Oh, thank you so much to the organizers for the invitation.

So the talk is a moduli of p-divisible groups. In a way, it's complementary to the last talk. It concerns more the supercuspital region of Schmoor varieties rather than the ordinary. Good job on the ordering of the talks, I would say.

This is with Peter Scholz. And the setup is as follows. So let's say k is a perfect field of characteristic p. And let's start with a p-divisible group.

And I'll call it H0 over k, p-divisible group. And we'll be interested in deformation problems related to this H0. How do you deform H0 to an H, let's say, over some local ring of residue field k? So the work of Rappaport and Zink, they study this problem rather intensively.

And they consider a certain deformation problem of H0,

which they show is representable by a formal scheme over the ring of bit vectors of k. So I'm not being perfectly precise here. So I'm not being perfectly precise in order to head off questions about

what sorts of rings and what category it's on. Yeah, okay. Representable by a formal scheme, M. So I'll write Roman M H0 for this formal scheme over the ring of bit vectors. And the next thing to do is to take this formal scheme

and pass to the generic fiber to obtain a rigid analytic space. So let me do that. And I'll write with a script M sub H0. So let's pass to the generic fiber. This is topologically a finite type, so you get a rather well-behaved rigid space right here.

And then what you do is you add level structures to this problem. You consider deformations of H0 equipped with a basis for its P to the N torsion. So let's denote by this.

So it's a covering of this rigid space, and they're arranged in a tower. And this is the space you get by adding a P to the M level structure. So again, a rigid space. This covers M H0, and these guys are structured in a tower.

This guy individually has an action of the group G, L, N, Z mod P to the M, Z. And what's the N? N is the height, height of H0. And the goal here is to construct a space

which is going to be the inverse limit of all of these guys at once. So that could only reasonably be called M H0 infinity,

and it should be the inverse limit of these individual spaces. And this, oh, so I want to show that this makes sense. It can't possibly make sense as a rigid space. It's just too big.

But it does make sense in Peter Schultz's new category of perfectoid spaces. Space. And it even admits a simple, well, simpler description in terms of linear algebra.

And that idea is going to be the goal of the talk. And furthermore, the reason I might be interested in this space

is that it admits an action of the group G, L, N, not Z, P, but actually all of Q, P, this whole group. And the cohomology of this space should realize the local Langlands correspondence for this group. And by now, that sort of thing is known

by what we know about these individual spaces. I mean, the cohomology here is the direct limit of the cohomologies here. But I should also remark that you can adapt to the situation to get Rapoport-Zink spaces at infinite level for other reductive groups,

G over Q, P. Okay. All right. Very good. So I guess I have two additional remarks.

So one is that I'm going to spend a little bit of time at the end considering the case of Leibn-Tate. So by the Leibn-Tate case, I mean the simplest possible H zero.

So that means that H zero is connected and one-dimensional. Let's call its height N once again. I'll remark that this M H zero without any level structure is a disk.

Sorry, it's a ball. It's an open ball of dimension N minus one. N minus one. And the tower of covers M H zero N, each one is et al over the last. So I'll focus.

All right, components, okay. So it depends on what your deformation problem is. If its deformation is up to isogeny, then you actually get Z copies of this open ball. Actually, that's maybe a better point of view. I like when things are symmetric and then you actually do get an action of the whole thing. So I like that comment enough that I will even adapt.

Z copies of the open ball. So that's just one example here. All right, good. Excellent. Oh, and then, yeah, I guess one thing I'd like to say. So in order to try to convince you that this can't possibly be anything reasonable like a rigid space,

the analogy that I like to think about is another inverse limit. Let's take a manifold like S one. So this is a manifold with the structure of a group. And let's take its inverse limit under a map like X goes to X to the P.

And this is an et al map from S one to itself. But the inverse limit is not a manifold. It's this rather awful thing. So it's R cross Z P modulo Z. So a neighborhood of a point really,

no matter how small a neighborhood it is, doesn't look like anything reasonable. It still has this fractal nature to it. And in a rather similar vein, you expect this M H zero infinity to have a fractal nature to it. And that's what perfect spaces look like. They don't have tangent spaces, by the way. Good. All right.

