 instinctively or without any ‫בספר סיפור לדלת לנקודם, ‫בספר סיפור לדלת לנקודם, ‫בספר סיפור לדלת לנקודם, ‫שצלוי מלך קצת, ‫אז אולי פסט דריכת, ‫אבל לא חשוב, ‫אבל יש עבוד חבר ‫ששמר, ‫האנחנו לא יש סיפורים כפה כדי דלת ‫לספורת את זה, ‫אגב, כמובן דלת, ‫אז נראה יש הרבה סיפורים, comes from arithmeticity. But so maybe being a lattice in product of three groups already is so rigid, so it implies arithmeticity, we don't know it. That's an open problem. For a product of two groups there are examples of it, which are not arithmetic. First example that I am aware of was constructed by Danny Wise. And independently by Brogier Mozes, הוא פרוב סימפליסטי של איזה קרובים עד פרידות, של איזה קרובים עד פרידות של קרובים, ויש כהלוי קצמודי קרובים שבארפינת פילס שתתעשו על רמי וכפרס רמי. גם אחד יכול לנסות סימפליסטי של איזה קרובים כך, אז זה מאוד ערוץ. ובכלל, כפול, תרשו על אל שארפ понравוף הרבה הקרובים שקרובים כך acho LOL ולכבison, גרוע זאת equivalent וכividade יש dale כמובש tap אבל אני ח gebru הכניכה זאת שאתה מעט קירורה baru ח Aku וכן היום אני מנסה לך מגולי סופריגיטי של קורסים סופריגיטי של היראנק. אבל כאן, אני already, אני already gave some an attempt for proving this, ואתה רואה כשאנחנו יכולים לדבר, סוברל, נונטריבר סטפס דווקא, כדי להתכנת את כתובילים של כתובילים שאנחנו התעשבים כך, אבל אני תל exped Will add a new layer of complexity of this category and by this I will get actually a very simple proof eventually. So the title is now a new category of representation. So again I ask you to bear with me with some general nonce and stuff, So, previously, we represented a narcotic space X. Now, we will not take arbitrary X, our space that we represent, will be the locally compact group S itself, thought of as a space. So, we start with a locally compact, second countable group S, and we consider it as a space on which groups act on left and on right simultaneously. So, I'll denote for a reason, a left-acting group by gamma and the right-acting group by T, closed subgroups. And I will fix also an algebraic K-group, G, K is a local field, as usually, and a continuous homomorphism from gamma to G. That's a representation of the group gamma and I will study a representation of this feature in the following form. Now, I will vary as before with suggestive notation, a KG space V, that will be the recipient of the representation. So, I will take a group of automorphism of V commuting with G, which is a K-group by itself. I mean, the automorphism group of V in general is not an algebraic group. I mean, this could be some sort of an infinite-dimensional feature, but I'll take an algebraic subgroup of it and I will have a representation, theta. So, gamma and T are not symmetric. For gamma, I fix the homomorphism and for T, the homomorphism to L will be part of the varying structure and V will be a G times L space, as you see here, and S, that's a measured space for me, it acted upon by gamma and T, and the heart of this construction I'm giving is this map phi, which should be equivariant with respect to the G times L action via rho times theta. This is a measurable map only. So, it does not... So, phi does not, a priori, respect the S-group structure only in parts, it respect the gamma structure on the left and the T structure on the right. So, it sees some pieces of the full equivalence that I might want to gain. Somehow, the game will be that I want to prove super-agidity result, I'm starting with rho, I want a map from S to G, and I will have it by pieces, I'll start with maps from T groups, which are small in S, to some algebraic groups, and somehow I will enhance this structure using these tricks of commutativity that we played before. So, all these are objects in this category of representation that we study, and morphisms will be given by this more or less same diagram that we had before. I replace previous guy X by S, and now I'll have V and maybe another U, and here I have just a reminder of the action of this one, G times L1, G times L2, maps, and diagram is a variant, and a morphism is this alpha, which is a KG map, morphism of varieties, as we had before. And of course alpha should be gamma times T, I take variance by this, by commutation, and, okay. So here I didn't assume much yet on this thing, I didn't inject ergodicity just to get this, but now I'm about to make some assumption and get some theorems. But first of all, is this world of representation clear? Yes, please. So then what's the relation between L1 and L2? הפריורי, no relations, I mean, I can always find, maybe I'll put it into, I can always find map. I mean, I can always pretend that there is a group here, and it's a reception of the product representation, and in fact I can take the zericy closure of the inside, but I don't insist on any a priori relation. So let me erase this, so just not to confuse us with the picture. But this diagram needs to obey the T-comotativity as well. Yes, take a variance. Thank you for asking. Okay, so now again, we should, I want to emphasize that this is a space for me. And I mean, of course, it has an extra structure that we will use. And in principle, I could develop this for other spaces on which I have by actions, commuting actions, and this is reasonable, but I restrict myself to this for simplicity. Also, one can play with definition and make theta part of the invariant structure, et cetera. I mean, one can play many games, but this is the game that we found useful eventually. So we stick to this category for presentations. Okay, now, a theorem. If gamma is a lattice, so S-mode gamma is a PMP space, and the action of T on S-mode gamma, I'm writing it left to right, but it should be somehow, conceptually, right to left action, is Me, קלאסיקלי, what we call weekly mixing, then you should remember that we don't have a representation of this X, T representation of this X. That's what we gained from weekly mixing. And it will follow that there exists an initial object, maybe gate in this category. And it's of the form S goes to, it will be a corset space. And here G, and as usual, I'll have this one here, and gamma times T, rho times theta here. Of course, for some H0 and G, קלאליבראיק, and that's what I claim. So we need a bit stronger, go this to the assumption that we had before, we need weekly mixing, but sometimes you remember our discussion in semi-simple league groups setting, we have weekly mixing for free, so this is not an issue. And I will give you the proof, but you already know the techniques, it's the same all over. So since I expect, maybe you don't see it from the very start, but I expect the reception to be just a corset G space, so what I need is just to look at possible ages that can be put there. So I look at all age, K-group, too early, full, such that there exists, and stands for the normalizer of age, I will not write down in details that you understand this, such that I can't put this structure on. That's a big collection, and of course G is inside it, it's a non-empty collection because I always have the trivial representation. And I'll take H0, a minimal element, and I will