 What I was going to say, I'm going to explain a little more. I wanted to explain the remaining terms in this analytic torsion formula. And then, so far, I've discussed what we expect and some heuristics for it. And I'll say, I'll discuss a bit what we can prove. I just wanted to make a little remark before that, which in response to a question that Paul Nelson asked last time, which is that I've been talking about imaginary quadratic fields all the time. There is a kind of parallel phenomena, which you can see for just classical modular forms. So that's the kind of thing we've been discussing. So the sort of thing we've been doing is you take this group and you take it to billionization. And what you find is it has typically a small free part, but it can have very large torsion part. And there's a similar phenomena for SL2, for just classical modular forms. But it occurs somewhere which maybe one doesn't think about so much, which is for weight 1 forms. So if you look at S1 of n, that is the space of weight 1 forms. Now this space, the weight 1 forms are in some ways the most interesting weight because many things one think of don't work here. As the simplest example is there's no dimension formula. The dimension is typically very small and it varies irregularly with n. We have some understanding of it, but anyway it's very irregular. But it turns out this has a very similar, you can interpret this as part of the classical space of weight 1 forms. You can interpret as, so you can make an integral version, I'll call it S1 nz of this space, so that the classical space is just what happens when you take the free part. Another way to say it is it's possible for a weight 1 form, it's possible to have a weight 1 form mod p, which doesn't lift a characteristic 0. And except for very relatively trivial thing instances, that phenomenon doesn't happen in higher weight. In higher weight, once you've worked out what the right notion of a mod 7 holomorphic form is, it always lifts the characteristic 0. But here this can really fail. And this was, but this story is much, I would say much less, so the first evidence of this was given some mod 2 classes that don't lift were found by Mestra. And then some more were found by Buzzard, but these were still relatively sporadic. And then on the basis of the kind of analogy is what I was saying, I had a PhD student who I asked to compute this integrally for many n, so George Schaefer. And he found that, in fact, this really does behave like this. That is to say, it has a very large, typically, it has a very large torsion component. So the classical space of weight 1 forms is like this small tip of an iceberg, but underneath there's this vast number of n very large primes where you have a mod p weight 1 form that doesn't lift the characteristic 0. This story is less well understood. I don't have. So in principle, this is an analog of everything I'm saying, but Schaefer's data is a little bit, so it will be nice to have a conjecture here that this is growing at some specific exponential rate and so on. There are some mysteries in this data, which I don't understand. But anyway, so let me just say for now this is less understood and it would be worthwhile to develop it. Yeah, so usually we don't think of weight 1 forms, but somehow as time passes, I sort of see them as the most. They have many kind of boundary behaviors, which are very interesting. OK, so here's what I'm going to say. So what was our setting actually? So gamma 0n, so now O is this imaginary quadratic order, so something like z squared minus d. And we take this group inside SL2O. Sorry, are there any questions about this story? So yeah, Kevin Buzzard has a nice paper about it. My student's thesis, Schaefer's thesis is online. It's interesting also from just a computational point of view that it's not trivial even to compute these things. OK, we go here and remember we have this subgroup. And again, I'm going to say that although I'm stating this setting, many of the results are proved only in the co-compact setting, although they should certainly extend here. Yes? What do you mean by computing these fun things? Yeah, so for example, I give you some n, like 500. And I ask you to give me a basis for the space of weight 1 forms over the complex numbers or mod 7. And even over the complex numbers, it's not a trivial question. Of course, it's clearly computable, but it's better than mass forms. But I mean, to say one word, you can embed this space in a higher weight space by multiplication, but then you have to figure out what its image is. That is, it's not clear that when you divide, you don't get a pole. OK, so back to, right. So what is the conjecture we've been discussing is that the size of the abelianization of this and the correct scaling is the volume goes to 1 over 6 pi. So let me state again this, all right. So as far as, so I spent quite a while giving you heuristic motivations for that. Now, as far as analysis goes, the main way we have to understand this is this story of analytic torsion. And I'll write the names again. This was