 Thank you very much and thank you for the invitation to speak here It's a pleasure to be in Paris and it's a particular pleasure to talk at this conference in honor of Fontaine and Vantaine-Bergé I Think most of the other speakers have a closer connection to both of them than I do but I just wanted to say a couple of remarks at the beginning about There were a lot throughout the talk about both of their work in some sense But in fact, they were both very important to me so very early on the first thing I did in my career was some work on companion forms and the weight part of Seir's conjecture and I realized I had to compute I was trying to do some lifting while talking about in a bit and at the time there was only a method of Ramakrishna for constructing lifts of global representations mod p to characteristic zero and One requirement of this which has been removed recently by patricus and kare and facudin Is Is that you needed local deformation rings to be formally smooth so you needed to know they were given by a power series ring and The context in which I was going to need to apply this result with some some potentially bustle He take deformation rings and you could try and compute these things using Bray modules But I tried to compute them and I just couldn't eventually some years later Arianna and Kristoff Bray Mazar computed these things completely and beautifully, but I look at their computations like I'd never have done them but I was kind of saved from having a disaster of just not being able to do these things because Karyan Vandenberg in around 2004 Found a new method for for producing lifts which I will talk about in the second half of the talk today Which was the Karyan Vandenberg method and this required far fewer hypotheses. It's basically no hypotheses at all. So Vandenberg a kind of Incidentally saved me by completely removing the need to do this horrible computation that I was stuck on for six months and The Karyan Vandenberg method in fact is something I've used a lot and I'll talk about a bit today So so that was kind of my connection with Vandenberg There's also another connection with some some work of one of his students, which I'll talk about in a bit and Fontaine in fact, so this this work I'm talking about today. It's all with Matt Emerton and this project started in I think March of 2011 and We finally finished the paper a couple of months ago So I think it's like by far the longest project I was ever involved in but I Think a month or two later sort of around May June 2011 We were both working at Northwestern in Chicago at the time and Fontaine visited for for a month or so and He basically kind of corrected all of our misconceptions about how modulo spaces of FIGAM modules and fontaine of five modules And so on would work So that was kind of crucial early on because we had some some early ideas and if you look back at them now They're just completely wrong and I think without him Patiently explaining to us where everything how everything actually worked. It would never have got going And then several times in the years in between. He was very helpful in particular Kind of the last time I saw Fontaine was last summer there was a special program in Leon and I came and gave four lectures there and it was somehow pretty intimidating because I had Rappaport asking lots of Questions and somehow I can never answer his questions But luckily and they were very basic questions out FIGAM modules often and sort of the behavior But I would get confused but then Fontaine would Helpfully answer all the questions and so he kind of saved me from a complete nightmare there So I'm kind of very grateful for all of that and on one sort of personal note there was one summer I don't remember when probably four or five years ago some conference in Lumini and I would I always go on to my wife about how great Lumini is for conferences and it's a great place to visit Once she came to a conference there and it was an extremely hot summer So hot that you couldn't go outside most of the day I think it was also that time there was some time when the cooking changed at Lumini And it was kind of the unique low point of these things and she's a vegetarian So like she would kind of have to sit inside all day or we were at talks and then come and have this food And it was not so much fun, but I think the one thing that saved it was that and and she always remembers this that We would go to lunch and sometimes also go into a conference and you're talking to mathematicians and sometimes this is not so easy But anyway Fontaine was always really great to talk to and Yeah, I miss him. I'm sure you will do Okay, so some mathematics, so Theorem let me start by stating the theorems so Theorem one there's basically a local theorem and a global theorem. So k over qp is going to be a finite extension arbitrary finite extension and Roba is going to be What dimension d-dimensional? mod p representation And always my representations Mod p or p-addict will be continuous with respect to and we hear the discrete topology and the profiler topology I Could instead of having fp bar coefficients put some finite extension that we valued in a finite extension of fp But I wherever possible. I'm just going to write fp bar to try and avoid Kind of making my coefficients big enough. So then the theorem is that there exists a representation row to GLD Zp bar again Continuous, I think this is the last time I say my representations are continuous And it lifts Rowbar so in the sense that this modular the maximum ideal is just this So in fact that would already I think be a a theorem that wasn't known before in fact we can say more about it We can arrange No Definitely not irreducible. I'll explain in a minute that irreducible is kind of basically trivial for this theorem So we can arrange the when you invert p That this is actually a crystalline representation with distinct Hodge take weights in fact, this is basically what comes out naturally from our construction But it so it doesn't in some sense seem to be any easier to construct lifts without this property, but You might imagine I mean it's possible. There is some argument that we don't know And furthermore, I won't explain this condition, but it's useful very useful in practice You can arrange that it's what's called potentially diagonalizable Maybe I can just say some some particular case of this is that if If row bar is the kind of the opposite of being Irreducible if it's an extension of of characters So all the this is some