 Well, thank you so much for the invitation. It's a great pleasure and honor for me to speak here. It's been amazing to have this great resource throughout the epidemic and I hope we keep it going long, long after and I can't believe it took us so long to figure out that we could actually do this and all get together every week and so it's nice to see a lot of familiar names and faces. Please ask at any time interrupt. I'd like to have as much as possible lively discussion. So, right, so I wanted to tell you about some things that I've been thinking about for the last many years about the interaction between these two subject sphere packings and arithmetic groups. But let me start at the very beginning. So don't look at these pretend you didn't see any of those pictures yet. So let's start at the very beginning since, well, a number of you have seen some version of something but there's always new stuff to there's lots of new stuff that I want to get to but I can't assume that everyone's heard me speak on this too often. So let's go back to the very beginning. So recall. Recall what an Apollonian circle packing is. Apollonian classical Apollonian circle packing. Let's do a little review. So this comes from the ancient Greek Apollonius. Apollonius was interested in a general problem about tangencies. He had a work called tangencies, which in fact was lost. But it spoke of taking three circles. Let's already start with them being mutually tangent. So if these are your three circles, the Apollonian problem is to with straight edge encompass construct a circle that's tangent to the three given ones. And its theorem is that there's exactly two ways to do this. So these are the two solutions to the Apollonian problem. And maybe I should mention, so this is construct, construct tangent circles by straight edge and compass, of course, straight edge and compass. His original method was lost. And a method that used only the tools available to him was rediscovered or invented anew in the 17th century or 16th century. Actually, very recently, so if you this is just a baby geometry, high school geometry. So Arthur Beragar, and I gave a construction gave the currently best known construction, so lowest number of moves, lowest number of moves, construction, and we conjecture that this is the best possible construction. But anyway, that's just that's just an aside. So once you can put in these tangent circles, you see you have another triplet of tangent circles, and you can put more tangent circles into these. And apparently this was first done by Leibniz. So Leibniz constructed what we now call an Apollonian circle packing, which is this fractal in the plane. Around 1700 or so. And then somehow people kept missing the number theory here. The number theoretic structure was uncovered by Frederick Soddy. So this is in the 1930s, maybe around 1936. He saw that there exist configurations, there exist configurations of these Apollonian circle packings of this with all what he called Ben's. Ben's are the curvatures of a circle. But since we want to speak of sphere packings, we'll actually use the term Ben, because curvature can mean one over the radius squared. So the curvature of a circle is one over the radius of a circle or a sphere. There exist configurations of Apollonian circle packings with all Ben's integral, with all Ben's integral. So there's some strange number theory hiding inside this. Sorry, was there a question or comment? No, I thought I heard something. Okay, so just one example of that is what's being, is the first illustration here. So this is a classical Apollonian circle packing. This is a circle of radius one over 18. So the radius of this is one over 18. The radius of this is one over 23 and so on. And so these numbers are the Ben's of the packing. Okay, any questions so far? So far so good. Okay. Well, there were sort of two natural questions, question one. So if you sort of see all these circles everywhere, if we want to draw all the circles with Ben at most 1000 or million, so what is the number of circles in a given packing p? So there exist configurations of this packing p. What's the number of circles in the packing with Ben bounded by some growing parameter x? In other words, so we like sieve methods, but this is literally sifting out, putting a sieve that throws out all the circles that are too small. So circles whose Ben's are greater than a million or a billion or a trillion. So what is the asymptotic growth rate of this picture? This question was answered in a paper of Oh, and myself already a decade ago, it's hard to imagine, and made more precise by lots of authors, including mean Lee and he owe and you have in a gradoff and many, many others. The asymptotic formula here is is some constant that depends on the packing. The the rate of growth is x to the house dwarf dimension. So Delta is the house dwarf dimension of the limit set of this packing. And this is some some number 1.30 and so on. It's a number greater than one, because the this fractal contains circles. So of course, the house dwarf dimension is going to be at least one. It's less than two because it sits inside the plane. And it's strictly less than two, which one can prove without too much work. And these days, we