 OK, so thank you for the invitation. It's very nice to be here again in Montevideo. And my talk will be about something which is not directly related to dynamics, but I will try to make the relation to a kind of dynamical system that appears in the middle. So the main object of my talk will be this kind of things. This is the famous Penrose tiling, which is tiling of the plane by using two pieces, two polygons. And the special feature of this tiling is that it is a periodic. So there is no translation of the plane that sends the tiling into itself. This was not the first examples of such a thing. But I think this is the most beautiful one because it uses only two pieces. And well, it is still an open problem whether you can build something similar using just one piece. This is the so-called Einstein problem. And well, this is an open problem. There have been some announcements in both directions. So in the recent years, but it is still an open problem. The only thing that it is known is that it is not very hard to prove that if you have a polygon that tiles the plane only in a periodic way, then this polygon has to be non-convex. So convex polygons, if they tile, then there is a periodic tile. And well, there are many special features of this tiling. One of these is that it has somehow a self-similarity property, so you can build this tiling by iterating some procedure. So you start with the so-called, this is the magic triangle. It is called the golden triangle. And you take the division of this triangle using following the golden proportions. And then you get some configurations that if you rescale, then at the limit what you get in one of these rescaling is the pendulum tiling. This is one way to produce the pendulum tiling. But actually, there are many others. And as I said before, well, this has been very interesting because it was discovered that there are real quasi-periodic tiling. I mean, in the real world. So the discovery of quasi-crystal in the 80s showed that this kind of geometric configuration actually appeared in nature. But the pendulum tiling was not the first in history. And actually, this is somehow an equivalent form of the pendulum tiling. And this was discovered only, I think, five years ago in Iran, in a city which is called Isfahan. This is the temple Dar Hifam, which is from the, I think this is from the 11th century, if I guess. And this is, believe me, this is really equivalent to the pendulum tiling in a certain way. But this kind of knowledge disappeared, actually. And nobody knows how they produced such a thing. But actually, there are many, many copies of this kind of tiling in Isfahan. Now, the mathematical object that I want to concentrate on is what I call the delon sets, or delone sets. So what is the delon set? So keep in mind the pendulum tiling and just keep in mind the vertices of the polygons that appear in the pendulum tiling. And the set of vertices with the Spanish accent, sorry, is what is called a delon set, which is uniformly separated and coarsely dense set. Uniformly separated means that if you take any two points, then there is a minimum distance or a positive infimum for the distance that separates any two points in the set. And coarsely dense set means that almost all the plane is covered by this set, which means that there exists a row, a positive row such that any point in the plane is a distance at most row of a point of this delon set. So this is the definition. This is a kind of discretization of the plane. And of course, you can define the very same thing for any metric space. So a delon set is uniformly separated coarsely dense subset of the metric space you are dealing with. It's a kind of discretization of the surface. And of course, there are obvious examples. You take C2 inside the plane. This is the delon set. And again in the plane, if you take the set of vertices of a tiling that uses only finitely many polygons, finitely many pieces, then, of course, the infimum of the, say, of the systoles of the polygon is going to be the row. And it's going to be the sigma. And the row is going to be the diameter of the maximum of the diameters of the polygons. And let me go very, very quickly to the main question of this. This is very interesting because it was formulated in an independent way by Forsenberg and Gromov with very different motivations. The motivation by Forsenberg was a motivation coming from dynamics. It was related to some recurrence times for some flows on the torus. And the motivation of Gromov was mostly geometric. And it came from geometric group theory. In geometric group theory, there is a very nice theorem which is not totally easy to prove, is that if you have a group which is, say, by ellipse equivalent to C2, then actually it is C2 up to some finite index of group. And so the question of Gromov in this view was very natural is whether every delon subset of the plane is by ellipse equivalent to C2. There is a by elliptic map, the finite on the delon set, which sends this set of points into C2. This question was answered by Burago and Kleiner, and again independently by MacMillan. And the answer is no. There are strange delon subsets of the plane. There are delon subsets of the plane that are not by ellipse equivalent to C2. And actually this is very recent work by Dmon, Calusa, and Copeca. They managed to prove that this is still true for elliptic equivalent, which is something very strong. They