 So I'll just remind you quickly at what point we were yesterday. So last time, the thing we discussed was that, OK, so O is an order in an imaginary quadratic field, so something like this, where B is less than 0. And so N is an element of this. And we defined the congruent subgroup, A, B, C, D. So these now belong to O. The determinant is 1 and N divides C. And so the phenomenon which we want to discuss is that when we take this group and abelianize it, so the size of this or the size of its torsion part, if it's not finite, grows exponentially. So this was with N squared. OK, so this is a I gave yesterday a more precise conjecture, and we'll revisit it later today. But that was the kind of phenomenon that we were discussing. And OK, so the plan of lectures is that I'm still in this. So after I set this up, I described a heuristic, which at least very crudely explains why this should be large. And so I'm going to spend more time talking about that. So I'm going to finish discussing that heuristic from yesterday, and then we'll talk about another heuristic. And probably by the end of the day, we'll start on this analytic torsion. So this first heuristic from yesterday, so what I called heuristic A from yesterday, suggested that when you take this group and make it abelian, you should model this as, so it can be modeled as modeled integers where, so for V1, VG, these are, let's say, random vectors with some, all right, so with entries, let's say minus 1, 0, 1. OK, so this came from looking at how a presentation for this group coming from geometry might look like. All right, and here G should be very roughly on the order of n squared. OK, so this is what we discussed yesterday. Now, I want to make a second, I want to explain one way in which this heuristic is definitely inadequate. So OK, so this heuristic, as I said, it's a crude heuristic, it very roughly, correctly predicts the size of this. But if you ask about anything finer, it's really wrong. So let's examine what this says about. So far, I just discussed its size. Let's look at its actual group structure. What this suggests about the group structure. OK, so if we're going to talk about, in other words, is it a cyclic group or a product of cyclic groups or so on, and I'll explain that this heuristic really gives you a wrong picture if you ask such a fine question. So there's something essentially missing with this, and it's sort of important to be aware of that. OK, so if we're going to talk about the group structure, we might as well look at it prime by prime. So if you believe this heuristic, let's just look at a single prime p, and I'll look at the p part. So in other words, I throw away everything. I throw away all the torsion that's prime to p. So I just keep whatever factors of Z mod p, Z mod p squared and so on. So if you believe this, you would think, all right, so I'll write this meaning this can be modeled as you take, now, patic integers to the g, quotiented by. So if you believe this, it's reasonable to leave it. Let's just sort of take p parts everywhere where these v1 up to vg, you choose random vectors in Zp to the g. Up here, there was some 0, minus 1, 1, and 3, but you don't expect that should have any influence on the group structure. It's kind of an Archimedean constraint. So if you believe this, you think maybe this is a reasonable model for how the p part of this behaves. You take some fairly large number G, you take Zp to the g, and you quotient it by this. Now, this is some way to produce a p group. And in fact, this way has already shown up yesterday in a different context. So in fact, you can very precisely understand what this process produces. So if you take, so quotienting Zp to the g by g random elements, vg, at least in the limit as g goes to infinity, it produces exactly the same distribution that you see in the Cohen-Lenzstrah heuristics. So I'll say this precisely, it produces the Cohen-Lenzstrah distribution. So by this I mean that if g is a finite abelian p group, then let's look at the probability that Z, all right. So take some integer g, pick, so to speak, a g by g matrix of piatic numbers at random and do this quotienting by the rows of that matrix. The probability that you get g will be exactly the probability that we see in Cohen-Lenzstrah, which is the product from 1 to infinity, 1 minus 1 over p to the i, divided by the number of automorphisms of g. Let me say it's not maybe so relevant for our story, but in a second I'll say why you expect to see exactly the same thing in the Cohen-Lenzstrah context. That's a separate but interesting story. But what I do want to say is that, in particular, just as Fouffry discussed in his lectures, the probability of getting a group like Z mod p to the k is very small. So Fouffry discussed Z mod 3 times Z mod 3. If you do Z mod 3 to the k, this probability will be on the scale of p to the minus k squared. And if you look at data, if you go look at a table of data of this, a bunch of different n, you look at the delinization, you see that this is not realistic. You see many factors of Z mod p much more often than you would expect. So this doesn't match the data. So