 Actually, usually when I have given this type of lecture series, the attendance goes something like this. So I'm happy that this time it's more like this. So I'll remind you what we're today. We're just going to just talk about regulators. So let me spend the first few minutes just recalling the story from last week, and then we'll talk about the regulators. So from last week, our situation was we had these subgroups, gamma0 of n inside SL2 of an imaginary quadratic field. Okay. So this is a matrices where n divides C, but now remember everything, all the coefficients and n line sides of this ring here, and the conjecture that we were discussing is that when you take the abelianization of this and you just take the finite part, so it might have some infinite part and you scale it by the volume that this has a definite limit, which is 1 over 6 pi. Okay. So what I said last week are the main tool in trying to understand this from an analytic standpoint. So I described some heuristics, but the main tool is this story of analytic torsion, and in particular the following formula. So the statement that I'm about to write is literally true, only the non-compact case, the size of this multiplied by a regulator, which I'll divide it by the volume, is equal to some ratio, something to do with determinants of laplacians. Okay. So what I said in the previous lecture was firstly, if you allow to pass to us, if you switch this for a different weight, this is the analog of weight too. If you go to some different weights, we can prove the analog of this conjecture. But for this conjecture, so using this analytic torsion formula, one can prove the conjecture in the non-compact, or at least in the compact case, at least, and as I said, this now close to being proved non-compact case. Also, one could prove that if we know two things. Okay. The first few small eigenvalues. So you can understand this object rather well, as long as modulo the issue that this Delta 1 might have some eigenvalues very close to zero. So Delta 1 has few small eigenvalues, and I wrote the precise meaning of that last week, so I won't write this again. This is something we certainly heuristically strongly expect to be true, but this is totally I have nothing to say about it. I don't know how to prove anything. But the second thing, which I'll call R, is that you need to know. So once you understand the right-hand side is large, and you can understand exactly how large, you need to know that that size comes from this finite group and not from R. So you need to know that log R, scaled by the volume, goes to zero. Okay. So I just discussed, let me give you the definition of R once more. So this is what I'll talk about today, and in particular, I'll say more about what the actual, so what is interesting about this R is not just it's trying to prove this, but it's actually a very interesting invariant. Okay, that is, it's related to many other objects of number theory, including L functions. Right, so here's the definition of R, and I'll, maybe I'll write it down here. So the definition of R, and I will just define it in the case when the free part has rank one, otherwise it's similar definition with determinants. When, okay, so I said this last week, but I'll remind you, so there's a pairing. Just as in the case of the thing I said at the very start of these lectures about classical holomorphic forms, there's a pairing between this and a space of modular forms. Okay, so this, I've written it S2 to be suggestive so it kind of reminds you of the case of classical modular forms, but here in our context, this is the space of harmonic one forms on this mod H3. Okay, if you were working, if you said make the same definition in the case of the upper half plane, the harmonic one forms are sums of holomorphic forms and their conjugates, F is Z, D, Z and it's conjugate. So this is really an exact analog here of the space of holomorphic weight two forms. And this pairing is given by the same formula as in the first lecture. So that is if you want to pair gamma and a form which I'll call omega, you integrate omega on any path between W and gamma W for any W and H3. Okay, so still not done. All right, so now this, yeah, by the way, let's, at some point in this lecture, I'm gonna switch to the non-compact case, but here let me make the statements in the compact case and one has to slightly modify them for Eisenstein series. So what do we do? So we can integrally normalize an element of omega in this S2 of N by requiring that, all right, so I'll write it gamma, what should I call it? Gamma naught omega to be one where gamma naught N projects to a generator of Z. Okay, so in general, you know, if you have, so you should think of this as a space of modular forms, we have different ways to normalize a modular form. Okay, so for analytic purposes, the two common normalizations that one sees if you think about holomorphic forms is you can normalize the L2 norm to be one or you can normalize the first Fourier coefficient to be one. Okay, but in the context we're discussing this is an equally natural normalization, you require that when you integrate it against these cycles that you get one on a generator for this. Okay, and once you've done that, then you define this regulator to be omega in a product. So maybe I should call this, I should just have a notation for this. But let's, I'll call it omega subscript Z to just to suggest this is a integral normalization. So now let me, okay, that's the def, so this I think up to here were all things I said last week