filtration by one dimensional Subquotions then you can arrange that row Is a rain can it can be arranged to also be of this form so it can rearrange to be ordinary So that's also sometimes useful Yes, I mean this is somewhat surprising if you haven't ever thought about this question, but in some sense This is the hardest case is Ordinary means that the lift is also an extension of characters Successive extension of characters and the hodge take weights are strictly increasing So it would be something like in Christoph's talk. There was this example of things that look like This where epsilon is the cyclotomic character. So something like this would be ordinary, but I wouldn't allow you to swap these two characters around Okay, and of course there so the flag the lift the flag. Yes. Yes And there is a general when the representation is reducible and you have a Jordan under series and you can lift it to a to one for the Lift I don't know that unless they're all one dimensional. I think Um Yeah, I don't I don't know if I've thought so much about that I might expect that was true But I'm not sure off to my head if that's what our proof gives so let me Let me be careful, but certainly in the rank one case. It's exactly what comes out I Mean in general what happens in the higher rank case is something to do with this condition But I'm not going to talk about about that today Any other questions about this theorem before I state theorem two Okay, so theorem two is some kind of related global results. Oh, sorry Is ZP bar complete No But it doesn't really matter what I choose. I mean in practice what the proof actually You replace fp with a finite extension you replace this with the ring of interest in a fire extension This is just a convenient way to make a true statement without yeah, I think it's remarked somewhere It's automatically automatically lens in gld of some ring of things right it follows from Bear category theorem and compactness Yeah, exactly now. It's just kind of irritating and I think you have to be a bit careful in that At least with our arguments I'd be a little nervous about if I just replaced it if I just wrote fp whether I could actually lift a zp For example, I think you have to be a bit careful those rationality issues It's sometimes convenient to have a ramified extension of the coefficients We'll come to that Any other questions Okay, so the global version Let me make sure I don't miss a hypothesis. Oh, I should also. Sorry. I should miss that that kind of remark This was previously known. Perhaps I should have raised this ordinary representation In the case that D is at most three So this is a theorem of Alain Muller In his PhD thesis supervised by Bantam Burje. He proved this and already here. It's not so easy to prove this Kind of by a bare hands approach in the case of an extension of three characters It's kind of for most choices of characters. It's kind of easy enough, but in general There's actually quite a bit of work to do this Various people sort of out this problem before and I think they're probably unpublished works where people maybe I think Did sort of D equals four maybe but even D equals five there starts to be a real problem trying to do it by hand Okay, sorry, so back to theorem two so assume now that Make a a mild assumption on P relative to D. So it's assumed that P is not two and doesn't divide D Then there exists. So let me I'm going to maintain my notation from up there F Over Q. So this is a fine. This is a number field. It's an imaginary cm field and And a global representation. So a representation Another more P representation now the absolute Galer group. I see I never defined that that gk is the absolute Galer group of k Apologies, this is the absolute Galer group of F So a D dimensional global representation Which is irreducible and it's automorphic in the sense in Christoph's talk Christoph's talk this morning IE It comes from some automorphic some customer automorphic representation on some ranked D unitary group Which is compacted infinity. So I'll just say it comes from an automorphic representation for Let's just say UD So so here it is that you're working with five Cartesian peak coefficient. So you're what kind of long-range I just mean it's the reduction mod p of something from characteristic zero. Yeah Yeah, I mean one point about using these these These units groups which are compacted infinities you only have an H zero. So there's no Question about whether you can lift a characteristic zero or not you just write down Your kind of algebraic modular forms and they they do just all come from characteristic zero I mean equivalently by a bunch of trace formula results and so on I could be saying that it comes from the reduction mod p of the representation associated to some Conduct self-dual automorphic representation of GLD all of these things are the same and in the if you imagine that D equals 2 You there's a there's a version of this work. Just be Hilbert modular forms instead just any kind of reasonable Kind of case where there's discrete series automorphic representations. Okay. So so far. There's no content to this theorem Such that so it's gonna have some relationship to this. So for all places V Dividing P of F F V is isomorphic to K the field I have above and Well, I would like to say that that locally at all the places this our bar is given by this this row bar I Can't quite say that so what I will say is that either Our bar restricted to G F V is awesome isomorphic to row bar or the same thing is true If I replace V with its complex conjugate and so why am I saying that? Well the point is that? The definition of automorphic implies That there's some conjugate self-duality. So something like our I'm I may well get a sign error here But the complex conjugate and the dual are isomorphic up to some twist by The mob piece of atomic character to the 1-n as I said, I apologize if I've got the The power the wrong way around but in particular this means that Sorry, what is our ah Sorry, thank you Sorry, sorry. Yeah, that was yeah. Thank you So what this means is that if you These two representations determine each other they're given by one is the dual twisted of the other so If you like what you can do is you can insist that This Cm field actually contains an imaginary quadratic field And what I can do is just make one choice in which P splits And I can just make the choice of one prime and above P in that and then I say for all the places above there My local my global representation locally looks