sort of discovered by Ray and Singer. And the proofs were given by Chigar and Mueller. And what they say is, well, we can access, at least in part, this thing analytically. So the size of this, again, right. So this statement is valid in the co-compact case. If I have time, I'll say some. I mean, we now know a fair amount in this case, although short of any complete formula. This, multiplied by a regulator divided by the hyperbolic volume, is equal to some ratio of Laplacian determinants. So as I said, this right-hand side is related to a Solberg-Zeta function. What I want to do right now, well, we've discussed this action hyperbolic space. So you know what this is. I'll tell you what this is. And then I'll spend some time saying what we can prove towards this. And then we'll come back to the extent there's time left. We'll discuss this right-hand side a bit more. But to say, for example, this delta 0 is just a usual Laplacian operator on functions. The 1 is means on forms or in different weight. So it has some eigenvalues. So 0. So in the co-compact case, it has no continuous spectrum. And this determinant is some way of making sense of the product of the eigenvalues. So we're going to talk about that just by example after at the end of the lecture, because as I said, there's some, you can also think of this in terms of Solberg-Zeta function. I actually prefer thinking about it this way, because this way, it's sort of clearer what the effect of the eigenvalue spectrum is on it. So that's why I say it this way. OK, what is R? This is what I call the regulator. And what it's meant to measure. So R is 1, as I said earlier, when the abelianization is finite. So in general, what it's meant to measure is it measures the size in some way of the free part. All right, so just as in the case of number fields, in general, it's some expression with determinants. So to keep life simple, I'll just talk about the case where it has rank 1. So I don't have to write down determinants. So let's say that suppose that this gamma naught and ab was actually a copy of the integers plus a finite. OK, so in general, the story will be similar to what I say, but with determinants. So what R is meant to measure more precisely. So in this setting here, you have, so you can make there is going to be the way I set it up, there's a unique up to sign, surjective homomorphism like this. So you can just project this onto z. And that's the only up to sign, it's the only surjection from gamma naught n to z. And what R is meant to do, it's meant to measure that in some way it measures the size of this homomorphism. All right, so I will give you the precise, I mean, it's a precise definition, but I want to just describe. So this quantity is interesting. So let me describe a couple of different ways that you could try to measure the size, and it'll turn out all these ways are from the analytic point of view the same. That is, they're all essentially the same order of magnitude. From the algebraic point of view, the specific definition of this is interesting, because its exact definition is related to other arithmetically interesting quantities. But analytically, so some ways we could measure the size. So the most naive way is you could just say, well, this is a homomorphism from this group to z, and I'll just let's look at a bunch of small elements here, and I'll just take the largest value it takes. So there's a group theory way where you just take, let's say, the largest, all right, so you pick some bound for how big your entries should be. So maybe you go up to, so a reasonable thing to do would be to go up to some power of n. And you just take the largest, and you see how big it gets. But by the way, because you're multiplying matrices, if you made this n to the 20, you don't really expect it to get that much bigger, because when you multiply two matrices, the entries multiply. So the matrices get larger without fee getting. Roughly, you expect this to just grow logarithmically. So exactly what you put here is not so important. By the way, what we are concerned about for our applications or what we're concerned about is, will be does this r grow exponentially with n or not? OK, so if we're just strictly thinking about this problem, we need to understand it only at a very core scale. OK, there's a topological way of thinking about it. So all right, so I'll say, again, a fact from topology, which is true exactly as stated only in the compact case, and otherwise you'd have to modify it slightly. But the space of homomorphisms from this group to the integers is isomorphic to the second homology. So I said I'll try not to talk too much about homology, but this will actually be quite explicit, so I'll explain. So this homomorphism fee, you can represent it by something in the second homology of this manifold. So I'll say what the map is. So suppose that we have S, which is a surface inside this three dimensional manifold. How do we make, I want to say from it, how we get a homomorphism like this? OK, so I'm just going to make this explicit. So this S went overboard