 The next speaker is Professor Srinivas from Tata Institute of Fundamental Research. And he's the title of his talk, the final presentation of the theme fundamental group in positive characteristics. Please. Well, I'd like to thank the organizers for inviting me to this meeting, a nice meeting. I'm happy to be back and see us there in a beautiful place. Come here a couple of times at least. I think lectured in this room also previously. One very different topic. So I'm going to talk report on some joint work with LNNO, Mark Shusterman. This work done during the COVID lockdown period. I've never met Mark Shusterman or talked to him. All correspondence has been entirely electronic. Because LNNO knows everyone in mathematics and everywhere. So one of those, air dodge number type things. So I've, during the COVID period, met in some fashion more than one math foundation through LNNO. It's been very interesting. So background is this. Let X be a connected algebraic variety to start with, or an algebraically closed field K. Then I want to talk about et al fundamental group of X. I'll just write it as phi 1 X. Strictly speaking, I suppose to write a base point. I don't want to get into that. It's not so important for what I have to do. So some properties of this thing. This is a profinite topological group. So et al topology in particular. And finite quotients to finite groups G correspond to finite Galois, I should say G Galois et al, covering spaces Y to X. So this is et al covering space. That's the corresponding extension of function field is Galois, and G is the Galois group. Now, if K, the ground field K, is the complex numbers, there's another fundamental group. X is also a topological space in a natural way. So it has a topological fundamental group. And there's a relation between this algebraic fundamental group and the topological. And let me write like this. There's a group homomorphism from the topological fundamental group to the et al fundamental group. This is a profinite group. This is just an abstract group. So it will factor through the profinite completion. And it's a theorem that this is an isomorphism. Sometimes this is called generalized existence theorem. Mathematicians from Germany tell me this is the wrong reference. It's Hebbar-Kite-Zerts, is the correct one. Removable singularities is actually the proper reference according to some people. Anyway, whatever it is. In the literature, it's always called Riemann existence theorem. I'm just using that. Now, the point is, we know that pi 1 topological of Xc is a finitely presented group, abstract group. This can be proved in many ways. One is to argue that the space Xc as a topological space has the homotopy type of a finite Cw complex. So in particular, if you take such a model of the space up to a homotopy, you have one cells and two cells. The one cells give me generators, and the two cells give relations. That gives a presentation. I mean, there's nothing canonical about it. There are many ways to do it. But anyway, the finite presentation is clear. Therefore, this implies that the et al fundamental group of X defined algebraically is a finitely presented profinite group. Then there are some theorems comparing telling you what happens to the algebraic fundamental group if you change the ground field from one algebraically closed field to another one. So you are able to, from this, deduce that if characteristic K is equal to 0, pi 1 X is finitely presented. And the proof is compared with. The natural question is finite presentation in the profinite sense. Also, in characteristic P, it's a natural question. And I'm going to assume, get to that later. It is a conference related to characteristic P geometry. So it's natural to bring this in here. Seems a bit tough to find unbroken chalk. So let me state, OK, I need some terminology. A good compactification of a non-singular variety X is a pair X bar D, where X bar is smooth and proper over the same ground field K. And D in this X bar is a simple normal crossing divisor. That's why I've identified X, the original space, with the complement. So of course, naturally, then I think of X bar as a compactification of this X. So I need this terminology. And then let me state the first main theorem, that X smooth variety over K such that there exists a good projective compactification of this X. Then the tame fundamental group, I'll explain later what this is, but is finitely presented as a profinite group. In particular, pi1X is finitely presented if X is smooth and protected. In some sense, even this was a new result in the characteristic P case. Let me write down a corollary to work out a little later with Jacob Sticks and Marcin Lara. If X is over K, it's proper and connected, I suppose. Then pi1X is finitely presented. I say corollary because in some sense, there is a machinery which is there in SGA. I'm sure Dale knows it very well. Which allows you to write a kind of van campan theorem if you have some sort