 Already half the definitions I'm going to give are there. Thank you so much, Sarah. Look, it's been 60 years since I first came to the IHES, which is pretty much when the IHES was founded, or at least the first time. It moved to Bure. Yeah, it was the first year that it moved to Bure. The residence de l'Ormaille just didn't exist, but all the visitors, or most of the visitors and professors, were in residence class. The air was electric. It was, at least for me, it was just extraordinary. There was a René Thaum understanding topological singularities, structural stability, morphogenesis, and Grotendieg transforming algebraic geometry and, on passage, a good deal of the vocabulary of the way in which we deal with mathematics. Grotendieg would very often like to say, when he talks about something called x, he would say, x n'est rien que y. And by that, he would mean there's a change of viewpoints that you have to make and you have to try to understand. And I was trying eagerly to do that and to learn some mathematics. I knew almost no mathematics, actually, when I came first. And I'm so really grateful that I learned so much and felt the inspiration of the IHES from its very beginning. I also want to thank Will Hurst for his generous and just overwhelming spirit in thinking of various ways of how to put it, of making the mathematical community more congenial and more open, more capable of doing things. He is equally generous in the sciences and the arts and literature. Of course, I don't and neither does Gretchen deserve to have our names as named entities in this visiting professorship. I'm sort of speechless and don't know what to say about it. But it seems like fun. Anyway, the other thing is this is the fifth lecture today. And I can imagine that people are exhausted. We've seen wonderful things. I'm so happy that Sasha Gontroff is the first visiting professor in our named chair. And we learned enormous amounts of things in many different directions. So my lecture today will try to be as untechnical as possible. And that's pretty easy for my subject because what I'm going to talk about is a project that's an ongoing project with Carl Rubin, which is more experimental mathematics than theoretical mathematics in some sense. It involves a lot of computation. In fact, I will not overload you with computation on a screen, but I do have one page. Many copies of one page if anyone wants to take a look at it. At the end of this lecture, there are various distributions you see here. And I'll try to make some sense of what these distributions are and why we think they're vastly interesting. And the other thing I want to say is by making such a lecture, I want to emphasize that we're doing computation. I just say machine, ordinateur, computation. And neither of us are really professionals. And so if anyone sees something that might be of interest to them and they have essentially any expertise, they would have more expertise than we would. And I'd love to chat with them about it. OK, what do I want to talk about? I want to talk about a variety. v will be a variety over k. k is always a number field here. That is, say, a field of finite degree over q. And what I'll be interested in is it's a set of rational points, v of k, k rational points. But I'm interested in it from a kind of a relative standpoint. Let us say I want also to be considering always a finite extension field, extension field, l over k. And in some sense, the connection between v of k and v of l. Let us say, do we get, I don't say I didn't like new. My mirror is saying new without introducing extension field. But do we get new rational points over v of l given the extension field? So this is a relative question. And by the way, most of the time, but not always, I'll be interested in when l over k is Galois. So there's lurking in the initial data. There'll be its Galois group. This will be around if l over k is Galois. And I'm going to ask various questions. In fact, it's very interesting to change the quantification. You could fix the variety and ask for all l over ks with a given g. You could do various things. And in certain contexts, you could even be more explicit. For example, if v is an obedient variety, a, you could just forgetting torsion. You could take Mordell-Vey group tensor q and consider the inclusion in the Mordell-Vey group of a over l tensor q. And in the case with the Galois group is, where the extension is Galois, you have a g representation, g acts on a of l over k. And instead of asking about new rational points, you could be even more specific and choose an irreducible representation with a character chi. You can ask, does chi occur in this g representation space a of l? Because to have a certain amount of easy language about this, I want to make a definition. I see. OK. This is going to go up all the way. I