here. So what you do is you intersect S. You send S to, so the homomorphism, so send gamma in gamma 0n to the number of intersection points, so of S with the path from w to gamma w. So for any. OK, so in other words, this path from w to gamma w, it's really a loop here, and you see how many times does it intersect this S, counted with a suitable multiplicity. OK, now given that what you can do, so over here, by the way, one reason I'm mentioning this, I know it's a little bit of a feel of always, but is that it's actually very useful in estimating R. OK, so I'll say something about that if we have time. Right, so the question of measuring the size of this, it translates to the question of measuring size there. But over here, there's a famous norm due to Gromov and Thurston, which is that you can just measure the minimal complexity, or the minimal genus, of such a surface S. OK, so you look at all surfaces S that induce this, and you see how, in other words, this right down the simplest possible surface I can hear that induces it, and you measure how complicated that surface is. OK, now I'll finally come to the definition which is the correct definition for this, which is analytic. I'm sorry, are there any questions so far? Can you please explain what is this way, Matt, for the last one? Sorry? For the homomorphism group to homology group? Yeah, so it's sort of clear to write just this way. So it's a fact, not that this is an isomorphism, but it's not. OK, so analysis. And for analysis, let me just quick first remind you of something that we said at the very start of the lecture series for SL2Z. OK, so at the very start, something I said is for subgroups of SL2Z, we had this pairing between gamma 0n, abelianized, and the space of weight 2 forms of level n. And so how this pairing went was something like if you had some gamma here and an f here, you just send it to the integral of f of z dz from w to gamma w. OK, so there's a similar pairing here. And I'll just say what it is. And what we're going to do is basically transfer this phi over to something here, and then we can measure its L2 norm. OK, so I'll write this up. But so that's what we're going to do. Phi is represented by a harmonic differential form, and we can take its L2 norm. So there's a similar story to this. So I'll use the same notation to be kind of suggestive. So let S2 of n be this. So over here, you had weight 2 holomorphic forms, which you can think of as holomorphic differential forms. In our setting, there's no holomorphic, but we can still say harmonic. So harmonic differential forms on gamma 0n mod H3. So you can think of these as being mass forms of a specific certain eigenvalue and weight. And then there's an exact, this story goes over word for word. You can pair these by sending it to integral exactly the same. So same formula. OK, so there's a pairing from this. In both cases, the pairing from this to the complex numbers. All right, now having said that, so now what happens is there'll be a unique, there's a unique omega such that phi of gamma is given by pairing with that unique omega. OK, so what we've done in other words is we've taken this discrete object, which is a homomorphism, and we've kind of spread it out over the manifold to get this representative omega. And we define our regulator to be the L2 in a product. So sorry, this symbol is used in two different ways, but this is the L2 norm on the manifold. So it's like a Peterson norm. This object is very delicate. I'm going to be saying more about its algebraic properties in the next lecture. It's not so hard to try to see what it is up to rational numbers. But if you're interested in the size of something, it's not so great if you know it only up to a rational number. And that's very subtle. So next time I'll also try to mention some problems in the next lecture also. But the point is I've spent the time saying these three things because they're actually all essentially, from the size point of view, they're all essentially the same. That is so these all have essentially the same size. So the most useful of these is the, and this is not a hard fact, but maybe it's nice to have all these. The most useful is the relationship between B and C. And if you want an exact statement, you can see my paper. I'll refer to it again. So I'll say this is my paper with Nicholas Bergeron and Haluk Sengun. And there. It's certainly not a hard fact, but we just write down exactly what I mean when I say that these have. And we don't write the relationship with A, but I don't think it's very hard to check. Unfortunately, A is almost useless. Somehow, in the sense that computationally, this phi is easy to compute. But theoretically, it's very difficult to get a hand on. Theoretically, what's easy to get a hand on is this omega because it comes from automorphic forms or physically constructing a surface, which you could also do by geometry. All right, so that is what r is. And next time, we're going to talk more about r. And especially, how is it related to L functions and so forth? So it's very. All right, so now let