 Well, I'm very happy to be at this conference, even though I'm virtually there celebrating Thomas' birthday, Thomas inspired the whole generation, current generation of geometric group theorists in France, both by his work, I mean, there were a lot of things that we're trying to understand. And Thomas' exposition was brilliant and gave us a really new insight. And he was also very generous, sharing his ideas with many people. And finally, he has a great passion of mathematics that was contagious, I think. So it was really great to talk to him. So, and in fact, one of the topics somehow goes back to a lunch I shared with Thomas 30 years ago, maybe I will go back to this later. But, okay, so let me start. So this talk is kind of two talks, but I would like to show you an instance where ideas that we use in course geometry, apply also in differential geometry or in systolic geometry anyway. So let me start with this. So, there will be two halves in this talk, both dealing with some kind of dimension. The first half is about a synthetic dimension that I think is now familiar to people that are doing geometric group theory. So we know it's a large-scale analog of topological dimension. Now, the other type of dimension I will talk about is macroscopic dimension or more specifically q-eurison width, which you have a space, say a manifold, and you see how close it is to something lower dimension, to a lower dimensional simplicial complex. So I will go back, of course, and give definitions of this, but this is what this will be about. And I'll try to convince you that, in fact, we can use similar arguments for both types of dimensions. So let me start with the synthetic dimension. I just recall the definition. So the synthetic dimension is less than n if for every m there is some constant d of m and the covering by sets of diameter at most d of m says that every m ball in x in this x at most n plus one elements of the cover. So just to illustrate, let's prove that the synthetic dimension of the Euclidean plane is at most two. You might have seen this proof. You just have this cover by bricks. Say if you take a ball, an m ball, you take a brick of size 10 m by 10 m by 20 m or something. And now you see that this m ball is going to intersect at most three of your bricks, which shows that the asymptotic dimension is at most two. And it's easy to see it's not one. And again, you have to play this game to draw this picture for every m, right? It's always the same picture. Another example is trees. Trees have asymptotic dimension one if they are unbounded. Now, the main reason we care about this is that if G is a finitely generated group of finite asymptotic dimension, then the novel conjecture holds for G as a result of U. So generally, we try to prove, so this is a way to prove the novel conjecture for a group. It's easy to see that this is a quasi-esometric invariant, the asymptotic dimension, and one can take a finite notion. So if we can take d of m to be linear, so d of m, remember, was just a constant depending on m. So if we can take it to be linear, then we have the strunger notion that the sub-negada dimension of x is at most n. So now I will be interested. So again, as I said, generally, we want to know if this is finite, but in this talk, I will be interested in asymptotic dimension in itself. So I see it as a geometric invariant and I want to calculate it. And there are some results in this spirit. So the most important perhaps is this result by Bialo Labedeva. So if G is a hyperbolic group, then we can give an exact formula for the asymptotic dimension. The asymptotic dimension is equal to the topological dimension of the boundary plus 1. So for hyperbolic groups, we can calculate the asymptotic dimension. And then there are some results by Gromov and Lang Schleschenmeyer about Hadamard-Mannivoltz. So Hadamard-Mannivoltz, the dimension n, and pinched the asymptotic dimension exactly equal to n. But if we pinched from this, I don't think we know that the asymptotic dimension is even finite. So we need pinched curvature to calculate. Now, another result, planes have asymptotic dimension less or equal to 2. So when I say planes, one can take planar graphs or planes with somebody's money and metric. And now this might appear trivial. Of course, we just said that R2 has dimension 2. However, here we allow any metric. And now if you try to do this with R3 and you put any metric in R3, you can easily construct a metric with infinite asymptotic dimensions. So in dimension 3 and above. So the way to do it, for example, take R3 and embed the sequence of expanders there just by tweaking the metric. So in this case, we can give a more or less exact formula. And this is the result together with Fujiwara. We gave the bound 3 and then Jorgensen-Lang and this bunch of people, Bonamibus, Kesper, Greenland, Lupiro and Scott remarked that, in fact, one can push our proof because what we are proving is we are proving actually that the SODNA gathered the dimension of spheres is at most one. They are which suffices using some material results to prove that, to give the bound of 2. So you can also remark that a plane can have asymptotic dimension and, for example, think of a cylinder, right? A half cylinder and cap of a half cylinder. This is a plane. Again, it's a homomorphic