6/6 On the local Langlands conjectures for reductive groups over p-adic fields
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Transcript: English(auto-generated)
00:12
So let's recall the setup on Wednesday. So we have G over QP, a reductive group.
00:31
And at some point I was assumed for simplicity that it's quasi-split. Also, I fixed some algebraically closed base field.
00:45
Work in a geometric setting. I was considering the stack bungee. I defined on the category of perfectoid spaces over K. Taking any test object S to the groupoid of G torsers on XS,
01:15
where this was the relative Fock frontend curve.
01:34
And so I stated last time a theorem to the effect that
01:42
in the suitable interpretation of these words, this is a smooth Artin stack of dimension zero.
02:03
And so then I fixed my coefficients. So coefficients were OE model L to the N, where E over Q are also finite extensions.
02:20
Of course, L is not P. And so we had a VIG duality function in this category of chiefs of lambda modules on bungee to itself.
03:03
I can't believe that Kontsevich asked this question about GABA.
03:29
So it maps it into the dualizing complex of bungee.
03:40
And then we had the definition that F is reflexive if F maps isomorphically to its W where G is dual.
04:13
And this was our substitute for the notion of constructible sheaves on the sky, where the usual notion of constructibility would not be a good thing to consider.
04:26
And the main theorem I stated last time, which still had a star because it still depends on one conjecture,
04:45
says the following, that you can completely understand when a sheaf is reflexive. So F of bungee to lambda is reflexive if and only if for all points.
05:03
And as I said last time, points on the stack are classified by Kontsevich's set B of G of isocrystals. And all come logical degrees i. So this B somehow corresponds to a geometric point xB bar to bungee.
05:31
For all such guys and all integers, if I look at the stock of this complex really, at xB bar, there is a complex, so I can look at all the cohomology groups.
05:47
These are all representations of the corresponding automorphism group of this B, which is this, well at least of G's quantum splits,
06:03
it's this inner form of a levy of G, GB of QP, if this is an admissible representation.
06:30
And so today I want to explain the proof of this theorem. And so the proof is by induction on bungee,
06:53
which is somehow bounded by some fixed B. So you somehow start proving a similar result
07:04
just on the semi-stable locus, and then you go further and further out into the non-semi-stable locus, one stratum at a time. And so... Is this sort of an abstract theorem behind the set of these stackers?
07:23
This discrete set of points? OK, the question is whether there is somehow a special case of a more general theorem applying to more general stakes, and I don't think so. I think it's really something very special for bungee.
07:40
And yes, so I proved the same theorem for all open subsets of bungee. And what about things which are in subsets et al over bungee? Things which are et al over bungee...
08:17
So for things which are very close to bungee it might still be OK, but...
08:25
OK, I can't tell. Certainly the proof makes use of very special properties of bungee. So you start on the semi-stable locus.
08:42
And so what I've already said last time, implicitly, is that it's this following theorem,
09:09
which for GLN is due to Gaussian value, and then for general G is due to phi.
09:28
It says that if you look at bungee, and then you look at the semi-stable locus, then this decomposes into a disjoint union over all basic b of a classifying space
09:50
for the corresponding inner form of G.
10:13
And so we have to understand what this d-add of such a guy is. And for this let me state the following proposition.
10:25
In fact, it holds in the following generality that if you have H in some GLN QP closed subgroup,
10:50
so for example G of QP for any linear group G, or an open subgroup thereof, or whatever,
11:01
then you can try to understand what this d-add of the point mod H is. And so the proposition is that the d-add of the point mod H coefficients in lambda
11:28
is really just the derived category of smooth H representations on lambda modules.
11:52
The ARAR was late. Smooth H representations on lambda modules.
12:10
And under this equivalence, and with a fixed choice of hard measure,
12:39
the idea duality gets identified with smooth duality.
13:01
And which takes any pi to the homomorphisms from pi into lambda, but then you take the smooth part of this. And then this passes to the derived category.
13:31
I mean this sounds very reasonable, but actually verifying this is a little bit in some trouble, but okay, you can do it.
13:44
In particular, the fact that you really get the smooth part here only is related to some fineness results for the cohomology of smooth diamonds. So there is something hidden, some work hidden in this statement. Also it looks very reasonable.
14:01
Not true that the abelian categories are different. Which abelian categories? I mean this is not a priori, the derived category of an abelian category, although somehow a posteriori it is here. Because here we're not in the setup where, so if this was some diamond, some locally spatial diamond,
14:23
then this N test some final cohomological dimension, then this D ed would automatically be the derived category of the tau sheaves. But here we're not in the setup. In particular, sheaves on this are somehow,
14:40
You have to first find the smooth cover of this guy by a diamond and so on. A priori unraveling what this really is, this is a bit of a trouble. Anyway, once you have checked this, you find that in particular if you try to understand what reflexive sheaves are,
15:01
then those are the ones where passing through the double dual and double smooth dual is an isomorphism. It's reflexive if and only if, for all i and z,
15:24
I locate h i of f at a somewhat geometric point, which then becomes a representation of h.
15:51
That's because it can translate to the same question about smooth representations. And this functor is exact, so understanding the radial duality, you can understand it in each degree individually.