So I'll just... That's fine. Say that again. So when you do this limit of, let's say, smooth things with finite et al,

in the limit, does the tangent bundle make sense? It should be the pullback from the tensor, so it doesn't make sense. Right. So the question is about the tangent bundle at the top of this tower. And it doesn't really make sense. It has to be zero, if anything. But think of this... I mean, just think of this example. What's the tangent space to the origin of that space?

It just doesn't look like any open subset of R around the origin. So it really can't be a tangent space. I want to push this up even further. Well, I think that's okay. All right.

So we need to first talk about P-divisible groups over... Yeah, there it goes. That's what I feared would happen. So I can't ever leave anything in the middle.

No, you can leave something in the middle, but if you put it here, you should stop it in the middle. Oh, I see. Oh, I see. There's some imaginary line. All right. So let's have the next section be P-divisible.

P-divisible groups over K, where K is still a perfect field in characteristic P. And here, the situation is extremely well known. So I'll just do this to fix notation.

So the category of P-divisible groups over K is equivalent to a category of Dieu Donné modules. And for this talk, the Dieu Donné modules will be covariant. So this is an equivalence, not an anti-equivalence.

And on the other side, if you have K, the Dieu Donné part means that they're equipped with an F and a V, and F times V equals P. And it's rather aesthetic that I'm choosing the covariance formulation, but it just makes my life a little bit easier

for the rest of the talk. But it produces this... Well, maybe you're not used to this kind of thing, but here, the constant P-divisible group, it's Dieu Donné module. Its F is P, and its V is 1. So maybe the opposite of what it is

for the contravariant module. So here it's the opposite. So the multiplicative P-divisible group works the other way. All right, so there's different equivalent formulations of how to construct the Dieu Donné module.

So Professor Fontaine has a whole book on the subject. But I wanted to do the point of view of messing in terms of universal vector extensions of a P-divisible group. So I will construct the Dieu Donné module in terms of universal vector extensions.

Okay, the idea is this. So if R is any... If R is a K-algebra, I can just take...

And H is a P-divisible group. I can take H and just evaluate it on R, and I get an abelian group. And what does this mean? So it means... H is composed of HNs, which are group schemes over K. I can take any one of them and evaluate them on R,

and now I just take the forward limit, okay? Okay, so H is going to be a functor from K-algebras to abelian groups. So... All right, so now I need the hook. So maybe that's close enough. All right.

So the idea is to take H... I actually wanted to notate by H0, my P-divisible group,

so... Imagine it's H0 in both of those places. So let's lift H0 over K to H over W of K. And then this H fits into an exact sequence like so.

Ah, no, I'll just put V here. V for vector. So this is the universal vector extension like this. And so what are these things? So these are also functors from R-algebras to abelian groups. But this one is isomorphic

to some number of copies of the additive group. So you can consider the category of such extensions, and then the initial object of such a category will be the universal vector extension. And we can just consider this as a black box, but then the Dieu Donné module of H0

is defined to be the Lie algebra of EH. Yes, yes, so that's the black box I'm referring to. So this is M of H0, not M of H. This does not depend on the lift that you chose.

So this only depends on H0. And in fact, you get a functor from H0 to Dieu Donné modules, and this is what the Dieu Donné module is, the covariant one. And for all this to work out,

the only things that were necessary about K and W of K is that this map between them, WK to K, is a PD thickening.

So the kernel has divided powers. So we can actually place this whole discussion in a more general context, and when we do that, we get the notion of the Dieu Donné crystal. So...

Yeah. Yeah, periodically complete, separated, nilpotent PD thickening. I hope I haven't missed any adjectives. No, not nilpotent. No, not nilpotent. Oh, sorry. Sorry. Sorry. Sorry. Sorry. But what I'm next going to say,

I'll introduce the crystalline period ring, so... Okay. Okay. Hmm. All right. Oh, okay. So let's suppose R... Hmm. All right, so what we do now is replace K

with a ring R, and R will be a ring in characteristic P. So it's an FP algebra, which I assume to be not perfect but semi-perfect, meaning that the Frobenius map from R to R is surjective.