plane. It has a strange metric and it has asymptotic dimension 1. But if you have a 1-handed plane, the asymptotic dimension is exactly 2. And now some further results of exact calculation. If we take one related groups, then they have asymptotic dimension 2. This is a recent result by Tsilekidis. And now the same bunch of people showed that minor restricted families of graphs have asymptotic dimension 2. So this, so a minor restricted family of graph is you take a family of graph that is not allowed to contain some other graph as a minor. And this is an important class of graphs for graph theorists and computer scientists. So that generalizes planar graphs and it turns out that the result from planes can be generalized in this family. I mean, it's a different proof to show that these graphs have asymptotic dimension 2. Maybe a couple of comments. First of all, as I said, this is an important theory, the theory of minor restricted graphs. It is developed by Seymour Thomas and it has a very geometric flavor, but it is unknown to geometers outside a graph theory, which I think is a pity because it's really geometric theory that should be widely understood. So another peripheral remark is that it turns out that computer scientists also have studied asymptotic dimension. This is something I didn't know. So they have some results because they are also interested in the composition of space. So maybe that's why some of these names are not familiar when we talk about asymptotic dimension is that it turns out that asymptotic dimension is interested, is interesting beyond geometric theory. So finally, another example, Hadamard manifolds of dimension 3 have a asymptotic dimension equal to 3. So this uses the previous result on planes and it's done by Georgiansen and Lang. As I said in general, for Hadamard dimension n, we don't even know the asymptotic dimension is finite, but for 3 we have this result without restrictions. Okay, so I would like to talk of another natural class of spaces which are graphs of polynomial growth. So a specular antiquities in 2018 asked the question if graphs of polynomial growth can finite asymptotic dimension. So of course one might be interested in this from a geometric perspective. A main example of spaces of polynomial growth are manifolds of non-negativity curvature. So one could ask this for manifolds of polynomial growth, but to get an equivalent relation question we need to assume also bounded geometry. Okay, so this question was answered in 2020 by the same bunch of people that were in the previous slide monomy and co-authors. They showed that if the growth function of a graph is polynomial of the form r to the k, then the asymptotic dimension is O of k log k. So some constant times k log k is bounded, I said equal, so it's less or equal to this. And now their proof uses an aerial result of class grammar and leaf from 2007, which was conjectured by linear. And the result is the following if gamma is a graph polynomial growth, then there is a function for gamma to r to the power n where n is omega of k log k such that, well, this function is Lipschitz, so the distance of f u f v is less than 10 times distance u v, but also it's coarsely one-to-one say. So the distance from f u f v is greater or equal to one. Now if we had a stronger second inequality, we would have a coarse embedding, right? But this is a much richer result on coarse embedding. And now if you just consider maps with this property, of course, they could arbitrarily increase dimension, for example, you take one dimension thing and make it 10 dimensional by such a map, right? But what Bonamy and co-authors observed is that it doesn't go the other way, you can get a lower bound from this, even though it decreases, it cannot decrease dimension. So this is a conjectured bilinear. And of course, it's interesting to ask whether, in fact, you can have a coarse embedding. And I don't know that. I mean, so a coarse embedding would give some bound immediately. But yeah, this is much richer than coarse embedding, but still it suffices to give a bound. Okay, so now I would like to talk about how to get a sharp bound, right? So this is a bound, but we don't know any examples of of graphs with growth, say, n to the k that have asymptotic dimension k plus 1, right? Or 2k, which this allows. So first of all, the growth function of a graph is the supreme of the cardinality of a ball of radius r, where I allow v to vary over all vertices. So some some people call this the uniform growth function. But this is the growth function, we consider that this is the natural growth function to consider here, right? You take the maximum possible volume of a ball where the volume again is just the cardinality of the number of vertices in the ball. And now the theorem is if the growth function is strictly smaller than r to the k plus 1, so if gamma to the r over r to the k plus 1 tends to zero, then the asymptotic dimension is bounded by k. So, so the difference with before is that we just get a sharp bound. So in particular, there is no graph with growth smaller than the k to the graph of z to the k and bigger asymptotic dimension. So let me, are there any questions maybe so far? Is the statement clear? Yeah, excuse me, I just have a question. It's Indira, do you know the what's the asymptotic dimension