managed to prove this dealing with a problem coming from this discrete geometry, from pure discrete geometry. And a way to restate this theorem is saying that there is no uniform bound. So let me write it. So there exists no C such that for every subset of C2 with finite cardinality, there exists a elliptic map into, say, let me take a subset of n square points. So there is no uniform constant for which there is a elliptic map, C elliptic map from S into the square of length n. And this is very, very impressive because you can think that this cannot be true because if you have a set S, then this is a very sparse set on the plane. But in order to make this set coming into the square, you have to do many things to wrap around on the plane so that the elliptic constant has to go to infinity. So this is a very interesting work. And actually, I will try to say a few words about the proof because the proof is not very different from the one of Burago-Klein and Agma-Mulen. It's a kind of technical improvement but which is very, very clear. Anyway, so let me give some concrete features about the loan set. So this is an exercise. If you take the loan subset of a plane, then it is by elliptic equivalent to a subset of C2. This is very easy to see because if you have a loan set, then you make each every point on the set to correspond to the closest point in C2. Maybe you have to rescale in order that this operation makes sense. And doing so, you get something which is by elliptic equivalent. The second exercise is not that easy. It is not tautological. One may think that this is just a consequence of the definition, but it is not. So the image of the loan set by a elliptic of the plane is at the loan set. This is not obvious because maybe by taking a elliptic map, you can create kind of holes in the set. But actually, this is impossible. But it doesn't follow from the definition, I mean. You have to work a little bit too. Yeah, so OK. OK, thank you. So this is for C2. So actually, it's correct. Because you could start creating holes in the image. And at the end, the set you get is not coarsely dense. This cannot happen. But you need some. I have some argument somehow tautological argument to do that. But thank you for the question, because all the things that I so I forgot to say something. So the question of first seminal gromo makes sense in C2 because in the one-dimensional case, it is an exercise to show you have the loan subset of the real line. Then it is equivalent to C. This is just moving points along the line. So this is quite elementary. So it makes sense on the planar. Actually, it makes sense on CD for every D larger or equal than 2. But the proofs are mostly the same starting from dimension 2. So I will only concentrate in the case of dimension 2. And actually, there is a nice open problem for coming from analysis. So there are many partial results on this problem. Is whether there is every by-lipschitz map defined on, say, on C2 or any other, the loan set. But actually, it is already open for C2. Whether these maps can be extended into by-lipschitz homomorphism of the plane. So whether you can, starting with something which is defined on a discrete seed, to extend it into a homomorphism of the plane. This is an open question. This is a question coming from analysis. It is known that the answer is yes, if the by-lipschitz constant is very close to 1. But in general, it is open. So here I will only concentrate on by-lipschitz maps defining on discrete sets. And I will say a few words about this general problem for particular cases. Well, what happens with the penrose tining is that it is not only at the loan set, which is a periodic, but it has a complementary property, which is very nice, which is almost periodic. Which means that somehow every pattern in the tiling can be appeared. So it is copied in almost every part of the tiling. And so the precise, the technical notion here is what is called repetitive. So the definition is there. You take the loan set, and you say that this repetitive is for every small r, there is a capital R, such that every pattern of diameter small r appears in every ball of diameter capital R. And this is what happens with the penrose tiling, and this is what in general is referred to as being repetitive. And this is a theorem that we proved some years ago with Maria Isabel Cortez. Is that there are many loan sets that are not by-lipschitz equivalent to C2 in the plane, but which are repetitive. So they are something like the penrose tiling, but they are not equivalent to C2. And the point is that the penrose tiling, it is equivalent to C2. And this is a theorem by, so I forgot the name of the guy who proved for the first time. But this follows from a more general theorem by Alistair Coronel and Gambodot, which says that if you have a loan set which is linearly repetitive, then it is by-lipschitz equivalent to C2. And this is the k for the penrose tiling because of this self-similarity property. And linearly repetitive means that in this definition above, the capital R depends linearly on the small r. If you have this property, then you know immediately that the loan set is by-lipschitz equivalent to the standard lattice. And the proof is very nice because it is a mixture of many, many arguments coming from different fields. There are even partial differential equations which are solved in the middle of the proof. This is really very nice. And