try to convince you that this one will be good for us as before. So as I did at some point in the past, I will now try to illustrate a diagram of S representation, I will not write S and the arrows inside it, just the receptions of the representation in a natural way. And it will be basically the same diagram that we used when we discussed the existence of an initial object in a previous category. Only, of course, I will have to argue further about some things. So first of all I have a fix, I need to fix, I'm sorry, some representation with the extra decoration that I'm not putting. I'm fixing some representation V and I'm trying to explain that G mod H0 will have M up to V. Is this clear? I mean, what I'm trying to do? We've seen this before. So here is V. And here is G mod H0, and again our task is to construct an arrow, right to left. And as before, I'll put here G mod H0, the product space, and I will have the product representation into them, because the L-reception of T will be the product of Ls. Again, I'm not putting it into the picture, but that will be the thing. And now here is the non-trivial part, maybe. So you see T act ergodically on S mod gamma, it act weekly mixing, which implies ergodically. So T times gamma act ergodically on S. So this is an ergodic thing that I'm representing into V, which is a G times L variety. What I'll get here is a G times L's corset space, inside which receives a mark from S. The T times gamma ergodic measure on S, when sent to this variety, will hit one orbit. My problem is that this is not a G orbit, it's a G times L orbit, but this I will resolve by this extra ergodicity assumption that I put there, okay? But so far I'm having this diagram, and now I need then some discussion. So let me then put the map, which it's not explicit here, into G times L mod M forward. And again, M here is a subgroup of G times L, and take pi, the map to L. So in here I have pi M. And now let's project on this L coordinate. So the result of the projection will be this thing, L mod pi M. I just somehow took the space of G orbits in here. But G was the reception of the gamma action. So this will factor from, maybe I should try it on the left just to be clear, this will factor from S mod gamma. And this will be a T-rep, you see. Now this is T acts on here, weekly mixing. And weekly mixing implies every rep is constant. I was emphasizing it very much in the past. So this means that this is, the image is just a point in here. A T invariant, T invariant point. And this is a trick that we were using many times and I will use in the future, but let me remind you. The point is T invariant, so also it's invariant under T bar. Now I guess that I missed something. Excuse me, forgive me. I want to assume always that the map theta is a risky dance in T. This is really not a restriction. I'm just, my L guy is kind of a free player. I'm just taking an algebraic representation of T and the ziric closure of it as my L. Sorry for not writing it from the very start, but now after having it, you see that, I mean, I used here the ziric closure of T and this point now is L invariant. Now this is, this justify this equation. So this means that pi M is T. Pi M, excuse me, is L. I have an equation here. Okay, I will. And this means that the G action on L times M G times L mod M is transitive. So this space, as a G space, this is just G mod H for some H. So I plug this information in here. This is G mod H. With this extra information, I'm coming back to the basic diagram and somehow now I can complete it in the way I would and I did in the past. So we found that this is a G space. This is G mod H and this H is one of these guys that I was looking at here. H, I'll give you a call, now H is in C, in the collection C. And now I'm getting this map here and by minimality, I'm getting that this is an isomorphism. That's what we had before. And then I can climb this isomorphism back and project myself onto V and I find myself this map. This is the proof. Any question about it? It's not so hard. There is a trick here played by some algebraic game, but really what's beyond it is that there are no algebraic representation mod gamma. This means that somehow the G part sees it all and we have transitivity and this collection, I mean, if I would try to minimize over the ms in G times L when L is arbitrary and its dimension could grow arbitrarily, I would fail. So I really need this extra ergodicity assumption, but now I have a restricted subset of ages all living inside same space G, for which I can apply an ethereality. And we win. Okay. So, again, we prove this theorem. And now I want to apply it in a suitable context, but before this I want to remind, we need sometimes an engine to our, I mean, a starter to our, engine, so here is a theorem. If father T is amenable, the action of gamma on S mod T is then amenable. That's for free. I want to some father that is ME. Again, something that I told you that I get for free in the semi-simple world, in the semi-simple setting, and if G is K-simple, my map raw is unbounded, which means the image of gamma is not pre-compact, then we get non-triviality. These are all assumptions to get that this H0 here, I mean, G mod H0 is not a point. The proof is very easy, given what we've done before. I mean, those many assumptions are just the assumption to treat the representation of S mod T as a gamma space. For L-trivial, we have S mod T, sorry, I like it. Consider S mod T as my X, and consider it as a gamma representation. This is a minimal and metrically ergodic, so it implies that there exists some H and a map X, S mod T, to G mod H. I want now to view it as a map from S, and here I will act by G times the trivial group. I have in this complicated category for representation, I have these two theorems, and now, from now on, I will assume, I will not even give a name to the local field, you can think of S to be a simple Lie group, I want it to be a non-compact simple algebraic group, maybe over another field than K, on which G is defined, maybe it's over the reals. So this will be a standing assumption for me now. קרלו. אני חייבה. כן, תודה. לבקשה, איך זה נכון לגמר של גמר, ואיך זה נכון? כן, מה אנחנו עובדים, אני חושב, כן, מה אנחנו עובדים הוא, שזה S-mode גמר כתריבי אלגיבריקה, כתריבי תספי, לכתריבי תספי. So I know I want this property, the only way I know to guarantee it is weekly mixing but I suppose some other gothic assumptions can be used just as well. Yes, but you need metric erodicity for example, I mean I don't even need gamma to be, sorry, metric erodicity by itself is not enough. Okay, I mean we can discuss this but I suspect that weekly mixing is more than enough or maybe I'll add here before I will always assume that gamma is a lattice. And maybe a remark, many of the things I'm about to say also works for semi-simple groups but we don't care about semi-simple groups anymore, we prove the super rigidity result for product of groups without any linearity assumption on S1 and S2 in the past. So we had this strong theorem for product so now