What's up everybody Jason for jazby's case breaks calm 2021 flawless collegiate football just sold out This is a one box brief case break picker teams number seven fresh case guys with this one And again, we are going by our college rules guys So teams they are currently on or teams they play for the longest if retired So examples drew breeze will go to the Saints currently on is active example Tom Brady go to the Bucks And again, we had a couple of random number blocks. We needed to sell out which we did But thank you guys to everybody that bought in straight up first of all We had two random number blocks first one was for four teams including the Jaguars 49ers Bucks and Cardinals. There's the customer names there And then here are the customer names for the 10 teams that have the Bears Dolphin Chargers Lions Packers Colts Rams Raiders Steelers and Titans Which will be right there and Again, as you can see us as flawless collegiate football Little pick your team on it so that we know it takes it from us Now what we're gonna do here guys is we're going to And again, we'll go top or bottom Which one we'll use here. So again guys what I'll probably do is after this after print out the labels I blow the videos put the break away if nobody's taken any teams in the next pick your team What we'll do is we'll just Put the rest of those teams in a random number block So one two three four five six for a dice roller. I see we get a good roll here Nice one three one two three.  So the bottom will be for pick a team number eight There you go. Here we go three Alrighty first one we have is Des Fitzpatrick 12 out of 25 A little two color and I believe des Fitzpatrick was a Tennessee Titan. I want to say Yes So Tennessee times of course is for the random number block I believe yeah 10 teams To Shane Thorin there you go. Next one we have is Najee Harris three out of 25 Little patch there for the Steelers Pittsburgh Steelers that is part of the R&B for the 10 teams as well Three that is going to Joshua who got that straight up or sorry. Actually, that's your giveaway spot Another one there for the random number lock Next what we have is a four out of five Cornell Poel Which I believe that's Kansas City Chiefs little green four out of five And if that is Kansas City Chiefs, that's going to straight up to Tom I'll confirm that right now really quick though Yes, there you go Next one we have here is a five out of 25 Kalio Herbert Which also I believe that's gonna go to the Bears and yes Chicago Bears five out of 25 That's part of the random number block as well spot five going to mark last spot mojo We have a Niko Collins 10 out of 20 RPA Niko Collins always forget right away where he plays for Houston Texans I believe Yes, Houston Texans That is straight up Ryan. Oh, look at that beautiful patch JK Dobbins and that is a update actually from 2020 Update JK Dobbins six out of ten It's like a little house they bowl patch Baltimore Ravens going to Aaron He'll be back and last one is a dual autograph to a tub of Iloa and Jerry Judy which I believe mark bought the Broncos straight up So if you win the randomizer by the Broncos do that'll be yours mark If the Miami Dolphins for to a talk of Iloa win the random number block or sorry when the randomizer Then it'll be going to the 10 team R&B Which will be going to spot six Matthew. So I have to randomize at the end of the break So I know the corners are usually the big hits the middle is usually like something else So let's go with the middle first and see what we got here. Oh Very nice Devonte Smith Five out of 25 little diamond from Alabama fly Eagles fly Philadelphia Eagles Logan There you go. All right, let's go And I feel like the left side might have the hits today. Yes, it looks like it Yes, Calum on six out of 15 That is a little sapphire, which would be going to the Minnesota Vikings straight up Steve Locke All right, what do we got here guys? There's got to be something crazy here, right? Oh snap, I think that's green It's also a rookie. Oh, it's an RPA three out of five Trevor Lawrence Wow three out of five RPA Trevor Lawrence and that is gonna be going to the random number block Gale with spot three Wow Look how beautiful that it looks Very nice There you go Gale I'm good man. Thanks. Yes, sir. Have a good night. See ya. Hey, that is a beautiful RPA Alrighty guys, and there you go That was the break So let's quickly do that randomizer. So again, it's a dual autograph Broncos Dolphins Broncos when it goes straight up to Mark Dolphins when it goes to the random number block spot six which will be Matthew s switch scenes now dice roll here Dolphins, let's click roll Three to one four times. Good luck one two four Stays with the dolphins after four times boom So dolphins when that ran up when that randomizer which means it goes to the random number block So there you go Matthew. I sit right there man six out of 25 or 16 out of 25 So there you go guys that was the break that was flawless collegiate football number seven number eight on the website We'll random number block the teams up again If you guys want to grab your team straight up while you can guys appreciate it. Jasper's case breaks calm


So, I remind you, so end of the proof, so we want to prove that, so we have a variation of abstractions on a smooth, quasi-projective ITS, and then we get a period map, phi from S-an to some S-gamma GM, which is a quotient of some period domain D by gamma, which is a group in G, and G is a direct number 40 group of your variation of abstractions. And so we explained last time that you are reduced to the following, to the studying local period maps. So, this means that we look at the restriction of phi after compactifying S by a strict normal crossing divisor. So, we are reduced to prove that when you look at what happens locally at infinity, so we get this period map, from the star to the n, to S-gamma GM. And we want to prove that this is R and X, definable. Yes. So, delta is a disk, delta star is a punctured disk, and I take this product. But you could also take this, no problem. Yeah, but I don't need it, because in some sense I can admit that some of them are trivial. Yeah. This is enough. Okay, so this is what we want to prove. And as I explained last time, the situation is as follows. You look at the universal cover, so you have the upper half plane to the n. And inside this, we have the Z-gel set that I defined last time, right? And so you have this period map in restriction to the Z-gel set going to the universal cover D, which is G-mod M. And then, so this is the lift of my period map, and here I have my period map. And as I explained last time, we reduced to proving. So using the nil-potent orbit theorem, it's easy to see that this map is R and X, definable. And then to conclude that this one is, we had to prove that this phi of the Z-gel set is contained infinitely many Z-gel sets here in D. And then, as I explained last time, let me repeat it. So for n is equal to 1, this was proven by Schmidt, and we will use it later. In general, and this is a consequence of the SL2 orbit theorem, in general, this is more tricky and this is what I want to explain today. So the idea is that you use the fact that your variation of structures give you a rational representation of your generic map for T group in GLV. And so this induces the following picture. First you have D, which is G-mod M. Then you can project it to the associated locally symmetric space, G-mod K. K is in fact the unique maximal compact containing that M. And then this guy can be embedded inside the symmetric space associated to X to GLV. So this GLV-R, this way, modulo maximal compact, I can even choose SLV if I want to simplify, SLR, which is this way. And so what is this map? If you have a point Z in this period domain, really what you are associated into it is just the odd norm seen as a real quadratic form, HZ. Yes, I use the nature of it. So defining the VHS, V on S, right? Because this group is a generic map for T group. So it has a canonical representation. It is a generic map for T group. So it has a canonical representation in, yes, it's just because I don't want to have to provide the center here, okay? So symmetric space of GLV-R, okay? So this is just a set of norms up to a MOTC on that vector space of VR, okay? And why are we doing this? Well, I recall you last time these properties of Z-Gal sets. So we want to prove that the image of sigma H to the n is contained in finitely many Z-Gal sets here. But this is, what I claim is that this is enough to prove that this is contained in finitely many Z-Gal sets here, because the pre-image of a Z-Gal set here is contained in finitely many Z-Gal sets here.  