 Thanks to the organizers. Been a great workshop so far. So yeah, what I'm going to talk about today is joint work with Grant Lakeland. He's another former student of Alan Reed's. I'd like to say part of Team Penzole. So what I want to talk about today. So let me start with let's take, for the rest of the talk, just so I don't have to keep writing PSL2C, because that's a lot more letters. Let's let G be PSL2C. And gamma in G, a co-compact lattice. Ooh, dusty chalk. What does this mean? This means gamma is a discrete subgroup of G, which means it acts properly discontinuously on H3. So we identify this with the isometry group, orientation preserving isometry group of H3. And then the quotient, M gamma. So we take hyperbolic three space in quotient by gamma. This is a closed hyperbolic stream outfold. And so one of the very interesting things that happens in dimension three and higher is that you have Maastar rigidity. So Maastar rigidity says that this hyperbolic three manifold is rigid. So what does that mean? That means any other hyperbolic structure on M gamma differs from the given one by a homeomorphism. What do I mean? So a hyperbolic structure, so that's a like GX structure. That's a GH3 structure. And then the group of homeomorphisms of M acts on the set of GH structures, GX structures, by just taking your atlas of charts that defines the structure and precomposing with that. And what we're saying is that any other hyperbolic structure that you put on M gamma differs by such a deformation. In fact, the result of goodbye says that, in fact, this homeomorphism is isotopic to the identity. So all right. So maybe that seems like the end of the story for sort of deformations. But there's another interesting quotient that you can take. So another quotient you can take is G mod gamma. So I can take the quotient of PSL2C. Gamma is sitting inside of there and acts on the left. And I can take the quotient of G mod gamma. And in 95, maybe before that, she showed that this is not rigid. So I need to explain what I mean by that. And that's going to be the some sense that the focus of this talk is the kinds of structures that I want to describe now. So observe G cross G acts on G. So if I take two elements, A and B and G, and then I want to act on some other element, well, I can act on the left by A and on the right by the inverse. And this defined the left action of G cross G on G. And for subgroups of G cross G acting properly discontinuously, we can take a quotient, get a nice power store space, should say we're acting freely, we get a cross G structure. So I'll say more about this kind of structure later and why it's interesting. But let me just, if you haven't thought about this before, make a remark. So one nice thing about this situation is that if I have a discrete subgroup of G, it acts properly discontinuously on H, on H3. However, what does the stabilizer in G cross G of the identity? Well, that's the diagonal subgroup. And this is isomorphic to G, which is non-compact. So the stabilizer of the identity is non-compact. So discreteness is not enough to guarantee proper discontinuity of the action. So maybe just look at some examples. So let's say, so maybe from now on, let's let J denote the inclusion of gamma into G. So what are the kinds of subgroups I want to look at? So I want to look at subgroups. So we have J. And then we have some other representation, rho from gamma, necessarily injective, any representation. And then we can look at J cross rho of gamma. This is a subgroup of G cross G. So it acts on G. And we can ask, when is this J cross rho of gamma acting properly discontinuously? So maybe just a couple examples, I said. So let's say, suppose rho, this other representation, is just J. Well, in this case, now I have J cross J of gamma. So I've just put this into the diagonal subgroup. And this fixes the identity, so definitely not properly discontinuous. On the other hand, if I take the trivial representation, so rho, say rho naught, rho naught of any gamma is the identity, well, then when I look at J cross rho naught of gamma, this is just gamma cross the identity. And in this case, J cross rho naught of gamma, so this does act properly discontinuously. This is just the original quotient by the action on the left that we saw at the beginning. OK? That make sense? What's this non-rigidity that I'm referring to here? So the theorem, jeez, let's see, for all rho sufficiently close, J cross rho of gamma acting on G is properly discontinuous. And in particular, if you have non-trivial deformations of the trivial representation, then you get sort of new G cross G G structures. So the quotients turn out again to be difumorphic to G mod gamma. So these are new structures on G mod gamma. OK? So let me just say a little bit, so this is definitely outside of my area of expertise, but this is really cool. Let me tell you just a little bit about, if you