the point is that if you have points, then you have to transform this point into weight, into somehow into density functions on the plane. And then you have to move a density function into another one. And this is encoded by a partial differential equation. And it is proved that in some cases, these partial differential equations have solutions. And then you put all the things together and you get the theorem of Alistair, Coronet, and above. A small r and capital R. No, linearly it's bounded above by a linear. So it's something like a small r is something like this. But if you change the condition by, say, 1 plus epsilon, so the theorem is no longer true. Actually, the theorem of Gambo-Doc, Coronet, and Alistair can be improved. And for the case of, you take a linearly repetitive the long set of the plane. And actually, there is a binitism amorphous of the plane that sends the long set into the standard lattice. So the open question, which is in general open for the long set, can be solved in this case. And the proof is somehow the same as the one of ACG. Except for that, you have to use all the things that are already proven and point to remark something which is quite technical in the middle of the proof. There is no new idea in the proof of the last theorem. But the point is that we wanted to add the repetitivity to the construction of Vurago-Klein and McMullen. And we managed to do that. And actually, we managed to do that in many different ways. It is here where dynamics appear. The point is that you can restate the property of being repetitive in dynamical terms. First, you take the family of the long sets of the plane. And it is not really a metric space or a compact space, if you wish. And not in general compact space, but there is a natural metric, a natural distance in the set of the long sets. So you say that, or a natural topology, you say that two long sets are closed. If when you restrict to a very big ball, then what you see are finite sets of point which are very close, one from the other, in the house of distance. And then if you want this set to be closer and closer, then you pick larger balls and you select an even smaller house of distance for the trace of the long sets on the ball. This is a classical construction that I think goes back to Chaboti. It's in general referred to as the Grom of Distance, but actually this is a much older idea. And the point is that if you take in this, this family is very large, it doesn't give a compact space. But if you remember that in the definition of the long set, there were two parameters, the sigma and the rho, which are corresponding to the final properties of the long set. And if you fix this constant, so at least you make them to move in compact sets, then what you get are compact sets of the long sets. In particular, if you take a single long set and you translate this long set, then you get new long sets. But all these long sets move in a compact family of the long sets. And the property of being repetitive is equivalent to that the action by translations, by R2 translations on this set of the long sets is minimal. This is exactly the same. It's really tautological. You just take the definition and you will see it's really the same. So the point of our work with Maria Isabel was trying to relate the original question, the question by Gromov, the geometric question, to dynamical properties for this action. But actually what we prove is that there is no relation at all because you can manage to do almost whatever you like. So you can construct repetitive long sets that are not by literally equivalent to the standard set and for which the action is minimal because they are repetitive and for which the action is, for instance, uniquely ergodic. And even more than that, you can prescribe any set of invariant measures. So any Chaget Simplex can be realized as the set of invariant measure of repetitive, non-rectifiable the long set. So you can do whatever you like. And somehow there is no relation between the geometry of the long set and the dynamical properties of the translation action on the closure of this set. This is the point. OK. But what is nice here is that you can formulate this question in another way. So remind that if you have the long set on the plane, then you can move this plane. You can promote this set into a subset of C2. So that the question makes sense only for subsets of C2. And the question is whether a subset of C2, which is coarsely dense, is by which is equivalent to C2. And the answer is no, because of Burago-Kleiner theorem and McMullen theorem. But this question, C2 is a group. So this question can be asked for any group. So the question is, what are the groups that contain a coarsely dense subset? So coarsely dense, you take a group, a finite regenerator group, and you take the word distance on the group. And you take a subset, which is coarsely dense, which means that every ball of radius, say, 11, contains a point in the subset, whether such a subset can exist, which is not by which is equivalent to the whole group. And this is the case for C2 and for CD, because of Burago-Kleiner theorem. And actually, this is the case also for some solvable groups, like the Bauller solitar and sol, and some nilpotent groups. And actually, the conjecture is true forever in a potent group, which is not essentially Z. And this is mainly the word by Tulia D. Martz and collaborators. And for the case