That's what semi-perfect means. So that simplifies the discussion of Dieu Donné crystals considerably, because you consider the category of PD thickenings, and it has this initial element due to Fonten.

So there exists a universal PD thickening, a crystalline of R to R. And then you can copy everything here

or if you have a PD divisible group over R, you can lift it to a Chris of R, consider the universal vector extension, take the Lie algebra of the center, and then you get the Dieu Donné module. So I'll just write that as M of H zero.

And this is a projective module over this ring with... And it has the F and the V. So that's the more general Dieu Donné module.

All right, so... All right, now I need one of these boards again. So I want this in the middle and this here. But... Yeah, okay.

All right, so now we have this functor from PD divisible groups to Dieu Donné modules over the crystalline period ring. Unfortunately, the situation is not as nice as for perfect fields. That functor is not going to be an equivalence of categories in general. But we can ask for something

a little less strong, namely, is this functor fully faithful? And so now we have our first theorem. So I assume the ring is slightly... It's a slightly stronger condition than R being semiperfect.

So the condition is F semiperfect. And that means that R is the quotient like so, where S is perfect, perfect ring in characteristic P, and I is a finitely generated ideal.

Okay, so such a ring is always semiperfect because it's a quotient of something perfect. But the quotient is by a finitely generated ideal. And an example of this would be the ring OCP modulo P.

So it's a very large ring. This is the quotient of... Well, now this is known by OCP-flat modulo some uniformizing element like P-flat. So this is a good example to keep in mind. All right, then the theorem states

that this Dieudonné module functor is fully faithful. H zero F H zero is fully faithful. That is, as a functor from P divisible groups over R

to Dieudonné modules over a Chris of R. Good. Up to... oh, yes, yes, there I really do. Up to... Yeah, okay. Yeah, okay, with some stronger conditions,

we can get it on the nodes, but I must say up to... So that is Ham's from H zero to G zero. Transited with QP. This maps onto Ham's, which can be with Frobenius and of H zero.

All right, so I want to specialize a little bit to the situation where the first P divisible group is constant. And we'll see what the consequence is.

So this will release into the top. Yes. All right.

So now the role of the first P divisible group will be played by the constants, QP mod ZP. So... and the second one, well, okay,

so this will be my H zero base change to R. So, yeah, so I should have said this before. So let's say H zero is a P divisible group over a perfect field K. And let's say that R is a semi-perfect K algebra.

So under those conditions, now we have two P divisible groups. I'm going to consider them over R. The first is the constants, and the second is H zero base change to R. So this... oh, yes, of course, isogenies, yes.

This will be the same as Ham's, which can be with Frobenius. Of M of QP mod ZP. M of H zero R. Tensor QP.

Yes? R is F... thank you. R is... I forgot the F. F semi-perfect so that I can apply the theorem. And let's evaluate the left-hand side first. So here, well, one over P has to go to a P torsion point. One over P squared has to go to a P squared torsion point.

So what you get is a tower of P torsion points. So you get an element here of the inverse limit of H zero with torsion and P to the N evaluated at R. And this is still tensed with QP.

But this is quite the same thing as the inverse limit of just H of R itself. Remember, you can just take... I erased it. But remember, you can just take H and apply it to R. It's essentially the union of all of the P to the N torsion points.

And so if you have a tower like this, well, because R is P torsion, every element is eventually torsion for some high enough power of P. So everything here lives in here. And then conversely, it's pretty easy to see that every element here lives in here. But then this is a QP vector space. So this is the map from X goes to PX.

You can take any abelian group you like and do this to it, and then you get a QP vector space. So, of course, it extends from this ZP module to this QP vector space. So this is an isomorphism. And I'm going to give this a name. I'm going to call this functor. When you send R to this guy evaluated at R,

I'll call that H tilde of R, and I will call H tilde the universal cover. Cover just considered as a functor on K algebras. I forgot the zero. I forgot the zero everywhere.