16:03
And then the condition that the double smooth dual is the same thing, means that on all the fixed factors under open subgroups, it must be finite dimensional. It must be reflexive London module, which means it's finite.
16:22
There's no restriction on degrees, right? Okay, so that's good.
16:43
I want to say next. So that somehow deals with the semi-stable locus. Okay, so now for, so it finishes semi-stable case.
17:13
So now we need to, so for the induction,
17:21
let's fix some b and b of g, not semi-stable, not basic. And so then we have this inclusion from some b bungee, which is less than b, into bungee, which is less than or equal to b.
17:48
And let me also already give a name to the complementary closed. So we have some i bungee b, meaning equal to b.
18:05
And so we assume that the theorem is known, I mean really the obvious variant of the theorem, known for bungee less than b, okay?
18:43
So what do I have to do? So we need to prove the following results.
19:06
One is the finiteness result, which says the following. So assume you have some f,
19:25
which lives on bungee less than b, which is reflexive. So then if I take, push this forward to the whole guy,
19:48
and then restrict to the new closed stratum, so then I get a sheaf on bungee equal to b from there, and we need to know that this is reflexive.
20:09
And you know that it is et al in D et al by general thing that's been confirmed? No, because all functors are by definition functors on D et al. Ah, so you modify them to be D et al if they are?
20:22
Right, for non-father compact immersions, I said that I define this rj lower star to be the adjoint of j upper star. So rj lower star, if I did it on v-sheafs, would not lie in the D et al, but I just reflect it back into the D et al. By using the right adjoint to the inclusion.
20:40
This means it's complicated to calculate it. This means it's complicated to calculate it, yeah. You must do it this way.
21:11
But still commutes with smooth space change. And the second thing you need to prove is a duality result.
21:26
So again, if f lies on this D et al bungee less than b and that's reflexive.
21:52
Okay, formally the statements look very similar now, but I will try to explain that the first thing really is a fineness result, whereas the second thing is some kind of local duality result.
22:02
That will become, I think, clearer when I go into the proofs. So why is this what we want to know? So does this imply, why does this give the induction step?
22:40
Well, you already understand reflexive sheaves on the two strata.
22:47
So we understand reflexive sheaves on the strata.
23:03
Bungee less than b by induction. And this bungee equal to b is this point modulus group, which are these kind of automorphisms of eb,
23:22
whereas this was an extension of this periodic group gb of gp by a connected group.
23:40
And so you can prove the usual thing that connected groups don't actually interfere with the analytic sheaf theory. So actually the D et al on this bungee equal to b lambda is equivalent to D et al of just the point mod gb of qp lambda.
24:04
And so this means that on the stratum you again can apply
24:22
the result that says that these are given by smooth representation. Do you mean the connected group has no cohomology or something? Right, the connected group has no other cohomology. So yeah, connected, some kind of unipotent group.
24:47
Yeah, so it has no cohomology. You don't use the semidirect model structure, you just need an extension. Do I need to semidirect product structure?
25:01
I would think I just need an extension. Okay, I have a semidirect product in case I need it. Okay, so we understand on the strata.
25:27
So then say for the theorem we need to prove two directions. So which direction should I do first? So assume first that say g in D et al on g less or equal to b lambda
25:55
has admissible stalks in the sense of the theorem.
26:05
And then we have a triangle like so.
26:21
And this is reflexive by induction. And then the whole thing is reflexive by two. So this is reflexive by the case of one g equal to b.
26:54
And then the whole thing is reflexive as i is proper. It's a closed immersion.
27:03
And we had seen that inclusions by proper maps preserve reflexivity. And so extensions of reflexive shoes are reflexive and so we see that g is reflexive.
27:22
And so conversely, assume that g is reflexive. Or maybe.
27:48
Okay, so what do we want to see? We want to see it has good stalks. And so you do the same thing.
28:07
So the left thing... Okay, so the same exact sequence implies that this thing is reflexive.
28:31
The same triangle. And... But then because this is a closed immersion, this actually implies that i opposite of g is reflexive.
28:52
I mean, so... On the open part, we already know that it must have good stalks, admissible stalks.
29:14
And so this means it has admissible stalks.
29:23
So for the argument, the fineness didn't quite appear, but it will be a step in getting to the duality.
29:47
But so because of this identification of reflexive sheaves here again with admissible representations of JB of QP. This statement about this reflexivity is really just saying that something is an admissible representation.
30:00
So in this sense, it's really a fineness result. Whereas this reflexivity on this stack here is a much more geometric statement. It's much more like an actual duality statement. Like a duality.
30:32
And so... I think I'm on the wrong page here.
30:46
So... We want to do these things, and for this we use a chart. So to compute anything.
31:06
Chart is MB, which was an MB tilde mod g of QP. And so where MB...
31:42
So B... I said this somewhere, but let me now say it explicitly again. I assume that g is quadric for simplicity. So... Then B is actually the image of some BM, BM basic.
32:11
Where M is contained in P, is contained in g. So this is a levy. And this is a parabolic encoding some slope filtration.
32:23
And what is MB tilde? That classifies P torsers. E plus an isomorphism of the reduction to M with the bundle corresponding to this basic element.
