 But maybe just to save time, I will read loud a summary of what we had last time, because it was already a week ago. So we discussed a category of representation of ergodic actions. So main players is an ergodic action, S on X. And again, I reserved letters like S and T for locally-compared groups, with not algebraic structure a priori, and G and age will, for us, identify themselves as algebraic groups. So this is a typical notation that we adopt, but we have this connecting homomorphism, rho, from S to G, continuous homomorphism. And now an object of our category of representation is this equivalent map, phi to V, where V is a G action in the category, in the world of algebraic geometry. And the morphism is a commutative diagram of this sort, when the right-hand side, again, is algebraic, algebrao-geometrical object. So this is a nice category of representation, but it's not, I mean, the tension between complicated dynamics or ergodicity in one side and very simple dynamics that we have in the algebraic world create the existence of this initial object in this category that I didn't introduce the name. Usually we call it gate, because this is the entrance gate for X into the algebraic world. Another metaphor that one can use is that S on X is a naked body, and representation is some algebraic clothing that you put on it, and maybe this is the underwear that is the first clothes that it is able to get, and it's behind everything else. It's the father of all other representations. I'm just summarizing what we had. H0 is a risky clothes. And H0 is a risky clothes, yes. K subgroup. By this I mean a risky clothes in the K category, in the file of the K. And again, things I say pretty much works with very little modification over arbitrary local field of arbitrary characteristic, and in fact arbitrary complete field with absolute value. Nevertheless, just to avoid fine things, I mean, difference between groups and the K points and such things, I typically think of my local field K as being of characteristic zero. Okay, and now we had two theorems that we already proved about existence or nonexistence of non-trivial gates in the weakly mixing case. So classical PMP action and metric ergodicity, which is equivalent to the classical weakly mixing assumption, the gate is trivial. This means that this space is a point, H0 is G. And all actions or all maps for X are just coming from fixed points. And the non-trivial theorem, not so hard to prove after machinery is made, but it's a non-trivial. And this is kind of the engine of the starter of many of the techniques that we will encounter is the non-triviality of the gate when we have amenable action, which is also a metric ergodic. ME here stands for metric ergodic. And with this extra assumption that G is simple and the map is unbounded. And the remark here is just if you think how could these theorems stand together, then you should know that, I mean, typically those actions, PMP actions are not amenable, unless S itself is amenable. So these theorems do hold simultaneously. OK, and today I intend to discuss mainly, the main object is lattices in product, and I will prove a rigidity result about those, but there is some leftover from last time that I wanted to do. So I'll start with this. And this is a simple but very important notion of functoriality, very important observation related to functoriality. So what happens if I change the main player, S on X? So I still have, still I am fixing this part of the equation. But now I will consider different S actions. So I consider, in fact, if you wish, the category, I will use these fancy terms. I think it's useful, of S ergodic actions. These are the possible X guys. And to each one, the gate provides an element in the category KG varieties. In fact, coset spaces for X. Not only that we found a space, in fact we find a map between them. But now just don't care too much about this map, just we assign to X a guy g-mode age. And if we add another guy Y, then we would get some g-mode, let me number it, right? And the functoriality is the following observation. If I have a map here, pi, then I could view g-mode age 1 as an algebraic representation of X. So I will get a unique such map, which I could, I will denote theta of pi. Oh, in fact, maybe fancy annotation would be the gate of pi. I'm applying the gate functor. This is how it arcs on morphisms, but let me just denote it theta of pi. And I proved for you this functoriality, it's very easy, but it has a remarkable consequence. And it is the following. If we take Y to be X itself, but pi here to be a non-trivial automorphism, or a priori non-trivial, then we look at an element in the automorphism group of X. And also these spaces will be the same. Actually, to be very formal, there is a choice made. To each object, I can define this in many equivalent ways. I make choices when constructing this functor. But once this choice is made, then I will get an element theta of pi in the g-automorphism group of g-mode age 0. Now, this automorphism group could be identified naturally with a familiar object. And this is the normalizer in g of age 0, modulo age 0, which acts on the right, on this corset space. G acts on the left, and whatever commutes with g is a multiplication by a normalizer element, a normalizing element, where age 0 itself acts trivially. So this is an identification. And usually, now this is just a notation. Usually, I will just denote it n0 mod age 0, n mod age 0, and I hope things will be clear for notation. And moreover, if you think about it, you realize that we gain an extra equivalent. This phi 0 of ours now is x is acted upon by this product of groups. And this is acted upon by this product of groups. And this map theta provides extra equivalence of this action. So this little observation will be most important for us. So are there any questions about this? OK. So now I will fix this, and I will move to another subject. I will introduce the notion of and discuss a bit lattices in product and discuss a relevant super rigidity result. And later on, I suppose, after the break, I will prove the result. So but now the remaining of this time is a little introduction to the subject of lattices in product. So I will forget about this theory of gates for a while. So if there are any questions about the picture, now it's a good time. So let me give a little introduction. Take the ring, z square root 2, and I can embed it into the reels, and embedding of rings in two different ways. Usually embedding and it's going to conjugate. This field is totally real. So these are indeed two embeddings into the real. And the image is a lattice, is a ring lattice. A similar phenomenon happens if I take the ring, z1 over p, and I embed it as usual in the reels. But also I can take a different completion, p-addict one, which same p, otherwise the image will be compact, and get, again, a lattice. And for these two injections, I can plug the sln functor and get a natural map, which I can identify as follow. And this injection now, again, that's not entirely trivial, but it is correct, and I think we all know it here. This is a lattice. And similarly, this embedding is a lattice. And these lattices, I mean, not just lattices in groups, but also lattices in a product of groups. And they have the extra property. I mean, if I project z2 over 2 into just one factor, just one r, it is dense. And same for each of the other four maps that are on the level of rings on the board. And it follows that the same goes for