of solvable groups, it's not that easy to conjecture anything, because for the case of the potent group, what happens is that they are mostly realized as lattices in the group. So somehow, you can repeat the arguments of Burago-Kleiner. But for the case of solvable groups, there are many groups that are not lattices. And then it is not clear what happens in general. But what is interesting is that these were examples of very small groups. And this is a necessary condition, because there is a very nice theorem in the thesis of Kevin White, which it says much more than this. But if you translate into this framework, the theorem of Kevin White says that if you have a group with coarsely dense subset, which is not by this equivalent to the group, then the group is non-aminable. No, sorry. Then the group is amenable. Then the group is amenable. And the reason, well, this is somehow natural, because in non-aminable groups, what happens is that if you move from one ball to the next ball, then there are many other points that appear. So you have a lot of, somehow, volume that is being created. And moving very strictly the points, then you can manage to actually achieve this by this equivalent, just by translations, in a bounded concept. This is an illustration of the theorem of Kevin White. So remember that in the plane, you take C2. And you take, somehow, randomly points in C2. And then you get a point, a set. You prescribe a condition so that it is coarsely dense. And then you get something which is not by this equivalent to the plane. And here, if you take this picture by, this is by Escher, you can do more or less the same. So there are some, these are the angels and devils. How it is called this? Yeah? So you pick, for every devil, you select one or the two ices, and you get at the long subset of the hyperbolic plane, which is coarsely dense because you are selecting one or the two. And for most choices, you will get something, we will get sets. For every choice that you make, you will get a set that is by each equivalent to any other shoes. For any choice that you make, you get something which is, which comes from a universal model. So, for instance, the universal model could be that you pick the two ices of every devil, okay? Which is something that doesn't happen in the plane. So there is really a difference here. Okay, and this is, this is another side of the history. And this is the key, the key side, at least in the view of Browclain and MacMillan, because this is not a theorem on geometry, this is a theorem on analysis, okay? And I will try to explain why. So there is a classical theorem by Radimacher, which says that if you take a elliptic map, elliptic function actually, then it is almost everywhere differentiable. So this is a famous and classic theorem on harmonic analysis, which is somehow, again, this is easier in the case of dimension one, but in higher dimension is more difficult. But what they prove Browclain and MacMillan is a theorem on analysis that translates into the theorem on the, on these, the long sets. And the theorem is that on the plane, there are positives, sorry, there is a mistake there, positive and continuous function, which are actually bounded away from zero, okay? There are not the Jacobian of a bilithismomorphism, okay? And this is the theorem. And this is actually equivalent to the existence of the long subset of the plane, which are not bilithism equivalent to C2. And the reason is that when you have a delon set, what you, what you can create is somehow a density function of this delon set. And then you rescale, you rescale the set and you rescale the density function. And at the end, you get some function which is continuous because it's a limit of density functions. And this function is not the Jacobian of a bilithismomorphism, and then you come back and you get, as a conclusion that the original delon set is not bilithism equivalent to C2. This is the way how they prove the theorem. So the proof is not constructed. At the end, they don't get, at the long set, what they get is a density function which is not realizable as the Jacobian of a bilithismomorphism. And this is why they prove. If you go into the paper, what you will see is lots of lemmas about analysis, okay? And this is a very, very nice theorem. And actually it's a very technical theorem because if you, for instance, if you, instead of having a continuous function, you get, you have a holder function, then this is no longer true. This is a theorem by Moser. If you have a holder function which is bounded away from zero and positive, then it is the Jacobian of a bilithismomorphism, okay? Yes, this is somehow related to this, okay? When you do the rescaling, you see, you translate the holder condition and density function into a condition like this one, okay? So this is a very technical theorem, but what is nice again, this is a theorem by Rodolfo Vieira, is that this is, when bad things happen, they happen generically, and actually this is a generic property of continuous function. If you take a positive continuous function, in general it is not the Jacobian of a bilithismomorphism. And actually because of the Calusa and the other guys, recent theorem, this, you can replace bilithism, by ellipses and you still get a theorem here, okay? Well, the point of comparison is the theorem of the fundamental theorem of calculus. If you have a positive function on the interval, then you