should have practiced this before. Here's a definition. We'll say that v over k is diaphanthine stable, all abbreviated as ds, for the extension l over k, if there are no new rational points. OK. Now, there are certain varieties that have sort of very clear characteristics regarding this notion diaphanthine stable. For example, if v is a projective line over k, there are no non-trivial extensions that are diaphanthine stable for it. So even if v is a variety that contains not only an image of a non-constant image of a projective line or a non-constant image of an open in a projective line, there are none. One might ask whether the converse is also true. So there are all sorts of questions you might ask about this. In the case of v to be an abelian variety, you might ask, does chi occur in a of l over k? And the standard conjecture would have it that chi occurs in a of l tensor q, in the case where chi is an irreducible representative of the Galois group of l over q. Standard conjecture would say that that's true if and only if the hoss of a l function of a twisted by chi at the natural point s equals 1 is 0. So this is conjecturally that. And after all, we have, therefore, if you want, we can think of this question as an arithmetic question up there and an analytic question down here. And it's fun to sort of try to play one of these against the other. So let's take a specific example. What about the l function analytically? What? How much do you know about the l function? Not much. Not much. It's s equal 1. Yeah. So this is also the conjecture includes the fact that l of a chi at s equals 1 is defined. They're to say l of a chi s has an analytic continuation, as Ofer points out. So that's part of the conjecture if you want. Of course, in terms of the quantification here, you could ask, for example, fix a group g. You could fix the group g to be the symmetric group sn and fix the character to be the standard character. And ask, for example, does the standard character occur in A of l tensor q for a fixed abelian variety over a fixed field k for infinitely many fields l? That's already an interesting question. And it's also interesting that we can prove this. We can prove it. It's not that hard. We can prove it where A is an elliptic curve. And we can almost prove it when A is an abelian variety. Let us say we can prove the analogous statement, but at least for n sufficiently large. So there are all sorts of questions that you might phrase from the data of the v over k and l over k and the Galois group. So I'm going to now talk about v, an elliptic curve, and k cube. When v is an elliptic curve and k is q, the Schontal-David, and Jack Fernley, and Hershey-Kosolowski have pretty interesting conjectures about this where g is a cyclic group of order p. Namely, the conjecture is quite simple. If p is greater than or equal to 7 for a fixed p, fix it. There are only finitely many l's over q, cyclic order of degree p, that are a diafantine unstable. That is, say, that you get actual extension of points when you pass from k to q. In other words, all but finitely many are diafantine stable, the conjecture in this range. And the way they do it is by using a rather beautiful but curious use of random matrix heuristics. And the random matrix heuristics will also give asymptotics. The answer is false for p equals 2, 3, and 5. But they give asymptotics for p equals 2, 3, and 5. And I might return to this a bit later. It's false, but with asymptotics. And proved false or conjecturally false? Proved false. Proved false. Proved false for, the thing is, I'm worried about being over q. But if you allow me to make a slight base change, it will be proved false for any of them, 2, 3, and 5. 5 is really interesting. It involves a certain curve called Brings Curve. I don't know whether anybody's heard about it. Does that mean the conjecture will work over another number field or not? Will the conjecture work over another number field with other p's? With, in other words, if you change the number field, you change the 7. The heuristics is about the L function or about the? Effectively, the heuristics is both about the L function. Or I'm the matrix heuristics is about the L function. And I should say, also, that Ruben and I, Carl Ruben and I, have heuristics as well. But we call them naive heuristics. They're dependent only on certain distributions having a property that we think they have in various pieces of data that we've collected. So if you know the tensoring is q, that is the stability from the functions, then is it easy to get back to without tensoring is q but you know that tensoring is q, it doesn't change. But you define stability. Oh, tensoring with q. Yeah, no, I mean, the thing is, if you have an elliptic curve over q, the only difference, that's a good question. The answer is yes, but that's a good