me, I still have to say something about this, but let's now pause and I'll tell you sort of what we know how to prove in the direction of this conjecture. So the tool, the only real tool, we have to prove things like this is this formula. But even here, there are still many obstructions, which I'm going to discuss. Are there any questions? So the definition of phi really comes from part b. Can I think of that? So I'm sorry, I might not be understanding your question, but I'm assuming I'm in a setting where this happens. OK, so there's a unique such thing up to sign. OK, and what I'm doing in these different parts is I'm saying, well, here are different ways to measure the size of phi. You can do it directly, or you can translate phi to being a surface, or you can translate phi to being a differential form. And this one is kind of the maybe the most canonical one, but they're all size-wise similar. By the way, I should say, yeah, this thing, you can also sort of talk about similar things just for SL2z. You can ask how big are these homomorphisms. And there's a nice paper of Goldfeld where he discusses this question. But in that case, you have a much more sort of concrete way of understanding it because you have these modular uniformization. So our problem here, our basic problem is that we have removed ourself from the world where you have kind of real algebraic varieties to play with, because everything is purely analytic. OK, so now let me spend some time telling you what can we prove towards the conjecture. Are there any other questions? All right, so let's look at this formula for a moment. So let's call this formula star. Let me sort of list the difficulties you have. And so one can try to apply star. So one can analyze, let's say, the analytical behavior of the right-hand side. So this question, all right, I haven't precisely defined this for you, but as the level goes to infinity, this is related to something Valentin was saying, you understand what the distribution of eigenvalues converges to very precisely. And correspondingly, it's fairly easy to understand the asymptotics of this, except for the following painful issue. Maybe you have an extremely small eigenvalue. Maybe this operator has an eigenvalue very, very close to 0. And then this becomes small and there's nothing you can do about it. So if there are no small eigenvalues, or maybe a better way of saying it is few small eigenvalues for delta 1, OK? So the exact condition, so let's call this condition no f for few small eigenvalues. So if you go through this, you find everything is fine. You understand exactly what's going on, before you have to drag out the small eigenvalues by hand. And when I was thinking about this, and now I still know absolutely 0 about this, absolutely nothing. But the nature of the statement that you need is the product. So we would need to know something like this. The sum over all eigenvalues lambda, so these are eigenvalues of this Laplacian operator here, basically what you need to rule out is that there are a whole lot that are exponentially small in the volume, which is absurd. So it's like sort of clearly will never happen. But so let's say I'm going to call v the volume of your thing we're looking at. So let's just say that we're always assuming a co-compactness, but I will comment about what we know about the non-co-compact case. But for my initial setup, let's not worry about it. What you need is for any positive delta, you take all the eigenvalues less than volume to the delta and sum their logarithms. So you'll worry that there's a, and you scale it by v, this should go to 0. So if you think about what the limiting distribution is, you see this is very, very unlikely to fail. But it seems extremely difficult to prove. So as a toy case of the kind of issue here, you can see, ask for SL2Z, do you have any understanding of how close the eigenvalue of a mass form can be to a quarter without actually being at a quarter? So if you think about a moment, this clearly should be true, but I see no way of, anyway. So this is, that's one thing. So the second thing is that you need to know that R, I'll call this R for regulator. You need to know that the regulator is also growing at a sub-exponential rate. By the way, the only problem, yeah, it's very easy to bound R from below. It's not a problem. And anyway, if R was small, it would only increase this. But it's, OK. So the point is that given, about, so I know Andrea Strumbergs, and I talked to him about this many years ago, as probably, but certainly looking at that, the spectrum near zero didn't show any peculiar behavior. That is, it behaves as you would expect from the limiting point of view, and then certainly you would, yeah. I would love to see even any much weaker, I basically know zero about this. This is, the situation here is a lot better, I'll say about in a second. But so what do we know? Given F and R, so here's a fact. Given F and R, we get the conjecture. Now, all right, this is, so I'm going to talk, as I said, I know nothing about this. We know quite a bit about