32:46
Then this maps via the GB of QP, which is the same thing as the MBM QP torsers. To MB, which is a set of P torsers.
33:01
E such that E times PM is fiber-wise isomorphic. This point-wise isomorphic.
33:24
And then this maps to bungee E via sending E to the push out.
33:59
And the conjecture, which is this conjecture that explains the star, its main theorem,
34:07
is that this map from MB to bungee is equal to B, which is open. Bungee is l-comologically smooth.
34:28
And we can prove this for GLN. And MB itself is also a smooth arting step, as a consequence.
34:44
And because the map is smooth and bungee itself is smooth, this guy needs to be smooth. This is actually un-conjectural, so you can directly show that this is smooth.
35:04
Now we have the following diagram that this maps to this bungee. It's equal to B, plus MJ tilde. Let's call this J prime here, actually.
35:32
The pre-image here I will denote by MB circ. Here you have I, bungee B.
35:43
Let's call this I prime. What you get here as a fiber is actually just a point modulo GB of QP. So effectively what you're doing over this stratum is you fix the splitting of the hard anaerysmal infiltration. And then you get this as the automorphism group.
36:11
And you also have on top of this a similar diagram. Let's call this point XB.
36:22
It's just a copy of spark K. Is it conflicting with standard usage? Yeah, this is the open part, this is the closed part. I mean, I use I and J in the usual way, I think.
36:49
We have this picture. Let's start with the finiteness.
37:17
So by using the conjecture, smooth space change, plus some unraveling of the formalism,
37:42
plus this part of the unraveling of the formalism, that this MB tilde is some kind of strictly handzillion space. So it behaves a little bit like power series over an algebraically closed field.
38:06
What you can prove is the following, that if you look at the sections on point mod K, where I say K and GB of QP is an open pro-P subgroup.
38:23
Pro-P just for simplicity, essentially, but makes things slightly easier.
38:40
So, I mean, I fixed here my F and D at all, on G less than B lambda, and I should put the pullback on the pi B here.
39:03
So if I pull it back on the pi B here, I get a shift here. And then whether I do push forward here and pull back here or upstairs, it doesn't matter. And so I want to understand what happens if I push forward here and pull back here. And then I want to know that there's an admissible representation here. So what I want to know is that these kinds of global sections,
39:20
if I go to the point mod K, that this is finite in each degree. And you can compute this just as a cohomology of this open part mod K.
39:45
So recall that MB tilde over MB was this torso and there's this group, so you can particularly divide by this group. That's still some kind of stacky object, but anyway. With coefficients, where this is still reflexive.
40:13
A different way of writing the same thing is actually to write the cohomology of this guy. Let's call this pi B tilde F plus star, takes the K invariance.
40:28
Where pi B tilde is the composite.
40:46
This description is better because here we still have a reflexive sheath. Here we're pulling back too far, this is not reflexive anymore.
41:02
But what's nice about this formula is that it tells you that it's actually, if you do this thing, you push forward and you pull back, and you know that what you get is just some smooth representation of this group JB of QP. The smooth representation is just this thing here, because we've identified the K invariance for any K with the K invariance here.
41:24
So what you want to prove really is that this cohomology here is an admissible representation. The thing with the not smooth, you have the conjecture that the map is smooth. Well, okay mate.
41:40
What does it mean? Pi is smooth. Pi B is smooth. Not pi B tilde. Ah, so the torso is not smooth? Yeah, so coefficients on a profane group are not smooth. Because profane sets, the fibers are not smooth. Profane sets are not smooth. There was this other thing that if you have something smooth,
42:02
and then you divide, then the quotient is still smooth over the base. But that's a different statement. Okay, so the torsos on the profane groups are not? No.
42:33
Okay, so this gives you a way of computing what you want to compute. And so now we need to prove, so we need to show that this is finite in each cohomology.
42:55
And so these are the two ingredients.
43:12
By the way, when you write k acting on this R gamma, this requires some foundation. So in general, if a group acts on the top,
43:22
it's not a discrete group acting on the top. So you need to know that it takes the other definition. Okay, so I will use two things. Well, actually, no.
43:41
I claim that in each cohomology degree, the map is an injection. This I can just say. And the image will always be just the invariance. Forget about any topology of this group acting on this set. I mean, I think it's just... And is it also a coherent couple? Everything inside is a...
44:02
Okay. Let me say what these objects are. So the open part of this MB tilde is actually a quasi-compact separated spatial...
44:25
Well, spatial implies quasi-compact, but diamond. Whereas this quotient here would again be a stack. In this sense, this is nicer here.
44:44
And so this means that you can compute cohomology... I also find it dimensional.
45:08
Compute cohomology as a direct limit of chesh cohomologies
45:31
over finite, over covers, over covers with finitely many.
45:57
If it was not quasi-compact, then to do this, you would have to use infinitely many things.