like this Okay, so Why do I care about these theorems? kind of lots of reasons one one remark is that if P doesn't divide 2d so under this assumption here, then this theorem here is implying this one Because essentially by definition if this gal representation is orthomorphic It means it comes from orthomorphic form so it lifts the characteristic zero so in particular the global representation is lifting Then so is the local one of course in the proof I'm not of course, but in the proof we the proof of this is actually purely local and it's an input into this So it turns out that it's not as far as we know In any method that's known for lifting global representations I'm going to say global for representations of the absolute Galer group of a number field all techniques that we know for lifting Mod p representations of a global field to characteristic zero you need to know the existence of lifts everywhere locally There's not there doesn't seem to be a way to go in the other direction. So it seems like you always need Things like this to to get things like this I should also say it's kind of the the analogous statement locally if you're at If you're looking at mod l representations of l not equal to p It's pretty easy to prove by hand that everything lifts So the only kind of ambiguity in in this local questions is exactly at P One reason to care about this kind of theorem is this this is giving you some local to global principle so this is giving you some way of taking a purely local mod p representation and then viewing it as Something local in a global situation and using this it's not so hard to then If you if you look at lifts of this robot to characteristic zero say you can then kind of see them Globally as well as living inside things corresponding to some space of pietic automorphic forms So when you're trying to do things in in the langans program It's often useful to even just the first proofs of class field theory were given local class field theory was proved by putting it their global situation And similarly the only proofs I think at the moment of local langans is for GLN is to put it in a shimmer of variety situation so one motivation for this kind of result is that You can use this kind of thing together with with techniques from the Taylor-Wiles method and so on to give candidates for pietic local langans and things like this and also in particular, it's kind of proved useful for this this broim as our conjecture that Is kind of one of the central problems in in trying to prove better modernity lifting theorems It's a purely local question but a lot of the recent progress is due to an idea of kissing has been to prove global results and deduce local ones from them and Theorem's like this of what led us kind of take a local situation and make it global Final thing I want to say about this theorem before I start talking about proofs and so on Is that probably this assumption here is not so crucial? It's kind of fairly orthogonal to the method it comes up at some technical point when proving some watermorpi lifting theorems I think at least the assumption that P doesn't divide D could be removed using some recent work of Brian Herman and Shren and Quite probably you can remove the assumption that P is not too But they think anyone thought so much about it But it certainly you should think of this as being basically orthogonal to the to the methods we're using here Okay, so what I want to do in the rest of time is talk a bit about the proof of theorem one a bit about the proof of theorem two Any questions before I do that Yes It depends what you want to do so so You In some sense, yes It I mean at the very least you would probably want to either have the same representation or a representation that you understood really well I mean the advantage of this is if you if you have some you're trying to prove some Some statement about some local representation It's kind of advantageous to just have just that local representation you're dealing with and not have to think about other ones at the same time Certainly, there's one instance of this kind of result isn't there's an earlier result of mine and mark Kesson's in the case D equals to where we use this to prove The bromazard conjecture for potentially butterly tape representations and there it was certainly extremely useful to have just copies of the one representation Yeah, so here it's implicit that Congress when regression act to self-exploit on the place above P or Don't you don't care? alcohol Well, it's an imaginary Cm field so Yes, sorry. Yes, you're right. I should I should probably have said that so if I've got f over f plus If this is the maximum totally real field then implicit in what I'm saying I mean is coming out is that P is going to split I Think you can prove versions of these kind of statements in which it doesn't split, but I haven't thought too much I should be careful. Yeah, thank you Yes, yeah. Yeah, sorry. So yes That should have been that was not so very clear. So that should have been V splitting as WWC or something but in particular there's no I'm not I don't have any control on how big this field is so in practice I think even probably in the construction Maybe the first step is that we choose an imaginary quadratic field then everything is an extension of that So it's an imagine imagine there's just an imaginary quadratic field to start with in which P splits And then everything is lying over that Sorry, thank you. Yes, Ariane We can construct lots of it so we can construct lifts where We don't have complete control over them But for example, we can make the whole state where it's arbitrarily spread out or we can make them kind of all lie within some range It's like Kind of D times P minus D minus one times P or something whatever the kind of best thing you expect is We certainly don't have production of all of them and in fact the construction involves I guess the construction naturally produces a kind of modular space of lifts of some fixed weights and so on Yeah, I mean certainly in general you expect and I guess it follows from this that you have lifts to infinitely many different Sets of hodged eight weights and so on. There's a there's a lot of lifts Yeah okay, so So maybe yeah, let me say a bit about the proof of serum one so kind of Sort of case zero as it were is the kind of the easy case as I already said is that If the case that row bar is irreducible So this is easy because there's a kind of a simple classification of these things which is also quite easy