 Last time, I finished the discussion of a talk homology of diamonds, the abstract setup. And so let me, some of the whole program that I'm trying to pursue, some of us consistent in some kind of three step process, the first one has to understand the abstract theory of a talk homology of diamonds, which is something that I pursued. This is what I've talked about in this, written. Then the next step is the study of the stack banji that I will talk about today. And chiefs on it. This is something that is work in progress with Laurent Farge. And then once one has understood this, one can try to really go to things one is interested in. And so on the one hand, one can try to define on some local lengthness correspondence to study it. That's what I sketched out in the very first lecture. And one can also try to prove that this local lengthness correspondence is realized in the homology of some proper point sink spaces and generalizations. So in fact, the formalism allows for the base generalizations of proper point sink spaces and. And does it include proving that it is the same as the previously known local lengthness correspondence? It includes for GLN, it includes proving that it's the same thing as the previously defined correspondence. The problem for other groups is that in the homology of a specific variety, you can't really see an L parameter. You can only see a map to some GLN. And so you can usually only detect the image of some symplectic parameter or whatever in GLN, which sometimes is enough to determine the parameter. So that's the kind of picture. In the abstract to the whole series, I included talking about something here. I'm afraid I won't be able to really get there. And so I've needed three lectures here and I will probably need three lectures to really discuss this. So I will spend the last two lectures talking about this part of the picture. Okay, so I want to talk about this bungee today. So for this, as was already done, I fixed g over qp reductive group. And for some reason, it's actually better to be in some kind of geometric setup in the sense that one works over a fixed algebraically closed base field. And I want to fix this algebraically closed base field from the start. So p, for example, k, p. My test category will from now on be the category of perfectoid spaces over this, which have a structure morphism to the eddy spectrum of k. So the algebra will be k algebras. Okay, let's call this perf k. And also an important notation will be this field l, which is the fraction field of ring of it vectors. And so then bungee will be the following thing. There will be a stack on the category of perfectoid spaces. So let's say, well, there are slightly different perspectives on what a stack is. It's a function, a function of fiber in group points or something like this. The category of fiber in group points, so let me pretend it was actually a function to group points, which takes any, let's call the test schemes now S. It maps it to the group point of G torsors on something that's called XS. And I need to recall what XS is. Your XS is a relative of ten curve usually said to be over S, but actually there is no map to S. So it's kind of lie, which is the same thing as some other space YS. Divided out by the action of the Frobenius of S to Z. Where YS is a space that I called last time on like this, which doesn't make sense. But I find it intuitive way to think about what this should be. It should be the product of my test space S was the kind of curve I'm considering. Which means the curve is some of this punctured formal disk of Z around P. Meaning by QP, or maybe I changed to an algebraic closed space field. But it's essentially QP here, which is small punctured disk. So it's a family of small punctured disks. And so I will give a precise definition in just a second. Which has a Frobenius phi S, which is the Frobenius on S times the identity. And because this is a product with spa L, this thing should sit over the edict spectrum of L. So it's an analytic edict space, automatically analytic. And actually I should have said somewhere, which I maybe didn't. I said sometimes I just write spa R, but really this spa needs two ingredients. And if I don't write the second ingredient, always mean that I put circle. So if you take Frobenius on S, S is what, is a perfected algebra of LK. Perfected to S. So the Frobenius on it is like for schemes. If you want the map to spa K, it could be a covariant. You have to put not identity. How can you put identity? Sorry, yes. But there are two kinds of Frobenius you could put on spa. Well, so that's the same thing as S times spa FP, spa QP. And then it makes sense. So there would be two, well, on spa L there's really only one Frobenius. Okay, so let me tell you two things about YS, what YS is. One thing I can tell you is that if I pass to the diamond, which I'm allowed if it's such an edict space there, it really is true that it is S times, so absolutely. So as diamonds this formula is true. And well, for a general space, it's slightly tricky to write down what it is. Because I mean, so what you do is you define it for a phenoid spaces and then you glue, and you glue so that this formula stays true. If S is the edict spectrum of some r plus, then YS is the following thing. You take the edict spectrum with vectors r plus, with vectors r plus. With the intuition that this is something like taking the product with ZP, but then you want to do a product with QP. So what you're going to do is you remove the locus where P is zero. But you also, I mean, you get R by inverting a pseudo-uniformizer here and you just need to do the same thing here. So you also need to remove the locus where Pi is equal to zero. Where Pi and R plus is a pseudo-uniformizer. And you check that this is independent of the choice. So this is just for the usual case of tape? Well, yes, yes, yes, yes, yes, yes, yes, yes, yes, yes. I mean, I asked the perfectoid in the sense I defined in the lecture, okay. And so you see that this has a Frobenius, which is just a Frobenius on R plus. Yes, it's induced from the Frobenius where R plus is some perfect ring, okay. So it's an automorphism and you check that this acts totally discontinuously. And so it makes sense to pass the quotient. So some kind of picture of this is that you have here is a ZP direction and here's the S direction. And then maybe this device I would be or maybe let me put here some of the R plus direction. Then here's somebody has the device of Pi is zero and the device of Pi is zero. And then