46:01
And then if you write down the chesh complex, there's some infinite products and they become very big. But if you use this information carefully... But why is there no type of cohomology? Okay, so there's this thing that f might be unbounded, but actually for the statement, you can reduce to the case where f is bounded
46:20
because you know by induction that canonical truncations preserve reflexivity. By induction, canonical truncations preserve reflexivity because the condition of being reflexive is a conditional cohomology sheaves. And so this means, and each individual cohomology group of this guy by some finite dimensionality
46:42
only depends on some bounded piece of the complex. So you can reduce to the case where this is actually just a sheave, in which case you are allowed to just use a direct limit of usual chesh cohomologies. I don't know what I'm saying. If there is a result on the coherent top of the cohomologies completed in general by hyper cover and all that. If you want, you can also do hyper.
47:01
Do you have an analog of Artin's theorem on joints of NZ rings that allow you to do check cohomology? So maybe, yeah, so maybe, let's say hyper. It's all hyper covers.
47:20
But still, we're finding many terms in each degree. And also, you know that this sheave f is not too big. For example, it has countable dimensional stalks. And then you have to sit down and do it carefully,
47:40
but it implies that all the cohomology groups, they are at most countable. So in particular, you have to check that there are not too many et al. things and so on.
48:09
But the critical part that makes everything work is the quasi-compacity here. You have a countable, a countable system. Yeah, you have a countable, a countable system of this kind. I think so.
48:27
And then the second thing is you apply Poincare duality. So because f is reflexive,
48:42
so in particular, it's a dual of something else, of some f prime. And then this implies that actually the cohomology of this guy with coefficients in this guy
49:05
is dual. See, our home of the compact support cohomology of this guy, of the single footage is a dual of into lambda.
49:43
And OFA will again complain because I'm using a compact support cohomology for something which is not representable. The way to get around this issue is to not try to do this directly for the projection, absolutely, but for the map to the point mod k.
50:01
And then you use the description of sheafs on the point mod k. So you can justify this. But this means that actually the cohomology is a dual space.
50:33
Lambda module. That's lambda module, yeah. The structure theorem for lambda modules. Right, and then, I mean, if you have something,
50:42
the dual space which is countable, then must be finite.
51:05
One can also do this argument in slight variance when one doesn't need this countability here, but argue slightly differently. Then you need that everything is stable on the pullback to four finite sets. But anyway, so that's the essential idea that on the one hand, it's automatically dual,
51:22
and on the other hand, by this quasi-compacity, it's somehow more of a direct limit nature. And so playing this off against each other implies finiteness. All right. And so the other part is reflexivity.
51:53
It's a duality.
52:02
So again, we start with our f, which lives on this less than b part. And it's reflexive. And we want to prove that the extension by zero.
53:11
So let's try to unravel what this means. So what is the dual of j lower shriek?
53:21
Well, by the general formalism, that's rj lower star of the dual.
53:42
So we have an exact, we have a triangle that j lower shriek of the dual of f which rj lower star of the dual of f maps to the stuff concentrated at the point for the
54:07
push forward. And so by the way, this is a step where we come here in a second, where we really already need to know that this kind of gadget here is finite to know that this extra,
54:27
this extra term that we get here is not too big. And then we apply duality again.
54:42
So then first of all, the order of the terms change. And so, okay, so here in the middle, we get the double dual of f, which is a dual.
55:09
What do we get on the right? We get the dual of this guy. The dual again turns the j lower shriek into the rj lower star. And then we get this off the double dual, but it's just rj lower star of f.
55:36
And so then, here's the duality commutes
55:40
with the i lower star. And so what we get here is the i lower star of the dual of this guy.
56:08
And we have our natural map here from j lower shriek star.
56:40
Okay, so what is clear is that
56:44
if I pull back this map under j, then this is an isomorphism. For example, because then this term vanishes, this term will be f again.
57:00
This map is an isomorphism, this map is an isomorphism, then this map is. So it remains to check it on the complementary closed. So i alpha star.
57:26
But if I pull this back under i, then it's zero because I took the extension by zero. So I need to prove that this is zero if I pull it back under i. So in other words, I need to prove that these two things become isomorphic after I pull back under i
57:49
with a shift. Equivalently, if I do this,
58:06
the i alpha star, then it's isomorphic to the dual of what I was getting for the dual sheaf,
58:22
which is some kind of local from gradient lt.
58:40
And let me actually go one step further and then we have to break. As a proof of finiteness, we figured out how to compute this procedure.
59:01
Maybe instead we do it here, and then if we do it there, then the sections are given by these guys. go and get one step further. You have to prove the following. That for
59:21
all k in kv of QP and open for p-subgroup, the sections of
59:47
this guy, this is the left hand side, isomorphic to
01:00:01
the dual of the sections for the dual of pi upper star f.
01:00:21
Well, maybe just destroy some degree shifts. So in other words, this is a Poincare duality on the stack, which is this open part modulo.
01:00:57
You also need that the map is the correct map.
01:01:02
And so let me say one last thing before the break. If you want to prove the statement for potentially unbounded f, there is again a way to formally reduce to the case that f is concentrated in one degree. So we can assume.
01:01:31
OK, and so then I will explain how to prove this after the break.
01:01:41
No, there's no, it's not a bounded giraffe category, sorry. When you say d, it's already there. No, it's potentially unbounded. I mean, what you do in this case is that you see that, well, if you formulate the statement for the dual, then both statements take direct, filter direct limits to limits.
