 Thank you very much, and thanks to the organizers, the two heroes here for and you, for inviting me to give at this very lovely place a talk about commensurability of hyperbolic coxsetter groups. Let's start. So let's look at coxsetter groups. How can we think about them? We can think about them geometrically. This is what I want to do basically today. So you take a hyperbolic polyhedron, a triangle, a simplex, a cube, a hypercube, a simplex of higher dimensions, and you assume that all dihedral angles between bounding hyperplanes are integers sub-multiples of pi, so of the form pi over m, say. This is a very nice polyhedron in general, so we do not care about the combinatorics, and when you now take the reflections in these bounding hyperplanes, these reflections generate geometrically defined hyperbolic coxsetter group. So we have a group of isometries, but we also have a very nice fundamental domain, a polyhedron. Abstractly seen, it is a subgroup of the isometry group of hyperbolic n-space, real. You could also speak about the complex case, the quaternionic case, or even more generally. Let's greet, and I assume always co-finite. That should mean finite volume of the simplex, or your polyhedron, whatever it is, co-finite. Then the fact that we have reflections in hyperplanes meeting in phases of co-dimension in two, and these phases contain vertices, that means that two reflections in two hyperplanes intersecting one another non-trivially give, when you compose them, a rotation about this ridge. So we have, because of that, very nice, oh, very nice general relations. This is the rotational condition. This is the order of your rotation, m is an integer bigger than or equal to 2, and this means that the two hyperplanes intersect under the angle pi over m. In hyperbolic geometry, we have the nice feature, which you do not have on the sphere. You can ask the same questions, of course, in another space of constant curvature. We can have hyperplanes which intersect at the point at infinity, and then we do not have a finite nice relation order of the composite, and we can also have hyperplanes which do not intersect at all. They are ultra-parallel, or hyperideal, and inside hyperbolic space, the two hyperplanes emit a common perpendicular, that we can also have when we have a cube. So here we are with hyperbolic coxet groups and hyperbolic polyhedra. That we should slowly absorb. Now the problem is, which I would like to address today, we take a particular family of coxet or polyhedra, or reflection groups. A particular family defined through a condition on the combinatorics, and let's take a family which consists, which is not empty, which has a lot of family members. And in hyperbolic geometry, they are quite rare, actually. In fact, the classification problem for co-finite hyperbolic coxet groups is not yet done, and I think during my lifetime I will not leave that. So we do not have a list of all co-finite hyperbolic reflection groups, contrary in contrast to the spherical and euclidean cases. There we have a complete picture. So again, I am too fast. I'll take a family which is particularly rich, also finite, we do not like that too much. I'm looking at coxet or polyhedra, which have at least one vertex at infinity, on the boundary at infinity. And when you are at infinity and you look at the structure close to this vertex, I would like that the structure is a pyramid over this vertex, and when you look down you see close to the vertex at infinity a product of two simplices, of positive dimension, meaning maybe a segment times a segment, so a quadrilateral, then we are in dimension three, or a segment times a triangle. This gives something three-dimensional around the vertex. We are in four dimensions, or two triangle times triangle, or whatsoever. So the structure around this apex at infinity is a product of two euclidean simplices of positive dimension, and that gives us a pyramid or some open part in hyperbolic space, and then we cut by one hyperplane opposite to this apex. So this is a very nice family. They are not co-compact, these groups, polyhedra are not compact, finite volume, at least one vertex is at infinity. And Pavel Tu-Marquin at Durham, he classified them, I'm too stupid today, I'm sorry. Not so intelligent or smart, right? So Pavel Tu-Marquin classified this family, I can say they are generated by n plus two reflections in hyperplanes with a particular behavior, and he listed them up and it turned out in each dimension we have finitely many, the number decreases with dimension dramatically, but the dimension goes up to dimension 17. And in dimension 17 we have one single example left, and this example I love so much that I like the whole family. Why is this 17-dimensional example so particular? Actually it is a pyramid with 19 facets, it gives us a group, and in that sense it gives us a quotient space, a space of orbits. This orbit fold of dimension 17 is actually