and now I'll try to say something about the nature of this thing, are there any questions? All right, so the only, okay, so now I'm gonna state a result from this paper with Bergeron and Hallux-Sangoon. So a result from, so after I state the result, I'll say a bit later about the proof, but the main thing I'll try to communicate is the difficulty here, okay, it's not of maybe a difficulty one is so familiar with in the analytic theory of modular form. Difficulty is trying to, so to speak, find such an element, okay? So in other words, this pairing is not so hard to compute, but if I just take an element, you really have no idea where it projects to in here. Okay, you could have an element of very modest size, write down a physical matrix here, very modest size, but yet its image here is very large, okay? And that's the basic problem which one faces when you're trying to bound this from above. But okay, but this, right, this result is actually in a way more algebraically suggested than analytically. So suppose now this really depends on, suppose that, so we're really in a non-compact case, okay? So in other words, that means at various points I should put in the word cusp forms rather than all forms, but that's all. This is important and you'll see why it's important, okay? Now, according to the Langland's program predicts, okay, so yeah, I should have said in this setting that we are where this is z plus a finite group, maybe it was implicit in this, but this is one-dimensional, okay? This has just one copy of these, just a one-dimensional complex vector space, okay? So just one, this really pins this down. Now over, if you were in a setting over in classical modular forms, where the space of weight two forms is one, then there's an elliptic curve associated to it, okay? This happens at, the first time it happens is at level 11 over q and then there's a, it's associated to an elliptic curve. So the same thing is predicted, although not proven here. So Langland's predicts that associated to this, this modular form. By the way, actually I said it's a harmonic one form, but if you write out what that means, it means that the mass form of a certain eigenvalue and weight, okay? So associated to this, there is an elliptic curve, E, over the field we're dealing with, q squared minus d of conductor, N, and there are other proper, a matching between Hecchi eigenvalues of this and point counts on this. It's, so I'll just say, same condition as over q, okay? So if you look up the story for weight two forms over q, it's exactly the same. Now I should say this statement is actually not, there's a subtlety which happens over an imaginary quadratic field, which is that even when the dimension of the space is, so over q, for example, when the dimension of the space of moduloforms is two, you might not have two elliptic curves, but you could have an abelian, this could come from an abelian surface. In this setting, something similar can happen even if the dimension is one, this can come from an abelian surface with quaternionic multiplication, which is sometimes called a fake elliptic curve. So I won't, I'll just ignore that, but I think exactly the same result should hold with the same proof in that setting. Okay, so Langlund's predicts this, so we will suppose this to be so is true and also assume the Birch-Sunin-Dyer conjecture for E and its twists. Okay, when I get to it, I'll say exactly, but this is, then this regulator R divides, all right, so there's an explicit fudge factor, which I won't, so let me just call it M, I'll say time, the more important part of it is omega E to the minus two, so this is some explicit fudge factor, explicit and involving only the torsion on the elliptic curve and the level N and also actually the adjoint, the adjoint L value of this modulo form. But this is the part which is interesting. This is the area of the fundamental parallelogram for E. Okay, so in other words, what you can do, you can take this elliptic curve and you take its complex points and you can integrate over it, you take a integrally, a neuron, an integrally normalized one form, wedge its conjugate. Okay, so the absolute value of this. If this doesn't have class number one, there's a slight issue in defining this, but again, I won't, I'll skip over that for simplicity. Okay, now, so what I want to, so in fact, I think that there should be an equality of this form. So let's say I think that this is really an equality, so R is should be equal to some essentially explicit factor times this. Okay, the proof says it divides it but I think in truth, it's an equality. Now, and let me say that another way to make it, to try to say why this is an interesting thing. This omega E to the minus one is essentially the faulting's height of the elliptic curve of E. Okay, so in other words, if you write out explicitly an equation, if you write explicitly an equation as E is a minimal equation, which is Y, which if E is given by this, then this is going to be approximately, I'll say what that means in one second, it will measure the size of these coefficients. Okay, I hope I got the powers right. This approximate sign means it's not that they are, it means the logarithm of the two sides is the ratio of the logarithms goes to one when something goes to infinity. So in other words, so this statement, okay, what it says, so this thing here is some measure of the height of E, so it's a measure of the arithmetic complexity of E. And this thing, as I said, it measures the complexity of the module of form associated to E. Form