01:02:02
And so you can reduce to the thing where you bounded. Comlogically bounded above. But then by some finite dimensionality, you can forget about stuff which is too far on the other side. Anyway, you can do this. So break for 15 minutes.
01:02:20
So maybe I should have stressed that. I mean, there's one commutation of duality with J lower shriek, which is that it transforms J lower shriek into J lower star. But this other junction one might know is actually failed. So it's not in general true that this R J lower star would be taken by duality J lower shriek. That's what you want to prove.
01:02:40
For constructible sheaves, you know it's true, but only because you know by duality for constructible sheaves. Okay, so. Wait, wait, wait. This time it's not the R here.
01:03:05
Okay, so. Actually, something slightly funny happens here with this Poincare duality. Namely, it's a Poincare duality where the pairing is in odd degree. So let me actually tell you about an analog.
01:03:22
So it is like a punctured spectrum. Right, it's the analog of the punctured spectrum exactly. So I assume you have an a2 over a k and you have the origin.
01:03:43
And let's say m is a strict generalization of a2 at x. And then it has a closed stratum, which is x.
01:04:00
And it has a punctured spectrum at zerk. And then, well if you get inspired by the complex picture, kc.
01:04:21
Then you would think that this is a very small neighborhood of this point. And so it's like a 3 square and then there is a coordinate which gives you the distance to x. And so this has a three dimensional punctured duality.
01:04:43
Because for cohomology this factor doesn't matter. So that's the kind of Poincare duality which we want to prove. So we have the strict generalization of our space at this point.
01:05:05
And then there is some homotopically trivial direction which gives you the distance to the origin. And then you have some kind of Milner sphere around this point or whatever it's called. Which is this compact manifold of one real dimensionless.
01:05:25
But now we do a trick. So let me explain the trick in this example first. So assume you have a self-map which is contracting towards zero.
01:05:53
For example in characteristic p you might take xy goes to x to the p.
01:06:16
And assume it's actually an automorphism. So let me also assume that f is an automorphism.
01:06:25
So for example after passing to the perfection rectified. Which doesn't really matter as regards a telekomology.
01:06:40
So then what I can do is I can divide the punctured spectrum by f to the z. Where you would imagine that in the analogous situation what you're somehow doing here is you're contracting. And in regards to the distance you're contracting to the origin.
01:07:02
So the idea is that this is something like this. But then on this distance factor you're somehow folding this together with multiplicative period p. So this would be in this three times as one.
01:07:20
And this satisfies some four dimensional concuriality.
01:07:44
And one can, if one knows finiteness of cohomology for m-circ, one can deduce the three dimensional duality.
01:08:11
From the four dimensional duality m-circ mod f to the n times z where n is sufficiently large.
01:08:31
So the idea is that the cohomology of this quotient is some of the group cohomology of z acting on the cohomology here. But then if n is sufficiently divisible, the cohomology is finite.
01:08:41
This will actually act trivially on cohomology. And so the cohomology of this quotient will actually just be direct sum of two copies of the cohomology of m-circ. And then you can deduce the desired duality here.
01:09:23
So in algebraic geometry it's slightly tricky to make this work. You can, I think, actually do something like this if you replace a strict generalization by this formal completion and then pass the eddic spaces. So then this m-circ is some eddic space itself and you can make sense of this quotient and it will behave like a proper smooth object.
01:09:52
And so you can also make sense of these objects in our case. So what you do is the following. So in our situation, pick some kind of up.
01:10:13
It's just called up. It's an element of gb of qp. Which is central, but is some x with positive slope on the unipotent radical.
01:10:49
So let me try to say in an example what I mean. So if g is gl2 and b corresponds to this bundle O of 1 plus O, then gb is this levy, it's gl1 times gl1 sitting inside of the Borel.
01:11:17
And then I look at the element p comma 1 in this guy.
01:11:40
And so then I can take the quotient by the action of up.
01:11:55
So the idea is that up is such an operator as this f up there. I'll take some power of it.
01:12:03
This is an analog of this operator f up there that is contracting mb's tilde all towards the origin. But if you remove the origin, then it's acting freely. You can take the quotient, so the action is free.
01:12:24
And now the map to the base is proper and smooth.
01:12:45
Let me give you an example of what's happening here. In this gl2 example, what was this mb-circ?
01:13:05
I said last time that this was the eddic spectrum of this Laurent series field, modulo a profinet group.
01:13:24
It was a free action. So this is a quasi-compact space, but the map to the point is not quasi-compact,
01:13:44
because whenever I base change this to a non-archemedian field, it becomes a punctured open unit disk, modulo a profinet group.
01:14:17
In these coordinates, the p corresponds to the map which takes t to t to the p.
01:14:30
And so if I take this guy, modulo p to the z, then it's this guy modulo for b to the z, and then also modulo a profinet group.
01:14:57
Forget about this profinet group, it's not the important part of what I'm trying to say.
01:15:01
So you get this quotient. And so again there is this crucial difference of perspectives between looking at these just as a diamond or looking at the structure morphism. If you look at this just as an object, this is very bad, it's not quasi-separated.
