Okay, so we talk about Thame Geometry in Hot Steery. Okay, so the plan of the talk today, there will be almost no proof, so it will be kind of a survey of what I want to explain later. So we start by a review of Thame Geometry, and then I will explain the main questions and some of the results in Hot Steery that you can get out of it. Okay, so let me start with Thame Geometry. Okay, so first about the idea, so I think it was first mentioned by Grotendick. In esquise en programme, in 1984. And there, he argues that general topology was invented by analysts for the use of analysis, but not for studying the natural topological properties of natural geometric forms. And so that it should be replaced by some kind of Thame topology. And so what is the basic idea? The basic idea is you want to discard pathologies, let's say a wild topological phenomena. So let me give some examples. So of course, you don't want to have cantorsets or piano curves of this kind of thing. But much more basically, you want to discard some kind of usual monsters. So what is a usual monster? I guess the best example is given by the gamma to be the graph of X give sine 1 over X, for X a positive real number, okay? And of course you have this picture of accumulation to the vertical interval 01. So gamma is not Thame for at least three reasons. So what are the main reasons? First if you look at the closure, gamma bar of this graph in R2, then you get this vertical segment 01, and then you have gamma. So this is the boundary of gamma. And well, this is bad because this is connected, but not R connected. So if you believe that real geometry has something to do with drawing curves, then this is some kind of thing that you don't want to have. So what does that mean? The world's boundary, actually some remark to me that there are different notions. Okay. So I do not define it. So let's say this is definition of boundary of gamma. It's just in this very special case. Now I agree. Yes. Closure minus interior if you want to, yeah. But it's not. Okay. So whatever, so Arthur will complain because I do not give a definition of dimension in this setting, but I think this is very intuitive. So if you look at the dimension of the boundary, so this is this vertical segment. So clearly this is dimension one, whatever the definition you use. But then this is also the dimension of gamma. And so it's well known that if you have this kind of properties, you cannot have any good stratification theory. So this is bad. So no stratification theory. And really this idea of having a stratification is the main argument developed by Grotendig in his Esquizon programme. And much more basically, you can just look at gamma intersected with the real axis. Right? And then you see that morally this is a dimensional zero object, but it's an infinite collection of points. And so this means that in some sense this is not a finite type, and so you don't want to have this. So this is the kind of heuristic, you don't want to have this kind of properties. These are no good objects. Okay. So what is the prototype of tame geometry in the mind of Grotendig? The prototype is semi-algebraic geometry. Semi-algebraic, let me just write SA for semi-algebraic geometry. So notice that we are doing a geometry of R here, right? So I remind you of the definition just to fix ideas, everybody knows it. So I remind you that if you take a subset of Rn, then it is semi-algebraic if it is a finite union of basic blocks of the form x in Rn such that f of x is equal to zero and gi of x is larger than zero for a finite collection of functions gi, which are polynomials with f, g1, gk in R of x1, xn, okay? And so why is that considered as being a good category? Well, what are the tame properties of semi-algebraic sets? Well, if x is semi-algebraic, then you don't have this kind of pathologies in the sense that if you just define the boundary this way, x bar minus x, then this is still semi-algebraic. So the dimension, well, of course I should define precisely the dimension, but here this is very intuitive again, is strictly smaller. The number of connected component, okay, I will never discuss the empty set here, okay? I don't want to enter this kind of problem. I don't know, this is not a serious problem, okay? You agree with me about that. So you have only finitely many connected component. So this is an important finiteness property. And any connected component, let's put it here just for offer, then any connected component of x is also semi-algebraic. So we have this kind of stability property by taking connected components. Okay, so this is nice, but the problem is that this is too rigid. This is too close to usual algebraic geometry to be really useful to do new things in some sense. And so we would like more general and more flexible-themed geometries. And in fact, those other-themed geometries were developed exactly the same year as Gotendic published his manifest, namely, so model theorist developed-themed geometries exactly the same year. So basically there are two papers, one by Van Dendris, where he defined structures of finite type, and the same year, Pele and Steinhorn, where they defined all minimal structures. And this is now the accepted terminology. So this is the one I will use, and these are essentially equivalent, so all minimal structures. And the idea of model theorist is to have an axiomatic approach, right? So you start with the properties that you want, and then you prove after that that such same geometries, interesting such same geometries do exist. Okay, so there are two ways of presenting all minimal structures, one from model theory, which is in fact the most efficient one, and the other one, which is the geometric one. First, I'm not an expert in model theory, and second, it would take a long time to write all the definitions, so I will not do that. So I will choose the geometric presentation, so let's do that. So I remind you a few things about structures first, and then I will talk about all minimal ones. So what is the definition of a structure? So the structure expanding R is a collection S equal SN for each integer N, where SN is the subset of the power set of RN, satisfying the following properties. So the first property that you want is that the algebraic sets of RN are in SN. Second is that you want that SN in P of RN is a Boolean subalgebra. That means that you are stable, i.e. stable, and the union, finite unions, finite intersections, and complement. So you see that up to now, this axiom two is pure set theory. The third one, two, you want stability in the product. So if A is in SP and B in SQ, then you want A times B to be in SP plus Q. And then the fourth axiom is the geometric one. You ask that you are stable under linear projection. So if P from RN plus 1 to RN is a linear projection, and if A belongs to SN plus 1, then you want that P of A is in SN. OK, so this is the definition of a structure. This is just a collection of sets you are allowed to play with. And you just require those very basic properties. So you don't insist all the algebraic sets are not semi-algebraic? Yeah, but by 37, because of 4, you will get all the semi-algebraic. So this is the definition of a structure, so additional definition. So these are the bricks with which you are allowed to play. So the elements of SN are called as definable sets. And once you have the sets, the spaces, you have the map. So a map F from A to B is said to be as definable if A, B, and the graph of F, which is contained in A cross B, are as definable. So now you know what are the sets you are allowed to play with and the maps. So let me give one example. The trivial example is R-alge. So, well, the definable sets are exactly the semi-algebraic sets. So they satisfy all these axioms because the only non-trivial axiom is 4, but this is starts with the second axiom. So it's well known that a linear projection of a real algebraic set is not necessarily real algebraic because you will get some inequalities, but you get exactly the semi-algebraic sets. And the other remark is that this structure, by definition, because of axiom 1 and 4, is contained in any other structure. So R-alge, any other. The second remark is that if you have S1 and S2 two structures, then S1 intersected with S2 is also a structure. I hope the meaning is clear. This means that for each N, you take S1N intersected with S2N. And then it still satisfies this. And so, as a result, this implies that if you give yourself any collection, if f is a collection of functions f from Rm to R for different m or of subsets of Rnm, then one can define the structure Rf generated by f. So this is a smaller structure such that in the first case, all the functions f are definable. Or in the second case, all the subsets of your collection f are definable. All right, so just to make a few comments about the structures, you have a few facts that are easily provable. So first is that if A is as definable for S some structure, then automatically the closure, the interior, and also the boundary are definable. B is that if you have f from A to B as definable, then f of A and f minus 1 of B are as definable. And C is that as definable map can be composed. So if you have f from A to B and G from B to C as definable, then g composed with f is as definable from A to C. So these are easy facts, but I want to give the proof of this implication that A definable implies that A bar is definable so that you get some feeling of what happens. Otherwise, I would just give you a long list of properties. So let's consider A bar. So of course, you define it as being the set, the usual definition in topology. So you have quantifiers. So this is a set of x in Rn such that for any epsilon larger than 0, there exists y in A such that sum of xi minus yi square is smaller than epsilon square. So it's not clear on this that this is definable because this is not written in the terms of the axioms. So what I want to do is to try to write this as using the force axioms, the projection formula. And then you see that this projection axiom is used to cancel quantifiers. So let me give you the formula, and you will see that this is a pain. So if you think about it, this is the following thing. Now I have to erase. So I claim you can write it in this way, where what is b? Well, so you have to go to a higher dimension. Of course, you take the intersection in r to n plus 1 of this product with the set of x epsilon y in Rn cos r cos rm so that the sum of xi square, xi minus yi square is larger than epsilon, is smaller than epsilon square. So I hope my formula is correct.  