can prove this with a power savings error term, let's say plus big O of X of Delta minus epsilon for some small epsilon. Okay, so so these counting questions I sort of consider settled, much less settled are questions about, okay, we see these numbers, we see these bends. So if our packing is integral, so we'll say a packing is integral, if all the bends are integers, which numbers arise, which numbers, which numbers arise. So in our example, above the set of bends, this is the set of bends contains, let's see, what numbers do we see 18, 23, 27, 35, 62, and so on. So this is just the list of numbers without multiplicity that we see. So this is 18, 23, 27, 35, and so on. So there's this list of bends, infinite list of bends. And in a beautiful series of papers by Graham, the late on Graham, Ligarius, Malos, this was an AT&T group, Wilkes, and Yan at Bell Labs before Bell Labs closed down. So it's starting about 20 years ago, maybe 02 or 03. It was a long series of papers that was studying the integral properties of bends. Rishka is asking, is this unique? No, so there's a different set of bends, the set of bends depends on which initial configuration you started with. So there are infinitely many, there's a, so apollonian packing, there's one apollonian packing up to complex structure. It's conformally rigid. Any one could be moved to any other by a Mobius transformation, but over Q, there are infinitely many rational structures. And so each different apollonian integral, apollonian circle packing gives rise to a different bend set. Yeah, so this is the set of bends of the packing, set of bends of the packing P. So in our example, these were the bends that we saw in the example up above. So Graham, Ligarius, Malos, Wilkes, Yan observed that there were some local obstructions here. There are local obstructions. You only saw certain residue classes. It turned out mod 24. So, so let's, we define, we define N, we say N is admissible, admissible for a given, for a fixed packing P. If N is in the set of bends, mod Q for all Q. Are we interested in characterizing B for a fixed packing? No, we want to know for a fixed packing. So given fixed, fixed a packing, and then what set of bends will you see for that packing? Yeah, thanks. Please keep the questions coming. So definition, a number is admissible if it's in the set of bends, mod Q. Of course, if you're not in the set of bends, mod Q, for some Q, then obviously you're not in the set of bends. And the question is, to what extent is there a local global principle? So their conjecture, their conjecture. So this is the, let's, let's give this a name. This is the local global conjecture. This is the local global conjecture for Apollonian circle packings. Is that, well, what else could go wrong? For example, if you have a, let's, let's say the number three, mod 24 is admissible for this bend set. But there is no three here because a circle of radius three would be way too big, a circle radius one-third would be way too big on the scale of this packing. Okay. So there's also a local obstruction at infinity, which the, so the full local conjecture would say fixed P, fixed an integral circle, Apollonian circle packing, Apollonian packing P. Let's, let's give a name. Let's A, which is A of P, be the set of admissible numbers, admissibles. Then every sufficiently large, large admissible arises in the bend set, arises in the bend set. Okay. So these should be the only obstructions. You have the local obstructions, mod Q for some Q, and then you have the admissibles. In the case of the classical Apollonian circle packing, Elena Fuchs's thesis showed that you could take Q to be 24. So 24 being two cubed times three, these are the only primes at which strong approximation for the Zariski dense group, which is the symmetry group of the Apollonian packing, whatever all that means. These are the primes for which you do not, you're not fully onto the, the mod P, or the, or better yet, the piatic orthogonal group preserving a certain quadratic form. There's some stuff that's going on here that I hope to get to a little bit later. But anyway, so we know it seems like admissibility is a hard condition. You need to check that you're in the bend set for all Q, that you're in the bend set mod Q for all Q, but actually it's a, it's a a finite check and a very simple quick one for any given packing. So admissibility is easy, sufficiently large. Well, that depends on the packing. Again, you could start with packings that have better, have smaller and smaller initial bends, initial curvatures. And so, but in principle, this is a, I don't know, the question is how, the question to me is how falsifiable is this conjecture? Maybe just haven't gone far enough out to see what's going on. But again, given the previous theorem, whereas the previous theorem here, the number of circles, in other words, bends with multiplicity is X to a power greater than one. And so, the reason you might expect this kind of thing if we look at all circles, all circles with bend less than X. So this is, this is some set. And the number of the number of these is X to the 1.3. And then we take the bend. So we take the projection to the integers. Here's the integer line of size X. And the question