 Well, thank you for inviting me here. It's a great honor and pleasure. I met Marcel in 1981 in fall when I came to Paris as an exchange student, being completely naive and ignorant, I have to say. But due to not only Marcel, but the atmosphere that he created around him and his young people. So when I came, I had some vague idea that maybe geometry was something that I could like. And after one year, it was kind of clear that I would never escape geometry again. I was completely captured and fascinated by the subject, and in particular also by Marcel's way to discuss mathematics and introduce concrete problems that had difficult, but perhaps concrete, answers. So I hope I can give you today a little bit of lecture in the spirit of this philosophy that you might be interested in concrete questions. There's been a lot of progress ever since. But somehow, maybe the progress leads now to us being able in the hopefully near future to solve problems that were not reachable back then. But maybe there isn't reach right now. Let me start with something so very simple related to Bruce's talk. So Bruce said, now that we understood geometrization, we can ask sort of more advanced questions that follow up. And let me just go back to something that is familiar to all of you. Name, let's look at a closed surface where somehow you could think of everything as understood. And for me, everything is oriented. I don't write this down. All my manifolds will be oriented. And let's take a closed surface of genus G bigger equal than 2. And we all know that this surface has a hyperbolic structure. That means a constant curvature, a metric, a constant curvature of minus 1. And furthermore, there exists a 6G minus 6 dimensional parameter space of such metrics of curvature minus 1. And we, in addition, have Gauss-Bonnet theorem, which says that for each such metric, the integral or the volume of the surface is equal to 2 pi times the Euler characteristic, the absolute value of the Euler characteristic. So it's a fixed number. And this has a kind of very easy consequence that for all x and s. So now I think of it as equipped with such a metric, the injectivity radius at x is less or equal to some constant times logarithm of G. Where does this come from? It comes from the fact that if you have a disk in the hyperbolic plane of a radius r, then the volume is roughly e to the power r. It's not quite correct, but up to some constant. And so if we had a large injectivity radius around a point, the volume would be very large. But the volume is not very large because the volume is uniformly bounded. Of course, we can have points where the injectivity radius is arbitrarily small. We can produce hyperbolic metrics on any of the surfaces such that the system, that means the length of the shortest closed geodesic is arbitrarily small. That's not a problem at all. So now, all these sort of things related, we can look at the rank of the fundamental group of s. What is this? This is a minimal number of generators of pi 1 s. That's a number. And this is actually the number that we can easily compute. It's 2G. So what we can say is that injectivity radius here point-wise is bounded from above by a function which only depends on the rank. That's a kind of very satisfactory result if you look at understanding the geometry of the surfaces. And you immediately observe that there's nothing much to do with having a hyperbolic metric here. We could have a negatively curved, pinched metric. And we would get not the same constant, but we would get very similar statements here. So there's no problem with that. Let me give you just to make an advertisement for my student, Bram Petri, give you a kind of, I think, nice result that he proved last year. So we can have, actually, this is sort of sharp in a way, a strong sense. There exists a sequence of surfaces, hyperbolic surfaces, GI of genus GI going to infinity, such that S, say Si, GI, GI going to infinity, such that Si, GI has a systolic pens decomposition. So what is a systolic pens decomposition? A pens decomposition is something that I might wish to draw on the right-hand side here. So you decompose your surface into pairs of pens. A pair of pens is just, for example, like this. A pair of pens is just a three-fold sphere. And now you can decompose your surface into such three-fold spheres. I forgot one corner here. And now what he proved is that there is a pencil composition of this particular surface, which is systolic. That means each of these pens curves from the decomposition has the length is equal to the smallest length of any closed geodesic on the surface. So the metric is hyperbolic. And the length of each of these curved systols is bigger equal 4 over 3 times the logarithm of GI minus some constant. So in other words, you can produce things where, I mean, you have really have your sharp estimate. I showed you here the point-wise injectivity radius is less or equal than c times log g. And now he produced something where you actually have a pencil composition. And all these pens, these curves here are systolic. So that means injectivity radius on each of these curves is what you get from the number. And you have plenty of them. So no, no. It came out of the proof. Yeah, it's just a number. I mean, here. So it has to do with the construction. But I mean, you may or may not be able to improve this. I don't know. So now the