arithmetic, it comes with a lattice. It is related to the unimodular lattice of type 2, which is pretty well known. And this orbit fold, arithmetic, has among all arithmeticly defined hyperbolic orbit folds of any dimension, and they exist in each dimension until infinity, the smallest possible volume. That is fantastic. And so this family is a family, these are not simplices anymore, a simplex has a vertex, the boundary, the vertex boundary is a simplex and not the product of two simplices. They exist in higher dimensions even than coxsetter simplices, a coxsetter simplex exists at most up to dimension 9, here we are 17. But okay we have to pay a price, the combinatorics is not so nice, and if you establish the gram matrix, what is the gram matrix? Well, you have n plus 2 hyperplanes, to each hyperplane you have a normal vector in our Lorenz Mikowski vector space model for hyperbolic space, and all these vectors, when you take the bilinear form of signature n1, we get a real symmetric matrix of signature n1. And this gram matrix is not invertible anymore, it is singular. That is also a price we have to pay. So the aim is now to classify them up to commensurability, what does it mean? You take two groups, say in the group of hyperbolic isometries, but this is a general notion. You can take just a bigger group, two subgroups, and they are called commensurable, here we take the wide sense because we are geometers, all of us I hope, yeah, okay. If there exists an element in the ambient group such that group G and this conjugate of H, this group, this intersection group has finite index in both G and it's conjugate in the conjugate of H. What does it mean geometrically? If you pass to a subgroup of finite index of a group of hyperbolic isometries, this means that when you have your orbifold, you pass to a orbifold, a finite sheeted orbifold, cover. You can even, we know by cell back, you can even take a subgroup of finite index which is torsion free by passing even lower. That means you can get the cover by a manifold. So commensurability is something we like to study. However, there are not so many results about commensurability. When you look at low dimensions, what does it mean, PSL2R, PSL2C, we have a beautiful picture coming from Takayushi in dimension two and in dimension three by Colin McLaughlin and Alan Reed and others. There's a whole book about that by McLaughlin and Colin and Alan Reed and there are also quite a few results. In low dimensions because there we have, we understand the isometries of these spaces very well. These are two by two matrices where traces give us a background about the geometry and trace is a conjugacy invariant, so if we are studying such problems when you look at trace fields when we have a group, that is a good, that is a very good tool. But okay, so here is this problem where we make one step further in a more complicated context, non-invertible matrices, isometry groups, which are maybe Clifford matrices or not, things become a bit more complicated. What I tell you today is, I would say, one of the first steps to do that in higher dimensions and it could be the methods which I present to perform this aim can be generalized. They are, the methods are universal in some sense, but one vertex must be at infinity. It does not work so well in the compact case, but okay, maybe you will do it. Let's continue. So what are these? These are coccet graphs and I invest maybe two minutes in order to encode or to let you know how to read or to understand what is coming next. So one node in this graph is a reflection. Sometimes people write also S for generators. So before I had an S, I do not quote it there. Two generators, so two reflections in our group, are joined by an edge with a weight M. If the hyperplanes HS and HS prime intersect and their angle is pi over M, which gives us this nice relation I showed you on the blackboard. So here, there you have an M, N, K and L. M is an integer bigger than or equal to 2. But then there are also these bold faces at these bold edges. That should be bold. I could also write infinity. This means that these two hyperplanes join one another at the point of infinity. So they are parallel in the Euclidean sense. And when you look at these in the upper half space, when you look at these hyperplanes, they do not have a hyperbolic common perpendicular, but a Euclidean common perpendicular on each horror sphere. And the composite of these two reflections gives us a translation by the double distance. So you can read these coaxial diagrams. Each one gives us a group. And what is this node over there? So this node over there. When you take it away, together with the emanating edges, you have here this situation, so something infinite and the translation. And here, if you omit this, you have that connected component of a graph. And those who know a little bit about Lie algebra theory, you know this is a Euclidean tetrahedron or