 for an invitation and let me start. So I'm going to speak about the single conjecture in dimension three and now this is going to be revisited and these extensions. Okay. So as I mentioned in the abstract, this is a joint work with Mike Hull, NC State, and Mark Stern at Duke. And I have to say that everything I'm going to present today is an application of a theory which I've been developing with Mark's since 2017. So this is going to be an application. Hopefully interesting to you guys application of a theory developed with Amsterdam starting in 2017. Okay. And this is ongoing. So I would say let's start with the statement of the single conjecture. So what is the single conjecture? So if I take an n dimensional Riemannian manifold and for us, the manifold will always be closed. So compact without boundary and orientable. And I'm going to add a topological assumption, namely that it's going to be as spherical. Then the single conjecture is a vanishing statement for the L2-bit numbers of this manifold. So then single conjecture in the, I think in the 70s that the L2-bit numbers of m are always zero for the index of the bit number different from the middle dimension. Okay. So this is the statement of this conjecture. Now let me try to tell you what. So as spherical means that if you take the universal cover, let's say from m tilde to m, then the total space m tilde is contractible. Okay. So in particular, if you, if you think, for example, of torus, universal cover is at n and that's an example of an spherical manifold. Now I have to give you a definition for the L2-bit numbers. Now here it is. It might be a little bit complicated, but I'm going to translate it in very elementary terms in just one second. So if you're better with me, you will see the light at the end of the tunnel very soon. So what does, what are these L2 invariants called L2-bit numbers? So these are defined as the full Neumann dimension of the gamma module of L2 integrable k-harmonic forms on the universal cover of m. So let me just say this. And here I have to tell you where I consider now the Riemannian universal cover. So I have m g. So by these I simply means that pi is nothing as the topological universal cover and the g tilde is nothing as the pullback of the metric on the base manifold. And what is this Hilbert space of L2-harmonic k-forms on m tilde? Here this Hilbert space is nothing as, so here I'm going to define the, oh and sorry, this index here has to be i, the same as the i, the i-bit number. So what is the space h2 i of m tilde is nothing else that the omegas in the space of i-form on the universal cover such that they are in the kernel of the Hodge-Laplacian. And moreover they have an L2 condition, an L2 growth condition. Namely that if I take their L2 norm, and the L2 norm is simply, maybe I'll express it this way, if you take omega wedge the star of omega then you get something of finite L2 energy. So this Laplacian here is the Laplacian on i-form and so this is going to be d star plus d star d. So in particular this, in order to define this Laplacian you need the metric. Okay? And I have a quick question. So just to confirm, so these are always smooth forms, right? Yes, because by elliptical value diffusion. And then the question is, so does any of this make sense for compact? Oh no, universal cover is never compact. Okay, got it. For a second I thought that the integral was on m. Okay. Yeah, no, it's on m tilde, yes. And then I noticed, and yeah, I noticed, sorry, I noticed that since I'm assuming that the base manifold is a spherical, the universal cover is always not compact. Got it, got it. And the last quick question, you said von Neumann dimension, could you? Yes, I am. Yes, I'm getting there. Okay, got it. So notice that here I have this gamma here. So that gamma there comes from the way these objects are defined, namely that, so you have this contractible universal cover, your base manifold and m can be expressed as a quotient by an infinite group of isometry acting on the universal cover. So, and then you have m tilde. So there is a way, and this was defined by idea, so you have in principle, this inverse space of L2 harmonic id form on the universal cover. This might be a Hilbert space, might be infinite dimensional, but what happens is that you can define this phenomenon dimension where you take this quotient by this very, very big group, and you get always a real number, which I call the L2 beta numbers, and this real number is always finite. Okay, so you can arrange this with some, it will take probably half an hour to go through all the functional analysis, but you can always do that. And this number here, which is this quotient of this potentially infinite Hilbert space, but you're taking a quotient by an infinite group, this thing always exists and is finite and real. Now, so this guy here is the guy acting on these Hilbert spaces. Now, though, it's not terribly important that you go