So in fact, and now the second step is to say that, okay, I understand the Z-Gal sets here. Basically, this is just a property of being reduced, of a reduced form with respect to some norm, right? So this is what we did last time. And so the theorem that we have to prove, as we control Z-Gal sets, in X, it is enough to prove the following result that let sigma n be a set of elements in H to the n, such that the real parts, so Xi is the real part of Zi, are between 0 and 1. And Y1 is larger or equal to Y2, larger or equal to Yn, which is larger than 1. So as I explained last time, if you take into account all possible permutation of the coordinates, those guys will cover your Z-Gal set to the power n. And so to prove that you are contained in finitely many Z-Gal sets, it's enough to prove that the image of each of those guys is contained in finitely many Z-Gal sets. And now Z-Gal sets are contained in those set of reduced elements. So you want to prove that there exists a basis E of VQ and a constant Nc larger than 0, such that for any Z in that guy, sigma n, the norm Hz is EC reduced. So I introduced this one. I discussed the Z-Gal sets for X. So let me recall the definition, maybe, because obviously if you are like me, you forgot. So I remind you the definition that I gave last time. So let E being E1 up to En basis of VQ, and let's see a positive constant. Then an element B in X is EC reduced. If, well, you look at your quadratic form, then you have this property on the vector of your basis. So anyway, this one is positive definite, so I don't need an absolute value. So this is the first condition. The second condition is that B of, you have some, to respect some order. And at last, you have to bound the product. So you take care of the determinant. The product for I is equal to 1 to N is smaller than C times the determinant of B. Okay, so this was the definition. And these are the conditions classically given by a reduction theory. All right, so this is what we want to do. And I introduced the basic notation last time. So let me maybe remind them. So I introduced the ring O, which was the ring of functions on sigma N. But obtained by put back the projection P from Hn to delta star to the N of restricting analytic function on this delta N of restricting analytic functions on delta N. Then you can take polynomials in x, y, and also y minus 1. So this is a ring of polynomials in z1, in x1, xn, y1, yn. And y1 minus 1, yn minus 1, which is coefficient in O. Yes, and we have its fraction field O of x, y. And I introduced a class of nice function into those rings, which was roughly a monomial and polynomial function. So the definition was let f in O of x, y. Then it is said to be a roughly monomial. If y is equivalent to y1, s1, yn, sn for some integers, possibly negative one, of course, for some integers si. And it is said to be a roughly polynomial if it is of the form yf, sorry, if f is g over h, where h is roughly monomial and g is a polynomial. g is in O, x, y, y minus 1, and h is roughly monomial. Yes, and maybe I should have introduced the notation that I'm using, which was that the equivalence was that f, so there is a constant such that. And f is equivalent to g, f, if f. Just in the neighborhood of when you shrink delta star to the x, when you go to the y goes to infinity? Yes, when y is good to infinity. So this is for y bigger than something? Yes, you are always working anyway here. So all the yi's are larger than one and you asked that there is this constant. So do you have this when you go to infinity? Okay, and you take absolute value also? Yeah, sure, yes. And the restricted analytic functions means that they are analytic on some larger? Restricted analytic function is in the sense of last time. This means that basically you are zero outside a compact and on a compact your restriction of analytic function. Okay? But again, you don't react to this. All right, so this is it. And why was it useful? This class, it was a claim. So the claim was that f is roughly polynomial. And suppose that g is roughly monomial. So you see that basically the kind of conditions that you want to test are exactly this kind of being much smaller than. So these are exactly the conditions that would be here. And the claim is that for functions satisfying those conditions and you can test it, then f is much smaller than g on sigma n. If and only if it is true in restriction to curves. In restriction, this is an easy computation, to curves tau of the following form. Tau was given by a system of equation. Alpha 1 z1 plus beta 1 is equal to blah blah is equal to alpha n0 z0 plus beta n0. And then you fixed the next variable.  And zn is a constant of the form zeta n for some alpha in q positive. And beta i in r and zeta i's in sigma n. So this is the interest of these functions. And basically you see that we'll be reduced to proving that our functions, our norm functions are of this type. Okay, so let me now explain the argument. So the first proposition is that your norm functions are roughly polynomial. The structure of the proof is the following. Proposition a is that for any uv in vc, the function of z, which is az of uv. So this is a function that belongs to my ring. And it is roughly polynomial on sigma n. And second is that if you restrict to the diagonal, so you look at az of u, then it is roughly monomial. And this proposition will tell you that you are in business because then what you want to prove is this kind of inequality between those functions. So this one, for instance, will be roughly monomial and this one will be roughly polynomial. And so you will be able to tell this kind of inequality only arguing in restriction to curves. And then in restriction to curves, we'll use Schmidt's result in dimension one. How do you know this is live in O x y round? No, this is a claim. This is part of the proposition. Is that this function belongs to that ring and that it is roughly polynomial? Okay. Okay, but when you take the fraction field, you get six which are not defined somewhere. It could be infinity somewhere. Yes. So this means that the denominator is not... It's not. The denominator is not vanishing. Okay, it could also be re-deconcelation. Okay, it doesn't... I'm just catching you. I'm not giving you all the details. So this is proposition a. Why does it have to consider real beta in the equations? Q? I already said the difference between alpha and Br. Well, I wrote it that way because I think in fact this is enough. But no, you see, you're right. I mean, I should take... I should take... Sigma H, right? I mean, but I want my point to... The total points have to be in sigma n, right? So this is like, yes, this is not really sigma n. I mean, let's write it in H, okay? I think this is fine. Okay, so suppose you have proposition a, and I will come back to this proposition a bit later. So what is the second point? So the second point is suppose B that we have basis, E is equal to E1EN of VQ, such that there exists a constant C1, there exists C1 larger than zero, such that you have only the third condition here, such that the product of this Hd of EI is smaller than C1, for I is equal to 1 to n, for all Z in sigma n. Then I claim that if you have this proposition AEN, if you are able to exhibit such a basis satisfying only the third condition, then you are done. So that is condition C for Hd to be EC1 reduced, right? Okay, so how does it work? Now, for any curve tau as here, so for any curve tau as above, it was above on my sheet, but of course as in the claim. Then what happens? Well, you apply Schmitt theorem for n is equal to 1, so you can apply it because you are working with a curve.  So it tells you that the other applied to phi tilde of Z lies in a set of E tau, C tau reduced forms, right? Schmitt theorem tells you that if you apply to curves and you know that you lie in finitely many single sets, you can find a basis E tau, depending on tau, of course, and a constant C tau depending on tau, such that you are in that set of E tau, C tau reduced form, okay? And the problem is the uniformity. You want to exhibit an E independent of tau and a constant independent of tau. So let's do that. So what was the crucial claim that I also stated when I gave that example about EC reduced form? Let me recall what was the claim. So this is after some shrinking because you are working on an open disk, so you don't control near the bottom. You control it when you shrink slightly a bit. No, I don't have to shrink anything. The constant will depend on everything, but that's it. The constant that I choose depends on tau, but I don't have to shrink anything. Yes. So the claim that I stated also last time was that if you have B in X, which is E prime, C prime reduced, and if you know that you satisfy the set condition and E is the basis for which condition C is satisfied and holds for some constant, then in fact there is another constant depending only on E, C, E prime, C prime, such that B is also E C double prime reduced. So this was the crucial claim last time about this reduced form. So you apply this here. You are in the situation where by my assumption B the condition C is satisfied and here I have exhibited this form satisfying exactly this E tau C tau reducedness. So this implies that there exists another constant C tau, but let me write it in the same way, such that yotta phi of tau lies in a set of E, so now E is fixed and only the constant depends on tau, of E C tau reduced form. So these nice properties of E C reduced form tells you that you get the uniformity in the basis. Wait, this is what I am saying. I started assuming that I found it such a base, satisfying the condition C. Suppose you have a basis. I start with a basis satisfying condition C. Then I apply dimension one to all my curves, Schmitt result, I exhibit another basis E tau C tau, such that in restriction to that curve I am E tau C tau reduced. Then I apply this claim and I say that because E satisfied the condition C already, then I know that changing maybe the constant C tau, then in fact for each of those curves there are E C tau reduced. We know the question is why don't you have the determinant on your inequality. I think he is asking why do you have the determinant in inequality. Ah, okay, because... So the determinant of the original symmetric form is known. Okay, so... You change it using the value of... You have to put it here. But anyway, in fact you can scale everything so that you reduce to forms of determinant one. This is not a big deal. Determinant of each z. Z and sigma, okay? Sorry, just because... This was the original way it was written. Okay, in particular, once you are here, you know that... You know that you have this first condition A. In particular, this tells you that H z of E i E j in absolute value is much smaller than H z of E i for z in tau. Well, of course the constant here depends on tau.  