OK, so just say, OK, everyone can hear to tell me if you can't. So I tried to make this the level of this. I know there are people of different levels and different backgrounds. So I tried to emit it at, say, graduate students as long as with some knowledge of holomorphic forms and mass forms. OK, so I'm going to start. I'm going to go quite slowly. And the other thing is, I don't know how familiar people are with homology. So I'll try to, by and large, stick to the simplest case where I don't have to make any use of it. OK, so let me start for a while with SL2Z. And then we'll, so for the first half an hour. So I'll just get to the question which we're going to discuss for the rest of the series. So first, just a simple notation of gamma as a group. We can, gamma ab means for us the abelianization of gamma. In other words, that's the largest abelian quotient of gamma. So it's the quotient of gamma by all. So you take gamma quotiented by the subgroup generated by subgroup generated by all elements x, y, x inverse, where x and y are in gamma. So what we're going to be talking about is what happens when we do this to groups like SL2 over an imaginary quadratic field. So if gamma comes with a presentation, this is very easy to compute. So for example, if gamma, if you happen to be given a presentation, let's say it's something like it's generated by two elements, I'm just making this example at random, beta to the 5 is the identity. Then, so the abelianization will be, so it will have two generators, so Z squared. And then you quotient by relations which come from this. So in this case, you quotient by the row vectors 2, 3, and 4, 5. And I guess that's the group Z mod 2. All right, so what I'll start by doing is we'll do this for subgroups of SL2Z. And I'll explain why you might be interested in doing that from the point of view of holomorphic modular forms. And then, so what I'm going to talk about is so we'll discuss this gamma ab first for subgroups of SL2Z. And then after we do that, then I'll make this Z into a Z square root D. And we'll see how the situation changes. So I'll stick to the usual. It doesn't really matter which family of subgroups we choose, but I'll stick to the standard ones that we talk about in number theory. So gamma naught of n will just as usual be the elements abcd in SL2Z. That is determinant 1. So determinant is 1 and n divides C. And in a second, the same definition also makes sense. So right now, abcd are integers, but this also makes sense if we did Z square root 2. So what happens for subgroups of SL2Z first? So for subgroups of SL2Z, what happens when you abelianize this is very closely related to weight 2 modular forms. So gamma, if you take this and you take its abelianization, what you get, so I'm not going to prove this, but I'll say in a second where it comes from, is you get, actually, I guess the statement I'm about to make is literally right only when n is prime. So it's Z to the 2h plus 1 plus a small, in a sense, I'll say, small finite group, and where h is going to be the dimension of the space S2 of n. So this means weight 2 level n holomorphic cost forms. Holomorphic cost forms. So I'm going to assume people are familiar with that definition. When I say a small finite group, in this case, what it means is that every element has order at most 6. So it really is very small. But in particular, it's at most 6 torsion. So I want to explain where this comes from, because this maybe gives you some motivation about why you would be doing this abelianization thing in the first place. So the point is that, in fact, this is very closely related to the space of weight 2 forms. And maybe if you didn't know the definition of a weight 2 form, you would notice many of the same things if you were studying this. So I'll kind of explain where this relationship comes from. So in fact, are there any questions? Just shout out if there are questions. Because this relationship between these comes from, there is a pairing. OK, a bilinear pairing. Let me say bilinear pairing. All right, this plus 1. Yeah, as I said, this is right if n is prime. And this has to do with Eisenstein series. So it's a little bit of a, if n is not prime, it's plus something a bit bigger. But anyway, I'm going to sort of not worry too much about it. There's a bilinear pairing. So if you have, from this to the complex numbers, and I'll define it for you, so given an element, so let's call it gamma bar in this abelianization. OK, so this is an abelianization. So it comes from some element gamma. So the image of some gamma in the actual group. The way you define the pairing is you just integrate the form. So gamma, and you have a form f in here, you define this to be the integral from w0, w, gamma w, f of z, dz. OK, here w is any element of the upper half plane. So this integral is independent of the choice of w. OK, so you could, so this means you take any path from w to gamma w. It doesn't matter which, the integral is independent. You take any w, any path from w to gamma w and integrate this. And this gives you a complex number.  