of Heisenberg group? Do we know the sharp bound or? I guess it's three, but I might be wrong. I mean, my guess could be three. Well, first of all, it is known. Okay, for important groups, it is known. Now I don't want to give a wrong formula because you have these exact sequences and you have an exact sequence like that. You have an exact formula. So for every, and I think it's three. Okay. Okay. Okay. Great. So, so yeah, I should say, of course, yeah, groups theoretically, my talk is not interesting because groups of polynomial growth are nilpotent groups. And since we understand them, that that was a good question. Actually, I mean, to say that and since we understand them algebraically, and you have these exact sequences, we have a formula which gives us the exact asymptotic dimension in all cases for groups of polynomial growth, right? But for graphs of polynomial growth, that's something, yeah, that is not so obvious, right? We don't have this algebraic description. Any other questions? Okay, so let me try to say something about the proof. So first of all, I would like to define to give some sort of finer definition. So I will say that their dimension is less or equal to n, right? If there is a cover so that my definition works for R balls, right? So it's the same as a definition for a simple dimension, except I fix them or now I call it R, right? So I say, oh, if can I play this, this game that I played before, but for R balls, not for every size balls, right? So for example, let's take a grid with size 10. So the size, so I take the infinite grid on the plane such that the edge side, it has length 10 rather than one, right? Now I claim that if, for example, I take the one dimension of this, the one dimension of this is less or equal than one. Of course, the asymptotic dimension, since it's a grid, well, the asymptotic dimension is two. But if I take this one or even two dimension of this, the two dimension of the grid is less or equal than one, right? And the proof is obvious. I think you just, since you look at this at such small scale, at the one scale, it's easy to find this, the composition, right? I just decompose it into two sets of different colors. If you like this red and the green sets, and if I take every one ball, it can get set at most two of the sets, right? One green and one red, it could be entirely set at red or a green. Okay, so this is a somewhat finer definition, but it's clear that the asymptotic dimension of a graph is less or equal to n if the r dimension is less or equal to n for all r, right? Okay, so if I can play this game for every r for a given space, this, in fact, is exactly the same as the definition of asymptotic dimension, right? It is just a technical device. And we have a lemma, for example, just to get used to this. So let's say I have a connect graph index, a subset. So I would understand the dimension of the subset if there is some n such that the growth of the graph, so just for this specific n, right? Just a fixed n satisfies that the growth at n is less than one over two r times, and then the r dimension of fakes is zero. So, and now the reason is, so the proof is immediate, so it's just to digest this notion if you'd like, the reason is that if I take an r-connected set whereby r-connected, I just mean that, oops, sorry, that distinct dot-sided distance at most r, then r-connected sets have diameter less or equal to n, right? Just by my growth assumption. I do not have big blobs of these r-connected sets or I would contradict my assumption. But now these blobs are going to give me, these r-connected blobs are going to give me my cover, and the diameter of the sets of the cover is bounded, which shows that the r dimension of x is zero, right? Because if I take an r-ball in the sets, at most one of my blobs. Okay, so now in general, I'm going, I would like to apply some sort of induction, so I'm going to define what it means for a subset to be dr-separating, so maybe you can look at the picture, a dr-separating set, so this is my green set in this picture, is a set that breaks the space into r-connected components, no, r-components, maybe they're not connected, but which have diameter, oh, maybe, yeah, let's consider r-connected components, which have diameter at most d, right? So I just break up my space, so this is a dr-separating set. Okay, so let me give you an example of a dr-separating set, right? So let's take the plane and now take a thickened grid, right? So a thickened grid is a dr-separating set, so I take, I take, say, I can take a thickened grid so that the whole wide space in between has diameter at most a hand and r, and the grid has thickness r, right? So this is an example of dr-separating set. And now if I go back to my proof, that the dimension of r2 to the classical proof is less than 2, you might say, okay, this is a simple proof, but it's a bit bizarre because it uses these bricks, right? So when you, when somebody asks you about the dimension of r2 or something, you tend to think of the grid, right? So, and I'll show you this is the right way to think about it. In fact, forget the bricks, we can prove it using the grid. The asymptotic dimension of r2 is 2, so this is my proof. So here is a proof. We take this thickened grid, right? And now since the r dimension of the grid is 1, the r