is, do we hit, what's the fiber of this, this map from, from circles in the packing to bends? And so a typical fiber, typical fiber would, would, you would expect you is expected to have multiplicity. Well, if you're going to hit every number from, from one to X, or at least every admissible number, which again is a finite set of arithmetic progressions, then a typical fiber should have multiplicity. If there's X to the 1.3 circles, then each bend you would like to have multiplicity X to the 0.3. In other words, X to the delta 0.3 here is meant to represent delta minus one. Okay, this is just from the most naive box counting type argument. Any questions so far? So far so good. And so, so this is the local global conjecture. Delta does not depend on the packing. Delta doesn't depend on the packing for classical Apollonian circle packing. So, so this number delta is, is, they're all, as I said, there's one complex structure on the Apollonian, on the classical Apollonian circle packing, which has its own house dwarf dimension. In a minute, I'll show you that this is just one of infinitely many such things, and they will each have their own unique house dwarf dimension. So hopefully that answers that question. Okay, and so, so the best known today is an old theorem with Bergan from, I guess, seven years ago now, which says almost, almost all admissibles arise. In other words, if we look at how many bends there are up to X, relative to how many bends there should have been up to X, in other words, a number of admissibles up to X, this ratio goes to one. In fact, with a power savings rate, but never mind. They said B can miss some elements, congruent mod Q, some small elements that are congruent, that are in the admissible set mod Q. Yes, there'll be congruence restrictions. Although there are some, there are some universal packings, packings for which the set of admissibles is the entire is all, is all integers. Okay, so, so this is the, the kind of the best we currently know in this setting. Right. Let me say just a word about what goes into the proof of these things. So this is by the circle method or what we call the orbital circle method, the orbital circle method, which is to say the circle method but applied to group actions instead of just polynomials giving you the points on some variety and you're trying to, you're trying to understand some kind of local global principle that major arcs, the major arcs are handled by so infinite volume, automorphic forms, so automorphic forms, automorphic forms and representations, infinite volume spectral theory, and, and the theory of expander graphs. So expanders, expanders and super approximation, infinite volume analogs of the remodeling conjectures, these are all synonymous things. Um, does the set of admissible bends have density zero and Z? No, the set of admissible bends is just a finite list of arithmetic progressions. So there's, there's some arithmetic progressions, there's some numbers mod, there's a finite list of numbers mod 24, and the, the, the admissibles are the numbers that are following to one of those classes mod 24, in the case of the classical packing. Yeah, so it's a finite, it's a, it's, uh, yeah, by the way I should say the positive density, the fact that, um, let me aside, the fact that the number of bends up to X is at least some constant times X was proved by Brigham and Fuchs also about a decade ago. Okay, so we, we've known for some time that a positive proportion of integers arises, the fact that 100% of admissible numbers arises is currently the best, the best we know about this. For the minor arcs, for the minor arcs, we use maybe more standard analytic techniques, so analytic number theory techniques, analytic number theory, cancellation in exponential sums, exponential sums, Klostermann and so on, exponential sums, and cancellation. Okay, so this is kind of, uh, old news, I want to get to new news, we're not ready for this picture yet. Okay, any, any questions so far? Doesn't the positive density follow from the existence of Ford circles? Uh, no. So, uh, the Ford circles, great question. The, the Ford circles will have, um, the Ford circles are the analogs of circles, in fact this will be important to what I'm about to say. The Ford circles will be the circles, the analog of Ford circles in this picture are the circles that are tangent to two, uh, circles. So the values of these, this 278, 203, 135, and so on that you see here, uh, 47, 27, 23, 35. These are the circles tangent to two circles, and they are the values of a quadratic polynomial. So you'll only get square root the number of, uh, you'll, you'll, you'll, in this way, and in fact this is, this was observed already by Graham Lagarious-Mellos-Wilxian and implied, so I should put what was previously Graham Lagarious-Mellos-Wilxian. Exactly, you use the analog of the Ford circles to show that the number of bends up to x is at least square root x. Did I answer that question? Uh, or does that guarantee for the standard packing? No, it doesn't, yeah, it doesn't guarantee. Even for the standard, by the standard packing, Sanya means the, the, the packing that's the strip packing. I think that's what you mean by the standard, the standard packing. So the Ford circles