So that means, in fact, so in fact, we can take a basis. Let's say gamma 1, I'm just going to write this very explicitly. Gamma 1 up to get, there is a basis for this. Let me put the bars for this z to the, so such that these gamma 1, it's just the functionals you get from them, or at least the first 2h of them, gives a basis for the dual to the space of cusp forms. OK, now something a bit strange about this is cusp form, this is h-dimensional. It's an h-dimensional complex vector space. So this dual space is also h-dimensional. This is an h-dimensional complex vector space. But this actually gives a real basis, OK? It gives a basis, so this is h-dimensional over c, but it's 2h-dimensional over r. So that's where the 2 comes from. OK, so this is the real, this is kind of the reason that you see this 2h here. And do you pick any 2h of them? You have 2h plus 1 of them on the top one, and then 2h. I said you can choose a basis in such a way that this is true. Yeah, OK. So aside from the plus 1, this pairing is tensed with the reals is perfect. OK, so a bit even more is true. So one can also define the action of the Hecchi operators, define an action of Hecchi operators. So for every, let's call them T sub m, where m is relatively prime to n, let's say, on this group in a way which is compatible with here, OK? In such a way that if you apply a Hecchi operator on one side, it's the same as applying it on the other side. OK, so what this means is that in other words, we could so can recover all of, if you want, everything we want to know about these way 2 modules together with the action of Hecchi operators on them, you can completely recover if you know the same thing about this. OK, this is the start of a lot of the algebraic theory of modular forms. But why is this useful? For example, this is a very efficient way of actually computing this. So if you take n as a few hundred and you want to write down what are the weight 2 modular forms of that level, this is actually rather computable because, as I'll mention again later, it's actually very easy to write down things like a presentation for this, which is not how people exactly do it. But so this is computationally useful. And also, for example, it tells you things like, so this is a fact of fairly basic importance. For example, it tells you the eigenvalues of the Hecchi operators or algebraic integers. Because if you think solely about holomorphic things, there's no reason why this should be the case. But here, over on this side, these Hecchi operators are represented by integer matrices. So there. Why do you consider weight 2? Well, if I considered weight, what do you mean? Why do you start with weight 2S2M? Why do you start with weight 2S2M? So if it were not to this, literally speaking, this would no longer be. You can't initially come to this. No, I would have to. OK, so there's no problem making an analog of this theory. But instead of the abelianization, I would have to take homology with some twisted coefficient. So it's not a, it's just for, sorry? If we show that, why this is important? Maybe I don't. The generalization is straightforward, but probably leading to no, I guess. No, no, so all the statements that we make about this, there'll be parallel statements in higher weight, which would be the only differences. They're more annoying to formulate. Because instead of this, I would have to write the homology of this group with coefficients in something. And I was hoping to avoid that. You mentioned that 2H plus 1, 1 is for eigenes. 1 is for eigenes in a series, so it would be more. That's 2H in a series, and the thing is correct. Yeah, that's true. But so for example, even for n is 1, so you lose 1. I don't think we really want to go there. Yeah, but in fact, you make a good point. At some point, I'll put a permanent disclaimer that I'll make many statements that are exactly correct in the compact case. Just because it's quite OK. So this is, sort of speaking, an introduction of why you might want to do this process of abelianizing it in the first place. So I say that it has the same information as the space of weight two forms. And now I'll do the same thing, but for other. So it's gone, all right. Yeah, so now I'll do the same thing, but I want to replace sl to z by something bigger. So now let's, now we study the same question for, OK. So now, so let's say replacing z by z of square root d, or by any quadratic order. Replacing z, z square root d. OK, so in other words, let me call o z square root d. And now, this definition, as I said, still makes sense. But let me just write it out so we're clear. So gamma n is still. It's now determinant one matrices in o. Now these are, OK, the determinant is still one. And now this element n is allowed to belong to o. In fact, n could be an ideal, but just to keep it as similar to the previous case, we'll just take an element and n divide c. All right, so let's first take the case where d is greater than 0. By the way, everything I say in these talks has an analog for any group, OK, that is to say if we're talking my gl5 or any. But I'll try as far as possible to stick to sl2 over quadratic fields just because the statements are clear, and I don't have to talk about higher homology groups. But this is not the analogs of everything I say. There are analogous things to everything I say, OK. Right, so first, d bigger than 0. So again, I'll state this somewhat imprecisely then explain it. What you will find if you do exactly the