question that arises is, what about three manifolds? Now, we know that due to the work of Perlman, there is geometrization. So what we know in particular is that if m is closed, again, my manifolds will be oriented for simplicity. So I don't write this down as closed three manifold, but a spherical, meaning that the second fundamental group of m is trivial. And a to order, meaning that their pi 1m, so I'm slightly informal here, does not contain z cross z. Then Perlman showed, so this is one of his consequences of his work. So this statement is weaker than geometrization. But it comes out of Perlman's work on geometrization that m admits a hyperbolic metric with curvature minus 1. So now, most of the rigidity, as Bruce explained to you in the last talk, most of the rigidity implies that this metric is unique up to isometry. We are now in 3D. So we're not in the two-dimensional case anymore. But of course, the diffeomorphism group acts on this metric. And so it looks, if you try to describe this metric, it's maybe slightly complicated due to the fact that the diffeomorphism group acts. But up to isometry, this metric is unique. Now, once we have understood this, we somehow can ask now refined questions as you see here that are known for surfaces. So one of the refined questions would be that was a question that was raised by McMullen, actually in the early 90s. So the conjecture by McMullen says, let's m be closed hyperbolic, that means. And again, I mean, this has a friend. This conjecture has a friend. So this conjecture was stated before geometrization. So now, in the view of geometrization, this conjecture should have a friend for variable negative curvature. And it says, if for all k bigger than 2, there exists a constant u of k bigger than 0, such that if the rank of pi of pi 1 of m is less than equal than k, rank in this sense. So you look at the minimal number of generators of the fundamental group, then the injectivity radius of x is less than equal than c of k for all x. So meaning that although there are definitely examples with arbitrarily small injectivity radius at some point, we require that at every point, the injectivity radius is bounded from above. So and again, if this were true due to geometrization, we assume that the same thing is true also for variable negative curvature provided that you have some pinching assumption. Now, this would shed a light on the geometry of those three manifolds, which would lead to a better understanding. Let me now give you some overview of what's known towards this conjecture. So first of all, there's a CO by white. Unfortunately, there are lots of people whose name is white. So and I forgot this first thing. I never met this person. He's a student of McCullough. So he did in the early 2000s. He proved that for all such m, x and m such that the injectivity radius at x is less or equal to, I'm saying, c prime of k. So the global injectivity radius for any such manifold whose rank of the fundamental group is less or equal than k is can be bounded from above by a uniform number. But of course, this is very far away from saying that the point was injectivity radius is bounded. That's much, much stronger. So there is a serum which somehow solves the conjecture, which is due to a Birner and Zutow. And they show that they prove the conjecture. So this is 2017. This is maybe 2005. This is 2017. That proves the conjecture for a manifold with injectivity radius at x, bigger or equal, some constant, a nu, bigger than 0, nu is universal, plus some small additional assumptions. I confess that I did not read this paper. This paper is available on the archive. And it's 170 pages long. So it's a very difficult technical work. But I mean, that proves this conjecture under these assumptions. So now let me give you a sort of weaker form of this conjecture of McMullen that is maybe geometrically more accessible than which I will discuss for the remainder of my talk. So the rank is controlled by something out which is called the Higa genus. And let me explain what that is. There is a fact. Any clues? And this is easy to prove. This has various ways to do it. I mean, if you believe that manifolds can be triangulated, just take a triangulation and you take tubular neighborhoods of the one skeleton you get going. That's an easy result. Any closed three manifold can be glued from two handlebodies with a different morphism of the boundary. So what does that mean? So I, in some sense, draw already here a picture of a handle body. So handle body is a neighborhood of a bouquet of G circles in R3. So you think of it as this thing here bringing this surface here built in the interior. Now you can take a point here and you draw a bouquet of, in this case, three circles. And you flatten this up and you get this handle body as we draw here. And now what's assumed this is, if you take a closed three manifold, then you can find some G. So the genus of the handle body is a genus of the boundary surface such that you obtain this three manifold by taking the boundary surface here, taking another copy and gluing it with a different morphism. So now the smallest G, smallest genus, genus G such that this is possible. That means such that we can glue our manifold from these two handlebodies of genus G. It's called the Hagar genus. Now as a Hagar genus, understanding the Hagar genus is a bit more friendly if it comes to