question, yeah. Because you can bound the torsion. Exactly. Well, you certainly can bound the torsion, but you can do even better, yeah. Great, OK. So our heuristics, which I am going to give you in a moment, but before I do that, let me tell you a theorem that we proved a number of years ago. It's sort of just published in the American Journal, which got us really interested in trying to understand this whole thing in some serious way. So here's a theorem. This is with Rubin. I know you're supposed to put a dash there, right? I'm going to do me. OK, with Rubin and me, the theorem is let v be one of two things. It could be a curve geometrically reducible of genus greater than 0, or it could be an abelian variety, geometrically simple. That geometrically simple actually is important. And v is over a field k, and there'll be a hypothesis on k, which I'll tell you a bit later. It's a mild hypothesis. I don't want to, OK. And so here's the theorem, and the theorem is kind of loved by mathematical logicians, because there are lots of quantifiers, OK. There exists a set of primes, p, of positive density, positive density, so fix p, such that for all integers, n equals 1, 2, 3, and so on, fix now p to the n, I don't know, forget that fix p, yeah, but fix that one, such that for any fixed p to the n, there are infinitely many, what? Fix means for every n. Oh, choose, you choose any p in this set, and choose any n in this set, and then. For every n, p, n, you have that. Yeah, that's what I mean, for every n, p, n, OK. That's right. You don't need logic, if you write a quantifier, don't say. Right, right, right, right, right. OK, very good, very good. You're right. I just wanted to make sure I was not that the infinitely many wasn't confusing. Anyway, there exists infinitely many extensions, cyclic Galois extensions, l over k, of degree p to the n, let us say for fixed p to the n, there are infinitely many Galois extensions, maybe I should have said that, l over k, that are divantine stable for v over k. Oh, did I? So I have some various things I have to say. I have to tell you what the hypothesis on k is, which I'll do in a moment. Instead of positive density, it's what one might call a Chebotarov set. There's some finite extension, and you put conditions on these p's having to do with Frobenius for things lying over those p's. And you get positive density. We expect that this is of density 1. And in fact, we expect even better than that. Infinitely many, this is also rather an impoverished infinitely in the sense that if you go up to a conductor where you've forgotten cyclic Galois extensions of that degree, it'll be on the level of x over log to some power of x that we actually prove exists. So we don't even get positive density of such cyclic Galois extensions. But we expect. Could you say it's Chebotarov? But if it is entity 1, if it's complement of finite sets, you expect that. Oh, I expect complement to a finite set, exactly. Exactly. The hypothesis on k is simply we want of, I won't write it, but I want the Jacobian of c to have its endomorphisms. The endomorphism ring of the Jacobian of c, I want to be all defined over k. Let us say the endomorphism ring over k bar of the Jacobian of c is the same as the endomorphism ring of k over k. And with a, it's pretty much the same thing. All endomorphisms of a over k bar should be defined over k. So if k doesn't satisfy this hypothesis, you just, you have a single variety in question. You just raise a pass from k to k to whatever field you need to make every endomorphism rational, and you get this theorem. So the next thing I would like to do is go back to the question, the more general question, does k occur, and is it connected and do it in a manner that's connected to the Hasse-Vael function? But we're going to do it for a now being, not only a billion variety, but an elliptic curve, and k being q. OK. Well, perhaps I could tell you I was going to end with the conjecture that we have. But perhaps I'll tell you the conjecture that we have, which is, in some sense, inspired by the Dawey-Fernley-Kazlevsky conjecture, but is motivated by what we call our naive heuristic rather than random matrix heuristic. We conjecture that if you, if we have e over q, and if we consider a billion Galois characters, let us say, or Dirichlet characters, if you wish, chi from the Galois group of l over q to c star, and you consider the set x of all such chis such that, and now, Ofer, I don't need to use the conjecture that the l function extends because it's going to all such chis such that l e chi 1 equals 0. But I exclude the chis that have the order of chi. I exclude the ones that are of order greater than or equal to 5. And I also want it not to be order 8, 10, and 12. And our conjecture is this set x is finite. But try to give you some sense of how we come to it. And for