same thing for d bigger than 0, OK, so for example, so d should be square free. So for example, maybe z of square root 2, OK. So what you will find if you do this, again, I won't be entirely precise, but is that this is always a small finite abelian group, OK. So let me not say exactly what small is, but I can again give you some explicit bound. In fact, I can almost compute it in advance, but I can give some very explicit bound on what size and the order of elements are in terms of n. OK, so first of all, it never happens in this case that you have these factors of z, OK. It's always a small finite abelian group. So let me just explain why this case behaves so differently to z, OK. So the main reason that this behaves so differently is that there are units, OK. So it is forced to be small because O has many units. All right, so I just say, so when you form this commutator group, you're quotient by elements x, y, x inverse, y inverse.  I only see two pieces, and I dropped them both. OK, so if we take this, then x, y, x inverse, y inverse is 1, 1, U squared minus 1, 0, OK. So it might not be immediately clear why it's useful. Usually when you take this commutator of two matrices, you get a giant mess, OK. But if you have units, you can use it to produce a very simple class of elements of this type. And then you can try to check that they, where did I put the definition? That they generate a large subgroup, OK. So basically, you can use this fact to sort of force this abelianization to be small. Now, from some point of view, that's a little bit disappointing. But you can recover an analog of the theory above. So one can recover an analog of theory for z. If we replace this abelianization, so rather than work with the abelianization, there's another way to talk about abelianization, which is as the first homology group. So I'll briefly talk about homology, but it's just a passing comment about how we generalize this. So instead of this, we work with the second homology, also called the Schur multiplier, OK. If you were to work with this, you would see something very parallel to this, where S2 of n will now be replaced by, all right, 2, 2 of n. If you work over a real quadratic field, you can talk about holomorphic forms with two different weights, one for each of the real embeddings. So there'll be a very parallel story with a pairing and so forth, but with these kind of, what is sometimes called parallel weight 2. And here again, as before, there's a corresponding story for any weights, as long as they're both 2 or larger. Weight 1 is a completely different story. Anyway, so if one wants to generalize the things we're talking about, one needs to make this kind of substitution. But OK, we don't have to do it for, turns out we won't have to do it for Bianchi groups. All right, so now let's take d less than 0. So for example, z of i. And here we find actually, so I'll give you some, yeah. So here, these analogous groups are called Bianchi groups. And so this is sorry, something called Bianchi groups. Or at least maybe, well, maybe this term is usually used for just SL2 of O. So these were so called because they were studied by Bianchi. And these are very pretty. They act on, so just as they act on hyperbolic 3 space. And I'll describe this action later when we need to say more about it. But so they're very parallel to the SL2z, which acts on hyperbolic 2 space. And Bianchi actually used the geometry of these actions, so this is more than 100 years ago. He already figured out fundamental domains for these actions. And for certain small values of d, he essentially computed presentations for these groups. So Bianchi did quite a lot already. Anyway, here what we find a phenomenon that's quite different, so now there are no units. And now it turns out that gamma naught of n, ab, can be quite large. All right, so let me write some data related to this, and then we'll, are there any questions? OK, so I'm going to give you some data that was computed by Haluk Sengun. And it was computed in response to a conjecture that Nicholas Bergeron and I made. Now I'll say the conjecture in a moment, but let me show you the data first. OK, so this is 4, o is z of i. And I believe this data is actually not for SL2, but for PSL2. But OK, so here we go. For n is 9 plus 4i, disabilitization is z mod 5 plus z mod 3 plus z mod 2 to the 6. For n is, OK, sorry, I'm just going to pick this up so I can transpose it here. For n is 41. So here it's z mod 4, o, 7, 8, 7, 9, 3, 5, 1, 3, 6, 7, 1. z plus z mod 2, 9, 2, 3, 0, 6, 0, 3, 3, z plus z mod 2, 2, 0. OK, well, I'll stop there. It goes on, actually. You probably don't want to see another version of that. But I'll just to abbreviate if you go up a little bit higher the same thing, because z. So it can z plus t, where t is a finite group, t is finite of order. It's order is bigger than 10 to the 310, OK? So unless you really want to see it. OK, by the way, the primes and so forth that show up in these are essentially random. There's no sizes of these that look more or less like random numbers from a factorization point of view. OK, so this was actually in this kind of thing had been computed before, but somehow you only kind of see how this blows up when you go moderately deep. To do this computation is not so