of covering of the space X in some suitable sense. So here we would want some kind of hypercovering. So basically, you'd start with your X, which is proper. You'll have a Y mapping to X, which is, let's say, a Dijon style alteration. And then you have a Y1 lying over Y and Y2, and so on and make some diagram. Then somehow the properties of this diagram, and if you know finite presentation of the fundamental group for each of the terms appearing in the diagram, you can deduce it by some kind of abstract van campan type result, which is there in some sense. Tim, we're not saying here this X is proper here. There's no boundary. There's no boundary. I haven't said what is, Tim. I'm going to explain that. I said, in particular, X is smooth and protected. If there's no boundary, exactly. So Tim has to do with X, which is non-proper. And then I have to take a boundary and good compound, et cetera. This should have been obvious, I suppose, even to us. Anyway, somehow this came a little later. Styx is an expert on this van camper type descent stuff. So I'm not going to really talk more about this. It's a corollary of this with known machinery, but it's a little abstract. The real thing is the smooth, productive case. And as Dale is pointing out, there's something about Tim. Whether something like that, what is Tim in the general case, where you have singularities we don't really understand. And so I think we don't have a good equivalent of the Tim thing here in this more general context yet. I'll comment a bit about some of the issues with Tim. Now let me give you some background remarks about the theorem. First, there is some stuff related to this already in SGA. This says pi 1x is finitely generated if x is smooth and productive. This is an old result of growth headache. And in fact, as I understand, this proof of this is one of the triumphs of growth headache algebraic geometry, one of the early ones, which couldn't be done by other methods. And the style of arguments is to use two things. Use left sheds hyperplane theorem for pi 1, which says that if I take a curve C inside x, smooth, complete intersection, implies pi 1c maps on to pi 1x. Well, this part of the left sheds theorem is actually not so difficult to prove. It is essentially there in Hart Shorn's book in some fashion algebraic geometry. The lemma of Enrique's Severe, which says that an ample device was connected. Basically, that's the statement. So that implies things like this without much difficulty. So it's not very hard. But the second thing is, so now I want to just prove finite generation. And if I know this is finitely generated, I deduce that one is finitely generated. So I have to do it only for curves. And for curves, the second step is growth index specialization for pi 1. So the idea is that if you have your curve, so suppose you have some curve C over spec lambda, where lambda is some complete DVR. And you have then sort of its spec k, the quotient field, and then pass to the algebraic closure. So I have a sort of generic fiber of this thing. A geometric generic fiber. And I have the closed fiber, which is, let's say, my curve C in characteristic P. So this script C is some sort of lifting to characteristic 0 of the curve in characteristic P. Such a lifting always exists because obstructions to lifting live in H2 of the space which you're trying to lift. C is a curve. It has no H2 of any reasonable sort of sheet. So there are no obstructions to lifting. So you can construct such a diagram. And then growth index says that there's a homomorphism from pi 1 of this geometric generic fiber to the pi 1 of C. And it's rejected. This is called growth index specialization homomorphism. So if I want to get finite generation of this thing in characteristic P, I can deduce it if I know finite generation in characteristic 0. But in characteristic 0, pi 1 x is always finitely presented even by comparison with the case of complex numbers. So I deduce that pi 1 of a smooth, productive variety in characteristic P is finitely generated. But of course, it's only a surjection each time. So you've lost the control of the relations. You don't know finite presentation. SGA, in fact, has a comment that, quite possibly, finite presentation is false for pi 1 of curves in characteristic P. No, no, smooth, productive curve. There's a speculation. But of course, our theorem. Absolutely, next comment. Absolutely. I'm going to discuss that next. Remark 2. Dale has already anticipated me. Pi 1 of the affine line, not finitely generated. In fact, if you take the first and fancy language, et al-komalji with Z mod Pz coefficients, this is hom of this pi 1 a1 abelianized into Z mod Pz. Infinite dimension is an infinite k vector space from Arte and Schrodinger theorem. So concretely, I can think of coverings of a1 to a1 given by equations like y to the p minus y equals x raised to m. And I can take mi, some integers. Assume p doesn't divide any of the mi's. And m1 less than mn, let's