that, I'm going to move to the combinatorial way of thinking of l functions, namely theta elements. Or maybe one thing I might do is this is a digression. And if I have time, I might have time. Our conjecture that we give is more precise than that because it gives asymptotics for the chis that are missing there. And those asymptotics compare well with the asymptotics of David Fernley and Kicilevsky that they get from random matrix heuristics. But there are lots of wonderful things you can do for the missing primes, for example, p equals 2, 3, and 5. And perhaps I should, since I mentioned my conjecture, talk about 3. I'll try to do it fast because. So the question for 3 is, how many cubic cyclic extensions of q have the property that a given elliptic you have fixed your elliptic curve over q. How many cubic cyclic extensions are there where the elliptic curve picks up more points over that cubic cyclic extension? OK? In the sense of the rank. More in the sense of the rank, as you'll see, it's not going to matter because it's going to be governed by a pencil of cubic extensions whose total space is a curve of genus greater than 1. And so we can use mountain, Mumford, and faultings to show that you have a whole pencil of rational points over the p1. And for each rational point, you get anyway. So here's a, I like this example. I'm going to try to do it fast so it's not, it doesn't disturb the rest of my talk too much. But the easiest way to try to understand this is you take E cross E cross E, the elliptic curve that you're looking for, cubic cyclic extensions, for which it's not diaphantene stable. The symmetric group of order 3 acts on this. And so you can divide by a three cycle. And this is your very generous cubic cyclic extension that you're going to sort of milk. You can, if you want, sum to 0. Right, you can take the sums and you can take 0 in here and take the inverse image of 0 in here. So it's the set of cubic things that sum to 0. That would give you a surface rather than a three-fold. And you could go all the way. You could divide by the symmetric group of order 3, which gives you sim 3 of E, which also sums to E. And if you take the inverse image of 0 there, you get something that I'll just call p. It's really a projective plane. It's in some sense that can be viewed as the dual projective plane of the elliptic curve. This is a double cover. And it will be ramified at a curve of degree 6 with nine singular points. It's the dual curve to the elliptic curve. And it's a kind of k-3 surface want to be that this x is, if you blow up those nine points, you get 18 projective lines. And so this x is k-3 surface of Picard number 19, 18 plus the ample divisor. And so every time you find a rational curve in this, you just pull it back. So let's suppose I find the rational curve and take the pullback of that rational curve. I get a curve in here, which, well, the pullback, I want the normalization of the pullback. So this is a smooth curve now, which maps to E cross, E cross E. And here we are. We have a pencil. This is a cubic cyclic extension. We have a pencil of cubic cyclic candidates. We take any rational point there, go up there, and project to, say, the first factor and you get a cubic cyclic point in E. And as I said before, money and money and faultings tells you you actually get infinitely many L over k. The C is a genus greater than one. That's the three. The five is, as I say, even more interesting, but for a lack of time, I'll go on to the analytic story, which is this. But I'm gonna restrict it to elliptic curve and over k equals q. Okay, now there's a whole essentially mini subject that connects the values, L e chi 1, to more combinatorial objects like modular symbols, as it was mentioned in Sasha's talk in the beginning of the hour. And lots is known about modular symbols. In particular, lots is known about the statistics of the values of modular symbols. But what I want to pass to is not modular symbols, but things that are built out of modular symbols that we call theta elements. And they depend upon a field extension L over q, with cyclic Galois, of course. It's cyclic field extension. Save degree, call it's degree D. And I'll call it just theta sub L. And the thing about theta sub L is it's really more an arithmetic object than an analytic object that lives in the integral group ring of the Galois group of L over q. So when I write it out, I'll write it as sigma for gamma in that Galois group of some coefficient C L gamma times the element gamma in the group ring. So these guys will call theta coefficients. And the virtue of this theta L is that if you take any chi from that C L gamma is in what? What? In Z, in Z. Is C L gamma? In Z. Is in Z. Oh yeah, yeah, yeah. These theta coefficients are in Z. So they're integers. And if you have any chi from the Galois group of L over