trivial. This doesn't look like such a large number, but what's relevant is it's square, and then you have to do linear algebra and integer linear algebra things of that size. So it's not. And so he computed this because Nicholas Bershorn and I made the following conjecture. And I conjectured. All right, again, this is a general conjecture, but I'll specialize in this case, specialized to here, that this grows at a very specific rate. So if we take the size of this. Now, as you see, this could also have an infinite part. So when I say the size, I just mean the finite part or the torsion part. So what we conjectured is, and I guess maybe I should say that maybe the constant I'll write is, let's say n varies to primes for this to be, this should converge to lambda over 18 pi where lambda is this. So sum of alternating odd squares. OK, so in short, it grows exponentially with the absolute value of this number. But by the way, the agreement with this conjecture is very good, numerically speaking. So what I want to talk about is more or less discuss some of the things that are around this conjecture. So in a way, maybe the conjecture itself is not as interesting as some of the things which show up when you start to try to understand what's going on. So the goal of lecture series is to give some, try to get some understanding of what is happening here and what it's related to. Are there any questions? Yeah? The size of the torsion subgroup? Yes, I'm sorry. I said that, but I should have written it. Yes, it is the size of the torsion subgroup. So it absolutely, it absolutely can be infinite. And by far, the most interesting phenomena here happened in this case when it has an infinite part. So that's why I'll sort of head towards. Yeah, any other questions? The lambda dots are odd squares. Those are odd squares, yeah. Yeah, so this number is like. Is there anything about the rank of the group, the pre-part? No, the rank of the pre-part bathes in an extremely irregular way.  The rank of the group is, a lot of the time is zero. Let's put it that way. It's hard to even make any sensible statement about it because it behaves so irregularly. The previous example has zero free part. What is that? The previous two examples, they have zero free part. No, the first one doesn't have zero free part. These two, the last two. Oh, sorry, this and this? This and the other one. These two have zero free part. That's right, yeah. Yeah, so most of the time what you'll see is a large, simply a large finite group.  Let's go ahead. Any other questions? All right, so here's what I'll talk about. 305, OK. So, question? So the first thing I want to talk about is, I'll describe two heuristics. So this is kind of a plan. Heuristics as to why this should be true. OK, so when I say this, I mean this kind of exponential. This means exponential growth of torsion, exponential growth. OK, so I'll describe two heuristics for it. And they're quite different, and they both have some advantages and some problems. So I guess probably this will already take up the rest of the time today. The second thing I'll talk about is analytic torsion. All right, so let me just tell you roughly now what that's going to be. The statement I make here, I'll say it in the context we've discussed, but it's literally true only in the co-compact version of it. So let me say that the following, the statement I'm about to make is really due to work of Ray and Singer, Chieger, and Mueller. So this comes from a statement that has no number theoretic, is not of number theoretic nature, but here it is. It says that the size of this group, multiplied by a number r, so I'll call this a regulator, is equal to, well, there are different ways of saying it. For now, I mean, I'll state it exactly when we get to it. For now, let me just say you can think of it as a value of the Solberg zeta function. I don't really like thinking about it in this way, so I'll say something different later. But this is, again, in a way that I'll make precise, this is a sort of Donabillion class number formula. So this is a, and again, I have something more precise in mind when I say that, so class number formula. OK, anyway, what this r measures the size, so r is 1 if this is really a finite group. So it measures, in some sense, is finite. So r measures the size of the free part. So r measures the size of the free part. There are several ways you could think of measuring the size of the free part that they're all equivalent of the constants. So I think this is a good way to think of it. And I'll explain more about why later. But once you think of this object, it measures it's kind of a non-Abelian class number. OK, and again, there's something precise to be said. So in some way, it has certain features which are like usual class numbers. And in particular, it has this sort of analog of a class number formula. The third thing I want to talk about is I'll talk about this regulator r. This regulator r is itself extremely interesting. OK, I think for a long time, I somehow ignored it because I was focused too much on this. But it is something very interesting in its own right. And I'll talk in particular about its relation with l values and with heights of elliptic curves. Let me just say l values and heights. All right, so that's