would be the circles that are tangent to, uh, this bottom line here. Pardon my confusion, but does the set of admissible integers depend on the packing? Yes. Yes, so the, the set of bends, although there's a finite list, well there's a finite subset of the numbers mod 24, so of course there's a finite list of what the admissibilities can be. Okay. Good. Good. Yep. No worries. So, um, in what, so there's, there's a general, generalization of this. In fact, Sadi already observed. So this is kind of the end of the story as far as, as far as what I can say today about the classical Apollonian packing. And what about generalizations? Other, other packings. So already in, uh, 1936, Sadi observed that the sphere packing, that there was a spherical analog, what I call the Sadi sphere packing, which is the analog of the Apollonian packing, instead of starting with, um, three tangent circles and inscribing one more, you can start with four tangent spheres, which, uh, we can draw spheres as having a point at infinity. And so if we start with four tangent spheres, there's an obvious way, hopefully you can see, there's one ball that I can put, let's stick to the color scheme, there's one ball that I can put in front of these, and there's another ball that I can try to put behind them, so, so that you get this configuration of four balls. And that's the analog of Apollonius's theorem in this setting. And, uh, Sadi observed through explicit, uh, construction, he actually built these things out of, uh, metal, you know, steel ball bearings, he observed that there exist configurations like this. So this is a Sadi sphere packing. So it's a sphere packing, uh, of balls in space, which, uh, all have integer bends. So this is a circle of radius one over 28, radius one over 81, radius one over 85 and so on. And, um, and for this, uh, Sadi sphere packing, the counting problem is again, uh, the counting problem is now solved in, in great generality for the four spheres in the packing. But what about the bends? So, um, it was observed, I think by Boyd, I think Boyd in the 70s already observed that all the bends of this packing will be either, so instead of 24, now the modulus is three. And these are all one or zero mod three. So 28 is one mod three, 81 is zero mod three, 85 is one mod three. So the bends are all one, zero or one. In this, for the Sadi packing, of course, again, there's infinitely many, uh, different rational configurations, but, um, there, the, the bends are all zero or one mod three. That's the analog of the local obstructions. And here I was able to prove, I guess this came out last year, that every, so given an integral Sadi packing, Sadi packing P, every sufficiently large admissible, admissible now is the residue class zero or one mod three. Well, it might be zero or one or for some packings, it might be zero or two. Those are the two possible values of the admissibles. Every sufficiently large admissible is represented, appears in the, is in the bend set, is in the bend set. So the full local global principle can be proved for sphere packings. We still don't know it for circle packings. One of the reasons, it's not exactly the reason, but one of the reasons is that the house dwarf dimension is now contains numbers. Well, this contains spheres. So the house dwarf dimension will be at least two. And it's something like, I don't remember, 2.4, something like, but it's less than, it's less than three, since three is the ambient dimension in which this fractal will live. And it's greater than two. And, um, all right. So, so that's kind of all we know about this classical, these two classical objects, the classical Appelonian circle packing, integral Appelonian circle packing, and the Saudi sphere packing. Any questions before we move on to generalities, generalizations? Okay, so far, so good. So, um, there were a number of generalizations that were observed in work of Boyd. So there were generalizations by Boyd, by Maxwell, by, by a number of authors giving finite lists. So there were a number of finite lists, finite lists of packings, although it wasn't clear what packings even was supposed to mean necessarily, with these integral structures, with integral structures. In some sense, this might go all the way back to, as I'll show you in a second, here's an old paper of Bianchi. Bianchi, this is 1890s now. This is, this is from Bianchi's paper. And he's starting to construct what viewed in the right context will be what we now call the cube octahedral packing, which I've reproduced here, what these parts of circles are. So here's a little bit of this red circle is here, and so on. And this is how one can make a packing out of it. So there were a number of generalizations, sort of ad hoc generalizations without, this was, we had lots of examples, but no theory. And so what I've been working on for the last many years is trying to understand what is there a general theory to, so is there, is there a general theory, which explains the existence of these objects, these objects, and constructions. So that's what I would like to try to explain today. So, yeah, by the way, I should mention, there are more examples, more recently made by Baragar, Stang, and her collaborators, Sanya Shedvasser, studied by