groups. So we get the irreducibility, which now I'm giving you as an abstract definition. So a lattice is irreducible if, let me re, the projection to each factor is dense. That's a name. And again, these classical examples are examples of irreducible lattices in products. And in the theory of arithmetic groups, this notion is we find it all over the place. And let me now give an ergodic interpretation of this piece. If gamma, I will typically assume, let's agree that typically, I will assume that gamma does not intersect each of the factors. So when I project it, I can just identify the image with gamma itself. So there is a copy of gamma inside S1. I mean, I should have divided by the kernel of this projection map, which is the intersection of gamma with S2. But let's not bother about it here. So I will not use this cumbersome notation usually. So gamma is dense in S1. If gamma wasn't dense in S1, then, oh, no. Recall that S1 itself is in a big space. It has the hard measure. If gamma wasn't dense, the action of gamma on the S1 would not be ergodic. Because I could just factor the closure of the image of gamma, call it T, and have a map on S mod T, which is not, which, I mean, S mod T will be the space of an ergodic component. Or at least S to S mod T will be a gamma invariant map. So actually, I claim that the density of gamma inside S, whatever, is equivalent to ergodicity. Gamma act ergodically on SI is ergodic. So I try to explain why if gamma wasn't dense in SI, then the action would not be ergodic. But if gamma is dense in SI, let me explain why is the action ergodic. The thing is that the action of SI on infinity SI is continuous. So this is a bit, I mean, it is continuous. Hence, if I have a gamma invariant point, and gamma is dense, this point is invariant under S. So to show that the action is ergodic of gamma on infinity SI, I need to show that there is no invariant function, or every invariant function is constant. If I have a gamma invariant function, then this function is already invariant under SI, I claim. And it must be constant. Now, you might object, I mean, I think you don't object because you have seen it before, you might object because this is not really the case if I consider the norm topology on SI, on infinity. But there is another topology, which is the weak is continuous for the weak star topology. Coming from the duality between L infinity and L1. So there is a finite topology, sorry, a closed topology on this space under which the action of S is continuous, and therefore, every fixed point under a dense subgroup is fixed under the full group. Yes, please. Do you want SI or gamma I to have an L infinity? I'm saying that this is a phenomenon that I'm using in proving this equivalence easily. It is just a hint how to prove it. OK? I mean, OK. And now I want to rephrase once more or twice more or whatever. Now, gamma on SI is ergodic. So obviously, gamma, let's choose coordinate. I mean, just let's break symmetry here. It's easier for me. Gamma times S2, then, is ergodic on S1 times S2. Right? I just added one lame coordinate here. And now let's. So S2 acts on S2, and gamma acts diagonally. You forgot ergodic. Thank you. And now let me divide by gamma. So this is S. So now I assume that gamma acts diagonally on these, and S1, maybe gamma acts whatever. And this is like S1 acts on S mod gamma ergodic. So this irreducibility, again. So let me repeat. Of gamma on S is equivalent to SI on S mod gamma is ergodic. That's the finding. So this is a rephrasing of this group theoretical statement in ergodic theoretical fashion. So this is, again, very, very standard, but very, very useful as well. And of course, we will take the ergodic point of view and we'll use it. Now I want to. OK, we have this now. And so we understand to some extent the irreducibility. And I want now to elaborate a little bit about this notion. And maybe why is it so important in ergodicity? I will form some sort of an inverse, a converse to the notion of a irreducibility of a lattice by means of common durability. So let me do this. So let me start with, again, this setting. So this is, I mean, title here is discussion. This is an irreducibility. Assume that S1 is a TDLC group, totally disconnected locally complex group. So it's connected component of the identity is trivial. So this implies that there exists in S1 a subgroup K1, which is a compact open. Now, a very nice feature of a compact open subgroups inside a locally complex group is that they are commensurated. Shall I define the notion of commensurability? K1, K2 in S are commensurated, commensurable, commensurated? Commensurable, I think. If the intersection inside each is a finite index, K inside S is commensurable, commensurated if the conjugation, if for every conjugation of K, we get a commensurable group. Now, if K in S is a compact and open, then these two are compact open subgroups. And it is easy to show that every pair of compact open subgroups are commensurable, because the intersection is, again, open and finally cover each of these compact dice. That's why we get to. So somehow, in fact, I will explain in a minute, or I will recall in a minute, that commensurability is always related to this phenomena of group theory, of compact open things. Now, if I have a totally disconnected group, then Vadansig, the theorem tells me that there exists a compact open subgroup in it. And these guys form a commensurability class in S, a canonical one, in fact, like the commensurability class of finite groups inside countable groups, which actually is an example of such a thing. So I'm having this thing. So you say that K in S is commensurability. If S is a commensurator of K, that's clear. I mean, there is commensurability of two groups and commensuration of a subgroup in a group. Mean that the commensurator is full. Commensurator is full, yes. Standardization, not my fault. But now, let me go back. OK, so we have this commensurated group. Also, I can intersect this with gamma. And I will define lambda to be the intersection of gamma. Now, K1 times S2 is commensurated in S1 times S2. And its intersection with gamma then is commensurated in gamma. It follows, is commensurated in gamma on one hand. But also, you see, this is an open subgroup of S. And I intersected a lattice with it. So also, this is a lattice in S. So E is a lattice and E is a lattice in that line. This board, right? So now, I want to ignore K1, which is just, now I'm working in this group, which is S1, sorry, which S2 itself is a factor of it by compact kernel. So I want to project things into S2. And do this identification I do. So my notation will not be cumbersome. So please allow me to write that lambda is a lattice in S2. So again, totally, this is a projection of lambda to S2, which is commensurated, I will not write the word, commensurated by the dense group gamma inside S2. So this is a very nice situation. And I want to claim that it is a typical one. So I will now discuss a converse. So assume having lambda inside gamma, countable groups inside T. This is S2 for me. But now I'm taking an abstract setting. Well, we have these conditions. This is a lattice. This is commensurated. And this is dense. I can sort of reconstruct S1 from this data. Let me call it T prime in this context. There exists T prime, T DLC. Now this is locally compact, second countable. So I can find locally compact totally disconnected group, T prime, and embedding or a map from gamma to T prime with dense image such that the image of lambda is