Okay. All right, so that's the left side. The right side. So M of QP mod ZP I said was this. So the thing is I want to kind of put the QP inside. So this is going to be A Chris of R tensor with QP. So that's B Chris plus, plus of R

times some basis element. And F acts on this basis element as P. So that's what this part is. This part. All right, so this arises from base extension from something that lives over K, the perfect field.

So due to name modules work well with respect to base change. So this is just M of H zero, which is a W of K algebra tensored over W of K with B plus Chris. Okay, well, this is a little misleading. I mean that this tensor with QP is this. And so HOMs from this into this,

they have to respect Frobenius. So the basis element here has to land somewhere where Frobenius acts as P. And so the whole thing just goes down to M of H zero tensored with B plus Chris. I should really put the R there.

Okay, R. So it lives in the part where... So I'm going to write this as phi equals P. But the understanding is that phi acts on the left factor as F and on the right factor as phi. Okay, yeah, so far so good. So as an example, when H naught is constant,

you recover a familiar result, ZP. Well, the universal cover is actually just QP.

When you take the inverse limit of this with respect to P, you get QP. And now this is telling us that QP is isomorphic to... All right, so there, the due to A module has Frobenius acting as P. And so divide by P and you just get

Frobenius acting as one. So phi equals one, okay? So this is known when R is OCP and now we know it more generally. And then for H0 equals mu P infinity,

so that you get mu P infinity tilde of R, which is sequences of nil topologically... No, just nil potent elements of R compatible under P power maps. This is the same as B plus Chris phi equals P

because now the due to A module has F acting as one. So phi equals P just puts phi equals P there, okay? All right, so these results are maybe not such a huge surprise. So in certain situations, for instance, when R is like OCP mod P, isomorphisms like these,

and more generally, like even when phi is P to the n, you get a P divisible group of height n here. So these were investigated by the recent paper of Farg and Fontan as applications to their fundamental curve of Pia de Koch theory.

All right? All right, so that's my discussion of P divisible groups over rings of characteristic P. I now want to lift the situation to characteristic zero because in the end, we really want some object which classifies

P divisible groups over characteristic zero rings because I want something on the generic fiber. I want a rigid space over QP. So this goes up.

So P divisible groups over characteristic zero rings, all right, okay?

So once again, K is a perfect field. And the kinds of characteristic zero rings I want to investigate are, they fit into pairs like this, R plus.

So this is meant to evoke the notation of attic spaces. So here are R plus. So R plus, I want to be a P torsion-free W of K algebra,

which should be piatically complete. And I want R plus mod P to be F semiperfect, semiperfect,

which is, by the way, the situation you get when this is a perfectoid to affinoid, okay? But just as an example, certainly... Oh, and I didn't say what R was. R is merely R plus tensor QP. So R plus is a subring of R. And so CP OCP is certainly an example of this.

So under these circumstances, once you have something in characteristic zero, then the Hodge theoretic objects have filtrations. So first example, so we have this map.

No, it's R plus mod P to R. And so this is the map called theta. So familiar from piotic Hodge theory.

And let's let H zero over K be a P divisible group. And let's let H be a lift of it. Well, the right way to say it would be a lift

of the base change to R plus mod P. So H naught is over K. H naught base change to R plus mod P is over R plus mod P. And let's lift that to just to R plus. So now it's really in characteristic zero. Then, well, it makes sense to define

the universal cover of H applied to just R plus. So again, it's this inverse limit under the P, multiplication by P map of H applied to R plus.

Now this should go in the middle. But I realize I can't really recover the back one. Is the back one just stuck there forever? I can't, well, suggestions anyone?

You have to lower every, yeah. It's like the Tower of Hanoi problem. Just undo.

So by what I just said, ah, right, okay, okay.

So one little lemma is that when you take the universal cover of H and apply it to R. So you can just, plus. You can, of course, you get a reduction map like this, but this is an isomorphism.

So, right, and I won't prove this, but it's really reminiscent of various computations you do in Pieta-Koch theory.

So for instance, you can take OCP and map this to OCP mod P. That's a reduction map. And you take the inverse limit of this guy under Frobenius and the inverse limit of this guy. So this is just a monoid, but this is a ring, but this is an isomorphism, right?