when actually considering league groups I can just as well assume simplicity. So we'll assume this and recall for every T in S closed non-compact gamma on S mod T, T on S mod gamma, RME. So this is the assumption that I had in those theorems, I get it now for free when now consider only the sympathy group setting. Okay, question so far, what I'm about to do next is discuss functoriality as we did before in this setting. And again, I remember how powerful was the simple functoriality that we got out of this abstract setting. So let me be clear about what are the categories that we consider. First category, so we'll have two categories, categories of ergodic theoretical setup and category of algebraic geometrical setup. Previously, the category of ergodic theoretical setup was gamma act on various spaces X and Y. Here I have the fixed data S and gamma, what will change is the T, is the right hand player. So fix S, C, gamma as above with all these assumptions and consider gamma S actions. Objects are closed non-compact subgroup T in S, maybe I'll write it from the right, more suggestive because they will act on the right, gamma will act on the left. And morphisms, maybe this is harder to guess now. So let me put it in the aromatic fashion first. So I'm having S, now I consider only ergodic theoretical sites, not representation yet. So I'm having S and I'm having the gamma action here and the T action here. And I want to move it, maybe this is T1 and I want to pass to T2, another set of action. So here I will take S in S and will act acting on the right. So when little S will act on the right, that will be an equivalent map for the gamma action, which act on the left, it commutes with this. But it will not be equivalent for T, but these are two different players, but it will conjugate the T1 action into a T2. If T1, so just to be clear, this is X in S, move to XS inverse here. So it's S in S such that S T1, S inverse, what I will call T1 to the S, is in T2. This is the space of morphisms and this is how they act. And this condition really says that this is an equivalent thing. So this is the natural thing to consider here after looking at it for enough time. So this is now a category, if you wish, if you want to use this language, of representation of gammas and partners on S. And again, the objects now are just the different partners that gamma might have. The category of G spaces is much simpler, so just put it to be clear. Objects are Vs on which G act, G is fixed, so it's the V here, and with L commuting action. This and that are the varying data. And morphisms, of course, you know it, but write it down. V to U, KG morphisms, without caring about electrical variance, as I said. So this is completely algebrao-geometric category. And now the functor between these two categories. Okay, so we have one object here. Let me put gamma times T1, and I will assign to it, of course, as you might guess, G times... Because I have here the map phi1, I don't care about it. The functor is the association to S of D space, right? But of course I have this extra data, the map connecting it to phi. Okay, that's the gate, nothing about it. But this diagram that I'm trying to now illustrate on that board explains functoriality for maps, which is very important. So assume we have another object, gamma times T2, and a map here, given by a little s, and corresponding conjugation map. Now this guy comes with another G-mode H2, on which G times N, N2-mode H2. Act, everything is connected here in a nice way. The thing is here that this guy really is a representation of that object. And this is an initial object, so there must be such a map. Because with the color, this is because of the gate property of the first object. That's what being an initial object is all about. So let's call it maybe theta of S, that's alpha of S, that was used here. I guess that's rho times theta 1, rho times theta 2, and I used alpha previously for those morphisms. But now this alpha depends on this little s. So, or maybe also you can call it, if you are a formalist, you can, and if you think of this as the gate, so this is the gate functor. Just to be clear, this is the gate functor applied to the morphism little s. But of course, I will use the symbol alpha. But now, as we had before, we had this very neat, in particular section. Is this clear? It is clear. It is formal, it is nothing too hard, but now we get a non-trivial thing. So we get a map between the automorphism of this gamma ST. Now, fix one object, we get a map in particular from the automorphism of this one to the automorphism of GMOD H0 of the gate. Now, this is in particular for a given object. I'm missing some words here when I write. I apply this for this equals that, for one equals two. You get this for three, and you get this connecting morphism alpha. But now we have interpretation of this. This, okay, this is not entirely, I mean, this sits inside N0 or the H0. I mean, these are all maps that, this is all KG morphism of V. What is this guy? These are all maps that, where is this condition? It should be here. All the s's such that when I conjugate T, now T1 and T2 are the same. When I conjugate T1, I'm inside T2. I mean, principle, this is a bit less than normalization, but with whatever, hope, hope or whatever. I mean, for those groups that we study, this is just normalization. This is the normalizer in S of T. I maybe I'll emphasize this, because now we have too many normalizer. Normalizer in G of H0, mod H0, and we get this map alpha. Now, a bit more explicit than what is written here. This is the non-trivial data that we got, and as before, this is the reception of this homomorphism. Now there is a funny thing. I didn't divide by anything. This is a true normalizer that I'm getting here. For any element normalizing T, I'm getting them up here. And this is a non-trivial data. And actually, I mean, the proof was now very, very easy, very, very formal. The ergodic theoretical part in it was this weekly mixing thing. That we worked in the first lecture to explain its effect on algebraic representation. I mean, the fact that weekly mixing cannot be algebraically represented. And of course, this fact here, which is related to how more. These two things together gives me, by pure formalism, this non-trivial connecting homomorphism. And now I will explain how can one use it and get a non-trivial result. Questions? When you say, so what's behind it is how more perfect it is. Yes, I mean. Okay, yes, I need to do something. I try to avoid those things. Yes, this is how more. And maybe let, to avoid this, let me stick to the group setting. I mean, yes, there is some subtlety here, which I'm putting under the rug. Thank you. I mean, how more doesn't work for S not connected. So you need connected, but also connectedness in what sense. I mean, there is the, the risky connected is not enough. Sometimes, typically you have something called S plus inside S, the group generated by unipotent, which is not S itself. Over algebraic closed fields, things are