understanding, trying to understand the topology of such a manifold. So what we know is that if the Hagar genus is of M, is less or equal than G, then the rank of pi 1 of M is less or equal than G. And this comes from the fact that the fundamental group of such a handle body is a free group into NG generators. So that's an easy statement. But now if you replace in McMullan's conjecture the rank by the Hagar genus, then it becomes a little easier. So let me, however, state that the rank is not equal to the Hagar genus because there is a theorem by Tao Li. It was conjectured to be the same by Waldhausen, but this is false. There is a theorem by Tao Li such that the Hagar genus of M minus a rank of pi 1 of M can be arbitrary and large. So some care has to be taken into account. So now if you think about, however, using the Hagar genus to understand our manifold, then it reduces to understanding how the gluing works. And the fact is that the manifold only depends on, so as a topological manifold depends on a differentiable manifold on the element, the gluing map or the class of the gluing map. In what Bruce already introduced, it was a mapping class group, mod S. So this is a group of different morphisms of the boundary surface S divided out by the different morphisms, which are isotopic to the identity. So in other words, if you take two different morphisms that are isotopic, you get the same manifold back. And actually, it depends on much less on a double coset, but this will not be important with the reason being that an element, if you have a different morphism of the boundary surface, which extends to the three manifold, then we are actually in the setup of Bruce's theorem that we deal with pulling everything back by different morphisms of the three manifold. And so this is not what we will be interested in. So now the mapping class group here. This is a nice group. It's finally presented. It's finally generated, was proven by Lekorich finding, sort of recovering of an old theorem by Dain and Nielsen. And then it's finally presented. This was initiated by Hedger-Sressen, the same Hedger as before. And then there is work by Weinweb. So this was done in the early 80s. Everything was done in the early 80s, Lekorich's earlier. So now what we can do is we can choose now a probability measure nu on mod S. So for the sake of sort of making things simple, let's assume that it's finally supported, and that the support generates mod S. For example, you could take a simple probability measure which just takes a final generating set and gives each of these generators an equal weight, such that the weight sum up to one. And then you get such a probability measure, and then you get a random walk on mod S. And what you can ask now is, I mean, what is a random manifold? Which random manifolds, in some sense, may be easier to study than other manifolds with other properties. Now I have the challenge here that I have to somehow maneuver the blackboards. All right. So what I mean a random walk, so what we do is, I mean, we have a probability space. So the probability to go from the identity to one of my generators is exactly what is prescribed by this measure. So it's a total mass of this generator. And then the probability, so we go here, say, from the identity to some point G, with probability, say, epsilon. And so now we start our process again. So we have this weight epsilon. And we go to the next step. So we attach a generator here. And then we ask ourselves what is the probability to go from the G epsilon to the next element, which is of the form G epsilon times S, where S is one of these generating things. And this will be given according to the weight of this S. And then we have to sum up overall ways to get to this point in order to get the probability. But now we can ask ourselves what is a random property. Let me define this formula. A property P holds for a random three manifold of rank G, say, if the probability for P tends to one with the step lengths of the random walk tends to infinity. So now what does that mean? We fix the genus. So we fix the genus of this handle body. Now we look at all elements in this mapping glass group. And we ask ourselves, I mean, we produce now using the elements in the mapping glass group, we produce three manifolds by gluing them together with this corresponding diffeomorphism. And diffeomorphism will present in this glass. So we glue together to this manifold. And we ask ourselves, what are the expected properties of such a manifold as a step length scores to infinity? Now I assume, which was proven by Dunfield and Thurston, maybe 2000, about 2000. And it says a random, so random three manifold in the sense of rank G. So it's not clear that the rank is actually equal to G. If you start this process, a random three manifold of rank G has rank G. So the rank is actually equal to G of the fundamental group, and not smaller, not bigger, has HECA genus equal to G. So the HECA genus is actually this G and not anything smaller. So you could do some sort of cancellation here that produces, for example, this fear. And the HECA genus of this fear is definitely zero in the way you construct it. So the random manifold is actually HECA genus G. And it is hyperbolic. So now let me state a theorem that I proved in joint work with my student Gabriele Biacchi. So what we showed is that a random, so we cannot prove