to prove The point is that wild inertia is This is the you know the pro-P subgroup of inertia and that has to act we're acting on a Space in characteristic P and that means it has to act with some fixed vector just because a group of p power order acting on Something of p power order has some fixed thing and since we're assuming it's irreducible This just tells you that wild inertia actually acts trivially because so that implies that Row bar is tame and then the tame thing is is also easy You've got the Frobenius and you've got tame inertia that you know explicitly how this works And in particular what's happening? It turns out is that you just have what inertia some list of characters and Frobenius is permuting them. So if you raise Frobenius to a big enough power It's acting trivially and your representation splits up as a sum of characters. So the conclusion Messing around a bit is that row bar is actually induced From some unramified extension unramified Degree D Okay, but then then life is nice. You can just lift chi bar to account You can lift characters very easily even with complete control over making them crystalline with hot state weights and so on and so you lift Chi bar To some character chi and you can arrange the hot state weights of chi to be Well, you can arrange Sorry, you can arrange chi to be Crystalline and you have kind of control over the hot state weights. You can kind of choose The hot state weights of chi to be distinct and then you set Row to be the induction of chi okay, so the the case of an irreducible representation is It's somehow no no not much harder than the case of a character in the case of a character is easy so so then there's a there's a natural strategy at this point, which is The one that's been employed. I think anytime anyone's tried to solve this problem is just you do exactly this You you lift you look at the irreducible sub quotients You lift each of them and then you just try and lift the extension classes so strategy is Just Sort of write Row bar isomorphic to row one bar Row n bar with the row I irreducible Choose row I Lifting row I bar and then try and lift To Row one Row n Okay, and I already said that I can I can choose my My row eyes to be crystalline and if I choose that the hot state weights of kind of row I are bigger than the hot state weights of Row I plus one maybe even plus one So by this when I write this I mean that kind of the minimum of these is bigger than the maximum of these plus one for example If this kind of condition holds then then in fact this representation, let me just call this row This implies that row is Crystalline so this this kind of statement I Kind of know at least two references by people in the audience for this kind of statement one is that At least for gel for k being qp these these statements are kind of analyzed carefully in in primary use article on ordinary representations in prior pedeek and you can also compute these These things You can prove this just by your so you're trying to prove that every extension of these representations is automatically crystalline and you can read that off from the formulas for the dimension of h1f in in yann's article on pedeek height pairings Probably these things were very well known to people in the 1980s and not my history But yeah, this is that this is a reasonable strategy if you can lift these irreducible things As long as you can lift the extension classes, then you get crystalline representations I'm sorry when you say the odd state place it's of course there waits for each embedding of k yeah Yeah, so so just make the statement for every embedding and you impose this in equality of me All I really need is that the hodgetate weights are strictly the hodgetate weights here a bit in every embedding a bigger than here and All of them are bigger and in at least one embedding that's bigger than bigger than bigger by at least one I basically need to avoid the case the ratio that there's there's a twist by the cyclotomic character. That's what I'm trying to avoid Yeah, that's the only thing I mean because there are of course Extensions of the trivial representation by the cyclotomic character that are semi stable and not crystalline If I wanted to have I mean I should also say that these methods are quite flexible and all these lifts and you can produce Potentially crystalline lifts of some type I could produce semi stable lifts and you know, but this is just To kind of make this all fine Okay, so Let me just show you what's kind of I Mean this sounds very good And I think anyone who ever thought about this problem as I say like pretty much had this idea and kind of checked that for gl2 You could do this and then kind of assumed that you could do things more generally I mean, I think several people I know got this set This is a PhD problem and you know, you get quite far You guys say Miller managed to do the gl3 case quite nicely, but then it's it kind of gets very tricky when you try and do The high-dimensional cases and I think there's a good reason for that that I'll try and explain So, okay, so what about the gl2 case? So let's just imagine D equals 2 Chi 1 bar Chi 2 bar and maybe even fact There's no loss of generality in twisting this This kai chi bar away. So let me just set Chi bar is Chi 1 bar Chi 2 bar inverse and So let's just take row bar is Some chi bar zero one Okay, so then then I choose some chi Lifting chi bar, which is going to zp bar star and Then the question I have to ask myself is well the extensions here are parameterized by the h1 of this character so I can ask does h1 of Chi Subject on to h1 Chi bar and of course the answer is Usually yes, but not always because there's an h2 that can intervene So so if you choose a random character then then the s this is fine And as long as this holds then I win because I just take my extension class given by row bar here And I just take any lift and that gives me my representation But it's not always Sugjective so maybe let me do the kind of the hard case for k equals qp is The case that chi bar is Epsilon bar the mod p cyclotomic character So this is the case where you have a non trivial h2 so let me just imagine that That in fact just for ease of writing. Let's just imagine that I'm actually Temporarily just actually having characters valued in zp rather than zp bar just so I can write mod p and Have it be fp and then you have the kind of long exact sequence in commodity So I have a short exact sequence sort of zp I copy of zp with chi acting on it. I have multiplication