through all of this, because of the following fact, if you look at a single conjecture, they say that these objects, which are complicated to define and they require some functional analysis, they always have to be zero. And now you understand that it's reasonably intuitive to believe that these objects are going to be zero if and only if the Hilbert space is zero itself. And so I could have rephrased a single conjecture by saying, if I take a closer spherical manifold m, m, I pull back the metric to the universal color, I look at the L2 harmonic height form in the grid outside the middle dimension, I have none of them. Okay, so now the fact, which will simplify things for us, is that bi2 of m is actually equal to zero if and only if the space of L2 harmonic height form on the universal cover is a vector space, the empty vector space. Okay, so now you see this is a really easy now statement in geometric analysis. So this is telling you that if you take a closer spherical manifold, you look at the height L2 harmonic forms on the universal cover, they should not exist unless you're in middle dimension. Okay, it's very clean statement, very easy to state. The bad news, maybe before I do that, let me also remark that one can show that these objects, which in principle are defined by using a metric on the base and pulling them back, they're actually homotopy invariant. So remark, the bi2 are actually homotopy invariant. And this is due to Atiya and Dodd-Juke. Okay, so single conjecture, closer spherical, you don't have L2 harmonic height form on the universal cover with a pullback metric unless you are in middle dimension. Are there any questions? I guess what fails here is basically all the time what fails is the L2 condition. Sorry. I guess this is more of a comment. I just want to make sure I'm following. So what fails in this, I guess there are usually a lot of harmonic forms, but none of them are integrable. Yes, for example, on negativical spaces, you can, on the universal cover, for example, the hyperbolic space, you could have bounded harmonic forms, but your bounded harmonic form now is not going to have an integral, which is an L2, because this underlying space is huge. Okay, it has a lot of memory. Okay, so here the harmonic forms, they have to decay at infinity. Okay. Maybe I can also have a quick question. Sure. So, in the compact case, so harmonic forms are, the dimension is the same as the dimension of the romcomology. So I would assume that fails here, but still, do you get some information from the single conjecture about the romcomology? Yes, and we're going to see regular Betty numbers. Yes, we are going to see the connection in a second. Okay, good. Yeah, that's a famous theorem of Luke, and we're actually going to use it in the proof. Okay, but yeah, that's a very good point. Okay, so this, since this is an informal seminar, let me, if there are students in the audience, let me give you a little bit of examples so that I'll convince you that the class of a spherical manifold is very interesting for geometric analysts. So, for example, if I take mg, such that the sectional curvature are all non-positive, then we know that by Hadamard theorem, the universal cover is actually diffeomorphic to Rn. And so, in particular, Mn is going to be a quotient of these Rn by a group of isometries. Okay, so, and then this guy is going to be co-compact inside the isometry group of these Rn equipped with a pullback metric, GTU. Okay, and let me add torsion free, so we want to get a manifold and co-compact. Okay, in particular, hyperbolic space is there. There are more examples which do not come from non-positive sectional curvature. Other examples, for example, if I have an object like this, such that the universal cover is Rn, again, this is contractible. And examples, for example, infranil manifolds or infrasalt manifold. And those do not support metrics with sectional curvature, let's say, M0. Okay, so, for example, if you take the Eisenberg three-dimensional group, which is the upper triangular three-by-three matrices where you have all ones on the diagonal, and you have the, then you take the lattice where the non-zero entries are all integers that the universal cover is the Eisenberg group, which is the unique non-abelian, important, non-solvable group. And that's a copy of R3. And if you take the compact quotients, non-trivial S1 bundles over Tori. Okay. So, if you never heard about the single conjecture, let me add that, first of all, it has a long history. It is of interest for many people also outside geometric analysis. Like, for example, if you do geometric group theory, this is something that those people are very interested in. And it has been studying a lot, a lot, but unfortunately, definitely results for this conjecture are kind of hard to come by. One result that we have, which I don't know, maybe I labeled as fact zero, is that Lot and Luke have shown in 1995 that the