McMullan's conjecture. But we can prove the following. So a random three manifold of rank G McMullan's theorem holds two. And we showed something else which is related to some other geometry quantities that you associate to such a manifold. And maybe show that for random three manifold. So McMullan's conjecture holds two, meaning that there is a constant only depending on the rank here, that a random three manifold has satisfied McMullan's conjecture for this particular constant. And for random three manifold, we have that of rank G in the sense that we have that lambda 1 of m, the first eigenvalue of m, is contained in and now comes a uniform constant b over the volume of m squared and a constant depending on G over the volume of m squared. So we somehow understand at least part of the spectrum. That means we understand the smallest eigenvalue for such a manifold. Now, let me add some remarks on this lambda 1 of m. So where does this estimate come from? This comes from an old theorem of Rick Shane. What he proved in, I have to check the date for an 82. What he proved is that regardless for all closed hyperbolic three manifolds over under some suitable curvature conditions, so if say n, let me give a different symbol, so make sure that we are not in the realm of hyperbolic manifolds anymore, but n of uniformly bounded curvature, uniformly bounded curvature closed say dimension. I mean, this theorem works for all dimensions, but we have looked at dimensions v. So the constants depend on the dimension and the curvature. Then lambda 1 of m is bigger equal than some constant b. And again, this constant depends on the data that I was discussing. So in the case of the pinch negative curvature, b is actually explicit. That's nice. Now, this was generalized to finite volume manifolds of negative curvature per random. And let's see, Randall, sorry, and Dodger, in 86. So let me discuss briefly what is known for the upper bound. So here, what we say is for random three manifold, this is essentially shop. So this is essentially what we expect. Of course, this c of g is not really explicit, unfortunately. But I mean, this is all what we can say. Let me discuss a little bit where upper bounds for the first eigenvalue come from, because it will somehow show you how things are a little bit related. So the first upper bound comes from the celebrated Chica constant. So what is this? This is something that is kind of very easy to write down, which makes it so attractive. So the Chica constant is the infimum. Probably all have seen this before. Overall, set E, so submanifolds E, that are hyper surfaces dividing my manifold into two pieces. And now we look at the volume, or the surface area. If in our case, it will be a two dimensional manifold, then we can look at the surface area of this E divided over the minimum over the volume of A and volume of B, where these are the two components that I obtained by cutting along E. So in other words, I have A disjoint union of B is equal to M minus E. This is a way to write it down. So this is a celebrated Chica constant. And now, what the Chica inequality says at lambda. Yeah, that was funny. So the Chica constant is nowadays, so very celebrated. I'm very sort of honored that now Jeff just ended the move. So the Chica constant, this lambda 1 is bigger equal to H2S squared over M over 4. All right. And so this gives a lower bound for lambda 1. This looks the same as here, but I mean now we also have under some curvature assumptions and dimension restrictions, we have an upper bound that comes from the work of Boozor. And Boozor showed that lambda 1 of M is less or equal. And these constants now depend on the dimension and the curvature, but I mean here everything is bounded. It's less or equal than B times H of M plus H squared of M. So this was proven in the 80s. This gives an upper bound for lambda 1 of M and let me now sort of quote a theorem that allows us to use the upper bound. Oops, now this was to use the upper bound coming from the Chica constant to give us, in the case we are interested in, to give us a bound on lambda 1. And this is work of Mark Lackenby. So Mark Lackenby proved in 2006 the following beautiful film. So it's very explicit. And he showed that this Chica constant of M, so now we look at M a closed type of Bollack's V manifold. So this is less or equal than 4 pi times. And so in order to formulate it in the sharp form, I don't use a HAKAR genus. I use a HAKAR Euler characteristic. So this is given by 2 HAKAR genus minus 2. And you divide out over the volume of M. And now if you put this together, we get immediately that lambda 1 of M is contained in universal constant B over volume of M squared and some constant over the volume of M. And this always holds true in the case of hyperbolic V manifolds using this theorem of Lackenby and what we saw before. Now this turned out to be conjecturally somehow not sharp in our case, the upper bound. The lower bound is, as I showed you, up to constant, but conjecturally, so is the conjecture, if the bank. So let G be the bank of pi 1 of M. Then lambda 1 of M should be less than some constant over the volume of M. And now comes an explicit exponent 2 to the power 2G minus 2 over 2 to the power 2G minus 2 minus 1. That looks a bit funny. So the exponent is 2 to the power 2G minus 2 over 2 to the power 2G minus 1. Sorry, 2G minus 2 minus 1. So this is