haven't seen this before, maybe why you would care about these G cross G G structures. So one, actually PSL2C is a complex league group, and the quotient is a complex three manifold. And actually, G showed something stronger. He showed that these deformations, so these deformations of the theorem are precisely, well, he was working in SL2C. Anyway, precisely the deformations of the complex structure. So when you deform the complex structure on G, a sufficiently small deformation of the complex structure actually corresponds to a deformation of this G cross G structure. Let me also mention that the killing form on G is invariant under both left and right multiplication and defines, let's see, G defines the holomorphic metric, in fact, of constant negative curvature. So let me just say a little bit about this, because this is not something I'd ever run into before. So what's a holomorphic Riemannian metric? That's just a holomorphically varying non-degenerate bilinear form on the tangent spaces. And so Susan doesn't have anything to do with Hermitian forms, sort of the direct analog of the real Riemannian metrics. But you can do sort of Levy-Civita connections, curvature. And so anyway, this is an example of a holomorphic Riemannian metric of constant negative curvature. And so these have been studied quite intensively, especially in low dimensions. So Dimitrescu and Dimension 3 show that these are quite a bit more rigid than, so holomorphic Riemannian metrics are quite a bit more rigid than their real counterparts. So in Dimension 3, they're always locally homogeneous. And then Dimitrescu and Zagiev proved kind of uniformization theorem. And so these G cross G comma G structures are one of the sort of models that you want to study in Dimension 3, OK? Yeah, that's a good question. I don't know, somebody, I don't know what sort of, I mean, I think they're not Kaler. I'm sorry. For co-compact lattices in SL2C, that's right. I'll say more about this. This has been generalized quite a bit, and I'll get to that in a little bit. Any other questions on the sort of structure? OK, good enough time. All right, so now, so you get these properly discontinuous actions, there's something that you have to do to actually get that. And so let me describe condition for proper discontinuity. Let me state the theorem, and then I'll talk about a little bit because it sort of relates to lots of different things. So theorem, Gerito Castle, which says, we'll look at the inclusion of gamma into G. And this is true in much more generality than what I'm going to talk about. For now, let's think about gamma, this co-compact lattice. Rho from gamma to G, any representation. Then, J cross rho gamma acts properly discontinuously on G, if and only if there exists a J, rho, aqua-variant contraction, a strict contraction. It's called F tilde from H3 to H3. So here, I've got two actions of gamma on H3, one by J of gamma, just the given action from the inclusion, and the other one rho. And so we want a J rho, aqua-variant strict contraction. So in the case where rho of gamma is, again, discrete, rho is not going to be injective in general, but rho is discrete, then this map, this aqua-variant map, is the same thing as a map on the quotients. So from M gamma, the quotient of H3 by gamma, to the quotient of H3 by rho gamma. And so let me, this has a name, this is sometimes say that rho is strictly dominated. OK, so let me pause for a few minutes and just say something about this theorem. So this was proved first in the case, actually, where G was PSL2R. And in that case, that was due to KSL. The reason that's particularly interesting, I don't have anything to do with this, but rather, in that case, as PSL2R cross PSL2R, comma, PSL2R structures, those are exactly ADS3 structures. And so a lot of people have thought about discreet, properly discontinuous actions in that case. Going back, let me just say some people, Goldman, Kulkarni, Kulkarni-Raymond. I think the first way of thinking about this kind of condition due to Salaam, right, Salaam. But a lot of other people thought about this, KSL, Gary Toe, KSL, Gary Toe, KSL Wolf, Danseger, Gary Toe, KSL. There were a lot of sort of PSL2R results about these sort of actions. As I said, this is true in more generality. So this theorem, they proved, was actually for G being the orientation-preserving isometry group of hyperbolic n-space. And gamma just needs to be geometrically finite. So, and now it's been generalized further by Gichard, Gary Toe, KSL, and Veinhard. This condition about this strict contraction might look familiar. So Jeff talked about something very similar. He was talking about actions on the Lie algebra. And there, he had these vector fields that were contracting. So that's sort of the infinitesimal version of this strict contraction condition. And there's another version of this theorem where one thinks about lengths of hyperbolic