single conjecture holds true when the dimension is three. So, in the case of Riemann surfaces, it's a nice exercise to show that the conjecture is true and it holds and it follows from the uniformization theorem. But that's kind of an easy case. So, let's go to the first non-trivial case. So, the single conjecture holds true when m is equal to three. And now there is a big if, if tarstone geometrization holds true. Okay. So, the approach which I'm going to present to you today provides an alternative solution of the single conjecture in dimension three, but unfortunately it does rely on tarstone geometrization as well. Okay. The nice thing about the approach that I'm going to explain to you today is that it does generalize to large classes on manifold and higher dimension quite easily. Okay. But let me point out that to give you an idea of how complicated this conjecture is, is that even in dimension three, at this time, there are no proof of the single conjecture which do not use the tarstone geometrization. Okay. So, it's a tough one. So, since I want to use a tarstone geometrization, let me summarize it. This, this result of Perman now is a theorem of Perman. What do we know after Perman? Well, we have a picture of three manifolds which is kind of similar to the case of surfaces in many case. So, first of all, we know that if I start with an M3 closed as spherical, then I know that the universal cover is the thermomorphic to a three. Now, this is not a trigger result because there are a spherical three manifolds which are not the thermomorphic to a three. The point is that they do not cover anything compact. Okay. So, and now once you know that M, the universal cover is the thermomorphic to a three. In the language of three manifold theory, I know that M is the one in the base is irreducible, which means I recall you that if I take a two sphere inside M3, then this bounds a ball B3. So, three bounds a B3 inside M3. Okay. Okay. So, irreducible and then tarstone tells you that any irreducible three manifold can be split along to a right and each piece that you get is a geometric manifold. So, moreover, so I'm going to draw for you hopefully an interesting picture. Moreover, for any irreducible M3, there exists a collection of finite disjoint embed that Tori, Tori say, I don't know, t1 up to tk in M3 such that each component of M cut along the union of this Tori, cut along the union of tk is geometric. Now, let me define what geometric means i.e. it admits a complete finite volume metric modeled on one of the eight torsion geometries modeled on one of the torsion geometries. So, let's see if I remember them. So, the most common one is the hyperbolic one, then you have the Euclidean one, then the spherical, which is not very important in the case of spherical things, but then I have SL2R. So, SL2R is a nice three manifold. If you put a left invariant metric, you get a nice complete three manifold, take the universal cover. Then I have the nil manifold, which was the Heisenberg group. Then you can take the solve. Now, there are a ton of solvable three-dimensional groups, but only one admits a compact quotient, and so you get that one. And then what else is left? I am a 2, 4, 6, and then I have the products, the products geometries. So, something that looks like a hyperbolic Riemann surface cross R, and then a spherical S2 cross R. Okay. Now, when I say modeled, let me try to be not too sloppy here. So, this means locally isometric. Okay. Okay. So, I guess I'll give you a picture. So, if you take, then for us, we are interested in a spherical manifold. So, what we're going to have is maybe you have a finite volume hyperbolic manifold. You know that this object is a manifold of phonetoblogical type with finally many casps. And those casps, in the rentable case, have to write cross section. So, let's say this blue manifold here is a quotient of H3R. And then you can glue in some pieces, other geometric pieces. And for us, really, especially if you require the splitting to write to be essential in P1, this is going to be in most cases H2R cross R. Okay. So, each of the red piece, what you have to have in mind is a puncture Riemann surface where you truncate the casps and you take a product with an S1, then you get a boundary which is an S2, and you glue that boundary S2 to the boundary of the casps of the hyperbolic three manifold which you chopped. Okay. So, this is a great theorem, because it's really a structured theorem for any three manifold. So, given any three manifold, now we know that you can split it this way. Okay. Are there any questions? No? Okay. Okay. Okay. So, as I said, the proof I'm going to present to you today, it relies on certain geometric inequalities for harmonic forms which have been developing with Mark's turn since 2017. Now, I haven't lived the time, so I'm just going to give you in a nutshell what we can prove in the three dimensional case. And so, I would say the