you're looking at the remaining part here. And then the Frobenius is something which shifts everything in this direction. And then if you want to take the quotient, you can somehow take some annulus here and then identify the outer ends to form some kind of torus. So under which condition I can pass to a quotient in attic spaces? I think here it's totally discontinuous in a very strong sense. It's totally discontinuous in the strongest possible sense, I mean. So you think that maybe it is joint? Yeah, yeah, yeah. So I mean the quotient map is really a local isomorphism, you have local sections. Okay. So okay, so one important warning is that there is no map from, not even from YS, but in particular not from XS to S. Because this is a characteristic P, this is a characteristic zero. But there is such a map of diamonds, yeah? And because diamonds determine topological spaces in the tall sides, you also get a map on topological spaces and the tall sides. So in this sense it is over S, but only in some vague sense. What else do you want to say? I mean, also some slightly funny property of this curve is that say if S is some sparse cc plus, where c is algebraically closed, not a convenient field, then you can compute the global sections of the structure sheave on this curve. And it's just qp. So usually for a relative curve of S, you would expect that the global sections would have something to do with the field in some kind of usual coherent setup. But here it's always qp. And this is very important for what's about to happen. So this is because phi invariance and summing of width vectors. So I think I believe it's farther from ten defined such curves from more generally you can have at any local field, so they use the width vector relative to sum. Right, I mean, so as I said, in everything I'm saying, you can replace qp by any local long-line communion field. And then you just replace width vectors by a corresponding generalization. And essentially everything goes through in the same way. Finite extension of qp or fp long-series. But if you call it the equal characteristic case, you will have a still characteristic. Yeah, in the equal characteristic case, it's still characteristic p. And so in this case, it would actually be such a method. So in this case, this would also be a valid formula. Okay, but anyway, so, okay. So what's important is that there's a function from isocrystals over k, which I recall are finite dimensional L vector spaces. V plus phi linear isomorphism phi V from V to V towards vector bundles. So why? So that's because vector bundles on Ys on Xs are the same thing as vector bundles on Ys plus the same as phi s, the covariant vector bundles on Ys. And you can take some V phi V and map it to V tensor over L with O Ys. And the Frobenius, which is a tensor product. All the vector bundles on Ys should be. All the vector bundles on Ys, well, not for general S, if S is a geometric point. S is a spectrum of maximally complete field maybe. Okay. At least in the equi-partistic case, it has its force if it's not maximally complete. Spericality. So Laurent-Farck says that let's put ourselves in the equal characteristic case and assume that S is your geometric point. Then they are not usually trivial. So there are not examples where bundles are not really even for line bundles, I think. There is a thing about the Picard group of these things. Even for line bundles, they are not trivial. But it might be true if S is spherically complete. I mean, there's a theorem of Lazar apparently. For general vector bundles or just line bundles? Lazar for line bundles and gruson for any vector bundle. Lazar for line bundles, gruson for any vector bundle. And so more generally, you get a functor from what's called G isocrystals, which are just tensor funtors, exact tensor funtors from the category of representations of G towards isocrystals towards bun G of S. And so there's the following important theorem, which for GLN is a theorem of Farck and Ponten, and for general G is a theorem of Farck. And let me also mention that in the equal characteristic case, this was open for a while, but it's not also resolved. So this is one instance where the equal characteristic case is actually harder because reductive groups and positive characteristics are more difficult. This was recently solved by Johannes Anschutz. It says that if you have a geometric point, then actually it is true that G bundles over this geometric point up to isomorphism classified by isocrystals, G isocrystals, bijection, with a set that called C nodes by B of G, which is just G isocrystals. So over a geometric point, you can completely classify these guys. So this is just a bijection, so you have a map, what is it from? Right, so there's a functor, but it's not an equivalence of group odds. And I will say to what extent it fails to be an equivalence of group odds in a second. Yeah, so let me just stress this, so this is S sets. So in these last two lectures, I want to, on the one hand, discuss the case of a general group, but on the other hand, I want to give the example in the case of GL2 to make it a little more concrete. So I will start discussing this example of GL2. And so say in some connected component, maybe here, you have O squared, which which specializes to something called O of 1 plus O of minus 1. So they are the picar group of this curve is given by the integers as a consequence of the theorem, for example. And I denote the corresponding line bundles by O of n as usual. And so you have this kind of specializations here. Then there are also vector bundles that are called O of a half. So this is not a line bundle, this is a rank two bundle. So in general, the simple guys are classified by rational numbers, S in the case of isocrystals, and that you know it's a simple guy of some slope by O of this slope. So this is rank two. And this will specialize to O of 1 plus O, will specialize to O of 2 plus O of minus 1. And then here you have maybe O of minus a half, specializing to O plus O of minus 1. And then here you have O of minus 1 squared, specializing to O plus 1. And so this looks roughly similar to the picture you see for the stack of G bundles on P1. So for P1, you also just have a discrete or countable set of points and some funny specialization relations. It looks more like a list of rational numbers or two conspicuations. But for elliptic curves, I think you have a modular, right? Yeah, so there's some philosophy that if there is any genus of the far from 10 curves, somewhere between 0 and 1. So the