 So thank you very much for the invitation to speak here. So I want to speak about, yeah, so what I'm going to speak today about is joint with Colmez Dospinescu. Right? And Dospinescu is going to speak after me. So the second talk this morning, and I'll explain a part of the argument, and he'll explain a different part of the argument. Yeah, so what is the argument about? So let's see. So g is gl2qp, okay? And so p is any prime. So the fact that this p is allowed to be anything is essential. Well, it's kind of the main point of this talk. And then we prove the following theorem that if I look at pi, so these are pun admissible gl, so these are absolutely irreducible, non-ordinary. Yeah, so I'll come back to these words. And look at isomorphism's classes. They're naturally in bijection with rho gqp gl2l. So this is continuous, absolutely irreducible, and also take isomorphism classes. Ah, so what is l? So l is coefficient, so l over qp is some finite extension. Yeah, so I have some ring of integers, a uniformizer, and then the residue field is called little k. And then, so this missing, these are admissible, unitary, albannac space representations of g. Okay, so now absolutely irreducible. This means that if I take a bannac space and I extend scalars to a finite extension, but this is irreducible for all l prime over l finite. Okay. Yeah, so this is what absolutely irreducible means. And then non-ordinary means the following, that pi is not a sub-quotient of ind bg chi continuous for any chi b l prime. Okay, so such representations came up in the second talk yesterday, and sub-quotions of such representations will call them ordinary. Okay, so this is, the name is motivated by this emittance function of ordinary parts, which was explained yesterday. So we look at non-ordinary things. Ah, so now this is just not, although this will not play such a big role for my talk, but this bijection is just not any bijection, but the way it works, you take this pi, and there is, to this pi, you can apply so-called Colmeses Montreal Functor, which is kind of somewhat magical, and you get a Gallo representation. So this will be explained in more detail in Dospinescu's talk. Yeah, so it's really great giving a first talk. You can claim all sorts of things. It will be explained in the second one. So now, ah, so I guess I did it wrong. So what I should remark is that this was known for P greater or equal to five, but now the proof, this proof is different. This proof is different. Right, so maybe, yeah. So before I go into details, the proof for P greater or equal to five is kind of a characteristic P proof, and this proof is a characteristic zero proof. So maybe this will become apparent. So now you see what I want to explain today is a part of the argument, and so let me call this theorem two, which is the following. So I have pi, such Banach space representation, and I assume that it's absolutely irreducible, and then I choose some theta inside pi. So this is unit ball with respect to some G invariant norm. Norm on pi, right? Because this representation is unitary, so I can choose, there exists, I mean by very, the word unitary means that there exists a G invariant norm which defines a topology on pi. So I just pick one and I look at the unit ball, which gets called capital theta, and then the assertion is either one, when I reduce this module of a uniformizer, this unit ball in a Banach space representation, I get a smooth representation in characteristic P. So many assertions, this is isomorphic to pi, irreducible, super singular, or two. So if I reduce this module of a uniformizer and I semi-simplify, this is contained in the following. So I have NB chi one, chi two, omega inverse, I semi-simplify, NBG chi two, tensor chi one, omega inverse, I semi-simplify. For some, chi one, chi two, QP times, so these are smooth characters, okay, right, so now, so this notation means that I take a borrel, so these are just upper triangular matrices. So this is a character of borrel which sends, well, a matrix A B zero D to chi one of A and chi two, omega inverse of D. Now what is omega? Omega of X is equal to X modulo P for all X in QP times. So this is what I would like to explain how we kind of prove this, but before doing this, let me also mention how this relates to the next talk. So, you see, when I apply this functor V to pi, it's essentially the same as applying this functor to theta and inverting P, okay, but now, so now V of theta is some module, some O module of actually finite rank, and you can read off the rank by applying V to the reduction of theta modulo to the uniformizer. Now because you know this explicitly and you know what is V of pi and what is V of this principle series things, you can conclude the following. If pi, so this is absolutely irreducible, then dimension L V of pi is equals to two and then equality if and only if pi is non-ordinary. So a priori, you have no control on the dimension of this space, so a priori could be say infinite dimension, but once you get some control on the reduction, you can control, well, on this thing and once you have this result, you can then apply a different construction due to Colmez and then developed by Colmez and Dospinescu, which two-dimensional Gaul representation associates, well, representation of GL2QP, right? Okay, so let's see. So I think I did this in the wrong order. General properties of what? Yeah, so this is either irreducible, this thing, yeah, so I should have said this, this thing, for generic chi and chi too, this thing is irreducible, this principle series is irreducible. It can happen, but the whole thing has either two, three, or four. No, I'm not asking this specific, just the general theory