associated to E. So at least it came as a great surprise for me that these two things might be the same. I mean, of course, they both measure complexity of something, but the nature of them is so different that I can sort of, as you see, the proof will be sort of very computational, so it doesn't really explain why such a thing should be true. Moreover, I think that a statement of this form might be true more generally. Okay, that is to say that the complexity of a, in some normalized sense, such as faulting's height of a variety should have something to do with this analog of regulator of the associated module of form. But that's very speculative. So, okay, anyway, so this is the, as I said, what's interesting about the statement is this relation between R and the height of the elliptic curve. Now, let's see what that does for us in the context of our wanting to bound it. Okay, so the point is that to prove that, so to show that log R over the volume goes to zero in this case, then you need only to show that the size of the coefficients of this elliptic curve are bounded, I mean on a logarithmic scale, basically, by the level. So, this is a standard conjecture of difantine geometry, the size of coefficients of the elliptic curve is bounded by the level. So this follows from any, from many standard conjectures, for example, the ABC or the Fray-Spiro conjecture. These conjectures are much stronger than this, okay? These conjectures are about polynomial bounds. We only want bounds at a much, much weaker scale. But anyway, the point is that, yeah, so in this way that this gives at least assuming BSD and sort of very safe difantine estimates, okay? If I mean far weaker than what we had, just than what we actually believe, this implies that the regulators indeed implies this slow growth of the regulator. All right, so I'll talk about where the proof of this comes from, I wanted to first have a little aside about this type of regulator in some slightly different context, but are there any questions about this? Yeah? Is this an integer or? What? No, no, no, no, nothing is an integer. These are all transcendental. I mean, presumably. When you say R divides, you mean? Yes, when I say R divides, a good question, yeah, I mean that R is just divided by an integer, yeah, yeah. You'll also see where that integer comes from and why it really should be one, but certainly the proof doesn't say this. And you use the full form of BSD? I use the full form of BSD and in fact I use, when I say BSD for E plus twists, right? So you use something that's actually not, surprisingly not written in the literature anywhere, which is that if you take an elliptic curve and you twist it by a Dirichlet character, you need a form of BSD for this, okay? And anyway, so that's what's being used. Not just quadratic twists, but old twists. Why everything in non-compact case? Yeah, I'll also explain that, yeah, yeah. In fact, so I don't know how to prove this in the compact case, okay? In the compact case, let us say you make the same assumption, all you get is that R is a rational multiple of the same thing, okay? So we're going to use the non-compactness in some essential way for the proof. Well, I might as well say that we're gonna make some use of modular symbols, okay? And that, and you don't have that. So I would love to have such a statement in the compact case even, yeah. Any other questions? All right, I'm gonna have, oh, well, I don't know if this is, all right, I was gonna make a slight aside about how this relates to some problems related to K2, but maybe I'll, okay, maybe I'll try to sketch what goes into this kind of thing, and then we'll come back to that. So I'm gonna sketch some ideas of proof, and this will sort of show how, well, all right. All right, so the basic point is you need, in some way, of course, to be able to evaluate this pairing, okay? So there needs to be some class of examples where you can evaluate this. So let's call, so let omega, let's call omega, let's call plain omega the L2 normalized form, okay? So it's inner product with itself as one. So this is not going to be, so, right. So what we're gonna try to see is how to adjust it to get this. So take any, so suppose we can produce some element, delta in gamma naught n, for which we can evaluate this pairing, such that we can explicitly evaluate, and by that, this will mean in practice in terms of L functions, okay? So that's the meaning of explicit. Evaluate omega, which way did I write it? Delta first, okay, delta omega. So somehow we can explicitly work this out, let's call it E, so it's, this will be in practice, something to do with L functions, right? So now this delta, so let delta in, so you can take this and you can send it to this gamma naught ab, which is as I said, z plus something finite, and so let's say it goes to q times the generator, okay? So let's say delta goes to q in this variable, and the finite part doesn't matter at all when we're doing this type of computation, the integrals against differential forms annihilate that, okay, so there's no, so delta, so q, i.e., okay, so delta is sent to q times the generator. So then what this says in other words is that, so this is q times the generator, omega over e is one, or in other words, this tells you what the integrally normalized form is, omega z you get by q times omega over e, okay? So when you're in this position, when once you've explicitly evaluated just one value of the symbol, you can explicitly, you can understand what this integrally normalized form is. By the