dimension of the thickened grid is also 1. If I take an r ball, I can have a cover of my thickened grid that intersects at most 2 of my sets of the cover, right? It's exactly the same thing as we said before, that a grid has r dimension 1. Okay, but now if I think of this as the whole plane, so I have a cover of the green red pieces and also the white pieces, right? The space in between. Then you see that my r ball intersects at most 3 of these pieces, which proves that the r dimension of the plane is at most 2. But we can play this for every r, right? So this shows that the asymptotic dimension of the plane is 2, just using the standard grid. Okay, so in every mathematical talk, there should be one proof. And that was my proof that the asymptotic dimension of r2 is 2, a new proof of this. Okay, but now what is good about this proof, while one couldn't play with bricks when you have general glass of polynomial growth, we can play with these dr separating sets. So this is the strategy and I might not say much more about the proof. So the strategy works in general. If we fix an r and take a dr separating sets with the much greater than r, then the growth of x at the scale n comparable to r is of the form cn to the k minus 1. So this is the idea, right? As the grid, you know, our grid had a small r dimension because it had a small growth at some scale, as we did in dimension zero. So by induction, we have a bound on the growth of x. And this gives us a bound for the r dimension of x of these dr separating sets, which gives us inductively a bound for the r dimension of gamma. So I will say, I will go through the technicalities a bit. But this is the main idea, exactly what we did with the grid in R2. We took the second grid, right? And we show, okay, the second grid has a small r dimension for a fixed r, but then we could play this game at every level, at every scale r. So this gave us the r dimension of R2. And this is exactly the same argument that is going to work for graphs, for graphs of polynomial growth. So now we need... Hey, I'm sorry, Panos. Can you go back to the dr separation, separating set definition? Yes. You want the complements to be a bunch of connected pieces of diameter less than d? R connected pieces, yes, of diameter. You need to have enough of them, so that some neighborhood of it covers the whole space, or something like this. Otherwise, I could just take a hold. Yes, yes. I mean, we have to play, so this is a definition, so now we will need minimal side sets, right? I see, I see, I see. And it's only if you have a good dr separating set that you can actually finish your proof. Okay, thanks. Right, yeah. So I need to take, yeah, that's a good remark, right? So yeah, this is an empty definition. I could take the whole space, right? It does nothing. But now, if you, now the game you're going to play is you want to look at this set, and you're going to take the smallest possible side set, right? Now the smallest possible side set for a fixed R, right? Okay. So the smallest possible side set is going to have smaller R dimensions, so you have to work at each place, at each scale separately. And this will do, okay? So thanks, thanks, I got it. Okay, so we have, again, we want to look at minimal the R separating set. And now this is where it gets a bit technical because your minimal separating set might be minimal when you intersect it with a ball of radius n, but not minimal anymore when you intersect it with a bigger ball and a bigger ball and a bigger ball, right? So for each ball, you can say, okay, this is a minimal D as a parading set. Minimal in the sense it has the smallest number of vertices, but now when you change the size of the ball, maybe the sets change and you might not have exact convergence, right? So that it works. So, but this is not a big obstacle, right? It just says that you have to define the minimal separating set carefully. And this is what this definition does, right? It says, okay, we define it so that it works for every ball and so on. And then we take a limit, okay? And then if you take this careful definition for this to work, this minimal D as a parading sets do exist, right? And this is what we want to work with. And then the main technical lemma is if we have, again, a bound of the growth, so we have an R, so we fix this R that we want to play with. We want to bound the R dimension for this R. And now we assume that for some n0, that is much bigger than R, 4 to the m times R, we have this nice inequality for the growth. So since this coefficient 1 over 4 to the R to the k is quite small, you can think that n0 is much bigger to get this n0 is much bigger than R. But then if we have a subset that satisfies this, then its R dimension is less than k minus 1. So if we have this small growth, the R dimension is less than k minus 1. Okay, so how do we prove this? We'll take a minimal separating set now, 2n0 R separating set of x, right? And the central inequality, the central thing to prove. So again, now I'm taking separating sets not for the whole space, the R separating sets not for the whole space, but for a sub space because to carry the induction, I need this more general setting, right? To take this separating sets for subsets. But I can do that. And now I claim that when we do that, the growth drops, drops farther. So