the plan of the lectures. And for the rest of today, I want to talk about heuristics as to why we might expect this kind of phenomenon. So I'm going to say two different heuristics as to why you expect this thing to behave this way. So let's say first heuristic. So I'll say the heuristic, which is pretty simple-minded. And then I'll discuss it for a little while, including some ways in which it's not right. So I haven't yet described how when we come to this analytic torsion store, by the way, when I say I'm going to discuss this, I think the main point I'll get to, I'm not going to. Well, maybe I should say. When I talk about two, I will say two and three, I will talk about some, like we have several theorems in the direction of this, which I will discuss later on. But I'm not going to go into the proofs of anything. I'm just going to try to give some sense of the story. So the first heuristic is so we can write down a presentation for gamma naught n. So from now on, when I write gamma naught n, we're always dealing with gamma naught n inside. So here, this is inside SL2 of O, where O is now in imaginary quadratic order. So from now on, I never mean inside SL2z anymore, always inside the imaginary quadratic thing. So we can write down a presentation for it using the geometry of a fundamental domain of the fundamental domain. So in other words, so this gamma naught, this guy now acts on hyperbolic three space. And I will say more about that. And using the geometry of the quotient, you can write down a presentation for it. So I won't go into that, but what does this presentation looks like? So this gives a presentation of the following type. So it gives you a presentation with some generators, x1 up to xg. And these generators have to do with you take the fundamental domain, and you look at fundamental domains adjacent to it. And then you impose some relations. And if you do this, there's more than one way to do this. But if you do this carefully, you see you can get the same number of relations as generators. So this group, it has a presentation. Some generate the same number of relations. And g is some measure of the size of the fundamental domain. To get the size of this fundamental domain is going to be essentially, it's a measure of how small this group is. So it's more or less proportional to the index of this group gamma naught n inside gamma naught 1. So very roughly, that's going to be of the scale n squared. I'm going to this approximate sign. It means really approximate. It doesn't mean n squared n squared log n. It's all the same. I'm sorry, when you say we can do this, you mean we can do this fairly explicitly, right? Yeah, I mean, if we did it naturally, because firstly, there's some issues because it's non-compact. And that naturally gives you fewer relations. But yes, we can do this explicitly. Sorry? Yeah, if you have something written as an explicit complex, you can always write down a presentation for it. Which is not to say it's easy to algorithmically write a program for it, but it's in principle, it is no issue. OK, so now I'll say something more about the nature of these relations in a second, but OK, so it's still there. So if we abelianize this guy, what we get then is we get g copies of z. So this is just like the computation that I said at the start. The group with an explicit presentation. To make it abelian, you just have one copy of z for each generator. And then you quotient by some vectors, which I'll call v1. So the subgroup generated by some vectors, v1 up to vg, where vi is, if you look at this example, it's kind of the exponent vector of this relation ri. So for example, if ri were something like x2, x5 inverse x2, x3, then the corresponding vector vi would be 0 in the 1, 0, 2, 1, 0, minus 1, then 0. OK, so these numbers here, the 2 means that x2 shows up twice. It doesn't matter where the order doesn't matter. Minus 1 is x5 shows up with x1, minus 1, and so on. So OK, now given that this presentation came from geometry, so these relations come from something like edges of the fundamental domain. The generators come from something like faces. The relations come from edges. And you can see these vi's, most of the entries to the vi's of vi will be in either minus 1, 0, or 1. OK, so that's not, it's something about how we can explicitly construct the presentation. OK, so now we have to address the problem of how to bring down that board. And right, so let's for a moment forget about our exact setting, and we can consider the following problem. So if we take, if we choose, let's say, random vi in z to the g with entries, you just choose all the entries at random, we can, so that's some sort of naive model of this, we can ask how large this group is going to be. And the size of this, OK, so you take z, g, and you take the quotient by the group and you enter by these vectors, the size of this is going to be the determinant of the matrix you get whose rows are given by these vectors. OK, the absolute value of that determinant. And if you take a matrix, you just start filling the entries in with, say, 0s, 1s, and minus 1s at random, the size, the determinant you expect, so depending on how exactly you model it, you expect this to be of size exponential in g, maybe g factorial, so I'll just say one