way, I should say this, maybe I, this is related to the aside that I was, I'll try to get to at the end. This story works very generally, okay? That is in the setting of the cosmology of arithmetic groups, you can always ask about this kind of, you can always integrally normalized forms, and it's always actually very subtle, okay? It's very difficult, this integral structure is very interesting and very subtle. Okay, anyway, this is the integrally normalized form, and so the regulator R is going to be the inner product of this for itself, which is q squared over e squared, and the entire problem will be that we can't bind, is that we cannot, we cannot bound q, okay? When I say we cannot bound q, of course, I don't mean it in some, you, in typical, it's more like, you can give some naive bound, which is of enormous size, let's say, so you need some, okay, so let me explain, for example, how you can try to do this in the co-compact case, okay? As we'll see in the co-compact, I don't know how to solve this issue in the co-compact case, so there'll be no full answer at the end of this, but it will just show how do you explicitly, so the next thing I'll discuss is how to produce such delta, okay, how do you, how do we have anything for which we can explicitly compute this? Okay, are there any questions? So you, if you, do you need to show me that the free part is one dimension? No, no, no, I'm just saying that so we don't have to write determinants. If the free part is enormous dimensional, when you run into a separate another issue, the free part is as large as the volume you, so let's say when you make this estimate, you would have another problem that the free part is a dimension that's very close to the full volume, but other than that, if it's significantly smaller, it's no problem, yeah, any other questions? Do we have some power theory bound for the free part? Okay, yeah, on the free part, here's what we know. Just from topology, for example, you know that it's bounded linearly in the volume. If you think a little bit more, you can bound it by something which grows slower than the volume, like volume over the logarithm of the volume, but in, and there are two kind of very interesting papers, one by Caligari and Emerton, and then one by Simon Marshall, where they give some specific instances where they can do better, but by and large, we don't have a robust way of showing. So, oh, by the way, I mean, certainly in terms of what should be true, a way for this, in this setting, the free part, the dimension of the free part should grow as a part of the volume less than one, but we lack very general ways of showing that. Yeah, I just had a question. Is finding this delta, is this in any sense, shape or form analogous to what people are trying to do, like explicit, false, perjury type formulas? Yeah. Like instead of functions, is it any sort of an analog? Yeah, no, it is. In fact, I'll say it's from this, do you find delta by the false perjury formula? But in fact, I mean, not just, false perjury himself, that that was his application, one of his applications in his original paper for his formula was exactly in this setting to understand this up to a rational number. Okay, so the story I'm about to say is not, it's gonna be the false perjury formula, but it was, in fact, already proved by false perjury in the same paper where he shows that formula. So yeah, so as you've asked, the way one does this, okay, this is not the main, as I said, the main problem will come later when we discuss how to bound Q. But in terms of just producing such delta, the answer is given by the false perjury formula, which I think Philippe mentioned now at the end of his talk. So it says, so if pi is an automorphic representation is an automorphic representation for GL2 over this field K is Q squared minus D and L is a quadratic extension of K, then this, and phi is a vector in the space of pi, then Walsh Berger, in his paper, he gives an explicit formula for this, okay? And in fact, he gives a formula for any vector at all. It's very beautiful, but anyway, what is important for us is that up to some fudge factors, what we get is the central L function of pi times the central L function of pi twisted by the quadratic character associated. So this is the quadratic Dirichlet character associated to L over K, yeah. Okay, now, all I'll say is what happens when you make this work out explicitly, what this says, it exactly computes some value of this pairing, okay? So in our case, so we take pi to be the automorphic representation associated to this to omega and we'll take phi to be the vector in pi that corresponds to, corresponding to omega, be a bit careful about this, but corresponding to omega. And then if you work out what this says, so you have an integral over this adelic space, but when you project, so the projection of this to gamma naught n mod H3 is a union of geodesics. It's a finite union of closed geodesics. So let's say I'll call this direct union of GI. Okay, so let me just write out what geodesics look like, okay? So each of these closed geodesics, this is just a computation, okay? You just have to unwind the adels, you project it down to here and you'll see this. Each geodesic GI arises from some hyperbolic element from hyperbolic gamma and gamma naught n in gamma naught n in the following way. So you take, so we're in hyperbolic three space, but this gamma fixes, so alpha and beta kind of fixes. It's exactly the same as if you know this story for the upper half plane, it's the same story. Gamma fixes two