a little smaller than this one. So you have a little higher exponent. So this is a conjecture that this upper bound should be OK. And this is essentially this is true and sharp for manifolds, which fiber. So hyperbolic V manifolds, which fiber over the circle. And I guess that the same should hold true in general. But I mean, we are very far away from proving this. So this is due to myself and to work with Hong Kong bike and Ilya Gekhtman independently due to Lensen and Suto. So that's the picture. And now let me spend the last maybe 10 minutes so giving you some idea on how to prove the theorem that I wrote down for the random three manifold and a little bit what kind of tools go in. So the tools is actually a combination of things that were discovered somehow after Perlman and tools that were developed before Perlman's work and the Ricci flow does not enter. So Perlman does not enter. So what the starting point is, the starting point is foundational work of Jaya Minsky and then collaboration of Jaya Minsky together with Jeff Brock and Dick Canary. And so what Jaya proved is this is an amazing theorem. And then so Brock, Canary, Minsky. So they proved the following. So for M closed hyperbolic as before, I don't write this down for the sake of time saving, fibering over the circle. So if I have a closed manifold which fibers over the circle, then this admits an infinite cyclic covering whose fundamental group is the fundamental group of a closed surface. So M hat, the infinite cyclic cover, pi 1 of M hat is equal to pi 1 of some surface, Sg of some genus. There exists an explicit model manifold. So that means a manifold that you can write down by hand, where you can write down the geometry by hand explicitly. And the curvature is a bounded curvature. The curvature bounds do depend on the Gs or Gs fixed, a bounded curvature, negative curvature. This can be arranged in such that this model metric is uniformly by-lipschitz to the hyperbolic metric. And of course, this model manifold satisfies McVown's conjecture. That's something amazing. I mean, so I discussed at the beginning of this lecture how you can describe surfaces explicitly. So in the case of surface, there's no big deal. But here we can write down, for those particular manifolds, we can write down explicitly how the geometry looks like. Let me recall also a theorem, a celebrated theorem of Egan-Egel, so he proved that any M, so M closed hyperbolic admits a finite cover, fibering over the circle. Now you might think that you are set with these two results, but you are not very far from being set, simply because the degree of this cover cannot be controlled. So I mean, once you can't control the degree, the injectivity radius is going to grow up, blow up. And so we don't know. We cannot argue using the injectivity radius. But now, so the goal is somehow to try to understand somehow how you obtain an arbitrary manifold using the information, or at least a random manifold, using the information that you get from this theorem. Now what the idea is, the idea is very simple. And then comes some tools that were sort of paper almond, which I'm going to explain to you. So the idea for random three manifold. And let me now give you the idea for the control on lambda 1 because this is a little easier. The idea for random three manifold is somehow you have information on the gluing map. You don't have information on the geometry, but you have information on the gluing map. And now how you would like to think about a random three manifold is as follows. So you have your two handle bodies. And you glue them together with an element in this mapping class group, which is very large. Here you have a handle body. Now what you think of it, this is sort of a precise piece. Here you have a handle body. And now just pictorial you think of as if we didn't glue them directly together along the boundary. But what we fit in is a large collar. And this collar will be the surface, the boundary surface cause some enormous interval, say minus kk. And now you think of this manifold being decomposed into three pieces. You have these pieces. This piece that is, I mean, a different morphic to S-class this interval. You have here the bottom and the top. You have here handle body where you know nothing much about. And now you try to use Minsky's model for this piece somehow. You think of this piece as being part of this manifold or piece of the manifold that you obtained by taking a manifold fibering over the circle by taking this infinite circle cover. And you like to import this piece somehow into such a manifold. This is actually what you can do for the random manifold. In the following sense, you can identify gluing regions where you produce the following picture. So it's possible to decompose this manifold M into three components. Say V1, V0, V1, V2. And these components are as follows. So V0 is stiffenmorphic to a handle body. V1 is stiffenmorphic to the surface cross, say, an interval. Let me take an open interval. And V2 is stiffenmorphic to a handle body. Now, in addition, in this case of, so you use some information on the random manifold to do this in such a way that you can equip these pieces with some hyperbolic metric suitable. And then you glue the pieces. So the metrics on the pieces, glue them to along the overlap. So you think of it as these pieces overlap. These are open submanifold. They