elements. And you require that the length of, so you could equivalently just require that the length of every hyperbolic element in row is strictly contracted compared to the length in gamma. And that version is related to the infinitesimal version of this condition on lengths due to Goldman, Laverie, and Margulis. So Jeff, I think, mentioned that also briefly. So there's a lot sort of going on here. I'm just picking out one particular aspect of it that's relevant for what I want to talk about. So OK, well, if you haven't thought of this before, seeing why this is true, let me just sort of, the only part that I'm actually going to worry about is if you have this strict domination, why do you get proper discontinuity? So let me explain that picture in it. Again, this should look familiar, because this is similar to what Jeff was talking about. So one way to think about why discreteness, so if you act properly discontinuously on H3, then just the usual left action of the group on G is properly discontinuously, because you have an equivariant proper map from G to H3. And so it turns out that you also have that when you have a strict contraction. So let's suppose F tilde from H3 to H3 is J row equivariant strict contraction. And so there's a really cool construction here, just building a proper equivariant map from G to H3. So define map from G to H3. So how am I going to use the map F tilde to find the map from G to H3? Well, let's just take a group element and let's send it to the unique fixed point of G composed with F tilde. So F tilde is a strict contraction, so it has a unique fixed point in H3. And if I compose with an isometry, it's, again, a strict contraction. So it's not too hard. You can check that this map, this is equivariant. So on G, we have the action of J cross row of gamma. And over here, we're just looking at the action of gamma. So this is equivariant with respect to that action. And the fibers of this map are all copies of SO3. So again, if you're used to thinking about the case of G of a group acting just on the left, so the trivial representation on the right, the strict contraction is just a constant map. And that's you pick a base point and you look at the orbit of the base point. That's exactly what this map is in that case. So all right, so now we have a proper map from G to H3. So we're acting properly discontinuously here, which means we act properly discontinuously on G. Does that make sense? OK. So in this case, the equivariant, so this is really a vibration of G over H3, an equivariant vibration of G over H3. And so we get out of this an SO3 bundle. This is actually a vibration with fibers that are this compact SO3. Yeah, I should say it's actually a vibration. So we get some SO3 bundle, G mod J cross rho gamma to N gamma. And in Jesus' theorem, that's what we're seeing. Now we're seeing different G cross G structures on the given SO3 bundle. So let me give this set of representations a name. Let's write DOM gamma G. So these are the representations. Rho is strictly. So equivalently, these are the representations that exactly give you proper discontinuity for the action. OK. So action of G cross G on G also preserves a volume form. And so you could look at the volumes of the quotients. So there's some normalization involved. I'm going to say anything about it. Let's all as on gave a formula for the volumes of these quotients. Again, in general for quotients of SON1 by groups acting properly discontinuously in SON1 cross SON1. But let me just, again, stick to this case. So suppose J is the inclusion, some representation, so that I get a properly discontinuous action. And then the volume of this quotient, let me just write it J cross rho. So that's the volume of these are acting properly discontinuously. I can take the quotient. The volume of this quotient is given by some normalization constant. So this is the volume of SO3, again, with respect to certain normalizations. And then you just see the two volumes. You have the volume of the original compact hyperbolic manifold minus the volume of the representation row. So what is this volume here? Let me just say. So we've got this, because we're strictly nominated, we've got this Equivariant Lipschitz map from H3 to H3. We can pull back the volume form. So we can take F tilde star of the volume form on H3. And then we can integrate that. We push it down to the quotient. The map is Equivariant. That volume form is invariant. We can push it down to the quotient and compute the volume, OK? So I want to make some comments about this. Oh, there's one more bit that's useful for us. So moreover, this function from these representations to R given by sending row to the volume of this quotient is rigid, is locally finite. So let's see a few words about this. So this theorem is also true, again, not just for G PSL2C, but actually for SON1 for all n greater than or equal to 3. And gamma can be non-compact