I think there's a clear guess as to what is missing from this heuristic. But let me just have a small aside about why the same thing shows up in the Cohen-Lenzstrah context. Are there any questions up to here? Yeah. So you choose a metronome with the trees minus 1 to 1? OK, so for the ease of analysis, let's say we choose them at random with respect to the Haar measure. However, in practice, I know there are theorems of the sort, but I don't know what the state of the art is. But in practice, if you simulate this, you find pretty much any way you pick them, you will get the same distribution. OK, that is minus 1. Any reasonable distribution, it seems it's very insensitive to how you pick them experimentally. But when I make a statement like this, let's say that I mean that VI are chosen uniformly with respect to Haar measure. Any other questions? OK, so this is a slight aside, but I thought it was worth making because we're also talking about the Cohen-Lenzstrah heuristics in another course. So why should this show up? Why does this show up when you model when studying class groups? Oh, one thing I wanted to say is disequality here. Oh, oops, so much should have said. Sorry, limit as G goes to infinity. The convergence here is extremely fast. So as long as G is slightly more than the number of generators for G, this would already be very close. And the proof of this is an elementary exercise in algebra. OK, it's not much harder than knowing the number of invertible matrices over Fp. OK, so why should this same distribution, which I've said is not correct in this case? But why does it show up in studying class groups? The point is that you can also, so if you take k as an imaginary quadratic field, so the class group of k is by definition the group of all ideals quotiented by the group of principal ideals. Now, this is the quotient of two infinite groups. So but you can replace this by something that's kind of finite and easier to analyze as follows. So let's say let S be a large set of prime ideals. So for example, maybe you take all the ideals with norm, e.g., all with norm less than some very large number. This should depend on k, so some big number. And so if S is large enough, you can replace this by kind of finite version where we just look at those ideals generated by primes in S. OK, so you look only at those ideals who have all their factors in S. And you quotient only by the principal ideals intersect the set. OK, so principal intersect the same set. OK, now this top thing here, so it's what you can get from all the primes in S, it's clearly a free-ablean group of rank equal to the size of S. OK? The bottom thing, it comes from all elements of k. Principal ideals are just things generated by k, but they should have all their factors in S. OK? So this, in other words, comes from this, sorry about that, this comes from the S units. OK, the elements of k, all of whose prime factors lie only in S. And Dirichlet's unit theorem shows you this is also a free group of rank S. OK, that's where the imaginary quadratic is used. If it were real quadratic, it would be S plus 1 at this point that you would get a new. So the point is the class group of an imaginary quadratic field is naturally presented as a quotient, in other words, of some free-ablean group modulo the same number of relations. OK, so in some sense, so that kind of explains heuristically why you should expect this distribution. OK, there would need to be some conspiracy among these relations for it not to happen. So this model of group shows up, I'm not saying it also shows up in other contexts. It's a very natural model that you take some number of, a large number of generators and quotient by the same number of relations. OK, now what was I going to say? I was going to say, yes, what has gone wrong here? Are there any questions? So yeah, I think the short thing which has gone wrong is that this, so let me say it this way. I'm going to say something with, oh yeah, so suppose we looked not at this, but we looked at a gamma that was similar, a group that was not arithmetic. OK, so you take something like a hyperbolic three-manifold, which is not arithmetic, and you take its fundamental group. And you go through the same story. Well, in fact, Nathan Dunfield and Thurston showed that at least under a certain model of random such groups, you really do get this Cohen-Lenz distribution. OK, so the fact this doesn't seem to hold is related to the arithmeticity of, so it's somehow related to, so this is related to the failure of the model to account for arithmetic symmetries, and by that I mean the Hecke operators. OK, this model knows nothing about the Hecke operators, and the Hecke operators greatly change the, all right. So let me just finish, there's a very similar phenomena to this failure, so I'll just comment on that, and then we'll be done with this heuristic, OK. But so in other words, yeah, this heuristic, this is the main point, this heuristic doesn't account for the Hecke operators, this is one way in which it becomes visible, it gets the group structure wrong. Right, so what we did here is I took a random g by g piatic matrix and