of representation of something, like if you start from the bottom, what do you call a admissible unit? Nothing is known. No, I mean it's not known. But it is admissible, okay. Yeah. It is only known to be admissible. Well, it's known to be admissible, but you cannot deduce this off, but it's finite length. Okay. Basically, I mean, if you do representation theory over C, you know that admissible plus finely generated implies finite length. In mod P, you don't have such a statement in general, in the representation theory of PI, the group's modulo P. But it is fairly generated. Well, you can, so you know that it's admissible. Let's see, so at least you know that it's locally finite. If you take, well, I'm getting confused between the particular case I'm dealing with, GL2QP and the general case. So in general case, it's not known that much, right? It's only known that it's admissible. So for example, the notion of irreducibility, you start from a vector, you know it generates, you start from a vector, not zero vector. And then what do you know about this thing generated by it? You know that it's admissible representation and that it's finely generated. No, no, but irreducibility is just algebraic. So for mod P representation, it's always just algebraic notion. For the Banach space representation, it's a topological notion. So if you start from a non-zero vector and take the all-module generated, the closure of the all-module generated way, is it okay? Maybe I can say that. No, I mean, I think even basically all the troubles, I think, well maybe, I think even if a group is ZP, when it's already you kind of see that nothing is that straightforward, right? But okay, so let me proceed with this. Okay, so what do I want to say now? So in fact, it makes me feel better, you know? Let's see. So what I want to explain is how to put, how to bound multiplicity of pi in some error. So I take an irreducible smooth representation over K, so let's say absolutely irreducible, and I want to, this is what I want to explain and theta, like in this theorem. So basically, so to prove this theorem, I want to bound multiplicity for each irreducible representation, so this is one, and two, I need to be able to say that given that pi occurs in sub-caution, I have to basically have a bound on which other irreducible sub-cautions can occur. So this last part is done by a theory of blocks, and it works, I mean, pretty much the same way as in representation theory of finite groups. So I don't want to explain this. What I want to explain is this boundedness argument. Okay, so now, so let's see. So I have this smooth representations, G, on all torsion modules. So this is anti-equivalent to a certain category, mod-profinate-augmented, G-o, and this is, this anti-equivalence is done like this. I take some representation, and I map it to its Pontriagin dual, so this is from O tau L O. And on this thing, I put a compact open topology, okay? So one can describe exactly what these things are, but for the purpose of the talk, I mean, I just take a discrete thing, and I take its Pontriagin dual, right? So I get something compact. So in here, I have this category of more locally admissible representations. So this was explained in the second talk yesterday. So this category is due to emittent, and this just means that for every V, in appreciation, every vector, if I look at the representation generated by it, this is admissible. So now, so this gives me some category C, okay? By definition, just like whatever I get by taking Pontriagin dual of this thing. So what I... Right, so now one can show, well, actually I can, for my purpose, I can actually assume this, but it automatically holds that this pi is admissible. So I guess... So this is, I guess, due to better share. But pi is admissible, so this means that if I take Pontriagin dual, so this lies in this category C. So now I look at P, so this is projective envelope, of pi V in C, and then I let E to be and Cp beyond the morphism ring. Okay, so... Right, so now, this instead, so now if I take pi, an admissible Banach space, and then I can take some, well, a unit ball as before, G invariant unit ball, and then I define M of pi, this is hom C in this category, P theta D, and I invert L. So let's see. So I have this theta, when theta D, I simply take hom O, theta into O, I equip this guy with a weak topology, so this is so-called Shikov dual of theta, and you can show that admissibility of pi implies that this guy actually lies in this category C. Because this is periodic, I don't have to do this. This has been pointed out by several referees of my papers. Right, so let's see. Okay, so now, so let's see. So what I have is a following proposition which says the following. So one, that this M pi is a finely generated E1 over P module. So this condition of admissibility translates into finely generatedness. Two, now if I look at dimension L of M pi of this module, this is the same thing as dimension K of P theta, I reduce this module of a uniformizer, so I get something discrete, and I take the, it's one-tragon dual, and this is the same as, ah, yeah, oh, sorry, yeah. Okay, and you see now this P is a projective hull of this irreducible, well, of this pi V. So now, so this guy you can actually determine this multiplicity of pi in theta module O semisimplified. So you can actually read off a multiplicity if you know the dimension of this module. So this is one, and then part three says if this pi is absolutely irreducible, then part A, this M P is absolutely irreducible E1 over P module B, and G quant pi. So this M is a functor, so if I have an andomorphism, a G-equivariant andomorphism of a Banach space, this induces an andomorphism of this