theory of pricing equalities for harmonic forms in a nutshell. Okay. So now, for the one of you in the audience which are pure geometric analysts, here is the old geometric analysis is condensed here in the sense that there are a lot of integration by parts of geometric integrations. So, curvature plays a very important role. And moreover, the most reiteration arguments for harmonic forms does play an important role. In particular, the key aspect of this geometric and analytical argument is the fact that you can bound the L infinity norm of harmonic id form in terms of the L2 energy of this harmonic form in a bowl. Okay. But anyway, if I had to present to this theory, it would take just one seminar by itself. So, I'm just going to condense it. And I'm going to tell you what kind of inequalities I can get by this theory without providing, of course, a proof. Okay. But if you're interested, we can talk at the end of the seminar and I can refer you to the relevant literature. One interesting thing is that I think some of you study Young Mills theory, some of the audience. So, these pricing equalities where are extremely well known for people studying Young Mills theory. Apparently, these ideas will never apply to other harmonic sections of other bundles. And so, what we have done is to develop this theory systematically in the case of harmonic k-forms on a manifold. Okay. But definitely, if you work in Young Mills theory, you have seen some form of pricing equality in that context of it. Okay. So, what wasn't developed was the case of harmonic k-forms. Okay. So, let me just give you a nutshell of this theory. So, in the three-dimensional case, what you can produce is the following. If I give you a closed M3G, and now I'm going to assume from now on that injectivity radius is bounded below by a constant epsilon naught, which is fixed. And moreover, I'm going to assume that the section of curvature in absolute value is less than one. Then what you can do by using the odd theorem, as Tamash was mentioning before, you can express the B1 as an integral over M3 of, let me call it a density, a B1 density, which is going to be a smooth, a non-negative function over your manifold. And I'm going to tell you in a second how you do that. So, this is going to be, let me call it for the sake of this talk, this is a density function for B1. So, in particular, it's going to be bigger equal than zero other points, and moreover, it has only isolated singularities. So, how can you do that? Well, you know that the space of harmonic k-forms, the dimension of it, computes the beta number. So, you can take, you know, if the B1 is k, you can take k-th, linear independent harmonic one-form, and you can normalize their L2 energy to B1, and then if you integrate their sum over B1, you get the beta number. Now, the interesting part, which uses all these price inequalities ideas and most reiteration, is that one can say the following that there exists C1 bigger than zero. This C1 is going to depend on the lower bound of injectivity radius and on the bound of a section of curvature such that, well, this density function, which in this case is the B1 density function, is bounded from a bow by C1 for any x in m. So, let me call this one. This is just going to be one of the two ingredients I'm going to use in the proof. And then two, what else can I say? Well, if x inside m3 is such that a bowl of radius, the injectivity radius at x of this bowl centered at x is isometric to a bowl inside the hyperbolic three-space with the hyperbolic metric, whose section of curvature is normalized to be minus one, then there is a very nice upper bound on this density function that goes as one over the injectivity radius. So, here is the main content of the pricing equality in the hyperbolic case. And now here, again, for some positive C2, and this C2, once again, the important thing is that it depends only on epsilon or not and on the bound of the section of curvature. So, let's call this fact two. So, what I want to show to you is that by using just this two fact plus geometrization, one can prove the single conjecture in dimension three. Therefore, recovering, they will no result of lot and look. Are there any questions? So, in the beginning of your slide, you mentioned that B1, the petty number, is integral of some function. So, in the beginning of the slide, you mentioned that the first petty number is integral of some function. Yes. And I told you how to get that function. So, you use the Orch's theorem. Right. So, similar statement will be true for higher petty numbers as well, right? Yes. That is correct. So, we have similar estimates for any petty numbers. For example, if you are in higher dimensions, now what it does change is that the first estimate is the same in any dimension. The second estimate might be different. For example, if you are in the fourth dimensional case and you have a bowl of injectivity radius, which is