I formed a group out of it, and that's its distribution, and the story has some parallels, this has some parallels with the kind of classical random matrix theory. OK, so just to give you an example of such a parallel, here I wrote that, so this probability that you get z mod p to the k, I said it's very small, why is it so small? It's because this g by g matrix is most unlikely, it doesn't want to have many eigenvalues near zero. So this is actually exactly the same phenomena that if you take, I think if I compute right in the GUE, matrix ensemble, so random Hermitian matrices, the probability that you have k eigenvalues in a window of scaled length epsilon has exactly the same behavior, it's epsilon to the k squared. So there's some, the sort of Cohen-Lentzter story has some reflection of what you see in random matrix theory, it's much easier than random matrix theory, but you see some parallels, and in particular this failure, there's a parallel failure to do with random matrix theory, so the failure of this naive model to, well, above, so it's kind of similar to an analytic failure, which is that to the failure of random matrix theory to model the eigenvalues of, okay, so if you don't know what I'm talking about here, it's not going to matter, but I'll just say this, to model the eigenvalues of gamma mod d upper half plane if gamma is arithmetic. So something physicists noticed about 20 years ago is that typically, if you take a quotient in the upper half plane and you take its Laplacian eigenvalue, the gap statistics look like random matrices, but they don't if it's arithmetic, and that's sort of presumably because the presence of Hecchi operators forces it to behave in a non-generic way, and so this failure is kind of parallel to that one, okay, where there's some similarities to this. Okay, so that I think that's the end of the first heuristic, and now I want to, so now I want to talk about the second heuristic, which is much more specialized, but it also gives me a chance to talk about some of the reasons one might be interested in this story in the first place. Are there any questions? All right, so for heuristic B, I have to describe a little bit the relation of this to the Langland story. So let me start by saying one, let me start by briefly saying something which happens in the classical case over Q, and then I'll describe the parallel phenomena in this setting and why it's in a way more interesting. So over Q, all right, so if we have F, which is a weight two, so F is a weight two holomorphic form of level N, so with rational coefficients and a Hecchi eigenform, and so let's say, okay, so it looks like some A n, Q n. So Eichler and Chimura made this wonderful discovery that they could associate to F an elliptic curve, so this is due to Eichler and Chimura, E, it's called E sub F, which is an elliptic curve, and it has this amazing property that this AP here is related to the number of points on this over Fp. Okay, so number of points of this over Fp, so this is due to Eichler. And Chimura, okay, so that's just P plus one minus AP. And in fact, you don't, there's a way you can just kind, if you don't even have to remember the full elliptic curve, one can just remember the Galois representation on the tape module. Okay, so if you don't know what that is, don't worry about it, I'll, so Galois representation. So I'm just saying this so that if you've seen it, it'll make contact with what I'm about to next, but it doesn't matter if you have not seen it. And this has some property, trace of row, that it's traces on Frobenius elements are related to these APs. Okay, so that's the story over Q, now let me now say the corresponding story over our imaginary quadratic. Okay, so over our ring of integers in an imaginary quadratic field. Now, yeah, maybe I should just remind you that, as I said at the start of yesterday, if we didn't know about the space of weight tube forms, we could nonetheless extract all the information in it from the, just abelianizing this group. Okay, so as far as the Hecchi operators go, they have exactly the same information. So over O, all right, so now I'm gonna, for example, here, there's also a statement that the coefficients are not rational, but I just stuck to this because it's simple. So again, here, let me make the simplest statement I can and then I'll say something to try to make it a little more palatable. And then I'll talk about the, so it's involved the work of many people. So let me start, so the simplest, I'm restricting to a simple case to make this simple, to make this, so let's suppose that we are in a setting where the P part of this, okay, so gamma naught n here, we're in this, now I'm over it, it's imaginary quadratic field. Let's suppose that the P part of this just consists of a single factor of Z mod Pz. It's not important, but it just makes the statement as clean as I can make it. All right, and then in this setting, all right, so let me pick an element, a non-zero element here, which I'll call alpha. So here, you can make a representation, so it goes from the Galois group of, so K is going to be the corresponding quadratic field. So it goes to