nice, but sometimes you get some extra complication. I'm not dealing with it now. This could be settled. But yes, I want to focus on the ideas and the simplest setting for this. Okay, and maybe I just now add one more line. Maybe it's obvious, but now we get extra invariance. This guys act on, this act on S and this act on G mod H0. And we add the map F0. This is the gate map, which was supposed to be gamma times T equivalent. But now I claim that it is, well, it is gamma equivalent, but also it is equivalent under this extra thing. So this bigger. So this is extra equivariance. I want to emphasize. And I'll put a corollary now to this. If T1 and T2 in S normalize each other, then they have the same gate. S to G mod H0 gamma rho times theta i. I mean same guy F0 could be used to illustrate both gates. Why is it so? Say assume T1 is included in T2 and normalized by it. Then what I showed here is that the T1 gate is actually a T2 object. I mean the map is invariance with respect to T2 as well. So and obviously every T2 object is a T1 object because T2 is included in T1. So again, so if T2 contains T1, maybe this is easier to see, but then apply to T1 and the group generated by T1 and T2, which normalizes T1, and they have the same gate also T2 is contained in this and normalized by this group generated by T1 and T2. It has the same gate and you get this. Okay, so it's the same map phi. It's actually the same. Yes, it's the same because that's the phenomenon we found here. It's the same map phi0 that is actually more respect for the invariance than intended by its universal property. It's actually the same map and this is very, very important. Thank you for asking this, Friedrich. Okay, so we've been working and we got some theorem. So we defined this fancy categorical setting. Maybe leave this. This might be useful at some point. We found that under this standing assumption, we always have initial objects and the initial objects have this extra equivalence for the normalizer. That's what we found and also now I manage not to erase if T is a manabelle, this initial object is non-trivial and this is an important starter as I emphasize many, many times. And now it's time, even before the break, we managed to discuss Margolis' superigidity, the simple case. So I'll set the theorem. Let S be higher rank. I think everyone here knows what it means, but yes, please. I have missed something. So this theorem, have you provided a proof of it? Yes, I think it is... I did it very, very fast. Let me repeat it. If T was a manabelle, then S mod T was a manabelle gamma space and we view it as a representation of such in the setting that we studied earlier and we find such a non-trivial reception here and we view it as actually a map from S, which is trivial on T. So that's a very easy way to do it. Maybe this is not the most efficient way to do it. Maybe I can go down and take another edge with the price that T will not be invariant. I mean the T morphism will not be trivial anymore, but this is an object in the category of such representation which is not trivial. Then the gate, this is... Maybe I'll write it down. This is a non-trivial object and it implies that the gate is non-trivial. It's just a deeper object. Thank you. And this is really... Okay, I said this many times. Now let S be a high rank. I will just take this setting. Setting simply group. I need to explain what high rank is. Again, I think everyone here knows, but I will give a brief introduction to this and gamma inside S a lattice. And since I'm being simple, I don't need to assume irreducibility. And let G be K simple, K a local field. So really I do want to emphasize that the reception is quite arbitrary. I mean, it's defined over an arbitrary local field. And maybe I should say, I mean, this could be generalized for a local field of any characteristic and also not necessarily a local field, but a complete field with absolute values. And here one should use a work that was done with Jean de Courailles and Bruno Doshan for this generalization. In my exposition, typically K was a characteristic zero local field just for simplicity. And assume we have this row, which is a risky dance and unbounded. Then row extends uniquely to a map from S. That's the formulation. And now, again, you know what higher rank means. S is simply group. In fact, you can think of it as an algebraic group over the reels. This theorem actually applies in a more general setting, but I avoid discussing this. Whenever we have an algebraic group defined over a field, we can look at the maximal split source in it, whatever that means, and look at its dimension. And that is, by definition, the rank, the split rank of the group. But let me give you nevertheless a higher rank. A higher rank is just a convenient way to say that the rank is bigger than two. It is not rank one. Rank zero is compactness. Rank one is a very special feature and rank two is more generic, a stabilized feature, but here, this fact now is an equivalence formulation of higher rank, which is what I am about to use. There exists t1 to tn, such that we have both property. Together, they generate S, ti commutes with ti plus one. They sequentially compute. So this property, maybe, is equivalent to higher rank. So higher rank is an adjective describing the group S, and it is equivalent to this thing over here. Let me illustrate this. Take S to be S3R, and take our groups to be S1, T2, and T3. I'll take them to be the group consisting of parts. I mean, here they are root groups, and these three are the ones that generate together the standard Heisenberg group. Do you know what ti is? Yes. Billion, yeah, yeah, yeah. Thank you. That was too trivial so far. Non-compact subgroup. I mean amenable will be enough for applications. In fact, amenability of t1 will be enough for applications, but definitely. And also, normalization will be enough for applications. But just to get, but really, higher rank is equivalent to this one, and I'm again continuing to illustrate this. We have T1, T2, T3. This is part of a six groups of root groups in S3R. And this triple generates in the Heisenberg group. This is the center of this Heisenberg group. So it commutes with this one, and it commutes with this one. But now, I'll add a fourth group. It will be that one. This triple is the same as this triple up to conjugation. Just renumerate your columns up to conjugation by an action of S3, conjugation by permutation. So also, this is the center of that relevant Heisenberg group. And you have commutation here, and here, and same holds for the next. T5 and T6. And in fact, also, this one commutes with that one. And if you haven't seen this before, just check it. So this kind of trick, this is a game we played with root groups. You can do always, if you know. I mean, in fact, you can take this as definition of high rank, if you haven't seen this. If you saw it, definition of high rank, definition of rank, you can prove it. Another candidate will be not to use unipotent groups as I used here, but take singular