an affine coaxial diagram. And so this is exactly Euclidean times Euclidean. This gives us the vertex structure of the apex at infinity. So you leave away the hyperplane associated to the apex at infinity, and you see the product of two simplices. This is a one-dimensional segment, and this is three-dimensional. So the node always tells you there is the apex at infinity. And we have two Euclidean simplices of positive dimension, giving us the vertex neighborhood. Okay? So this is the classification of two marking in dimension three. This is one group in dimension five. It has one, two, three, four, five, six, seven, n plus two. And what is this here? This is, one could call it the width symbol or the coaxial symbol or whatsoever. This is a graphical, another symbolic notation how to read this coaxial diagram from, not from the British side, from the continental side, from the left to the right. So here is a four. We start with a four. Here is a three and a three, but a bifurcation. That means we write the three, one leg here, two legs here. Then we come to this point, and we have a circuit, a circle, which means we put a parenthesis. It says we have to go three, infinity, four, okay? So now you can exercise on next slides by understanding how we write coaxial diagrams. And I show you now very quickly the full classification of two marking. It consists of 200 graphs. So I do it quickly. That meaning you take one graph from the left side, one graph from the right side, and you glue the circled node together. So we fix the combinatorial structure with a vertex at infinity, and we cut with a hyperplane. That gives already a whole lot of coaxial graphs. Then the next, you glue one of these two to any one of these. This gives higher dimensional ones. N plus two nodes gives us a n-dimensional pyramid. This is one, and finally I end up with the last one. So there are 200 groups up to dimension seven. I call it the queen. I could also call it the king or emperor. It's an important group. And it is this group here with 19 nodes. It's this pyramid group. It is arithmetic. It is the reflection group generating the hyperbolic lattice which has at infinity, which is related to the unimodular group of type two. And so if you look at the set of all arithmetic hyperbolic oriented n or before, it's quotients of hn modulo gamma, gamma discrete in the direct in the group of isometries conserving the orientation. And arithmetic, I will just come back to this word later, among this huge set which exists for each n, no bound, this group here or this group which is the fundamental group of a particular or before it realizes in a unique way the smallest volume. And this is the volume, a certain rational multiple of set of three. And that was shown by Vincent Emry not so long time ago. Now what is arithmeticity? I could now give a talk about not just one hour just to explain what is arithmeticity and to make the bridge. I will not do that. I will give you the working definition which is a deep criterion and an equivalent description for arithmetic which works quickly in our context. So let's look at the group, discrete group, finite core volume in this group of hyperbolic isometries and let's look at the commensurator. So you look at all conjugates of gamma such that these two groups are commensurable. So up to finite number of sheets, they somehow give us the same or before. Now by a very deep theorem of Margoulis which he proved in an even more general context, one can characterize arithmeticity in the following way. This group, it is not difficult to show that this is a group because commensurability has also very nice properties. I did not mention that but my time is limited and the physicists are in my back, right? I have to hurry up. So the commensurator is a discrete subgroup, is a discrete subgroup of this group and contains gamma of finite index if and only if gamma is non-arithmetic. Or in other words, gamma is arithmetic. You could take this as a definition if you believe Margoulis. Gamma is arithmetic if and only if its commensurator is a dense subgroup of the group of hyperbolic isometries, whatever. But I would like to see once a proof which is not so difficult to understand. Not so good. But okay, we are looking at hyperbolic coxset groups. They are distinguished by having very nice generators and relations. But in that case, Wienberg, the king of hyperbolic coxset groups, provided us with a very efficient and linear algebraic tool with a bit of number theory of course to decide whether such a group is arithmetic or not. It turns up or turns down, boils down to look at the gram matrix of such a group to look at cycles of coefficients and then to check a few things also concerning embeddings of your algebraic number field which you get by looking at the coefficient of your matrix. But I cannot go into this. But I want you to know there is a big separation of discrete groups of isometries, the