01:15:27
Well, just a sheaf on the perfectoid space in characteristic p. It's not quasi-separated because if you take a quasi-compact object over it like this eddic spectrum and then take the fiber product,
01:15:40
you get z-many copies of it because you're dividing by the z-action. But if you look at it relatively, it's still quasi-compact but not quasi-separated.
01:16:06
But this map is wonderful, it's quasi-compact, quasi-separated, it's even proper and smooth. Relatively represented by diamonds, by proper and smooth guys.
01:16:23
Because whenever you base change the picture to an algebraically closed field, what you get is a punctured open unit disc modulus reaction of Frobenius, which is a wonderful object of the base change to some sparse C.
01:16:46
I get punctured open unit disc modulus Frobenius. And this is a wonderfully nice object.
01:17:10
And so, and because it is proper and smooth, we can apply punctured duality on this space.
01:17:22
So, if you then just use a formalism plus our assumption that f is reflexive,
01:17:42
we get this kind of even-dimensional punctured duality on this stick.
01:18:08
And then, okay, let me write the argument down again.
01:18:25
Yes, there is to check some compatibility of maps. So if n is sufficiently divisible, the cohomology of this quotient is the same as,
01:19:05
I mean, up to the nz acts trivially on the cohomology of this guy by finiteness.
01:19:29
Because if you have any automorphism of a finite object, then some big power of it will be trivial. And then, once this happens, this means that the cohomology,
01:19:43
the cohomology of this quotient is just two copies of the cohomology of.
01:20:02
You have got the spectral sequence, but then you need to know that it acts trivially in the derived category. I want that the action is trivially in the derived category, and then I think it really splits as it acts.
01:20:24
And then you need to check that the even-dimensional punctured duality, which you get on this space, really comes from two odd-dimensional punctured dualities, relating this with the dual of this and the other way around.
01:21:05
So that's what I wanted to say about the proof of the main theorem.
01:21:34
Let me make some concluding remarks about a few things that one can deduce from the formalism.
01:21:50
So this is somehow continuing the discussion of the first half of the first lecture.
01:22:19
And for this, I would like to pass to analytic coefficients.
01:22:25
I don't want to claim that I can make sense of this in as good a way as would be desirable, but you can make sense of some objects.
01:22:40
So you can just make sense of some D reflexive on G with OE coefficients. That's essentially by passing to the limit of the story modulo powers of L. So if you would have done everything in infinity categories, you could literally pass to the limit.
01:23:06
There's a way to do it without infinity categories. Then I can just formally set.
01:23:21
And so this is a canonical object where any possible definition should give you the same answer in the end. For the next thing, that's not true. I define this with E coefficients to be just the idempotent completion.
01:23:42
There is a problem in the limit process, because then you go from OE modulo powers of L to one to another. I don't want to do this just under the derived category. I don't take a limit of derived categories. If I do a limit, I do it of infinity categories. No, but I'm saying that because of infinitely many cohomology groups and infinitely tall dimension,
01:24:02
when you pass from OE mod L to them to OE mod L to their small power, you don't necessarily go from a reflexive guy to a reflexive guy when you take mod, when you pass from one ring to another, because you have infinitely many possible cohomologies at a given number. And then you take toes with a smaller ring, and you can have many things on the left and on the right,
01:24:25
and then you have many toes, so you get possibly will. So you can work with bounded maybe, reflexive. Bounded, let me do bounded.
01:24:44
And maybe I also want the idempotent completion of this thing tensed with E.
01:25:14
And then I don't claim that this satisfies any kind of descent anymore, or that you can check.
01:25:23
I mean, this is some kind of not good thing, maybe.
01:25:45
You need idempotent completion, you don't need your... Yeah, let me do the idempotent completion, because I might want to use some operators, some excursion operators to cut an object into pieces.
01:26:02
So I do want an idempotent completion. But is it necessary? You ask, yeah, I think it's necessary. Because for DPC of nice schemes it's not necessary to ready idempotent, No, but I would expect that you can have representations over OE of such periodic groups,
01:26:27
which have some congruences, but then generically they decompose, and then if you want to somehow get the decomposition that you have generically, if you want to get it integrally, I don't think you can do it. So I think you need an idempotent completion there.
01:27:13
At least for the part of the derived category that I'm eventually interested in. I'm confident that this is going to be defined.
01:27:23
And so let's now fix an L-parameter. Fix my simple L-parameter.
01:27:42
One can define a full sub-category, which is somehow the phi part of this guy.
01:28:02
So there is also a question like in Latic shields, that you can have for various eigenvalues which are not Latic units, then you need values. I don't know if this occurs in this. Well, I'm working over an algebraically closed field right now. So right at the beginning I fixed an algebraically closed space field,
01:28:22
and I'm working over the space field all the time. Oh, but this W of QP, so you have the... Ah, yes, so it might be, well, I can define this for anything. It might be that it's just empty. But I mean this way of group, it's not related to... The fact that this way of group appears here has nothing to do
01:28:41
with this algebraically closed space field that I had. I'm asking if the Frobenius element is not a Latic unit. So yeah, it's probably true that if I have a guy here whose image is not compact in some sense, then what I'm just defining just now will be empty.
01:29:01
But I can just still define it, okay? LG is a Langdon's tool, yeah. The phi part.