elements in Torai. You always have Cartan groups, which are two-dimensional or higher. And I can take singular things in them, and the singular things will commute with others and commutation, by such commutation, you can go all over the place. If you want the symmetric space or the corresponding building, you can move on it via singular elements and change your apartment, change your flat, and go all over the place. So the smallest n, given s, the smallest n is some, that's relatively the root number. I suppose, I suppose if you take, if you take singular, singular semi-simple element, you can be maybe a bit more efficient I'm not so sure. If you want to go over root groups, this is some root space, root system computation I never carried about. Can you go down to 2? Sorry? Can you go down to 2? No, 2 will generate an abelian subgroup. Maybe you can go down to 3. I, typically I think, no, 3 will have center, the second will be a center, it will be some Isabel group. Maybe 4. Okay, we are well prepared for the proof, it will be very, very easy now. After all this preparation, maybe you can see it by yourself. Let's take a break, and after the break I'll give it to you. 12. I keep this one. Okay, welcome back. So we stated Margolis superigidity under the assumption that S is a simple group of higher rank, I explained what higher rank means here, and now we're ready for the proof. So the thing is that all the TIs, so I'm using the TIs that I presented there in the fact, and I will use this corollary that they normalize each other, they centralize each other, so they have the same gate, each pair. But of course this goes all over the sequence, so they all have the same gate. All the TIs have the same gate. Let me name it. So it's S, G mod H, gamma times TI, G times N mod H, connecting automorphism here, the rho times TI, same thing here. And now I want to explain that when I mod out N here, so N mod H act here, and when I mod out N, of course I'm getting G mod N, but then all the TIs group cancel, so this is S mod the group generated by TI, which will be mapped here. But of course I have the assumption that TI generate S, so this is just a point. So it means that N is G. I mean this point, the image, I'll just say, the image is a gamma invariant point in here, and it is invariant under the risk enclosure of gamma, which is G itself. So this is a G invariant point in here, which implies this cos of space is trivial. Okay, so this is a trick that we've seen many times before, so I've got that one. But I haven't used amenability, in fact I'm using the fact that T1 is amenable, as Jean said, and this is enough to show that H is non-trivial in G by this theorem. But this one implies that H is normal in G, right? And it's a normalizer of H by definition. This is not written anywhere on the board, but this is our standard convention regarding the notation. Hence we got now that H is actually a trivial group. Okay, so now I'm in a position to rewrite this with H being trivial group. And I call this part now already pretty soon we are in the endgame and I want us to remember this because this endgame we will use again in a different context data. So let me summarize what we have. We have this map phi is in fact a map from S to G itself. Remember, this is a map between two groups, but it is not a homomorphism, a priori. It is just a measurable map. And here we have gamma times TIs and here we have the G times G action and the map rho times theta i that in the background we have all these maps Ti to G itself. And remember, the Ti's do generate S. So we have like a partial homomorphism I mean many partial homomorphism which are all over the place. So we just have to integrate them together somehow into one homomorphism from S to G. And this is what I will do next. Okay, so in order to do so I mean that's now it's kind of easy but I want to show how we actually can do that. I'll look at K-valid polynomial functions on G and algebra and I will take the map into measurable functions from S to K. That's an algebra of classes of functions to be precise. That's phi star. And so we have this homomorphism of algebras with the same equivalence as we had before. And I want to observe that the Z support so if you take the measure here on S and you map it to G I can take the support of the image in G. That's a measure class. Somehow, I mean that I can push a priori but I have a well-defined support of this measure and it will be gamma invariant. Again, gamma for a subset invariance means G invariance. So it is everything. This means that this map over here is injected. If you have a polynomial here which vanishes on the support which image is zero here it means it vanishes on the support and we cannot have such a thing. So it's a nice thing now that I can actually identify this algebra as a subalgebra of this piece. On this algebra I have left and right actions of G and on this algebra I have the gamma and Ti actions but also I have the S-action on the right. So somehow by this injectivity I can recognize the S-action here as an S-action here. So phi star KG if you want to be formal and again I think of it as a subalgebra of KG itself as a subalgebra of this guy is Ti invariant for all Ti hence it is S invariant and this gives an S-action on KG itself extending the Ti actions simultaneously Here they act on G on the right. So I'm getting S acts on G on the right. I'm getting a map S to G which is theta and actually this diagram here S to that's I have maps actually this is a homophysics and I had this phi this was not a homophysm a priori but now I'm having gamma times S and G times G and this is a homophysm Rho time theta and we've been here before we've seen that if we have such a situation Rho extends as before we've seen that actually Rho extends to a certain conjugation theta we use the same lemma somehow we didn't frame it as a lemma but we used the same lemma several times in the proof of the super agility for products and this is it we've proved it any question? so that was a bit anticlimactic it's too easy but now we have a super agility for higher rank and we might ask for more what about rank one? so from now on I will assume that S is a simple non-compact P group so everything is just the same only we don't have that fact that we use dramatically but now I want to discuss what can be saved is there any question about the setting? so I want I said this is the end game and I want to use it so maybe I'll just now refer as a lemma if S is generated by a bunch of TIs they need not to commute at this point if I get such a diagram then I get extension so in my mind I'm under I will not write it completely I'm under the assumption of Margulis super agility I'm assuming having Rho with all this property I wish to get extension of Rho but I do not assume a higher rank anymore that's the game and I'm trying to see what I can save so what extra assumptions do I need to finish so I assume this maybe let's do recall it I have this the dance I have this setting target is G now and I have G