isometric, blah, blah, blah, then in that case, the upper bound is better. It's going to be exponential in the injectivity radius. So, yes, this theory of price inequality that we have developed systematically. First of all, the constants and the upper bounds depends on which geometry you are talking about. In the case of hyperbolic geometry, you have certain exponential decay, or in this case, one of the injectivity radius. If you are doing complex hyperbolic geometry, then you get always exponential decay. Then if you have metrics with pinch section or curvature, then you have exponential decays that might be a little bit weaker, but that's the idea. But you're completely right. This theory works for any petty numbers. Now, I'm not going to talk very much about it, but if the key is always to get a bound here on the right hand side with decays on the injectivity, which decays with the injectivity radius. If we could do that for any metric with just no positive section or curvature, then we will have part of the single conjecture. But that is, again, unproven at the moment. I have to say, this can be done always in any dimension for any petty number, but you don't always have some very useful upper bounding too. With that said, for hyperbolic, for metric with pinch section or curvature, for complex hyperbolic, you have all of these things. Thank you. Now, I can finally talk about answer the question of Tamash. What is the connection with the regular petty numbers? I'm actually going to use this in my proof. So the way I want to show this conjecture here is by using another fact. So now let me, this fact is a summary of two results, one by Luke and one by Hampl. So it goes as follows. Let's say I give you a closed spherical three manifold, then Hampl show in the 70s that the pi 103 is residual finite. For simplicity Rf, I think I'm going to use it more later. So now what does this mean? That is, there exists a sequence gamma K of nested subgroups in gamma, which is just going to be the pi one of my three manifold such that, well, gamma K is normal in gamma and the index of gamma K inside gamma is finite. And then if I take the intersection of all these nested subgroups, I get the identity. Okay. So moreover, a result of Luke tells you that whenever you have a manifold which is a spherical was a fundamental group is residual finite, then you can compute the L2 petty numbers defined by idea with this functional theoretical arguments as the phenomenon dimension of the gamma module given by the Hilbert space of i dimensional L2 harmonic forms just in terms of the regular petty numbers. Okay. Moreover, we have, and this is a famous result of Luke, that if I take the limit as K goes to infinity of bi of mk, the degree of pi K, this is equal to bi2 of m. And now I have to explain who pi K is where pi K is the regular color associated to the normal subgroup of finite index gamma K is the cover associated to gamma K. Okay. So here I can then finally formalize the connection between the regular petty numbers and the L2 petty numbers. So the L2 petty number encodes the asymptotic growth of homology, homology in this case, of towers of the given manifold. Okay. Sounds good Tamas? You're good. So I don't know if this is what you were thinking about, but you know you have this manifold which has a huge fundamental group. So it has a lot of colors. Then you take colors that converge to the universal color and you take the petty number of the color, you normalize by the degree of the color, you take this limit, and then Luke tells you that first of all, the limit exists and this is not completely obvious. And second, that this magically computes these L2 petty numbers defined in terms of L2 harmonic forms on the universal color. Okay. And so this is something which is of interest for many mathematicians. A lot of geometric topologists, they look at this kind of questions and interestingly enough, this is connected with L2 homology on the universal color. Okay. So then you see that if I want to prove the single conjection in dimension three, then I can do the following. So in order to prove Singer for n equal to three, then it suffices to show that the limit in three is zero for i equal to one. Okay. Now notice that the case is i equal to zero and three are trivial because the degree is going to infinity, but the numerator is always zero. Okay. Are trivial. And if I have the case i equal to one, I can cover the case i equal to two by Poincare duality. So can be covered by Poincare duality. Okay. This is kind of interesting that if I can control this limit, just in the case i equal to one, and I can show that this limit is going to be equal to zero, then I have the single conjection dimension three. Now, similarly, if you go in higher dimension, in that case, you have to work with more indices, but a similar strategy will apply. So as long as you have an spherical manifold, which has dimension n, residually from