But of course, the formula has become very complicated. You have to go to a higher dimensions. So I hope this example convinced you that formulas are much better than projections. And so this is where we go to model theory. I will be very light. So definition? If I just look at epsilon bigger than 0, it seems to me that maybe you need to put it inside the description that you have to put to use r bigger than 0 somewhere in some fact where it is a small detail. You mean here? Because you changed the quantifier for every epsilon bigger than 0 to the? But first, I want a to be in a bar. So I guess I need this. But you can take for any epsilon, right? Yeah, I could take for any epsilon. Because if you do it for every epsilon, it's not true. I am putting epsilon squared. Let's do it this way. No, it's not true because epsilon equal to 0 is not true. You cannot do it with epsilon equal to 0. So you must change the description. OK, I will leave it this way. And you just correct it so that this problem does not happen, OK? That's OK. Sorry, I shouldn't have been wanting to say that. So yeah, what I want to say is that you should move from this language using only projections to language of formulas. So let me recall what is the first order of formula. The first order of formula in the language of s is a formula. So this is not very precise. I'm sorry for again. I think you get the idea. It's a formula constructed out of the following rules. First is that if you take a polynomial that's real coefficient in n variables, then the sentence p x1 xn is equal to 0. And the sentence p of x1 xn larger than 0 are first order formulas. Second is that if a is an element in sn, then the sentence x belongs to a is also a first order formula. Three is that if you have two first order formulas, phi of x1 xn and psi of x1 xn are first order formula, then phi of psi, phi of psi, no phi, and phi implies psi are first order formula, OK? And the fourth is that if phi of yx is a first order formula, where y is in rp, so to be in rp and x is in rn, then if a is in sn, then the sentence exists x in a such that phi of yx. And for all x in a phi of yx are also first order formulas. And now the claim is that any set defined by a first order formula in this language will be definable. And so you see that thanks to this four, you are allowing quantifiers on elements of sets. So lemma, this is easy, this is by induction on the construction of formulas, is that if phi of x1 xn is a first order formula in the language of s, then the set of x1 xn in rn such that phi x1 xn is as definable. So basically, 1, 2, and 3 are more or less trivial, by definition, using the axioms of 1, 2, 3 of structures. The only non-trivial one is 4, but you just have to do the same kind of manipulation to write those sentences in terms of projections. So I'll let you use this as an easy exercise. OK, so it was a quick reminder on structure. So now let's stop one second. So what is the non-trivial geometric axiom? This is axiom 4, right? Because it tells you that if you choose a structure and in dimension 2020, you add a new set and you decree that this new set is as definable, then in all smaller dimensions, in particular in dimension 1, you would create a lot of new definable sets by taking linear projections of those guys. And so this means that very fast, if you are not careful enough, then this will be pure third theory and counter sets will appear. So actually, this level holds in a slightly stronger form that the smaller structure containing a given collection is exactly those things that you need to know. That's correct. Completely agree. But as I said, I don't want to enter into even basic model theory. But what you are saying is completely true. OK, so what I want to do now is to counterbalance this wild axiom 4 in order to do geometry and not pure third theory. So this is a notion of a minimal structures. And so what is the definition? A structure S is all minimal if you add the fifth axiom, which is that any element in S1 is a finite, is semi-algebraic, namely, it's a finite union of points and intervals. So finite intervals are always non-reduced to a point. So you are just asking that in dimension 1, there is nothing more than semi-algebraic sets. Yes, so the offer is right. You allow things of this form and things of this form are allowed. And so you see that there is a tension between this axiom and this projection axiom that creates a lot of new sets if you start from something random in high dimension. So of course, we still have a trivial example, which is RL. This is trivially O minimum. The second thing that you can say is a corollary of this definition. So first remark. And this remark is kind of strange if you think that we want to use it to do some hot theory and maybe some arithmetic geometry is that Z is never definable. Z is not definable in any O minimum structure. So it does not exist, because this is an infinite set, an infinite union of points in R. So another second remark is that periodic functions are not allowed, periodic functions like sine are not definable in an O minimum structure. Of course, because the pre-made of a point would be an infinite collection of points again. OK, non-constant. Thanks. OK, so one remark that I will not write, because I will have no use of it. I won't have time to talk about this, is that what is so special about R? Can we work with other fields? And the answer is that if you want, of course, there is nothing here you can start basically with any ring. There's no problem. But the O minimality condition needs you to have a notion of interval. So the best thing to do is to start with a totally ordered ring. And then it's not difficult to see that if you start with a totally ordered ring and asking this O minimal axiom that the ring itself should be O minimal, so satisfy this property, then you will see that you get a real close field. So in fact, you can replace R by any real close field. But I won't talk about that. OK, so we have the definition. So now we have the definition. We have only one trivial example up to now. But let me first prove that at this state that this axiomatic definition give you some nice tame properties of O minimal structures. So I will state a sequence of theorem and maybe tomorrow I will give you some indication of them. They are not difficult. Everything is completely elementary here, but it takes time. So the first thing that you can ask is basically what is the structure of a definable function from R to R. And so the first theorem is called the monotonicity theorem. Monotonicity theorem. And it says the following. It says that if you start from a map from an interval A, B to R, which is a definable, higher. So now I fix S by definition an O minimal structure. And S and definable means S definable. So I start with this definable map in some O minimal structure. And then the claim is that you can conclude that there exists a finite partition of this interval A, B such that F in restriction to the open interval A i, A i plus 1 is continuous, is continuous, and either constant or strictly monotonous. So now you feel a bit better. Those functions are really very tame. You have this very nice property. So one remark is that you can replace a continuous for a finite p, but not by c infinity. Of course, if you replace c0 by cp, then you will have to change this finite subdivision. So you will have a refinement, but usually it will not be possible to go to infinity. Yes, there are examples. There are examples of O minimal structure where you can't do that. So of course, there are kind of pathological from my point of view, but they exist. So this tells you what are one-dimensional functions. So now what about higher dimension? Well, the second, and this is the main topological result, really, is the decomposition theorem, where I think, no, not that one, decomposition theorem for O minimal structures. So this means that if you start with a finite collection of definable sets in Rn in some O minimal structure, then there exists a very strong kind of decomposition. There exists a cylindrical definable cellular decomposition of Rn. So I will call this a cdcd such that each ai is a finite union of cells. So you can decompose the full Rn such that each ai is a finite unit of cells. So now I have to explain what is such a cellular cylindrical decomposition where, OK, let me do this here. I guess I can now erase the properties of structures. So the definition of cylindrical decomposition is by induction, because you will see the cylinders. So for n is equal to 1, so you just have the real line. So what is a cdcd of R? Well, this is what you think. So this is just your finite partition a1, a2, aR in R. And what are the cells? Well, the cells are just the points. And the open interval, yeah, so where a0 is minus infinity and aL plus 1 is plus infinity. What is this notation? OK, so this is what happens in dimension 1.  