times G and various TIs which are homophisms so I have partial homophisms from S but global measurable map alright that's the the data and we got we had a little computation showing that Rho extends to S so now in the rank one case I don't have enough commuting Unipotence but still I discovered it I have this whenever I have an amenable group I can try to start something on so I will now try to discuss what happens for amenable subgroup in particular for the parabolic subgroup in S but before this let me add a piece of notation denote for T and S the T gate because I will play with some I will use S to G mod H sub T and that will be T sub T and here I will have gamma times T to G times N mod H T N T mod H T and the map here I will use I will call T sub T okay just so I can play with various T and study the relations and the most prominent T in particular I will consider the parabolic or a parabolic they are all conjugated we have only one on trivial but also I will use T to be A or U etc and we understand what those are right this is the splitorus this is the maximally important subgroup inside P etc standard notation and here is a lemma that I can squeeze from what happened if for the parabolic this age is trivial so I'm sort of in this situation then raw extends is my notation clear is it clear what I'm proving now so I'm assuming that the gate map for P when T is P the gate map is to G itself G is the reception of the gate map consider P A the thing is that so I'm having S to G mode HA I'll also have S to G mode HP which is just G this is this is P P this is P A but since A is in P this is also a P an A map and this is the gate so I'm getting this so HA is inside HP and this forces HA to be trivial too but now P P did not have a normalizer P is its own normalizer so applying this corollary that we had before for P didn't give us anything but when I restricted to A I get just a little bit more I get the vile element which normalizes it and this means also but now P A is P of the normalizer in S of A that's the corollary over there H the normalizer of A then is trivial and what we get really is that this map here I mean they all end up here it's the same thing P all over again but I have gamma and T I for T 1 which is P and T 2 which is the normalizer of A and these two P and the vile element they generate so by this extension lemma I'm done so if I'm lucky and I consider the P representation and it's as big as it could then I'll be good you need to be lucky here you need some miracle to happen you will not get it for free because we know that there is no super agility in general in rank 1 groups but sometimes miracles do happen and I want to discuss such a situation so I will discuss a certain condition which you might think is a bit unnatural it is the miracle I will just discuss it formally first and explain it geometrically later if I have time assume there exists W inside S which is a simple group by itself typically S is S O N 1 in our game maybe W is S O M 1 for M smaller than N assume that by this miracle I'm considering I can find such a representation of a gamma representation of S mod W for H non-trivial that's the thing that's a gamma rep as we discussed last time and let me call this condition star these condition stars imply that U is not in the kernel of the P representation should be technical what I'm writing I consider it a certain condition and I got myself a technical result I'm having the P representation that we discussed before and associate with each representation where did I put the notation associate with each representation I had 13 maybe I didn't emphasize it enough this went from T to NT mod HD in P I have the unipotent radical and I claim that it is not it cannot be fully in the kernel the proof is easy I mean once you understand all the technical notations so let me give it to you I'll consider U prime which is the corresponding unipotent radical of W so I'll take W W and intersect it with U W is a is a simple ego by itself so this is a non-compact TETA P is a map from TETA P is a map from P to a now I don't know what it is now I'm not under the assumption here it's for something here and I ask myself this question I really am curious by this question and I find myself very complicated the star condition answering this question sometimes in the negative NP is the normalizer of NP is the normalizer of HP yes yes it's P is an index here yeah sorry yeah the notation gets a bit cumbersome here but okay let me let me move on sorry I wish to clarify questions I don't want to give the wrong impression okay so now the thing is it's a very strong assumption that W is simple and it means that it has unipotent by itself and the unipotent radical of U is just of parabolic intersection of W with P will be a parabolic with W maybe there is a conjugation that I need to do here and U prime will be the unipotent radical of parabolic in W so this is non-compact and the thing is now that if I'm in SON1 by the way U is just a vector space it's a billion U prime is normal in U and U is normal in P if I'm in SUN1 maybe U is more complicated it's a two step unipotent in fact this is the case always but the thing is if I go for normalizer of normalizer of normalizer of U I think three steps should be enough always maybe four I will just get P so P is the normalizer of the normalizer of the normalizer the normalizer let me be safe and all these guys all have by I guess it's here these guys all have the same gate so it's this is why this I emphasize so much this HP by the way so this goes for U for A we don't get it for free A is not normal in P so we need this strong assumption maybe it's here we need this strong assumption here of triviality to get it for A and P we get the same gate but for U and for subset of U it's free Do you need a specific property on U prime to get this P quality iterating normalizer? No Because I need a construction which is taking normalizer on the plane for the unipotent radical maybe maybe you need to go down to groups normalizer by I'm not sure now okay this is an approximation I'm not sure now but the thing is that I claim that U prime by taking normalizer and maybe go down to things normalized by but non-compact things and go up again and go down if necessary eventually you get P thank you for this clarifying question the thing is that you can get P actually I'm having in mind SON1 and now just two normalizers will be enough but this is the important thing that you get out of pure algebra but the thing is you should I want to emphasize it in rank 1 somehow the thing is that you cannot go out of a parabolic unless you are on A A took us out for the parabolic it gave us one extra element which is the vile element which is enough now to generate everything but usually if I'm just unipotent and I'm playing this game of going to normalizer and getting something which are normalized by etc I'm stuck in a parabolic if I could move out I would get full super rigidity it's this algebraic technique which is completely the opposite of