this separating set, if my original head growth n to the k has even smaller growth. So a similar, satisfies a similar formula by, but it has a smaller growth. So then by induction by the same lemma, right? So the case n equals zero of the lemma is proven already, right? It was my previous result for zero dimensional sets. So the n equals zero case is done. So now by induction, if I have this bound on the growth, I am done, right? So now, so the crucial thing is to prove this bound for the growth, for the separating set, for the separating set y, but the idea is very simple. So again, I have drawn y as a thickened grid. If the growth of y is bigger than expected, then just look at surfaces around the point V, right? A sphere, a thickened sphere around V. Because of the polynomial growth, I have a control of how big are the spheres. So if the growth of y is too big, I can just cut by this sphere, by this annual, you say a thickened sphere. So I shouldn't say spheres. When I say spheres, I mean our neighborhood of spheres, right? Because we want the set to be d r separating. So I take this thickened sphere, I cut by it, I throw away what is inside the sphere, and they have a better, a smaller separating set, d r separating set, which contradicts minimality, right? So d is here two n zero, right? Okay, so this is a proof, right? So this proves this inequality. And by induction now, I have that the r dimension of my effects is bounded by k. Was it k? Yeah, by k minus 1. Yeah, it doesn't matter. So again, by induction, we have this inequality. And since the r connected components have bounded diameter, right, of x minus y, I have the dimension of x is bounded by k minus 1, sorry. So I get the bound I wanted for x. And of course, this implies the theorem now, right? Because if the growth satisfies this, gamma to the r over r to the k plus 1 is equal to zero, then for every r there is an m such that for this big n zero, I have the inequality, the exact inequality I had in the lemma. So by the previous lemma, the r dimension of gamma is less or equal to k. Okay, so this, this ends the proof, the proof, the theorem clearly follows by the lemma, right? Okay, are there any questions? So this gives us a sharp bound. So maybe some remarks, of course, we get the same theorem for manifolds of bounded geometry and polynomial growth. Now, if you, if you don't assume bounded geometry, of course, it's false because you can just take a graph and thicken it and make a two dimensional manifold of volume less than one, right? You can always do that. And then, or do I need an extra dimension? No, you can, yeah, you can have a two dimensional manifold of volume less than one and infinite asymptotic dimension. So, but I'm not sure, for example, if we assume non-negative verity curvature, if we can get rid of the bounded geometry assumption in this. So maybe some geometry in the audience knows the answers to this. Anyway, this is the end of the first half of the talk. So I will move on unless there are more questions. Okay, so this is about the Stoic inequality in this course back to 30 years ago. So I was discussing with Toma, and of course, the only thing we shared about that time was understanding hyperbolic groups. And at some point, he told me that, you know, there is this filling paper of Gromov that is not well understood by the community. At the moment, I didn't make anything out of it, but I still remember this conversation. And then 30 years after I came back to this, I tried to understand this filling paper. So here is a definition. Let M be a non-simply connected closed Riemannian manifold. The c-stall of M is the length of the shortest non-contracted loop in M. So in this example of the torus of this metric on the torus, I have drawn the c-stall. And now there is a result of Lerner from the 50s which gives a bound for the c-stall of the torus. So the c-stall of the torus is bounded by this constant times the square root of the area. And this is optimal, and it's one of the very few cases where we have an optimal, an exact, sharp bound for the c-stall. Now, some other remarks. So I gave you an upper bound for the c-stall, in the case of the torus at least. There is no lower bound in general, right? Because you can always pinch. I mean, we are looking for bounds where we are allowed to pick any Riemannian metric, right? So the bound of Lerner is valid for any Riemannian metric you put on the torus, and it is an upper bound. Well, we cannot hope for a lower bound because we can always pinch our curve, right, and make it as small as we like. And the second remark is we cannot hope to get an upper bound for all manifolds in terms of the volume. For example, if you take s2 times s1, you can arrange so that the volume is equal to 1, and the c-stall goes to infinity, right? Just take a very small s2 and a very big s1. So there is no upper bound for every manifold. And now the main theorem is Gromov's from 85. So for any n-dimensional closed as spherical Riemannian manifold, the c-stall is bound in terms of the volume. By the formula, the c-stall is bound by a constant that depends on the dimension times the volume to the 1 over n. So this 1 over n shouldn't confuse us because, of course, any such formula should be invariant by dilation of the metric, right? So if we multiply