01:29:32
By the condition that all excursion operators,
01:29:44
as I defined them in the first lecture, are given by phi, or are determined by.
01:30:17
And so let me discuss what one would expect about this category.
01:30:28
So let me assume that this phi is actually cusp at all. So assume that the group S phi, which is a centralizer in the dual group of phi, is finite.
01:30:49
This implicitly also says that G is, well, the central torus has no split part.
01:31:05
So GLM would be disallowed for this state. G hat is a dual group. So LG is the semidirect product of G hat with a real group.
01:31:23
So in this case, you would expect, and these expectations are results that I would believe to be within reach, okay?
01:31:41
You would expect that all these reflexive guys in this component are concentrated on the semistable locus.
01:32:11
So actually, what this guy is,
01:32:21
this lump, actual bar phi part, it should just be direct, well, it doesn't matter, there's only finite number. So under this assumption, it's automatically true that there are only finitely many basic elements, only finitely many connected components.
01:32:49
Direct sum over all basic B of the derived category, of derived category with admissible cohomologies
01:33:04
of this group JB of QP, of all these inner forms. And then you take a phi part of that in the sense that you only allow those admissible representations, all of whose constituents have this fixed L parameter phi,
01:33:30
where the L parameter is as defined in the first lecture. So this means that this thing which is a priori
01:33:40
defined in terms of some very complicated geometry, boils down to some very concrete representation theory. Why would I expect that this can be proved?
01:34:01
So if you would have a sheaf which has this L parameter but lives on some other stratum, then this other stratum is given by the automorphism group there as an inner form of a levy. And you would expect that if you have representation of an inner form of a levy, that then the associated L parameter should be induced, which is ruled out by this hospitality condition.
01:34:22
So you would only have to prove some compatibility with parabolic induction to get this statement.
01:34:43
So that's the first expectation you would have. And the second expectation you would have is that all hack operators are T-exact
01:35:03
for the standard T-structure. That is not perverse, but... Well, it doesn't really matter, right? Because these are just of points here.
01:35:21
So the perverse is the same as the usual T-structure. Ah, but you are speaking about just... Well, on this part and on this part, some of the perverse and the usual T-structure are the same. It's certainly not true that hack operators are
01:35:40
exact for any kind of T-structure on the full derived category. You need to make some assumption. And for these cusp parameters, it doesn't matter which one you take. And so you would actually this way get an action. But was perverse ever defined?
01:36:00
Perverse was never defined anyway. So that one talks about something else, not this one. Right now, everything is a little undefined, maybe. Let me try to explain this picture. So you actually get some kind of action
01:36:24
where you have here the algebraic representations of the odd group times... Okay, let me call c, the category for phi to be the direct sum
01:36:44
of all b and b of g basic. So the category of admissible QL bar representations of gb of Qp takes the phi part.
01:37:02
So it's some kind of representation theory at a category attached to this cusp parameter phi. And so this would mean that whenever you have a representation of the odd group, you can act on this category
01:37:21
and you get another object of this category. But actually, you get an object with a variable action.
01:37:42
And the action is continuous.
01:38:02
This action satisfies some conditions. These are objects of c phi together with the action of w. So continuous means it's like in a smooth representation
01:38:21
that every vector is an open step. Every vector is an open step. Well, a priori that might intertwine the analytic action, the analytic part of the way group with the analytic topology of this QL bar.
01:38:42
It will actually not happen. And so this gives you some purely category theoretic setup here where you have a category,
01:39:02
you have some category representations of an algebraic group acting on it, where after you act, you get an extraction of some other group and satisfying some compatibilities. And you can write down how the structure there is. And then there's the following theorem, which seems to be well known to the experts in geometric language.
01:39:20
So for example, Gates-Gury and Winston-Lafork and so on, they seem to know this. It seems to be well known. But anyway, somebody who actually wrote down a proof that I understood is Johannes Anschutz.
01:39:44
So yeah, so assume that S phi is finite.
01:40:19
Then there is actually an,
01:40:29
also tensor category of representations of the centralizer group S phi
01:40:43
on this corresponding isotropic component, which under these assumptions,
01:41:02
under these expectations, would actually be an action on this category C phi, such that the action up there,
01:41:28
actually this action is unique, if you write down correctly all the compatibilities.
01:41:49
It's given by the following procedure. So if you have a representation of the L group,
01:42:01
then in particular, you get a representation of S phi times the real group, because you can restrict via phi
01:42:20
to a representation of the real group, but then this action of the real group commutes with S phi by the definition of S phi. So you get a representation of the product group, which is the same thing as representations of S phi,
01:42:43
which are real-QP equivariant. And then if you have a representation of QL bar over S phi, then you have this action. So you get an endomorphism of this reflexive subcategory. What do you mean by action of the tensor category?
01:43:03
Just on the usual Dirac category, or there is some higher? No, no, on the usual Dirac category, just on the category. So this means there is some drives of opposites and some compatibilities? Right, so for every object, you get an endomorphism,
01:43:21
and then you have two objects, then the action of the tensor product is the composite of the head chance. And when you have three, you have a compatibility. And when you have three, you have a compatibility. But it's only on the drive category, so it just stops at some point. Let me not try to remember at which point.