this. But still, actually what got Berger and I started on this was the observation that, in fact, in this closely related context, we can completely sidestep both of these at the same time, and then we get unconditional results. So let me say a bit about, so I have to change the problem slightly, and then we'll really get the analog of this conjecture, and then we'll come back to what more we can say here. So, Berger and I, this is in a paper, so, OK. So there's another paper of just with Nicholas Berger and I from a few years ago, but not with Sengun, OK. So you see, can avoid both problems by changing the weight. So now, here I, right, so in this, in this board I'll say, in a sense, the most, the really unconditional things we know. So when I say changing the weight, let me go back for one moment again to SL2Z, OK. So I'm not going to, I'll state the result but without defining all the terms, because, again, I don't want to go into homology. But suppose for SL2Z, as I said, there's a pairing between this and weight two forms. Now suppose you want to know, as Henrik was asking about in the first lecture, well, what's the analog for higher weight? There's no analog for weight one, OK. Weight one, there's really nothing. But this works perfectly well for higher weight. You just have to be willing to. So for higher weight, the analogous statement is rather than this, you have to take the homology of this, the same group with coefficients in some module. So this is a module depending on k, OK. So all right, and the same thing is true here. The analog of changing the weight is looking at homology with coefficients in a different module. And so what we show is that, so we prove the analog of the conjecture. So for in this context, so in the context of a co-compact group in SL2C, for what we call, for what we call strongly acyclic weights. And I won't say what this is, but the point is that, in fact, so the weights here, in this case, the weights are parameterized by a pair of integers and strongly acyclic means that the two integers are different. So it's actually a generic condition, not a special one, OK. When I say the analog of the conjecture, there's a different constant, but otherwise the conjecture is exactly the same, OK. So this is very satisfactory in the sense that, from the point of view, say, of the Langlund's program, there's no one weight as as good as another. So it certainly gives you pretty strong evidence for the underlying phenomenon. But in fact, so we did this for all groups. And when you do this for all groups, something very pretty happens, which I will say, OK. So for the exact definition of this, the exact statement, you can see our paper, but it's that type of statement with a different constant and with abelianization replaced by homology. One thing I wanted to say is that at almost the same time as our paper, Müller proved a similar statement, but essentially the same time, but as the weight goes to infinity rather than the level, OK. So it's sort of strange that we were, yeah. So the case, so the proof is by following the same strategy. Yes. So OK. So the strongly acyclic case, I mean, the point is, and it's a little bit surprising, is that these strongly acyclic things exist in the first place. But given that they exist, what happens? The eigenvalues are all forced to be away from 0 and r is 1. Yeah. So it's a complete surprise in a way that they exist because they don't exist for more varieties. So it's not, OK. Anyway, so this is interesting. I mean, as far as I know, no one is really thinking about this before. And then Müller somehow was thinking about it exactly the same time as we were. Let me say something about general groups because there is something to be learned from there. So we prove a similar statement. So Müller means Nicola, Berger, and I. With SL2C replaced by any group, so similar statement means exponential growth of some similar object, by any group G satisfying, OK. Some condition, which I'll, the rank of G minus the rank of a maximal compact subgroup is 1. And gamma ab was replaced by some higher homology group, OK. So for some higher Q. So I'm not going to larger Q. Now, this condition here is a strange condition. Rank G minus rank K is 1, OK. So for example, it includes the groups SL3R, SL4R, maybe SL, OK, SLPQ with P and Q odd, but not SLN for n bigger than 5. Bigger than or equal to 5. SLNR. So these are groups. So if you don't know what to do, it doesn't matter. Almost, there's only maybe one other example, if I remember rightly. But OK, so you go through this method. The point is there's all sorts of computations to be done here depending on the group. And you find here disappointment that you get a result of this nature only in this case. But here something wonderful happens is that according to heuristic B, so you can work through this heuristic with the non-Abelian Cohen-Lenzra heuristics, which I mentioned earlier, and you find this wonderful thing that these should be the only groups, only G, with exponential growth of torsion homology. OK, so here's where this heuristic B