So as z gives H z of E i E j is roughly polynomial and as z gives H z of E i is roughly monomial, then the criterion by restriction to curves, tau, gives that in fact you have uniformly, now independent of tau, you have H z of E i E j is much smaller than H z of E i uniformly for z in sigma n. Okay? That is, there exists a constant independent of tau such that... This I want to keep, this I want to keep, and this I can erase. That is, there exists a constant c2 such that this function can be compared like this for any z in sigma n. Okay? So now you take c being the max of c1, c2 and you get that yota phi of zeta is easy, easy reduced for all z in sigma n. Yeah, here there is no difference. You can treat A and B in the same way, yes. So you have to treat now really because you have these conditions in 30. This is a clever claim. Yeah, but it's written here when I say that, basically, when I wrote this one, right? Okay, but anyway, if you have open account of things, then you have some inequality. Okay, so this is the ID and without too many details and then it remains to show A and B. Those two statements. And this is asymptotic hot theory. So you really look at the degeneration of hot structures. So let's try to do that. So A and B follows from classical study of degeneration of hot structures. So if you have a variation of hot structure and you degenerate it at infinity then you will go to limiting mixed hot structures. So let me recall a few things about mixed hot structures. So I remind you the definition. I think I gave it last time, but maybe I will write it. So you want to have your increasing filtration on VQ, your weight filtration. You have your hot filtration on VC and you want that this thing to be satisfied that the Gros WL VQ is a pure off-weight L. Right? And as I explained, the same thing as a bi-grading. And this is important for what follows. The sum of the IRS where WL is the sum for R plus S smaller than N then Fp is the sum for R larger than P of IRS and the condition is that IRS bar, the complex conjugate is ISR, not exactly, but mod the sum of the IAB for A smaller than R and B smaller than R strictly. And I explained that you say that you are split, so sorry, yeah, sorry. This is badly written, this is ISR, yeah. And the condition of NA and P as well, so is R and S for R? Yes, yes. So you want, you say that this thing is split, is R split if IRS is ISR. Okay? And then I said that the link proves that you can, if you start from any VZWF then you can associate to this canonically a split one of the form VZ, of course you keep the same weight filtration but you change the arch filtration by some endomorphism so this is R split delta is canonically chosen in this sum. Okay, so now let me recall the generation and this is where things become to be complicated in many variables. So you start with VZ, you have weight filtration, so your variation of structures you have the allomorphic vector bundle associated to it V, NABLA the flat connection and F, your hatch filtration and you are on delta star to the N you have your variation of structures on delta star to the N. Where is delta? Delta is here, so this is a way of... So you choose the pieces of your filtration for the l-algebra so that you are negatively good. This l-algebra is the structures and so you take those pieces and you assume that you have unipotent monorhomy so with unipotent monorhomy you write them as exponential of Ni and Ni is nilpotent. And then what I claim is that out of this data you will produce mixed structures so to do this maybe let me recall the standard tool in this which is the monorhomy filtration so the classical remark of the lean is that if N is an unipotent endomorphism of VQ those guys are rational so this is important. Then associated to this you get a filtration W of N as a monorhomy filtration on VQ and it is characterized by the following properties that N of WNL is contained, it shifts this filtration by two and if you look you have a hard left sets integrated in this data namely the grue for this monorhomy filtration L is isomorphic when you apply NL to the grue minus L WN so this is SL2 triple business as usual so now of course the problem is that we have many filtrations and the order will consider the variables will give different results and this is why you have the introduction of this sigma N for the asymptotic so let's do that so let's see the cone, the open cone generated by the monorhomies lambda j lambda j positive strictly so this is an open convex cone of gr the l algebra generated by the log of the monorhomies then I guess I will erase this what is the classical theory the classical theory tells you the following that for j any subset of one N then you can look at the cone Cj contained in the closure of this cone C so the facet corresponding to j so this is the set of linear combination but only now for the parameters in your given subset with lambda j positive and so this is a facet of you have this cone and you have finitely many facets in the closure and the basic result of Catanie and Caplan which was really the beginning of the business in many variables is that all if you take any element in that cone any N in Cj determine the same filtration W Cj on vr so the monorhomi filtration that you get depends only on the facet not really on the true elements that you are considering and in particular because this guy contains rational vectors rational elements you see that this filtration is defined of a Q in particular for the elements which are not defined of a Q of course but you get only something of a R but as you know that this is independent and you have rational vectors everything is fine and it is defined of a Q okay so this means that for each facet but you see that this is really the facet the subset of one N that you are considering that plays an important role then what Schmitt SL2 orbit in one variable tells you SL2 orbit in one variable so this is the only SL2 orbit we use but of course it is crucial it tells you that if you take the underlying module to the variation of our structure then as a weight filtration you take the monorhomi filtration given by one facet and then you have to shift it so that this is centered in zero K so K is the original weight yeah I think I was confusing the notation at some point it was N but now N is the number of variables so let's call it K so this is the weight of your variation of our structure so you center everything at zero and then you take this F infinity that I defined last time which is the limit hatch filtration obtained by the nilpotent orbit theorem then this is the mixed structure and what is the next step the next step is that those weight filtration are compatible when J runs through a descending set of indices so you have a complicated compatibility between those things so let's try this so let T be really one T and this is where you fix the order of the variables otherwise if you fix a different order you get a different result then you can write, I will write the nice notation introduced by Javier in his survey which are much better than the one in our paper so you define it as being the weight filtration associated to this facet shifted by minus K for each T between one and N so T is now underlined to remember that you really look at this as being that subset and then the claim is that all N in CT in this facet they preserve the previous filtration W T minus one and you still have hard left sets for free and to the L in that facet CT maps the grue L plus J W T grue J W T minus one as an isomorphism to grue minus L plus J W T grue W T minus one J so more or less speaking what this means is that the nilpotent orbit in many variables will be sharply approximated by an SL2 to the N orbit once you fix the order of the variables and you restrict to the corresponding sector sigma N that I considered in the statement this is morally what it means what is proven later in a Kaplan-Schmidt paper but I don't need so now