the Galois group of K bar over K to GL2 Fp, let me, GL2 Z mod P, so it's the same, okay, same P here and here. So I'll write it, and then I'll sort of, in a special case, I'll try to make it a little more friendly. So, and it has the following properties, it has many properties, I'm not listing all of them, but two important ones are, it is un-ramified outside n times P, and for any prime Q, the way that the Q-teche operator acts on this alpha is by multiplication by trace of the Frobenius of Q. Okay, if you've never seen this kind of thing, this looks like sort of incomprehensible, so let me just pick a special case and try to make this a little more friendly. And so, by the way, this is a lot harder, okay, this, I'm gonna talk about the history of this in a second, but this is not due to actually, this is much harder. So let's just illustrate the kind of what this means, let's take the case when P is, yeah, it should be rho, yeah, let's call it rho, let's just call it rho, okay. In this setting, I just have this one, yeah, thank you. P is prime to n, I don't think it matters, but maybe this is only one n, but I don't think it matters that P is prime to n. Oh wait, what did you say is prime to n? Oh yeah, okay, but maybe Q should be prime to n, yeah. Okay, so let's just take the example when P is two, okay, so I just wanna spend some time saying that this, you know, how should one think about this, okay, because it, so when P is two, I mean, in practice, so let's say P is two, so you do bilinize this and you see that the two part is just a Z mod two. In practice, if you actually do this, a lot of the, you'll have this annoyance, this won't be surjective a lot of the time, but let's say it is surjective, so say that rho is surjective, so what I'm gonna try to do is translate this right-hand side into something more down-to-earth, so rho goes from the Galois group of K bar over K, it goes to GL two, Z mod two Z, and this is a group with six elements, it's isomorphic to the symmetric group on three letters. Okay, so to give this map, to give the map from the Galois group to S three is exactly the same as giving a cubic field, okay, so rho determines, so to give rho is equivalent to giving a cubic field, which is not Galois, okay, so a cubic extension, and yeah, if it's not Galois, you will get a S three extension from it. And now I'll translate these other two properties, A and B. Okay, so A says that this L, the discriminant of L is divisible only by, only by primes Q dividing N times P, and condition B says, if you translate, if you just go through the translation, what it says is that if you, okay, so I have this, here's this alpha, one thing I've sort of said here without explaining is that there really is a way that the Hecke operators act on this, okay, so I sort of, I said this at the very start when we were discussing weight two forms, I said the Hecke operators act here, and it's kind of, they compatibly act here, okay, so if you sort of think through this and figure out how to translate the Hecke action here to here, then the same, exactly the same thing works over here, okay, but I never wrote out, I never actually wrote it, wrote down the definition, but whenever you do this, take an arithmetic group and abelianize it, the Hecke operators act on it. So the alpha is an ideal form, this is automatic? It's automatic in this case, because I assume that the P part is just C mod P, yeah. So, right, and this other assertion says that, what does B say, it says TQ of alpha is non-zero if and only if Q, so Q is a prime of K, is inert in L, okay, so when you, so something that, at least at some point I did to try to make myself feel better about this is for small p, you can translate these assertions to things about explicit number fields and the way that prime split in them, and then you can go and test it numerically and it's very satisfying to see you can, so as you see for small p is really these statements, the content statement is something very explicit, okay. So, I'm gonna talk a bit about the history, is there any questions? All right, so this was, so on the one hand in some way you expect it, perhaps by analogy with this, but there's something, so let me first say, so this is recently been proven very recently by Peter Schultzer, but I wanna say some of why it's, it's so much more subtle than this, okay, and some of the history of it. When we look at, in the case over Q, as I said yesterday, this thing is essentially, it's more or less a free group, okay, so I said yesterday it's some mini copies of Z plus a small finite group. Now, the problem you have here is some, a situation like this, this Z mod PZ is not detected by anything in the classical theory of automorphic forms, okay, the classical theory of automorphic forms, it has things in objects that exist only in characteristic zero, like Laplacian eigenfunctions and holomorphic forms, and so it has no hope of seeing this. And actually, so the reason I got interested in this problem in the first place is, somehow all the standard techniques of automorphic, so this thing, this alpha behaves just like a modular form, it has every right