pre-compact, its closure is supposed to be k1. I mean, this is supposed to be S1. And the image of lambda is supposed to be k1. So the image is pre-compact. And gamma inside T times T prime is an irreducible lattice. So this is a general fact. I will not reprove it for you, but it is rather easy. Rather easy. The buzzword for this is the schlichting completion of the pair gamma lambda. Whenever you find a group and a commensurated subgroup, I mean, if lambda was not just commensurated but normal, which is a pretty much closed notion, then I could have defined gamma mod lambda. And actually, this would be my T prime in this setting. Then those maps will have kernels, but it doesn't matter. But in general, there is sort of an analog of a quotient group thing. And this is this T prime, again called the schlichting completion, which actually it's very easy to construct. You look at the symmetric group of gamma mod lambda with natural point one convergence topology. And you embed gamma in it for its action and take the closure. That will be T prime. So there are things to prove that all these properties do hold. But it is correct. So let me stop and breathe and say something. If and only if situation for lattice is in product. I started with an arbitrary lattice in a product. Only I assumed something. I assumed that S1 is totally disconnected. And then I tried to convince you that this is an equivalent picture to this picture of a commensurated for lattices commensurated by dense subgroup somewhere. And this is typical from number theory setting. This example is of this sort. We have this factor. And in fact, I want to say just by passing, because I think this is a very important phenomenon, not enough understood in my mind, that lattices in actual groups for locally compared groups are sort of, I mean, by simplicity assumption or something, they are either totally disconnected or simply groups. And irreducible lattices in simple groups are very, very special. They are not of this kind. They want somehow to be of this kind. They want some commensurability phenomena. But we cannot find this K1 inside here. So we cannot construct it. And these are very complicated objects. And somehow lattices in these groups are somewhat simpler. Or we have a dual way to view them. So always a dual way to view things is an extra tool to play with. Any question? So now I will state theorem. Let me just state it here. OK, so take an irreducible lattice and consider a K-simple group, a map, an algebraic representation of the lattice, which is generic, in the sense that it is a risky dense, and unbounded. Maybe I already said it many times. Over the rails, unbounded is included by the risky dense. So compact groups are also always algebraic. So if we assume also that G itself is non-compact, the risky density implies unboundedness. So this is a non-assumption. But in this business, piadics and other fields are very important. So let me stress. Nevertheless, the unbounded assumption. Then let me complete a diagram. Then we have, let me also give it a name. Superigidity, which is code for the following that I have gamma inside S, and rho extends uniquely into, in fact, a continuous homomorphism from S. And now that's the main part of the theorem. Let me also just stress that if I map a product of groups in a risky dense fashion into a simple one, then the risky closure of each factor will normalize the other. In fact, the full group, which is the risky closure of each, will be normalized by the risky closure of S itself, which is G. So they cannot coexist somehow. One of them should disappear. And it means that rho bar factor from S to one of the factors. So this is just a byproduct. But also when we prove that this map extends in a minute, we will prove, in fact, that either it extends to S1 or it extends to S2. Now, the following corollary is an immediate application of this theme that I was discussing here. If lambda inside gamma inside T as above, I guess this is the second line over there that I'm referring to. And also, GK simple, and I have Z dense unbounded thing, and then rho is supposed to be risky dense. But the unbounded assumption I will assume on lambda. Lambda unbounded. If I have a situation of a dense subgroup inside T, which I map into a simple group in a risky dense fashion, and now the only assumption about lambda, which is a commingerated subgroup of gamma, which is a lattice in T. If this is an unbounded image, then we can complete this diagram. So a word about this corollary, it follows because of this discussion that we had over here. I can always embed gamma inside T times T prime and apply this theorem now and have either that this map rho extends to T or to T prime. But I rule out the T prime extension because if T prime is mapped to G, then the image of lambda in it is bounded and the map is assumed to be continuous. So the image of lambda will be bounded inside G, contradicting this assumption. So the factoring here must be from T itself. OK, so this is just a sketch. If you see it for the first time, it could be seen arbitrary to you. But actually, this is a very, very important theorem by itself due to Margulis. Should I write it down? Maybe I'll write it here. I have to read here. Application that's a criterion by Margulis take T to be G itself, a lattice gamma inside G is arithmetic. Now, this gamma, think of it as lambda. And we applied things in the case G equals T in the corollary. It's arithmetic if and only if it has a dense comm-mangerator in G. So look at all elements in G which commangerate gamma. That will be a sub-group of G. In fact, the sub-group I want to think as gamma in the theorem and I want to think about gamma as lambda. Sorry for this. And I want to apply this. And the thing is the following. If gamma was arithmetic, if gamma is arithmetic, then it is comm-mangerated by the Q-points, which is a dense sub-group in G-R. I'm explaining briefly why you should believe it. The fact, the general fact that an arithmetic lattice is comm-mangerated by a dense sub-group is a result due to Borel. But you can believe it. You can check yourself that S L and Z is comm-mangerated by S L and Q inside S L and R. So this should explain the implication left to right. For the implication right to left, if it has a dense comm-mangerator, then we are in the setting here of super-rigidity. And there is a general explanation by Margulis, which partially I hope to explore in the next lecture, that explains why super-rigidity implies arithmeticity. I will not go into the full explanation of this, but presumably you have seen it, or you will see it somewhere else. So it's super-rigidity. It's this theorem that is in the heart of this criterion. And this is a very, very useful criterion for arithmeticity. OK. Any question about this? So now we have a goal. We have a theorem to prove. And this will occupy much of the next minutes. And let's see. I will erase this. So now I will make some preparation. So we're focusing on proving the theorem. I will not start with the proof. I will do some preparation. So we have gamma. We are under the assumption of the theorem. We have S, which is S1 times S2. We have gamma, which is an irreducible lattice in it. We know that for each group S i, I can find an action. I will call it bi, not xi, because I think of it as a boundary action, which is amenable and ME. Amenable and metrically ergodic. I claim that we can do that. So I'm using the claim. Next claim, easy one, but I will do it for you, is that the