how does the asymptotic work well you start with N so let W N and F be the R split so each time you have to come back to R split stuff so this is really painful so B is the R split mixed structure associated to the first mixed structure given here associated to W N and F infinity so you start with W N and F infinity then you R split it and I call this R splitting the new hatch filtration is F then using the compatibility that I have written above it would give you a multigrading in fact of VC so using the compatibility of the filtration W T one can refine one refines Dulling splitting for one weight filtration into a multigrading into a multisplitting into a multigrading VC can really be written as the sum for all R S1 and SN of I R S1 SN such that the filtration Fp is the sum for all other than P of the I R SN and WLT now for all T is the sum for R plus ST smaller than L of I R S1 SN so if you look only at the weight filtration associated to T you take only into account the T's variable ST okay so you get this sorry so this F is the hatch filtration that I have introduced here and I am stating that you get a multigrading that gives you in fact all the weight filtration okay so now I need one more piece of notation for introducing the Kashiwa normate estimates so as G of Z acts transitively graphically on the check we can write everything is canonical unique multigrading satisfying those properties no but of course I have to give all the compatibility with the complex conjugate for each variables and I don't want to do that this is the same conditions as for doing splitting but this is true for all there are additional conditions so that this is uniquely defined once you have fixed those conditions so the complex group acts transitively on this dual flag variety so we can write so recall that we had this twisted PN map that descended to the disk and we can write G as being just an element G of T in F and for some holomorphic function for holomorphic function G from delta N to G of C and in fact if you want more details there exists a unique holomorphic V function from delta N to this direct sum A and B negative of GAB such that G of T is exponential V of T so you can choose uniquely this G in some sense is also it's not canonical but you have a natural choice let's say PSI I did not introduce this notation so you have PHI which is your period map at PHI twilder which was the period map at the level of the universal cover untwisted by the monodromy to kill the monodromy and you get a function PSI tilde but now this is invariant under the monodromy so it descends to a function PSI from delta N to D check no from delta N it extends by this is delta star to the N but the result of Schmitt that explained last time tells you that it extends to delta N and it goes to the dual flag variety for the twist that I introduced last time alright so with all those notations what is what is the result of Kashiwa and Katanika plan Schmitt about the Hodge norms so the theorem due to Kashiwa is 85 and Katanika plan Schmitt give a similar proof in 86 is that for all the in HN well we have this Hodge norm then and for you sorry sorry sorry okay all due respect the theorem is that for ah I forgot this maybe I shouldn't mention it yeah sorry I need one more piece of notation that I forgot is now you can glue together this by grading so notice that everything is only over R but after all those weight iterations are also defined of a Q so you can glue these things together to obtain we know that all weight all WT are rational and so you get a multigrading VQ which is really the sum for all S1 SN in the N of JS1 SN JS1 SN is really the grue SN for WN then the grue SN-1 WN-1 and the grue S1 WS1 W1 and such that the WT is the sum for ST-1 SN JS1 SN so I'll let you write with this description, I'll let you write the expression of the JS1 SN in terms of the I RS1 SN you have to make a sum I mean you just write what it means and then you get it yeah once more because it is deduced from this I guy can't modulate the conditions that I did not give okay so now the theorem of Kashiwara so I can put many K right and then Kapani and Kapani Schmidt is that for you in the piece so you want to understand the asymptotic of the odd norm of course it will depend on the piece of the weight filtration where you are so if you are in that piece then as a function of Z in sigma n so you will get an asymptotic on that sector then the result of Kashiwara is that this function is of that form then you also need to understand what happens when you twist by the model of me and then this is equivalent to the same exactly the same expression of course not necessarily for the same constant but and then C is that HZ now of this gamma of Z what was what did I do with the gamma of Z was this function here sorry I have to twist again by the model of me because this gives you psi and now you want to go back to phi so in some sense this function is a function that controls the distance between the multi SL2 orbit and your nilpotent orbit so and then you also have the same kind of asymptotic in the sense that this is asymptotic to the previous one EZ and U so this is the term on the norms now if you look if you look at the proposition A which was the fact that these functions are roughly polynomial then you will deduce directly from this estimate so I will not make the detail but it follows from that so proposition A follows from that result and now for the basis E so for B you need to exhibit the basis E when you choose any basis adapted to this multi decomposition so the C the left hand side is not dependent on S1SN and the right hand side does sorry no it does this is as a function U is in part C sorry that is U and then I am just saying that those two functions are essentially the same sorry excuse me so what is the basis that you choose you choose any basis compatible with this decomposition choose any basis of VQ adapted to the composition in terms of the J S1SN and then you have to check that the condition C is satisfied which was the product of the values and then the crucial thing is that everything here is centered around 0 so this means in particular that J S1 SN and J minus S1 minus SN is the same dimension because the weight iterations are centered in 0 and then the estimate A that you get here then estimate A so when you make the product you will see appearing that space will compensate that one in those products you will have Y1 over Y2 S1 but there will be also Y1 to the Y2 to the minor S1 and they will appear in the same numbers because they have the same dimension so you will get the estimate from estimate A you get the condition C on the product of the value of HZ of the EI forget the determinant ok so this is roughly the proof so if you are not disgusted let's move to something less technical so but the right idea is that the original idea you wanted to prove you knew the theorem in dimension 1 for n is equal to 1 and what you are missing is some kind of uniformity so at the beginning I was hoping that in some sense some nice statement like suppose you have a holomorphic function which is definable in some minimal structure in restrictions to each holomorphic curve then is it true that this is globally definable in that minimal structures so for real algebraic maps you have theorem like that and here I don't see how to prove this and I don't think this is true I think probably you can construct an example but here you see you go a little bit further you really need precise estimates more precise estimates in dimension 1 and this works but of course I would prefer a cleaner statement is there a step for real algebraic maps? I think if you have a real algebraic function which in restrictions to all let's say even lines or real algebraic curves is it real algebraic then this is globally real algebraic and how the restricted algorithmic functions appear in this A for position A well because anyway I mean they come everywhere here you see I mean in this case of business the untwisted thing descends and extends to delta n so here you have typically the kind of things that appear those functions are of this kind are they just a homomorphic function these are all the functions that appear that can be extended to delta n and they appear many times they appear in the nilpotent orbit theorem and they appear here again so they are real analytic? they are real analytic in the statement I don't care about anything homomorphic it's of no interest they are actually on an open containing zero containing zero this is why it has to be defined on something compact something compact okay but it is contained zero alright so in the first lecture I stated the direct application to Cateny doing Kaplan theorem and I gave roughly the proof so I will not repeat it just see that your Hodge-Locas is a pre-image of something definable and homomorphic so it has to be definable and homomorphic and so this is algebraic in S and then there was this theorem by Baker and Brunmar the image of the period map as a canonical structure of quasi-projective variety I have prepared this but it would I mean in some sense I don't think the proof is really enlightening so you