the fundamental group, which is residually finite, then by studying the asymptotic growth of homology or homology, you might be able to show the single conjection. Okay. With that said, it's very hard. Okay. So this is not a trivial problem at all. In fact, has been open for many decades. So are there any questions regarding the strategy that we are going to follow? No? It's clear. Okay. Okay. Now, I have a little bit more than 50 minutes. I'm going to work out an example, which hopefully you will find interesting. And this is an example where the geometric trim manifold, which you are looking at, has two geometric pieces, and those pieces are all real hyperbolic. So now, one thing that you'll have to take my word for it is that manifolds, which have hyperbolic pieces are the hardest to study from the single conjecture point of view. So if you have, for example, a graph manifold, while all those pieces, you didn't have any piece which was hyperbolic, and it was a spherical, then the vanishing on the L2-bit numbers is relatively easy to accomplish. For the following example, which I was going to discuss, at first sight, it might look like just, okay, you're solving this conjecture. As an example, but it encodes all of the possible difficulties that you are going to encounter when trying to solve this problem in dimension three in full generality. So I would like to start now with, so we have seen that if you have a hyperbolic piece in the third-stone geometrization, then that piece, it's a diffeomorphic to a finite volume hyperbolic manifold with the casp truncated. So you're going to have, we're going to start with a hyperbolic three manifold of finite volume. And for simplicity, I'm also going to assume that we have only one casp. So now the argument that we're going to do on the casp are bluing arguments. And so you will see that once you know how to do it on one casp, nothing is blocking you from repeating this for more general hyperbolic three manifold, which possibly have more than one casp. So if you have just one casp, your object looks like this. And moreover, it has torus casps, something like this. So also you have a parameter s, which tells you how deep in the casp you are. Now let's say if I consider the cross section, cross section of the casp for the value of the parameter s equal to zero, then I get a certain two-dimensional two-dimensional tori. These torus here, which is smaller, let's call it ts. And so I have that a casp is always going to be the theomorphic to the half-align cross t2. And moreover, the minus one hyperbolic metric has a very nice form. So you can actually write it down explicitly. And this is going to be ds square minus e plus e to the minus two sg on t2. Now the metric on t2 is a flat metric. So this is flat metric on t2. So now how can we construct a closed three manifold out of it? So if you remember the picture I've given you for you at the beginning, what you do in the geometric decomposition of torus, you chop the casp here and then you want to try to double this manifold. Now you can double this manifold by gluing the boundary tori with a fine diffeomorphism. So let me just give you a few examples. So first of all, given n, I'm going to say that n bar is the manifold with boundary obtained by truncating the casp. And then if I want to construct a closed three manifold out of it, well, I can do some things like this. So let's say that t0 is a two model of the integers. Then I can consider maybe the identity from t0 to itself. And I can double the manifold and I get a closed three manifold m, which is the manifold which I obtained by gluing the boundaries of n bar by identity. And this is sometimes called the double of n. On the other hand, maybe I can consider a nanosoft diffeomorphism of t2. So let's say I consider the matrix 2111. This matrix does preserve the integer points in z2 and gives you a nanosoft diffeomorphism of t0 to itself. And then maybe I can glue the boundaries in a more complicated way. So I can get another three manifold m, which is obtained by taking n and n bar and n bar, two copies of n bar, and gluing them together by the nanosoft diffeomorphism. So both those examples are a spherical and they are metat geometric decomposition with two pieces, those two pieces are hyperbolic. And in the case that I'm doing here where we have just one cusp, then we have just one tori where we have to cut. Now, since I'm talking at the University of Maryland, let me show you that one of the good students from Maryland, Bernard Leeb, studied those kind of manifolds in his University of Maryland 1992 thesis, I guess. And he also showed that, for example, both those two examples which are presented here, they actually admit metrics with a non-positive section of curvature. In the case, in the first case, it's trivial. In the second case, it's not completely trivial. And moreover, those manifolds are, they have geometric rank one in the sense