product action preserve these properties. The product action is amenable and ME2. So amenability, take non-empty S compact convex set. So I will use the notion of amenability, what I call baby amenability here, the existence of maps into compact convex sets, which is what I need here, really. So I'm taking an S compact convex set, and I want to show that there is a map from B1 to B2. Sorry, from B1 times B2 to C, as a grand map. So think of it as an S1 space, so I can find a map which commutes with S1 from B1 to C. Now this guy is an S2 compact. It has a natural compact convex structure. So I can map B2 inside this one. So there is an S2 map of this sort. And for B, it tells me basically that there is a map to C. That's a very quick explanation of the amenability of the product action. And let me discuss the ME aspect. So for this, I take U and S isometric metric space. So a metric space on which S acts isometrically. And I want to assume having a map, an S map, B1 time, B2 to U. Now fix B1, generic one. Use the notation. For almost every B1, I'm getting a map B2 to U, which is S2 equivalent. So since this is metric lyricodic, this must be constant. By this, I'm getting a map taking an element B1 to U, taking an element here to the constant image of that map, which is S1 equivalent. And this one is constant, too. So on the space of B1, B1 is your problem. We are in amenability part? They are compact spaces, or? Yes, OK. I didn't really explain this. I just wanted to understand why this space was compact. Yes, I actually didn't say it. You can view it as a functional analytic standard technique. You view this as a subset of L infinity B1 to C, which is a dual space, or maybe to be more precise. I can embed C inside a dual banner space and embed this thing then in a dual banner space as well as a closed subset. So I mean the game to play here, which I do not explain really. Now, I'm just giving you the ideas. I claim that you can put a natural compact typology on the space of maps from a big space into a space which is convex compact in a topological vector space, in fact. I said banner space, but it's slightly more general. Yeah, I didn't explain. And I used that, indeed, in order to find fixed point and proven amenability. The ME part is even easier because I don't need to use any tool which is more sophisticated than Fubini. If I just fix the first coordinate and look at the map from the second coordinate, it must be a constant by metrical density of S2. And look at the constant image here. I'm getting a map from B1 to U, which is S1 equivalent. And it must be constant again. So this is it. So this is the proof of this claim. And where am I? OK, so corollary, I already said that if I take a, OK, it's a general fact that if you have a group acting amenably on a space, then also each closed subgroup act amenably as well. That's a general fact, due to Zeeman, in fact. It's rather easy to show, but I will not go into it. And I actually proved to you, or sketch approved, why metrical density goes from a group to a lattice. Now this one doesn't go to any closed subgroup. Here I'm using the assumption that gamma is a lattice in S. So this is an important fact for me, because as you remember, amenability and ME is the starter of our engine, of the proof of rigidity result. Now another claim I will need is that gamma act, so the action of gamma on B1 times B2 is, in particular, it is ergodic in the usual sense. Now the action of gamma on S1 times S2 is far from being ergodic. Gamma is a closed subgroup, the action is proper. But if I just change one coordinate to the group, and one coordinate I keep to boundary, this is something I will use in the proof, then this is ergodic as well. This is not metrically ergodic nor amenable. But I will just use ergodicity. Somehow the starter will be already given. But very soon I will replace this space with this space. And this space I already give you the punchline. This space has extra symmetries. If gamma act diagonally on the left on S1, then this space has symmetries of action of S1 on the right. And with this in mind, I will be able to start getting homomorphism from S to G, or an approximation of. So this is very, very useful. So this is an important fact. So let me, before proving this claim, let me say some general remarks about similar settings. So the thing is that this space on the right is not just a gamma space. It's an S space. S1 acts here, S2 acts here. So given the space I give you the punchline, S2 acts here. So given, here is a general discussion that actually does not apply necessarily for product of goods. If I have an action of S on X, now you may ask whether this restriction is ergodic. And the claim is that this restriction is ergodic if and only if the S action on S mod gamma times X is ergodic. I want to replace this problem by such a thing. And let me explain this, because this involves some very, very easy, but very, very useful tricks of ergodic theoretical nature. So let me sketch some things. Ergodicity of a gamma on X is equivalent. I already used this very, very stupid trick with ergodicity of S times gamma on S times X. I just add a lame extra coordinate. And here this is, now so far I don't use the fact that X was a priori in S space. This is true for any gamma space. And now you can mod out the gamma action, modding out an action, taking the space of orbit is something that we are not allowed to do usually in ergodic theory. We do allowed when the action is proper. But the action of gamma on S is proper. So the action of gamma on S times X also is proper. We have a nice space of orbit. So I can take now a space of, what do I want to write? Sorry. I want a bit more room. So I will take the space of gamma orbit, and then I will denote it by this. So this S times X over gamma is the space of gamma orbits, which is well-defined and nice in the setting of the gamma action, diagonal gamma action on S times X. And this ergodicity is equivalent to that one. And this space is important. It's called the induction space from gamma to S. It's a very, very useful construction. So we replace gamma ergodicity with S ergodicity. This is a general remark. It has nothing to do with this extra assumption that X was an S-space to begin with. Now I will add this assumption. If X is an S-space, I can also define the map, the stupid map, that take S comma X. Maybe I already used this map. I will use it several times. It's a very useful observation that this map is, in fact, invertible. It's easy to invert. But it conjugate, I will not write down because it's very complicated to write down. I will explain. On this space, I have a gamma action. Let's say I'm acting diagonally on the right on X and on the left on X. And on the right on S. When I move along this map, the diagonal action on the second coordinate disappears. Remember that when I act on S from the right, I act by gamma inverse. So multiplying this by gamma inverse and this by gamma, they disappear over here. So this map intertwines this diagonal gamma map into a gamma map, which happens only on the left coordinate. But also here, the S-action, S does act on the left on S, commuting with the gamma action, because I assume the gamma act on the right. So this is also an S-space. And the same map, now don't take the S-action on X. So I have the action of S on this object only on the left factor from the left. And this map intertwines it with the diagonal S-action. So this map takes this space, the diagonal gamma space, into a space on