have to develop a gaga formalism for definable complex analytic spaces this is nice but I prefer to move on to transcendence okay so of course we can discuss it outside the lecture alright so part 3 by algebraic geometry and transcendence so let me start with some philosophical remarks so the way I see variations of attraction in some sense is just a way of algebraically controlling universal covers so the identification of complex algebraic varieties right so this is a basic fact that over C you really have existence of universal cover so if you start with something algebraic then you can look at the associated complex analytic space and then you get the following picture that you have a universal cover which is not true in general if you work in a more restricted setting of any field and it's very classical that you can use this picture so you have a group acting on that complex space and such that that guy is a quotient and the first remark is that usually if this group is infinite then you really leave the category of algebraic space right I mean this guy will just be complex analytic without any algebraic structure but nevertheless I mean since the start of complex algebraic geometry that picture has been much used to study complex algebraic varieties I mean basically it was used by Euler to study elliptic curves by Abel to study a billion varieties and whatever so it gives you completely in some sense a bit different perspective on complex algebraic variety which is not really algebraic anymore so let me make some remark so those pictures are two basic questions the first basic question is that what are the finitely presented groups that can appear here that gamma that can be pi 1 s n where s is let's say was it projective or even projective and almost nothing is known about that question it's a classical question of ser and we know almost never nothing about that the second question is that complex analytic geometry is kind of wild I mean complex analytic space are much more complicated than complex algebraic varieties and so you can ask what are the geometric structures that you can expect for universal covers and here there is only one kind of conjecture so let me mention it so this is a Shafarovich conjecture yes yes yes yes sure there is a classical theorem of the limb that if you do not fix any restrictions then you can realize any of course the proof is very easy so what is Shafarovich so from now on s is smooth what is Shafarovich conjecture at least if s is projective it tells you that if you look at this universal covers it has to be allomorphically convex well I don't know if you have to call this a conjecture in these books this is written as a question rather than a conjecture but people now talk about a conjecture so let me remind you what it means in terms of geometries this means that you have a proper vibration of a Stein space so there exists phi proper vibration over r and r is Stein so something compact over something fine if you think in algebraic terms so as I said the main remark is that if by one of s and is infinite which is the case which is of interest because if it is finite then this guy is just finite at a cover of s so it is also algebraizable by the classical result of I don't know Riemann so for this picture to be of interest you have to work with infinite finite groups and then usually s and tula has no algebraic structure so of course there is a clear exception which is the case of a billion varieties which is a striking example but basically there is a recent result telling you that you have no other so there is a TRM to Clodon, Kola and Horing ok this is conditional to the abundance conjecture but I think everybody in this topic believe it so let s be even normal projective over c so here I am talking about the projective case there are results in the quasi-projective case but not as clear so assume the abundance conjecture which is the last step missing in the minimal model program at least for a no fit like me so it tells you that if s is not unirold then its canonical bundle is eventually base point free, sufficiently large so assume this conjecture then the canonical bundle is just which is q quartier yeah q quartier ok anyway I am interested in this smooth case then what is the TRM the TRM is that if s n is biolomorphic to quasi-projective variety if and only if it is really a product of c m by a projective one connected so essentially this is a bundle in a billion varieties of something simply connected I didn t check the paper and the paper is written with quasi-projective but I am sure Kola can answer your question so basically this result tells you that the only guys with algebraic universal cover up to phenomena of vibrations the basic pieces are really a billion varieties ok so now the idea of bi-algebraic geometry and transcendence is to emulate, to force in some sense the existence of an algebraic structure on the universal cover so this is a definition of what I call a bi-algebraic structure I think this terminology was introduced in a paper with Ulmo and Faf so you take s irreducible algebraic over c and maybe let's take it smooth for simplicity although you can give a general definition so a bi-algebraic structure on s is a diagram of the following form so you have s n you take the universal cover s n twilder then give yourself holomorphic map to the identification of an algebraic variety where f is holomorphic and x is algebraic f is holomorphic and you want the link between this f of course and your base so you ask that this map is equivariant under an action of pi 1 so pi 1 of s n equivariant for some morphism from pi 1 of s n to the algebraic automorphism of x so what is the idea, the idea is that you try to get some kind of model for the algebraic model for the universal cover so this is a very formal definition of course this is interesting if f is big enough for instance you do not lose too much in terms of dimension, if f is an embedding this is even better and so on and so forth so you are trying to test the structure of the universal cover in terms of various algebraic models and once more this is interesting only if the image of pi 1 is infinite because otherwise there is not sufficiently many link between the base and here and then the transcendence problem will appear in trying to compare this algebraic structure that you force on the universal cover by that map and the original algebraic structure on s so let me give a typical example how we came to this definition let's take an obedient variety or you can take a torus but let's take an obedient variety of a c then the picture is the following this map is just the identity to this the example 2 is Shimura varieties so of course this case is trivial because here really the universal cover is algebraic but for Shimura varieties this is more interesting so suppose that s is s gamma gk with my previous notation assuming that d is a permission type and so this is a connected Shimura variety and then the picture becomes the following so now the universal cover is not algebraic it's an emission symmetric domain but as we have seen so this embeds as an open set in something which is algebraic which is this flag variety and this map this map is even g equivalent and gamma is contained in g so this is so h from gamma to g of r to g of c and g of c is the algebraic automorphism of of the check okay and so still here this is a nice example because as I said I mean this is really an embedding but now you see that the variations of abstractions are just a marginal example than example 2 and where it might be possible that you lose some dimension I mean this period map is not an immersion anymore so you start now with any smooth quasi-projective of a c and suppose that you have a variation of abstractions so now you want to test the algebraic city of the universal cover by looking at variations of abstractions so you fix one a non-trivial dvhs in particular you assume that the p1 of s1 is infinite and then you have a European map to d which is g mod v and then the map that you are really looking at your f for the definition is again you put it in the dual flag guy but of course now this is more complicated you don't really understand that map s1, phi, twilder has no reason to be an isomorphism and of course in general it is not but still this is a algebraic structure in the sense that it is different okay good so to which extent can you understand these algebraic structures as really putting an algebraic structure in the universal cover at least you can define algebraic sub-varities so we say that v an algebraic erodici-ball, ah, okay, sorry analytic erodici-ball sub-variety of the universal cover is said to be algebraic, algebraic with respect to the bi-algebraic structure that you have fixed bi-algebraic structure on s if, well, it is a pre-image by that map f of something algebraic here if v is f minus one of something algebraic but now you know what is something algebraic it has to be the Zarski closure of the image so you can even say this and while there is a problem of erodici-ball components so if it is an erodici-ball component of that pre-image, okay and second now what is