lattice, non-compact lattice. And in dimension 2, oh, sorry, locally constant. Sorry, that's, thanks, sorry, locally constant. Yeah, it's also locally finite, locally one. So right, so whether gamma is compact or non-compact, this is true in dimension 3 and higher. In dimension 2, it's also true for compact, but it's false for non-compact. And so this rigidity, this sort of volume rigidity mostly follows from the Besson-Cortois-Gallow volume rigidity because of this formula. The non-compact dimension 3 case requires a little more work because there you don't have that volume rigidity. But it turns out that whenever you have one of these dominated representations, they always have to send parabolic to elliptic elements. And in that case, the volumes are locally constant. So let me just also say that in the non-compact case, you have to be a bit more careful with this. There are various things you might try to do. And the volume that you get turns out to be the same as all the usual volumes. So Frank Avila, Kim and Kim, Buker, Boucher, Buker, Berger, and Yotsa, all versions of this volume. And this agrees with that in those settings. So they said that the two-dimensional cases, so when G is PSL2R, it's not what I'm thinking about here, but let me just mention, and gamma is compact, so long it gave lots of values for the volume. In fact, he constructed what turned out to be all the possible volumes that you could get for a co-compact lattice. And using this formula, Tholeson ensured that that was all the possible volumes. And then Tholeson in the non-compact case in dimension 2 showed that actually not only is this not locally constant, but you get an entire interval of volumes. OK, but again, that's not the situation I'm talking about. I want to think about dimension 3. And there, well, if you think about the construction of Gs, those were deformations of the trivial representation, which has zero volume. And so this term for all of those representations is zero. And so then the question that Tholeson asked was, actually, it was here last fall where I heard this question, does there exist gamma, G, rho, or these dominated representations? Such that the volume isn't just the volume of this guy, such that the volume of rho is non-zero. So we started looking and we found an example on math overflow due to Yenegle. So let me describe that example. Math overflow, isn't it great? So let me just say these are what I'm thinking of as these exotic G quotients, not just deformations of the trivial representation. OK, so how's this example go? So you start with a pair of compact hyperbolic tetrahedra. So let's say this is 2, 3, 5, 2, 3, 4, and 2, 3, 5, 2, 3, 2. So let's call this T in this D prime. So what do the numbers mean? So there's a hyperbolic, compact hyperbolic tetrahedron whose dihedral angles are pi over 2, pi over 3, pi over 5, and so on. And the nice thing is, if you, the group generated by reflections, here is the coccider group. So let me write down what the group is. This comes from these numbers. So we have, let's say, let me call it gamma bar. This is the group generated by reflections in T. So this is, there are four faces. So it has a nice presentation, tau i squared is one. And then, well, if I do the reflection in one face, let's say this face and this face, that's going to be a rotation of order two. So the edge adjacent to the pair of faces tells me what the order of rotation is. So tau i, tau j to the E ij is one. This is the label on the edge adjacent to face i and j. And then, there's also one for T prime. Looks very similar. Let me put primes on them, tau 3 prime, tau 4 prime. OK, same sort of thing. And the first thing that you notice is that all the labels here look like the labels here, except this is a 4 and this is a 2, and 2 divides 4. That's nice. So there's a homomorphism row from gamma to gamma bar, or sorry, gamma bar to gamma bar prime that just sends tau i to tau i prime. And you just need to check that all the relations are satisfied. That's pretty easy. They're exactly the same, except one of these is a 4 and one of these is a 2. But if the second powers the identity, so is the fourth. OK? All right. There, she puts some primes. OK, so we get a homomorphism from gamma bar to gamma bar prime. And now you just need to check a few things. So let me not do it, but let me just say what you can do. So you check. What do you need to check? So we have our homomorphism. So I need to build this equivariant contraction from h3 to h3. And what can I do? Well, I'm not going to diss the Klein model. I'm going to use the Klein model so we can take a linear map. So let's put one of the vertices. Let's put this vertex at the origin. And this vertex at the origin, actually, if you do that, then this guy actually sits strictly inside of here. And you can find a linear map, let's say, in the Klein model, a really projected linear map from t to