And so what happens in higher dimension? Well, you work by induction. So a cdcd of Rn is a cdcd of Rn minus 1 plus for each cell c of Rn minus 1. So let me make a picture, of course, in dimension 2. So I have a cell. And above that cell, I give myself two functions, one strictly larger than the other. So this is fci plus 1. So fci plus 1 is greater or equal to fci. And so what are the cells in Rn plus 1? Well, you have the graphs of those functions. And you have the bands here. And you're allowed to take only finitely many functions. I hope this is clear. I will not write everything. You give yourself finitely many functions satisfying those inequalities. And then you consider the corresponding graphs and the corresponding bands. And the function is strictly less than the next one. Yes. And are they continuous or some say? No, not necessarily. Just definable, yeah. But of course, it will be part of the theory that you can refine the decomposition so that everything becomes continuous in restriction to the cell, right? So you use the monotonicity to prove this. So then you get also that maybe the band, as you say, the types are monotonic in some sense? You can also refine so that you satisfy those properties, of course. Yes. So you can refine so that those fci satisfy the conclusion of this monotonicity. OK. What else? Is there anything else? So you see, this is a very strong cellular decomposition. And as a corollary, what you get automatically already is that if you have a set which is definable in an all-minimal structure, then first, it has finitely many connected components, right? Because it's composed out of finitely many cells, so. But you need some continuity to verify the co-activity. OK, so there is a notion of definable connectivity. But in fact, over R, this would give you the same usual connectivity. But you're right. So this is not completely proven. But this is true, is that if A is definable, then the number of connected components is finite. And any connected component of A is definable in the all-minimal structure, because it is also finite union of the cells. And the same way, if you define the dimension as being the largest dimension of a cell appearing in such a decomposition, then you get that dimension of partial A is smaller than the dimension of A. So once you are here, you guess that you have a stratification theorem. And this is true. Yeah, so I'm cheating again. In fact, first, you have to prove that what I call the cell is really a cell in the sense that this is homomorphic in the definable category to Rk, for some k, any cell. And then you define the dimension as being the maximal k, so that you can embed definably Rk inside your guy. And you prove that this is the same thing as a naive dimension in terms of cellular decomposition. And it is also the dimension in the sense of topology that our history expresses some critical definitions. All of them coincide here. So there is no problem. OK, so what is the stratification theorem? The stratification theorem tells you that if you take some guy which is definable in some all-minimal structure and you are OK. And once more, again, as I said, there is a notion of Cp decomposition. Everything that you can do, you can do it C0 and then Cp. And so this is no problem using this kind of monotheism theorem. So here I will fix the p, because sometimes this is important for this theorem, is that there exists a finite stratification of a. And by this, you can write a as a disjoint union of strata, where a stratum is what? This is a Cp manifold, but a very particular one. This is a graph manifold. So first, it is a Cp manifold. It is definable. And this is a graph of f from u to Rn, which is, so u here is an open definable in some Rp. And f is Cp, and with bounded derivatives up to p. So this is a very nice stratification theory. And in particular, and I will use that later, the corollary of this stratification theorem tells you that if you have a in Rn, which is definable in some minimal structure and of dimension k, and which is bounded in Rn, then the k-volume, the k-euclidean volume of a is finite. Can you put the community of a is finite? Yes, this very strong decomposition theorem to give you a very nice, simple decomposition. And, well, let me mention one last result to give you one more tamedess result, which is a trivialization theorem. So in the stratum, it is not so clear what was the state. It's a graph of something. Yes, it's a graph of a Cp function from Rp to Rn. To some open set of Rp to Rn. And this map is definable, of course. OK, this map is in Rp plus n. But do you then have to use some inner transformation of the rotations of the coordinate or what to put it in? You want some kind of rectilinearization of this thing? Because it's not the same n. No, OK, OK, OK, OK, sorry. Rn plus p. Minus p. Minus p. Minus p. And but then you would need also to change something. Because if you take the graph of y per x squared, the derivative is not bounded, so you have to take the inverse. And so you have to do the rotation of the coordinate. So we need to be more precise in the statement. I would need if I really used it. But I guess it gives some kind of idea of what happens. So I will not try to correct this. But in fact, I think OK, I will leave it like this. But I agree with you that you have to be a bit careful about reordering the variables. OK, so what is the trivialization theorem? Well, it tells you that if you have x from x to y continuous and definable, then in fact, you can trivialize it in a very strong sense. There exists a partition y into yi, where yi is definable. So this is a finite partition, definable, such that, well, if you restrict yi, then here in x, you have f minus 1 of yi. Then this is trivialized on the full yi by yi cross the fiber, some z, where z is definable. Yes, so z is definable and this is definable, yes. And of course, you can do it with CP if you want. But you have to refine the partition. And as a corollary, I will not write it, because I'm already tired, is that if you have a definable family, then there are only finitely many homomorphism type in the fibers using this trivialization theorem. OK, so this is nice. This is completely axiomatic. This would be a bit sad if the only example were alge. But this is not the case, so now let me try to say a few words on this. Well, in some sense, it was already known at the time that there are other kind of tame geometries, for instance, subanalytic sets. So let me say a few words about that. So the main problem is given f a collection of functions, when is the structure rf au minimum? It's clear that if f contains a sine function, then this will not work.  I will not say much, but most proof work like this. So definition, a structure s, not necessarily au minimum, is model complete if, in the definition that I raise of definable set, the operation of taking complements is superfluous. And then the result is that if a structure is model complete, then it is enough to constrain the connectivity of some very specific set. So definition, an f set is x in Rn such that you satisfy p of x, f1 of x, k of x is equal to 0 for some polynomial p. I don't understand the definition. I'm not surprised because I was not very precise. So by this, I mean that you can construct all the definable set without using complement anywhere. No, but a structure already includes everything that you can construct. It's not a structure but a collection. OK, so I will try to be a bit more precise in the theorem. I think it makes sense vaguely. And I will explain what I mean by that. So let me start with my collection of function f. And then I define what is an f set. So these are exactly the solution of such equations where p is a polynomial and the f i's are some of my functions. And then the proposition is that, suppose that the collection f satisfies that Rf is a model complete and 2. So the way I want to say it to answer a offer, by this, I mean that you start with the graphs of all your functions. These are your basic definable sets. Then I can project linearly. I can take product. I can take finite intersection and finite unions. But I do not consider out of this complement. So now does it make sense? And then I say that the structure is model complete. If you get doing only those operations, you get all the definable sets in your structure, including the complement. Exactly. So this is what I mean by this vague sentence. Does it make sense? Yes. So suppose that your structure satisfies that property and that each f set has finitely many components. And then the structure is O minimal. In the usual to pluralical sense? Yeah, I'm working over R. So if you work over a real close field, then you have to define a definable notion of definable connectivity. But yeah, this is fine. So if you have these strong properties that you do not have to use taking complements, and if those very peculiar sets are finitely many components, then your structure is O minimal. In fact, what you prove here is that any definable set is a projection of an f set. So in particular, if the f set has finitely many connected components, then its projection also has finitely many connected components. And so in particular, in dimension 1, you have only finitely many connected components. And so you're done. This is important because this is basically the way you prove. The O minimality of our alge correspond to the fact that you have quantifier eliminations. But this is not true for all O minimal structures. On the other hand, model complete is much weaker.  