points on this plane at the bottom. You join them by a semicircle and what happens is gamma kind of moves you along the semicircle. So if you take a point W here and you keep applying gamma to it, it moves you further and further along, okay? And the way you get this geodesic GI is you just project this path, the piece between piece from W to gamma IW. You take this piece and you project it to this quotient. Okay, so in other words, you just take one of these, this piece here and you project it to the quotient and if you took any other such piece, you would get the same closed geodesic. So if you, okay, so in other words, when you integrate the form omega over GI, that's exactly the same as this, by definition, that's the same as what you get when you pair it with omega, okay? So the GI, the definition of that pairing is exactly, so this is true. And so then what Walsh Briget's formula says, what it translates to is that you add up all of these over all the geodesics, sum over I of these pairings and you square it, well, in this case, it'll actually work out to be real. So you just square it and you get L of, up to these fudge factors, L of one half pi, L of one half pi times this quadratic character, okay? So in other words, this gives you an explicit evaluation. So this, you can think of this as being, comparing the sum, how do you say, let's say the product of all the gamma I, gamma one up to, so this is an explicit formula, okay? So this is how you produce explicit things for which you can compute that. And as I said, this was already done by Walsh Briget, okay? So to deduce statements, what he showed using that was he showed that the different L values or different twists are related by rational numbers, okay? So this, now, by the way, you also need to do something a little bit, if L one half of pi is zero, this is useless to you, but you can always find a quadratic twist with non-vanishing central value. So this allows one to compute this regulator up to rational numbers in terms of L values. In either the compact or non-compact case, okay? But this is useless for us. I mean, at least it does tell you that this regulator is something interesting because it has the same, sort of it's algebraic properties are the same as the central values of these L functions. But for us, the entire subtlety is the integral story, that is how do you bound this Q, okay? So, and I don't know how to bound, so let me say I don't know how to say anything more, which is very, just very problematic. I mean, because you don't have a huge architecture, I mean, I think of Prasanna's paper. Yeah, so yeah, but even Prasanna's papers are really used something, right? They like you find many imaginary quadratic fields with, yeah, oh yeah, no, that's right. No, that's right. As Philippe says, maybe I'll just restate what, there are similar problems even in more algebraic settings and even there, they're very far from obvious what happens. Okay, this, so, yeah, Philippe mentioned this thesis of Karthik Prasanna, which is dealing with quaternion algebras over Q and already there the compact, these questions are difficult. Here it's much worse because nothing is algebraic. All right, so now what can we say in the non-compact case? Here we have available, fortunately, this trick of modular symbols. Yeah, any other questions? So here what we can do, evaluate any pairing. So let's actually take, let's, so let's choose, we'll just choose an element in gamma root n which projects to a generator of z, one in z. Okay, so over, in that notation over there, so Q is one. So in other words, the regulator will be given by, if you look at that formula over there, is just one over this pairing delta with omega squared. All right, now here we have the following thing we can do. So to evaluate, which is the integral from w to gamma w of omega, you can take this point w to infinity. Okay, so this idea goes back at least to Manin. Okay, you can choose any point w in this and as long as omega comes from a cusp form, so it decays rapidly, you can replace this integral. So what happens when w goes to infinity? This, so w is going off to infinity, gamma w is down here and this becomes a geodesic. You can take this to be a geodesic. So i.e. integrate omega over geodesic path from infinity. All right, so we'll write this in terms of L functions and there'll still be a problem there, which I'll discuss. Okay, so this alone doesn't, these are in terms of the, you can see this as the, the special case in fact of this Walsh-Bichet formula where you take L to be the split quadratic extension. Okay, so this, but then these have another feature which will be useful later. All right, so let me discuss what these integrals look like. So just to keep it more familiar, I'll discuss the analogous story for the upper half plane. Okay, everything works exactly the same. So for familiarity, I'll compute the analogous story for the upper half plane. Okay, so it really works out exactly the same, but if I were to write it out here, the here I can write it explicit for expansions where here would have a Bessel function and so on. So imagine you have a weight two cusp form and this element delta let's say is A, B, C, D. And what do you want to do? You join infinity. So delta applied to infinity is to point A over C. So you join infinity to A over C and you want to integrate f of Z, D, Z over this. All right, and again, this will be easy enough to do but the issue, there'll be an integrality problem which we have to handle. So how do we do this? Let me first take