So the proof follows exactly the same logic. And then, of course, this is one of the intermediate steps involved in the construction of the period ring. So what we get as a result of this, so we get a map from,

so H tilde of R plus goes to H tilde naught of R plus mod P. But this is the same by the full faithfulness theorem as Dieu Donné module of H zero,

tensor B Chris plus of R plus mod P, phi equals P. And then I have this map theta, so I'll just apply it. So that's the map one tensor theta.

And this now lands in M of H zero, tensor R. So now, by doing this, I've certainly introduced denominators. So this map from H tilde of R plus to M of H zero, tensor R. So this is all called the quasi-logarithm map, H zero.

So Q log stands for quasi-logarithm. And I've labeled it by H zero because this map does not depend on the lift from H zero to H. Yeah, yeah, okay.

I mean, of course, R, it depends on the lift R of R plus mod P, but not on the lift H here. So that's an important point. I have an alternate interpretation of this quasi-logarithm map, which will be important. So I'll need some room for this.

So the other interpretation, quasi-logarithm, okay.

So just remember what this Newtonian module was. So we have this lift H of H zero to R plus. And so it can be put in the center, so I can put it into an exact sequence like this.

But I want to evaluate it on R plus. So this was the universal vector extension of H over R plus. Here's zero.

Okay, I want to put another exact sequence above it. This will be the universal cover of H applied to R plus. And the map between them, so this is a set of P-compatible sequences

of elements of this, and I just project onto the first one. The kernel of that is the set of sequences where the first element is zero. So the following elements are P torsion, then P squared torsion, and so forth. So that's just the Tate module. So I'll write just T H zero.

It's not guaranteed to be surjective. And then there's a third row here. So from H of R plus, there's a logarithm, log H, which lands in the Lie algebra of H. But the Lie logarithm introduces denominators.

So now I really have to tensor with R. And then here's the Lie algebra of E H, but that's, by definition, the Giardinier module of H zero, tensored with R. And this map is the log that comes from E H.

Okay, zero. This thing I'll call fill one, zero. All right, so here's this map from H tilde of R. So the fact that there exist... Well, maybe I'll put a dotted line here. Well, dotted line here,

because it's not totally obvious that it exists. But I'll sketch a proof just orally. So this is an element, a sequence of elements of this guy, which are P-compatible. Lift them all haphazardly here. And now we have a sequence of elements here. They're not P-compatible. But you do this smoothing process

involving a limit procedure, which is what you do over and over again in p-adic Hodge theory. And then you get an element here, which really does lie above the element here that you started with. And this is actually going to be well-defined because the ambiguity you get from lifting from here to here is measured by an element here. And that limit process involves multiplying by high powers of P.

But since this is additive, that kills the element here. So there's my short oral proof of why there exists such a map here. Okay, so this map exists, and this composition is the quasi-logarithm, H0. So it does not depend on the choice of lift.

This whole center column doesn't depend on the choice of lift. It's just the filtrations that depend on the choice of lift. All right? All right, so this will bring up...

Oh, I know. Yeah, yeah, I'll actually do the thing I said I was going to do. So that goes up. I'll bring this up too. I'll bring the back one down. I have to do it sometime.

So this quasi-logarithm map, so it's defined whenever you have such a pair, R, R+. So I mentioned something about attic spaces. Attic spaces are built out of attic affinoids

of this form, SPA, R, R+. So that's a set of continuous valuations on R, a valuation less than or equal to 1 on R+. So I'm not going to go into a diversion on attic spaces, but I just want to mention

that because of what we've just done, that is, for every pair, we've constructed a map from H tilde of R+, to M of H0 tensor R. What we do is we get a morphism. Well, it's like this. So H tilde is something like a formal scheme,

but you can take its generic fiber, its attic generic fiber, and then once you've done that, then you can go ahead and apply this quasi-logarithm map coming from H0. And it lands in M of H0 tensor R. So I'll write it this way.