module, so this induces a map, and the assertion is that this is an isomorphism. Okay, so now the statement looks kind of a little bit scary, but basically the proof works like this. You imagine that G is a finite group, and then you can just prove this by saying using the book of Sarah on representation theory of finite groups, and then you look at this proof and you realize that it's actually the same argument works, except that to deal with issues of duals, continuity, and periodic function analysis, you have to use a paper of Schneider Entitlebar. Let's see, so okay. Okay, so now, so this is maybe not so hard. Well, given... What did you write in this statement? Pardon me? What did you write? The anti-equivalence goes to mode, no energy, and then what was the single both? Yeah, so this is profaned-augmented. So if G was a compact group, a compact periodic analytic group, so this profaned-augmented would be simply compact, I mean linear topological modules over the group algebra. So in my situation, G is not a compact group, so one has to do a little bit of extra things, yeah? But essentially this is what it is. So let's see. Right, so now I want to explain one easy corollary of this, this theorem. If E is commutative, dimension of L m pi, so pi, yeah, so pi is absolutely irreducible. Yeah, so I want to explain how to deduce from this proposition. So first of all, I let script E be the image of E1 over P in this end L m of pi. So I mean, I may assume that m pi is non-zero. I assume, I mean, if it is zero, there's nothing to prove. So now I know, so what do I know? And this is commutative. This is the image of which, yeah? Well, it's just, I have, I mean, this is a module over this ring, and I just take a linear and a morphism. Oh, I see, okay. Right, so I know that the image is commutative because this ring is commutative by assumption. So now, so this implies that this E is contained in end E1 over P of this m pi, okay? But this is isomorphic by this proposition of m pi 3 b, b to end g pi op. Now, here, I can use Schubert's lemma, lemma. So the Schubert's lemma is not a triviality, but so it is a theorem of Dospinescu and Schren building on work of, recent work of Adlerkoff and Wadsley, that the endomorphism ring of this thing is simply L because pi is absolutely irreducible. Okay, so then, so basically I can conclude that this E is equal to L, but now I know that because pi is absolutely irreducible, implies, again by this proposition, that this m pi absolutely irreducible script E module and then, because the ring is simply L, this has to be a one-dimensional vector space. So from this, I conclude that dimension L m pi is equal to 1. And I mean, once I conclude this, I can then further conclude that pi occurs with multiplicity 1 in theta by using pi 2 of a proposition. So this kind of is also an easy consequence. Okay, so now I don't want to write this down, but in this proposition I assume that the ring is commutative. So this commutativity can be expressed that for every ordered tuple of elements pi 1 and pi 2, pi 1, pi 2 minus pi 2, pi 1 is equal to 0. I mean, there's no way to argue with that. But now we can generalize this to the following identity. You take, instead of taking two elements, you take 2d elements. And then you require, so this is remark, if for all 2d, so now I take 2d elements and I take sine of sigma and here I compose them. 2d is equal to 0. Then one can conclude in the same way. Well, okay. So this is, in this thing, I have to use a theorem of Kaplansky. So this would mean, I argue similarly, but now I look at this E and from this identity I can deduce that E is a ring of polynomial identity. And so then I can argue that the center of this ring is L and then, I mean, basically Kaplansky's theorem, this session follows from Kaplansky's theorem. Okay, so far, so good. Not too bad. Now, okay, so how to, so I have this abstract proposition and I know if a ring is commutative, then I can conclude something. So now, how about, so how to prove this commutativity of a string in some situations? Yeah, so how do I prove, how do I get a hand in this case? And the argument is kind of Bajarisky density. So this doesn't mean anything yet, but hopefully, right, so now density. So now, there is a family of representations where I can control this MPI. So these representations have been, basically been studied by Bergeron-Broy. So now, I look at, I give myself, say, I have some finite field extension and then I have some delta 1, delta 2, QP times to L prime star, some unrhymified, some unrhymified characters. And I guess I want to assume that delta 1 times the norm is not equal to delta 2. So then, to this thing, I can define a smooth, so the usual smooth principle series representation. So this is just a smooth principle series representations on L prime vector space. And then, additionally, I take V of a form similar A, say L squared times B in algebraic representation. And I let pi, okay, so I can tensor these things. So now, so this is a locally algebraic representation and I can consider its universal unitary completion. So this is universal unitary completion. So this universal unitary completion, this means that if I have a G-equivarian map of psi tensor V into some Banach space, then it automatically factors through a continuous map of this Banach space. So now, so far this is just introducing notation, but now, so what is the statement of a theorem when pi is admissible G L prime representation in absolutely irreducible and theorem two holds. You are happy? I wish you were saying, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah. Okay, but of course this doesn't really play a role in anything. I mean, this makes the