expects this, randomly speaking, to be exponential in g, OK? And you could just, if you want, you could just model this yourself, you'll see immediately that it has its, at least to the scale, e to the g or e to the g log g. So this very naive thing is at least compact, so remember I said this quantity g is of size roughly n squared, so this is at least, I mean, such a heuristic is certainly not, makes no precise conjecture of this nature, but at least it suggests to you that, so in other words, you can, in some way, make a random model for this group, gamma naught n, from a geometric point of view, and if you just think about that geometric model, it does suggest that, it does suggest that the abelianization should be large, OK? This model, so if this is, as a first heuristic, this is not bad, there are several problematic things about it, I'll discuss a couple of them, but it sort of gives you the right idea of what's going on already. Are there any questions? Yeah. Well, SL2Z doesn't have such a, SL2Z and subgroups are basically free groups, right? So I didn't say, I mean, specifically I didn't say where this fact of presentation is balanced came from, but for SL2Z, and there's subgroups that are up to a small phenomena, they're free groups, so there are no relations at all. Yeah, this is, another thing about this heuristic is that it's not very robust, like if you're analyzing SL3, it's much harder to figure out from this naive point of view what happens, yeah? Can you make a conjecture from the value of the subgroups about the size of this camera? Yes, and I did, that's the conjecture which is there. So this number comes from thinking about this. No, no, but I was wondering, can you just use some of the, make a conjecture about the central value and then transfer that to a conjecture of camera? That's essentially where this comes from, yeah. Yeah, yeah. And the problem in this story is not the analysis of the Selberg's data function, okay? I have to say exactly what that is, but that's not the issue. The issue, well, there are two issues, but the most interesting one has to do with this R. Can you say why most? These are heuristics, sorry. Is the heuristics scaled by the free part? Sorry. Is the heuristics scaled by the free part? Is there a free part? So, okay, I'll talk about that right now, how the free part affects the heuristic. There's another question. Can you give an idea of why most of the entries should be minus one, zero and one? Yeah, you see, this comes from the geometry. There are different ways to construct it geometrically, but these relations look something like this.  So, as long as not too many things meet at an edge. Okay, so, I forgot what I was gonna say. Oh yeah, okay. So, yeah, let's, we'll talk now. In this whole story, the most interesting thing is what happens when there's a free part as well as this. Okay, so this is a statement of this heuristic. Now, let me say remarks. Let me call this heuristic, yeah. Heuristic A, okay, because there'll also be a heuristic B. So, remarks on heuristic A. I guess I'll probably only have time to make one of these. So, there are two comments I wanna make about it. So, first, what happens if, so what does this say about the case where Gamonaut and AB is infinite? It has probability zero, but nonetheless, it happens, it certainly has probably that's exponentially small in G. For one thing, that's already a point is that it happens much more frequently than this. Okay, so it has in, if I just vary this number N, it happens far more than you would expect from literally this, okay? Right, so, but, right, but then, so let's look at a case. So, if this, suppose this group were actually infinite, we're infinite, then, right, then this matrix, this happens when this matrix is, whose entries are these VI's, this happens when this is singular, okay? It has rank, rank at most, well, it has rank less than G. So, let's suppose just for sake of discussion, now that it has rank G minus one. Let's say it's rank is G minus one. So, firstly, this, now where's the equality? Certainly, this, right, so there's a modification of, so let's say a modified version of star. Ah, what do I call A, okay. So, the following modified version of star still holds. Okay, in what I'm about to write, let me assume that A is symmetric, okay? Just because there is a reason why this is a sensible thing to assume, but it's not, mainly it just makes the formula a little bit simpler. So, what you still have is the size of the torsion part of this group, okay? So, now we just take the finite part and now we multiply it by something, which I'll call R because it's going to be similar to the regulator that will occur later. And this is, I'll call it debt prime of A and this debt prime of A is the product of the non-zero eigenvalues, okay? So, the determinant is zero, but one still has something like this and you still expect the product of non-zero eigenvalues to be pretty large, okay? One expects it still to be of exponential size. I have to say what R is. So, here, this R is given by, I'll write, so it's given by the sum of xi squared where x one up to xg is a generator for, is an integral generator. So, let me say a primitive, meaning all the coordinates are relatively prime, primitive integral generator