a very simple case. So if let's say A over C is actually zero. Okay, so maybe if A were actually zero, how do we compute this? Then it's, so if we just, okay, let's write the Fourier expansion, this sum of A and Q to the N. And what we have is just integral from infinity to zero, sum of A and U to the minus two pi and Y, D Y. And this just gives you one over two pi times the standard, the central L function. Okay, now the central L function, so let me see, let me now, oh, where did it go? Maybe it's gone. Okay, of course, this, so if we are actually in this case that you get the central L function, then you can apply Birch-Spintern and Dyer and Birch-Spintern and Dyer says the central value of the L function for, so let me just ignore that one over two pi. It's a product of several factors, the K-chaff-Ravage group of the elliptic curve a Tamagawa number divided by the torsion part squared times a period in the imaginary quadratic case it's this complex area, okay? So in this unrealistically simple case where A over C works out to zero, i.e., in this case, we would get this pairing works out to be this, so some integer, which comes from the size of a Tate-Chaff-Ravage group divided by size of torsion squared times omega E and then this regulator comes out to be the reciprocal of this, so E torsion to the fourth times omega E to the minus two, maybe I should keep this as modulo sum integer. Okay, so that's the nature of the statement I made. The regulator, it divides, okay, you don't know what this integer is, but anyway, the regulator divides a fudge factor, which is this torsion size of the torsion group, which can be very easily bounded, times this, this is the interesting part, this thing is the height of the elliptic curve, faulting's high to B. Okay, this however, so that's the way this argument is meant to work, but the analog of this essentially never happens. Okay, so if A over C is not zero, then you have to do something more complicated. Okay, your entire worry is in fact that, okay, I chose an element delta projecting to one and the entire worry is that this element delta is extremely complicated. Okay, so your worry is in fact this A over C has some very large height. So in that case, is that okay? Yeah, okay, but so if A over C, if you're not so lucky, what happens? Then you can still write this as a combination of L functions. So we have no other source of understanding of the integral properties of things besides Birch's foot and dire, okay? So if you don't have something like Birch's foot and dire, you're completely lost in this. Okay, so we need to go through L functions because in this, in the, when you talk about modular curves, you can make direct arguments sometimes using algebraic geometry, right? Here there's nothing, there's no algebraic structure. So the only, that everything must be reduced to L functions because there's no other way, no other handle we have. So what happens here? The integral from infinity to A over C, F of Z, D, Z is now sum over N, same, okay, similar thing. If you work this out, you get one over two pi and you should, this is what the sum I write is in the sense of analytic continuation. And then you have this factor E to the two pi I A N over C. So this is not an L function, but you can write it as, so this can be written as some linear combination. So by Fourier analysis, you can rewrite this in terms of Dirichlet characters. So you can written as a linear combination of things like L of one half F, F times chi, where chi is a Dirichlet character of conductor dividing. So this is sort of a sum. A chi is a Dirichlet character of conductor dividing C. Right, now you can hope to now do the same thing. You can apply Birch-Sudder and Dyer to understand the kind of integral properties of this. Now you have another problem, the A chi's are not integral, okay? And in fact, up to here, you can kind of make analog of this kind of story also in the compact case, but you run into a similar problem. These A chi's are not integral because when you write out this, for example, if C is prime, and when you do some expansion of this, you get a sum over chi, you get Gauss sums, it looks something like this, but you'll divide by C minus one, which comes from when you do Fourier analysis. Okay, if N is non-zero, mod C, you have something like this. So you get a denominator. So you get denominators related to C. And this is again, okay, so you're the, but now what you can do is, so this is where the sort of integral difficulty gets ends up, these denominators. But now you can play the following game. So to avoid these denominators, you take your original delta, rewrite delta, you take this delta and you write it as a product of two other elements. Okay, so delta omega is delta one omega plus delta two omega. Okay? And if you, you can do exactly, you can play exactly the same game for these, and what you find if you do it carefully is you can get a different set of denominators for the right hand side. Okay, so it can arrange to get different denominators. And then, okay, and then, so thus these denominators, which potentially occurred here, couldn't actually occur in the first place. Okay, so it's just this final, this is a similar story in the compact case, but the denominators become related to class numbers and I don't, then I don't know how to play this kind of game effectively. Okay, so this last step is the point where you kind of avoided the kind of win something. Okay, so that concludes the, so what have I done to here? So right, so regulator divides some