statement wrong, but doesn't evaluate the statements to come afterwards. Yes? Yeah, so this has been pointed out by Laura that because I don't put any conditions, except for this one, the universal unitary completion might be zero. But this doesn't really matter for me. I mean, this universal unitary completion is just done like this. I take some... What are you asserting then? So this has been... I have to assume that pi is non-zero. Yeah, sorry. Yeah, sorry. Maybe if pi is non-equal to zero, then... Well, hopefully it's true for zero as well. Yeah, so I wouldn't want to spend too much time on this. Actually, you're right, you're right. That's what I was thinking about. That's terrible. Okay, so now I still have 20 minutes. I hope. Okay, so now... I know there's a theorem 2 holds for pi, so it means that if I choose a unit ball and I reduce the semi-simplify, well, it satisfies kind of one of these things. And so to illustrate the argument, I want to assume from now on, assume pi is super-singular. So this basically means that this pi is not a sub-cautions of these principle series representations. So then... So the theorem of Berger-Bruy then implies that dimension L, well, M' of Mp is less than or equal to 1. Because pi occurs in the reduction with multiplicity at most 1. Okay? So as a result, we kind of deduce that image, E1 over P in end L' of Mp. So this is L' is commutative. Right? So this is a one-dimensional L' vector space. The action of E1 over P on this module commutes with the action of L'. So I get this map and this is just a field itself, so the image is commutative. Okay, so this is one point. Ah-ha. Well, I guess I don't need this anyway. So now basic idea, idea, try to show that the map E1 over P into the product. So now I take the product over all such unrhymified principle series and which I define over say different finite extensions of L and all their weights V. And to this, I have a universal entry completion and here I have end L' Mp. And then you try to show that this is injective. So this implies E1 over P commutative. Okay? Because well, all of these rings are commutative and the product is also commutative. So if you can show that this is an injection, you're done. So now, so I should say that motivation is Emiton's proof of the following fact. So here in a global setting, so Emiton says that proves that classical crystalline point are dense in the big Hicke algebra. So this is as a part of this Fontaine-Mezer paper. So we kind of try and run a similar argument. Right, but it gets somehow complicated, a bit more complicated. Maybe with a little luck, I'll explain why. So let's see. So now I have 15 minutes. So now, right. So you see, and because I assume that Pi is super singular, V of Pi is irreducible to the dimensional Mp representation. So this kind of basic idea, morally, well, that this injection should hold, morally says that crystalline points are dense in the universal deformation ring of V of Pi. So this is, but now, okay, so now I want to sketch. Okay, so this is going to be a sketch. And step one, first of all, we know that P, when I restrict it to K, so K is GL2ZP. So this is projective okay module. Okay, so this is something that one has to prove that okay, but we know this, so I guess this is, so this is maybe due to Emmett and myself. So once you know this, you kind of prove the following, using this, you prove the following statement. So if for all V, so now you look at algebraic representations of well, of G. So, yeah, so these are irreducible algebraic representations of GL2 over L, which I view as representations of GL2 of L. So explicitly these are sim A L squared, tensor that B A is a natural number and B an integer. Right, so if for all such representations, wait, sorry, sorry, I'm kind of getting excited. Okay, so have this, I take some phi in this E, one over P, okay? Now if for all V, phi, this phi kills HOM continuous, okay, P V dual, then this phi is equal to zero. Okay, so how, so this argument essentially already appears in a paper of Emiton, so in this front-end major paper of Emiton, so basically because you know that this is a projective OK module, you know that this is isomorphic to a product of projective, indecomposable OK modules. It turns out that it's enough to actually prove the same statement with replacing P with a projective indecomposable OK module and then it turns out that it's actually enough to prove the same statement with replacing P restricted to K with simply OK, the completed group algebra. And then the statement basically says that polynomial functions are dense in the space of continuous functions. So I mean, yeah, so as polynomials are dense in, well, you end up looking at the Banach space KL. So continuous polynomials or say polynomial functions are dense in this Banach space and this is what the argument reduces to so let's see, so yeah, let's look at this again. Right, so now now, as a corollary of step one to step one you deduce that this ring injects into a product of this algebraic representations E E 1 over P A V where this A V is an annihilator of this module of this HOM OK continuous P V duo. So if I can show that every single every this ring is commutative then I can conclude that this is commutative right, so now in step two fix fix H V G right, so now OK, so now I let this Pi P be HOM O P L continuous homomorphisms so this is a compact O module so this is in three Banach space where I take the supremum norm OK right so I should say that in emittance in analogy to emittance work this is a local analog of emittance completed cohomology but now if I look at this module HOM continuous well say K P P duo by Shikov's duality so which is so due to Schneider