by itself is certainly something you should be somewhat skeptical about. But it is a very nice thing that it meshes exactly with what you can prove. So the proof for a reason that I won't go into, but you prove something here and then otherwise you get nothing and you're left wondering what happened. But fortunately, heuristic says is that the cases you can prove are really, I think, the only cases where you have this. So in other words, we conjecture that we can't prove, thus, that there is no exponential growth in any other case in the cases where rank G minus rank K is not 1. I think we can't prove it's maybe not a strong. Again, this is the sort of thing about which I know is 0. That is, I have no way besides completely trivial ways of giving you upper bounds, interesting upper bounds for the size of the sort of things we're discussing. So it's OK. So all this is under this switch that you switch to a different weight where you can avoid these issues. And now I'll come back to what kind of things we can say about, especially the second thing. So now we'll go back to the original question, trivial weight corresponding to abelianizing. Oh, what about in the non-co-compact case? So now, at least for SL2C and the non-co-compact case, there's quite a bit of work. And maybe there's work of Jonathan Faff and Jean Rambeau, which give now similar results in the non-compact case. By the way, the non-compact case here is really painful. It's much worse than just the horror of the trace form. I mean, when you try to regularize this and you have a continuous spectrum, it's really, really awful. This strongly acyclic thing makes life easier. But anyway, I just want to say it's not, I put it like a small, I think the number of pages involved in this one line is probably greater than the number of all the other papers mentioned put together. So it's a substantial problem. It's an even more substantial problem to do the same thing in higher rank. And this is really not just a technical problem. I've thought about this for SL3 and I really, in fact. I mean, in rank one, at least, you can work hard and sort of see how to do it. But I actually don't even see how to do this for SL3. All right, so now, OK. So let's talk a little bit about what we can, let's go back to the original problem, what we can say about this. Are there any questions? Yeah? Is there absolutely anything known about F? Is there anything like trivial resolved or is it just? I personally know, oh, trivial resolved. One could give a bound for this. I'm not sure how big. I mean, it's certainly not sure it goes to 0, but you could give some bound. The point is, in this case, I mean, for example, in some cases, you don't even know how to give any bound on how close an eigenvalue could be to 0. Here, there's some a priori bound you can give over how close an eigenvalue could be to 0. Because somehow, you can switch the eigenvalue problem for a discrete problem with integer matrices and you have a lower bound on how close the eigenvalue and integer matrix could be to 0. So that would give you some bound, but it's totally useless. OK, so yeah, any other questions? All right, so nothing about F, but R is better. And so today, let me say a bit about the analysis, what we know about some things about the size of R. And then tomorrow, I'll talk about the kind of, or whenever the next lecture, I'll talk about the little more about it's an algebraically interesting object. OK, so what we want to show, so we'd like to show this rewrite it over here, log R over. So R, the regulator should be bounded by the exponential of epsilon times the volume. So if this fails, if you look at what am I saying, I'm saying that if this were to fail, it means you have this homomorphism to z, which takes exponentially large values on small matrices, which seems absurd. So it might seem that this is something which, again, this might seem like something which, so to speak, you should expect. But that's not at all clear to me. So I don't know the exact current status, but I'll say that Brock and Dunfield have at least have produced maybe not counter examples to this exactly, but closely related counter examples for non-arithmetic gamma. So they don't produce a counter example in the exact setting where, say, Bursar and I make a conjecture, but they produce something very closely related. So it's completely possible that this is always true. It really has some special feature of arithmetic and not a general group theoretic or geometric property. So let's study it. I'm going to say a bit more about what this, so maybe I'll just say in words, one case where we understand R well, one relatively easy case, so easy from the point of view of having a good upper bound on R, is when all the cosmology is when all of this space, which I called all of the, what did I call it? Well, let me just say when all of gamma naught n ab comes from base change, comes from q via base change. So I'll say via a lift in the sense of Valentin's talk. When I say this, I just mean the characteristic 0 part. So let's say tensor would C. All