by p fp bar Just I can write chi bar or epsilon bar. So now I take homology. Sorry another bar. Thank you. Sorry And So I get that this mod p maps to What I care about But there's a co-carnal given by the p torsion in the h2 and this this is genuinely not zero so I mean in particular this is As long as chi is not actually the cyclotomic character No, maybe even then then typically this this thing here is one-dimensional over fp. This is a rank 1 zp module This this thing here is two-dimensional. So there is unsurprisingly this co-carnal But I take duality you can turn the h2 into an h0 So this becomes h0 of Gqp with coefficients. I always get my twists wrong. So let's see if I can get this right qp mod zp Probably want Epsilon times chi inverse and I'm writing too big And this is mapping to h1 of Gk and now the trivial character, which is just hom Gk, so this is this is kind of the the dual of this and What we need to see is that we can actually Well firstly you can compute what this co-boundary is it's something easy which I'll explain in a second But you kind of need to see that you're gonna have to see that by varying this character chi you can fill out everything here so This is the kind of first place you see that there's something subtle is that because The h1 for chi is going to be one-dimensional and the h1 for chi bar is two-dimensional You need to allow yourself to vary your choice of chi So well, I said that the strategy was to kind of choose a lift if I've literally fixed a lift I can't I can't make things work so we need to allow chi to vary to Kind of fill out all of all of the possible classes here So it turns out Is everything okay? Cool. Thanks. It turns out that We just have to allow some unrammified twists to twist unrammified characters so let me kind of I'll sort of say explicitly in this situation Where my mod p character is the mod p cyclotomic character. I'm gonna take Well, if you think about what I was doing over there I said I wanted the hodgetate weights I wanted to choose my lifts so the hodgetate weights were sort of increasing by more than one So the obvious way to lift this this more p cyclotomic character would be to the cyclotomic character But then the jump in hodgetate weights isn't big enough So I sort of take chi to be instead the p power of the cyclotomic character Then all my extensions really will be crystalline and I will twist it by an unrammified character here so lambda a is some unrammified character taking the Frobenius at p to a and Zp star and then it turns out that the That you could also take p equal 1 and take an unrammified character I could indeed do that. It's just slightly cleaner for me not to do it right now, but yes As long as my my unrammified character was non-trivial then then that that would work It's just slightly easier for me to just sort of say I'm gonna do that and then And it also makes the kind of calculation you do here slightly easier Not to have to worry about what what's going on when you're taking the h0 if you because again This is kind of bad if I actually had the cyclotomic character this h0 is Really a Qp mod Zp, but as long as this this chi is not literally the cyclotomic character then this is of finite length anyway, what happens is that this Sorry for switching back and forth between the board so much, but the the kind of co-boundary here is basically this map If chi is whatever it's in there lambda a as long as to the p this maps to Kind of a minus one over p As long as a minus one is it is exactly as wide as long as this is non-zero this is giving me some scalar which is And this is this is I think the unrammified Map sending for being is to this a minus one on p. Let me just kind of put this in quotes But this is I mean you can imagine how co-boundaries work You I mean you know how to do this in short exact sequences you do some lift and and it turns out that what you have to do is you have to kind of consider Sort of the smallest power of p modulo which this character is not the trivial character And then you you take the difference in the trivial character and divide out by that power of p That gives you something non-trivial and that's how you do this kind of computation And this is something you think that extends to a very general situation So anyway, what happens is it turns out that you need to kind of vary a in such a way as you're varying this quantity a minus one over p and As you vary a this does indeed vary and you manage to span all the classes you need to so this is this is some Indication of how you make a proof like this And you can kind of do this in a sort of very very concrete way So this is something you can do for this This this gl2 case now Now you try and do gl3 and it turns out that the kind of problem that you can have is If you look at cases that look like Having basically the problem is always having lots of characters with ratios being the sick atomic character So this is already kind of bad I Think what happens is that maybe I Forget the exact details but the kind of thing that happens when is that you start having trouble not when you're kind of in the generic situation where lots of things are All the all the things are non-split as possible, but you really start having problems when you actually have some non-split things so This thing kind of thing is not so bad is not so bad I think although if I now take if I write down this formula, but now k is qp Z to p for example Rather than qp then things are suddenly a lot worse because this this character is trivial and Suddenly all my characters have ratio which are both trivial character and sick atomic character And I have h2s and h0s and things are kind of nasty You can still do all this kind of thing by hand and this is what Muller does I mean not quite by hand with some cleverness And you can prove that you have lifts what you have to do is you you lift this to sort of Epsilon I'll get my numbers wrong sort of 2p times lambda a epsilon It's the p lambda b. I Guess I only need to do one you can you can always make lifts like this But the other kind of scary thing that happens is you it's hard to do an induction so what seems to be hard to do is to either Choose your be here make a lift here and then try and kind of lift to this or to do it the other way around So how have you kind of split things up you get into some trouble? You basically find that when you when you come to make lifts up here You maybe need to go back and choose this be a bit more carefully As saying the rank three