the transcendence problem that you want to study you start with y algebraic now in s and erodici-ball then it is bi-algebraic so the idea is that if you look at something algebraic in the universal cover of course it is algebraic with respect to x to this morphism but it has nothing to do with the algebraic structure on s so when you project it back usually this is something disgusting not even complex analytic and so there should be very few things which are algebraic at the same time upstairs and downstairs and those guys are called bi-algebraic so this guy is called bi-algebraic still with respect to the given bi-algebraic structure if one equivalently any erodici-ball component analytic component of pi-1 of y-analytification in s and tilde is algebraic in the first instance is algebraic in the sense of 1 okay and the idea is that the collection of algebraic sub-ities will give you something interesting so let me give you examples of this example 1 so you started with your abelian variety and you ask yourself what are the bi-algebraic guys in your abelian variety and the claim is that this is the same thing as being translated for an abelian variety so really you are capturing something out of the geometry so if you are bored you can try to prove this this is a nice exercise about the universal property of the Albanian map it's very easy but okay example 2 so this was the case of shimura varieties so the result is due to ulmo and ff is that if you start with some algebraic erodici-ball sub-variety of a shimura variety of a connected shimura variety then it is bi-algebraic even on af in terms of group theory y is what is called a weekly special so let me not define this because I will define it in example 3 which is more general but you can just think that this means that shimura variety if you look at it as a complex anatic space it has a canonical localisimetric emission metric and then the claim is that this is the same thing as being totally geodesic and this is due to moonen well this equivalent let's write it this way this is due to moonen that this is the same thing as y totally geodesic for the canonical metric so you see that these notions of transcendence in fact they capture classical stuff about localisimetric spaces like shimura variety and well example 3 let's go back to the case of general variations of architectures this was properly done in paper by myself and Odvinovskar is that so suppose you have your D variation of architecture and you have your period map to D then you are in this situation then the claim is that y in S irreducible algebraic is bi-algebraic once again if I leave y as weekly special so I define the notion of special variety was a component of the hodge locus weekly special is a bit more general it is the following so let me tell you what it means this means that that means that y is of the form so now notice that I am restricting myself to the variation of a structure where I really have this and this is the main problem in the theory I will come back here at some point to this I am asking that why is the preimage of S gamma 1 g1 m1 cross a point x where you have one special variety S gamma 1 g1 m1 cross S gamma 2 g2 m2 embedded in S gamma gm so in S gamma gm you have special sub varieties which are exactly the guys of the same shape it can happen that the guy of the same shape corresponds to a product group g1 times g2 in that case this special sub variety will be a product and then a weekly special is when you keep one of the non-trivial factor of the product but the second factor is just replaced by a point and if you think in the case of S gamma a total geodesic will always be of that shape ok so now that I have given you the basic definition why do I think this is interesting? well there is a full heuristic about this kind of structure which is very rich so further is a functional transcendence heuristic which has proven to be extremely useful for solving a number of theoretical problems so let me explain this now functional transcendence heuristic for theme and this is where theme geometry comes into the picture, bi-algebraic structure so I did not define what is the theme bi-algebraic structure and the reason is that I don't have yet the right definition but I will give you some indication of what I mean by that but let me write first the heuristic suppose that S is endowed with the theme bi-algebraic structure this is just a heuristic so of course you can find counter-examples this is no problem but it's used as a guide and describing some similarities between different problems so let me not to brutal and explain first what is really most geometrically speaking to me so the first thing is that so this is called Axlindemann I tell you that for any y in s n algebraic in the sense of the bi-algebraic structure and irreducible ok so you have this guy which is algebraic in the universal cover then you push it down by the projection to your original variety this is not even complex analytic but you take the Zeissky closure of that thing then the projection of y and then you take the Zeissky closure and the claim is that this guy is bi-algebraic I think this is a very meaningful geometric statement and there is a more general statement where you see that this is really a functional transcendence which is called Axlindemann and which says I suppose that you have you let you in the product so this guy has naturally an underlying algebraic structure because s is algebraic and I'm considering the algebraic structure in the universal cover coming from my bi-algebraic structure and now I take some guy here which is algebraic for the product bi-algebraic structure bi-algebraic so this means that with respect to the universal cover it is bi-algebraic for the nature of bi-algebraic structure on that product the one coming from the algebraic structure here and the bi-algebraic structure on the universal cover so you require that when you push down to a certain process arm you get something algebraic and when you go by the morphism to your variety process arm you get yes let w be the intersection with the diagonal where the diagonal is a graph of pi and you take an analytic irreducible component of this where delta is a graph of pi then there is a very precise dimension estimate which always is me but then the conjecture is that the co-dimension in u of this w so the thing you started with is always larger than the dimension of the smallest bi-algebraic guy containing the projection of w so this guy is the smallest bi-algebraic sub-variety of fs containing pi 2 of w p2 of w w is living in that product, a project with a second factor and I take the smallest bi-algebraic guy that contains it okay and then Axlinderman is a special case of action for u of the form y cross pi of y zar but you take this guy of very specific form you take it as being a product with then 2 implies 1 easy exercise so examples of this heuristic 1 and 2 are theorem due to Ax so for example Axlinderman tells you that if you take an algebraic sub-variety of c to the n and you push it down to your abelian-varity you take the zaricic closure then this is a trans-state of an abelian-sub-varity okay and Axlinderman will give you a more precise statement about dimensions when you do this kind of procedure so example 2 well Axlinderman is due to myself in Mo and Faf and Axlinderman was proven more recently by Pilar Mock so I think we proved this in 14 and Pilar Mock and Zimmerman so I'm not sure when it was published the first draft appeared around 17 so this statement tells you that you start with a bounded symmetric domain and you take the intersection of your bounded symmetric domain with something algebraic in the flag-varity then you project it to the Shimura-varity you take the zaricic closure then this is a weakly special sub-manifold so it gives you a very strong information then example 3 so the case of a variation of hatch structures so it was conjectured by me in a paper that will maybe someday want to appear and manage to make the proceedings no pressure and this was proven last year by Baker and Zimmerman so I think the conjecture is something like in 16 and now we are back to the first lecture because the main ingredient for proving all this kind of functional transcendence result is the other algebraization theorem due to Pilawilke so let me think a bit about how I want to proceed maybe it's useful that I don't want to give something completely empty for today, for the second part so maybe I will try to prove axiom demand for obedient varieties so this is a baby case but still you see how it works so you see all the main ingredients the main ingredients are a mixture of arguments on the monromy and the application of Pilawilke theorem and tomorrow I will go probably to try to explain some axiom demand or axiom well in hyperbolic settings so how does it work? so what is the theorem? I remind you you start with an obedient variety so you look at the uniformization map and so you have a map from cg to the unification an which is cg mod lambda where lambda is some z to the power 2g and now as I said one way of saying it is that you take something irreducible algebraic here you project