say, take n of these guys. That gives me n Z mod Pz galore coverings, one for each of the integers mi. So this equation defines a Z mod Pz et al-komalji. But by choosing different exponents mi, I'm able to ensure that the corresponding field extensions, k a1 to k a1, right? So I have n of them with n different choices of exponents. I have n field extensions. They are linearly disjoint. So I'll get that if I take the compositor of all of those fields, the galore group is Z mod Bz, Z mod Pz to the n. n was any number. So this is certainly not a finite rule. And why do the different exponents? Well, you can do it directly in some fashion, if you like. But there's a fancy way to see it using the so-called swan conductor. There's something called swan conductor which measures wild ramification. If p doesn't divide mi, basically this mi is the swan conductor in these things. And they're different. So since the swan conductor are different, they must be disjoint. OK, so in any case, as Dil therefore remarked already, if x is affine, there's no chance whatsoever that the fundamental group is finite. It's not even finitely generated. So if you want to get any decent statement for non-compact x, you want to put some restriction on the type of ramification you have with the boundary, which is why our theorem in the end has something about tame. And I have to explain what is this tame. Right, so how does one think about tame? So here I've got a non-proper x. And I'm assuming, for example, I'm in the situation of my theorem, x is non-singular and it has a good compactification x bar. So suppose I have a suggestion pi 1 x to g for some group g, finite group g. And then I have a diagram like this. This is a g Galois eta cover, right? Because quotients of pi 1 correspond to such things. But now my x came with an x bar, a compactification. So I can normalize this x bar in the function field extension here and get a normal variety y bar with a map to x bar, a nice finite g Galois. But now this may be ramified at the boundary. So let's choose y, just notation. y0 contained in y bar minus y, a boundary component. And x0 equals this map that's called f bar, compactification f, f bar of y0, the image component below. Then I have discrete-valuation rings, o x0 x bar, o y0 y bar, these are DVR. And because both of these are normal and these are divisors, so these local rings are DVR. So I can look at the ramification in the following sense. I can take the maximal ideal here, the discrete-valuation here, and look at how much it's ramified there in the classical sense. That is, you take the maximal ideal, m x0, and then multiply it by o y0 y bar. This is m y0 to the e for some power e because this is a DVR. And this e was sort of ramification index. But now, where in geometry, where in possibly higher dimension, so the residue field of these are function fields of subvarieties. They may not be perfect. So when I want to talk about tameness, I have to take that into account also. The residue field not being perfect. So I'll say that this extension here of DVR is tamed if and only if p doesn't divide this e, this exponent, and ky0 over kx0 is separable. In general, there could be a purely inseparable part. So I call something tamed if that bad phenomenon doesn't happen and p doesn't divide. So no p isn't interfering with anything. That's what you want. So now, I can define the original covering y to x. So we say, is tam relative to our compactification x bar if every such DVR extension is tamed in this sense. This is good in some sense. It's a finite number of things to check. But it depends on the choice of a compactification. It's not obvious that it's intrinsic to the space x in any fashion right now. So it's natural to ask, what about you change the compactification and so on? There are subtle issues there because I want to stick to the situation where my x and my x bar are non-singular. Luckily, we are saved by a theorem of Kurtz and Schmidt. They showed that if this appeared in 2009, though I must say this is the published reference and they have studied other notions of tameness and funcular reality and various things. But the particular thing I'm going to say maybe was not to other people. I wouldn't be surprised. They've looked at other aspects of tameness. People want to know a little more about that after the talk. I'll be happy to talk more. So their theorem is, if x has a good compactification, tameness does not on this choice. I'm saying it a bit loosely. What it means is that if you have two good compactifications, tameness with respect to one of them is identical with tameness with respect to the other one. So at least in that sense, it doesn't depend on the choice. And the way they prove it is there's another notion of tameness to do with restriction to Kurtz. That doesn't depend on any choice. And he shows the same as what you get with any given good compactification. Anyway, so if you're interested in issues to do with this