So what are the examples that you can get this way? You get Rn. So what is Rn? This is a guy generated by the collection of restricted real and net functions. So what are those things? These are functions from multi-cube to R, which are restriction of real and netic on the neighborhood of this cube. And you extend them to Rn by putting 0 outside of the cube. Right much more. Then what we can prove is that in this structure, so this means that you are allowed to play only with real and netic functions defined on compact sets, basically. So what are the definable sets? It's not hard to show that the R and definable sets are exactly what are called the globally subanalytic sets. Time is passing and I don't want, if you have questions, we can talk about the precise definition afterwards. I don't want to make a lot of reminder about semi-analytic and subanalytic geometry. So what is crucial here is a Gabriel of theorem. So basically the proof of the hominimality comes from Gabriel of theorem, which tells you exactly that the complement of subanalytic is subanalytic. So all this theory was developed by basically Gabriel of Wojaciewicz and Ironaka. Let me move to another, at the time, more exciting structure, which is Rx. So this is just the expansion of the real field by the full real exponential function. So the proof that this is hominimally due to Wilkie and it uses a model theoretic techniques. So let me maybe say a few things. So an exponential set, so that you get an idea of what happens, is the guy of the form x in Rn, exactly of this form, so p of x1 xn, exponential x1, exponential xn, you can take the same number of variables in this example, where p is a polynomial. And a sub exponential set is a linear projection of such a thing, p of exponential sets. So pm of an exponential set in Rm plus m. And then, basically, Wilkie was inspired by this Gabriel of theorem and he proved using model theoretic techniques that the complement of sub exponential is sub exponential. And in the analytic case, it's kind of clear that the globally subanalytic have only finitely many connected components. And so this makes the proof. Here, this is absolutely non-trivial that the sub exponential finitely many connected component. But this is a classical result of Hovansky. So exponential sets and also sub exponential sets have finitely many, finitely many components. You look at this kind of equation, this is far from trivial. So as I said, I should write that nowadays, in fact, you don't need model theory at all to prove these results. So you can look at a paper by Lyon and Roland, where they have some kind of what the, ah, sorry, there is only one L. Preparation theorem, what they call preparation theorem for log exponential functions and the same in the analytic case. So this is purely, everything now is completely geometric. Example four, and this is the one I will use later in the course, is usually if you take two minimal structure, then the structure generated by those two is certainly not or minimal. But in this particular case, this is true. So you can take the structure generated by all restricted analytic functions and the real exponential, and this is still all minimal. So here, the Hovansky was that geometry didn't model? No, this is purely geometric. So the model theory was here in this proof. And this is removed by some geometric preparation theorem that gives you an explicit kind of parametrization for those sets. I don't want to enter this at all first because I didn't read in detail. And OK, so I've given the definition of Omeano structures. I have claimed that there are many of them. And of course, what is remarkable and interesting for algebraic geometries is this appearance of the exponential that usually algebraic geometry does not consider. But notice that this is only the real exponential. Of course, you cannot put the complex exponential because of why the sine and the cosine would be. So this is really real geometry. OK, so now what I want to do is to talk about, just say a word about globalization. It's nice to work in our end, but it's better to. So here you use the standard real numbers. You simply thought you could use any real closed space. Here I am doing the standard real numbers. It was just a remark of that. And as I said, I would not use it. But the logicians can go see this. Yeah, of course. Logicians are much more powerful. I know that about that. Any question? Is Tavsky's theorem used increasingly or not? No. Well, I don't know how to answer this question. It's not true that those structures have quantifiers elimination if this is your question. OK, so globalization. Well, it's just a usual business. You construct many-fold spaces out of gluing. But what is important is that you keep the finiteness. So you want at less of charts with only finitely many charts. So an as-defineable topological space or many-fold m is topological space m n-dode with a finite at less of charts phi i from vi to, so vi is in m, to ui open in rn, such that where it's defined, equidimensional stuff. So n is fixed multiple. Such that one for all ij, ui, and uij, which by definition is phi i of ui intersected with uj are definable. So you want your charts to be definable. And the change of coordinates is definable. So for all phi ij from, so by definition, this is phi i composed with phi j minus 1. So this goes from uij. OK, the way I did it is you have to take phi j, sorry, from uij to uji are definable. So this gives you a notion of, and of course, if you want the many-fold, you have to be sure that those change of coordinates are c0 or cp, if you want to have cp differentiability. And then what is a morphism? A morphism, f from m to m prime of s definable space is, let's say, is a continuous definable, a continuous nap, definable in the charts. Let me not try to look at everything in your finite collection of charts on m and m prime, and everything is definable. So you get a nice category of s-topological spaces, or if you want s-many-fold, I guess, s-topological space is enough for what I'm going to say right now. So of course, what are the important examples? The important examples, you start with x and algebraic variety over r, then you can equip x of r with its Euclidean topology. But it gives you also an r-algebra manifold, right? You cover x by finitely many affine charts, and then those realizations give you those charts there that you want when you take the r-point. So this is naturally an r-algebra manifold. Likewise, and this is the case I will be interested in, is x over c algebraic. Well, you can do it without any assumption on smooth. Here, I didn't ask things to be smooth. You said open is a red. OK, I don't have to put it. In fact, I just want a homeo if you want here. I can also do it. But anyway, with all the stratification business that I've explained you, you will see that this is no problem. So you can do it also for x of c. Well, just, for instance, proceed by restriction of scalar alave. And so you see that if you want, you have algebraic varieties over c, and then you have s topological spaces. So you can call this a definability functor. And then here you have the usual realization in topological spaces if you want. And here, this map factorizes. All right, so this finishes what I wanted to say about the geometry. I'm very late, as usual. And so now I want to kind of survey what I want to talk about in the course. So two applications, two, let's say, algebraic geometry, and in particular, and arch theory. So when I say algebraic geometry, this is, of course, of s c. So why should a complex algebraic geometry care about this kind of business? Well, let me start as a motivation first, because so we see that at least if you think of the c points, you get definable topological space in an O minimal structure for free. But this is a much larger class. So you can hope to try to work with spaces which are almost algebraic, but not really. And this is to prove algebraization results. So this is the first motivation. So there are two main algebraization results. One I will not talk about today and tomorrow. I will just mention it here, stated. What it will be very important next week is diaphanthine criterion. So diaphanthine geometry, usually you start with a very, very, very defined number fields. And you start, you want to understand the q bar points. Let's say, the q bar points. Here, you do the reverse. And this means that, so this is a fantastic result by P-line-Wilke that tells you the following. Suppose that you start with a subset of Rn, which is definable in some O minimal structure. So from now on, so let me write it for the last time in some O minimal structure. And from now on, definable will always mean in some O minimal structure. In some O minimal structure. Then you consider its algebraic part. You define it as being the union of all positive dimensional semi-algebraic sets of Z. And then you would like to know, well, is it empty or not, for instance. And also good because you can have a union of the point and the positive dimensional. And you don't want to count the points. Yes, but you know perfectly well the answer to this, right? I mean, pure positive dimensional stuff. Then, but in some sense, you're right. I mean, as I would say, I mean, a Pila and a Bombayian Pila proved the result concerning curves. And then they needed 10 years to see that the right