integrate this function, you get a diffimorphism, and the derivative of this diffimorphism is the original function, okay? So what this theorem says is that there is no analog of the fundamental theorem of calculus in dimension two. This is one way to say, to restate the theorem, okay? But the point is that it is this theorem that is proved by Burago and Klein and also by MacMillan. The proofs are somehow similar, but they, there are some tricky differences, but I will retain the proof of Burago and Klein because it is the proof that is closer to our view of this thing. Well, the proof is very technical. It's really very technical, and what we did with Maria Isabel was just to restate the proof of Burago and Klein, which is a theorem on analysis into a statement on discrete subsets of the plane. So we are trying to restate at least the key lemmas of the proof into lemmas about functions defined on discrete sets and doing some course differentiation and things like that. These are things that are well known in geometric group theory, but it becomes technical, and this is an example of a lemma that you can find in our paper. I just took a picture of the lemma and this is the statement, and I know it's very hard to read, but I will try to explain a little bit what is in this statement without any notation here. So there are many proofs of Radimacher theorem. Radimacher theorem is the theorem that says that you have something which is Lipschitz, then it is almost very well-differentiated, okay? And but one of the proofs that I like a lot, which appears in the appendix of Gromov's book, I think it's approved by Stephen Cimes, says that if you have something which is not differentiable, then when you take the, I think this has a name, the increment, what is the name for this expression? The increment, it has a name. The increment quotient. So the point is that when something is differentiable, the increment quotient converts to the same limit at every scale. So you can take here T, you can take one over T, one over N, or you can take one over E to the E to the E to the N, and you will convert to the same limit because it is differentiable, okay? And the key point in the proof of Radimacher theorem is that if there is no derivative, then there are some sequences of scales where if you pass from one scale to another one, then these increments will explode, okay? This is a classical idea for Radimacher theorem. And this is somehow what is happening here without saying what is exactly this condition, so you can see that there are some, let me see. So here you can see that you are adding some M over P, and here we are adding some M, which means that there is a change of a scale, okay? And here we are saying that if moving from this scale to this other one, from one scale to another one, it doesn't change too much the increments, then there is an almost derivative, okay? This is the statement in this theorem. So let me draw a picture to explain a little bit what are these sets here. So here you have a rectangle, you have many squares, and you will compare the increments between two points here and two points here, which are at different scales. And what I am saying here is that if this is somehow M over P and this is somehow the M in the statement of the lemma, and if the increments, so what is written here is the increment at this scale, and what is saying this is that if the increments do not increase too much from passing this from this small scale to this larger scale, then at the end what you get is an almost derivative, okay? And when you get an almost derivative in a discrete sitting, what you are saying is that somehow your map is like an affine map, okay? And since the map is somehow not an affine map, then you do the last trick, which is the following one. Now you will select your delon set and you will put many points here and a quantity of points, which is much smaller here and many points here. For instance, you select all the points of Z2 in this square and this square, but only a half here and so on. But what you get is that in some set, the map is almost affine. So when you look at the image, the image of this has to be very similar to the image of this, but this is impossible because here you have a quantity of points which is much larger than the quantity of points here, okay? So this is another point which is very important in the proof is that at the end what you are doing is to do a very large variation of the density of the function. And this is why at the end the density function cannot be holder, for instance, but it is allowed to be continuous, okay? So this is somehow the idea. This is the translation into a discrete setting. One of our main jobs here was, first of all, trying to understand but I would like a paper because the proofs are very short. At the very beginning we were thinking that it was wrong, but actually everything is correct. Except for one sentence which is in the paper, they claim that everything can be done in the Lipsy setting, but actually that way it cannot be done. And this is the recent work by Calusa and his co-authors. They managed to do that, but there is a very clever remark in the middle and the clever remark is that the density function you are getting at the end is bounded away from zero and when you integrate such a thing, what you get is not only a Lipsy function, but it is a Lipsy function with some supplementary conditions