of Bauman, Breen, and Eberlein. So I guess Breen is retired from your department now, but so those manifolds have geometric rank one. So in many ways, they resemble hyperbolic manifold, which also have rank one. But anyway, this is just an aside remark. Okay. So let's say I have this object now. I want to construct a sequence of metric in the spirit of Shigel Gromov, and I will add also Pugaya, which does certain things for me. So next, once I have this move three manifold, which are obtained this way, I want to consider a sequence of metric. So next, I want to construct a sequence of metrics GN on, let's take a more complicated example because otherwise the double is too easy to see as follows. So I have this manifold, the casp here, and then I have this another copy on the other side. And what I want to do is I construct a manifold, a metric, sorry, as follows. So here I have somehow to glue the boundary to right in a non-trivial way. So this is where the Anosoft-Diffie-Morphis act. So if you recall, so at this height, we have the T0. So this is the casp parameter S. Then I'm going to, I don't know, stop here. This is N. This is N plus one, and this is N plus two. So, and here I have the cross-section TN. So first of all, what I can do is I can make the warping factor in the metabolic metric going to constant. And so in this transition regime here, I can keep the section of curvature GN to be less or equal than zero. Here I keep the warping factor constant. So actually in this case, I had in this region here, I have the section of curvature identically equal to zero. And then I perform my gluing. So that gluing, what I claim for you is that you can do that gluing by keeping the section of curvature bounded between minus one and one. So such that I'm going to say a few words so that you believe me. And if you need more, maybe I can go through the example in detail. So I want to keep the section of curvature bounded by one in absolute value. And I also want that the limit as N goes to infinity or the volume of GN of my manifold here, what am I recovering is just the volume of the two original finite volume hyperbolic manifold, which I truncated them then double. And so this is two ball G minus one of N. So what is the metric structure of the gluing? So imagine you have a cylinder, which is a torus cross an interval, and then you take the two boundary to write and you glue them by Anosov diffeomorphism. What you get is a non-trivial T2 bundle over S1. In the case where the gluing map is Anosov, you actually get the solvable geometry. So in that case, you get something which is modeled on an R3, modular lattice, co-compact and torsion free, which acts by isometries on the solvable geometry. In that case, the metric, the model metric is just a left invariant metric there. And the rescaling, because in this case, what is happening is that the fibers are actually collapsing. They're making the fibers smaller and smaller, but they are still claiming that you can keep the section of curvature bounded. And this is an easy exercise in the manual geometry, because if you take the solvable three-dimensional geometry and you rescale the fibers will correspond to the enormous subgroup inside sol, which is just a copy of R2. And you rescale the fact metric on R2, and you get a one-parameter family of left invariant metric, which are actually collapsing to the interval of the S1 with bounded geometry. So the gluing can actually be done. And if you want, I can be more explicit, but it's just a computation with a left invariant metric on the solved geometry. Now, let me remark that the same construction holds true in a higher dimension. And here I have to cite some certain names. So Chigar Gromov, Foucaille, Farrell-Johns. So they are all a storybook at one point. So I noticed that this is not a completely trivial matter, because what you can do in higher dimension, imagine you do the same thing with a nine-dimensional hyperbolic manifold and truncate the cast, and then I glue the boundary to write by an affine diffeomorphism. But then let's say I glue an exotic sphere on the neck, the resulting manifold will not admit a sequence of metrics like this. So when I say the same holds true for n bigger or equal than 3, this is always if the affine diffeomorphism of the boundary are affine. The same holds true in n bigger or equal than 3. Let me put it in red if the diffeomorphism are affine. Okay, it doesn't always hold true. And I can tell you more about that business of summing up exotic spheres. Okay, it's very interesting and it's due to Farrell-Johns. Okay, so I'm ready to go ahead with the proof. So let me define the following disconnected subset of m. So I'm going to define mn to be the following subset of my manifold m, which is disconnected. So I chop the part where the gluing is happening and actually I chop it at, so if this is the cast parameter s, I chop it at n equal to s equal to n over 2 and same