overlap somehow along V0, intersected V1, and V1 intersected V2. We will have V1, V0, and V1, V2, which is joined. We glue them along these intersection components to metrics of bounded negative curvature. Now, we don't quite know what's going on on V0, but on V1. But we can produce this model geometry on V1, where we have a lot of control. And the model geometry is imported from the work of Jai Minsky and his collaborators. So we can arrange, we can assure, that a large piece, a large portion of V1 is uniformly bilibchitz embedded in the hyperbolic manifold. So what does it mean is we have this metric which has bounded curvature. And we have this large region where we have essential perfect control. And we can assure that this large region, though we haven't basically the identity map, which maps this region into the hyperbolic manifold, this original manifold has a hyperbolic metric. This is a uniform bilibchitz map. So the geometry is different, but nevertheless, we have a uniform bilibchitz map in between. And now, we could use the control on this V1 if we knew somehow that we have some control on V0 and V2 as well. And this is a little more difficult, because this work of Minsky and his collaborators is not valid in this case. But we, however, can assure also that we can assure for the metric of variable curvature, the volume of V0 union V1, V2, sorry, is much, much less than the volume of V1. So almost all the volume is eaten up by this region where we have control. But now what we are going to do. So what comes now, what we like to say is this geometry here is not too far away from the geometry that we're aiming at, it's a hyperbolic geometry. So the entire summer, I stared at this problem and tried to use a Ricci flow to get this settled. And I never succeeded. I asked everyone that I knew who works on the Ricci flow to help me out. And I didn't get any answer. And then eventually, I remembered that there was another fantastic tool, and that's called the BCG map. All right. And I'm very happy now to say that this was, came to my rescue, the work of Besson-Coudreaux and Galovic states, is this prepairman. So what that proof is, that now we are in a situation where we can apply it. So what that proof is, and so this is some souped up version of the work is that, in the case at hand, there exists a map. Now, we are talking about the same underlying manifolds, but the geometry is different between the variable curvature. And here comes to the constant curvature. And this essentially, so I mean, I'm slightly cheating because I'm running out of time. It's essentially decreasing volume. It's a point-wise decreasing volume, essentially. So point-wise. It's not literally true, but I mean, so let me cheat a little bit, point-wise, decreasing volume. And now, we use this, and so now we analyze this map, what it does on the part that we have a perfect control. And we realize there are nothing much happens, so the volume essentially remains the same. This uses precise information on this piece in the middle. And on these pieces here, v0 and v1, where we do not have much control, it decreases the volume. So now, we use a volume information to get lambda 1 going very easily. So this allows you to show, to deduce, that the volume of this v0 and union v1 for the hyperbolic metric is not bigger. Not bigger is not literally true. There is a small constant. But this is, I mean, we had arranged this for the variable curvature metric that the volume of this union is much, much smaller. It's not bigger than for the variable curvature metric. Now, even more, this approach by Besson-Côte-Var and Galoux is so amazing that it can be souped up and proved. So I don't use exactly what they do, but I use a slight variation of what they do. And it can be souped up to also prove McMullen's conjecture in the situation where I have to analyze a little bit what these great minds did in their work and then using my souped up version that get me going. So that's, my time is over. Thank you very much. I'm not sure it makes much sense, but could one hope to get some average value in this Ago theorem you quoted? And could one hope this number to be one and then get a proof you want? Here's a theorem that we can prove. V meaning me and seven, it was a paper that I wrote with seven other courses, so I don't name all of them. But what we proved is that for the random sweep, so when you fix a number, any number of 10,000, for the random sweep manifold, you will not find a cover of degree 10,000, which does a job, so you have to go a higher degree, yeah. So this, I mean, all this can be sort of made a quite, so this isn't the sort of the power of random method that you can make sure that you don't try to do something that's not going to work. Is there any geometrical understanding about the bone on the ramp that you get in the random case? Sorry, what? Is this C of K on the ramp? Is there any geometrical understanding of these constructs? Yeah, I mean, you can basically, this can be given an explicit interpretation. Actually, the C of K can be more or less calculated, more or less explicitly. So you have to do a lot of refined analysis. Depends a little bit on the random work, however. But you have to, this kind of refined analysis has to do with dynamics of the mapping class group, actually, that can be made quite explicit. So I'll question you. So thank you again.