t prime. Well, it's a linear map from all of r3 to r3 that sends t to t prime. And actually sends, you just need to check this, some computation, you can explicitly calculate where all the vertices of this thing are. It sends hyperbolic 3 space, the Klein ball, strictly inside of h3. Here's h3 with one of these tetrahedra. And then here's h3 again. And then let's go sort of strictly inside of here to get to the other one. So here's t. OK, and then use the Hilbert metric. And you check that means that you get a contraction. So this means the map from t to t prime is a contraction. And now there's another important property of this map, which is that this is sort of the key point. So let's call this map f0 tilde, sort of the building block for the map. If I look at the image of a face, let's say the ith face, this is fixed by rho of tau i. That's tau i prime. OK, if I want to build an equivariant map, what do I do? Well, I have my map here, that's f0 tilde. And then I'm going to look at all the tetrahedra that are adjacent to this one. So how do I map that guy? Well, first I do a reflection in this face, I map over here, and then I do the reflection according to the image of that reflection. And so in order for that to give me a well-defined map, I need to make sure that the thing over here that I'm reflecting also fixes the image of that face. And that's how this homomorphism is defined. OK? So that's sort of the key point here is to get a homomorphism, a map between these two tetrahedra, and then ensure this property. And then you just start reflecting and you extend the map to all of h3. Now, OK, so why does this give us a representation with non-trivial volume? So if you look at the index 2 orientation cover. So gamma bar is not inside of G, right? It's inside that it has orientation reversing. There's reflections orientation reversing. So let's take the intersection with G. This is the index 2 orientation subgroup. And this is the kernel of the determinant homomorphism. The determinant sends each of my reflections to minus 1. And the image also does that. So this map actually row. I can restrict row now to gamma. And I get a homomorphism from gamma to gamma prime, which is just the same thing. I intersect. So gamma prime is gamma bar prime intersected with G. That's the index 2 orientation cover there. I get this map, this homomorphism now between these two groups. And I can take the quotients. So I look at the map. M gamma. So this map, f tilde. Oh, I guess I didn't write it down. So from this, we get our map f tilde from h3 to h3. It's a real equivariant. That descends to a map here between these orbitals. And well, what is this thing? This is just the double of this thing over its boundary. And the other guy is the double of that over its boundary. And the map is actually, if you forget about the fact that there's this singular look, this is an orientation preserving homomorphism. So this is a degree 1 map. So the volume of this representation is just the volume of this guy, which is twice the volume of this. So the volume of row restricted to gamma is twice the volume of t prime. That make sense? OK. OK. So what we wanted to do was, well, to see if you could do this in any other situation. And it turns out that it's pretty easy to do this for lots more examples. So let me show you the setup. So let's start with, so if q is a compact, all right. All-Hedron, h3. Let's let gamma, we'll say gamma q bar, be the group generated by reflections. So for each face, we have a reflection. And this has a nice presentation. So it says tau f squared is the identity. And tau f tau f prime, so the two reflections and two faces commute if and only if f intersects f prime non-empty. OK? So this is a right-angled coxeter group, an example of what Jeff was talking about. So it's interesting that these same groups are showing up here, even though the construction is a bit different. OK, so well, we can also take the orientation subgroup called that gamma without the bar. So gamma q, again, gamma q intersected with g. This is our index two subgroup of gamma q bar. So what do we do? Well, we do something like this. So the theorem, tensile twins. So for any q and q prime compact all-Hedron, there exists another compact polyhedron. So the group generated by reflections in this one is commensurable with that one. So this is compact right-angled polyhedron, such that, well, so first, the group generated by this is a finite index subgroup of the original guy. Two, these are reflection groups. Actually, sorry, it's the same for the rotation groups. There exists a homomorphism, rho from gamma q1 to gamma q. And if I think about this sitting inside of g, then this representation is strictly dominated by the inclusion of gamma q1, g. And well, just like over here, the volume is going to be exactly twice the volume of q prime. So the volume of this representation, rho, is twice the