to be called a modular form, but none of our, none of the existing story of the classical techniques apply to it. And in fact, so the idea that you should, you should be looking at these, so that one should think about the possibility this has torsion and that you might have a sort of story goes back, so I think it was predicted and maybe Closel knows more about the history, I think Abner, Ash, Closel and Taylor already knew about this from the 1980s, is it fair to say? I think that's fair, but Sam knew about it, whatever. Okay, so there's something to be said here, it's not obvious, this is something you should be looking at, and it's not obvious that there is this torsion phenomenon here at all, so I think it's, insight to realize this is something worthy of study. Anyway, this goes back to one of the main motivations for studying this conjecture I made with Nikola Bergeron was to show that this kind of phenomenon, this torsion is actually complete, it's not some sort of strange exception, but it's, once you go away from groups like SL two inch Shimura varieties, it's actually the norm. This is what you see is primarily torsion and you see very little in characteristic zero. Okay, so all this was just to give you some context, so this story fits in some, must be a part of the Langlund's program, okay? And still, so now Schultz has proved this analog of the Eichler-Schimura story, but the amount we know about this is still vastly, vastly less than, okay? So very few of other tools that we know about characteristic zero modular forms can be applied here. There's also a converse to this assertion, okay? Which is not known, so let me say, I'll just put in quotes as Sarah's conjecture. Sarah's conjecture is usually understood as a statement about q, where it's, the rational where it's proven, but so this is a should be, so it's a conjectural converse. And I've said this in the kind of a very simple case where this is Z mod p, but even if this is Z mod p squared, but whatever this is, there is some precise statement you can make whereby you can exactly predict this in terms of Galois theory. All right, so that's a very short description of how this relates to the Langlund's program and now I will come back to the question of its size. Are there any questions? I mean, if you have three parts on your group, the representation in principle can come from some characteristic zero morph, or you always predict it should come from distortion. No, okay, so sorry, but when I mean this notation, let's say I literally meant it Z mod p, but suppose, as you said, suppose this lifted to characteristic zero, then correspondingly this would lift to a piatic and presumably even geometric representation. Yeah, any other questions? So all right, so now back to the question of size, size. All right, so if you have some description of this in terms of Galois theory, you might try to figure out how large this is by seeing how likely it is that such row exists. So I'm gonna say this very, this deserves more explanation than I have time for, but so by a non-Abelian version of the Cohen-Lenster heuristics, so I explain why I say this in one moment, the probability that such a row exists should be roughly one over p. Okay, so let me just compare here. Let me say why I'm mentioning Cohen-Lenster. So the usual Cohen-Lenster heuristics say that for an imaginary quadratic field K, they say so for K imaginary quadratic, the probability that p divides the size of the class group is this number which is one minus that number up there. So one minus the product from one to infinity, one minus one over p to the i. And yeah, I think this is a, so I put a statement above the g equals one and then take them. All right, now this is roughly one over p to first order. But we can reinterpret this in the following way. This is the same as the probability that there's a non-trivial homomorphism from the class group to z mod pz. And by class field theory, this is the same as there being an un-rammified homomorphism from the Galois group. Okay, so this way you can think about the Cohen-Lenster heuristics as telling you, measuring something about how likely is it that you have a a billion extension of a number field and this context the answer is supposedly one over p. So when I say a non-abillion version of these heuristics, what I mean is something that describes for you the chance of you having a homomorphism in the Galois group is a non-Abelian group. Okay, and this exists, I would say so, largely due to work of Bargava. So let's say Bargava formulated this for the symmetric group. And then if you take what he did and then you apply it very recklessly, you come to this claim. Okay, so suppose you believe this for a moment. Is it the same as what don't you have a question yet for Cohen's with my opinion? Yes in some, I think yes in some idealized situation when there was no Archimedean story. The Archimedean story is very problematic because some elements are forced to be of order two. And so if there were no Archimedean places, it would be like that. Yeah, but sorry, maybe I should say that. So another way, how would you guess what these heuristics