Xinjiang. And there's forthcoming work of Edna Jones on analogs of this local global theorem in sphere packings. Okay, so what's the, what's the general theory? So let's give a definition. So by a sphere packing, what's a sphere packing? First of all, a packing, a packing P of, we can say, Rn union, the point at infinity, what I really want, this is our, this is space plus, this is the Riemann sphere, I mean, this is the Riemann sphere, which we can also think of this as the boundary of hyperbolic n plus one space, as you'll see in a second. A packing, by a packing P, what I mean is, is an infinite collection of spheres, so these will be Sn minus one spheres, or n minus one spheres, with two properties. One, I want those spheres to have disjoint interiors. So with disjoint interiors, disjoint interiors, so they, in order to have interiors, they need to be oriented. So oriented spheres, which means if you like, if you prefer, you can think of as balls, or discs. So spheres with disjoint interiors, and the, I want those, the discs to be, balls to be dense, the balls, balls on the spheres, balls bounded by the spheres, this, the set of balls are dense in Rn. In other words, if I take all of the spheres in the packing, and I take the balls that the spheres generate, these will have closure equal to all of Rn. So there's no space missing. All right. And again, a packing is integral, P is integral, if every sphere in the packing has integer bend, the bend of the sphere is an integer. And we can generalize this to number fields in certain ways, but not, not certain other ways. Page. Now, as stated, there's going to be no structure. There's nothing we can do with this because you just start, if you want to study integral packings, you just make a circle with radius one over here, 27, and another circle of radius one over 100, and another circle of radius one over a million, and you can fill up space in this way. And there's nothing that one can say about which bends arise, how many spheres there are, and so on. So the structure, so without more structure, without more structure, there is little that we can say, little to say. So the structure that I will impose, so key definition, key property of all of these things is the following. A packing, a packing P is called Kleinian, if there exists a group, gamma, which is called a symmetry group. Gamma is a subgroup of the isometries of hyperbolic space in one higher dimension. So packing P is called Kleinian, if there exists a Kleinian group, a group acting by isometries, which is discrete, geometrically finite, this means there's a fundamental domain whose, so that an epsilon thickening of it has finite volume. The convex core has an epsilon thickening of the convex core has finite volume. So if there's a group that's discrete, geometrically finite, such that the limit set of gamma, so the set of limit points, is exactly the set of limit points of the packing. So let me give you a sort of a definition by example. So again, let's go back to, let's go back to the very first thing, which is the uploading packing, and I drew here a coxeter diagram. So I'm going to redraw this coxeter diagram. This is meant to be the grand matrix of the corresponding coxeter diagram. Let me show you how to, how to interpret these things, what they're supposed to mean. So example, so here's the diagram again, there was a thick wall, a single, a single and a thick again. So this is the coxeter diagram, and let's label these walls. So this is wall one, two, three, four, five. So a thick wall means multiplicity infinity, which means that the two walls, the edges are meant to be dihedral angles. So the nodes correspond to walls or roots, simple roots, if you prefer that language. Walls of reflective walls. So coxeter diagrams are supposed to tell you about reflection groups, reflective walls, and the edges, edges tell you the dihedral angles, the hedral angles between walls. So this single edge is meant to tell us that the walls meet at the hedral angle pi over three. The fact that there is no edge from one to three, there's no edge, no edge means that they're orthogonal, means that the walls meet at a right angle. And so, and what is, so how do you meet an angle infinity, these two are tangent at infinity. These two are tangent at infinity. In other words, they're parallel. So let's start with walls one and two. So wall one will be this wall, wall two will be this wall. Walls four and five are also tangent because they have an infinite line. And wall four is orthogonal to walls one and two, and so is wall five. So it's pretty easy to guess that walls four and five will be vertical lines. So so far, all we have is a square. Walls two, walls one and two, and walls four and five generate this, this square. Now what about wall three? So wall three will be orthogonal to walls one and five. I can draw that as a circle centered at the point of intersection of wall one and five. And I want it to intersect two at angle pi over three cosine of pi over three is a half. And so if I go twice the distance to wall two, and I draw that circle, this will be circle three. This is wall three. Okay, so wall three is orthogonal to walls one and five as desired. There's no edge between walls one