ramification in higher dimension, please do take a look at this. And in particular, they haven't fully resolved all the issues. There are interesting things left there to think about. And if good results come out of that, I think arithmetic geometries will be able to use it. So now that I've got a notion of a tame covering, I have automatically a notion of tame fundamental group. So that's some sort of quotient, pi 1 t, is the quotient such that pi 1 x to g factors through this pi 1 tame, this quotient, if and only if f y to x, the associated cover is tame. So tame coverings are a proper sub-collection of all covering. So there's an associated Galois category and so on. There's a fundamental group. So that's the tame fundamental group. So our theorem is about the tame fundamental group of a variety smooth, variety x, which has a good predictive combination. Okay, there's another theorem. If we start trying to follow the growth of the extractory, it says, well, you want to prove something about x, let's try to reduce the dimension to simplify the geometry as much as possible. So that too is a relatively recent, not so old result. I would have thought it's there in growth of the equation. This is published only in 2016. So let's take a good predictive compactification of a non-secular x. Let z bar in x bar be a general hyperplane section. z equals z bar intersection x. Then z bar meets transversely, whatever that means. Since it's general, I can assume this by burtoning. And then I'll have a map from pi 1 tame of this z, this open z, which has a good compactification z contained in z bar. And that maps to pi 1 tame of x. So this is like the sort of setup for doing a left shift theorem. And the theorem is that exactly what happens in the non-segular predictive case happens here with the tame pi 1. This is a surjection dimension x is greater than or equal to two and an isomorphism dimension x is greater than or equal to three. Now the tame fundamental group does have the feature which you expect with fundamental groups in topology that a two-dimensional part of the space actually determines completely the fundamental group. And that's just like in CW complexes, you need only the one skeleton and the two skeleton. Whether this, if you don't put tame, it's false. It's false for the affine line and the affine plane. Things like that. Just Art and Schroder theory again just says nothing like this can work if you take affine and don't put tame. You don't put some condition, just completely wrong. For any affine variety and any divisor, pi 1 of the divisor to pi 1 x is never subject to a characteristic peak. So left shift is hopelessly wrong without some condition of the boundary. While the ramification seems to really mess everything up. But in case with tame, picture looks good again. So this says in order to prove our theorem, without loss of generality, you can assume x is a surface or a curve. And I claim also the same proof which growth and leak give for finite generation of simple modification of that implies finite generation for the tame fundamental group. So this is basically growth and leak. The sense that once you have the left-shift package, growth and leak allows you to carry through everything. Growth and leak has looked at tame in the case of curves and compactification of curves. Of course, curve has only one compactification, which is auto handling good. And he has constructed his business of specialization and stuff like that. Yeah, this sort of picture works even with a boundary with a divisor. And this is there in HGA somewhere. And he has constructed such a home awesome and even putting, well here you don't put anything, but here you put tame, it's rejected. C is non-proper and put tame, this package still works. And in characteristic zero, it's the same fundamental group which you had. So finite generation works the same way. That's what, HGA has all this and says, possibly finite presentation is going to be wrong. That's the speculation. The last remark, which I found kind of provocative, but we do not know in any explicit form, saying this badly, let me explain it again after writing it down. In other words, we do not know a single example of a curve of genius, at least two, for which we can write down our presentation. That's what I'm trying to say. It's then no example, characteristic P. Because in characteristic zero, we have a standard presentation. If you want to prove finite presentation, therefore clearly it's not by writing down any kind of presentation or doing any calculation, there has to be some other indirect method. And there is one, and that's somehow what was unearthed by Shusterman, colleague. And he proved in particular finite presentation holes for curves using this method. And then higher dimensions of what Ellen and I have joined the product. So there's a theorem in group theory, Lubbocki 2001, this is very interesting, that gamma be a finitely generated pro-finite group. Then gamma is