that are well studied in harmonic analysis and then you can make the matching of Bravo-Glenus work again and they managed to prove the theorem for Lipsy maps. But actually it's an 80 page paper in GAFA which has been published this year, but anyway. Now the point with Maria Isabel is having this lemma which is the discrete analog of a lemma in Bravo-Glenus paper. We start with these two patterns and then we put these patterns together along this rectangle here which is somehow the analog of this rectangle and then you make the variation of densities here, but then you have to do this again and again. So using this you have this rectangle and then you build a square which is going to be the analog of this square but in the larger scale. And then you do the same, but you put a square with this density in a larger square and then you repeat the construction. But the point is that again, the proportion of this length to this length is going to be unbounded in the limit because otherwise you will get something which is linearly repetitive for which the Coronel Gambodou Alice theorem applies and so it would be rectifiable. Okay, as I said before, we proved this some years ago, but actually what we would like to do is something more interesting, but this is still an open question and this is the natural open question. Okay, but it is still open and somehow we think that we have the ingredients to prove that it is such a thing, such a set exists. So there is a classical way to produce the long sets which are called the Catam project in the long sets. So for instance, well, this is a picture in dimension one, so in dimension two, but actually it projects into something in dimension one. So you take C2 and then you put a line such that the slope of this line is irrational and then you project the points that are, let's say at this and at most one over two to this line into the line here and doing so you get the long set. Of course, here we are in dimension one, so this is the long set by which it is equivalent to C, but you can do the same in any dimension, okay? And the natural thing to do in dimension two is you take a plane in dimension three, which is totally irrational, so that which means that the angles do not have any kind of linearly dependence among them and then you project into the plane, you get at the long set of the plane and the question is whether this subset of the plane can be a non-byelift is equivalent to C2, okay? So this is a very concrete open question, whether for every alpha and beta, which are linearly independent, well linearly independence is a condition which means essentially that the long set that you get is not has no periodicity, so it is a periodic, okay? It is a quite natural condition and this question is open, but what is very interesting is that it seems that the answer to this question is yes, I would like to say it is yes, but it is a very, when both are rational, you have something, yeah, yeah, when one of these is right, yeah, yeah, of course, of course, of course. So, yeah, because if one of them is rational, then you have in the other direction something in dimension one, then it is violated. And, but the point is that you have rational independence to get a periodicity, okay? Exactly, exactly, exactly. Okay, so for this problem, there is a theorem by, again, by Bouddhaug and Kleiner, which says that if the angles are geofantine, actually if one of the angles is geofantine, then the long set what you get is by which is equivalent to C2, okay? And this is a very nice result. And actually, you need less than geofantine, but you need some kind of arithmetic conditions on the angle to make this proof work. But it is totally open for newville angles. And the key, the idea would be trying to produce such a long set using somehow an approximation method, like a Katoch, a method, something like this. So you start with the, with sets which are actually periodic, associated to rational angles, which are very bad, which converge to very, which converge in the limit to newville angles. And then you will produce this, the long set as, as a limit of the long set which are actually periodic, but for which the periodicity is being broken in the limit, okay? And in the middle, one would be to be able to reproduce the Burawan-Kleinert estimates in order to show that this is not by which it's equivalent, but this is not that easy. And the point is that repetitivity comes for free, okay? This is an exercise. You have something which is, if the angles are rational and independent, then the long set what you get is non-periodic. So the only thing that remains to do is to select the good angles and to re-prove the estimates of Burawan-Kleinert to produce such a thing, okay? We are somehow optimistic, but we don't know. Maybe we are wrong on this, but anyway. So I will stop here. Thank you very much. You still be projecting on the assumption that it's aligned over the plane to rational reproducers, right? You project on. So instead of projecting on planes which are linking objects, you project on the surface, right? And maybe it's not being one of your... Ah, let me try to remind. I think we tried to do that, but it didn't work. But it is not equivalent. No. It is not an equivalent problem. And we tried to do that. Because after projecting into this surface, then you could project this surface onto the plane. But it is not exactly the same, but we don't know, okay?