right. So I'm not going to, so without going into details, suppose you start off with m, which is an integer, and then you can also think of m as being in O. So you can sort of form the subgroup gamma naught m in SL2z, and you can form the corresponding subgroup gamma naught m in the big in SL2O. It turns out that there is a lift that goes from the sort of free part of this, the free part of lift, which is called base change. OK, so this is one way of producing examples where, in fact, it's a fairly common, at least not all uncommon, that all of the free part of this comes from base change in this way. Because it's sort of not here, unlike SL2z, this free part is somehow usually small except if explained in some way. So in that case, we can give the correct upper bounds on the regulator. And that uses this topological interpretation, part B. So here what you do is you produce enough totally geodesic surfaces and use part B and use part B. Now, there's a reason why we wanted to study this specific case is that for numerics from the torsion, this case actually looked pretty bad. So we were a little bit worried that this would be, this was a case that had some reason to, might have some reason to believe this would fail. But in fact, we can give a rather good bound. The actual bound we get in this case is something like, so we never wrote down that we just proved it's sub-exponential. But I think what we actually prove is something that it's exponential in square root of the volume, something like this. Can you give a criterion on M which you can check if it's valid for infinity many times? Oh, no, I can't. But I certainly expect, so this case and the other case I'll mention, I expect them to be valid infinitely many. In the numerics we did, they were both valid, say, around half of the time. Yeah, up to the, of course, some of these things change for very large values of M. But when I say this happens a lot, I mean I don't remember the numbers we wrote in the paper, but maybe half of the time in some range. OK, so a more interesting case is when, well, OK, let me just say next time. So I did want to say at least a few words about this Laplacian determinant. So all of next time will be devoted to studying this R in the case where you have a single form that comes from an elliptic curve that's not base change. So related to an elliptic curve over O. And what we'll see is that this conjecture is then becomes related to questions of difantine geometry in a sort of very surprising way. All right, so in the last few minutes, I'm just going to go back to the analytic torsion formula and talk and just give you a quick, very brief flavor of what these Laplacian determinants look like. Are there any questions? So I was planning to, I would have liked to spend more time on this Laplacian determinant thing, but it's probably the least number theoretic part of this. So I'll just, so how do we define this sort of determinant of delta? I'm going to, as I said, I prefer this way rather than the Solberg zeta function because this way, it's easier to understand this effect of small eigenvalues. OK, you can really see if there's a small eigenvalue, it causes a problem. But the definition actually makes sense for the Laplacian operator for any compact Riemannian manifold. And again, this definition is due to Ray and Singer. They introduced it for the purpose of stating their formula. And so what you do, so let you have the eigenvalues, be the eigenvalues. And a basic fact true in this context is that the function sum, so it's called z of s, sum over i bigger than or equal to 1 lambda i to the minus s. So Weiler's law tells you that this is convergent when the real part of s is large. And it actually has an analytic continuation to the complex plane, an analytic, meromorphic. Sorry, meromorphic. And you understand exactly where its poles are and so on. So this result is obtained by, it's not very hard. It is obtained by taking the Mellon transform of things we know about the heat kernel. But anyway, it's true for any Riemannian manifold. Now, formally, so ignoring issues of, well, OK. So in the region of convergence, if you differentiate this with respect to s, what you get is negative log lambda i over lambda i to the s. So formally, if you put s equals 0, so I'll put this in quotes, log of lambda i is equal to, in quotes, the negative of the derivative of this is 0 sum, if you just formally put s is 0. And this is what we take as our definition. So we define the determinant of the solution to be the exponential of minus z prime of 0. That's the definition. And it works for any compact Riemannian manifold. So let me just do, to give you a flavor, I'll just do a single example and then we'll stop. Are there any questions about the definition? Do you think you have a functional equation? No, no, that's not. Yeah, this has no sort of, none of the other algebraics. It just has a meromorphic continuation. And it's a very, it's a soft fact. All right, so let's do the example when M is a flat torus, so R2 mod z2 with the usual metric. So it looks like