are is related to what Emmanuel said. If you didn't know them, one way you can do it is you can think about a function field case and there is sort of a clear guess of what the answer is. So that's how you might discover this if you didn't know it. Okay, where are we? So if you accept this, and I admit we're somehow deep in the realm of heuristics here, but if one accepts this, that says that the size, so how big is this group gonna be? Well, it's gonna be at least for every p, every prime occurs with the probability of one over p and this is infinity. So you might think it's problematic that this heuristic predicts infinity for an integer, but I think, so if you were to, you could apply the Cohen-Lenzter heuristics the same way to predict, try to guess. If you just went prime by prime and applied the Cohen-Lenzter heuristics to guess the size of a class group, it would predict that imaginary fields have infinite class group and real fields have finite class group. Okay, so you should take this with a pinch of salt, but at least this suggests that it should be large. Okay, one advantage of this type of heuristic is it works very well when you replace, when you go to a general setting, that is you replace SL2 by a bigger group or you replace your imaginary quadratic field by another field. This always gives an answer and that answer seems as far as people have been able to test it to really be experimentally seemed so far to be right. Okay, I'm not gonna talk about that, but that's the main virtue of this heuristic is that it is very, it's much more generally applicable in the previous one. Okay, so that's I think the end of heuristics and now I'm going to talk, we will talk about next about analytic torsion and to start that I'm just gonna talk about, I'll introduce just hyperbolic three space. So I'll just be making, I think definitions for today so that we can look at the formula next time. Are there any questions? Yeah. So common order and I've been raised, common order and you speak, that means it has to have a three part? No, it's unclear, it's somewhat open to interpretation, but I interpreted it in the way that I want to, which is that the torsion part is getting very large. Yeah, so it's, yeah, like for example, if you literally apply this heuristic, it predicts that the probability that modular forms exist is zero. So you have to be a little bit careful about the realms where you apply it, but anyway, I'm not saying I found this to be quite reliable as long as you're a bit cautious about how you use it. All right, so now, okay, so now we're gonna talk about analytic torsion, but well, maybe I write the formula now and then our job will be to define the quantities that are in it, okay? So this formula, well, again, let me attribute it after I write it. So what I'm about to say is true, so true if, oh dear. Okay, it would be true if this quotient were compact and if not, it's much more delicate story. So I'll write a formula which is literally true in this case. The size of gamma naught n torsion part abelianized times R, this is this regulator, which will be the thing we, which I now feel is kind of the most interesting part of the story, this is equal, oh, now divided by the volume, hyperbolic volume, so I'm sorry, I'm gonna talk about this hyperbolic space and we'll define the quantities one by one, is equal to determinant of, so over here, there'll be two, okay. These delta naught and delta one will be Laplacian operators and we'll, I'll define them. Laplacians on gamma naught n mod H three, this is the Laplacian on functions that is usual mass forms, this is the function of one forms, so mass forms with some weight and we need to discuss how you take the determinant of such a thing. All right, this is really, it's a remarkable formula because it says that this object, which is something of discrete group theory, you can compute in an analytical way and so it was conjectured by Ray and Singer and I think, so in particular, when Ray and Singer conjectured this, in particular they defined this in the first place. Okay, so as far as I understand, that the definition of this, how you take a determinant of something infinite dimensional was not, was due to them. So I think it's tremendously insightful that they guessed such a formula could exist. I'll say something next time about how you might guess that but it's an incredible thing just to guess it's there and it was proved maybe by Chieger and Mueller independently and also I'll say a couple of words but this formula has nothing to do with number theory. So a version of this is true for any a version is true with gamma naught n mod H three replaced by any compact Riemannian manifold. So I said yesterday another way you can think of this right hand side is as a special value of a Selberg's data function. Okay, so I'll also say more a little bit about that next time, if there's time. Okay, but for now, I haven't, I want to define all the quantities here. This one we've seen. Let me define hyperbolic three space in the volume. Oh, and one thing to say is that this R, this R has to be defined too, but for now just say R is one if this