volume of q prime. OK, so that's what you can do, more or less, following this kind of idea. And it's the same sort of elementary hyperbolic geometry construction, anything fancy. So as a corollary, you get lots of volumes. So corolla, coroll, coroll, coroll, two L's, two L's. How about now? Oh, is this being recorded? So it's the corollary, say, for any angle, compact, polyhedron, I just abbreviate that. I'm done writing it. Q, there exists finite index subgroups. So we start with gamma q. So let's call that gamma naught. So inside of there, we have gamma 1, and gamma 2, and gamma 3, and so on. Such that for all n, there exists rho 1 through rho n representations from the nth group into g, which are strictly dominated by the inclusion, gamma n, g. And the volumes are all different. Volume of rho i is different than the volume of rho j. So you get lots of different values for the volume, in fact. OK, so this is a really easy corollary of this construction, so let me just tell you the proof of the corollary quick, up proof of, I must have been on my coffee killing the chalk, proof of the corollary. OK, so let's use this construction. I'm going to take two polyhedra. Let me take both of them to be q. The theorem says if I have two polyhedra, I can find a finite index subgroup of the rotation group of one of those with a homomorphism to the rotation group of the other one that strictly dominates the inclusion, or strictly dominated by the inclusion. So out of the theorem, we get gamma q1, finite index subgroup of gamma q. And this rho 1, call it from gamma q1 to gamma q, which is strictly dominated by the inclusion of gamma q1. So this is in gamma q1, g. And what's the volume? The volume is just twice the volume of q. Let me write it like this. It's the volume of m gamma, but this should be rho 1, the volume of the original one. So this is the volume of rho 1. Yeah, so q prime is just going to be q here. Oh, sorry. Yeah, let's see. This is q prime. This goes from q1 into g. Commas look like ones. And this is the volume. OK, so they go from gamma n to, well, they're going to go from gamma n to one of these other ones. I'll tell you in a sec. So I've only done sort of the, this is a sort of recursive procedure. So I just did sort of step one. I produced one finite index subgroup, and I do another one. So now I just repeat. So let's take, so now we take q1 and q1 and apply the theorem. We get gamma q2 finite index and gamma q1, and this homomorphism, rho, let's call it rho 2, from gamma q2 to gamma q1. And if I think about this inside of g, maybe I should just to remind you, this is sitting in g. So this is now a representation that's dominated by the inclusion of q2. And the volume of this rho 2, well, it's the volume of m gamma q1. But I can also compose this rho 2 with the rho 1. So now I do rho 2, rho 1. This goes from gamma q2, then to gamma q1, and then all the way down to gamma q. And the volume of that guy, rho 1, rho 2, is the volume of the first quotient. I just keep doing this, so you repeat. So this sequence of representations, they're actually just homomorphisms from one group to the previous one in the sequence. Any questions on the corollary? All right, let me, in the last seven minutes, I tell you the idea for the proof of the theorem. It's also pretty straightforward. OK, so we start with this q and q prime. Let me draw a picture. So here's q prime, sorry. And now, well, q prime is some compact polyhedron. Let's put that inside of a ball. There's some ball of some radius, the hyperbolic space. And then let's put this inside of a bigger ball. And now, I haven't done anything with q yet. So we're going to use this trick that goes back to Scott and was used by Egel Long and Reid, a kind of discrete version of taking convex hulls. Basically, what I want to do is I want to take q and I want to tile some convex polyhedron by copies of q that contains this big ball. And really, the right way to do it is that the tessellation of h3 by translates of q gives you a bunch of hyperbolic planes. So the planes parallel to it that contain all the copies of the faces. And then I can take this ball and I can look at the half spaces whose boundaries are not arbitrary planes, but those planes contained in there. OK, so long story short, what you get is something like this. So this is some gigantic polyhedron. So this is our q1. And it's a union, the convex, sorry, I'm writing up here and no one can see. Convex union of gamma q bar translates of q. So in particular, the group generated by reflections in q1 is commensurable. It's a finite index subgroup, the group generated by reflections in q and the orientation subgroups inside q. OK, so, sorry? q is maybe like, here's q. I don't know what it looks like, but I'm just going to start reflecting this if you want. And keep sort of tiling this out until I've eaten up this whole big ball. So, all right. So now we need to