q to C star, the virtue of this theta is the following. Apply chi, you can apply chi not only to the group, you can apply it to the group ring as a homomorphism to C of algebras. When you apply it to the group ring, you get chi of theta L is equal to something which is visibly non-trivial, non-zero. Non-zero. And more elementary well involves a period and gauss sum maybe, but it's just non-zero times L e chi one. So if we're interested in is L e chi one zero or not, we're really asking is, if I replace this a by e, we will be asking is chi of theta L zero or not. And these are equivalent not by conjecture but they are equivalent. Oh, I don't want that. Let me get rid of this. Well, just to give you a sense of what's going on, like my L over q was of any degree D, I'm gonna, and now let's suppose that the degree of L over q is a prime, although everything I say here with sort of more argument and notation has sort of qualitative analogs for any D. Okay, if D is a prime, take a look at that chi of theta L, chi of theta L is a sum of integers, C L gamma, chi's of gamma, and these guys are p through its immunity. And so this is in the brackets theta p. So you learn that, hey, we have a certain element in, a psychotonic element of you wish, element in the Z brackets theta p, we're asking is it zero or not? And the answer is zero, if and only if all the C's are equal. So chi of theta L is zero, if and only if all of the C L gammas are equal to C L gamma primes for all gamma and gamma prime in the Galois group. Okay, so this is the sort of thing that might get you interested in asking statistics about these numerical values. I mean, how often are they equal? All of them equal for a given theta element. So for example, before you do any statistics, you should try to figure out what regularities these C L gammas have. So we want to understand for every L over Q, cyclic Galois extension of degree p and for the gammas in their Galois group, we want to understand these integers C L gamma. And already we're specifically interested in whether, what happens when they're all equal. Well, there's some regularities. Before you do any statistics, you better take care of the regularities. And the first is the sum, we have one has a very clear expression for the sum of all the C L gammas. Let us say in the case where I take M to be the conductor of L and at least to say what I'm saying without the extra terms, I'm going to assume that this is square free and prime to p. If you take the sum of all these C L gammas, you get the following interesting thing. It's the product for all primes dividing the square free number M of the L Fourier coefficient of the modular form for the elliptic curve, I'll call that AL minus two times some number, a rational number. And that rational number depends on nothing except for E. Well, in fact, this rational number is non-trivial multiple of the L function of E over Q at one. And so therefore this is zero if the L function over Q of E at one is zero and non-zero if not. In any case, we see that in order to get such a such an effect, perhaps I'll put it again on this board, this is going to happen at least in the context where I make the hypothesis of square free prime to p that each every C L gamma, all of them have to be equal to one over p times in some sense, well, I'll call it the right hand side of that equation. What is at the AL minus two? What? In the equation, you're right, what of the AL minus two where L runs through all the primes dividing M. And AL is? AL, oh yeah, good question. In fact, I want one more requirement with this prime to p and two prime two the conductor of E. AL is the usual thing. Usual thing, yeah, yeah, yeah. And there's a similar formula, it's more complicated in general. Okay, so under these hypotheses, we get not only for a chi of L to be zero, that is safe for us to get a zero as a. AL is the trace of Fremini or minus the trace of Fremini? It's the trace of Fremini. Yeah. Okay, so we get this is equivalent to that. So that does suggest that maybe one should take a look at the distributions related to these values of these theta coefficients. I mean, after all, for fields of degree p, in order to get a zero, you have to have every one of these theta power coefficients to be equal to this specific number. And so what's going on here? So the natural thing to do is to produce data, I'll consider it data. And by the way, I'm gonna do this data not only for, oh, I should also give you one more regularity before I give you this data. Namely, the Atkin-Lehner or the functional, or what I'll call the functional equation, identifies essentially one theta coefficient with another in the following way, especially I'm still under the hypothesis that M is prime to p conductor of E and square free, in which case the identification is fairly simple. Namely, if