arithmetic ones and the non-arrismatic ones. Certain tools work only for the arithmetic ones. That's often the case for McLaughlin and Reed's approach. And then there are other problems where arithmetic tools cannot be used anymore. And we are confronted here with both cases. Among our 200 coxset-pyramid groups, quite a considerable part of them are non-arrismatic. Non-arrismatic. Okay. So how to check commensurability? How can we check that a group, a subgroup, is a subgroup of finite index in another one? Don't read this. Look at me first. So if you have a group gamma and you have a group H which is a finite index, it is clear that gamma and H are commensurable. So the first thing we do is among these 200 groups we check which one is a subgroup of the other. This is the first thing one does. And for this, this is already a task. But for coxset groups, George Maxwell from Canada, he provided us with a criterion which seems to be, my God, difficult to read, you get tired, don't read this. What it says is if we have such a group here, I just give you this example. And we start here from the left with a subgroup of type AK. Type AK means linear graph K, K minus 1, 1. This is a famous finite coxset group arising very, very often also as subgroup. So here we have a A2. If we have at one end, for example, here at the left, an AK, like here, or AL minus 1, then there is a procedure to study subgroup relations in other ones. That is the condition. And among our coxset pyramid groups, this happens from time to time. So this helps to decide, ah, yeah, this group is a subgroup of index 3 in here. It is not difficult to prove, but yeah, there are not many general results of that type. The index is also given. Then another one, just a useful criterion is we work on the graphs. Here we see a part of the graph which is of no interest to us. So I call it delta, just some part of the graph could be very long. But here we have one end of the graph which has a high degree of symmetry. And if we have a high degree of symmetry on the level of the fundamental domain of the fundamental coxset polytop, this means we can cut it into two pieces. In different ways even. When the graph is symmetric, we always have some bisecting hyperplanes. It's not difficult to prove. Just to say, aha, let's study symmetry on the graph, and then we take here an associated bisecting hyperplanes, for example, by choosing this reflection or this hyperplane and this one and we cut into, we cut there into two pieces. Okay, a last useful result which I want to mention. Then we go into the dirty work, but this is already the dirty work. So you could see on these different slides there were these cycles, these triangles. And geometrically this means, so here we have such a graph with this cyclic part here, so we have infinity or bold edge, three, three infinity, five. This is the coxset, simple. When you look at this geometrically, actually turns out geometrically, we have, we take this one, one, two, three, four, one, two, three, four nodes, but here we have five nodes. That's actually a pyramid. These are not pyramids, there is lacking one node, but one can complete them. The first one you can complete it like this, infinity, three, three infinity, and the second one in this way. That means when we are here in dimension three, this is a tetrahedron, which is not compact, and we add an additional hyperplane so that it becomes a finite volume. And we do this in this way. These are pyramids. And we could do it in such a way that by adding this additional hyperplane, we can glue them together. And the gluing of these two finite volume pyramids gives us this one. That is something one can see in low dimensions, and then one has to prove it algebraically. This is here very nice to explain. These two groups inject into this coxset group, and actually this coxset group is an amalgamated, it's a free product of A and B, of that group and that group, of A and B, amalgamated along a subgroup, and this subgroup is infinity three. This subgroup here is a well-known group, it is the modular group, which in the upper half plane is given by this coxset diagram. And so it has, this subgroup here gives us this plane along which we glue two pyramids together. So here is an algebraic description, and due to a result of Karas and Solitar, I hope I pronounce correctly Karas and Solitar, the Spanish people, we have a criterion how to decide when do we have a subgroup of finite index. So if we compare two groups, which are both amalgamated free products, and we want to check whether they are commensurable, we have to check a common finite index subgroup has to have a non-travel intersection with this subgroup for each conjugate group of it, and this also a finite index. So we just have to work on this connecting hyperplane. It's a very nice result. Okay, now all these criteria, quickly, quickly, previous slide, okay, so quickly, okay. Now I kind of, okay, now