generalization was taking this Z algebra. So you're right that it's not completely obvious what you have to do in terms of dimension. So what is the statement? The statement is that for any epsilon larger than 0, there exists a constant depending only on epsilon and Z, such that if you look at the set of points which are not in the algebraic part, and then you intersect them with Q to the n. So now I'm using the natural rational structure underlying Rn. And I want to count those points. So what I do, I take the naive height of those points. There are points in Qn, so I can take their naive heights. And the claim is that if you count such points, then you have very few. You have very few. This is smaller than C epsilon T to the power epsilon. So this grows in the height sub-polynomially. So in other words, if you have a definable set in Rn in some amino structure, and when you count the rational points, then they grow in the height, at least polynomially, then you know that you have a semi-algebraic positive dimensional component there. OK, so this is an incredible result. Let me make a picture. So the simplest case is just you take your favorite box 0, 1, 0, 1, and you take a real magnetic curve here. So C is real magnetic. Now you count the points on that curve, which are in Q square. And you count them with respect to the height. And if you have polynomially many, then your curve is real algebraic, OK? So this is really a very strong statement. And we'll see next week that this is crucial for studying functional transcendence properties in Hodge theory. So these are axianual type statements and then a typical intersection. But today is not what I want to talk about. So what I want to talk about has to do with complex geometry. So complex analysis, if you want. And the motto is that, so let me not write it. The motto is that the pathologies of complex analysis are not compatible with hominimality. This is really the motto. So let me illustrate this in dimension 1. So this is the following lemma, I think, which is kind of clear. So let f from a punctured unit disk delta star to C be a holomorphic map, which is also definable. So by this, I mean that I look at the underlying real structure. So C is definable, and this guy is semi-algebraic. Defineable in some immunostructure, again. Then, of course, you ask what kind of singularities can you have at 0? And the answer is that then 0 is not an essential singularity. In other words, you are meromorphic at 0. And the proof is nice. This is easy. Modulo is a great picartier. Otherwise, if you have an essential singularity, then by the great picartier, you can consider the graph of your function. And you take the closure. And then you know that when you remove the graph, then you will get vertical C. This is the content of great picartier. But then you get a contradiction, because the dimension then of the real dimension now of this thing is the dimension of at least 2. So this is 2. And this is dimension of gamma of f. So this is a contradiction to the different ability. Contradiction with definability in some minimum structure. Exponential 1 over z is not different. Sorry? Exponential is definable. No, real exponential is definable. Complex will never. So this is in dimension 1. So what can you get in higher dimension? So this is what I will use later. In higher dimension, then let me recall the classical result in complex analysis due to Rammert and Stein. So the problem is extending complex analytic subsets. So you start with S, a complex manifold. You can think that this is C to the n. This is not very important. This is. And then you remove some closed complex analytic set. And so now you suppose that in this open thing, you have X, which is C analytic irreducible subset enclosed in this complement. Then the question is, what happens when you take the closure, right? Then of course, in usual complex analytic geometry, the closure has no nice geometric structure. It's really disgusting. But the theorem of Rammert and Stein tells you that under some kind of timeness assumption, namely under this just dimension condition, if C, if X is sufficiently big, in other words, if the dimension of X is larger than the dimension of E, then when you take the closure, then the complex analytic structure propagates. Then X bar is closed inside S, and it's closed in C analytic and of same dimension, of course. OK, so you can use this kind of statement, for instance, to prove ciao theorem. Of course, this is over killing. So now what is the corresponding O minimal statement? Well, so I don't know if you read it. So there is an O minimal version due to Petrazine and Sachenko. So now what you add as a hypothesis is that this guy is definable in the sense that I stated before. You don't make any assumption on E, but you assume that this guy is also definable. And then you erase this, and then you have the statement. So now we can work with any dimension. But the strong assumption is that, of course, you need an ambient definable space and that the sub-thing is definable. Do you mean definable globally, not just locally? Defineable in some amino structure, globally, of course. Yeah, but before the assumption was local, it's analytic, it's very different. Yeah, I agree. I'm not saying that it implies it. But it's similar in spirit. But did you know already the closures are? No, you know nothing. Are definable. Why? Because you said before the general thing for a minute. Yes, I agree. But there is no reason, even if the closure is definable, why should it be complex analytic? Ah, OK. Let's make sure that's your question. Do you need to ask S to be smooth? Sorry. Yeah, here I'm asking smooth. I don't think this is really very important. But so the corollary of this is still by Patterson and Starchenko is a version of Chaotoram or minimal Chao, which is really nice because you can get rid of the assumption, the usual assumption of projectivity. So suppose that you have X, which is closed C analytic, and now definable in S n, where S is quasi-projective. Quasi-projective, you just have to think that S is A and C. So this guy is just C to the n, and you take any complex analytic subset of C to the n, which is definable in some minimal structure. And then the claim is that X is algebraic. The important point is that you see, I mean, if S is compact, if S n is compact, if you are in the projective case, then anyway, this guy is automatically definable in R n. And this is classical Chao. But here, what is important is that you can have a version of Chao in something which is only quasi-projective if you control how tame it is at infinity. And this is exactly what this definability assumption makes for you. OK, and I'm so late. That's crazy. So I can give you an idea of the proof, in fact. So I'm cheating. There is a proof using only hominimality, basically, a bit of complex analysis, but not much. And there is a proof where you use much more serious complex analysis. So I would give this proof. The other one is longer. So proof of the theorem. Locally, so it's a rough idea. It's not a detailed proof, so this is to prevent an offer to complain. So locally, so you look at the theorem. Locally, you have X in S minus E. And you can assume that S locally is open in C n. I'm working with a manifold. Then what you do is that you can assume also that X is pure dimensional. X is pure dimensional of dimension over C k. Now, as X is definable, the color is that I had from the decomposition theorem and the stratification theorem is that the 2k Euclidean volume in my open set in C to the n is finite. And then you apply Bishop theorem. Thanks so far. No, this is a remarkable result. So Bishop theorem tells you that in this situation, you have something complex analytic whose Euclidean volume stays bounded. Then when you take the closure, it's still complex analytic. And then to conclude, of course, it's important that you have only finitely many analytic shots. OK, so I arrived at what I wanted to talk about, namely, Hodge theory. OK, so in some sense, those results tell you that the first use that you can have for hominimality in complex algebraic geometry are some kind of algebraization theorem, which are really nice. But in some sense, I want to argue that Hodge theory is really at the border of algebraic geometry in the sense that, naturally, it produces transcendental objects, which are not inside the algebraic world. And that same geometry, to some extent, can deal with them. So more fundamentally, so what is the kind of transcendental objects that I have in mind? Well, these are objects which are constructed with algebraic recipe, to some extent. And nevertheless, they are not true algebraic varieties. So more fundamentally, some transcendental objects, there exists some quasi-algebraic transcendental objects produced by algebraic geometry. So I will come to periods. But let's start with something much simpler. So if you start with an algebraic group, so this is a nice real algebraic variety. Then if you take the quotient, so you look at all the metrics, the space of remaining metrics that you can put on Rn. So this is this quotient. So this is a naturally semi-algebraic space. But now what is important is you want to consider quotient of that form. And you have the feeling when you define this that everything is algebraic. This guy comes from equations. This guy is also algebraic. And this guy, you are just taking this quotient. And this thing, of course, so for n is equal to 2, this is a usual modular curve. But for n larger than 3, then this thing is not even complex analytic. And it has no natural