abelianization were actually finite. Okay, so this R is some measure of the size of the free part. So it's a regulator. So let's say R is one of the three parts. So it's a regulator. All right, so maybe just a few minutes on the hyperbolic space net, well, are there any questions? All right, so just like, so I've already written it up here. So just as SL2R, so okay, hyperbolic three space. All right, so when you study SL2Z, a basic fact is it and even the bigger group SL2R act on the hyperbolic plane, okay, which I'll write as H2 for the moment. So in other words, that's all as a space, it's just a positive half of the complex plane, but what's useful about one of the things which is important about it, this is that it has a metric that's invariant. So a Riemannian metric where, and that's invariant by SL2R, and then when we base the thing, so to some extent we try to study a group like this through the geometry of this. So the analog in this case up here is that SL, for example, SL2 of O, O is my Z's square D, this sits inside SL2C, and this acts on sort of an analogous story, one dimension higher, okay? So I'll just say the bare outline of it. It acts on a three-dimensional analog of this space, and in fact, everything is very similar, just one dimension up. So this space is now, you can think of it as three coordinates, X1, X2, Y, where Y is positive. The metric is exactly the same formula. The action, here the action is by fractional linear transformations. Here you can see it as being by fractional linear transformations when you think of these in some quaternionic way. But instead of writing that, I'll just show you how some a few basic classes of things here act. So for example, the element one. So sometimes I think it's, we can also think of this space, we'll think of it as being, yeah, so it's sometimes nice to think of this as being inside the complex plane times the positive reels where the complex coordinates are X1 plus I, X2 and Y, okay? And if you think of it this way, so one alpha zero one acts on, sends X1 plus, this thing here translates in the complex plane, X1 plus I, X2 plus alpha, Y. And this element here, this dilates both the entries. So it sends X1 plus I, X2, Y to T squared X1 plus I, X2 and then absolute value of T squared, Y. Okay, and finally the element, an element like zero one, one zero X by kind of inversion. So anyway, you have this geometry now to and one thing which you get even before you do anything fancier is that, so just the fact that this acts on this geometry gives you something already, so the volume of, you can form the quotient, okay? So let me just say in the simplest cases, just like the fundamental domain for SL2Z looks like this in the small discriminant cases, the fundamental domain for that just looks like a sort of three-dimensional version where you have like something like that. Okay, that's a typical picture, fundamental domain for these things are kind of built out of things like that. So, right, so you can now measure this because you have this metric and just that fact is useful. This gives a kind of an intrinsic measure of the complexity of, let's write this gammon on end, complexity or if you want size, okay? The fact that this gives you some number which in some way measures how large this group is. So in terms of this number, I wrote a conjecture last time now, let me rewrite it in slightly more generality. So my conjecture with Bergeron, so the conjecture of mine and Bergeron states that and it's naturally phrased this way is that if you take the logarithm of the size of this torsion, so the last time I scaled by n squared but the natural thing to scale by is this volume, then you get this should go to one over six pi. Okay, now this is not the way I stated it last time. So to recover the form from last time, so the version from last time, you can use the volume formula which was proven by Humbert in the 19th century. Okay, and what that says is that the volume of hyperbolic three space modulo SL2 of O where O is the order of discriminant D is given by absolute value of D to the three halves over four pi squared times the zeta function of Q squared minus D at two. Okay, so in the case I had last time, this D was minus one, D was minus, well D was plus one and the zeta function of Q of I at two is pi squared over six times this directly L function. Okay, so this constant which appeared last time, really it comes from the value of a zeta function of this quadratic field. Okay, so it's really a measure of hyperbolic volume, nothing more. Okay, so maybe this is a good point to stop and next time we will discuss the other, what the other things in this formula mean and a little bit about what we can do with it. We have questions. Is there anything we can compare this to in the SLQZ case? What's that? To the formula? Oh yeah, the formula of an S. Oh, no, not that one, the trigger muller. Right, come on. Well, the problem is an even dimension this tends to is, at least if you don't twist it in some way, it's one for trivial reasons to do with duality. So it has interest only in the odd dimensional case, at least if it's not twisted in some way.