finitely presented if and only if there exists a constant C greater than zero, such that for every prime dimensional FL gamma module, I should put here continuous, there's a topology, M, we have dimension over the finite field FL of the second group co-mology of gamma acting on M. M is a continuous gamma module. So there's a notion of group co-mology, continuous co-mology, continuous co-chain and so on. This is a finite dimensional vector space because M, the underlying vector space is finite dimensional. This should be bounded by constant times dimension. So claim is if the constant C doesn't depend on the prime L or the module in characteristic L, such that this estimate holds on second group co-mology, this is equivalent to finite presentation. So this is something to do with the theory of the Schur multiplier, if you've looked at those kinds of things in group theory, this is the criteria, okay. So of course our main theorem is going to be, we're going to check this. This is the condition in co-mology, we'll try to understand co-mology spaces and prove it. But I want to explain a bit what is this criterion about? What is it trying to do for you? Why do we believe it? So quite shocking that something, this looks like very little information somehow. And how do you get something out of nothing is what you might wonder. And that's what I like to explain next a little bit. So remarks for profiling groups only. No such, no similar abstract groups. So it's using the profiling structure in some fashion. And two, let's write gamma as F mod R. F equals pro-free group on a finite set, closed, normal circle. And the relations. So what Lubowski's criterion is doing for you somehow is the following. He's really looking at R mod its commutator as topological F mod R equals gamma module. This is now an abelian topological gamma module. And I'd like to know, so Lubowski's criterion gives an estimate for number of generators of topological generator of this R mod its commutator. So how does that work? Well, since the thing is abelian, and we're profiling it, I can work one primer at a time. So I can just look at the l-addict part for each L. Then I can, and remember I know finite generation of gamma. So I know F is a finitely generated free group. So from that somehow I'm able to do a kind of Nakayama argument. And if you know it mod L for each prime L, you get it l-addictly. And then once you can get it l-addictly for each L, it doesn't matter if it's infinitely made prime for a profiling, I can do a topology type business and close everything up. So his criterion here is really estimating the number of generators not only of this, but he's saying for each prime L, you look at R mod RR tensor Z hat of Z mod LZ. And then he's trying to estimate the number of generators of this thing. And if it's not finitely generated, I'll be able to find a sequence of things getting bigger and bigger, the covalence sort of growing. Something like that, you can imagine the type R. Okay, this we might believe. But what about this whole big non-abelian mess here? We haven't said anything, right? The criterion isn't telling you anything about it. So there's a business which I found, it's cute, but it sort of explains what's going on. Let me just write an example. So let GI be an infinite sequence of finite, simple groups. Finite, simple, non-abelian groups. Let G be the direct product of this GI. This is a profilite group. So now choose, let little GI in GI minus the identity be any element. And let G be the infinite word in the direct product which in the ith component is little GI. So this is a single element of the group. So now, claim the closed normal subgroup of G generated by this little G is G itself. So I don't know anything about the group theory structure, but nonetheless, up to taking conjugates and closure, it's generated by one element. This is how it gets rid of all the non-abelian stuff, basically. All the non-abelian stuff disappears for abstract reasons. And let me explain why this is true. This is actually pretty obvious. So fix I zero and choose H zero in GI zero. Take HI equals one for I, different from I zero. Fix and index I zero and choose some element in that particular factor. And everyone else just choose the other elements to be one. Let H be this sequence HI, which is non-trivial only in the I zero's copper. And let's compute the commutator, GH. So I claim this is basically the commutator of GI zero, HI zero in the I zero's factor and identity in every other factor. Obviously. But now if I let this HI zero vary, I get all possible conjugates of this GI zero inside the factor capital GI zero. But capital GI zero was a finite simple group which is non-abelian. So taking commutators gives me everything. So I can get the entire factor just by taking this element one, this particular element conjugating in and fiddling around. If I can do it with one, I can do it with finitely many. Once I can do it with finitely many, I can take