which a gamma act only on the left coordinate and S act now diagonally. So this here is equivalent. Now when I divide by gamma, I'll just get this one times X, diagonal action. So nothing here is complicated. The top line is a general statement of induction in ergodic theory. And the bottom line, or the fact here, the fact I was given here in this remark, is saying that, in fact, induction, when I started with an S-space and I did induction, I ended up with this trivial operation of taking product with S gamma, S mod gamma. So this is, again, some general remarks. And now I'm back to this claim. Any question about this? So the induction space is isomorphic to the coefficient S mod gamma times X. This is isomorphic under this assumption. Yes. And they're isomorphic as S spaces. Thank you. OK, those are trivial things. But I think they should be noted. And now I want to go back to this claim and prove it in view of this remark. So now I want to show that this is ergodic and using the remark, this is the same as showing that S acting on S mod gamma times, sorry, time is ergodic. I use this as my X, of course. But now this is S1 times S2. And it acts on many things times S1. Let's mod out the S1 action. So this is the same as S2 acting on S mod gamma times B2, being ergodic. OK, in fact, I mean, this is, it's very easy. But somehow I use the same trick that I used here. I change a bit the coordinate of the action. Whenever S1 acts, whenever S acts on S times X, I can absorb also. In fact, I went in the reverse direction. Considering the S2 action, it was acting, S2 was a part in here. It was acting on S mod gamma and on B2 diagonally. This piece it acted trivially. And so think of this as an X. And I inverted the action. I changed coordinate. So the action will be only on the S1. The action of S1 will be only on the S1 coordinate. And then I just erase this action. When I consider ergodicity. But this is implied by the MECT of the action of S2 and B2. We already showed that whenever you have a metrically ergodic action, it implies that the product action with a PMP space, with a PMP ergodic space, is ergodic again. That was something we discussed last week. And I used somewhere, I explained, I guess, I already raised this. I explained that irreducibility of gamma in S1 times S2 is equivalent to the ergodicity of the S1 actions on S mod gamma. So I do know this. And I do get this. And I prove that one. Because this is an assumption. Any question? OK. So remember, we are proving the theorem. We are in this setting. And we proved this claim. And that claim. We made many further remarks which are useful. And maybe I will use them, but this is the essence. Let's take a break. And after the break, let's do the proof. So our task now is to prove the theorem on the right upper corner. So we are under the assumption there. And we found BI amenable and ME actions. And we found then that the action of gamma on the product is amenable and metrically ergodic. And we know what to do with such a thing. Beatec asked me during the break. But of course, we could have used the theorem that we mentioned for arbitrary group and take a space which is gamma amenable and metrically ergodic a priori. But we want very much to keep the product symmetry that we have here. So this is why we bothered to prove this claim. So we know what to do with amenable and metrically ergodic space. We use it to generate a gate. So there exists non-trivial gate. And emphasize that h is a proper k sub-loop of g. And this is a gamma-equiviant thing. And now this space, it is a product, but yet it does not have extra symmetries. Maybe I'll rewrite it here. We found b1 times b2 to g mod h. But the thing is that I can generate extra symmetries somehow by replacing this space with that space. And this is how I'm about to do it. I'll view it as a map from s1 times b1 times b2. This is not an ergodic space by the gamma action maybe. I mean, I don't know. But the thing is that using this change of coordinate trick, I can assume that the s1 action doesn't see this coordinate. So there is. So s1 acts only on this coordinate by the same trick that we have used. I will not write it down. And so for a generic b1, we get a map from s1 times b2 to g mod h. And again, h is a proper sub-loop in g. And now we have a map. Now the gamma action on s1 times b2 is ergodic. That's the second claim over there. So I can find a gate, s1 times b2 to g mod h0. I didn't use the term h0 before, which is a gamma map. But also here, now, I have extra symmetries because I have s1 acting on the other side. So this is. So now I'm using this fact. We gain from factoriality. And I'm adding this extra equivalency for some, consult my notation, for some map theta, from theta goes from s1 to the normalizer. Again, I use the fact that this space is already still ergodic. It is not as nice as the previous space, b1 times b2, which was metric ergodic, also amenable. But ergodicity is enough, and we proved it. And these space have extra symmetries. And these extra symmetries give me already some algebraic representation of s1. This is already something very, very non-trivial. I started with a gamma action, and with almost no work. I mean, we did some work previously, but it was just pushing elements from one side to another. We get already a representation of s1, of the ambient locally compact group. But we want more. The thing is that now I want you to observe that n act on the right-hand side here. And if I mod out this action, mod the action of n, I'm getting a map to g mod n. And this map is already a covariant. I mean, it's still a covariant with respect to gamma. But now this map is s1 invariant, because the action of s1 here was via this normalizer action. So this action factor mod s1. And it gives me a map from B2 itself into here. And this is a gamma map. So now there are cases to consider. I mean, is this space trivial? Is it not? So let me do this. I don't know if I'm defining all this gamma map. Which one? So what do you use, paper? OK. First step, we took a non-trivial representation here. Then I took, in fact, a generic s1 orbit in here. I replaced b1 by the group s1. This is something I can always do. And got myself a map, a gamma map in here. And use the godicity. Now I turn this map into a gate map, a minimal one. So I replaced h by maybe a smaller subgroup, h0. Took the minimal one. And got a gate map. Now, this map is a priori gamma covariant. And again, gamma acts diagonally on this one. But this space has extra symmetries. It has an automorphism group, which is an s1 action acting just on the first factor from the other side. Gamma was acting from left to right. Now I don't remember. S1 can act on the other side. And I get some map just out of functoriality into the automorphism group of the target. And the map fee that I'm having is equivalent also with respect to this one. That's a result of the functoriality. So I got myself, afterwards I decorated this a bit more. But I got this square diagram here. So this map, and these are the groups just to indicate what kind of equivalency I have. So this is the skeleton here of this diagram. And then I elaborated further. Now I divided the action. I took this space, consider it with a g left action, and an action of the normalizer on the right. And I divided by the action of the normalizer. This is just taking the map from g mod h0 to g mod the normalizer. And here I lost, I mean, this is n invariant. And since now the s1 action over here is acting on this space via