picture you get is usually some mixture of the picture you get for P1 and what you get for an elliptic curve. Okay, so let me tell you what remains true of this picture in general. The first thing is that you can classify the connected components of bungee. They are in general given by what's called the algebraic fundamental group as defined by Voronoi. And then you take the covariance under the Galois group. Let me write Galois, to not confuse us with the group and here it takes P1. And every connected component has a unique semi-simple, a unique semi-stable. So these correspond to these generic points there. This is in general, yeah. And semi-stable is defined by some kind of idea about reduction theory. One way to say it is that if you push it out along the adjoint representation, that then it's a semi-stable vector bundle. Actually, the semi-stable vector bundles works as well as it doesn't characteristic 0, even in positive characteristic, where there are usually some subtleties that it's not completely functorial. For example, tensor products of semi-stable bundles need not be semi-stable. In general, it is true for this curve. So you don't get into this kind of trouble. And in fact, okay, let's denote this functor here by... So these guys are usually called B. And let me denote the associated bundle by EB. Then the condition that this bundle semi-stable is a condition that's known from the theory of G isocrystals. It's this condition known as being basic in the sense of Kotwitz. And in this case, it's actually true that if S is a geometric point, then the automorphisms of EB are the same thing as the automorphisms of B, automorphisms of this sec tensor functor if you want, which on the other hand are also known as a group usually called JB, but I kind of want to call it GB because it's just in the form of G. GB is an inner form of G. It's sensibly determined by this property. And so you see that the automorphisms... Let me give you an example in a second. So usually this is called JB, but this makes it look more asymmetric than the relation really is. So for example, you have that the automorphisms of O squared are just GO2QP. In fact, this remains true as you pass to some functors. So if this is a thing which sends any S to the automorphisms of O squared on XS, and this is a thing which I define for any topological space, then this is true. There's a continuous function from S into GO2QP. So it is not the algebraic group GO2. If you take the automorphisms of this bundle O of a half, what you get is D cross where D over QP is a quaternion algebra. And so in this picture there you get something of two-periodic picture where the even connected components you get GO2 and on the odd connected components you get D cross. On the other hand, if you pass to non-semi-stable guys, if you look at the automorphism group scheme or group diamond of some non-semi-stable guy, so say O plus O of 1, then the picture becomes more complicated. So what you have are diagonal automorphisms where you just act by a scalar on O of 1, which are the automorphisms of just a line bundle. There are no maps from O of 1 to O, that's one what maybe I expect, so you have a zero here, but there are a lot of maps from O to O of 1, it has a lot of global sections. So here in the upper corner you get the global sections of O of 1 and this is some open unit disk, so it's one-dimensional. Whereas QP cross is just a profiled set, so this is something zero-dimensional. So how does QP cross act? How is that? Well, just by multiplying by a scalar on these. Also just by, well, this multiplies and this divides. But then you want it to be in the open unit disk, so if it is open unit disk? Well, I mean it's like, sorry, which open unit disk is it? It's really it's the, you take the open unit disk inside the perfectified GM, which is the disk of radius one, open disk of radius one around the identity. And this is some kind of QP vector space object, because, well, because you're perfect, you can somewhat take piece roots also. It's one of these Banach-Kolmes spaces. Okay, and so there's a general picture, so in general, so if P is not necessarily basic, what you have is the following that you have, and maybe I should have used, these are not bigger either, what you have is the following, so you have the automorphisms of this bundle EB, and there's actually projection to this GB of QP, which now is not an inner form of G, but an inner form of a levy of G, at least if G is quasi-split. So for example, in this case, it would just be the trivial inner form of the diagonal torus, but then there's a kernel of this map, and the kernel, what is it? It's some kind of unipotent group diamond, which is a successive extension of some Banach-Kolmes spaces. What is O to B? What is O to B? That's over a point. Sorry? You said that you have a map of group weights, which is not the fact of physical right, so there's actually a splitting of this map, which is given by this functor, so you always have the automorphisms of this B, which if you do it functorially, we'll give you this GB of QP underline. It always maps to the automorphisms of this guy, but actually it's also a quotient because it's also parameterizing the connected components of this group scheme, so there's something connected here. Does it give, see what? Right, so if you look at G bundles, together with the splitting of there, so we can always define some hardenerism infiltration. If you fix the splitting of this, and then look at automorphisms which respect the splitting, then this always, if I would put this on the left-hand side, then this would be true as group weights, which somehow isolates the subspace here. It's the ones which fix some chosen splitting. Well, EB, if it comes from a B, then this EB has a chosen splitting, so it's inside. Good. So that's roughly what this space looks like, and so the basic theorem, I mean, as everything I will talk about this joint work with law and law, is that, first of all, this bun G really is a stack, it's a V-stack, so you can glue G bundles in the V-topology, but then you can ask yourself whether it's some kind of arting stack, which is true in the classical setup. So being an arting stack should mean that there is some smooth map onto it from something which is itself smooth, and this is also true, so there's some comatologically smooth diamond, with a comatologically smooth surjective map. Sorry, maybe I shouldn't call this X, just something. I don't think I will use C a lot, so I can use C. So you should think that bun