the metric by lambda, the c-stall is going to be multiplied by lambda and the volume by lambda to the n. So we expect this, so the only significant term is this c to the n, and Gromov gave the bound c to the n, c of n is less than 6 times n plus 1 times n to the n times square root n plus 1 factorial. So just to repeat myself, another way to say it is the c-stall is bounded by the volume. So what this theorem says, let's fix a dimension and now let's take any manifold of dimension n, just pick any spherical manifold of dimension n of volume 1. So you're allowed to pick any manifold, you don't restrict the topological type, and you're allowed to put any metric on this manifold, any Riemannian metric, so that the volume is 1. Now if you do that, the theorem tells you, okay, the c-stall, no matter which manifold you picked or which metric is bounded by a constant c of n. So this constant depends only on the dimension. Okay, so any questions? Well, I guess, I hope it's clear. And now Gromov's proof, so this is a very simple looking theorem, you would expect a very simple proof, perhaps, but this is not quite the case. So the main new ingredient that Gromov used to prove this was the filling radius. Now I have done a picture, so the filling of a torus embedded in R3, so you see the torus as a chain and now you take a homological filling of this and you see in which neighborhood of the torus you can feel, right? So the filling radius is a smaller star such that in this neighborhood you can feel. However, to work this out, of course, manifolds don't sit in Rn, so what Gromov did, he used the Kuratovsky embedding from M to L infinity of M. So you embed them to the space of L infinity functions on M. And then you look at the filling in this infinite dimensional space, and this is how you define the filling radius. And he proved the following two facts. The first is that the system is bounded in terms of the filling radius. And the second, he used isoparametric inequalities in this infinite dimensional space to show that the filling radius is bounded by the volume. So now the two together, of course, imply what we want, right? The system is bounded by the radius, by the filling radius, and the filling radius is bounded by the volume. So the system is bounded by the volume, well, by c to the n times volume to the 1 over n. So the first part is a relatively simple argument, while the second part was quite technical but was simplified relatively recently by Wenger. So there was another proof. Okay, however, it's still quite puzzling in order to prove such a simple geometric statement that the system is bounded by the volume to use the embedding in L infinity of M, right? So I would like to, so the second, in the second part of my talk, I will try to give a different approach to this systolic theorem, to this systolic inequality. So we need another notion, Eurasian width, which is, again, a notion of dimension. So x has Eurasian q width less than w, if there is a q-dimensional simple complex in the continuous map from your space x to the q-dimensional complex y such that every fiber, so every pi inverse of y has diameter at most w in x. So I have drawn some pictures of, say, two-dimensional spaces that, in a way, look one-dimensional. So they have small one width, right? So you see this torus on the left is close to a circle, and my strange sphere is close to a y, to a three, a y-shaped three. And now Gromov also showed that the filling radius is bounded by the Eurasian width. So we saw before that a bound on the filling radius bounds the systole, but now if we have a bound on the Eurasian width, we can also get a bound on the systole. And so now Gromov made several conjectures related to this work, and Goose proved one of them. So let's denote by, so v of r is similar to the growth I defined earlier, right, is the largest volume of an symmetric ball of radius r in a manifold m. And now what Goose proved is that there is an epsilon n such that if locally, if there is a radius r such that every ball of radius r in m has volume at most epsilon n r to the n, so if it has very, if locally we have a very small volume, then our manifold collapses to something lower dimension. So the Eurasian width, then minus the Eurasian width of our manifold is less than that. Okay, so again what this says is if locally the volume is very small, then we collapse, there is this map, this continuous map to a lower dimensional complex. So I have drawn an example with a thin torus, right, which collapses to a circle. And now it's a corollary. We get something that is similar in flavor to what we do in group theory. So if we have a closed spherical Riemannian manifold, then the volume growth of the universal cover is at least epsilon n r to the n for r, right. So it's bigger. So this reminds us of manifolds of groups of polynomial growth, right. So these covers grow at least polynomially, but it's a different theorem. It doesn't involve actually group theory. And the corollary follows easily from the theory, more to the fact that the filling radius of the universal cover of a spherical manifold is infinite, which is proven by a similar argument as before using homology, but I'm not. So yeah, there is an argument for the first line, for the right side of the, for the left side of the proof, but the