this, OK, with periodic boundary conditions. And the Laplacian operator is minus dxx minus dyy. And so this example, it's kind of a special example, but it, you know, we'll learn a little bit by doing it. The eigenfunctions are e to the 2 pi i nx, e to the 2 pi imy. And the eigenvalue is 2 pi squared n squared plus m squared, OK? So in other words, this function, the n and m are integers. So the z of s is, all right, I'm going to forget it. I'm going to lose the 2 pi. And I decide I almost certainly screw something up if I keep it in. So this function, z of s looks like this. So in other words, it's a value of an Eisenstein series. This is an Eisenstein series for sl2z evaluated at the point i. OK, so at the point i. All right, so to do this computation, we can actually compute this absolutely explicitly. And so I'll very briefly do it. So more generally, we can look at the, let's do it for the, so what we want, right, is we want to compute the exponential minus z prime at 0, OK? But let's do something more general. So we will compute this, but we'll do it by computing it for all lattices at once. So we look at es, which is the sum of y to the s. This is the standard Eisenstein series for sl2z. So we can recover this, look at, on the upper half plane, what sl2z is the standard Eisenstein series. And so our function we want is the value of this at i. And now, all right, so what we want is to differentiate it at 0, OK? It turns out you can do this basically by pure thought. So this Eisenstein series satisfies two properties. It's an eigenfunction. And its constant term is y to the s plus c of 2s minus 1 over c of 2s, y to the 1 minus s, OK? Where this is the completed zeta function. So now, if we want to work out the derivative d by ds es at s equals 0, let's call this function capital F, you learn something about this from both of these facts. The first thing you learn is that the Laplacian of f is equal to 1. Oh, sorry, this is not actually an equality. OK, it's this plus decaying. So this function differentiated at 0, it gets something of Laplacian 1. And the second thing says that f is equal to y, no, differentiate it, log y minus pi y over 3 plus decaying terms, OK? So if you differentiate this, and you evaluate the zeta, the pi over 3 comes from a zeta of 2. So what this says, all right, so I claim that in fact this function f is the logarithm of y times delta to the 1 over something, 6, OK, where delta is this modular form of weight 12, OK? And the reason I know this at this point is that the difference of these functions is now a harmonic function, which is going to 0 at infinity, so it has to be 0, OK? So in other words, what we've done, so in the end, if you go back to this x of minus z prime at 0, it's something like this y times delta to the 1, 6th. Maybe I might have missed an inverse. I'm not sure if there should be an inverse or not, but let me put minus 1, OK? So this is certainly a very striking calculation, because it shows disinvariant in this simple case gives you a number theoretically interesting thing. And I think that was, I mean, Ray and Singer did this computation. I think it certainly suggested to them it was a good definition. But actually, the important thing to learn from this computation is that this Laplacian, what did I do? This Laplacian thing, this series itself, I evaluated this at a specific point by looking at it as a function on the whole space, OK? And then I used harmonicity. And the sort of moral of the story is that this Laplacian determinants are very difficult to understand themselves. But you can understand how they vary, OK? You can differentiate them when you change parameters. And just to finish, so one of the reasons that Ray and Singer realized that they were onto a good thing is that the right-hand side of their formula, the right-hand side of this Ray-Singer formula, which involves the determinants, what they did is they said, well, here's something which depends on a metric. And then they changed the metric, and they watched what happens, and they found it to be metric-independent, all right? So it's very hard to understand those directly, but they could check that it doesn't change as you change the metric. So if you have such an invariant of a manifold metric, which is then independent of the metric, it's reasonable to think that it's something purely topological, OK? So that's the one takeaway from this, not the specific example, but that you can understand this Laplacian determinant varies when you make, when you change things. So sorry, I went a little bit over time. Are there any questions? Yes? Just a bit confused about your last formula. So you evaluate, you take the limit as 0, or something? I see my y on the left, and say, of x to the z prime. Sorry, where, where, where? x to the minus z prime of 0. Yeah, there's a y, but there's a y at the beginning. You see that for this? There's a y. Why are you going home? Oh, yeah, I'm sorry. That's right, I should have put, I should specialize this to the point i. That's right, sorry, thank you. So the y should be 1, you're right.