the limit, get all of it. So without understanding anything about the structure, I've managed to generate it by one element up to topology. And that's the kind of what Luboski has also managed to do. That's why his question works only in the profanity case. All the non-abelian stuff takes care of itself automatically in some fashion. So more carefully you want to argue this, you have to use the so-called Frattini subgroup which is the group theory version of the Jacobson radical. You know, take the non-trivial closed maximal subgroups and take the intersection left, right, doesn't matter. Just like usual Jacobson radical and so on. The quotient group is a direct product of finite simple groups. Some of them are abelian, some are non-abelian. The non-abelian ones are all gone, thanks to this. I'm left with the abelian ones. The abelian ones, well, he's giving you a criterion. Whether you have infinitely many or finitely many factors is a question, or to gamma action. And that's what he's saying, the criterion tells you. So this is behind the criterion, okay? That's why one believes the criterion. So then the question is, how do you check the criterion? That, of course, we get deeper into algebraic geometry. Let's try and do that. After all, it is the fundamental group of a space. If I want to compute some group theoretic property, I have to involve the space and its geometry in some fashion, right, so that makes sense. The only thing I know about the group is that it's pi one of the space, so, okay? Okay, so we want to apply the Lubotsky criterion in the case where pi one tx is my gamma. And I do know pi one tx is finitely generated thanks to the discussion earlier. So gamma is finitely generated as a profine group. Now, suppose I have one of these primes, L, and I have an FL gamma module, M, a finite dimensional model. Obviously, the case, L equal to the characteristic and L different from the characteristic are going to possibly be different. So let's first assume L is not equal to the characteristic. So P is a special prime. So let's take L not equal to P case first. Anyway, there is this question about continuous modules, by the way, if you take a general finitely generated profine group, it need not be the case that any subgroup of finite index is open. But if it is pi one of a space, this is correct. There's a property of the kind of groups which appear in Galois coverings. This property is apparently always correct. So I don't need to bother with continuity, it's automatically satisfied. Being the fundamental group of a space, this is correct. Right, so let's take one of these M's. Then this M is a finite dimensional vector space over a finite field, so M is a finite set. The map from gamma to the automorphisms of M, the kernel has finite index. So that is some finite quotient of the fundamental group. So let's choose one, any quotient like this. Let this be any quotient such that gamma through, there's a smallest possible G, namely the images that are present, but then I have other things which are bigger and in the end, if I want to prove something about the covalry of gamma, I have to take some sort of limit over coverings. What I'm saying is it's enough to take coverings of this kind that they are co-final among all coverings, okay? So let's choose one such thing. Then let's look at the geometric diagram I get out of this. And remember, I already assumed x is of dimension two because of the left-shifts there. And by the way, if x is of dimension one, this is h2 of something, you're able to somehow manage it quite well. That's what Shusterman did first-hand. He became aware somehow of this. So Watski criteria that it can be useful in algebraic geometry and that was his earlier work. And then higher dimensions, of course you have to do more algebraic. So let's look at this diagram, which we get in a natural way out of the picture. Then, of course, G acts on this space y-bar and x-bar is the quotient space. Here it acts without fixed points, but there it may have fixed points on the boundary. And M is a G-module, so I can think of M y-bar be the trivial et al local system on y-bar with G-action. Trivial local system is basically just the product y-bar cross M. G acts on y-bar, G acts on M, it, of course, acts on the product. So that's obvious, that's what it is doing. But now the part is I can look at MY equals MY-bar, restricted to y, it's the same, similar type of analysis. But now the action is free, and so there's a descent to a local system on x. And now this is some object appearing in algebraic geometry, I have a sheaf on the space x, I can talk of the co-mology of x with coefficients of the sheaf. And then I claim that h1 gamma M is in fact isomorphic to h1 et al of the space x with coefficients in the sheaf. And secondly, h2 gamma M, really the thing I'm interested in, sits inside h2 et al, so I can control the co-mology, I'm trying to understand the group co-mology by et al co-mology. Now, let me see, I