cases is fine But it's kind of easy to write down once you want to try this you can then kind of make up hard examples for yourself And you do some rank four or five example, and it's just completely hideous You can actually find examples where provably No matter how you break it up into an extension of two smaller pieces There's kind of no way that you could have chosen your lift for the two things independently and then made The lift of the whole thing even if you do some unamplified twist basically what you need to do is you need to If you're doing some inductive argument at every stage you need to be able to go back and refine all the choices you made in the earlier stages It turns out that it does always work and as I say we do always produce by this theorem these ordinary lifts But I think any kind of way to do it by hand is just basically doomed as soon as the dimension is at least maybe Five and certainly six. You're just the conditions you write down are too complicated Okay, so Morally though you can kind of see what you want to do What you want to do I Mean you don't want to make explicit choices what you want to do is kind of just consider instead like maybe fix here My powers of a sick atomic character, and then I want to just consider kind of the modernized space where I let Kind of a and b both vary and all these extension classes vary And I just want to show you maybe that that modular space is big enough that when I look at the reductions mod p I see all the mod p representations. That's the kind of thing that Seems more natural to be able to do and you can kind of when you do these calculations You kind of feel like this is all going to work basically What kind of happens is that the hard cases sort of involve some of these extension classes being Split rather non-split and that's imposing some condition that makes the kind of space of mod p things smaller and Correspondingly when you look at the space of possible lifts It's imposing some conditions, but these conditions are like in kind of quite high-co-dimension It feels like generically you kind of win if you just kind of chose at random possible lifts And then everything would be okay, but the problem is actually proving that But that's how we prove things in the end. We exactly do build some modular spaces Let me kind of state a precise theorem again kind of Just in this same kind of setting about what we what we prove and then let me say something about how we prove it So this is theorem 3 which kind of easily implies theorem 1 by induction so given a crystalline or given to Crystalline representations One is called row D. It's a d-dimensional representation and one is a-dimensional and it's just an arbitrary Irreducible crystalline representation so with Well, in fact, it's not just irreducible it's irreducible mod p so alpha bar Goes to GLD. Sorry. GLA FP bar is irreducible and I'll demand my same condition on hodge-take weights so It's assuming that the hodge-take weights of row D are bigger than The hodge-take weights of alpha maybe plus one Then what I'm going to show is that I can lift any extension of alpha bar by row D bar to an extension in characteristic zero so then any extension row D bar to Let's just say row D plus a bar So this is just my name for some D plus a-dimensional representation alpha bar to zero Lifts to I'm not have to be slightly careful I'm not going to lift it to an extension of alpha by row D because I can't literally do that like just like I can't lift every extension of the Trivial character by the more piece it because we're character to an extension of Of the trivial character by any particular given lift so I now I need to rely on some flexibility in In my lifting so what I do I have row D primed Row D plus a Alpha zero so alpha is the representation I fixed up there Row D plus a is crystalline and so of course row D primed is crystalline and With the hodgitate weights of row primed D being the hodgitate weights of row D So and this lifts this in the sense that this lifts this representation this this this one This is genuinely a lift. So in fact, we actually have more control than this What we really prove is not just these have the same hodgitate weights, but they lie on the same Irreducible component of the of the corresponding crystalline deformation ring So we have the same mod representation mod P. We look at all of the possible there's a modular space of the the lifts of that and What we're demanding is that this lies on the same irreducible component. So in particular has the same hodgitate weights So once you have this theorem, it's very easy to prove theorem one just by induction on D. You just adding an extra Representation at the end and just producing this Okay, so how do we prove this? bad management to my boards Well, basically what we do is we kind of employ the strategy I talked about here of just kind of considering I mean you just look kind of like consider all possible lifts So what we what we're going to do is we're going to consider all possible lifts of Row bar D to one of these row D primes and then we're going to consider all kind of possible lifts of the extension class and then just sort of show that this thing is well behaved so So given row D there exists Kind of a formal scheme that Yeah, let's call it spoof our Row bar D and what is this thing? This is going to be the universal I should say framed defamation ring for crystalline lifts of Row D bar With kind of hodgetate weights equal to the hodgetate weights of row D Okay, so I know that I have some crystalline representation. So I know this is some Non-zero thing and it's just literally it's a Modulized space for all the possible lifts which are crystalline have these given fixed hodgetate weights. So for example kind of in the case above If row D bar is One piece of atomic character and row D Was the six on the couch the power of P. I'd just be looking at Just a power series ring Corresponding to kind of to a minus one I guess Okay, so it's just the unramified twists in that case in general that it's something much more complicated I don't want to try and like make it explicit. I wouldn't know how to but it's something So this is kind of a sensible object to be thinking about because I'm allowing myself all possible choices of how to lift this This row D bar, so when I don't know much the theory, but when you define the model like problem So does it make sense to speak about crystalline representations? So no, so what you do is you can kind of construct the