real algebraic structure. So for n larger than 3, not algebraic in any reasonable sense. And the claim is that algebraic geometry naturally considers this kind of space. For instance, the theory of automorphic form is perfectly defined on this. So of course, this is the analytics side of the story. But suppose that you want to study Galois representations. Then it would be very nice to have that space to be a usual algebraic variety, such that you can take etalcomology and then make some kind of Lagrange correspondence, creating Galois representation in the etalcomology of this. And one of the main problems of the Lagrange program is that you have to go to groups such that analogous cushions are complex algebraic varieties. But you cannot deal directly with SNNR, for instance. But I don't want to. So this is a philosophical remark. But now what I want to explain is that Hodge theory really produces things of that kind. But are these spaces the same as right? So as I will explain, the first result that I will need is that those guys have a canonical Hodge structure. They are not semi-algebraic, in the usual sense of semi-algebraic geometry. At least, there is not canonically. So you see the main point is that on this kind of spaces you have all the echocorrespondences. And you want them to be algebraic also, in some sense. And I don't know how to apply Nash theorem or Tognoli theorem or this kind of thing to prove that there is a real algebraic structure, which is stable under echocorrespondency. So now let me make a short reminder on Hodge theory. So what is the Hodge structure, vz of weight n? So this is just a finite rank z module such that vc, when you tensorize by c, I mean the canonical decomposition vpq such that p plus q is equal to n and vqp is equal to vpq bar. And more abstractly, as explained by Deline, you can think of this as being just a representation of c star. So the restriction of scalar r from c to r to glvr. So s of r is just c star. So you are looking at a representation of c star. On the complex point, this is a representation of c star cross c star. So this gives you a bi-graduation. And the fact that this is defined over r gives you this condition. So Hodge theory is basically just a study of such representation, such star i. So already here, you see what will be important is the tension between the underlying z structure and the Hodge structure. The Hodge structure does not care about the z structure. It's something purely real. So now what is the basic of Hodge theory? Let's do it simple. So I would work on it with a smooth projective vitis. Then it tells you that if you take the associated topological space x of c and you look at the Betty Comunity of this, the usual singular Comunity, then this is the Hodge structure, of weight n. The usual construction for this is transcendental. It extends also to compact scalar manifold. Well, you use some parametric on that guy. And you take the harmonic forms that they compose in PQ types, and you get your decomposition. So this is very transcendental. So of course, you can argue that you have a much better construction in the algebraic case as follows. So you look at Hnx of c top. You take the rational Comunity. So what I want to explain first is that in some sense, Hodge theory is irrational and then that this is transcendent. So here you have a Q structure on this Betty Comunity. Then you use Durham theorem. So you know that this is the same thing as looking at the complex analytic manifold and the complex analytic Durham complex, Xn on the analytic space. So the hypercomunity of that complex by Durham theorem, this is the same thing as your Betty Comunity. Now you use Grotting-Dick theorem that tells you that this is the same thing. So now suppose that X is defined over a smaller field k. So we can take the algebraic hypercomunity. So this is a k vector space. And finally, how do you construct the Hodge structure? Well, here what you can say is that on this complex, you have the Hodge filtration, which is the filtration bed. So f, by definition, fp is the complex omega point larger than p. So you're filtered. Sorry? Yeah, I said Grotting-Dick, but OK. In that case, GaGa is enough. OK, I am not, sorry, I was thinking about the quasi-particular. OK, GaGa is enough. Sorry. Right, and here what happens is that this complex is filtered. So you have an associated spectral sequence with degenerates. This is the main content of Hodge theory. And then you get that this is isomorphic to the direct sum for p plus q is equal to n of hq of x of a k omega p x of a k. And then you can see. So this is another way of looking at this filtration. So now let me rewrite this. And I claim that, why do I claim that all this? So it seems that we have a natural definition, algebraic definition of the Hodge filtration. And this is true to some extent, but what is not true is the comparison that the comparison is algebraic. So by this, so let me recall, let me just write the full common g as being the hb of x. So this is the direct sum of all n. And then this is a q vector space, and I tensor it by c. So here, I will just keep this analytic well. So I get this period isomorphism with the Dirac homology. So this is a k vector space, and k is my subfield of c. And then I have this isomorphism, which is the Dolbo isomorphism with the Dolbo homology, the Hodge homology, depending on how you want to call this. It's just this direct sum of all n of these direct sums. And this is also defined as a k. Now, if you think a bit of those things, you can fix a basis of this vector space of a q and fix a basis of this k vector space of a k. So fixing basis, we see that omega is given. This period isomorphism is just given by a very big matrix is described by a very big matrix omega x in glnc, where n is the dimension of those spaces, which is the period matrix, which is just describing the change of coordinates. And you can see in glnc, it's intrinsic, but not in glnc. It's intrinsic in glnk, glnc mod glnk mod glnq. Yeah, but right now, I'm concentrating on the left side. I would go back to the right side after that. Then any z of a k inside x of a k, any algebraic sub-variate, defines what you can call Betty-Doram cycle. Or Doram, Betty, it's nicer. There is a cycle class associating to any such z, a class here without tensoring by c, and a class there without tensoring by c. So this is what I call, so this is a pair of classes in hd of x and h Doram of x, mapped one to the other by this pi. And then, glnc conjecture is that, suppose that k is a number field, then any Doram Betty cycle is algebraic, in the sense that this is a cycle associated, this is a comogical cycle, a pair of cycles associated to an algebraic cycle defined by k. So you see, I mean, the moral of this story, what I mean is that first, hd theory is irrational in the sense that it does not preserve the field of coefficients. You have to tenser by c, and this irrationality, but on the other hand, this irrationality is controlled by algebraic data. No day twist here. No, there have to be, but I don't want to write them. That's right, I agree. I agree. I'm already very late, too, if I put the day twist, I'm dead. And you can be more precise, and this is where irrationality goes to transcendence, is that you can consider cycles not only on x, but on the powers of x. So any cycle on x to the n gives via cunette homogenous polynomial relations of degree n between the coefficients of omega x. This is clear.  And then the Grotten-Dick-Pierre conjecture is that if k is a number field, again, then the ideal defining you take your matrix, omega x, then this is just a point. And you take the zicyclosure of a k. So now, this is an algebraic sub i t of j and k, right? Then the ideal is generated by the relations coming from algebraic cycle. So the model of all this is just to insist that Hatch theory that looks algebraic of a c, I mean, this isomorphism is just linear. So nothing happens. But if you are a bit careful about the fields of definition, first, this is irrational, and even more than that, this is transcendent if you look at all cycles in powers. But this transcendence is constrained by some very big conjectures. So here you take Hb is the direct sum of all the gradations. So the gradation is also a structure here that you are in school. Yes, yes, yes. In addition to. Yeah, I'm just going very fast. And of course, everything is wrong. But I guess you get the idea. So now, if you work on the same picture, but with the Dolbo or the Hatch relations, then, well, you get the similar kind of Hatch classes, similar kind of pictures. The Hatch classes are now the classes in Hbx with image in Hdolx RPP. So these are Q classes with image in the Dolbo comma g RPP. And the analog of the period conjecture is now the Hatch conjecture, that Hatch class is algebraic. And yeah, this is the same model. Hb is transcendent, but this transcendence is constrained by algebraic data. So let me let's. So here for Hatch, of course, Hatch is for any x over c. There is no, you don't need a base field. For the Glotendick period conjecture, the original version of Glotendick is of a number field. There is a version by André for a finitely generated field, which is not as precise as this. So you get an inequality, and I don't want to enter this. So let me be more precise. What are the algebraic objects in some sense that should control this transcendence phenomena? Well, let's do it at the Hatch level. So if you take a Q-Hatch structure of 8n, you can associate to it given by a representation of s to glvr. You can associate to it an algebraic group of