title bomb this is the same as HOM K V Pi P OK so now then by Frobenius reciprocity this is HOM G compact induction K V G Pi P OK, so now if I let H be the Hecke algebra endomorphism G of this compact induction KGV so so this is actually this is a very classical thing so this is the same thing as compact induction since this thing is algebraic this is the same thing as the trivial representation and this is isomorphic to TZ plus minus one so this is commutative so this is a nice ring and it acts on this guy and it acts on this guy so now well OK, it went away we do this step by step so if I take N a maximal ideal of this Hecke algebra and then I look P I look everything that is killed by this maximal ideal N so then if you unravel these definitions the same thing as HOMG kappa N the residue field at N H compact induction KGV Pi P this thing is the representation of the form I considered before Psi tensor V where this Psi is an ramified principle series in this finite extension and therefore this is the same thing as HOMG so now I can replace this tensor product by universal unitary completion OK and then if I apply this Shikov duality again I can deduce that this is the same as Psi tensor V with functor M so at the end I already know by Virgib Roy that this thing is one dimensional over kappa N so then I can conclude that dimension kappa N of this HOMG K continuous P V dual N but this is one OK so now using this so this is step 3 if I look at the morphism ring of H so I take this continuous P V dual but now I just don't take things by kill just by N but things kill by some power of N that this is that I have a I have a suggestion on N H E H kappa N so this is an injective hull of kappa N as an H module in the category of H modules so now I know what this thing is explicitly this is isomorphic to H localized N and completed I mean there is a theory called Matly's duality stand-ups in local homology quite a lot but this is commutative OK so let me see so I can actually start deleting some of the stuff here so now I can actually conclude from this thing that if I take N H HOMG continuous by V dual so now I take all the elements which I when I apply H to it generate a modular finite length OK so this is what the subindex locally finite means so this is isomorphic to N H this is a product over all the maximal ideals of H I have this HOMG HOMG K P V dual and here right so this is just Chinese remainder theorem but now because we have different torsion this is isomorphic to a product of N of these I mean I don't have sorry so this should be a direct summary I guess so I don't have yeah so I don't have a map between individual things right so now so this is the same as HOMG continuous K P V dual and infinity is commutative so this is step 3 in step 4 which just says that HOMG K P V dual locally finite is periodically dense in HOMG K continuous P V dual so this is compact and this is a final dimensional vector space so I can just take some K invariant norm on this guy and this makes it into a Banach space unfortunately I don't have enough time to explain the proof of this but if you believe this then you can at the end conclude that you have a map E 1 over P divided by the script A V injects into a product N well into this thing commutative and then you conclude that this ring is commutative so then right so then yeah it's still there very good thank you so then you can conclude that this thing is commutative okay so now I would only so this looks quite complicated right so what I'd like to say that in emitting setting because he's in a global situation he can already stop here because he knows that these rings is a product of fields okay so he can stop here and we have to work harder okay thank you very much yeah so now I already made remark about this this this Kaplansky's theorem right so now you see so now wait so let's see so where am I so now when I use this result of yeah so in this case if Pi is this Berger-Broyer representation then I know that M Pi can be at most and let's say can have dimension at most 2 and let's imagine it is actually 2 okay so then the endomorphism L of M Pi okay these are just just a matrix ring and the thing is that every matrix ring satisfies this identity okay so this is in this case would be S4 this identity S4 so if I take this I do this so this is a theorem of maybe I'm at Zulevitzky so okay so now so then I could conclude that the image of my ring E1 over P in there also satisfies this identity but then when I run the same density argument showing that that this E1 over P injects into this thing so then this identity is satisfied satisfied for this ring and I basically proceed then as in remark in this remark on Kapladsky rings Kapladsky's theorem rings of polynomial identity are there questions comments? the field is not Qt but a finite extension do you know today the length of all these reductions is bounded? no but this is actually a good question so because so there was a kind of a part of the argument right so this was part of the argument so this proposition that I wrote down on the board so this holds for any periodic league group but what I need is a family of representations for which I can control the reduction modulo P and the density so now somehow the whole GL2qP theory in a way works because that this universal unitary completion is reducible and this kind of corresponds to the fact that there is a unique admissible filtration and so on one hand the argument is very general but difficulty is finding such a family where you can control reductions and as far as I know apart from some ordinary cases ordinary Banach space representations so you can compute for reduction but apart from this not much is known