that one of IRANK that forces somehow the non-rigidityness of rank 1 groups but this I got and let me just consult my notes because I'm out of focus so FeU prime is FeU is FeP so I'll write it S to G mod HP and gamma times maybe I'll assume now that U is in the kernel of theta P and I'll show that star cannot hold and I'll assume star and try to derive a contradiction okay so just where am I so I have gamma times P here and I have G times N P mod HP again I apologize for cumbersome notation and here I have S mod W now I'm plugging here star and this is G mod H for some H non-trivial and U prime is in here so this is U prime invariant this map I mean this map here is U prime invariant in particular it's some representation of FeU prime and this is the gate of FeU prime so I'm getting that but now is this clear? I used sorry I didn't tried it out that's I consider this as a U prime map U prime map in a trivial way and I use this as U prime gate and by this I got this G map but now I have that U all U is in the kernel here and now I'm using this assumption this map is an equivalent respect to P and U is in the kernel so this map now factor from this and all together we get at this map from here to here S to G mod H is both W and U invariant but W and U again in W I mean it's W invariant because it's factor through here in W I have the vile element and with U and and also I have A so I have U and W and I don't need A U and with W I have the opposite of U they generate so I have S so again I have an S invariant mapping here it goes here and I show that this implies H equals G I mean this is our general theme we did it many times contradicting that one and I got a contradiction okay so it's nice it's definitely we proved something but what did we prove it's maybe not clear so can explain when this becomes a bit relevant before maybe let me give a very little observation if K is non-alchymidian then U is certainly in the kernel of whatever map I have right I mean because U is connected and NP mod HP is a group it's totally disconnected so I always agree this and I get that star cannot hold for free so I mean we have star we didn't discuss yet what is the meaning of this star but it cannot hold if S is real group and G is group so we already got something trivial maybe non-trivial but out of really general nonsense and also I'll add another definition then it becomes important I'll say that G is compatible with S if for every age which is non-trivial subgroup of G and for every theta from P to a normalizer of age mod age U is in the kernel of theta like in the non-alchymidian case this is trivial as I observed here so I'll give it a name I'll give it the name for every E of the sort and I get a corollary corollary that star plus compatibility implies the true extents why in the definition of compatibility I explained that I took the star means star cannot live with U in the kernel which is compatibility but in the definition of compatibility I allowed age to be trivial but if age is trivial we are here in this lemma and then we get extension okay so this might be look technical but sometimes it applied in nature so this is why we have this and I want now to focus I mean to freeze this picture and to discuss sorry change the subject now and I will discuss relations between superigidity and arithmeticity and then I will go back to this and explain how this could be applied in some situations so now new subject briefly relation to arithmeticity so S is SR it's the real point of an algebraic group in our consideration and I assume it is not SL2R itself now for this discussion S could be either rank 1 or higher rank I'll take a lattice so then irreducible lattice and here is a fact gamma is defined over a number field there exists a number field L which is embedded into the reels so gamma is a group of matrices up to conjugation I can put them inside with real coefficients up to conjugation I can put them inside a certain number fields called the trace field of gamma such that S first of all or maybe the group S is defined over this little field L I have already an algebraic structure for the algebraic group S over a little L and gamma is in the L points up to conjugation yes up to finite index thank you great now I will phrase something that should be called Margulis arithmeticity criterion gamma is arithmetic if and only if its image is pre-compact in any other place or in any place other than I0 I'll explain this means that whenever you take L and embed it into some local field real or not when this could be the reals complex, periodic numbers then either if I take gamma view it in here and view this in SK is pre-compact I call it bounded earlier or this J is this I0 that I started with or not only that K is the reals but it's actually the same way to embed L into the reals that I'm having here so you do it for a particular number field I do it for a particular number field called the trace field yes exist unique minimal yes thank you now I want to explain very briefly I don't have much time and I still want to say something new there connect some dots but I really need to one thing and this is how super agility implies arithmeticity given this criteria super agility implies arithmeticity the thing is that I will take I will take G to be S and just make it a joint basically I'm taking G to be S I know super agility I know things for every G and I want to prove I have gamma as in there it has this extension property I want to prove this to show that gamma is arithmetic okay so I test it overall embedding Ks basically so for every for every L to K I will do gamma goes to SL goes to SK and here I have SR that's the I0 thing that's the usual embedding this is what I called S maybe I'll go back to like this and I will get this by super agility alright and this one can walk out and see that this implies that sorry either gamma which is what I need in the criterion or I have this application but if gamma is not pre-compact and I have this then this implies now I wanted to expand not use too much time but by a little argument of Borel and Titz this implies that I is J so this is how a Margulis prouver arithmeticity out of his super agility thing and in fact this criterion I attributed it to Margulis I think is the first one to use it but there is a I was in Margulis birthday I think it was whatever birthday and most of let's not talk about age but most of was explaining there then he left but he had a session and he just interviewed all the big guys in the audience and asked them during his talk about the history to explain their point of view etc and he asked most of about this event when Margulis was supposed to I mean won the Fields medal but he couldn't let go from Russia to pick the Fields medal and then most of explained Margulis' work and he was communicated only very tiny part of the work and he was explained basically he explained why super agility by Margulis proves arithmeticity which I guess was known to people working on the subject so it was a very nice event I wish it