filling radius is less or equal than the universal width, which is infinite, right. So the volume cannot be epsilon n r to the n. So it follows. So the corollary follows by contrapositive from the theorem, right. Since we have infinite filling radius, the universal width is infinite. So the volume cannot be epsilon r to the n for any r, right. So it has to be bigger. Okay. Now, the stoichiometric quality also follows from Good's theorem by a cheap trick, right. We just choose r such that epsilon n, remember epsilon n is a constant, right. It's just a constant that Good's calculated so that epsilon n r to the n is equal to the volume, then the c-stole is smaller than the garrison width as we saw before, but the garrison width is less than 6r by Good's theorem, which is equal to cn volume to the one over n. So this cn just comes from the epsilon n of the previous line. So it immediately follows. So would you get another proof of Glomov's theorem, except Good's theorem somehow builds on techniques similar to the one in Glomov. It doesn't quite use a Kudatovsky embedding, but again we have an embedding in mysterious high dimensional space. Now, I would like to mention in passing the relationship with stellar curvature, there is a much stranger conjecture. So we showed that if we have the universal cover of an asphherical manifold, the volume growth at least epsilon r to the n, but the strongest conjecture is that if you have the universal cover of an asphherical conjecture, then the volume grows at least as in Euclidean space. There is nothing smaller than that, right. The volume has to grow at least as fast. And remember this volume growth is we take the supremum, so we move the base point around and we take the supremum, right. So this is the notion of volume here. And this is open, right. We don't know much about this, except one could imagine that this would imply the classical conjecture of Shenyao that MN does not admit a metric that such spaces do not admit metrics of strictly positive scalar curvature. Now there has been a lot of recent progress on this conjecture of Shenyao. So for n equal to 2, this conjecture it follows from Gauss-Bonne, for n equal to 3, it was shown by Gromov Lawson in 83. And now recently there have been breakthroughs, so Chodos List showed it for n equal to 4 and Gromov for n equal to 5 in a couple of months ago. So yeah, this is a very active field of research. And as I said, the shot version would imply these results for every n, except this sharp version above is not even for n equal to 2, right. So this seems to be a harder problem. Still it's still a beautiful question. Okay, so let's go back now to trying to find a different proof of Gromov's theorem. So Karrashev sent Gergo an email to Nobotosky explaining that in fact we can prove this bound that this is totally bounded by Eurasian width without the filling radius. We don't need the Kulatosky embedding, one can just use a direct homological argument and prove this that this is totally bounded by the Eurasian width. Still of course, so that's good progress, but still we have to have a bound for the Eurasian width. And now I would like to discuss this in a more general context. So I'd like to define the n-dimensional house of content of a subset of a metric space. So this is the infimum of the sum of r i to the n among all coverings of s by countably many balls b x i r i. So this is a similar definition as a house of measure and it's easy to see that it's then a house of content is bounded by the house of measure. If s is an n manifold, this is not, this hasn't been studied as much by analysts. One reason is that it's not quite a measure, but here it would play the role of volume. So if you have any metric space, so one good thing about this in contrast to the house of measure is that if you have any compact metric space, it's always finite for every n. The house of measure easily become infinite, but if you have a compact space, the house of content is going to always be finite. And okay, so again, I will use this as or rather good to introduce this as a substitute of volume in the very general setting of metric spaces. And he made the conjecture that his theorem, which was proven for manifolds holds in general for proper metric spaces if we replace volume by house of content. So now this is a purely geometric metric space geometry statement. And now this conjecture was proven by Lysiak Yokumovich in Nabutovsky-Rotman in 2019, but they use again goods argument, which is quite involved. It uses higher dimensional spaces and so on. And I would like to state instead the more direct proof. So now again, the idea of the proof, or maybe I should say something more before going to the proof, goods conjecture. So it's in a way, so the house of content has some advantages. So for example, if you take a graph and take a thickening of it, say an epsilon thickening, right, but n dimensional thickening, you see that this case, very small n content growth locally. So the theorem would tell you that or one content actually locally. So the theorem will tell you that it collapsed to one dimensional space that if you have an epsilon neighborhood of a graph, no matter in what dimension in our end, the theorem will still tell you that it's going to collapse something one dimension. However, if we