and five. And it meets wall two at angle pi over three. And similarly, wall four, it also meets at angle pi over three because this is a square. And so what you should what this image really represents is the following. If we take everything, so this is the complex plane, really the Riemann sphere, which is the boundary of hyperbolic three space. So this is the first Bianchi group, it turns out. If we take this to the Poincaré extension, hyperbolic three space, so I'll draw wall one here, wall two here. So this was wall one, this was wall two. Wall three, rather wall four is in front and wall five is on the, on this axis. So these are walls four and five. And then I have wall three, which is this sphere. So hopefully you'll, you can see that there's this intersection of these walls and then this familiar diagram, which is a double cover of, so the group, the group gamma, which is generated by reflections in walls one through five is commensurate with the classical Bianchi group SL2 of the Gaussian integers. So all of that, all of that information is hidden inside this Coxeter diagram. Okay, does that make sense? And here's the key observation for wall one is what we call isolated. This is isolated, which means it doesn't intersect any other walls except orthogonally or at infinity. And so the, the yoga is if you have an isolated wall, so let me call this gamma tilde. So let's take the group generated by the other walls, the non, the other walls, walls being wall two, three, four and five. And let's look at the limit set of this, of this group gamma. So I'm taking this sub diagram, which is the walls two, three, four, five. And I'm not taking the inversion in wall one. And let's see what is the limit set of this, of this thing. So hopefully this will work. So again, this is wall two, three, sorry, four, five, and three. So these are the walls that you see on the ground. Inversions in these walls correspond in the Poincare extension to inversions in the geodesic domes. These are geodesic hemispheres. And in the case of straight lines, they're just Euclidean inversions. So I start with some arbitrary points in the upper half space, and I start moving it by the walls. So here I inverted through one of these walls, maybe wall four or something. And then from here, I'm just at random going to do some move and invert, invert in, it turns out this geodesic sphere will invert me into there. Okay. And if we keep doing this, for a long time, we will get a limit set. So let's see if you can see this. So again, we started with this one wall, and we started moving it around more and more and more. And we're getting some limit points in the boundary. And those limit points, hopefully you see. So the limit set of gamma is exactly the classical Apollonian packing. My pen is having trouble, I think, because of the video that's running. So let me move away from the video, and hopefully it'll restore this thing. But any questions so far? So this is the key property of the Apollonian packing. It's that it occurs, it arises as the limit set of some geometrically finite group acting in hyperbolic space. Sorry, I must, I'm a bit confused. So do you mean the three dimensional picture? I mean, I'm not sure. Do you mean that I should be looking at the projection to R2? Exactly. The limit set will be in the boundary. The limit set is there are no limit points because it's discrete. Because the group action is discrete, the only place where you get accumulation is at infinity, at the boundary at infinity. So that's why we can draw these pictures in the plane, in the case of hyperbolic three space. Are we applying infinite words to some arbitrary starting point? Yes, we're applying, we start with some point and then you just apply letters at random. And when I apply those letters at random, what you really, that's what the image is illustrating. What the limit set is actually is take all the, take the entire orbit all at once and look at the set of limit points. There are no limit points in the upper half space itself. The limit points are only in the boundary. So where are they? These are all dots, but hopefully you see that the dots in the end make up, here's the uploading packing appearing after infinitely many steps. Changing the starting point changes the orbit, but does not change the limit set. Yes, exactly. Does the Hausdorff dimension have constraints other than great question. So in the case of Schottky groups, there's old work of Philip Sarnack and others that chose that the Hausdorff dimension will not be allowed to get close to n, can't get close to the maximal dimension. But for general packings, in fact, we can produce our sort of arbitrarily close Hausdorff dimensions to n, to n minus one, I'm not sure I never really tried. In fact, isn't that an open conjecture about the classical ablonian circle packing, being the packing with being the dimension closest to one, that one, that one can get. Okay, so I only have, let's see, I have about five minutes left. Is that, is that right? So let me try to state some, some theorems about this and then tell you what we can say. So, so what can we say? So we've defined these, these Kleinian packings. So a subclass of the packings, so definition packing P is not only