think in the interest of time and let me not go to the proof, it's not difficult for, again, people interested, I'll be happy to sketch it afterwards, for this audience just take it for granted right now. Or maybe I'll hand wave in the following fashion. If you were similarly working in topology, you'd have a universal covering space x tilde. When you pull back anything to the universal covering, it's the trivial local system, but the fundamental group is also trivial, the h1 of the universal covering is also trivial. So write a spectral sequence giving descent for the action of the fundamental group on the co-mology of the universal cover, converging to the co-mology of the space x. And then when you do that, but the universal cover has no h1, so some terms are zero in the spectral sequence and the correct term is zero in the right place to ensure these two things. And you write the five-term exact sequence or low-degree terms, whatever it is, this will map to that and then it'll map to the thing which I said is zero, it'll map to this, sorry, this follows. For those who've seen such arguments, that's a proof, basically. And then you have to make it et al, which you can do. But anyway, let's take these things. So this tells me that I want to bound this object, local co-mology of this x. So how do we try to do that? There are various tricks there, we'll just see. Okay, again, looking at the time and the context, maybe I should not spend too much time on this. Just say we can do various reductions. So you can take x and you can take x minus a hyperblown section and try to compare the co-mology of x and the co-mology of this open set, writing a long exact sequence of local co-mology. I claim the local co-mology terms are trivial for some obvious reasons, they're bounded in some stupid way. So you're able to replace x by this big open set. And instead of computing the co-mology of x, I'll compute the co-mology of this thing. But this thing has the advantage of h is a hyperblown section, so this is affine. Remember, our x had a good compactification with a nice Cartier device of the boundary. So we take a general hyperblown section. This new guy looks like a predictive variety minus the old divisor plus a new one. But then one knows this is affine for various reasons, x is a surface. So once this affine, therefore, I mean, I can assume without loss of gravity that x is affine. Not just a surface, it's affine. Once it's affine, I can say that it's et al co-monotical dimensions two, or at most two. By, for example, in characteristic zero, it would say it's a Stein manifold. So the co-mological dimension is bounded by the complex dimension. So the algebraic version of that is correct. It's called Artin's theorem, Anishin's theorem et al co-monotical. So I also claim because the fundamental group is, a tame fundamental group is finitely generated, there's no problem with the etch not and the etch one. Etch not of any local system is bounded by rank for stupid reasons. Etch one of a local system is bounded by rank times the number of generators of the group because an element of etch one comes from one co-cycle, a value of one co-cycle is determined by knowing the values of the generators. Of course, everything is topological and limits and so on, but let's not worry about it, those things work. So bonding etch two is the same as bonding the Euler characteristic. Bonding the Euler characteristic, then Deline has proved, now we have to use heavy things proved by Deline, he says that Euler characteristic of a tame local system is equal to the Euler characteristic of the trivial local system times the rank of the tame local system. This is for Euler characteristic and only for tame local systems. If it's not tame, then you have to worry about swan conductors, about where the nasty thing happened, but we are tame in our situation. Then the trivial local system, you say, well, okay, it's trivial, right? The rank is one. So it's just some invariant of the space x, but there are prime l involved. I have to check whatever the chi, which I got with FL coefficients. So x is a fi and we have chi of the trivial local, chi x, FL, the trivial local system. And I want to know that this chi is, there's a constant C bonding the chi. That's the type of thing I want. But in fact, this chi, if you think about it, it's the same as chi xql for homologically algebraic reasons because- This is time, yeah. In my time, okay. One more minute, okay. Fine, that's not it. So then you have to use Delain's proof of the veil conjectures, et cetera. This comes from the zeta function once you put ql and veil conjecture stick here. The interesting thing really technically is the p part, but let's leave that to experts too. So some people want to hear about it, I'll tell you. Thank you. So because of time, so we discussed, we asked the speaker after the- After the talk. So- Okay. After the talk of class. So we have two minutes for the preparation of the last talk.