for the kernel of A acting on z to the g, okay? It has rank, it's an integer matrix, it has rank g minus one, so it has a kernel and so you can pick a generator for that kernel and this, so this R measures the size of that generator, okay? So, another way to think of this, so another way to think of R is that it measures the size, it measures the size of a homomorphism from this group to the integer from a homomorphism from z to the g quotient by this v one up to vg. So, this is no longer a finite group, so it has a homomorphism to z and explicitly this homomorphism sends m, let's say it sends the m one up to mg is sent to sum of mi xi, i equals one to g. So, this sum of xi squared kind of measures the size if you actually try to write down in explicit coordinates, a homomorphism to z, the R measures how large that is. So, now you have this problem which is why should, so for gamma naught and ab, for this to be large, R should be small, okay? So, let me say this naive thing above will have a precise, you can think of it as a kind of crude version of this analytic torsion formula, which is literally right. But anyway, the fact remains is you have a competition between two things. One is the size of the finite part of this group and one is this R, okay? And this, for example, this happens in the context of class groups. If you take the class group of a imaginary quadratic field, the size of the class group times the size of the regulator is something you can compute explicitly with L functions. But what happens is in the real quadratic case, the R always wins almost for typical real quadratic field, the regulator is very large and the class group is very small, okay? So, this does not happen in the setting, okay? So for some, at least from an experimental point of view, most of the time that you see it, the R does not appear to be that large. And so, this was something, I mean, trying to understand this was one of my, some of it motivated me for a long time, understanding why it should always split up, so to speak, in favor of this, the finite group part. Okay, so I think I'll stop there and next time I'll start off, there's a second remark I have on this heuristic where I'll explain a way in which it's definitely wrong just to, like, this heuristic is clearly crude and I'll point to at least one way that it gives you wrong answers, so, thanks. Now there's a third question, sir, come on. Sure, so what's the heuristic part in the few signs? And can you prove that the size of the fundamental domain is very strategic or not? Oh, sorry, you're saying which, what part of this is heuristic? Yeah, the part of this is heuristic is that you have this, you have no actual idea that this determinant is as large, so you really could write down a presentation of this rough size, but that determinant might be one for all you know. Is there, say, any connection to, say, the class group of binary quadratic formula? No, so at the start of next time I'll, in particular, explain what I mean by non-Abelian class formula. Yeah, so there is a connection. So next time we'll talk about more what this has to do with algebraic number theory. I'm sorry? The number of cuffs is the class number, but this measures something else. The fact that you're putting the sum of squares here is again under the assumption that you have no random denominator. No, I think this is an, okay, so, I might have misremembered this, but I think this is more or less, this is an exact equality. Just a fact about quotients of. Would you expect a volume, except? This is, you can think of it as the volume of the kernel, if you want. Yeah, yeah, because, or the square of a volume. It was this confusion about squares that I was. Square, yeah. By the way, you might explain why you're at n square. Oh. In the main formula. No, Laurent, I said it makes a good point. Sorry, I should have explained this a little bit more. For SL2z, this index would be essentially of size n, but because it's over an imaginary quadratic field, it's the norm of the ideal generated by n, which is n squared. This is relevant. Can you realize there's something like, for a integer, I plot order of some division algebra, the same kind of thing? Yeah, as I said, the conjectures that I'll talk about, the conjectures are completely general. If you look at my paper at Bergeron, they are formulated for any group, isn't it? Yeah, yeah. What is not, the methods, okay, so later on I'll point to things that are more specific, but the conjecture that I wrote, for example, is completely general. What is the difference in, so if I understand, non-Abelian, it's with respect to the class number for imaginary quadratic field, which is A-B-R. Which is torus is A-B-R. It's there, non-A-B-R related to the class number, formula, computation of Tanaka number. No, no, I'll explain exactly what I mean. It has a precise meaning, which is not, but I'll say it next time. And so, you tell me, so the L value is at one. Yes. Oh, sorry, the Selberg? Point three and one point three. The L values at the center. At one half, so you could vanish. So, when I say that, no, this is something like a, more like a, it's up to a rational number and L value, which the rational number can be zero, but maybe it's better to, it's a height, or so it seems.


