says that it's at most it's less than or equal to three. I think so. It's the guy who worked on this kind of geometry I guess anyway, so you can so there's a reduction of the dimension by one here This is four-dimensional Okay And uh Then there's another kind of famous classical example where you have smooth cortex in p2 Modular reaction of pgl3. This has dimension six and then you can consider For each such quartic you can consider the lines in p2 which are tangent to the quartic at two points So called bi-tangent Okay, and this is a 20h there are 28 such lines So there's a 28 sheeted cover and the group of the normal closure is the vial group of v7 living inside s28 And it's conjectured that the resolvent degree here is six And so one thing that I want to explain is how to reduce all these Try to reduce these questions to another kind of covering where maybe one feels one has more techniques which is the modular space of principally polarized abelian varieties So let's consider ag the modular space of principally polarized abelian varieties And if I for any prime p I can look at the The p the p torsion of the universal family So then this group this covering has group sp2 gfp and So here's a proposition on which reduces the resolvent degree of these coverings from classical geometry to this setup Uh, in fact there are two so you see you have to concentrate here because it's g is the genus And then p is The prime so here I have 2 3 and here I have 3 2 Okay, so this is reducing the resolvent degree in this case to the resolvent degree of The kind of the three covering of the of the universal family of abelian surfaces and here I have The two covering for abelian three folds for the biotangent problem Okay, so let me let me kind of sketch the proof of this The point is that the two groups in this case are related. So let's just do the first one Here the covering that the vial group of v6 This has a degree to a subgroup which is nearly isomorphic to sp4 f3. It's ps When you define the resolvent degree, yeah for recovering again you You had some connectedness assumption, but of course the p torsion is not connected Because of the zero section, right? So Essentially you you have to take soup of all connected components or just here maybe yes everything is connected except for the zero right Yeah, right the thing you care about us is kind of the the Yeah, I mean the thing which is giving you kind of the monodermy the Right, I guess. I mean, yeah, that's right. It's the thing minus the zero section if you like but It's not the moderate space is the cost moderate space. Yes. Yeah, that's right. So I'm here. I'm ignoring Yes, so you should add some level structure away from p that doesn't matter You have minus one automorphism. So you don't have an universal family No, no, don't worry about that. You should you just add I mean I wrote ag Okay, but you just add some level structure That's not going to change anything Essentially independent. Yes. Yes, of which level structure five, right? Yes. Yes Okay, and we know that the monodermy then will be the full sp Right, okay So let's yes. We're doing the first one So remember here the monitor group was w v6 the bio group v6 This has an index to subgroup which is nearly sp of f3. So the covering group here Right, so this is this is for genus two So this would be sp4 of f3 Okay But we already saw that the resolvents degree is kind of insensitive to quadratic extensions So that that means the groups on these two sides are related And in fact what you show that both these families are versatile So you can imagine for any group g there's a notion of a kind of a versatile FAM a versatile g cover It's one which is kind of general enough that any other one comes from pullback From it. So for example families of polynomials, that's a versatile sn cover And for the other one there's a similar versality argument and the group theory looks like this again There's an index to subgroup which is actually sp6 of f2 okay, so at least for me this is this is kind of motivating because um I think for ag What you would expect is there's no reason for this resolvents degree in general To be any smaller than the dimension of ag And that's what we conjecture. There's one exception when g is two and p is two And that's because in that case the group sp2 g the In that case the group sp2 g fp namely sp4 f2 Is not simple. It has an index to subgroup And that allows you to reserve to reduce that that means you can pass to that quadratic cover for free And that allows you to reduce the resolvents degree by one. So in that case when g is two uh So if g is two this number here is three But in fact the you see you can show that the resolvents degree of that covers at most two In that exceptional case But I think otherwise there's no reason to believe That this is any smaller than the dimension of ag Oh, sorry, that's the remark to add an extra level structure. Yeah, so this conjecture is Yes for any But I have the feeling that's like not the main point. Okay, no, but but because It's true that it's not obvious that this is if I add a level structure Which is prime to p. It's not obvious that this is independent of p. Sure. Yeah, right, right That's right. Okay Any other questions? Okay, so one thing is that this agrees with the conjectures in this case for the two classical problems So remember in this case we said it should be three Okay, and then we're related this one to um To one degree related to To get three Sorry A two that's right a two three. Oh, yeah, right. That's right. Sorry. That's right No, I was confused because she was two and so I was still thinking about this example But then p is three, right? Yeah, good. So So in that case right so the dimension of a two will be three and we're conjecturing a three here And then in the other case it's a three two Which means the mention will be three times four on two. So it should be six As before Yep, I guess but We're not even can yeah, okay, I should uh, well, let's see if you mean if g is one G is one I get one So that seems to be correct Yeah, it can't be any smaller than that Uh, okay, uh, so at the moment we proved this at the level of essential dimension So that if I take this covering, this is