explicit multiple, some explicit and as you, as some extinct explicit times omega e to the minus two. Okay, as you saw this explicit thing really just came in the, from the torsion, the size of the torsion and the elliptic curve. So let me just say one word about why you would expect this to be an equality. Okay, the failure of this to be an equality comes from this integer. Okay, that integer if you trace back is the Tehchaf Ravich group of this elliptic curve. Okay, now the sort of thing which would need to happen for this to be a, if this weren't equal, is that you would need to have not just the Tehchaf Ravich group being on trivial, but there'd be some prime that divides all the orders of all the Tehchaf Ravich groups of all twists. Okay, so this kind of thing is, although it's very hard to prove there are conjectures of Caulibagin and others suggesting that that kind of phenomenon should never happen. Okay, so let's say to show equality would need to know, another way to say it, something like non vanishing mod p results. Okay, so which you could think of either as being about the Tehchaf Ravich group or being about the L function. Okay, but I do think for that reason I think this is really an equality. And yeah, I'll say one more time, I think this general phenomenon, this might be a general phenomenon of these regulators are related to heights. And if so, that's quite, as I said, it's quite interesting because somehow the two sides are of different natures. Okay, so maybe I stopped there. Questions? Could I give just a hint of the idea of how you do when the free part is a two, by the way? Is what? Is it a two when you have the two instead of that? Oh, actually to tell you the truth, I never wrote it down. So you can run this argument, so to speak, let's say there are two elliptic curves that contribute to it. You can run this argument for each one individually. Now then there'll be another factor about congruences between those two forms, which I think should, so if they, which I think you can bound in some explicit way in that case. I've never, so in other words, I've never written down this argument. I don't think there are any problems when it has bounded dimension, that is you analyze it form by form, but you have some subtlety because the integral structure doesn't break up form by form, but nonetheless, I think it's easy to bound. Without, I say that without having thought through it very carefully. When the rank of the free part, yeah, when it starts to, I guess it seems to me it should be fine even if the rank of the free part was as long as it's sort of volume to a power less than one, but I haven't really thought very carefully about the estimates for that. Yeah. So when you say you have a proto, this issue of eliminating the denominator is related to plus number of problems. Not really. No, you said this. Oh, oh, oh, oh, I see. Yes. So yeah, for example, in Prasanna's paper, it's Prasanna's thesis, right, he doesn't, when you do an analog of this, where in the Walsh-Pochet context, when you do Fourier analysis on this kind of, basically end up doing Fourier analysis on class groups and then you end up dividing by class numbers. Okay. Yeah. But I mean, you may have a, so when you say class groups, you mean class groups of, no, things of people that you think, so maximum. No, no, you lose, you don't have much control over things like that. Like, for example, here, there's this element where I call a delta. Now, this, which generates, I've lost it, but you pick an element which corresponds to a generator, you don't have much control if it, whether it corresponds to a maximal quadratic ring or not. Okay, so, but if you were non-maximal, you would have more problems, so. I'm not really sure, I don't, I don't know. So, is it possible that then you can translate your delta by some sort of a cooperator to make it non, so if you suppose it is primitive and it becomes less primitive, but then you are non-maximal? I haven't had much success with such a thing, but I don't know if, I don't know. Because this A over C looks like just non-primitive quadratic things over the split-quadratic algebra of probability to be. Yeah, but see what was important here in a way is that you could bound, this C minus one, you can think of it as coming from a class number but degenerate quadratic ring. And it is important in this denominator avoidance argument that you have such an explicit understanding of those. But so then my question would be to move the delta, try to move the delta to be in such a situation by some big amount, I don't know. Yeah, no one, okay, I don't know how to, I've had very little success besides this argument and I'm doing anything. Any other question? Yeah. To bound are you, why do you exactly need the equality, why not just integer time of the ring, is enough? I'm sorry. The R, the value of R it says integer times, the power factor times omega E. R divides this. Yeah, I mean, why do you need it exactly? You don't, you don't, this is enough to bound it. Yeah, because R is this divided by an integer, so it's less than this. So this is fine for the purpose of bounding, but I would like to know what's true and I think it's actually should be equality. Yeah, but strictly speaking for the application, this is fine. Yeah. Yeah, do you know of any G1 incarnation? Oh. Which this? No, I like, it already has interesting incarnation for GL2 over Q, but I don't know for GL1. I think it really might not have, I can explain why to you afterward because GL1 is a billion, I think it's.