G is a smooth arting stack, and I said that there is some slight confusion still about the notion of dimensions, but these confusions should go away in the relevant case, and it should be of dimension zero. And... So it's comatologically smooth for all l, and so whenever I will say comatologically smooth in this lecture, I will mean it's l-comatologically smooth for all l not equal to b. So this means that you can actually extend this formalism of this d-add, but actually this d-add you can always extend to v-stacks, small v-stacks, just because it makes sense to talk about the v-side of a v-stack, just saying all v sheaves mapping to it. So you can talk about sheaves of lambda modules on that, and then you can still isolate this d-add inside. So this is something you can always define, but you can also define a dualizing complex in there. So dualizing complex of bun G. For the second you need that the dualizing complex commutes with pullback under smooth maps, which was part of what I said last time. I actually don't really know explicitly how to describe the dualizing complex. I think it's an interesting question, but what I know is that g-bun G is locally equivalent to lambda and degree zero. So as you would expect if it's of dimension zero, but it's not globally. It's actually related to harm measures. So if you have say k, the subgroup of g of qp, compact open, then you can wonder what the global sections on point mod k of the dualizing complex of point mod k. And those are harm measures on k. And so if k is pro p, you can always choose such a guy. But sometimes if L is a prime which divides the pro order of k, you might not be able to find such a guy. It's not really important. Okay, so you can make sense of this d-ad of this bun G. And in particular you can also make sense of your g-duality. So let me actually assume from now on that my coefficients are the ring of integers of some e modulo L2 is the n, where e over z qL is finite extension. You said that there is no harm measure with the raise one. If you have a hopper group, even. Yeah, that's right. Take one of the smaller subgroup and put the translate the same thing. So what do you Yeah, maybe there is one. I mean, yeah, I'm not okay. I would have to think about this again. I think well, this is true. And then you can wonder whether there really is a global trivialization of this. Anyway, some of trivializing. Let's say it's not economic. It might be that there is a global. Okay, so let me assume from now on that we're in this situation, which is maybe the one of most interest. So in particular lambda is finite. It's Gorenstein. It's also self-injective, which will play a little bit of an implicit role in what I'm doing. So we have a value duality, which might not be a duality usually, but you can take any f to this r-home f into the globalizing complex of 1g. And the following definition will be critical. So I want to define a good class of sheeps of 1g that satisfies some good finiteness properties. And this notion of constructability from yesterday is not good. First of all, because the strata on 1g are not constructable. And secondly, because I want these sheeps to have infinite dimensional stalks because the stalks should be related to representations of these periodic groups. And so I need a different condition. And I make the following. I say that f and d at x lambda is reflexive. Sorry, yes, yes, yes, yes, yes. It's reflexive. It's a natural map from f to its double dual is an isomorphism. The usual notion. Reflexivity. So because d-1g is actually trivial, you don't have to know what's the cotangent, the dualizing thing is actually. So it's an equivalent to say that it's equivalent to the r-home. And the claim is that this is exactly the right finiteness condition to put. And so first, let me say that more generally, say working with, working with diamonds, let's say z over a fixed base s, which would be this bar k here. Let's say that f and d at z lambda is reflexive if f maps to the double dual of f. Where now the dual is defined to be the r-home into the p upper free lambda, p from z to s projection. And so why, why is the notion of reflexive sheaths nice? It's because it's easy to prove that reflexive sheaths are stable under some operations. So what will the condition to define this p upper free? Yeah, so let's say representable and locally spatial didn't find it. Okay, so shouldn't call these y and x. I mean these are now just abstract things and unrelated to the ys and the xs that appeared before. If you have a proper map and you have a reflexive sheath upstairs, then you push forwards automatically reflexive. And in the other direction for pullbacks, if you have a comodically, l-comodically smooth map and you have a reflexive sheath on the base, then the pullback is again reflexive. Okay, so this essentially will imply that reflexive sheaths are stable under the hack operators. I should say that there's still a little bit of work to do, but it's the essential idea that you can reduce the hacker correspondences which are given by smooth kernels. And then the pulling back first preserves our flexibility and then it's also a proper map. So push forward will again preserve our flexibility. And so it's a notion that's easily seen to be stable under the hack operators. That's why it's a good notion. Well, I mean it depends on those properties. That somehow is given by smooth proper maps. I can essentially completely prove this. So we always have the following identities. That the dual of r of lower shriek is r of lower star of the dual. So this is the duality on x and this is the duality on y. And also one for the shriek. So these are two on the full derived category. Well, assuming that f is good enough to even define what these functions are. I mean it needs to be more representable locally spatial diamonds with dim trug finite. No, I would only want the relatively representable maps. Yeah, right. So this is basically what This is relative to the content of vergeriality. This is vergeriality somehow, right? I mean this is essentially the definition of these functions. I mean x and y are the wrong way here, right? Okay. But if f is proper, then you have said r of lower shriek is the same as r of lower star. And so you also get the other one. So also someone talks logically, the other one is true. And so it's actually true that if you take the double dual of r of lower star, then it's the dual of r of lower shriek, the dual, which is double r of lower star of the double dual. And so if f is comodically