a Q. And this is the important algebraic invariant. So the moment for the group of v, by definition, this is you take the image of s, and you take the Zeissky closure of a Q. So this is a Q subgroup of glvq. And another way of saying it is that you can show that this is a stabilizer in glvq of all Hatch tensors in all tensor power of v and its yaw. And if you prefer, this is also the Tanaka group of this tank and category, which is a category of sub-pure Hatch structures generated by v and its yaw, depending on your taste. And then let me mention the last conjecture. Because I think this is the one that explains the best the link between the transcendence and the fact that this transcendence is constrained by algebraic data. So the conjecture, which is you take gothendic period conjecture plus moment for state conjecture, it tells you that if you work over a number field, then if you look at the transcendence degree of a Q of the field of definition of a K of your period matrix, then this is the same thing as the dimension of the moment for state group of HBT of X. So of course, you have some transcendence, but you control it. Of course, those conjectures are extremely difficult, and they are number theoretic by essence. So there is no hope to touch them using all minimal methods. But I wanted to insist, because I think this is really what is behind all these kind of transcendence results that you want to prove using hot theory, is that in some sense, you look at the functional analog of this kind of thing. So what does it mean? This means that you have to move to the case where you do not consider a single algebraic variety, but a family of them. So this is what I want to do now and state the result. So now I start with X over S, a smooth proper or smooth projective. If you want morphism, where S is quasi-projective and smooth. So everything is over C as usual. Then you globalize the situation. You look at the hot structures on the fiber. So you get the notion of variation of a structure. So you look first as a local system of the module of the S an. Then you have a corresponding holomorphic vector bundle. So this is VZ tensor over ZS an, OS an. So at the fiber, this is the common G of the fiber with coefficient in Z in degree n. And so you get a hot filtration of V. I have a question on because you go fast on many conjectures. So you had the period conjecture relative to the deram realization. Then you have the host conjecture. And the host conjecture shows that the host tensors are coming from cycles. Yes. But then you said the conjecture is period plus morpho. Did you agree? It's not very precise. It's probably period plus hodge plus mom48. I mean, what you want to put here is really the motivic Galois group. But if you just know the period and the object of state for this setup, does it imply this conjecture or you need extra? No, I think you need extra. No, no, that's OK. You're right. Maybe I don't know. No, you're right. I should rather put hodge here. Sorry, you're perfectly right. So I have the hodge filtration. And why is it important to work with the filtration rather than the hodge decomposition? Is that because it varies holomorphically. So fv is a bundle of v. And then I have the flat connection nabla. So this is the datum of a variation of that structure. And you have some condition. That is that there is some relation between this holomorphic filtration and your flat connection. Namely, it shifts the filtration only by one. So in fact, this gadget, if you forget about this geometric origin, is the definition of what is a variation of that structure. And the remark is that I will not write it. Everything is algebraic, in fact, in this picture. Even if you give yourself a smooth projectivity, just a variation of a structure not coming from geometry, then Griffiths proved that everything is algebraic. In other words, this holomorphic vector bundle can be algebraized. The hodge filtration, too, and nabla, too, in essentially a unique way. If you insist that nabla is regular. So you do the delinear extension. You do the delinear extension. So this gives you the extension of v. Then you have to prove that f also extends. This is the work of Griffiths. And for nabla, by delinear extension, you know that there is a unique extension, which is algebraic if you insist that this is regular. OK, so now what I claim is that the transcendence that we have seen over a number field, you see it over c if you work over the families. So what does it mean? This means the following is that here you have sn. Then you can take its universal cover. Then you have your holomorphic vector bundle v. Then if you take the pullback to the universal cover, well, here you have a flat connection, but it is trivialized because this guy is simply connected. So this implies that v is canonically trivialized as being s tilde n cross some complex vector space vc, where you have z structure. So you can identify either with the fiber of your local system or better with the h0 of sn. So I cannot ask again any small question. So the finite order of creation extends. Do you want it to extend as a sub-bundle or a sub-sheet? It extends as a sub-bundle. And this is the resultant variation of? This is the result of Griffiths, a variational structure. It is proven in the paper by Schmidt, but Schmidt says that this is due to Griffiths. OK, so now what transcendence I'm talking about? Well, you put this comparison isomorphism in family. So this means that you start from your universal cover and you think of your filtration as living on that space.  And you think of it as being the period matrix with respect to vz. And so you arrive in some flag variety gc mod p. And I did not talk about an important point, which is the polarizability of the h0 structure. So it will be assumed everywhere. And so in fact, you arrive in some open subset, which is an open g of r orbit. And m is compact. So this map is just a map classifying the hot filtration with respect to the flat trivialization on the universal cover. But now everything is equivalent under the action of pi 1. And so you get exactly one of the spaces I was talking about before, namely g of r mod m mod gamma. So this is the definition. So this is a period map. So giving yourself the variation of our structure or giving yourself a period map is exactly the same thing. So what are the properties? Well, this map is allomorphic because the hot filtration varies allomorphically. And there is a condition coming from this condition. So it is horizontal. I don't want to enter into the details. I will do that tomorrow. So what is, and gamma is an arithmetic group in g of q. So what is g? g is the generic moment for take group of a fiber of your variation. And gamma in g of q, an arithmetic group. So what is the basic fact that illustrates this transcendence I was talking about? Well, the basic fact is that usually, so fact is that, in general, as soon as v does not come by some tensor construction from weight 1, I'm not very precise here, but in the very general case, if you take randomly a variation of our structure, if you do not have a family of forbidden varieties, says, then s gamma gm is the complex space, complex manifold s gamma gm has no algebraic structure. So you see that even starting with a very algebraic situation, if you encode your variation of a structure by a period map, where implicitly you are using this period isomorphism, then you get some transcendental object. First that quotient, and then the map, which is transcendental. So now let me state the results. I have five minutes. Should be enough. Let's do this here. Now maybe I will keep this here. And there is here. So what is the first theorem that I want to explain tomorrow? So this is due to Ben Baker, myself, and Jacob Zimmerman. So it should appear soon. It will appear soon. So it's to say that, OK, you created some complex analytic guy, which is not algebraic, this s gamma gm, but nevertheless, they have some nice geometry. The sameness of arithmetic quotient is that, and this is pure group theory. It has nothing to do with Hatch theory.  So this is what I call, s gamma gm is called an arithmetic quotient. So s gamma gm, this double quotient there, has a natural r-alge definable structure. And more precisely, there exists a fundamental set in g mod m. So g mod m is always semi-algebraic. So there is a semi-algebraic fundamental set for gamma, such that the projection from this fundamental set to s gamma gm is r-alge definable. So the full map cannot be definable because it is periodic under the action of gamma. But if you restrict yourself to a fundamental set, then it becomes definable. And then two is that this is factorial in the sense that any morphism s gamma prime g prime m prime to s gamma gm of arithmetic quotients is definable. So a morphism of arithmetic quotients means that the morphism essentially comes from a group morphism from g prime to g following what happens for m prime to m and gamma prime to gamma. I'm a bit vague here. But yeah, I'm allowed to change the base point. So I will make that precise tomorrow, but it takes too much time to do it now. And the fundamental set means that it intersects only far too many. Yes. And this condition is independent of the fundamental set. Yes. It really characterizes an independent set. It's not uniquely, but you will see that these are exactly Ziegelsets that I take. I take Ziegelsets, or union of Ziegelsets. But you don't say that. Once more, I mean, this is not already. I agree with you that this is not a nice statement. I should be more precise about this. Let's say there