with all these criteria, and there are some others, there is one unpleasant pair, unpleasant. Those are amalgamated free products, but there is not a common subgroup or a plane which they have in common, yeah, here's three, here's infinity three, that group, but here's an infinity four, they are not the same triangles, and we cannot decide whether they are commensurable or not. Maxwell does not work, amalgamation does not work, nothing works, but it is in low dimensions, it has five nodes, it lives in dimension three. In dimension three, we can compute hyperbolic volume. In terms of dialogarism functions, so not just a number, explicit values. In terms of certain well-defined functions, okay, we come back to this unpleasant pair, it's a very nice pair. Now in the arithmetic case, I was speaking about arithmeticity, yeah. In the arithmetic case, and in dimension three, there are two invariants for commensurability, which in fact form a complete set of commensurability invariants. So if we compute them, things are done. And this is the invariant trace field, so you take your group, I call it here G, you take all the elements in your group as two by two matrices, they have a trace, but now you pass to the square of this element. We are working PSL2C, not in SL2C, because of that we have to pass to the square of the element, compute its trace, form the field of all these traces over Q. Turns out that this is a very nice real number field, and we compute the invariant quaternion algebra over KG, which is not so difficult to do, because we are in a non-compact case. In a non-compact case, this quaternion algebra becomes actually isomorphic to the matrix ring over KG, look, everybody can do that. Meaning in dimension three, arithmetic case, the commensurability classification is done due to Reed and McLaughlin, yeah. They form a complete set of commensurability invariants. Now Colin McLaughlin, one of the top experts in this domain, he generalized this approach by using the quaternion algebra to study arithmeticity of higher dimensional discrete groups in isom HN, as long as they are arithmetic. So there's a whole picture with quaternion algebras, which he wrote down in a beautiful paper which was published in 2011, and that was the last paper which he published before he died, unfortunately. So the arithmetic case is done due to McLaughlin. And I will give you now very quickly, quickly, quickly the list, just to show that we worked on that and that, so I give you now the list with all these commensurability classes. How should you read this? Well, it's not so difficult here, in dimension three, I have one, two, three, four classes, four commensurability classes, because commensurability is an equivalence relation, which gives us a distinction. In the first class, I have a number four, that's the cardinality, the number of elements in this group. And these are mainly subgroups of finite index, which we got by Maxwell, etc. In the second class, we have also four elements, this is a representative, this is a representative here also, and in dimension three, we have three classes. In dimension four, we have two classes, one is very rich, has 20 groups. Dimension five, dimension six, seven, eight, nine, ten, eleven, and in dimension seventeen, we have this one group, the isometric queen, okay? So that was easy. Now let's continue. This was the classification for a coaxial of pyramids. The simplex case, which was simpler, was already done with Johnson, Radcliffe, and Chance, we did that some years ago, and used the fact that the gram matrix is invertible, and developed a ratio test with the determinant of the gram matrix, and so the simplex case was done. And what we did here, in these classes I had before, we just tested, when do we have a tetrahedral group? So these are simplices, coaxial to simplex representatives. And this means, you have in dimension eight, in the first class, in this class, you have 16 groups of coaxial to pyramids, and all these 16 groups are commensurable with this simplex group. So apparently there is a common sheet, a common finitely sheeted cover of an orbit fold, which has a combinatorial model, which you can dissect into a finite number of coaxial to simplex, and you can dissect it in 16 different ways into coaxial to pyramids. If you want to do it by hand, this is not possible. Okay, home baby, voilà. Now the non-arithmetic case, I will not speak too much about these technical proofs, but I want to mention what we used. I think that is a new idea. Don't read this, don't read all this. It's a long text. So we have our coaxial to pyramid, and very important is, we have at least one vertex at infinity. That means we have a subgroup, which stabilizes this vertex at infinity. This subgroup is a Beeperbach group. It's a Euclidean co-compact group, a coaxial to group, and by Beeperbach we know it contains a translational lattice of finite index and full rank. That