smooth, then r of upper shriek is r of upper star tensors dual. And so, okay, this locally constant sheaf of rank one can somehow be, I mean commutes with everything tensoring with it. And so again you also get the other one that f upper star of the dual is r of upper shriek with the dual. And so in total you get that f upper star of the double dual, the dual of r of upper shriek is double dual f upper star. And so because you can pull the double dual through the functor, it means that this kind of condition also can be pulled through the functor. So this is essentially formal. Why did you say that some work, where's the work? I didn't quite say what kind of stability you need. The little bit of work is related to the fact that after you push forward, the hacker corresponds to actually get a sheaf on bungee times the curve. And you actually need to get rid of the curve factor. You just only do this reflexively after you pull back to a point on the curve. And I will say what I use for this later, but, and also you need to know that you can restrict to proper smooth hacker correspondence, because apparently you have these purpose sheaves as kernels. And so, I mean you have to make this reduction to demo zero solutions and so on. I mean, it's not serious, but a little bit of work. So it's time for a break for 15 minutes. Okay, so let's go on. So these reflexive sheaves are nice, because you can easily prove stability under the hacker operators. But on the other hand, it's not so easy to produce any example of a reflexive sheave, except maybe the constant sheave. We don't take a constant sheave and put it in many translates. Yeah, yeah, right. So it's, I mean, the direct sum of the constant sheave and all different comodic degrees is also reflexive if you want. But it's maybe not so easy. There's also this question in the scheme, so the analytical definition, the serial problem, you don't know. Right, so maybe as Lukas brings this up, let me ask the following question, which is, whose answer is not really important for anything here, but it's maybe an interesting question to think about. So if X is a separated scheme of finite type, maybe over another bracket close to field K, L is not equal to the characteristic of K, and F is maybe, let's say, in the bounded derived category of the atar side with FL coefficients, then is F reflexive, if and only if, F is constructible. And so there's certainly the implication that constructible implies reflexive, but it's unclear whether any reflexive sheave needs to be constructible. Okay, can I reduce it to the perverse case, for example? Okay, so, no, I don't have place for my main serum anymore. So I need actually to put an asterisk on this serum still, because it still depends on one conjecture. I will say what it depends on. It's something that I'm confident can be resolved, but let me be honest. So let's take such a guy, then F is reflexive, if and only if the following condition holds, that for all points, so for all elements B and B of G, which give rise to some geometric points, say XB bar mapping to 1G, and all integers. If you take this guy and restrict to this geometric point, so then after you pull back to your geometric point, this thing is just a complex of factor spaces. Sorry, of lambda modules, thank you. I take its ice core module group, and then the I have acting on this, this group GB of QP, and by the general formalism, this is always a smooth representation, but the condition is that it's admissible. No vanishing for large I, just because the direct sum of lambda and all degrees is reflexive. Maybe in practice you actually want to restrict to some bounded subcategories, but the serum is true in this part. There is an addendum, which is related to the small work there, and let me just add, write it this here, and the same holds true for 1G base change to some C, where C is okay, some algebraic closed non-accommonian field, and if I look at the reflexive subcategory on 1G and pull it back, then this is unchanged. So what do you mean by 1G? I mean this is just, yeah, the pullback. Just for perfect, I will see. And then I restrict everything to perfect, it's always C. And so also for the cotangent complex, I then take the, and not the cotangent, for the dualizing complex, I then take the one which comes from the projection to C, not to K. So this invariance under change of algebraic closed space field is necessary to get some applications running, really. But if a sheaf is a dual of something else, is it reflexive? No, because I mean, for any sheaf, which just might be very big, you might take the dual, then it's even bigger. I think the dual is always the direct summand of the triple dual, by general disability was referred to, maybe. So there is one. No, but if you have something which is, say, countable dimensional, and it's a dual of something, then it must be finite dimension also. So sometimes you can make such arguments. But you need to put some weak, if you have a weak finiteness, like some kind of countable dimensionality, plus it's a dual of something else, and sometimes you can deduce actual finite dimensionality. This will actually come up in an argument. Okay. So the corollary of the theorem, of the main theorem, plus the proposition, is the finiteness result for the corollary of Rapport-Sink spaces. But I don't want to explain this deduction. But I mean, essentially, these HEC operators are a way to encode the corollary of Rapport-Sink spaces. And then, particularly, you can plug in some admissible representation into this, apply the HEC operators, and you get some kind of isotopic part of the corollary of Rapport-Sink spaces. And this must still be admissible because the sheaf is still reflexive. So let me start by explaining the proof of the theorem. And so to prove the theorem, you need a very good understanding of the geometry of Banji. So you need very good local charts. The ones here need good local charts. And so they are the usual charts of Banji. So let me explain what the usual charts of Banji are that one would use, which are the ones that are also somewhat more directly related to how one views Rapport-Sink spaces. So the usual charts are as follows. So let me do the simplest case of GsGl2. But the general theorem is not as simple as GsGl2. No, no, the general theorem is general. I mean, for GLn, it's actually a theorem. So this conjecture is something I can prove for GLn, but it's open so far for a general group. But it must be true with this conjecture. The usual charts as follows say