it here and you take the zarsky closures and this is the transit of an obedient variety but equivalently you start and this is easier to prove in some sense just a matter of perspective but you start with something irreducible and algebraic in A and now you take something irreducible and algebraic in the preimage of v and you take it maximal for this property maximal for this property of being irreducible and algebraic contained in the preimage for these properties then you want to prove that the projection of y is the translate of an obedient variety this is the theorem due to ax and now let me give the proof or at least idea of the proof so that you see the ingredients so first thing that I do is that for simplicity I would assume that A is simple it's not a big deal but it makes life easier so assume A is simple y is maximal inside the preimage of v anitifier for the two properties of being irreducible and algebraic so assume that A is simple and then what you want to show you want to show that v is equal to A in that case as soon as the dimension right well let me let's say first reduce can you can assume replacing v by something smaller you can assume that so you can inside v you will have y zaiski closure so you can assume that this is v if not you replace v by something by this guy which is smaller and so you have the same statement okay so once you are there what you want to show is that in fact v is equal to A because you want to prove that v is a translate of an obedient variety but A is simple so v has to be A okay at least if the dimension of v which is what I'm assuming is positive just don't take just a point okay so this is what is enough to show this and this is important because it tells you that you really use a group of automorphism you want to show that the stabilizer theta y which is stabilizer in CG of y is positive dimensional because if you prove this then you know that v which will be the pi of y zaw will be stable under the zaiski closure of this group projected to A but the zaiski closure of this group as A is simple as to be A so you will prove that v is stable under A so it has to be A okay so this is really the idea is that to prove this kind of statement you prove that your y is very homogeneous so how do you prove this well this is why you use pilawilky theorem I remind you that pilawilky tells you that if a definable entity contains sufficiently many rational points then it contains positive dimensional semi-algebraic stuff so define sigma of y as being the set of elements in your group CG such that y plus z intersected with the fundamental domain of the action of gamma is different from the empty set and y plus z is contained in the pre-image of v so what is the picture the picture is that I have well first I have my fundamental domain so let's look at it as being a cube so I have my big cube which is my fundamental set F and then gamma acts by translation on this in the universal cover so I have pi minus one of v so let's imagine this is something really analytic of course I'm not able to draw this but I guess I get something like this pi minus one of v intersected with F and now I have my y which is algebraic so in some morally speaking it's much less curved because it is algebraic so it's living here and now I'm trying to translate my y so this is y and then I'm trying to translate y by some element so that of course this guy if I translate it's still algebraic because y was algebraic such that I'm still cutting the fundamental set and I'm still contained in the pre-image of v so I'm trying to deform my y into something still algebraic so of course now because I want to use pilawilki I take F as being semi-algebraic so in my case just a regular usual cube and then you see that this set sigma of i is in fact nothing else than the set of sigma in Cg such that the dimension of y plus z intersected with F intersected with pi minus one of v is the same thing as the dimension of y so now in the case of a billion variety of course the uniformization map is not definable in any or minimal structure because this is periodic but this thing from the fundamental domain is trivially undefinable because everything is compact so here there is no problem so this implies that if you look at this definition the dimension is a definable function and this guy by definition will be definable this is algebraic so this implies that sigma of y inside Cg is undefinable I don't care that this is Rn I just care that this is some or minimal structure and so what I want to show I just want to show that that guy contains some positive dimensional semi-algebraic set and to do that I use pilawilki so what is that enough so this is enough to show that sigma of y contains a positive dimensional semi-algebraic set w because in that case for any then by maximality of y then we'll have that y plus w is equal to y because otherwise if my w acts semi-algebraically and I look at what it generates with y then I will get something algebraic of higher dimension still contained in pi minus 1 of v no but you don't know that w contains the origin maybe it is a little bit shifted away so you know that y plus w maybe is bigger dimensional but maybe it doesn't contain y okay I have to be careful to shift everything to 0 okay let's be careful about not be careful about this it is but this is the argument I'm just saying that I'm creating a larger dimensional algebraic families and still contained in pi minus 1 of v okay no but you will see that I produce something that contains 0 anyway so okay and then I will be here and then I will be done because then I will know that this stabilizer contains that w so this implies that w is contained in theta y and I'm done okay so to do that we apply pilawilki and so we have to count rational points so of course the rational points will be counted in the group cg with respect to the integral structure given by the lattice lambda so what I count is really the intersection of sigma of y with lambda and then you see that this is the set if you make if you try to understand what are the integral elements here you will see that you are just a guy such that this problem become much simpler you are just counting the element in your lattice such that when you translate y by this element you are still intersecting the fundamental set right so you have your lattice you have your y which is algebraic in particular it's going to infinity and you are just counting all so you have your reference f and if you want you are just taking back y to your original fundamental set by an element and you are counting the number of elements okay and the claim which is an easy exercise is that now you count the set of elements that appear in pilawilki so we take this element there such that the height of z is smaller than t and then you will see that the height of z is essentially the infinity norm that is connected to lambda okay nothing else and so it is easy to see that this is larger for t sufficiently larger than t over 2 the idea is that each time you go to another fundamental set then the height moves at most by one it moves by one okay so and so you get this and then you apply pilawilki it tells you that this definable set sigma of y at least polynomially many in the height rational points in fact integral points and so it has to contain a positive dimensional semi-algebraic set okay and so we are done so this is the proof of ax in demand in that case okay so after so you have seen most of the ingredients you have some problem of timeness you have to prove that some maps restricted to fundamental sets are tame here this is trivial but in all the other cases this is deep you have to use the monodromy here this is very simple because the monodromy is a billion and you have to understand the height and once more here this is very simple because everything is flat the height is just in infinity norm with respect to the lattice of course in the case of shimua variety this is much more subtle this is related to the remanian distance you have to really compare what happens and to use non-positive curvature but this is basically the idea and for variation of a structure well you have to use the non-positivity curvature in the horizontal direction which are the only ones that appear at map C okay good so now the point of all this is that out of this functional transcendence heuristic you can transport this notion of biological structure of a q-bar and then you will get really a true transcendence problem and some atypical intersection heuristic so a definition a q-bar biologic structure is a biologic structure S and tilde as before such that biologic structure such over C such that S is different of a q-bar so that space X is different of a q-bar and the monodromy takes value in the automorphism of a q-bar X okay so you are just enriching the data of a q-bar and in particular this gives you this will define you a notion of q-bar by algebraic subvarities so this will be the algebraic subvarities of a C such that they are defined of a q-bar in S and there are the pre-image of something defined of a q-bar in X okay so this makes sense and what is the interest is that well with my naive definition