Kleinian, but crystallographic, crystallographic. If this symmetry group is finitely generated by reflections, which means that it's geometrically finite. So groups that are generated by reflections are the easiest to illustrate, and they're exactly the ones that one can write down a coxeter diagram for. For a general geometrically finite group, I don't know a good way to, to represent it. But in the case of reflection groups, we can really draw such, such pictures. And so that is what these things are meant to illustrate. So this is the classical ablonian packing when I take this vector to be isolated. Here's the Saudi sphere packing. This is a coxeter diagram, which gives a lattice, which is the group minus three. I'm running out of letters u squared. So this is the orthogonal group, the orthogonal group of this quadratic form, the integer orthogonal group of this quadratic form, which has signature four one acts on hyperbolic four space. So this acts on hyperbolic four space. And in fact, is this reflect, is the reflective group given by this coxeter diagram, it has an isolated vector. If I take the sub diagram and I take the group gamma generated by reflections in these walls, it will produce the classical Saudi picture. And more generally, so in a paper with K Nakamura, Nakamura, that came out last year, we, we studied packings generated by polyhedra. And the polyhedra packing, it turned out that this, this is the cube octahedral packing. This is the cube octahedral packing. And it also arrived, this is the coxeter diagram for the Bianchi group SL 2 Z adjoin root minus six, which is exactly the one I was showing you in Bianchi's paper. So this is the coxeter diagram. And if you take the wall one, which is isolated, then the inversion in the others is exactly what this picture is supposed to illustrate, which is this packing. And there's one more that I'll show you, which is this 14 dimensional packing coming from work of Wienberg. We have constructions of up to 18 dimensional packings. So this vector is isolated, isolated, and gives a 14 dimensional packing integral pack, not only integral, but super integral. So I have to give you one more definition. And then I can state the main theorem and maybe we'll end there. So one more definition. All of the packings that we've seen so far, not only are integral. So definition, I have to tell you one, two more things. A packing, given a packing P, let P tilde be the super packing. So this is called the super packing. What you do is you take the packing and you act on it by inversions in the packing itself. So in other words, if we start with the classical Apollonian packing, so here's the packing, in fact, this has nothing to do with the any other structure, actually, what we do is we take inversions in the spheres of the packing and let that act on the packing itself. So if I invert the entire packing through this sphere, I get an Apollonian packing inside of here, but I'm allowed to continue acting on all of the, take the entire orbit. So I just get circles inside of circles and so on. So this is called the super packing, the super packing of P. And definition P is super integral, not just integral, but super integral. If every sphere, not just in the packing has been, that's an integer, but every sphere in the super packing has integer band. So the classical Apollonian packing has this property, the Saudi packing has this property. Not all, so there exists integral, but not super integral, non super integral, packings. This was, this was given by Nakamura and myself. So here's, I can state the main theorem. Theorem, if a packing is Clinian and super integral. So this is joint work with Mishka Hoyich from this year. If a packing is Clinian and super integral, then the group generated by its symmetry group, so it's Clinian, so it has a symmetry group. And the group of inversions and all reflections of spheres in the packing itself. So this, this group is a lattice, acts on hyperbolic n plus one space with finite covolume. And not only is a lattice is, and it's q arithmetic. So it's an arithmetic group, arithmetic, defined over q and non uniform. So the, it's q arithmetic of simplest type, arithmetic of simplest type, which means it's defined by quadratic forms. And it's non uniform, conversely, conversely, every q arithmetic of simplest type, non uniform quadratic form, equivalently every O F Z, where F has signature and one F is defined over q has signature and one and is isotropic isotropic over q. Every, every such thing is commensurate to is commensurate to this this object is called the super group, super group, the super group of a super integral Clinian packing, a super integral Clinian packing. So this hopefully explains what I was trying to get at before, an understanding of where does the uploading packing come from where did this Saudi packing come from that we had these couple of examples. And in the greatest generality, it's just from the theory of arithmetic groups. So anytime you have q arithmetic, non uniform, hyperbolic group, then one can construct in fact, not just one but infinitely many infinitely many conformally in equivalent super integral Clinian packings out of it. So that's my time. So thank you for your attention.