GsGl2. So what you can do is the following. You have a map from Cp1 over this field L, say, and take the associated diamond and say you mod this out by some k where k and Gl2qp is open and Gl2x in the usual way. So you can map this to Banjil2. What you do here is you, let me explain this. So this is given by modifying the trivial Gl2, rank 2 bundle, at a point of the curve. The point corresponds to the map If you have a map to the ad-expectrum of L, this gives you an un-tilt and the un-tilt will automatically give you a section of the front-end curve. And so if you map to this p1 over L diamond, in particular you map to L, L diamond, using the given point. So if you want to modify a rank 2 bundle at a one point, the simplest kinds of modifications are the ones where you describe a sub-bundle, which somehow just modifies it by some rank 1 bundle at this point. And this is parameterized by a p1. So it's the usual kind of buvilasslu type of map. And so this map is smooth and the total space is also smooth. So this map is smooth, comodically smooth. So it's enough to modify just at one point? To get all bundles is enough to modify at one point. It's something that Laurel But of course you need to allow more general modifications than just by p1. Okay, and so what is the geometry here? So okay, so you have this map from p1, diamond mod k, and this maps to, it actually maps to the points isomorphic to either o of a half or o of 1 plus o of minus 1 inside of bun g l2. In particular there's an open stratum, which are all the bundles which are isomorphic to o of a half. This corresponds to, Dreamfield's upper half space in here. So this corresponds to you take p1 of l and remove p1 of qp, and then it didn't leave enough space for this. So there is a complementary closed subspace, which is the one where it's isomorphic to, sorry, what did I write? This doesn't make sense of 1 plus o. So this is then given by just the p1 of qp, well as a, sorry, I mean these are stacks really, there are some more to morphisms. Okay, so locally if you want to understand those two points you can think that smooth chart was given by this p1, where this closed subset corresponds to the rational points inside of this guy. And so a consequence of the main theorem, which is also a key step in proving it, is the following result. So if I look at I have the space which isomorphic to o of a half, this has an open immersion into the space where it's either o of a half or o of 1 plus o. And then there is this closed immersion from the space where it's isomorphic to o of 1 plus o. And if pi is an admittable representation of d cross, I can build from this a sheaf f pi on point modulo d cross, which is exactly this space. And then there are two ways to extend this to the whole bungee. There's the extension by zero, which obviously has a property that it preserves this condition that is satisfied for all b, because for all b other than the one we care about it's zero anyway. But you can also take the rj lower star. And so which is where j dual of what you would get if you would have started with a pi dual. So this is again a reflexive sheaf, because the j lower three is reflexive. And so consequence of being reflexive is that again if I restrict to any stratum, in particular to this guy, that must correspond to an admittable representation of in this case qp cross times qp cross. So you need to prove a finiteness result that if you start with something admittable on the open part and then extend by rj lower star to the whole guy, then what you see at the boundary must still be an admittable representation. But by smooth space change, one way to compute this is to compute it here. So let's call this j tilde here. So the smooth space change now there is no condition. I mean, so if you have the base usual base change theorem for lower shriek, it's always true. So if you always do a shriek and then upper star, it doesn't matter. And then if you pass to adjoints, you see that if you make lower star and then upper shriek, it's also always commutes, even without quasi-compacity assumptions. So there's an important issue that this inclusion here is not quasi-compact because, I mean, this is a quasi-compact space, but this is highly non-quasi-compact. And so it's really difficult to compute this function because, okay, I want to write this down. But smooth space change still holds true in this setup. Yeah, but in any case, for usual italic homology, they sometimes put this finite space condition. I don't remember the formulation of smooth space change, but probably it holds more generally, as you say, because you can... I think it just falls by passing to adjoints from the lower shriek base change. We can instead compute, obviously, f pi restricted to this, what's called omega 2, okay? So omega 2, I mean somewhere. It's p1 minus p1 of kp. So you take f pi on the strain-field space, you push forward to what's the whole p1, and then restrict to strata. I mean, I need that smooth things commute with both the lower star, and this is the argument I just gave, and then there's the other thing that commutes with i upper star, but that's what I wrote down last time. Well, it commutes with those functions. But the stalks of this are rather complicated. So stalks are some direct limit over all u in p1l, which contain p1 of qp, and are maybe k invariant of the global sections on u of this sheet. And so you have this p1 somewhere, then you have inside it this profile net set p1 of qp. And then if you want to compute this r gamma, you somehow need to cover all these guys, but a little disks, and then take a union, as is this string, or a direct limit, and it's pretty hard to control what happens here. No u, right? No, but u, I'm only supposed to make this. Maybe I'm taking k invariant here. Sorry, minus, sorry, the sections on p minus p1 of qp here, because otherwise I can't evaluate here. p1 of qp. So u contains p1 of qp, it's an open neighborhood of it, but to evaluate the sheet, I need to remove p1 of qp again. You can assume it's causal compact. There's a co-final subset of such guys which are causal compact. But the problem is that if I remove p1 of qp, it's not causal compact anymore. So already, for a fixed u computing those global sections is a huge inverse limit, because it's the inverse limit of the guys, of all the commodity on all the causal compact open subsets, and then you need to take a direct limit. So this is impossible to control, essentially. And the invariant under k is in the, is it in the right sense or in the? Well