n, the map from here to here is s1 invariant. Since it is s1 invariant, it's factors via just the s1 orbit space, which is b2. Maybe I'll write that this is the s1 invariant map. So I got myself a new map, b2 to g mod n. Now several things could happen. Maybe h0 is trivial. Or maybe the normalizer is everything. And this space is trivial. So now I want to discuss cases. And in fact, the proof of ours will be somehow by a case study, because we will show eventually either that the map extends to s1 or extends to s2. And we will not prove this right away. We will show that one. It will either extend to s1 or to s2 according to what h0 is here and what n is. So this is my next thing. I will now take an assumption, discuss it, and later we'll take the negation of this assumption. So I want to know that this is clear so far. So now I will assume that h0 is not trivial. So now h0, recall, h0 is a subgroup of h, which is a proper subgroup of g, right? So h0 is not g itself. If it is not trivial by simplicity of g, I will get that n is not everything. h0 is not normal. To be normal, it must be either trivial or everything. We know that it is less than everything. We assume that it is not trivial, and we get that it is not normal. So we get that this space here is a non-trivial space. And now this is a representation of b2. We got a non-trivial gamma acting on b2. Again, non-trivial is nice, but now I want to go down to get extra symmetries. Oh, sorry. And now we do the same trick. So we can take s2 times b2, just the action map, and replace b2 by a generic orbit. And we get a map s2 to g mod n, a gamma rip. So we have a map s2 to g mod n, and we take a gate, which I denoted by g mod h2, when h2 is some smaller subgroup than n, which by itself is smaller than g. So we get a non-trivial representation of s2 as a gamma space. So now we have a gamma action here and g action here. So we kind of reduced, I mean, we started with b1 times b2, then we take s1 times b2, then we move to b2, and then we replace b2 by s2. And just recall that this is ergodic. That was part of one of the equivalence definitions of irreducibility of the lattice gamma. So now we have this gamma ergodic space, and now we have this. We map it into g mod h, and we have gamma acting on one side. But of course, now we gain symmetry of s2, and that's h2, n2, mod h2, for this is rho times theta2. Theta2 is a map from s2, the symmetries of the gamma space here, to this space. And now this is just a measurable map with lots of equivalency. But I claim that this data now is enough to finish. Let me explain this one. Oh, maybe let me just ask before, maybe. Let me go to g mod n2, and this will be s2 invariant. I took the mod n thing, so s2 now act trivially. So s2 invariant just factors three point here, right? So I get a gamma invariant point in here. But of course, this gamma invariant point is also invariant under the closure of gamma. Gamma act via rho. I mean, I don't write rho, but you should understand that this is the image of gamma inside g. And this is g itself. So having such thing, this implies that n2 is g, if you have g invariant point in this coset space. And this implies that h2, which is normal in n2, is normal in g. So it must be the trivial group. So I don't know why I started writing on this side. Let me move back to the men's side. So let me rewrite this map phi. It actually goes to g, g mod h2. It's just g. And here the action, this thing, is just g acting on the right. And I have gamma times s2. And this is rho times some map theta. Did I use theta already? Theta 2. So now I get that the gate is really huge. It's g itself. And I get this extra variance. And again, phi is measurable. This is not homomorphism yet. But I claim that this picture forces having a homomorphism. In fact, it forces that rho extends maybe not to theta 2, but to a conjugation of theta 2. This is what I will explain. And what I'm about to explain now, a little piece, is very important lemma that I will not phrase as a lemma, but I will use this argument over and over again. So let me put this in a colored box. Whenever I have this yellow box, I will get the extension of homomorphism. Let me explain this. So instead of the map phi that I'm having here, take the map that take s. OK, maybe you want to know this s sub 2. This is an element in s 2. OK, I will take a new map, take it to phi s 2, and then take the inverse of theta, applies to it. And this is a new map, theta 2. This is a new map from s 2 to g. But now I forced s 2 invariance. So I get an element g. If this element was the identity, then rho is just a restriction of theta 2. Otherwise, it will be the restriction of theta 2 conjugated by this element g. So let me write a little equation. Maybe before writing an equation, let me write that we got that for almost every s 2, phi s 2, let me just change the sides, phi s 2 is g times theta s 2. And now let me use the rho invariance for every gamma in gamma, almost every s 2 in s 2. I will look at phi gamma s 2. And I have the rho invariance telling me that this is rho gamma phi s 2. But I know that I have this equation for phi s 2. So this is rho gamma g theta s 2. But also I can apply this equation right away and get that this is g theta of gamma s 2. And that is the homomorphism. So let me write that way, theta s 2. And I guess I need 2s. And this is for every s 2. Let me, I mean, I could have used, I could have write it with s 2 being trivial, but I didn't know it for the identity element. But just take one s 2 that it works and cancel it afterwards. I can cancel it from this equation and get that rho gamma equals theta 2 of gamma conjugated by g. So this is what I told you. I get that rho equals, what does rho equal? Rho equals the theta 2 conjugated by g, theta 2 conjugated by g applied to the projection on the second coordinate. And that's a complicated way to write that this is a, that theta 2 extends to s. I mean, this right hand side is defined for every s 2. So it's also by projecting from s to s 2, it is defined for every s in s. So having the right hand side homomorphism defined for every s, the left hand side rho, which was a priori defined on gamma is now a restriction of a homomorphism defined on s. So this is the resulting yellow box of this assumption. And maybe I'll put an extra. I was walking under this assumption. We are not done yet. I was walking under this assumption. And I got myself, by playing with symmetries, I got myself to this yellow box and said that, and I emphasize that such situation always implies extension. Actually, when I go on, the things will get a bit more complicated when I go on, this will be my goal. I mean, I want to prove extension. That's the end result. But now, after believing that we have such an implication, this will be our desire in the future. So that was the easiest case. So just to summarize, we are done under the assumption h0 non-trivial. So I'm now about to assume that h0 is trivial. Any questions so far? Now I'm assuming h0 is trivial. But of course, we'll have to recall where we are, because we took some way. So let us read the proof from the start again and see where we are. So we started with such a map. And soon we replace it with a map from S1 times B2, which was still ergodic. So we took a gate of this. This is where we are. And now we want to assume that h0 here is trivial. So actually, what we see here is g itself. Let me consult my note. Just OK. So also n, the normalizer of h0 is g itself. And this map, theta1 that we