was videoed I think it was not I'm sure it was not but now okay back to our reality you see that this proof of super agility implies arithmeticity is very not not tight super agility is just something about all possible targets and here we need only need only few targets G basically the G is just S or maybe you take a form S defined over L so you go down and then you extend to another local Field but it's those algebraic groups that pops here are just a very restricted class of groups that we need to have a rigidity with these kind of targets in order to achieve arithmeticity so there is a hope to get arithmeticity even if we don't have super agility and in rank one I remind you we don't have super agility usually there are some groups with super agility which are super rigid I will not have time to discuss this I see but what about SON1 for example so now I want to discuss in the last few minutes I just want to give you a very brief story and relate back to this condition star and compatibility where one can do proof of arithmeticity so so this is I'm talking about recent work with Fisher Miller and Stover and we proved the following proposition if on the group S mod K is maximal compact subgroup in S so S mod K is rank one C metric space and this is the manifold or an overfold or finite volume there exist infinitely many totally geodesic subspaces not geodesic real totally geodesic subspaces if this exists then condition star sorry immersed yes immersed and if we are being pedantic maximal I mean not everything is happening in a sub manifold then star holds for Gs relevant to arithmeticity so we proved this I will not discuss this proof surely not now but the proof here is is an application of Ratner style theorem Moses Shah and Danny Margulis so I mean it's it's not I mean it's an application of non-staff usually but here is an easy corollary of this right away from this observation over there it shows that our gamma cannot be represented when you prove Margulis arithmeticity criterion into piadi groups because then the compatibility thing holds so gamma is defined over a number ring so it's integral G not over Q or the analog of L and O I don't know however we want to write it down and and now really in the last minutes let me explain something related this is an exercise now in this theory for N for more and S being S O N 1 S C which is basically S O N F maybe some parity issue that you have to fix here is compatible if you take for G this group then you get compatibility I claim I mean and this is compatibility is something this is not related to Pagodic theory or anything or anything that we did here this is some leth theoretical thing that one has to check and here for S O N 1 it happens that you are always compatible when you try to apply Margulis superigidity and you get where is the corollary that I wrote at some point somewhere I wrote but you get that if there exist infinitely many TGs then gamma is arithmetic this is a theorem and it really follows whatever is on the board now I corollary up there thank you so this follows from the corollary on the upper left corner and and also maybe extra care needed for N equals 3 but also it works for N equals 3 just maybe we wrote a paper and my terminology here is not completely compatible with the terminology in the paper our compatibility criterion that we use in the paper is not the one that I put here the one, the compatibility criterion here is a bit simplification so we rework the compatibility criterion it's a bit more technical but it's something that still holds for N equals 3 and it's strong enough to get so it's a bit of a weaker condition still hold for N equals 3 but it's strong enough to get the corollary but let's say that there are only groups you need these are the only groups you need okay yes because you go down to L and then you go up to PADX but PADX is not a problem I explain or you go back to the reels and then you might get many complicated groups but if you go up to the complex numbers you get just that and this is and you get I mean it's enough to go all the way up to complex numbers here and by this you get compatibility for all possible groups so my time is almost almost up so I'll just say remarks about SUN1 or maybe very briefly now I will not write anymore very very briefly so rank 1 groups comes in 4 flavors there are isometric group of real complex coternionic or octaneonic spaces in the case of in the last case and we know that the 2 last families are super rigid and I mean lattice is there hence arithmetic and lattices in SUN1 I explained that arithmetic if you have infinitely many totally geodesics and for SUN1 it was open for a while and the 2 teams I mean us and also Ulmo and Baldi were here Emmanuel and Greg they prove it with a different proof we struggled a bit because then we don't have compatibility not always have compatibility we don't have compatibility for the complex numbers but then by Simpson's theory you don't need I mean your field is totally real your field L is totally real and you don't have complex numbers to deal with so we need compatibility for the reals only and for the reals you happen to work with SUPQ P plus Q equals N plus 1 and then also if the PQ are more than 2 2 or more then you do have compatibility and everything is fine and the only thing you have to play against is the target group G is SUN1 itself but with a different field embedding of L and then you have to squeeze a bit more from the theorems to understand to play a bit more those games that we played for 2 weeks now and to squeeze out of this extra information some invariance of some incidence relations and eventually we were able to use a result by Beatrice Posetti that was visiting here this week as well and about showing she showed that every map from every boundary map that preserved cartons incidence relation in boundary of SUN1 must be rational and we plug this I mean out of our techniques we squeezed eventually the condition to apply a theorem and from her result one easily deduced again the relevant super agility and this solves the arithmetic history altogether so we have this also for SUN1 and there is a companion paper by Emmanuel and Greg and this is the end of my talk so thank you very much so we have the full I mean for SPN1 no we didn't carry the detail for SPN1 no actually this is very interesting for me for SPN1 we should have super agility with no assumptions I mean with no totally geotasic manifold assumption I don't know how to achieve it I looked at it a lot and still didn't get the answer I believe that looking some more will reveal maybe extra structure but yes and I just to say the Baldy Ulmo paper is using also this arithmetic criterion but from then on this is a completely different techniques that you will maybe hear from others I'm trying to understand it it's very beautiful אינטגרלטי for SUN1 yes could be deduced from a previous result as well you don't need yeah how very recent for this yeah our techniques give it very easily but very differently somehow