I take the Galois group of L over Q, there's an involution, I'll call it I. It's a permutation of order two. It's not a homomorphism, needin' preserved zero, but there's an involution which gives a condition on the values of the CL gammas. Let us say if I apply CL, if I consider CL and apply the involution to gamma, it's equal to pretty much CL gamma up to a sine and the sine is the root number of the elliptic curve. So if we're trying to make data, which I was about ready to do on this board, I don't necessarily care to take all of the CL gammas, I will, but I should understand that there is a, there's a kind of correspondence between them in pairs. So here's what I wouldn't want to do. I'm gonna take all the data, I'll call it data of E and D, and this is just the set of all integers for the moment. I'll normalize them to make good use of them, where L over Q runs through all cyclic extensions of degree D and if D is even, if D is even I want it to be real cyclic, and I want gamma to range through all elements in G which are not equal to their image under that involution. These are we call generic, since the involution has a maximum of two fixed points and sometimes none. And that's my data, but if that were my data and I asked what's the distribution that's determined by them, it would just be, it would just flatten out to something horrible. So I have to renormalize it and this is how I renormalize C L gamma. I multiply by the square root of D and I divide by the Euler phi function of the conductor of L times log of the conductor of the square root of log of the conductor. So that's a whole lot of data. And the question is, does this converge to a distribution? If it does, what sort of distribution does it converge to? And how does it, how might it connect with the issue of how often is C L gamma gonna be equal to that specific thing? So here's our conjecture. So we begin, you're now on a computer. At this point, it's total experiment. But there is some heuristics why you put those factors. What? There is some reason you put those factors. Oh yeah, if you wanna know the C L, I will tell you just very rapidly. The C L gammas are really a sum of phi of M modular symbols. The modular symbols are move up into modular symbols with denominator M. The modular symbols move up by log M or they're bounded by a constant times log M in terms of the denominator. So in order to bring this to a reasonable range, I have to divide by this. And I multiply by this because I want uniformity for all D. That's okay. So that gives you some rapid reason why I do this normalization. But anyway, here's our conjecture. This converges. So the conjecture fixed E and fixed D. I fixed E, I fixed D. So the data of ED is this whole set. And I vary L through all real cyclic extensions of the BD. And I vary gamma through all, as I call generic elements of the gamma group of L over Q. And by the way, the other ones I've avoided, the ones the gamma is equal to i of gamma, we call special and they produce different distributions. But anyway, I'm gonna give you one and we'll discuss, we can discuss how that distribution relates to the special ones. Okay, this meaning that converges to a distribution which we call lambda E D of T. And you know what that means? That it means if I integrate this thing, if I think of it as a function or a distribution, if I integrate this thing over T in some range, that would give me the percentage of elements in this set that live in that range. Do you see what I'm saying? The measure in every function of the DT. Yeah, yeah. Well, I'm thinking of it, as you see, yeah. It'll be multiplied by DT. That would be the measure, exactly. But the distribution has the following properties. One, it's continuous. It's also, we have a sort of clear conjecture of its shape, but just continuous is enough, except possibly at T equals zero. Two, if D is large enough, it is even continuous there, simply continuous. Three, the limit as D goes to infinity, which is, oh, that's the reason for the square root of D there to deal with this, of these lambda DTs, EDTs is Gaussian with variance equal to some elementary term, it looks like a bunch of Euler factors, times the L function of the symmetric square of the automorphic form, modular form attached to F at the point S equals two. So that's our conjecture. And if anybody is interested in wanting to take a close look at one, we've done loads of these, and I didn't want to overload the lecture with a screen after screen of data, but I have an example here of one page. And if anybody's interested, they can take a look at it. I have many copies. We consider the only elliptic curve that people start their empirical journeys with, and Sarah can guess which one. Yeah. There are only two, you see, either it's 11 or it's 37. I mean, one or the other, 11's better. We did 