same story here. This is at n equal to 2. So this is a disconnected subset. So disconnected inside m. And moreover by construction, I know that the volume with respect to gm, the sequence of metrics which I put on my geometric manifold following trigger and glume off. This is the volume of gn of m minus a little piece and epsilon n goes to 0 as n goes to infinity, where limit as n goes to infinity of epsilon n is equal to 0. Okay, so basically this disconnected part, it contains most of the volume, it's the tick part. Okay, so now what I do is take, and I hope it's okay if I use just two more minutes, so take gamma k, a sequence of subgroups of gamma equal to pi 1 of m, as in the definition of residual effectiveness, as in the definition of residual effectiveness. And what I do is the following for fixed n bigger than 1, define the Riemannian covers which I denote by m with the index k up, which is not the dimension, the dimension is always 3. And then I have mk, I have this cover pi k, which is the regular cover associated to that normal subgroup. But now here I consider the metric, which is just the pullback metric of the metric gm on the base manifold. So where gmk is defined to be pi k star of gm. Okay, so in other words, this is the Riemannian cover associated to that subgroup. And finally, I have to define one more subset of mk. So finally, let's define mkn to be the pullback by the covering map of this disconnect set inside mk. So the picture is that you have your manifold at the base, you take this cover, and this cover is it's thicker. If you look on the cusp, you are unraveling the cusp. Okay. Okay, so in particular, I claim that by studying the sequence mk, gmk converging to the Riemannian universal cover m tilde with gm tilde metric, where gm tilde is the pullback of gm. Then we can show the following. We see that there exists a k0, which depends on n such that for any k bigger or equal to k0, and any p inside m and k, there exists a ball center of p of radius n divided by 2 isometric to a ball inside the hyperbolic three-space isometric to a ball in h3rg-1. Okay. And now it's really the last estimate. So what's happening here is that the manifold downstairs might be collapsing, but it's collapsing with bounded covering geometry. Okay. And here, as you go deeper and deeper in the tower, you're really converging to the universal cover, which is non-collapsed. Okay. And moreover, bigger and bigger pieces of this universal cover, the biggest part is hyperbolic. Okay. So as long as you stay far from the gluing region, what you are seeing as you convert to the universal cover is bigger and bigger balls of the hyperbolic three-space. Okay. So now I can do the final estimates and I show you that b1 divided by the degree of the cover goes to zero. So I promise this is just one slide. So if I want to compute b1 of mk, then this is, well, the integral of mk of the density, gmk. And now I can do the integral over the nk part of b1 of x, the mu gmk plus the complement. Okay. Nothing deep here, but now I can use price. Okay. And so this is less than 2c1 divided by n, the volume of gmk plus c2, the volume of gmk of the complement. Okay. So that if I take the normalized beta number, so if I take b1 and I divide it by the volume, what do I get? Well, I get 2c1 divided by n plus c2. Now these are remaining covers, so the depth volume is nothing else than the degree of the cover times the volume of g, with respect to gm of m minus mm, which we know is more. And then I divide by the degree of pi k ball of gm of m. So in other words, this is 2c1 divided by n plus a c2 epsilon of m divided by volume of gmm. But now as n goes to infinity, the volume of gm of m is converging to twice the hyperbolic volume of the hyperbolic pieces. So this is approximately for n large plus c2 epsilon m divided by 2 volume of these pieces. And this is as the volume of gm of m goes to twice the volume of g minus 1 of n. n was the finite volume hyperbolic manifold which we double. And so then basically I'm done. Thus, for fixed n, if I take the limb soup as k goes to infinity of b1 mk divided by the volume of gmk mk, this is nothing else than the limb soup with respect to k of b1 mk divided by now the degree. And this is what I want in order to use a luke approximation theorem, volume of gm m. And this is less or equal than now another constant, c1 prime divided by n plus c2 prime epsilon m. And then taking n that goes to infinity, I get the result that the limb soup as k goes to infinity of b1 mk divided by the degree and I'll close here. Questions? If there are no questions, I think I can close up here. I think I ran five minutes over time, but we started probably five minutes later. Yes, yes, you're fine, don't worry. Yeah. So are there any questions in addition to the ones we had during the lecture? If not, then let's thank Luca with our virtual round of applause. All right, thanks a lot. Thanks a lot. Thank you.