universal ring with no conditions at all And then you can take the Zariski closure of all of the crystalline points and that that's something that you can do So you can look at all the specializations of it at zp bar points and ask if those are crystalline And then you show that the Zariski closure of that is some sensible object and you can compute its dimension and so on Frames just means that you have to carry around a basis Because because there's automorphisms, but yeah, it's kind of it's harmless for everything Okay, so so then over this space I can think about There's kind of a sheath of the extensions the kind of x2 gk of Alpha by the kind of maybe I'm going to write row D universal for the for the kind of Universal galvanization living over this so I can kind of consider this this sheath of which is just the analog of my H2 above and it turns out that Basically, this is kind of This is where all the difficulty is if this thing just vanished there would be no problem and everything would you just lift or your X1 classes everything is it's easy. So we have to have kind of have some kind of control over this By our assumptions on The hodge-take weights and so on this thing is is At least set theoretically supported on the special fiber. There's no x2 in characteristic zero. There's no ratio being cyclotomic but I have to kind of Worry about how big this might be. I mean, this is exactly where this kind of a minus one on P thing is coming from So there's also a kind of a universal Something that's like a universal family over here So this is the kind of thing where that why I'm kind of considering all of my possible lifts to extensions This is kind of the universal family except it actually turns out that this is not Because this x2 is not zero. This is not a vector bundle. This is there's some some torsion here Exactly in this case So I was looking at this is exactly where this a minus one and P comes in So what we need to do is to kind of consider properly the universal family of extensions You have to be able to make sense of things like this a minus one over P and of course Here this this doesn't make sense. I'm not allowed to divide this by P But what I do how do I how do I make that make sense is it turns out? I mean just think about that particular case you need to make some kind of blow-up to make these these functions reasonable So what we need to do and what we do do is we replace This thing here we're calling this by blow up Kind of at the support of x2 and then you have to do some work But it turns out that the basically I mean You can make that construction and that doesn't tell you anything. I mean a priori you know nothing but it turns out that all you need to do to make Everything reasonable and make a make a kind of fairly straightforward argument about these instances of lifts is just to have some control About how big the support of this is So it turns out that what we need the kind of key input to make this work is Is Is the following bound we need that the co-dimension So for any r greater than or equal to one the co-dimension of the kind of locus of kind of points so that The dimension of this x2 Is greater than or equal to r that this co-dimension is at least are So for our is great r is one for example I'm saying that I don't have this x2 is not supported everywhere. That would be kind of like my disaster and Similarly, it turns out. I just need to know this in all dimensions that so the locus of things where This x2 is is very big is a very small locus So if you think about the kind of example I was writing down for gl3, maybe it still survives So by take duality like the I mean the loci where where this x2 is big is exactly to do with having having lots of Kind of characters here whose ratio is a cyclotomic character And having non-split extensions there so having sort of zeros here is making some x2 bigger and Kind of intuitively the only way you can make the x2 bigger is by putting by making you imposing a bunch of conditions Okay, that's not kind of meaningful in some sense though, and I'm kind of giving to the punchline of Why we need to do quite a lot of work to actually That key input turns out to be a lot of work to prove Because when I kind of saying oh, I'm like thinking about this and I'm putting a zero here And it's imposing some conditions and making things smaller implicit I'm talking about some Object where I'm allowing this mod p representation to vary within all of this this whole thing here This row bar D was completely fixed so So this x2 of Row D Universal Maybe I'll just write road Let me just write row D for a second just to be not carrying around too many symbols By alpha of I got my writing somewhere around alpha by row D. Thank you By take duality is isomorphic to Homs from row D To alpha twisted by the cyclotomic character So what I'm trying to do is I'm trying to say that inside this this Defamation space of all the kind of possible liftings. I'm wanting to Have some bound on on some kind of reducibility locus I'm kind of saying how often can I admit a map from my deformation to this representation this this fixed irreducible representation and and That's kind of some kind of question that seems quite hard to say much about in general so When when row bar is fixed. So what we do Is we build instead a? Modulized stack of Fee gamma modules Whose Zp bar points are Exactly the the crystalline Representations of GK I was fixed Hodge take weights so Fee gamma modules and gal representations are not the same thing in families, but they are when you when you look at the kind of points over fields or over Over kind of Zp or Zp bar or something and I'm not making any requirement here that the mod P representation is fixed So then we prove That this is actually a formal algebraic stack and so the special fiber is just An algebraic stack over Fp bar and then inside this This now sees all the mod P representations which come from crystalline representations of that given those given hodge take weights but now Now those those mod P representations are allowed to vary so then inside here so inside this stack you can kind of the sub-stack of representations with the dimension of the x2 of Alpha bar now that's just said I'm going to work mod P the whole calculation can now happen mod P This is great for me equal to R But you can compute this now Just by Galois homology. So now I'm like literally looking at Representations that kind of admit alpha bar as a quotient So I'm looking at representations of this form and it's