You get a map to the additive group. And, well, it's rather like this. I mean, I can even put this into a commutative square where I can project onto the first coordinate. So this is just H, attic. Here's a logarithm map coming from H,

and that lands in the Lie algebra, tensor GA. So, okay. So this is maybe this more classical thing. If H0 is connected, then this is a ball, an open ball, if H0 is connected.

So it's a map. I mean, it does what a logarithm should. It takes an open... Yeah. Okay. So this, again, once again, does not depend on the choice of lift. Nothing here does. But then this does depend on the choice of lift.

Okay, and one more little observation. So let's let N be the height of H0. And so originally this talk was about moduli of deformations of P divisible groups. So let's say one of those is given.

And that's our H. And now let's also add a level structure into the mix. So let's suppose that H is given together with a level structure. So what would that mean? That would mean a basis. So I want N basis vectors to be inside of... So I let VH of R plus be a basis.

So this is the rational tape module. And at most it could have dimension N. Well, I mean, assuming this is connected. So okay, that's dimension N. So let's suppose you have a full level structure.

Then... So you are starting H0 over your perfect field. Yes. And R plus is related to some affinoid in the moduli space. What is R plus? So R plus is like a...

Yeah, so again, it's like our test object. So... Oh, maybe it's underneath. So R R plus is some given affinoid W of K algebra. And I'm saying, let's suppose we have a deformation of H0 to that test object together with a level structure.

So this means that in R plus mod P it is H0 or R plus... I'm not saying it is H0. What was the question? So when you say deformation... Yes. You have made a map from R plus to K? Oh, okay, yeah, okay. So the right way to set things up is to have a map from...

Yeah. The right thing... I mean, I hadn't planned on going onto this. But somehow the right thing to do is to say something like this. P to H. So really a deformation should be

a pair H rho where this is a quasi-isogony together with an alpha. And an alpha is those n basis vectors. So a point of this moduli space at infinite level valued in R R plus really should be this kind of data.

There's even a white lie there too, but I could tell you about it later. Yeah, okay, okay. Really, you want to do these things locally on SPA R R plus. You really want to cover it and then... I mean, in general, this might not be a bounded subring. Well, okay, there are various white lies in this discussion, but I can tell you about them later.

So all right. So let's follow... Okay, so where do these alphas live? So they live in the Tate module, but maybe they have denominators. But if you follow them through this diagram here, they land in this filtration, which means that they can't really... So this thing is n-dimensional,

but our n vectors in here can't span this n-dimensional space. They actually have to lie in this space of codimension, whatever the dimension of H is. So because of this diagram, so this guy has dimension D, where D is the dimension of H itself. Pull this up. And this guy has dimension N.

And the observation I want to make is that when you take quasi-logarithms, they lie in a subspace of codimension N. D.

Where D is the dimension of the original p-divisible group. Just by that diagram. All right, so now we're actually ready to define the moduli space at infinite level.

So the deformation H disappears from the discussion. All that we have are elements alpha 1 through alpha N, and they are just some abstract elements floating in here, in the middle, in the part that doesn't depend on H. And the condition that we put on these n elements in here

is exactly the condition I just wrote down. Their quasi-logarithms span a space of codimension D, rather than filling up this whole space of dimension N. There's n of them. Vectors, okay. All right, so this here.

All right, so the idea now is to define a functor, which we'll call, I mean, so this is going to be the desired object,

M H zero infinity. And it will take, okay, so, it will be defined on such pairs, with the proviso that you know I'm telling a slight white lie about this. Yes, all right, so the correct, okay.

So to define an attic space, that's, well, that's one of them. Right. So if I say that this is an affinoid W of K, one over P algebra,

then part of that definition is that R plus is integrally closed in R, and it's open and contains this bounded subring. But I also want to assume that R plus is bounded. So there are, you know,

but it has to contain a bounded subring. So the white lie has to rather to do with that. So, but glossing that over, this functor will take the following value on this. So I want it to be the set of,

I'll do alphas, yeah, N tuples, which lie in H tilde of R plus. And by the way, even though I've written an H here, that doesn't mean that I've already lifted H zero,

because I could have just as easily written H zero tilde of R plus mod P. So, or you could just say, pick any lift and form its universal cover, and then that's where these alphas live. Okay, so I want to impose some conditions here.