Welcome to All Hands Update. Here's what's happening around your Navy. Sailors and Marines from the Boxer Amphibious Ready Group return to their home port of San Diego after completing their seven-month deployment. The crew had the opportunity to interact with the Republic of Korea counterparts and visit 13 different ports, strengthening relationships with partner nations. USS Boakleys participating in the multinational exercise Cutlass Fury 2016 off the coast of Halifax, Nova Scotia. The exercise focuses on ships from Canada, France, Spain, United Kingdom and the U.S., learning to work cooperatively to develop skills necessary in today's dynamic maritime environment. DoD officials confirmed that a U.S. air strike from the Dwight D. Eisenhower strike group killed ISIL leader Abu Muhammad al-Adnani in Syria. This is one of a series of successful strikes, hindering ISIL's finances and military planning, increasing difficulty for them to operate. You can read more about what's happening around your Navy at Navy dial-a-mill.


Hello people of the internet, my name is Johnny and welcome to what should be Five Nights at Chuck E. Cheese's episode I want to say four Four Yeah, four or five whichever one it is. Anyways when I say should be Five Nights at Chuck E. Cheese's I'm basically saying what you're gonna be watching is Five Nights at Chuck E. Cheese's, but at the same time isn't. You see when I recorded this episode, I actually didn't know it was gonna be an episode So I was just playing, practicing, seeing if I could get past night four with spoiler alert I do and then I check out night five and see what happens, but Me being the absolute idiot I am, I didn't actually switch over my OBS scenes to show Five Nights at Chuck E. Cheese's. So what you're gonna be seeing is just my face cam Now I'm only gonna leave in the part where I beat night four.  Whatever. So you're not gonna be You're not gonna see the actual Footage of me completing it, but I do have proof that I've completed it. If you remember There was no MMF applications for Five Nights at Chuck E. Cheese's meaning you cannot skip ahead nights The only way to pass nights is to beat them So what I'm gonna do is right now. I'm gonna show you the title screen and Then it'll say continue night five and then I'll show you the footage of me beating night four And then we'll return to me right now and I'll do the outro sounds good Hopefully because that's the only way this video can go. I don't know how loud it is But just in case it's so loud you can't hear me. I'm gonna yell So I'm gonna hit continue and it's gonna say night five. I Told you All right, you should be able to hear me properly now, but just in case I'm gonna talk slightly louder I Don't know why this game is so loud, but I don't care So basically on night, I had one go of night five and I'm gonna show you what happened. It happens, right?  Why is that not draining power? That's kind of strange I Wonder is that technically closed because if so I might be able to beat this It's technically closed But why is that cheated the system?  Are you just making who the who are you? Whoa? Whoa? Whoa? Wait? Whoa? I found a glitch I found a glitch I Found a glitch So if you if you that's a really useful glitch all right Anybody watching this right now if you are stuck on hard nights Just do this glitch the vents technically closed Who are you is that Billy Bob? Wait, so I get killed by Billy Bob, but I get tricky to use death screen a little bit confused All right, so I Could I think now that I know that glitch? I'm gonna end up returning to this game So I'm gonna quickly show you guys my night before footage and then we'll end it and then in the next episode We're gonna use this glitch to beat night five, but here is night for I should have opened the vent I should have opened the vent I What Come on Please please please please That was like my third try. Oh boys So that was me beating night four and showing off what sometimes happens on night five or five nets at Chuckie Jesus So thank you for so much watching again.  Goodbye


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