11 first, yeah. And so we did D equals three, and there's this, the distribution is sort of spiky. It's always spiky, so by the statistical jargon, the jargon of the statisticians, there's kurtosis, that is spikier than Gaussian, and we don't know why. Anyway, it's this big for D equals seven, it's this big for D equals 19, it's next. And by the time you get to D of 101, it's so close to that dotted line, which is Gaussian, that it's hard to imagine it's not moving towards Gaussian. Okay, and that happens with all the elliptic curves we've tried. And so we became fascinated by these things. We know no statistics. And when I have a friend who is a statistician, and I frantically call her every once in a while about some statistical thing, and she will always say, Barry, read a book on statistics. So anyway, so I'm reading books on statistics. Anyway, okay, whatever, we've become fascinated by these distributions, which in principle doesn't have anything to do necessarily with automorphic forms of, you know, on GL2 or any automorph. You can always twist by a Dirichlet characters and see what you get. And there are these statistics, and they seem to depend sort of somewhat, in fact, visually in our computations, they do depend a bit on the elliptic curve, therefore on the automorphic form. So they're interesting, and we would love to know, let's say a closed formula for them. We haven't found nor do we have any guesses. In any case, what this does, wait, am I, I have two more minutes? What do I have? What? Okay, anyway, here I go. I'm just gonna tell you our conjecture that follows, we think, from this and from also a conjecture about the correlations of theta elements of the given theta, of theta coefficients of given theta element. So our conjecture is let E be elliptic curve over Q, let M over Q be any a bian field of rational numbers. Let us say the interesting thing here would be of rational numbers, of algebraic numbers, but it contains only finitely many, finitely many subfields of degree two, three, or five, then E of M is finitely generated. I should say that this is actually true. This would be true thanks to Cato, Ribbit and Rohrlich. If M, for example, simply was un-ramified outside finitely many primes, but here we're not assuming that it's un-ramified outside finitely many primes. There are many a bian fields that are pretty interesting, that are contained in this. And one of the reasons why it might be interesting is that every time you have such a field, such an M, where E of M is shown to be finitely generated, Hilbert's 10th problem is full. So this is connected to recent studies of Hilbert's 10th problem where people don't know about the total, the maximal a bian field they don't know. Is it true or false? And the logicians don't know what to believe there. Which one will be there? What? Which one will be there? 10th. The ring of integers in that field would be what's called diaphantine. Let us say there's an algorithm that determines yes or no, does an equation with coefficients in that field have a solution in the ring of integers of that field have a solution in the ring of integers of that field. Okay, I think I've forgotten. I think this is my, okay. Thank you. Thank you. Thank you. Thank you. Okay, I'll have any questions. Why is symmetric squared 200 years? Ah, you're right. Why is, oh, why does it symmetric squared? Oh, no, that's a good question. The reason is that, I mean, we conjectured this and we conjectured this without even before, even before the data, because if you think of this metric square and you think of the Fourier coefficients, there'll be squares, A of p squares. And those, those Dirichlet series control the variance of modular symbols. And this is connected to modular symbols, so. Oh, and if you average such things, not for the theta coefficients, but for the modular symbols, you average it over all modular symbols and ask and normalize appropriately and ask for the appropriate variance. The variance will be, is proved to be L function of the symmetric square. It's AP square. You have to think of it that way. You did not collaborate with the relation with the analytic results, but as you mentioned on our list, if you twist by a character of all the p and you vary the character, those are sort of raw lists, the Murtis, vice-burge, and so on. They tell you that generically you have no zero, is that right? Yeah, yeah. And that's so how does it connect with that? It connects perfectly with it. Yeah, yeah. But they're always in a specific situation where there are only finitely many ramified primes. Yeah. Oh, that's the case? I think that's, yeah, that's the case. So what they prove, they prove this conjecture. If, forget this, forget this. They prove this conjecture if M over Q is unramified out by finitely many primes. Oh, I see. Yeah. Okay, thank you.