of a C all the points were by algebraic but here now you also have a notion of q-bar by algebraic points in particular you get the notion of a q-bar by algebraic point is a point S in S of q-bar such that there exists an S twilder in the unedification in the universal cover such that F of S twilder belongs to X of q-bar and pi of S twilder is S you have a point here of a q-bar, a point there and they define the same okay so example one of the algevarities of a q-bar here the story is a bit subtle because you have to understand what is the q-bar structure that you put on the universal cover so canonically the universal cover is the Lie algebra of A so you could put the natural q-bar structure on the Lie algebra of A and this so you would have Lie of A and here the map would be Lie of A of a q-bar then take C and then an 85 this would be your X with this q-bar structure and here this is AC and you have the projection to A if you do this there is a classical result of transcendence theory by Lang telling you that there are no q-bar algebraic points except the identity of the algevarity this is not and this is where now the things become tricky so if A has complex multiplication then the right q-bar structure is given by the lattice of period and then if you take this you prove that the q-bar algebraic points are exactly the torsion points if you want of course T.M. abilities are different of a q-bar so this is a particular case of what you want and in general it's more tricky this is the universal vector extension of your algebraic vitality to really define the q-bar by algebraic points and then they coincide again with the torsion points but you see that the better picture enter into the picture because the natural lattice is really the lattice of periods which is normal you are working at the level of the universal curve there is no reason to take the Lie algebra with its name the q-bar structure so I am not giving more details just that this one is not the right one example 2 so this is the case of Schimoire vitality you can just take the first case of the middle arc curve y0 of 1 so by the j function this is c and so here you are embedding inside p1c this is your model and here everything is naturally defined of a q-bar and so what are the q-bar by algebraic points you are looking for tau in each intersected with q-bar such that j of tau is in q-bar these are your algebraic points and it's well known by Schneider theorem that this is the same thing as asking that in fact you are not in q-bar but you are imaginary quadratic and this means that your point z which is pi of tau so your z here or if you want j of tau as complex so x is the same point so the point in your middle arc curve parameterized by j of tau is the point with complex multiplication so you are exactly parameterizing what you want the point with complex multiplication in your middle arc curve and in general you get exactly the same result so this is due to in general and in general you get the same so the q-bar by algebraic points well for Schimoire varieties of billion type where you have a modular interpretation this is the same thing as cn point example 3 global variation of abstractions then a complete mystery ok and this is related to a deep problem like absolute odd and so no idea this is open so what is the heuristics that you get once you go to q-bar so this is a typical intersection heuristic which is a q-bar analog of the previous heuristic so this is for tame q-bar by algebraic structure yeah so I did not explain exactly what I mean by tame but it means that tame geometry has to enter at some point so I can give you more details if you want later it's not yet clear I mean I have in all the cases that I'm giving there are obviously tame for some reasons but this is not enough for what I want to do in general so let s be endowed with a tame q-bar by algebraic structure than any irreducible algebraic sub-variant g of s containing zarisky dense zarisky dense set of q-bar by algebraic points is a q-bar by algebraic sub-variant so this is heuristic what does this heuristic give you in the examples well for a billion-varity of a q-bar this is renaut theorem, this is Manning-Mumpford conjecture so example one so this is Manning-Mumpford for a of a q-bar so a theorem of renaut and you take an irreducible sub-variant containing zarisky dense set of torsion points then the guy has to be a translate of a new billion-varity by a torsion point example two so if you have an irreducible sub-variant containing zarisky dense set of c m points then you have a special sub-variant which is the same thing as a q-bar by algebraic sub-variant okay so I guess I have two minutes remaining so maybe I try to give you the proof of renaut theorem using this business of tame geometry so what I claim is that basically once you go over a q-bar Manning-Mumpford in that case so basically what I claim is that the proof is first you use what you know as c namely a functional transcendence given by action demand or action well and then you have to use some kind of big galore orbits so you have to have an information about the galore size of your special points of your by q-bar by algebraic points so again the problem is that I don't know the characterization of q-bar by algebraic for not of a billion type I expect them to be the c m points but nobody knows okay so yeah I'm a bit late but let's take three minutes to do that so again I put myself in the simplest case where a of a q-bar is simple it's not a big problem to restrict to this so what was my picture of a c? I had my fundamental domain which was semi-algebraic in cg so just my usual cube and then I had pi and then I had the identification of my u-beam byte which was cg mod lambda and we know that this projection in restricts so this map here is undefinable and so now suppose that A so of a q-bar such that containing infinitely containing the ischid and set of torsion points and I want to prove that this is A because I'm in the simple case so this is the picture here so how does it work well the first thing is that by ax in demand we can suppose that pi-1 of V intersected with the fundamental domain does not contain any positive dimensional semi-algebraic set because if it were by ax in demand this would imply that V is already A otherwise this means that V contains dimensional semi-algebraic stuff so when I push down then necessarily I'm equal to my ischid closure because V is algebraic and then this would be A, otherwise V would be A and I would be done so this is where I use what I know of a c so now what is the second kind of statement that I need I need to say what is the size of the Galois orbits of torsion points and there I use pilawilki in the other direction so first by pilawilki this will tell me that I contain very few there exists a constant c epsilon such that the cardinal of Z in pi-1 of V intersected with F such that the height of Z is smaller than T is smaller than c epsilon t epsilon and it grows sub-polynomially by pilawilki because there is nothing algebraic in it okay this is good so now I want to prove the finiteness I have to prove that under the hand the Galois orbits are big so on the other hand if P is a torsion point so P is pi of Z in the final tool set then what is the height of Z well by definition this is exactly the order of P okay and so now you have a classical result of Masse for ablian varieties of a Q-bar that tells you that if you look at the Galois orbit so let's suppose that A is different of a K where K is a number field and then you take the Galois Q-bar of a K orbit of your torsion point and the cardinal of that thing which is the same thing as the degree of the field generated by K and P is larger than c times the height of Z so the order of the torsion point to some constant rho so what is important where c depends only on A and in fact rho it's even better, rho depends only on dim of A so in that case rho depends only on A okay so now you look at this equality 1 this inequality here and this inequality 2 and so from 1 plus 2 you deduce that the height of your torsion points or the height of the premise of the torsion point is a final set or if you want the order of the torsion points is uniformly bounded so any torsion points point on V as bounded order offer will not like this sentence there is a constant uniformly bounding the order of the torsion points in V so you have only finitely many okay I'm sorry I'm a bit late I will stop here and tomorrow I will talk about so you see the problem for variations of a structure is that I don't know how to characterize the cube bar by algebraic points because this is difficult transcendent theory so what I would explain is that I replace these notions by adequate atypical intersection conjectures for variation of a structure so I want to state this and maybe give an example then maybe try to give a hint how to prove the kind of action well but in a curved context and then the second hour will be devoted to the proof of the original theorem 4 which is not atypical intersection but typical intersection and where nevertheless you still use some functional transcendence result okay so I stop here for today