One of them is the condition I just wrote. So this lives in M of H zero, tensored with R over W of K. These span a space of codimension D.

So if you like, you get a matrix out of this guy. Actually, what I'll do is I'll put parentheses around here, and now it's an N tuple of elements, so it lives there. And then after having picked a basis for this guy, you get a, it's really N squared elements of R. And I want that matrix,

that N by N matrix to have rank N minus D over R. N minus D. Where remember, N is the height of H zero, and D is the dimension. This actually makes sense.

So I mean, you want certain minors to vanish, and that makes sense in R, and then you want other minors not to vanish, so you want the ideal generated by those other minors to be the unit ideal. Okay, so that, yeah. Or you could just say, like for every maximal ideal of R plus, of R really is a matrix of rank N minus D.

Okay, I need actually one more condition because for it to actually be a level structure, it should be linearly independent. I mean, yeah. So it should be linearly independent inside of a vector space. I don't think it's good enough to just say

that those things are, well, yeah, okay, that these are linearly independent on H tilde. I want to do it point-wise. So I'll say it this way. For all X, for all points of SPA R R plus, maybe the residue field is like so, K K plus.

When you take alpha and apply it to X, you get an element of H tilde of K plus, and this is a QP vector space,

and I want these to be linearly independent. So they're linearly independent everywhere.

All right? So, yes, yes, over QP, ah, yeah, exactly. They're QP linearly independent. So this, so H of K plus is a abelian group. H tilde of K plus is a QP vector space. H tilde of anything is a QP vector space. So I mean for them to be linearly independent over QP,

just like if you were defining a modular curve X of P infinity, it should be like elliptic curves together with a QP basis for the rational tape module. So, all right, so then next main theorem.

Oh, no, I can, I pull it all, yeah, I know, I know. Pull this down. So the next main theorem is just that this M H zero infinity does what we want it to do.

But I have to be slightly vague about that.

So M H zero infinity really does parameterize deformations of H zero, with level structure.

So a deformation with level structure is one of these triples H rho alpha over there. And, okay, maybe that's just part one. Part two is slightly technical. So if you have an attic space, Schulze has a notion of one attic space

being similar to an inverse limit of a tower of attic spaces. And this notion means a lot of things, but one of the consequences of it is that as a topological space, this is the inverse limit of these topological spaces. They're homeomorphic. And you also get an isomorphism,

an equivalence of et al sites. So as far as cohomology goes, this is really the limit of the cohomologies of these guys. And finally, well, yeah, I mean, you want this thing, I want to say something like this is a perfectoid space, but a perfectoid space requires a perfectoid field for it to be perfectoid over.

But when you base change to some perfectoid field, you get a perfectoid space. What? When you change it. For some. For any, actually, yeah. Yeah, for any field K.

And this last one, I mean, this is proved by first observing that the way that I've set this up, M H0 infinity lies inside of this space H tilde, well, n times generic fiber.

Because a point of this guy is n elements of H tilde. So yes, at least it's a subset as far as functors go. But as a subspace, it's locally closed. Why locally closed? Well, the conditions I've placed,

well, the linear independence condition is an open condition. And the condition about the vanishing of those minors is a closed condition. So it's an intersection of an open and a closed. So it's a locally closed subspace of this space. And then the last thing you've got to observe is that when you take this space and base change to some perfectoid field,

this is really a perfectoid space. And so for instance, in just the case where H is connected, well, then H attic is just a ball.

But then H tilde attic is something like a perfectoid ball. It's the inverse limit of balls under this et al map which corresponds to multiplication by P in the P divisible group. And that inverse limit has the structure of a perfectoid space. That's not hard to see.

I go till half past, right? Okay. So I planned something like an explicit formula. So let's see how, maybe I'll see how far I get here. So in the Lubin tape case.

So in the Lubin tape case. So the dimension of H zero is one.

It's connected and has height N. And I want some matrix which is an N by N matrix to have rank N minus one. And that just means the determinant vanishes. So as far as determinants go, what I can tell you is that the determinant map,

which I'll notate as delta, even exists on a level of formal schemes. So I haven't passed the generic fiber yet. Um, to UP infinity. Okay.

exists an alternating map, such that when you pass to generic fibers, well, you get this map.