got, or theta it was, is a map from S1 to g itself. We had theta from S1 to n mod h, which now is g. So to summarize, we have the following data, S1 times B2. I'm rewriting this. Goes to g. And this is acted by gamma times S1. And there is a map here to g times g. And this is rho times theta. So look, previously we had theta2, a map from S2 to g, where it's written somewhere. I had theta2 a map from S2 to g under one assumption. Now I got myself a map, theta, or theta1. You could call it from S1 to g. One of them will work. One of them will be the extension of rho up to a conjugation, as we saw. But I need to prove it. So it's a bit less simple than before. What I have to do now is I want to replace B2 by S2. But this will not be gamma. Now I want to take a generic orbit here and replace B2 by S2, as I did before. Same trick. But this is not a gamma ergodic space anymore. Under the gamma times S1 action, it is ergodic. So this is what I want to use. So now I will change my point of view. And instead of viewing my representation as taking place in g, I will view my representation as taking place in g times g. I will now take this gamma times S2 space. Gamma times S1 space. I'm sorry. It's the same acting group as here. And view this as a representation into the g times g space. I will identify this with g times g, more the diagonal. So again, maybe I'll rewrite this diagram here. Just to rewrite it in the fashion that we are used to, we have S1 times S2, that space. And we have g times g, more diagonal. And we have gamma times S1 acting on this measured space and a map to the acting group here. So this is an element in the category. This is an area of gamma times S1 to g times g. That's how I want to think about it now. So once I have an area, and this is a diagonal, diagonal ASR group, maybe one more, for the action for gamma times S1 acting on S1 times S2, which is ergodic. Once more, I'll use the gate theorem and take a gate for this action, for this representation. I mean, I want to emphasize. For this theorem, which is less trivial somehow, I need some simplicity assumption on g. Previously, this was complete general nonsense. And of course, I used simplicity over and over again. And now g times g is not so far from being simple itself. But it is not simple, and this is not a problem at this point. So we have a gate. I replace delta now by another. Basic object is S1 times S2 to g times g mod m, unknown group. Maybe I can call it h3 or something. But I take a different name because it lives elsewhere. This is a subgroup of the diagonal inside g times g. And here I have the action, and here I have the g times g action. Now, I want to claim that in fact m is delta. I didn't go any lower. So I want to assume that m is smaller than delta and get myself a contradiction. Not a contradiction, and explain how I'm done in this case. Then, now delta, if I add to it any factor, left g or right g, which are normal groups, so I can take the product of delta with any of the factor, and this will generate g times g itself. If m is smaller, it is not enough to generate. So this implies that if I take first factor and take the product with m, this is less than g times g. And now I can divide by it. So maybe I'll put the name. I'll give it a name. Okay, so now I can take from here to g times g, modulo this beast. I moded out the left g action. Should it have been the right g action? Sorry, I'm doing something stupid. So this is g mod something for the right acting group. But when I moded out the left group action, I moded out, maybe I changed, I moded out the S1 action here. So these factors, via S2 to here. This is a gamma. S, one of the factor, I think I interchanged. One of the factor was the reception of S1. So m modded it out and got something which now on the quotient, S1 acts rivially. But the quotient, because m was assumed to be small, the quotient is something non-trivial. So I got myself into such a situation. So actually this is not the situation I put in the box here. This is the situation I put in the box. I now put in the box here. I mean, we have maybe a stabilizer, but now I made a reduction using normalizing. Now g is simple and made a reduction and obtained no stabilizer here. And I explained previously that under such a situation, I'm getting this thing and then that box and get extension of the homomorphism. So what I'm trying to tell you now, again, let me repeat, is that if m is too little, then I can mod out one of the factor groups and lose the S1 action and from this get extension as before, as using a lemma that I didn't phrase as a lemma. Only now I got, now still it's an extension to S2. Is this clear? I know it's a lot to swallow altogether, but that's how life goes. It's a complicated proof of it. So if I got an extension under this assumption, then with outlaws of generality, I will assume from now on that we have an equality here. This setting, I'm here. I'm having this one. So I prove that I can assume that this is delta and this is the gate for this action and this representation. And now I will use the extra equivalence once more that I just erased. I'll use here the extra symmetry that they have an S2 action. So actually there is also an S2 acting here commuting with the gamma action and it will go into the normalizer in G times G of delta, modulo delta, which act on the other side here. I mean I could have done it before with m, but then I didn't have enough information and there is a piece of information and now I have that this is delta we're talking about and delta is its own normalizer in G times G. This guy over here is trivial because delta is the normalizer of delta. This follows from the fact that G is a simple non-ambient group. So the equivalence under the S2 action here tells me that this map factors via S1. We cannot see the delta. Oh, I'm sorry. Shall I rewrite? I'm not sure this is better. So now I have the normalizer of delta, which is delta itself, and I have S2 equivalence because S2 acts here and commutes with the gamma times S1 action we have. And because it's mapped into this trivial group, the map here is S2 invariant. So it factors via the S2 invariance. So I'm getting actually a map from S1 to G times G modulo delta, which recall this is just G with left and right action, and now gamma times S1 goes to G times G. We are back in such a situation that we know the map that rho extends, and this is for the last time. I mean, under all the situation, we are done, rho extends, and we are happy. Okay. That was long, but as you see, it's just manipulation with things and you do things that are forced on you and you use extra symmetries, and there are no deep ideas in it. Just using over and over again the idea of the gate map gains extra symmetries, and the product structure injects extra symmetries in our situation, and over and over again we also use the assumption of simplicity of G, which somehow forces things to go to, not to sub-portion, but to G itself. Okay. So this is it for lattices in product, and next time I will discuss super agility, super agility for higher rank lattices, but let me now give a brief, so now we are out of proof mode. You can breathe. We discuss again. I want to say something extra about higher rank super agility, preparing for next time, and also say something that tells you a little story which is related. So I'm raising. I don't need any of this anymore. Just starting with a new subject. I'm not starting. I'm making a little discussion in preparation for next time, so