k is pro-p, so it doesn't really matter. I mean k is sufficiently small for this. I mean I would also, I'm probably computing the invariance of this admissible representation under some compact open subgroup here as well, which depends on the k choice of k. But if you try to prove by some direct argument that the commodity of Rapport-Singspaces is finite, you run into the same kind of issue that Rapport-Singspaces are highly non-quasi-compact and you might have to balance it. What happens to the boundary doesn't really contribute much, but it's not so easy to see. I mean, so the moral of what should happen is that if pi is sufficiently ramified, or if pi is ramified, oh, there's some given ramifications, and on small enough this, it's somewhat too ramified to really have any commodity or something like this, but it's hard to formalize. And in the general case, what is P1 to be placed by what? By some Schubert variety, or the general resolution of it, if you want. Schubert variety, in which sense? In the speed-around-plus-cross-manion. Yeah, so there is a kind of provio-laslu in uniformization in general, which goes from the billions, from this speed-around-plus-cross-manion here to Banjee. And you can use that to prove this first theorem about Banjee that I stated that this is a smooth out instead of dimension zero. Okay, so you want different charts, and so the key here is to construct the following charts. And so what I will do is I will fix the point B and B of G, and I want a chart for the set of generalizations of B. So this case would loosely correspond to the case that B corresponds to O of 1 plus O. And let me assume for simplicity that G is quasi-split. Can make sense of this more generally. Okay, so then, well, if B is basic, there's not really much to do, because then this is actually an open stratum, and it's zero dimensional, so those guys are easy to understand. So the critical case is where B is not same as stable. And so if B is not same as stable, there's actually always a canonical levy inside of G, for which it is a basic subset. That's the analog of passing to the Hadanar Zeman filtration. So there exists a levy M corresponding to B and G, such that B actually comes from an basic element B of M and B of M maps to B. And actually, you also get maybe two opposite distinguished parabolic, and I want to fix one of them. So you also have a parabolic M contained in P contained in G, or as this corresponds to a slope filtration. And let me not try to say if it's an increasing or decreasing one, because I would screw it up anyway. And because actually in the function from isocrystals to vector bundles, there's a sign change in the slope. So it's anyway confusing. Let me give the example of GL2 in a second, and then we'll determine which sign I mean. Okay, so I will look at the following. I'll look at, let's call it Mb, the following space. It parameterizes p-torsors E over 5.10 curve, which are, if you push them to the levy, they are fibrous isomorphic to the bundle corresponding to this basic element on the levy. And so this has a map to bun G, which takes any such E and pushes it out to G. And you also have some torso over this, some GB of QP torso. So actually the way this works is that this GB of QP is actually the same thing as the form of M corresponding to this PM of QP. So BM is a basic element for M. So this is an inner form of this levy. And so, and so this is just trivializing E times p. Let's call this map here. Right. And I didn't yet tell you which one. And there's only one where this will be a reasonable thing to do. And I will determine this by giving you the picture for gl2 in a second, okay? And yeah, actually just now. So an example for what is this space. So if gl2 and p corresponds to o of 1 plus o, then the levy will be gl1 times gl1. And BM will correspond to or still somewhat the same thing, except now maybe as a tuple of 1 comma o. So, and in this case, what will this MB tilde be? It will parameterize certain p. Well p will be the braille here, which one I didn't say. So torsos under the braille, they're the same thing as the rank two vector bundles together was a filtration. So you have such guys. And the second condition determines the filtered pieces of this filtration. And if I'm going here, I'm actually trivializing them. So what I'm doing here is I'm considering the space of all extensions. And which parabolic I take is not determined by which way the extension goes. And the extension will be from 0 to e to o of 1 to o. It wouldn't make sense to do it the other way, because in the other way, all the extensions would be split anyway. So that would not be the good thing to do. It's the torsos under this, but it's the same thing. Okay, so then you can map this to MB, which will be extensions 0 to l e to l prime to 0, where these are line bundles. And then instead of saying that they are o or 1, you just say they are slope 0 respectively 1. Okay, and this maps to bun g l 2 just by forgetting about this extension. And the images again, so the images are all the vector bundles which you can write in as an extension of those two guys. So one extension is a trivial extension of 1 plus o. And then you can check by the classification of vector bundles that's the only non-spirit extension is all for half. So in this way it's s here. Okay, and so the conjecture, which is okay for gln, I mean true for gln, is that this map pi b from mb to bun g is combologically smooth. And mb itself is also alchomologically smooth. I mean actually you can check what this actually is doing here. And so you can check that these are actually, I mean for here it's just given by a class in h1 of minus 1, which is an example of a Banach-Kolmes space. In general they will, this thing will be some kind of extension of Banach-Kolmes spaces again. So these things, mb tilde is a reasonably explicit guy with an explicit action of this group until then this is the quotient. So you can understand this. But actually geometry is a little bit funny, so I will maybe end with explaining a little bit about the geometry of this space for, yeah, this space for gl2. But maybe before doing this, let me make one remark. Why is this true? I mean some are also true classically. So this statement would have an analog for usual Ban g and in this usual setting it's also true. And there you can just use the Jacobian criterion to check this. So this is actually an analog of a statement that appears in the work of Drinfeld and Simpson that a reduction to a choice of a generic borrel is