Here, there's a logarithm. It's the usual logarithm. Here, there's the quasi-logarithm applied n times. And so it lands in GA. The Lie algebra is one-dimensional, but there's n of them.

But I'm sorry. The Goudein module is n-dimensional, and there's n of them. So here's n squared. And then this is the determinant, so that's why I'm calling this thing a determinant. All right, I'll go over here.

And I even know what this quasi-logarithm is. Yeah, not eta, delta. Sorry. Oh, well, OK. Perhaps I'll just write some formulas, and if you have questions, you can ask me.

So I don't know what to say. I have three minutes. But the quasi-logarithms are all known when you read the paper of Gross-Hopkins. He computes the quasi-logarithms for H. And then you have to transfer them over to the universal cover. So H tilde, as a formal scheme, is actually w of k, double brackets, t, 1 over p infinity.

And the quasi-logarithm from H tilde to this Goudein A module, which is like, oof, eh.

So I want n coordinates, because this is like ga to the n. So it sends t somewhere. It sends t, in fact, to g of t, g of t to the p, g of p to the n minus 1. And g of t is a wonderful formula

that appears in the paper of Farg and Fonten. So sum i as in z, really over all integers, of t to the p to the n i divided by p to the i. There we go. So what I've written down is a power series

that converges on the perfectoid open ball. Notice that it involves extracting arbitrary p through roots of t. And then in order for this diagram to commute,

what's that? So i ranges over all integers, positive and negative. So for negative ones, you've got to extract p through roots. OK. So in order for this delta to really make sense, so basically delta of x1 through xn,

these are my coordinates on h tilde to the n. This is something like exp of the determinant of g of xi to the p to the j. And by an incredible miracle, this has no denominators.

Where g is that power series, which certainly has denominators. If you believe this theorem and it follows from the full faithfulness results, you get this delta out of working with period rings,

and you believe the commutativity of this diagram, then this delta is the exp of a log. So yeah, it's got to be true. OK. And then the luban-tate space is defined somehow by,

it's defined by setting delta equal to a system of roots of unity. That's all I'll say. Thank you. OK, so thank you very much.

Any questions? Yes. In the Drinfeld case, you have a tower, which is not on this side with Grn. Does this theory work also? Yes, the question is about the Drinfeld tower, which doesn't actually fit into this paradigm, but that relates to the remark I made

that you can adapt this construction to give you Rapoport-Zink spaces with EL structures. That's all you need for this. So the E means endomorphisms, and your endomorphisms could include an action of this division algebra. And so yes, the Drinfeld space at infinite level

works into this paradigm, and you actually get a proof of the isomorphism between the two towers this way. You get a new proof, a different proof? Yes, yes. I mean, it's natural, because both spaces now exist. So it actually says this thing, that there's an isomorphism between two spaces, two perfectible spaces.

It reduces to linear algebra. There's a question. In the case of a geo with the PDEs in the beginning variety, you have other interpretations of your quasi-log for example.

So the question is about an abelian variety? Yes, you can take the PDEs for example. So what are the other interpretations? Of the quasi-logarithm then? Yes, and the construction, and when you have also when the addition of the universal extensions and so on, so it creates an income of GH1 and so on, so I suppose, and then, so do you have further results

in this case? No, no, so I think you're getting at, is this quasi-logarithm map compatible with other constructions that come from the fact that it's an abelian variety? Yes. No, I haven't thought about that, but. So you're getting the idea of a quasi-logarithm? Of course.

It's more. Oh. Yeah, as Fontaine says this, so the abelian varieties also fit into this universal vector extension thing, so. Then you have the optimization, so. Yes, yes, oh, so you want to relate to this universal vector extension to the Hodge filtration of the H1

of the abelian variety? Yes, it must be compatible. Yes, of course, of course. As Jean-Marc says, it should be tautological. I'm serious. Okay, so any of us know how to think of this? Okay, so let's sum it up, keep going.

Okay.