means we have, when we are in dimension four, four translations, which are independent. But that means also that we have four translational lengths. And these translational lengths, which are Euclidean, Euclidean on a horoscope, we can compute them by using hyperbolic geometry, by looking at hyperplanes, which go up to infinity. They are connected, they are bounding hyperplanes of our polyhedron. And we have between vertices on these hyperplanes, a geodesic line of a certain hyperbolic length. And this hyperbolic length is directly related through cosine hyperbolic to the Euclidean distance. In this way, we look at this translational lattice, and we can check whether these translational sublattices, which we have in this group and in that group, whether they are equivalent or not by looking at the supergroup, which is a hyperbolic coaxial to group having this stabilizer. I hope this is somewhat clear what I say. Well, I want to come back to this annoying pair, which is a little bit a different story. But in this way, we can, in a geometric way, this is much more geometric together with Biberbach, but we use translations, decide whether two groups are commensurable or not. So difficult, but I cannot go into details, or you can ask me a question at the end when we have five minutes left, and I'll come back to that with a big pleasure. Okay, let's continue. So that is basically the idea. You have two hyperplanes intersecting at the point at infinity. Here we have vertices on the polyhedron, and this translational length, which is delta. We have half. There is this formula. There is this formula, which maybe many have forgotten or never used. You can read it in Cocksetter's book, but not in all books on hyperbolic geometry. That is a default. So in this way, for example, we can easily also treat amalgamated products, because they are glued copies. When one shares a vertex at infinity, the other one, we also have translational length. So this is not a drama. You don't have to read that. I continue, if I can. What's going on? Didn't I tell you that I have this unpleasant couple? I do not say unpleasant anymore, now I come more with interesting couple. So we have these two here, and this proposition above does not help to decide whether incommensurable or not. So what do we do here? Well, what is funny, I gave you here the theorem 2, which gives you a complete classification in a non-arismatic case. It's complete. So what about this unpleasant couple? Apparently, we could deal with it, of course, we could deal with it, but with another thinking. Do you want to go with me through the list? Yes? No? No. The lists are not so important. Nevertheless, it's some work. Okay. So this unpleasant pair lives in dimension three. Not in even dimensions, not in an even dimensional space. Because you know, in even dimensional space, if you have a polytop which tessellates space, then its volume is a certain rational multiple of a certain power of pi, always. That is due to the Euler characteristic. But in odd dimensional space, Euler characteristic is not there anymore, and what replaces the topological Euler characteristic are highly transcendental functions like dialogarism, Lobacevsky function, or hypergeometric functions. We are dealing now a little bit with analysis and number theory. So let's look at this pair and let's look at the Lobacevsky function. It is one-half of the imaginary part of the classical dialogarism function. Lea to z, sum, r to 1, infinity, z to the r, r squared. So you take this dialogarism function of roaches and others. You put for z a point on a circle, on a unit circle, and then you take here the imaginary part, one-half of it, this is called Lobacevsky function. And you can write it as a simple integral over the log, and that makes that this is a typical three-dimensional volume functional, because length in hyperbolic three-space is given by cosine hyperbolic, sine hyperbolic, tangent hyperbolic, something with hyperbolic, and that is always in coordinate something with logarithm, if you, with logarithm. And so integration over length gives you volume in some sense. So for these two, in our unpleasant pair, we can compute the co-volume in terms of values of the Lobacevsky function and which have arguments which are all rational multiples of pi. Oh, that is not so evident. That is not at all evident. And there are conjectures about rational combinations of Lobacevsky numbers at rational multiples of pi and commensurability of numbers which are unresolved up to today. Our approach has to do with that problem. Why? So if this pair of groups were commensurable, the ratio of their volume is a rational number, right? So if this number is irrational, the two groups are not commensurable. And those who started my list before saw, they live in different commensurability classes. So okay, a numerical check of this explicit quotient gives us a number which never ends here. We continue