use the classical notion of volume, goods theorem would only tell you that this epsilon thickening, the original theorem will tell you that this epsilon thickening collapses something n minus one dimension, which is not what you want, right? So this is another advantage of the content instead of measure, right? It gives you the right dimension of collapse. Okay, so let me try to give a stretch of the proof. So let's take a very specific case and now let's say we have a plane and we thicken it, say in R3, so we have a thickened plane. So the volume, so it's a three dimensional, my space now is three dimensional, so volume or even content for any notion of volume, but you can just think of the usual Riemannian volume. It doesn't really matter for this stretch. So the volume grows like R3 because it's a thickened plane, but epsilon times R3, right? Because it's very thin, right? So for some small epsilon, the volume is of the form epsilon times R to the three and now assuming this, assuming that we have this small local volume growth, we want to produce a map from M to something two dimension, right? A two dimensional, simple complex so that the fibers are small, that means pre-images of the point have small diameter. We want to collapse this to something two dimension. Okay, so we'll do something similar to what we did before. We'll take a thickened grid, maybe my picture could be misleading. So of course, since we have a thickened plane, this thickness in my grid goes, say in the vertical direction, but the whole point, again, is we take a thickened grid that cuts the plane in pieces of diameter delta of some small diameter. And now the crucial observation is that we can find such a grid, which has small local volume growth, right? So the grid, there is such a grid, which you can think of it as being two dimensional now, right? And it is actually in this stretch of proof we had this thickened plane, but now we cut by a grid in pieces. And of course, we only need two dimensional grid to cut it, right? This thickened plane, we don't need three dimensional grid. And I claim that we can find such grid so that the growth at scale R is some small epsilon times R square. Okay, but now let's believe that that we can do that. Then by the indexive hypothesis, we have a map with small fibers from G to a one dimensional, from this grid to a two dimensional simple shell complex. Okay, so we've got this map and we want to extend now to the whole space. But now if you use one piece, then the boundary, the image of the boundary of the piece is contained in the sub complex. And now of T, which was the one dimensional complex I mapped, and T is an absolute neighbor who retracts since is a sub complex. So it's implicit cone is an absolute retract. So we may extend them up from the boundary to the whole U. Now all this last part is just classical topology. So that I had to learn. The point is maybe you remember this extension theorem. So if you are not familiar with this classical topology stuff, absolutely retracts are spaces where extension theorem applies roughly, right? So you just need conditions that allow you to extend continuous maps. And it turns out that if you have a cone over N, N, N, R, then this is an absolute retract and you can extend your map. So this is basically just by classical topology. This inductive step allows us to extend our map to the whole space and we have a control on the fibers because remember I cut my space in small pieces, right? So my fibers cannot be bigger than the pieces that I cut cannot have diameter bigger than the pieces. And this is, oh, this is a proof. Okay. And now I see I'm, I went slower than I thought. So I won't have, I'll go through the technical details very, very quickly. I'll just go through the slides. So then equals zero case is this before, right? If the content is, is bounded, then the Euro, the zero Euro, which is bounded. And the reason is again, we cannot have big blobs, right? Otherwise, I would contradict my, my bound on the content. So we cannot have big clusters. And now the other ingredient of the proof is the core area inequality, which is, which is, we just say that if you want to have an estimate for the content, well, you just integrate the size of the sphere. So the side, the integrate. Yeah. So if you have a ball, you can estimate the content of the ball by just integrating the content, the lower dimensional content of the spheres. Right? So this is like Fubini. Anyway, so, so again, the proof just uses minimal separators. It tends, it, it, it turns out that we cannot have exactly minimal separators, but it's suffice to have epsilon minimal separators. And the idea for showing that the separators have small volume growth is exactly the same as I said before. If they did have big volume growth using the core area inequality, you could cut a sphere and contradict minimality of your grid. Right? So this shows that your grid has small volume growth, which is the crucial step needed for the proof. And now, and an application. So Nabutovsky used this proof to improve the constant in Gromov's historical inequality. So now we know that this is always bounded by not the same constant, but, but just by n, right? Times the volume. Okay, so sorry, I think I went a bit over time, but that's the end. Thank you very much.