a smooth thing. Generic reduction to the borrel. Well in the case where the parabolic is a borrel, choice of a generic reduction to borrel is smooth. Drinfeld-Simpson, so let me try to discuss the final five minutes, say, a little bit about the geometry of this mb tilde in the gl2 example. So it's quite a funny space. So mb tilde is the same thing as a function as the h1 of s of s with coefficients in line number o minus one, which contrary to the case of p1 is non-zero. So the analogy should have in mind that this is roughly analogous to the eddy spectrum of fp power series t1 over p infinity, or let's say k, because I'm working over k. So it's not itself a diamond, it has some kind of non-analytic point corresponding to the close point of this, the point t equal to zero of this kind of form of scheme. But if you look at the structure morphism to the base point, I mean these are both v-shefs, then this is representable in diamonds. So whenever you base change to the perfectoid space, say, then this thing will become a diamond because then it will just become like something like an open unit disk over this space, which is itself a perfectoid space in that case. In this case it will just be a diamond. Okay, so these are objects that far calls absolute diamonds. So there are some v-shefs which are not yet diamonds, but whenever you base change to a diamond, they will be a diamond, which is the same thing as saying that the structure morphism to the base is representable in diamonds. And so they are two strata, so they correspond to just zero and everything except zero, which are precisely the two strata corresponding to the two possible images inside of band GL2. I mean zero precisely corresponds to the non-split, to the split extension. As soon as this extension is non-split this must be over half. And so actually you have an inclusion just of this point zero, which as a function is just the eddy spectrum of k. Now the other end you have some j, and you can check that what this looks like, it's abstractly isomorphic to this long series field, but not quite modular per finite group. Okay, so in this sense it looks like this example of the power series ring, but then there's something extremely funny happening is that what, so this is some kind of punctured open unit disk. And so now you try to move in this punctured unit disk towards the origin. What does it mean? It means, usually it means that the absolute value of t has to go to zero. For this space it means that the absolute value of t has to go to one. So I don't know, and I don't know how to draw the space, like it says open unit disk, and then it has one extra point, but it's somehow it's a boundary of this disk. Is it punctured open unit disk? It's a punctured open unit, and it's this killer wrong series here, right? So it's like a punctured open unit. So whenever your base changes to some non-archimedian field, then this will be some punctured open unit disk, modular sub-profinite group. And with one extra point, but this extra point will be somewhere near the boundary of this open unit disk. And the profiling group is acting but like what you explained. No, actually, I mean this is isomorphism comes from this kind of map to bungeal two, and so this profiling group is actually related to the division algebra. It's maybe the OD cross or something like this. Right, so it's a kind of funny space, but well, it exists and it somehow behaves like a strictly Henselian space. And so in particular, okay, let me finish with one sentence. If you now want to compute with this chart of some sheave, then this is now some quasi-compact space, and there are no further other open neighbors of these points. Actually, all the subtleties go away, and this thing will just be the commodity of this mb tilde minus zero with coefficients in F. And so that's much easier to control. Well, that's the funny thing, right? So this is an affinoid edict space, modular profiling group. So it's quasi-compact. It's quasi-compact as a, it's just a diamond, but the structure of morphism to the point is not quasi-compact. So when your base changes, it's not quasi-compact anymore. But just absolutely, it is quasi-compact, and that's critical. And the map to mb tilde is quasi-compact J. The map is not quasi-compact, but that's not important. Okay, I mean there is some justification behind this, but in the end this formula is true. And so, because here you don't have this just for the compact, so computing global sections is easy and there's no more direct limit here, so everything is much easier now. Okay, but I'll continue next time. Okay, what is this doing? So if you look at this open subset where it is isomorphic to o of a half, then on this part you can fix the isomorphic to o of a half, which gives you some D cross torsor. And then there are some subtleties, actually it's more like an OD cross torsor in the end over it. But then if you look at such modifications from o of a half to this guy, you can also somehow switch perspective and do it the other way around. And regard, well, then some equivalent, this is o of a half and then you look at all possible ways to write as such an extension. But that's just parameterized by global sections of o of a half, which are not zero. And then maybe there's again some ambiguity whether I'm here or here, so there's again some profiling groups ambiguity here. But global sections of this o of a half, they are just given by such an open unit disk. Why you need to remove the origin. But this means that this open unit disk came about in such a way that t equal to zero would correspond to still having o of a half, but this extension collapsing. But that's not what we want to do. That's not the generation that happens in this space. In the space, what should happen is that the extension is still valid, but this bundle degenerates. And that happens at the other side. So the stuff you've just erased must be related to the Jacqueline's correspondence somehow, right? Right. So, well, actually, this is supposed to be basically always zero except for the trivial guy, because one general HEC operator say, take sheaves somewhere to somewhere else. And for example, it takes sheaves which are on the d cross part of bungee and take it to the gl2qp part of bungee. And then this will be doing some Jacqueline's correspondences. But that's essentially the known phenomenon that in the homology of the lumintate space, you have Jacqueline's correspondences. But then if you go to deeper strata, that's supposed to be some Jacket Fungus.