just the dimension of a g so again I take this covering so I can't contract it to anything smaller So that's what you would guess and I'll sketch a proof of that at the end In fact, there's a generalization of this Uh, so to explain this let me take the p the The seagull parabolic in sp2 gfp. So in the standard representation It just corresponds to upper triangular matrices which are in the symplectic group And let you be its unipotent radical so the The the structure of view it's an abelian unipotent group and it's just its rank is equal to the dimension of a g And then if I have any group in here I can take its intersection with you And I'm going to note by eg the fp rank of This intersection. Okay, so it's something Okay, and then what I can do Is I can start with the gala closure of this guy Okay, and I take the I take the the the um The quotient corresponding to the group g so I'm sorry So, uh, so then this this covering has group g here from from this ag tilde to ag And the statement is that central dimension is at least eg Okay, so when g is the whole group Then ag is of course Then ag a big g is of course equal to a little g And this is just the other proposition. It says just says that the the central dimension is equal to the rank of view Which is the just the dimension of ag Okay, so let's apply this To the problem we were interested in from before with alternating groups Uh So for this I need the following uh proposition So it turns out there's a family of in characteristic two There's a family of representations of an Which I won't describe but The property it has is that if I intersect it with this With this u Then this in fact the maximal elementary bill in two subgroups of an meaning the one of maximal rank And there's a formula a simple formula for it is two times the integer part of n over four And so in particular we get uh Yeah, we get an estimate for this an cover here for the central dimension um This is just to remark that these coverings here Uh turn out to be versal when n is five and six, but probably not in general So vers again versal means that the kind of the most generic kind of coverings And so one thing is that kind of in the literature There are there were methods for computing the essential dimension of versal coverings in this case But in general these are probably not versal. So it's kind of a new new result to be able to compute the essential dimension So in the literature I guess people Have already worked on this for eight like the essential dimension of a So the essential demand of am Yes, they know it. I don't know if they know what there's some law of bounds Okay, so yeah, and this is compatible So it's better be But that is exactly the previous result or you are improved You mean you mean there's the bound is is the bound that they got Just this bound somehow That i'm not sure. I mean, I know how to compute the I know I know how to compute a law of bound for the essential dimension of an it's obtained by computing what computing what's called the something else which is the called the essential p dimension which is Less than or equal to the essential dimension and that can be like using a result from a career that can be reduced purely by group theory But we haven't actually done that calculation. So you have to kind of look at the um I think you have to look at like the smallest something like the smallest complex representation of the p-sealer subgroup of of an You look at the dimension of that guy Yes, right Maybe even um, sorry Yeah, maybe you need to look at the dimension of the smallest faithful one or something like that Anyway, so for these covers you have this estimate And so let's look at these the first let's look at the first The values of the first the first few values of this ean. Well, it's a very simple function. So you can write it down here okay, and Let's compare this with hilbert's conjectured value for the result resultant degree of the sky Okay, well, we know for four and five. It's actually should be one. So let's forget about that Okay And then for the other values you see that this is nearly the bound that uh, he conjectured for Resolvent degree namely for a solvent degree. He conjectured two three four and four. So only in seven It's kind of off by one um Okay, and so the the conjecture is actually for n greater or equal six you have This is also a bound for the resolvent degree Of the of the of that cover and so particularly a bound for the resolvent law bound for the resolvent degree of a and itself so So we have some kind of progress towards this but it's still open Right, so this would imply the conjecture for six eight nine And the conjectures on cubics and by tangents that I explained before Okay, so let me to end let me sketch a proof of how you show this bound on the essential dimension Um, ah, this should be a capital g Okay, the reason I put a small g was psychological because I want to restrict immediately to this case Okay, so the general case is kind of similar but Okay, so we want to show that the essential dimension of this covering is just equal to the rank of this maximal unipotent, which is the dimension of a g and Well the idea is if this covering comes from something of smaller dimension Let's let's say a g mapping to v So first of all This has some integral nice integral model and then I can make an integral model of of v Over okay where now I switched okay was algebraically closed before But now I'm going to switch settings and assume k over qp is finite So there's some kind of specialization argument that allows you to do that Um, okay, so this is now an integral thing and then um What you actually want to do is pass to the piatic completion and then Over then this ag twiddle that I had before this was kind of this fp local system So now but now we're working integrally right So before I had this agp twiddle without the circle It was this fp local system his mondrem. He was the symplectic group sp2g fp Right, but now I put a circle there that means that Even characters to if I want to extend characters to p it doesn't extend as an atar local system But extends as a finite flat group scheme Okay, and what you what you show Is that actually you can choose this v hat zero so