next time I could... So I have an extra quarter of an hour, no? No? Ah, if you're too tired. Yeah, it will be light. I want to say something about mobility super agility, and I will come back to this, so I just want... So if S is maybe a real semi-simple group, or more generally, S itself is a group defined over a local field, and gamma inside S, an irreducible lattice, so classical situation, then if we have gamma to G, and this is K-simple group, all this assumption of ours, Z-dense map and unbounded of higher rank, of course, an axiom which I will discuss. So now this is just a brief. Then Margulis super agility says that under such assumption we have extension. Now, we have two cases to consider. S is a product, and this case is done. Or it is done by this more general theorem that I proved that did not assume any algebraic structure on S, only the product structure, semi-simple, non-simple, in fact falls into this picture, or S is simple of higher rank. And this we'll do next time. But now I want to tell you that we are very close to it and to say something about the relevant idea. So, but now an example here will be SL3 over a local field, maybe SL3R. So now let me make an attempt. Next time we'll approach this problem more scientifically and we develop new machinery, new heavy machinery, just an elaboration of this category of representation I explained so far, but it will be an extra layer of work. So now I want to explain how just using these things that we have over here already gives us a match. And so sorry Pierre, but not how things, only one idea. So consider inside S the group T, which is not a diagonal group, not the full diagonal group, but a singular one inside S. Now SL3. It is important group. I mean we know it is associated with the wall of the vial chamber. One thing about it is that this is amenable. It is a billion. Non-compact. So these two things put S mod T as an S space also, then as a gamma space as an amenable and ME space. Remember amenability here just follow for amenability of T, but the ME here assumption is a special property of simply groups that I know that the gamma action on S mod T is ME. This is something I solved you last time. So we have this, but also this group was chosen. I could have chosen the full diagonal group. It will have the same property, but this group has a huge centralizer. So by this you gain extra symmetries. Centralizer, also normalizer for that matter, is the full GL2 that you see here. I mean it's GL2 here because I put here that inverse of a matrix. And if you call this N, then N mod T is PGL2. This piece over here tells me that if I want to run an argument as before for proving super rigidity, I will get S mod T, take a lower level, S mod T into some G mod H and here I have the gamma action, but also a normalizer in S of T and here I'll have, I will have PGL2 acting commutingly on this and I will map it here somewhere. So in particular, I will get a map from PGL2 to something, non-trivial. So out of a gamma representation, I'm getting a representation of a big group already. Maybe it is trivial, but one can prove that it is non-trivial by a technique that I will take a different approach, but by such a technique I can show that, but there is a technique to show it is non-trivial and you see we are quite close to have an extension of a gamma action just by the simple technique. So this before I worked further, which I will next time and make an extra category of representations and now in a few minutes, I want to tell you a story, I will not write down things, how this idea helped us with Capras and Le Corée to prove a very, a theorem that I like a lot and this theorem that we proved is about non-linearity of a group acting on A2 tilde buildings. Now, SL3 over a local field, it acts on an A2 tilde building and the Breua-Titts building. If the field is R, we have a symmetric space, if the field is non-alchemidian, then we act on a building and this space S mod T is what one calls the singular carton flow. So we have an action, we have an ergodic flow that one can construct. I mean, now in my discussion now, gamma is not a lattice in a League of S. Gamma is, it acts geometrically on a certain geometric object which maybe does not have extra symmetries, but one can imitate this space and construct a metrically ergodic and amenable space, we call the carton flow, and study representation of these and out of these, find a representation of another group, PGL2, which now is not discrete anymore, but a continuous group. So if you have an A2 tilde building, and this is something I learned and I want to share this information because I think it's beautiful, out of it, you can construct, I mean, even if it does not have symmetries at all, a priori, as an object, you can construct a huge group called projectivity groups. At infinity, we have a projective plane. If the building is not classical, it's not above what it's building, this projective plane is not classical. So it's an exotic projective plane. Maybe it does not have symmetries. The classical one are acted by PGL3, or fulfilled. Non-classical maybe does not have symmetries, but they do have partial symmetries and a rich group of such, and maybe you saw it if you played with finite projective planes. I think now of a projective plane as an instance geometry feature, and for finite projective planes, one usually proves that we have the same number of points on every line, which, by the way, is related to an open conjecture. Is this necessarily a prime power, plus one. But the thing is that we have something happens for every line. And why is this? Because we have perspectives. If we have one line and another line, I can take a point out of this picture and to each point here, I have a unique line through this point and it must intersect this and I get a point here. So this gives me a bijection between these lines. So somehow these two things are related by a symmetry coming from here. And this gives us a group point of symmetries of this object. And if you think of the stabilizer in this group point of one object, of one line, we get a very, very rich group of symmetries, again, of lines. And if we have started with a projective plane which comes from the boundary of an A to T building, each such line, projective line, could be thought of as a boundary of a tree. And we get a very rich group, a street transitive group acting on the boundary of a tree just out of this geometric interpretation. And this group is that one. And just by constructing such a thing and working quite hard to show this property of this, we get a representation of the projectivity group and we show, in particular, that this group is linear if and only if this locally compact group, simple locally compact group is linear if and only if the building is classical. And this is much easier to show. I mean, it's not triviality, but it's much easier to show. And by this, we show that, by playing with such a game, we show that this gamma, a group of symmetries of an A to T building, is linear if and only if the building is classical. So I just wanted to share with you this in a seminar fashion. This is not part of the course. So I took this extra 15 minutes. Sorry for that. But this follows from the same idea of proving super rigidity. And next time, we are going to attack these questions. And again, I'm going to construct yet another layer of complication on the category of representation of ours, but it will be worth it. So stay with me. I'm done. Thank you.