pretty easy to compute just using Galois homology that the mention of this is Well that I want to prove there's a co-dimension of at least R, but in fact, it's much better than that you can Particularly if K is is ramified you can get at least a quadratic growth and maybe better Yes It's a revenue question sure this sub-stack now you write of representations, but do you really mean a frigate? I really mean a figure modules, but the point is that point wise I can it turns out that you can do all these computations point wise So yeah, I mean I'm about to be over time if I'm not already so I'm gonna not say so much But yeah, you use the the air complex long as That's a stack of a zp bar it turns out that the computation can actually just happen in a special fiber. So I have just an algebraic stack over fp bar and I literally get to do a computation just on its Points and I get to do a computation using Galois homology and and it's kind of easy to inductively prove and the dimension condition is Defined by you need some scheme structure by fitting right there or something like how the because usually they meant this defines a set and not Well, this is this there's no homology in degree bigger than two and so it's not so hard to check that I mean I'm just looking at the dimension of support of the h2 Yeah, we can we can talk about this after I mean I don't want to it's not this is not a problem I just wanted to say like 20 seconds on theorem two I'm sorry. I completely wrote a two-hour talk theorem one implies theorem two using This car event and verge a method that I've left myself no time to say anything about So What I can literally say is that for GL to Disimplication is basically literally there's an argument car in van to verge a found this great way of given a mod p global representation of constructing lifts with very carefully prescribed properties and They did this for GL to over Qp to prove says conjecture they Basically any time You would expect robot again from modular forms any kind of congruence you would expect from the modular form side They managed to construct purely on the Galois side and their argument for GL to If you kind of put our theorem in as an input this comes out So what we need for GLD is some some kind of version that in D dimensions Which has been basically built by the work of quite a lot of people on the automography lifting things for higher dimensions But the key argument is really this this argument here, but I'm definitely over time. So Apologize I'll stop Not that I know of It's a little harder to expect something so So mod p for example, there are at least you know You have all these mod p representations of the absolute Galo group if you wanted to do something globally that made sense You certainly want mod p there to be something so you would need some modernized space of mod p representations so if for example, you'd I mean it seems reasonably natural that you would Bound the conductor away from P or something and as soon as you do that then conjecturally there only we find out in many representations Like this is true by says conjecture for GL to if you're looking at odd representations. So yeah I Think in some sense actually what happens so we we have some of these stacks for crystalline representations We also have a much bigger stack that has all of the representations But that the special fiber of that is not actually an algebraic stack That's still formal and some other way the dimensions work out is that the the dimension of the the kind of crystalline part Plus the dimension of sort of global deformations adds up to the total dimension and somehow I think the reason so these crystalline things seem to be about the biggest kind of algebraic stack which is actually Sorry the biggest sort of thing which is which is genuinely algebraic mod p and not formal And I think in some sense the reason why you can't go any bigger than that is to something to do with a lack of global Families. Yeah, I mean, maybe there's some there's some analog, but I don't see what it would be something about geometry of this modernized stack and especially special fiber If it's in the structure relation to the x1 x2 Yeah, so so there's a lot you can say about it. So so so in fact yeah, so so essentially by definition the the infinitesimal deformations Points in this stack are recovering these these gala deformation rings of may zone and so on Either unrestricted if I if I kind of remove this word crystalline or in this case the crystalline things The global geometry we don't know so much But we did what we do know which is turns out to be very useful in applications and it's kind of actually important here Is that by thinking a lot about kind of families of extensions we can we can compute Exactly how many irreducible components There are and give a kind of description of what the generic representations on them Oh, so it turns out that in characteristic piece like you surprisingly for each irreducible component basically generically it is in fact Completely reducible so the irreducible things are actually a very small irreducible representations are just kind of some scattered points Yeah, well, yeah, so so generically these these are each irreducible component You get these and the the kai's bars are kind of fixed on inertia So if you think about how that works you can kind of work out all the possible things There's a natural way of labeling them by irreducible representations about the algebraic group, but I won't go into that It's probably possible to prove that the This thing is a complete local complete intersection. I'm not totally sure what's been proved. It's certainly expected to be No meaning the the special fiber at least of the of the of the stack Basically because I mean the again if I remove the word crystalline So the special fiber it turns out the special fiber of the whole thing even without the crystalline is the same as the with the crystalline things basically because of all the more peer representations have crystalline lifts just Part of how it's all proved But it's expected that if I look at deformation rings not with this crystalline condition just the unrestricted deformation ring Those are expected to always be local complete intersections and I think this has been proven It's certainly known for GL2. I think it's now close to being proved in general Yeah, so There's a lot of interesting questions, but I yeah, I I could talk about this at some length, but probably I should Not run too much into Lawrence talk