that this descends to v hat zero at least on some formal open In the ordinary locus of this guy right In any case when you're doing essential dimension you in any case have to restrict to an you have to allow yourself to A strict to an open so this is no not really In some sense this is cost free in terms of what you're trying to prove Okay, so you descend not just the local system, but in fact the whole finite flat group scheme And now you think about the defamation theory of these finite flat group schemes so if you take any If any close point in the ordinary locus Then in characteristic p so this isn't let's close point then you can think about in any tangent direction If you go in any tangent direction this gives you a defamation of this The the fiber of this finite flat group scheme at x Okay, let's just say over the dual numbers And then it follows well either from set a theory a growth in the messing theory or this set a theory works better for the for the generalization to an arbitrary g That in any in any tangent direction you get a non trivial defamation and that in some sense Well, there's no redundancy kind of if you go this gives you this gives you a every direction gives you a different Defamation And then the point is we get a contradiction because if you if this comes by pullback from some Something of smaller dimension you can't get enough defamations Okay Over the sky or that guy But how do you know that the integral model exists over p which one of of If it comes from something of smaller dimension, that's me Yeah, but then how do you get an integral model including p Of this one. Yes, you just I mean remember. We're always allowed here to restrict to an open So This I mean we can assume this is affine and I can just take the intersection of the rings Okay, and then you have to do a little bit more like here when you show it descends you can always replace You replace something by its normalization in here as well And then you can descend the finite flag group scheme Okay, this seems to be non trivial You know that the Essentials of this make it generically, you know that the covering comes from We uh-huh, but then the final group structure is yeah No, you have to I don't claim this is completely obvious, but maybe We can I can explain it to you later, but Okay, which part are you worried about the descent to v? Okay, okay, because there You have a final group scheme on a general five. Yeah, I'll say you yeah. Yeah. Yeah it's not The models of final extension the final flag group scheme is not Uh But you mean that you can descend reason the same day Basically you do it by I mean if you restrict things enough then this map has already flat and then um I mean in fact we do this just in a formal neighborhood But because all you that's all you need. I mean we're working on the ordinary locus here So this is an extension of an atoll part by a multiplicative part, right? And so each of those parts descend And certainly in the order in in a formal neighborhood because they're just kind of trivial and then you have to compute the extension classes And then you you get that like any extension class So basically if you have an extension class That extends to a finite flag group scheme here it already extends to a finite flag group scheme here Okay, so that's the end of the talk the end of the official part of the talk um and uh so I so I mentioned that uh that I benefited from kind of offer suggestions in In a few ways one story I wanted to tell was I think it was now 15 or more years ago. I was giving a talk in in or say and um After I afterwards I mean I should be a tea. I asked maybe over a question It was a kind of group theoretical question about Representations of fundamental groups of curves and if you have such a thing where every simple closed curve had finite image This implied that the representation find an image. Anyway, the talk was about something completely different afterwards. Um afterwards uh Ophrin Luca, Lizzie and I went up for dinner and then after kind of several glasses of uh wine Opha came up with a counter example and here is a picture of the Of the of the bottle of wine of the actual bottle of wine and uh So, you know when when my co-authors and I was kind of struggling with some of these questions I would some to say this is exactly the kind of thing you should ask opha perhaps over a nice bottle of wine Okay, thank you So there's one compliment Luke told me When you asked the question at tea, yeah, sir Heard the question You want to tell the whole story? I'll do it rapidly so Luke sent sarin email after the dinner telling him Opha settled it So I had asked say the question at tea is the yeah, so then said called Luke very early in the morning earlier than Luke normally gets gets up I said Gabbard to come and Luke said this is unlikely Why don't you think about this a little more? Say I called back a little later Gabbard our razor You have a story? All questions about the talk are also are also allowed Talk about Shumora Arnold You wrote the names there, right? So they have an article I forgot where it is it's in the In the in the 1960s where they just asked this question I think it's in some kind of proceedings Where they you know questions and all direct geometry being discussed Yeah, what about it? I mean Yes, you can yeah So there it's even somehow more complicated, right because You have like the mapping class group So implicitly you need some local kind of flattening model of because you are you are working with Well, maybe it's not so difficult, but you want to to pass on the characteristics of the situation to support this game Okay, at least in the neighborhood of your Generic point of social fiber. So you have kind of the disorient evaluation. I mean you needed a good model for the map in terms of Something which is smooth between smooth things over the BBR. This is this is okay This is part of the argument Yeah, I mean the the thing is in the argument. It's not quite as I think this is still okay like this, but But this is you see you really only need it in a formal neighborhood of a point. Yes, but Right. Yeah, sure. I don't want to just put the paper up Sorry, that's right So some other questions Yes We stopped at E7