01:43:48
So this means that you can describe the action of the L group, actually, in terms of the action of this S phi
01:44:01
and the L parameter. And so this actually, if you unravel what this means, this is a form of the Kotwitz conjecture. So this action of the representations of the L group, this encodes the cohomology of rapid port sync spaces and then take isotopic components under some guys in this phi part.
01:44:25
And then this computes the output of this functor in terms of the action of this category of representations of S phi, which is doing some combinatorial things like realizing some Jackie Langdon's correspondences. So this takes one representation with L point reader phi
01:44:42
and produces another representation. So this gives you the representation theoretic part of the picture. And then it also tells you how the VAR group acts, namely, well, it's also given.
01:45:10
This is a form of Kotwitz's conjecture on the cohomology of basic form of spaces.
01:45:25
Well, this is a form of the Kotwitz conjecture, except that it doesn't really tell you what this action is. So proving the Kotwitz conjecture would then come down to actually understanding what this action is. So let me say one last word about this.
01:46:00
And so there's a following conjecture, which is essentially due to Carlita, except of course, it didn't have this formalism to state it. But Carlita conjectured exactly how this category C phi
01:46:22
or the reducible objects of C phi look like. Okay, so let me assume for this that G is quasi-split. So then what Carlita effectively conjectures is that this category C phi is equivalent
01:46:42
to the category of representations of S phi, where he has a trivial representation, gets mapped to use a unique generic representation and fix the data.
01:47:02
Unique generic representation of G of QP, which is the one which corresponds to a trivial element. And then in general, he has a way to parameterize
01:47:21
all the irreducible representations of all these, what he calls extended pure inner forms, all these JB of QPs. And they should be given by certain representations of S phi, which have a certain behavior. And well, if you combine this over all possible different Bs, then they should be parameterized by all possible irreducible representations of S phi.
01:47:41
And so he relates the irreducible objects on both sides, but actually this should be an equivalent of categories. And this should also be true. What did I do? GQP equal what? G1 of QP. So I mean, if you take for B the trivial element,
01:48:08
then competitively with, so in other words, there should be this unique generic representation.
01:48:21
And then this action of the representations of S phi would give you a whole functor from the category of representations of S phi into here. And this should be an equivalent. So this isomorphism, it depends on the choice of
01:48:41
vitae datum, but some of the action of the representations of S phi on it, they do not depend on the choice of vitae datum. The unique generic representation here, so some of the basis elements here depends on the choice of vitae datum. But this kind of module structure over the
01:49:00
representation of S phi do not depend on it. So some base point free version of this conjecture, some of this category, some are free of rank one over the representations of S phi.
01:49:21
What Carl Leiter really made much more precise is he wrote down some explicit endoscopic character relations for this action.
01:50:00
And so this is actually also something that seems to be in reach. So there is work in progress of Carl Leiter and Weinstein where they use the Lefschetz trace
01:50:23
formula and check that it is compatible, and check that the bungee picture is compatible with these
01:50:52
endoscopic character relations.
01:51:05
So the way they phrase the argument is that they assume that Carl Leiter's conjecture is known in his form that the reducible objects are paired in the expected way and satisfies the expected endoscopic character relations, which Carl Leiter has proved in a wide variety of cases. And then they prove that what you see in the
01:51:22
cohomology of these basic robot thing spaces really is realizing this correspondence of representations. And so let me just say that this generalizes work of Faltings and Strouh for the Leibn-Tate case, or
01:51:50
Farak for some unitary groups. Farak, Chen, unitary groups. I think you want n minus one, right?
01:52:02
And also Mieda had some results of GSP for GSP4. So there are Faltings and Strouh proved by some purely local methods using the Lefschetz trace formula and Leibn-Tate or Drenfeld Tower.
01:52:21
And that in the cohomology of these Leibn-Tate spaces, you realize the Jackie Langan's correspondence. And this is some large generalization of this. And so my time is up. Thank you very much.
01:52:45
Very nice lecture. Any questions of course? I mean in this middle backward, of course, there's also the action to be a group, right? Whereas in the lower blackboard, there's no action to be a group, right?
01:53:01
Whereas this theorem tells you that the action of the very group is already as expected. And everything comes down to determine what happens on the periodic groups. And for example, for the Leibn-Tate Tower, which you
01:53:20
can't quite apply, so if you do PGL or something. Or proof of the version of the theorem which allows the center. The theorem would already tell you that if you plug in some supercast representation of GLN, what you get is some representation of the divisional algebra tensed with the correct L-parameter. And then if you combine this, tensed with the L-parameter
01:53:42
that is defined by these methods. If you combine this with Faltings and Stroff paper, which tell you that you see the correct Jackie Langan's correspondence, basically see that you get the correct thing, but actually the output of this is that actually what you see here in this picture is really the usual local Langan's correspondence.
01:54:01
Are you using some global thing to determine this? No, this is purely local argument. But I'm using that there is a local characterization of local Langan's for GLN. That is what's realized in Leibn-Tate Tower.
01:54:29
Well, I mean when... Yes. So that it is a local Langan's correspondence is proved by global methods.