with high precision, but it still could be a rational number, right? But somehow this suggests this terrible number. Probably this is not a rational number. Probably. Okay, let's assume it were a rational number. Now, this Lobacevsky function, which is related to this dialogarition, satisfies three beautiful functional equations or as properties which John Milner called essential functional properties. First of all, this is an odd function. It's pi periodic. And it satisfies this strange distribution law for which we unfortunately do not have a geometrical interpretation. We don't know what this means for on the level of volume of a polyhedron. We don't know. But it's not difficult to prove. Now Milner, in chapter seven of Thurston's notes, he conjectures the following, first, a very rational linear relation between real numbers with omega, a rational multiple of pi, is a consequence of the three essential functional equations. He writes and he says, he checked that for a small set of rational numbers. Probably this is true, but it's completely open since the 70s. And B is also interesting, fixing some denominator, like 24 before, 24 was appearing as denominator. All these real numbers with k relatively prime to n and because of the other relations, we can take them, k between 0 and n over 2, that they are linearly independent over q. That's conjecture part B. Okay. So let's assume that our number alpha, this quotient volume, is rational. What happens? Nothing. Doch. So alpha is rational, so if we take 2 over 3 times 1 minus alpha, beta is also rational. Yeah? This is just a rational, that is with our expression before, beta minus 1 times the Lopacevsky value at pi over 4 is this combination. That follows from the slide before, before. Now by the distribution law applied to two different cases, once with pi over 8 and once to pi over 12, gives us for these two parts, these two terms, this and this. So if you plug in everything here up, we get this. That means if beta were rational, these two numbers are rationally dependent. But then also we have, then we have a contradiction to Milner's conjecture part B and if Milner is conjecturing something, I'm pretty sure it's true, although it's still open. But okay, then we became confident, let's prove that they are not commensurable. This is not as yet the proof. And what is the proof? Now I come back to this commensurator. These two groups of this interesting pair, both are non-arithmetic. With Winberg's criterion, one can't decide it. Then they are non-arithmetic and they are commensurable. Then they belong to the same commensurator, which is a discrete group in isome HM, right? So they belong to the same commensurator. That is now here. Let's suppose that their volume quotient is irrational. This is what we want to show. This volume quotient is irrational. And we assume the contrary. So since both are non-arithmetic, this commensurator, which contains both of them, they are commensurable as we suppose. This is a discrete subgroup herein and contains both as subgroups of finite index, magullis. Thanks to magullis. C, it contains these two groups, which are both non-co-compact. C is also non-co-compact, but has finite core volume, because we are finite index. So for this hyperbolic group, which is discrete and non-co-compact, we know due to Robert Meyerhoff, we call it atoms in a non-orientable case, we know in Dimension 3, every hyperbolic 3 or before is bounded on the level of its volume by this universal number. This is the volume of this coxate tetrahedron 336, or you take an ideal regular tetrahedron of angle pi over 3, and this one dissects it 24 times. So this bound of Meyerhoff gives us this bound from below. This number is computed with high precision, and this helps now in the following sense. So we take alpha, this is this volume quotient, but this one is the finite index in C, so we just divide both by co-volume of 3. This is the same, but this here we can interpret as a quotient of two indices. This is an integer number, this is an integer number. And now, by the volume computation which I showed you before, we have an explicit value, which is this one, and that gives us this number must be at most 16, at most 16. And that allows, so you multiply alpha by this denominator, alpha times this number can be at most 16. But on the other hand side, it is this integer, which we can also, so now we just check, is it possible that by taking integers here, 1, 2, 3, 4, 5, up to 16, multiplying with alpha, we get our other number, and that is not possible. So they are not commensurable. This allows you to conclude that there is no rational solution alpha 2, this one, number 1, with alpha being of that size. I have two more minutes, but I think I finish here. Thank you for your attention. We should leave the room in time, and if you want to come back to one of the questions I put you in the mouth, please don't hesitate. Thank you very much.