construct this map. So we're going to construct a map from this big polyhedron down to the little one. And it's going to be built in some steps. So let me do, let me see if I can figure out the right distance. So the first thing I'm going to do is I'm going to do nearest point projection. So I'm going to do nearest point projection to this big ball. And now I've got two concentric balls, so I can take spherical coordinates on this ball, and I can just do a radial dilation. And if this was a ball of radius r and this was a ball of radius r prime, that dilation will actually be a contraction with ratio r over r prime. So that's an explicit sort of easy calculation. So I'm going to do some radial dilation, call this dilation. And then I'm going to do another closest point projection. Now, so what happens to this polyhedron? Well, the first thing that you want to show, actually, the diameter of these faces, even though I'm taking this ball to be really, really, really big, the diameter of these faces are actually uniformly bounded independent of how big this ball is. It's not so hard to see. And so when I do this closest point projection, these faces get mapped down. So because this sphere here is C1, and strictly convex, and everything's sitting inside, convex, this tells me that actually the map from this boundary of this polyhedron to here is actually homeomorphism on the boundary. So I get some new cell structure on the boundary. And then, well, this dilation takes that to some other cell structure. I guess I said it, but I didn't say. This can be a big contraction, right? By taking this sort of bigger sphere, big enough, this is a big contraction. So all these cells get sent to tiny, tiny little cells here. So now I've got some cell structure in this sphere. And they're sort of made of these tiny, tiny little cells. And then now I'm going to do this closest point projection. And things get a little messy here, because so what's the preimage of a vertex? Well, it's sort of this whole sector inside of here. And in three dimensions, it's something similar, but it's pretty easy to understand. So when I map through this final map down to here, these cells on the boundary, the images of the faces out here, they've got really tiny. And a bunch of them get collapsed down into the vertices, some of them. So basically what doesn't happen, though, is you don't see any faces that kind of overlap two of these faces. Every face goes into one of the faces down here. Now, because some faces go into a vertex or into an edge, you have to make some choices about where you're going to send a reflection if it goes into a vertex or an edge. But it turns out the sort of combinatorial argument allows you to do that. And what you end up with is, right, so you get a map, phi, that goes from faces of q1 to faces of q prime. And it has the property that phi of a face contains the image of the face under this map. So there's a map from faces to faces. So in particular, then, I can define a homomorphism from the reflection group here to the reflection group here by sending rho bar of the reflection in a face to the reflection in the image of the face. OK, so reflections go to reflections. Those are order two to order two. So that's one of the relations in order to get a homomorphism. The other thing is that if I have adjacent faces, those are exactly commuting, they better go to adjacent faces. But that's what this condition tells me. So if two faces are adjacent, their image is under f not r adjacent, and they're contained in these things. So these image faces, these phi of f, those also intersect. So you get a homomorphism. If you look at the orientation covers, you can sort of build everything just like we did before. You get a map from this orbit fold to this orbit fold, gamma q prime, a degree one map. So this f maps here. And so f star, this is our homomorphism rho. And the volume of rho is just the volume of the m gamma q prime. OK? So I'm out of time. Let me not write anything else, but just make some remarks. This is pretty flexible. You can make the optimal Lipschitz constant as close to zero as you want, or actually as close to one as you want. You have to work a little bit harder there, but it's not so hard. But let me also say that there are lots of constructions. So at the end of the day, we're constructing degree one maps, or in general, non-zero degree maps that are contractions. And there are lots of constructions for non-zero degree maps. So Soma, Balou and Wang, Siddharth Gettgill has a sort of general construction, sort of equivalent formulation of when you build a degree one map. And so trying to get examples where those are Lipschitz, though, seems to be a little tricky. However, Hongbin Sun has a construction of degree two maps, and I think now maybe degree one maps, which probably can be adjusted so that those are also strict interactions. OK, anyway, I'll stop there.