is an explicit construction of an F semi-algebraic such that this map is definable in our algebra. So what is the theorem 2, still by Baker, Micev and Timerman, is that tamedness of paired map. Those guys are wild. They are complex analytic. But now we have proved that the target is tamed. And the result is that if you start with any variation of a structure, polarizable, where S is quasi-projective, then the paired map phi S from Sn to S gamma gm. So of course, this guy has a real algebraic structure. This guy has an algebraic structure, too. But this map is not semi-algebraic. But it's still definable in R and X. I think this is a major result in that paper. This means that to write that map in nice charts, you need only to use restricted analytic function and the real exponential. Nothing else. OK, so it is tamed for this respect to this O minimal structure. S is smooth quasi-projective. Yeah, I just wrote Qp smooth. Am I allowed to take five more minutes? Or is it strictly forbidden by them? This is OK? You mean I'm already over? Ah, good. OK, so I think I like this result because conceptually, you understand the timeness of the geometry. It has a nice application. Application to hodge-locky. So you want to detect the points in S where you have exceptional hodge classes. So let's define this set. A set of close points such that there exists an exceptional hodge classes, hodge sensors for VS. So there is, in some tensor power of the fiber at S, there is some exceptional hodge classes that appear in that fiber, which does not exist on the nearby fibers. Then you can also say that this is a set of points such that the memphorted group of VS is not of maximal dimension. So now you can interpret. You want to understand the geometry of that lookie. So the remark is that, and this was done by Vail in 79, is that if you are in the geometric situation, in the geometric case, as the hodge conjecture implies that this lock S is a countable union of algebraic values of S. So the reason is that each time you see an exceptional hodge class, this means that there is an algebraic cycle appearing. But algebraic cycles have these nice properties that they glue together just because of the existence of Hilbert schemes. And so the lock S in the total space should correspond to a family of algebraic cycles. So when you project it, you would get something algebraic too, this is roughly the idea. So the question of Vail, can you prove this without assuming the hodge conjecture? So now in terms of the paired map, you see that this hodge locus inside S n comes from the paired map as follows. This is the pre-image of the union for all morphisms of hodge data, so g prime to g and some element g and gq, of spaces of the same kind.  And inside those ones, you have a collection of spaces looking very much like them, namely those S gamma prime, g prime, m prime. And you see that this hodge locus is the pre-image of that thing. And so the corollary of the theorem, well, the theorem tells you that this guy is definable. And the second part of the theorem one tells you that those guys seen as sub-vites here are definable sub-vites. So now the paired map is definable. So when you pull back one of those guys, you get some guy which is definable here. And because the map is complex analytic and this is a complex analytic sub-manifold, then the pre-image of each of those guys is a complex analytic and definable. Now you apply Petersil-Starchenko result. This is algebraic and you're done. And you know that the map from S gamma prime, m prime to S gamma gm is essentially a sub-variety or what? Yeah, it's essentially a sub-variety. But there is some properness, of course, to be taken into account here. And those are five-nets, so five, OK. Yes. So what you prove, corollary of theorem two, is you re-prove a famous theorem of Katanidine Kaplan with a bit less computations in the paper, namely that HL of S is a countable union of algebraic sub-vites. And without any assumption on the geometricity of the variance of our structure, what? Everything is polarizable. Everything is polarizable. Yeah, so now I extend really for, can I extend for five minutes? So what is another result which you can prove using this result is image of paired maps. So you see, I just erased it. You have this paired map five from the analytic space sn to that guy. And the question is, is there any nice structure on the image? Well, a first remark of Griffith is that you can always, extending a bit sn, assume that this map is proper. So the image will always be complex analytic. But the question is, can you really put a complex algebraic structure on the image? And this is non-trivial. In fact, this is kind of not so difficult if the image is normal. But a priori, there is no reason why it should be. And so you get this theorem of Baker, Brine-Barbe, and Jacob, and Timaman, which was a long time conjecture of Griffith. Let's say that, well, maybe let's try to be precise. Let phi from sn to s gamma gm appear in a map for a polarizable by a pure variation of a structure. Then there exists a unique quasi-projective variety, z, and a morphism g from s to z dominant such that you imagine such that sn, you have the identification of your algebraic morphism zn, then you have a closed analytic immersion, so closed immersion, and in fact, definable also, s gamma gm. I said that, but it was too fast. Griffith showed a long time ago that you can always remove the boundaries, the components of the boundaries. You take a normal crossing compactification, you remove the component of the boundary with the local monodromes trivial. You can extend through those ones, and then the map phi becomes proper. And so you know that the image is complex analytic. But your problem is algebra is this. And so this result of Baker, Brine-Barbe, and Timaman, it uses crucially theorem 2, plus the developer general, Gagaf formalism, extending the Ominimal Chow of Petersen and Sarchin-Clo. And to finish, let me state last result. So I want to talk about those results, theorem 1, 2, and maybe I should tell you I forgot to write 3. I will try to talk about this in more detail tomorrow. And then the last theorem that I want to mention now, I will talk about it, I think, the last Wednesday in one week. So you have one more result, which is a typical intersection statement. I will explain next week what it means, this typical against a typical. So theorem 4, due to myself and Anna Odinovska, states the following that if you have D2S polarized the variation of structure, assume, so I will make an assumption for simplicity, that the generic morphotic group, the adjunct group of that guy is simple. I don't want to have a product situation. I have a result in general, but it's more complicated to state. And then the question is, can you say more about the geometry of the hard lockers? You know that the hard lockers will be a countable union of algebraic subvarities. But for instance, you can ask, what is the Zajski closure of the thing? And then the result is kind of surprising to me, is that then either this hard lockers, OK, I cannot touch points. This hard lockers is a countable union of algebraic varieties of various dimension. I keep only the positive dimensional part. More precisely, what is mapped to some positive dimensional stuff in the period domain. Then the claim is that this guy, which appears as a countable union, is in fact either a finite union. So it is algebraic on the nose of special subvarities. In particular, it is algebraic, or it is dense. There is no intermediate, nice geometry that you can consider. So to really conclude, let me give an example. Classical one, take an algebraic subvarity of AG, moduli space of principally polarized bin variety. And then the claim is that either S contains only finitely many positive dimensional varieties, reducible ones, contained in special subvarities of AG and maximal for this property. Or the union of such varieties is dense in S. OK, sorry, assume S or generic. OK, I'm sorry. I'm late. I stop here for today. Is the assumption that the weight is not 1? Yes. Well, it's just because I'm not making an assumption. I'm just saying that when the weight is equal to 1, then this is a connected chimera variety. And it was known for a long time that I should mention that this map is already algebraic. So this is an old theorem of Borrel. Of course, we reproved it en passant. But it's not the same proof, by the way. In some sense, Borrel uses hyperbolicity properties. In some sense, in our proof, we also use hyperbolicity properties. But it's not exactly in the same way. And here, in the statement again, it doesn't look, it looks like some precision because a special subvarity of AG, I mean the focal special subvarity. Yes, that's correct, proper, contained in proper. Yes, there is always this ambiguity that the full space, of course, is proper, is special. Yes, so the plan is tomorrow. I want to discuss, try to give some details about those results. Then next Tuesday, I will talk about something I did not talk at all today, which is a really functional transcendence statement for this kind of period map, extraneous conjecture. And then on the last day, I would try to give some applications to atypical intersections and typical intersections. So I didn't say anything about atypical intersections. This is typically Andre Hort or Ziba Pink type conjectures. And a typical intersection is this one. Because here, you see that you do not put atypicality about the intersection. It's just a dimension question, which is not atypical. So that's the plan for those who plan to come.